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PostPosted: Wed Oct 29, 2014 8:09 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Puzzle rating table, with links to archive entries; each of these has a link to the puzzle thread.

Abbreviations used in Rating Table:
Est = Estimated rating by puzzle maker
Other = Puzzle posted from another site and/or newspaper
E = Easy
H = Hard
SS Score = SudokuSolver score, rounded to nearest 0.05; there are two columns, one for v3.3.1 and the second for v3.5.7.
In the v3.3.1 column these are the scores posted by the puzzle maker in the puzzle thread and/or in the
Assassin Schedule thread, except where indicated by * when I calculated the score using SS.
Thanks Ed for calculating the v3.5.7 scores.
+-------------------------------+-----------+---------+----------+----------+
| Puzzle | Made By | Rating | SS Score | SS Score |
| | | Andrew | v3.3.1 | v3.5.7 |
+-------------------------------+-----------+---------+----------+----------+
| Assassin 221 | manu | H1.50 | *1.85 | 1.55 |
| Paper Solvable 8 | HATMAN | 0.50 | 0.75 | 0.85 |
| Human Solvable 9 | HATMAN | 1.50 | 3.55 | 2.60 |
| Assassin 222 | Ed | H1.50 | *1.65 | 1.55 |
| Assassin 223 | manu | E1.50 | *1.35 | 1.25 |
| Assassin 223 V2 | manu | H1.25 | *1.25 | 1.35 |
| Assassin 224 | Ed | H1.50 | *1.75 | 1.55 |
| Double Diagonal 1 | HATMAN | 1.50 | | |
| Paper Solvable 9 Elevenses | HATMAN | 0.50 | | |
+-------------------------------+-----------+---------+----------+----------+


Some of the selected quotes in the puzzle entries have been edited to remove "spoilers"; the full rating comments are included with the walkthroughs. In some cases the puzzle makers gave hints; these are included in tiny text in the selected quotes.


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PostPosted: Thu Oct 30, 2014 1:45 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Assassin 221 by manu (September 2011) here
Puzzle Diagram:
Image
Code: Select, Copy & Paste into solver:
3x3::k:5120:5120:5120:3587:3587:3587:4102:4102:4102:3849:3849:5120:1548:11533:3342:4102:4624:4624:3849:2579:2579:1548:11533:3342:2840:2840:4624:2587:2844:2844:11533:11533:11533:3105:3105:4899:2587:2587:2598:2598:11533:2857:2857:4899:4899:5165:5165:5165:11533:11533:11533:4659:4659:4659:4150:4150:5165:4409:4409:4409:4659:4669:4669:2879:4150:9537:9537:4409:9537:9537:4669:1607:2879:2889:2889:9537:9537:9537:2638:2638:1607:
Solution:
+-------+-------+-------+
| 3 7 9 | 8 4 2 | 1 6 5 |
| 8 2 1 | 5 9 6 | 4 3 7 |
| 5 4 6 | 1 3 7 | 9 2 8 |
+-------+-------+-------+
| 4 3 8 | 6 2 1 | 7 5 9 |
| 1 5 7 | 3 8 9 | 2 4 6 |
| 6 9 2 | 4 7 5 | 8 1 3 |
+-------+-------+-------+
| 7 8 3 | 2 5 4 | 6 9 1 |
| 9 1 4 | 7 6 3 | 5 8 2 |
| 2 6 5 | 9 1 8 | 3 7 4 |
+-------+-------+-------+
Quote:
manu: As promised to Ed and Andrew, a more difficult one for the next. But no doubt you'll find a good path for cracking it. Enjoy !

Ed: Thanks manu for a really nice challenging puzzle. Took me a long time to find the key (step 14) but kept chipping away in the meantime. The chipping bits are not shown in this optimised walk-through. This level of killer really suits a two week cycle for me - can take my time and get a lot of satisfaction.

manu: Nice walkthtough Ed! I did not think about yours steps 14 and 15, which offer a quick resolution. My path was different. I wait for Andrew's WT before sending mine.

Andrew: Thanks manu for an interesting and challenging Assassin!
manu wrote:
I wait for Andrew's WT before sending mine.
Thanks manu. I feel honoured that you did that. :) I've been busy on other (non-sudoku) things for the past couple of weeks so only started on A221 after seeing manu's message.
I worked in the same areas of the puzzle that Ed used but saw several things in different ways so, although there's not a narrow solving path, the solution seems limited to particular areas of the grid.
I'm impressed and amazed that Ed spotted his step 14; then the next step was a neat way to crack the puzzle. Ed's step 11 was similar to my step 32, but based on a different number in the same cage.

Walkthrough by Ed:
Thanks manu for a really nice challenging puzzle. Took me a long time to find the key (step 14) but kept chipping away in the meantime. The chipping bits are not shown in this optimised walk-through. This level of killer really suits a two week cycle for me - can take my time and get a lot of satisfaction.

Assassin 221 19 steps
Please let me know of any corrections or clarifications needed.

Prelims
i. 6(2)n2 = {15/24}
ii. 13(2)n2: no 1,2,3
iii. 10(2)n1: no 5
iv. 11(2)n3: no 1
v. 10(3)n4: no 8,9
vi. 11(2)n4: no 1
vii. 12(2)n6: no 1,2,6
viii. 19(3)n6: no 1
ix. 10(2)r5c3: no 5
x. 11(2)r5c6: no 1
xi. both 11(2) cages n7: no 1
xii. 6(2)n9 = {15/24}
xiii. 10(2)n9: no 5

1. "45" on n7: 2 innies r78c3 = 7: no 7,8,9

2. "45" on n25: 2 outies r5c37 = 9: no 9
2a. no 4 in r5c7
2b. no 8 in r5c3
2c. no 2,7 in r5c6
2d. no 1,2 in r5c4

3. "45" on n69: 2 innies r58c7 = 7: no 7,8,9
3a. no 3,6 in r8c7
3b. no 3,4 in r5c6
3c. no 1,2 in r5c3 (h9(2)r5c37)
3d. no 8,9 in r5c4

4. "45" on n9: 2 innies r78c7 = 11 (no 1)
4a. no 2,3,4,5,8 in r7c7
4b. no 6 in r5c7 (h7(2)r58c7)
4c. no 5 in r5c6
4d. no 3 in r5c3 (h9(2)r5c37)
4e. no 7 in r5c4

5. "45" on n47: 2 innies r58c3 = 11 = [65/74]
5a. no 6 in r5c4
5b. r7c3 from (23) (h7(2)r78c3)
5c. no 5 in r5c7 (h9(2)r5c37)
5d. no 6 in r5c6
5e. no 2 in r8c7 (h7(2)r58c7)
5f. no 9 in r7c7 (h11(2)r78c7)

6. naked pair {45} in r8c37: both locked for r8 and 37(7)r8c3
6a. no 1,2 in r9c9
6b. no 6,7 in r9c1

7. "45" on r89: 3 innies r8c258 = 15
7a. = {168/267}(no 3,9) = [1/2..]
7b. must have 6: 6 locked for r8
7c. no 5 in r9c1

8. Killer pair 1,2 in h15(3)r8c258 (step 7a) and r8c9
8a. both locked for r8
8c. no 9 in r9c1

9. naked pair {45} in n9: both locked for n9
9a. no 6 in 10(2)n9

10. 8 in n9 in 18(3) or 10(2) -> 2 in 18(3) must also have 8
10a. -> {279} blocked (Locking-out cages)
10b. 18(3) = {189/369/378}: no 2

11. h7(2)r78c3 = [25/34]
11a. 11(2)r8c1 = {38}/[74/92]
11b. -> 2 locked for n7
11c. no 9 in 11(2)r9c2
11d. and no 2 in r6c1 (CPE)

12. 4 & 5 in n8 only in r7: both locked for r7
12a. 17(4)n8 must have both 4&5 = {1457/2456}(no 3,8,9)

13. "45" r789: 2 innies r7c37 = 9 = [27/36] = [3/7..](no eliminations yet)

14. "45" on r89: 1 innie r8c5 + 19 = 4 outies r7c1289
14a. 4 outies are from (136789): sums to 34 -> to sum to 20 must subtract 14, ie {68} only
14b. however, {1379} in r7c1289 clashes with h9(2)r7c37 (step 13)
14c. -> no 1 in r8c5

15. h15(3)r8c158 = {168/267} = 6 in r8c5 or 2 in r8c5
15a. no 7 in r8c5

16. 17(4)n8 must have 2/6 for r8c5 = {2456} only: 2 & 6 locked for n8

17. 1 in n8 only in r9: locked for r9
17a. no 9 in 10(2)n9

18. 9 in r9 only in n8: locked for n8

19. r8c1 = 9 (hidden single in r8)

cracked
Walkthrough by Andrew:
Thanks manu for an interesting and challenging Assassin!

manu wrote:
I wait for Andrew's WT before sending mine.
Thanks manu. I feel honoured that you did that. :) I've been busy on other (non-sudoku) things for the past couple of weeks so only started on A221 after seeing manu's message.

I worked in the same areas of the puzzle that Ed used but saw several things in different ways so, although there's not a narrow solving path, the solution seems limited to particular areas of the grid.

I'm impressed and amazed that Ed spotted his step 14; then the next step was a neat way to crack the puzzle. Ed's step 11 was similar to my step 32, but based on a different number in the same cage.

Here is my walkthrough for A221

Prelims

a) R23C4 = {15/24}
b) R23C6 = {49/58/67}, no 1,2,3
d) R3C23 = {19/28/37/46}, no 5
e) R3C67 = {29/38/47/56}, no 1
f) R4C23 = {29/38/47/56}, no 1
g) R4C78 = {39/48/57}, no 1,2,6
h) R5C34 = {19/28/37/46}, no 5
i) R5C67 = {29/38/47/56}, no 1
j) R89C1 = {29/38/47/56}, no 1
k) R89C9 = {15/24}
l) R9C23 = {29/38/47/56}, no 1
m) R9C78 = {19/28/37/46}, no 5
n) 10(3) cage at R4C1 = {127/136/145/235}, no 8,9
o) 19(3) cage at R4C9 = {289/379/469/478/568}, no 1

1. 1 in N6 only in R6C789, locked for R6 and 20(4) cage at R6C7, no 1 in R7C7

2. 45 rule on N2 2 innies R23C5 = 12 = {39/48/57}, no 1,2,6

3. 45 rule on N7 2 innies R78C3 = 7 = {16/25/34}, no 7,8,9

4. 45 rule on N8 2 outies R8C37 = 9 = [18/27]/{36/45}, no 1,2,9 in R8C7

5. 45 rule on N9 2 innies R78C7 = 11 = {38/47/56} (cannot be {29} because 2,9 only in R7C7), no 2,9 in R7C7

6. 45 rule on R1 2 outies R2C37 = 5 = {14/23}

7. 37(7) cage at R8C3 = {1246789/1345789/2345689}, CPE no 8,9 in R8C5
7a. 9 in 37(7) cage only in N8, locked for N8

8. 45 rule on N25 2 innies R5C46 = 12 = {39/48}/[75], no 1,2,6, no 7 in R5C6, clean-up: no 4,8,9 in R5C3, no 4,5,9 in R5C7

9. 45 rule on N4 2(1+1) outies R5C4 + R7C3 = 6 = [33/42], R5C4 = {34}, R7C3 = {23}, clean-up: no 1,2,3 in R5C3, no 3,4,5 in R5C6 (step 8), no 6,7,8 in R5C7, no 1,2,3,6 in R8C3 (step 3), no 3,6,7,8 in R8C7 (step 4), no 3,4,5,8 in R7C7 (step 5)

10. Naked pair {45} in R8C37, locked for R8 and 37(7) cage at R8C3, clean-up: no 6,7 in R9C1, no 1,2 in R9C9
10a. Naked pair {45} in R8C7 + R9C9, locked for N9, clean-up: no 6 in R9C78

11. 4,5 in N8 only in 17(4) cage at R7C4, locked for R7
11a. 17(4) cage = {1457/2456}, no 3,8

12. Law of Leftovers on N5 2 outies R23C5 must be exactly the same as 2 innies R5C46 -> R23C5 (step 2) = {39/48}, no 5,7
[It was only later that I realised that this step is done more simply as
5,7 in N5 only in 45(9) cage at R2C5, no 5,7 in R23C5
Law of Leftovers seemed an easier way to spot this elimination.]

13. R23C6 = {58/67} (cannot be {49} which clashes with R23C5), no 4,9

14. 1 in N7 only in 16(3) cage at R7C1 = {169/178}, no 2,3

15. 18(4) cage at R6C7 contains 1 = {1269/1278/1368/1467} (cannot be {1359/1458} because R7C7 only contains 6,7), no 5

16. 5 in N6 only in R4C78 = {57} or 19(3) cage at R4C9 = {568} -> 19(3) cage = {289/469/568} (cannot be {379/478}, locking-out cages), no 3,7
16a. R4C78 = {39/57} (cannot be {48} which clashes with 19(3) cage), no 4,8

[I ought to have spotted the next step earlier, after step 9.]
17. R5C46 (step 8) = [39/48], R5C67 = [83/92] -> R5C47 = [32/43], 3 locked for R5 (locking cages)

18. 45 rule on R89 3 innies R8C258 = 15 = {168/267}, no 3,9, 6 locked for R8, clean-up: no 5 in R9C1
18a. Killer pair 1,2 in R8C258 and R8C9, locked for R8, clean-up: no 9 in R9C1

19. 20(4) cage at R6C1 = {2369/2378/2459/2468/3458} (cannot be {2567/3467} which clash with R5C3)
19a. 1 in N4 only in 10(3) cage at R4C1 = {127/136/145}
19b. Killer triple 5,6,7 in 10(3) cage, 20(4) cage and R5C3, locked for N4, clean-up: no 4 in R4C23
[Alternatively hidden killer pair 8,9 in R4C23 and 20(4) cage at R6C1, 20(4) cage contains one of 8,9 -> R4C23 must contain one of 8,9 -> R4C23 = {29/38}]

20. R4C78 (step 16a) = {57} (cannot be {39} which clashes with R4C23), locked for R4 and N6
20a. 19(3) cage at R4C9 (step 16) = {289/469}, 9 locked for N6

21. Killer pair 5,7 in R4C7 and R78C7, locked for C7, clean-up: no 4,6 in R3C8, no 3 in R9C8

[Maybe some nibbling at the 19(3) cage in N6 will help?]
22. R5C46 (step 8) = [39/48]
22a. 45 rule on N25 1 innie R5C6 = 1 outie R5C3 + 2 -> R5C36 = [68/79]
22b. 19(3) cage at R4C9 (step 20a) = {289/469}
22c. 2 of {289} must be in R5C89 (R5C89 cannot be {89} which clashes with R5C6), no 2 in R4C9
22d. 4 of {469} must be in R5C89 (R5C89 cannot be {69} which clashes with R5C36), no 4 in R4C9
22e. 6 of {469} must be in R5C89 (R5C89 cannot be {49} which clashes with R5C46), no 6 in R4C9

23. Killer pair 8,9 in R4C23 and R4C9, locked for R4

24. 45 rule on N25 2 outies R5C37 = 9 = [63/72]
24a. Killer pair 2,6 in R5C37 and 19(3) cage at R4C9, locked for R5
24b. 2 in R5 only in R5C789, locked for N6

25. 10(3) cage at R4C1 (step 19a) = {127/145} (cannot be {136} because 3,6 only in R4C1), no 3,6
25a. 6 in R4 only in R4C456, locked for N5

26. 18(4) cage at R6C7 (step 15) = {1368/1467} [4/8 in R6]
26a. 20(4) cage (step 19) = {2369/2378/2459} (cannot be {2468/3458} which clash with 18(4) cage at R6C7), CPE no 2 in R4C3, clean-up: no 9 in R4C2

27. 45 rule on R6789 3 innies R6C456 = 16 = {259/457} (cannot be {349/358} which clash with R5C46), no 3,8, 5 locked for R6 and N5
27a. 8 in N5 only in R5C56, locked for R5
27b. 8 in 45(9) cage at R2C5 only in R235C5, locked for C5

28. 5 in N4 only in 10(3) cage at R4C1 (step 25) = {145} (only remaining combination), locked for N4

29. 20(4) cage (step 26a) = {2369/2378}, CPE no 3 in R4C3, clean-up: no 8 in R4C2

30. 3 in N9 only in 18(3) cage at R7C8 = {369/378} or R9C78 = [37] -> 18(3) cage = {189/369/378} (cannot be {279}, locking-out cages), no 2
30a. 6 of {369} must be in R8C8 -> no 6 in R7C89
30b. 6 in N9 only in R7C7 + R8C8, CPE no 6 in R6C8

31. R8C258 (step 18) = {168/267}
31a. 2 of {267} must be in R8C5 -> no 7 in R8C5
31b. 17(4) cage (step 11a) = {1457/2456}
31c. 1 of {1457} must be in R8C5 -> no 1 in R7C456

32. 16(3) cage at R7C1 (step 14) = {169/178}, R89C1 = {38}/[74/92] -> combined cage R78C12 + R9C1 = {169}{38}/{169}[74]/{178}[92] contains 9, locked for N7, clean-up: no 2 in R9C23
32a. 2 in N7 only in R7C3 + R9C1, CPE no 2 in R6C1

[At this stage I spotted a contradiction move which limited 37(7) cage at R8C3 to one combination, but this didn’t feel satisfactory so I’ll look for something more acceptable.
Then I spotted a forcing chain, so I’ll go with that.]

33. Hidden killer pair 1,9 in R9C456 and R9C78 for R9, R9C78 contains both or neither of 1,9 -> R9C456 must contain both or neither of 1,9
33a. 37(7) cage at R8C3 contains both of 4,5 = {1345789/2345689}
33b. 37(7) cage = {1345789} must contain both of 1,9 in R9C456 => R8C1 = 9 (hidden single in R8), R9C2 = 2
or 37(7) cage = {2345689}, 2 in R9C456
33c. -> 2 must be in R9C1 or R9C456, locked for R9, clean-up: no 8 in R9C78

34. R8C9 = 2 (hidden single in N9), R9C9 = 4, R8C37 = [45], R4C78 = [75], R7C7 = 6, R7C3 = 3 (step 3), clean-up: no 1,2 in R2C7 (step 6), no 6,7 in R3C2, no 7,8 in R8C1, no 8 in R9C1, no 7,8 in R9C23

35. R89C1 = [92]
35a. Naked pair {56} in R9C23, locked for R9

36. R8C5 = 6 (hidden single in R8)
36a. 17(4) cage at R7C4 = {2456} (only remaining combination)
36b. R8C258 (step 18) = {168} (only remaining combination), locked for R8

37. 18(3) cage at R7C8 (step 30) = {189} (only remaining combination), locked for N9
37a. R9C78 = [37], R2C7 = 4, R2C3 = 1 (step 6), R5C7 = 2, R5C6 = 9, R5C89 = [46], R4C9 = 9 (hidden single in N6), R5C34 = [73], R4C3 = 8, R4C2 = 3, R6C1 = 6, R4C1 = 4 (hidden single in N4), R5C5 = 8 (hidden single in R5), R8C46 = [73], clean-up: no 2,9 in R3C2, no 9 in R3C3, no 2,5 in R3C4, no 4 in R3C5, no 8,9 in R3C8

38. Naked pair {39} in R23C5, locked for C5 and N2 -> R9C5 = 1, R9C46 = [98], R1C4 = 8, R4C4 = 6, R3C4 = 1 (hidden singles in C4), R2C4 = 5, R4C56 = [21], R6C4 = 4, R7C4 = 2
38a. Naked pair {67} in R23C6, locked for C6 and N2 -> R1C56 = [42] , R6C56 = [75], R7C56 = [54]

39. 20(4) cage at R1C1 contains 1 = {1379} (only remaining combination) -> R1C123 = [379], R2C1 = 8, R3C2 = 4, R3C3 = 6

and the rest is naked singles.

Rating Comment. I'll rate my walkthrough for A221 at 1.5 to Hard 1.5. I used a fairly short forcing chain to achieve my final breakthrough.


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PostPosted: Thu Oct 30, 2014 2:06 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Paper Solvable 8 by HATMAN (September 2011) here
Puzzle Diagram:
Image
Code: Select, Copy & Paste into solver:
3x3::k:2825:2825:2309:2309:9749:9749:7188:2828:2828:2825:6670:6670:9749:9749:1284:1284:7188:2828:1287:6670:6670:9749:9749:9749:2050:7188:7188:1287:9231:9231:3341:5654:2050:2050:8723:8723:9231:9231:9231:3341:5654:5654:8723:8723:8723:9231:9231:5633:5633:5654:5654:8723:8723:3848:4368:4368:5633:8209:8209:8209:3602:3602:3848:3595:4368:3843:3843:8209:8209:3602:3602:4874:3595:3595:4368:8209:8209:2822:2822:4874:4874:
Solution:
+-------+-------+-------+
| 6 1 5 | 4 2 8 | 9 7 3 |
| 4 7 9 | 6 5 3 | 2 8 1 |
| 3 8 2 | 1 9 7 | 4 6 5 |
+-------+-------+-------+
| 2 6 4 | 8 7 1 | 3 5 9 |
| 8 9 3 | 5 6 4 | 7 1 2 |
| 1 5 7 | 9 3 2 | 6 4 8 |
+-------+-------+-------+
| 9 4 6 | 2 1 5 | 8 3 7 |
| 5 3 8 | 7 4 9 | 1 2 6 |
| 7 2 1 | 3 8 6 | 5 9 4 |
+-------+-------+-------+
Quote:
HATMAN: This on popped out on my work for the next Human Solvable. It's quite neat (note the number of large cages) so I decided to post it.
SS gives it 0.73 - but that feels high to me.
It is to be done without pencil marks or 45 marking. You might even wish to do it at speed.
As they say enjoy.

Frank: Well I did it without pencil marks or 45 marking. I'm not too sure about the speed :)
But I did enjoy it. Thanx HATMAN

simon_blow_snow: For what it's worth here is a "complete walkthrough", listing all 81 cell placements

HATMAN: My start was similar. Your point on the newspaper vanilla is correct - one reason I like zero killers is that they keep me in some practice on the vanilla stuff.
I'm not getting very far with Human Solveable 9, my attempts keep switching from easy to impossible. But I've got a good cage framework so I will persevere.

Andrew (in November):
HATMAN wrote:
It is to be done without pencil marks or 45 marking. You might even wish to do it at speed.
As they say enjoy.
Thanks for a fun puzzle!
I tried to solve it taking the instructions literally, apart from innies of cages. When I came back to it this week, I still couldn't see how to do that so I used outies on R123 (or one can use outies on R789). Is it possible to solve this puzzle without using any 45s apart from innies of individual cages?
Even after placing the innies for the cages, I found it fairly hard going to complete the rest of the puzzle. Maybe others are better at spotting hidden singles and pairs than I am? ;)

HATMAN: I've made no restriction on multiple innies and outies - just that you shoul keep them in your head.

Andrew (in November): Here is how I'd solved this puzzle. I think it's in the spirit of Maurice's recent message. The comment after step 4 was written a few days ago, when I couldn't see any other way of making progress.
My solving path is fairly similar to Simon's one.

Walkthrough by simon_blow_snow:
For what it's worth here is a "complete walkthrough", listing all 81 cell placements:

Walkthrough and solution

Innies N6: R4C7<>[1]
Innies N3, R2 5(2): R3C7<>[1]
R3 8(3): R4C6=[1]
Innies N5: R6C4=[9]
Outies R123: R4C17=5={23}
Innies N6: R4C7+R6C9=11=[38]
C9 15(2), Innies N9: R7C9+R9C7=12=[75]
R9 11(2), Innies N8: R8C4+R9C6=13=[76]
R8 15(2), Innies N7: R78C3=14=[68]
R6 22(3), Innies N4: R4C1+R6C3=9=[27]
C1 5(2), Innies N1: R1C3+R3C1=8=[53]
R1 5(2), Innies N2: R1C4+R2C6=7=[43]
R2 5(2), Innies N3: R23C7=6=[24]
[3] of R1,N3 locked in N3 11(3)=[731]
Hidden singles R4,C7: R4C5=R5C7=[7]
Hidden single C6: R3C6=[7]
N1 11(3)<>[7] (no [3])
Hidden single N1,C1: R2C2=R9C1=[7]
N9 14(4)<>[9]
Hidden singles C7: R17C7=[98]
N9 14(4)<>[6] (with [8]):R68C7=[61]
Hidden single N9: R8C9=[6]
N3 naked single: R3C9=[5]
C4 13(2)={58}
Hidden single N2: R2C5=[5]
N1 11(3)<>[9]
Hidden single R2,N2: R2C3=R3C5=[9]
N1 26(4): R3C23=[82]
R3: R3C48=[16]
R2: R2C148=[468]
N7 14(3): R8C1+R9C2=[52]
C1: R1567C1=[6819]
N1,C3: R1C2=[1], R459C3=[431]
C4 13(2)=[85]
R4: R4C289=[659]
N9 19(3): R9C89=[94]
R5: R5C25689=[96412]
R6: R6C2568=[5324]
R19: R1C56=[28], R9C45=[38]
R7: R7C24568=[42153]
R8: R8C2568=[3492]

To be frank I don't like my start very much (there could be an easier opening elsewhere), but I did solve this puzzle on a plain notepad without any candidate listing etc. After the initial placements the rest feels like a normal newspaper ("pen and paper") sudoku puzzle.
Walkthrough by Andrew:
Here is how I'd solved this puzzle. I think it's in the spirit of Maurice's recent message. The comment after step 4 was written a few days ago, when I couldn't see any other way of making progress.

My solving path is fairly similar to Simon's one.

1. 45 rule on N3 2 innies R23C7 = 6, max R2C7 = 4 -> min R3C7 = 2

2. 45 rule on N6 2 innies R4C7 + R6C9 = 11, no 1 in R4C7 + R6C9

3. 8(3) cage at R3C7 must contain 1 -> R4C6 = 1

4. 45 rule on N5 1 remaining innie R6C4 = 9

[HATMAN suggested that this puzzle should be solved without using 45s, but I couldn’t see how to continue without using a simple 45, so …]

5. 45 rule on R123 2 remaining outies R4C17 = 5 = {23}

6. 45 rule on N6 2 innies R4C7 + R6C9 = 11 = [38] (because [29] blocked)
6a. R4C7 = 3, R4C1 = 2, R3C1 = 3, R6C9 = 8, R7C9 = 7, R3C7 = 4 (cage sum), R2C7 = 2 (step 1), R2C6 = 3

7. 45 rule on N1 1 remaining innie R1C3 = 5, R1C4 = 4

8. 45 rule on N9 1 remaining innie R9C7 = 5, R9C6 = 6

9. 45 rule on N8 1 remaining innie R8C4 = 7, R8C3 = 8

10. 45 rule on N7 1 remaining innie R7C3 = 6, R6C3 = 7 (cage sum)

11. 3 in N3 must be in 11(3) cage = {137} (only remaining combination) -> R1C8 = 7, R12C9 = [31]

12. R5C7 = 7 (hidden single in C7), R4C5 = 7 (hidden single in R4), R2C6 = 7 (hidden single in C6)

13. 7 in N1 cannot be in 11(3) cage -> R2C2 = 7 (hidden single in N1), R9C1 = 7 (hidden single in N7)

14. 9 in N9 must be in 19(3) cage -> R1C7 = 9 (hidden single in C7), R7C7 = 8 (hidden single in C7)
14a. 13(2) cage at R4C4 must be {58} -> R9C5 = 8 (hidden single in N8)

15. 19(3) cage at R8C9 must be {469} -> R8C9 = 6, R6C7 = 6, then R8C7 = 1 (hidden singles in C7)

16. 28(4) cage at R1C7 = {5689} -> R3C9 = 5 (because 6,8 already placed for C9)

17. R5C9 = 2 (hidden single in C9, because not in 19(3) cage at R8C9)

18. R1C6 = 8 (hidden single in C6)

19. R5C5 = 6 (hidden single in N5), R4C2 = 6 (hidden single in R4)

20. 26(4) cage at R2C2 must be {2789} (because {4679} blocked) -> R3C2 = 8, R23C3 = [92]

21. 11(3) cage at R1C1 = {146} -> R2C1 = 4, R1C12 = [61], R1C5 = 2

22. R2C5 = 5, R2C8 = 8 (hidden singles in R2), R2C4 = 6, R3C8 = 6, R3C45 = [19]

23. 14(3) cage at R8C1 can only be {257} (because {347} blocked in C1) -> R8C1 = 5, R9C2 = 2, R7C4 = 2 (hidden single in C4), R9C4 = 3

24. R5C3 = 3, R9C3 = 1 (hidden singles in C3), R4C3 = 4

25. 17(3) cage at R7C1 = {349} -> R7C1 = 9, R5C1 = 8 (hidden single in C1), R6C1 = 1, R45C4 = [85], R56C2 = [95] (hidden singles in N4)

26. R4C8 = 5 (hidden single in R4), R4C9 = 9, R9C9 = 4, R9C8 = 9

27. R78C8 = [32] (hidden singles in N9)

28. R78C2 = [43] (hidden singles in N7)

29. R5C8 = 1 (hidden single in N6), R6C8 = 4, R5C6 = 4

30. R6C56 = [32] (hidden singles in R6)

31. R7C5 = 1 (hidden single in R7), R7C6 = 5

32. R8C5 = 4 (hidden single in R8), R8C6 = 9

Rating Comment. Even though I found this harder than Paper Solvable 9 Elevenses, because I found it hard to find some hidden singles, I'll rate Paper Solvable 8 at 0.5 based on Mike's original definitions.


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PostPosted: Thu Oct 30, 2014 2:38 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Human Solvable 9 by HATMAN (October 2011) here
Puzzle Diagram:
Image
Code: Select, Copy & Paste into solver:
3x3::k:5135:4627:1537:1537:8203:1794:1794:5134:5134:5135:5135:4627:8203:8203:8203:4884:5134:4884:2056:4627:4627:23:8203:24:4884:4884:2051:2056:7434:25:26:5645:27:28:7945:2051:7434:7434:7434:5645:5645:5645:7945:7945:7945:1799:7434:4113:4113:5645:29:30:7945:1796:1799:5394:5394:4113:7948:31:4374:4374:1796:5394:4624:5394:7948:7948:7948:4374:4374:5141:4624:4624:2054:2054:7948:1541:1541:5141:5141:
Solution:
+-------+-------+-------+
| 7 2 1 | 5 6 3 | 4 9 8 |
| 8 5 4 | 9 2 7 | 6 3 1 |
| 6 3 9 | 1 8 4 | 7 5 2 |
+-------+-------+-------+
| 2 7 5 | 4 1 9 | 3 8 6 |
| 1 4 8 | 3 7 6 | 9 2 5 |
| 3 9 6 | 8 5 2 | 1 7 4 |
+-------+-------+-------+
| 4 6 7 | 2 9 5 | 8 1 3 |
| 5 1 3 | 7 4 8 | 2 6 9 |
| 9 8 2 | 6 3 1 | 5 4 7 |
+-------+-------+-------+
Quote:
HATMAN: I've had the general layout for a while but it has taken me three weeks+ to actually get a workable puzzle.
SS gives it 3.57 and JS goes around the houses.
I've solved it of course, and I'm interested in seeing your solutions.

Andrew: Thanks HATMAN for your latest puzzle.
What have I done wrong?! It didn't feel like a Human Solvable to me, more like a very early Assassin. As the most methodical regular solver on this forum I started with methodical steps, expecting to have to use innovative ones later, but never reached that stage.
Maybe Simon or HATMAN will show us how to use "Human Solvable" steps to give a quicker solution, not that my solving path was very long?
Thanks Maurice (see post below) for pointing out the typo in one of my original steps which, inadvertently, led to a quick solution. I've deleted my rating comment and the steps after my error, and also added a sub-step which I missed earlier.
I've now had another try at this puzzle; see my later post in this thread which has my complete walkthrough.

(Archive Note): Since the faulty start wasn't deleted from the puzzle thread, it's been included in this archive entry as an example of what can happen when a careless mistake is made.

HATMAN: You are undoubtedly right and I have missed something.
Did you check the code in SudokuSolver, in case I made some mistake there?
I'll go over your walkthrough at the weekend.

HATMAN: I had a quick look at your walkthrough and your solution goes wrong at point 9.
You have:
9. 45 rule on N9 2 innies R7C9 + R9C7 = 8 = [35/61], R7C9 = {36}, R9C7 = {15}, R6C9 = {14}, R9C6 = {15}
It should be:
9. 45 rule on N9 2 innies R7C9 + R9C7 = 8 = [35/62], R7C9 = {36}, R9C7 = {25}, R6C9 = {14}, R9C6 = {15}
This makes the deduction in 9a wrong and hopefully reclaims the puzzle as HS.
Although the deduction is wrong the answer is right so the puzzle solution flows on correctly from here.

simon_blow_snow: The highlight of my walkthrough is step 7, which I don't know how to name the technique. :?:
Definitely a human solvable. Thanks HATMAN for a nice puzzle! ;-)

Andrew: Thanks Maurice for a challenging puzzle. Yes, it probably is a Human Solvable rather than a hard Assassin.
Hope I've got it right this time. :pray: (I was looking for a "fingers crossed" smilie but there isn't one).
My solving path is fairly similar to Simon's but I saw several steps in different ways, including the key breakthrough step.

Ed:
HATMAN wrote:
it has taken me three weeks+ to actually get a workable puzzle
Thanks for your efforts to make this very nice puzzle. I did it differently to the others. Managed to get it out quicker than most Assassins and some fun moves on the way. I started the same way so have pinched parts of Andrew's walk-through to save a bit of time. Hope he doesn't mind. My cracking step 12 is the only one that's any different.

HATMAN: Well done Guys - three different ways to break the ring.
My breakout was much less elegant:
Assuming 20(3) at r1c8 = {479} either:
gives a conflict on R1, N1; or
breaks uniqueness with r1c89 = {79} and r9c89 = {79}.

Both were unsatisfactory but I posted it as I was confident that there was a better solution and that you would find it.

Andrew's faulty start:
Thanks HATMAN for your latest puzzle.

What have I done wrong?! It didn't feel like a Human Solvable to me, more like a very early Assassin. As the most methodical regular solver on this forum I started with methodical steps, expecting to have to use innovative ones later, but never reached that stage.

Maybe Simon or HATMAN will show us how to use "Human Solvable" steps to give a quicker solution, not that my solving path was very long?

Thanks Maurice (see post below) for pointing out the typo in one of my original steps which, inadvertently, led to a quick solution. I've deleted my rating comment and the steps after my error, and also added a sub-step which I missed earlier.

I've now had another try at this puzzle; see my later post in this thread which has my complete walkthrough.


Here is start to my walkthrough for Human Solvable 9

Prelims

a) R1C34 = {15/24}
b) R1C67 = {16/25/34}, no 7,8,9
c) R34C1 = {17/26/35}, no 4,8,9
d) R34C9 = {17/26/35}, no 4,8,9
e) R67C1 = {16/25/34}, no 7,8,9
f) R67C9 = {16/25/34}, no 7,8,9
g) R9C34 = {17/26/35}, no 4,8,9
h) R9C67 = {15/24}
i) 20(3) cage at R1C1 = {389/479/569/578}, no 1,2
j) 20(3) cage at R1C8 = {389/479/569/578}, no 1,2
k) 20(3) cage at R8C9 = {389/479/569/578}, no 1,2

1. R1C67 = {16/34} (cannot be {25} which clashes with R1C34), no 2,5
1a. Killer pair 1,4 in R1C34 and R1C67, locked for R1

2. 45 rule on R123 4 innies R3C1469 = 13= {1237/1246/1345}, no 8,9 in R3C46, 1 locked for R3

3. 45 rule on R789 4 innies R7C1469 = 14, no 9 in R7C46

4. 45 rule on C123 4 innies R1469C3 = 14, no 9 in R46C3

5. 45 rule on C789 4 innies R1469C7 = 13 = {1237/1246/1345}, no 8,9 in R46C7, 1 locked for C7

6. 45 rule on N1 2 innies R1C3 + R3C1 = 7 = [16/25/43/52], no 1,7 in R3C1, clean-up: no 1,7 in R4C1
6a. R67C1 = {16/34} (cannot be {25} which clashes with R34C1), no 2,5
6b. Killer pair 3,6 in R34C1 and R67C1, locked for C1
6c. 1 in N1 only in R12C3, locked for C3

7. 45 rule on N3 2 innies R1C7 + R3C9 = 6 = [15/42], R1C7 = {14}, R3C9 = {25}, R1C6 = {36}, R4C9 = {36}
7a. R67C9 = {16/34} (cannot be {25} which clashes with R3C9), no 2,5
7b. Killer pair 3,6 in R4C9 + R67C9, locked for C9
7c. 45 rule on N3 2 outies R1C6 + R4C9 = 9 = {36}, CPE no 3,6 in R4C6
7d. 1 in R3 only in R3C46, locked for N2, clean-up: no 5 in R1C3, no 2 in R3C1 (step 6), no 6 in R4C1

8. 45 rule on N7 2 innies R7C1 + R9C3 = 6 = [15/42], R7C1 = {14}, R9C3 = {25}, R6C1 = {36}, R9C4 = {36}
8a. Killer pair 2,5 in R9C3 and R9C67, locked for R9
8b. 45 rule on N7 2 outies R6C1 + R9C4 = 9 = {36}, CPE no 3,6 in R6C4

9. 45 rule on N9 2 innies R7C9 + R9C7 = 8 = [35/61], R7C9 = {36}, R9C7 = {15}, R6C9 = {14}, R9C6 = {15}
As Maurice has pointed out this should have been
45 rule on N9 2 innies R7C9 + R9C7 = 8 = [35/62], R7C9 = {36}, R9C7 = {25}, R6C9 = {14}, R9C6 = {14}
Then my original step 9a was no longer valid.

I'll add one more sub-step now
9a. 45 rule on N9 2 outies R6C9 + R9C6 = 5 = {14}, CPE no 1,4 in R6C6 + R9C9
and see whether I can solve the rest of the puzzle.
Walkthrough by simon_blow_snow:
The highlight of my walkthrough is step 7, which I don't know how to name the technique. :?:

Typo fixed, thx Andrew.

My complete walkthrough (8 steps)

Step 1:
6(2) R1,R9 = {15/24} has [2/5]
--> 7(2) R1 <> {25}, = {16/34}
Innies N3: R1C7+R3C9 = 6 = [15/42]
--> 7(2) R1 = [34/61]
--> 8(2) C9 = [26/53] has [2/5]
--> 7(2) C9 <> {25}, = {16/34}
Innies N9: R7C9+R9C7 = 8 = [35/62]
--> R47C9 = {36} (C9)
--> 6(2) R9 = [15/42]

Step 2:
Innies N1: R1C3+R3C1 = 7 = [16/25/43/52]
--> 8(2) C1 = {26/35} has [2/5]
--> 7(2) C1 <> {25}, = {16/34}
Innies N7: R7C1+R9C3 = 6 = [15/42]
--> R9C37 = {25} (R9)
--> 7(2) C1 = [34/61]
--> 8(2) R9 = [26/53]

Step 3:
Innie-outies N1: R3C1 - R1C4 = 1
--> R1C4+R3C1 = [12/23/45/56]
Innies N3: R1C7+R3C9 = [15/42]
--> R1C4+R3C1 <> [12/45], = [23/56]
--> 6(2) R1 = [15/42]
--> R1C37 = {14} (R1)
--> R36C1 = {36} (C1)

Step 4:
6(2),7(2) R1: R1C46 = 6 + 7 - 1 - 4 = 8 = [26/53] has [2/3]
Innies N2: R3C46 = 5 = {14} (R3,N2)
Innies N12: R1C6+R3C1 = 9 = {36} (R1C123+R3C456)
Innies N23: R1C4+R3C9 = 7 = {25} (R1C789+R3C456)
8(2),6(2) R9: R9C46 = 8 + 6 - 2 - 5 = 7 = [34/61]
Innies N8: R7C46 = 7 = {16/25/34}
Innies N78: R7C1+R9C6 = 5 = {14} (R7C456+R9C123)
Innies N89: R7C9+R9C4 = 9 = {36} (R7C456+R9C789)
Outies N7: R6C1+R9C4 = 9 = {36} (R6C4)
R47C9, outies N7: R4C9+R6C1 = 9 + 9 - 3 - 6 = 9 = {36} (R4C123+R6C789)

Step 5:
8(2) C9, innies N9: R3C9+R9C7 = 8 + 8 - 3 - 6 = 7 = {25} (R123C7+R789C9)
--> 20(3) N9 = {479} (N9)
Outies N9: R6C9+R9C6 = 5 = {14} (R6C6+R9C9)
Generic X-Wing: N9 & Outies N9 each has a [4] in R68C9+R9C68
--> [4] locked in R68C9 (C9)
--> [4] locked in R9C68 (R9)

Step 6:
7(2) C9 = [16/43]
--> R46C9 = [31/64]
Innies N6: R46C79 = 14 = [6341/3614]
(Innies N9: R7C9+R9C7 = [35/62] --> R46C79 can't be [2381/8321])
--> R4C79 = {36} (R4,N6)
--> R6C79 = {14} (R6,N6)
--> R16C7 = {14} (C7)

Step 7:
R4C7 has no {479}
--> R456C89 has at least one of {479}
--> 20(3) N3 & 20(3) N9 can't be both {479}
--> 20(3) N3 <> {4}
--> Hidden singles N3,C8: R1C7 = R9C8 = [4]
--> R1C346 = [153], R46C79 = [3614], R9C3467 = [2615]
--> R3467C1 = [6234], R3C469 = [142], R7C469 = [253]
Hidden singles C9: R125C9 = [815]
19(4) N3: R2C7 = [6], R3C78 = [75/93]
--> R78C7 = [82]

Step 8:
16(3) R6: R6C34 = 14 <> [7]
Innies N4: R46C3 = 11 = [56]
--> 16(3) R6: R6C4 = [8]
Innies N5: R4C46+R6C6 = 15 = [492]
Hidden singles N6: R45C8 = [82]
Hidden triple N7: R7C2+R8C12 = {156}
21(4) N7: R78C3 without {56} can't sum to 14
--> R7C2+R8C1 can't be [61]
--> R8C1 = [5]
Hidden single C1: R5C1 = [1]
--> R46C2 = [79]
--> R16C8 = [97]
--> 19(4) N3: R3C78 = [75]
Hidden single N1: R2C2 = [5]
20(3) N1: R12C1 = 15 = [78]
--> R39C2 = [38]
18(3) N7: R8C2+R9C1 = 10 = [19]

Naked singles to finish.

Definitely a human solvable. Thanks HATMAN for a nice puzzle! ;-)
Re-worked walkthrough by Andrew:
Thanks Maurice for a challenging puzzle. Yes, it probably is a Human Solvable rather than a hard Assassin.

Hope I've got it right this time. :pray: (I was looking for a "fingers crossed" smilie but there isn't one).

My solving path is fairly similar to Simon's but I saw several steps in different ways, including the key breakthrough step.

Here is my re-worked walkthrough for Human Solvable 9

Prelims

a) R1C34 = {15/24}
b) R1C67 = {16/25/34}, no 7,8,9
c) R34C1 = {17/26/35}, no 4,8,9
d) R34C9 = {17/26/35}, no 4,8,9
e) R67C1 = {16/25/34}, no 7,8,9
f) R67C9 = {16/25/34}, no 7,8,9
g) R9C34 = {17/26/35}, no 4,8,9
h) R9C67 = {15/24}
i) 20(3) cage at R1C1 = {389/479/569/578}, no 1,2
j) 20(3) cage at R1C8 = {389/479/569/578}, no 1,2
k) 20(3) cage at R8C9 = {389/479/569/578}, no 1,2

1. R1C67 = {16/34} (cannot be {25} which clashes with R1C34), no 2,5
1a. Killer pair 1,4 in R1C34 and R1C67, locked for R1

2. 45 rule on R123 4 innies R3C1469 = 13 = {1237/1246/1345}, no 8,9 in R3C46, 1 locked for R3

3. 45 rule on R789 4 innies R7C1469 = 14, no 9 in R7C46

4. 45 rule on C123 4 innies R1469C3 = 14, no 9 in R46C3

5. 45 rule on C789 4 innies R1469C7 = 13 = {1237/1246/1345}, no 8,9 in R46C7, 1 locked for C7

6. 45 rule on N1 2 innies R1C3 + R3C1 = 7 = [16/25/43/52], no 1,7 in R3C1, clean-up: no 1,7 in R4C1
6a. R67C1 = {16/34} (cannot be {25} which clashes with R34C1), no 2,5
6b. Killer pair 3,6 in R34C1 and R67C1, locked for C1
6c. 1 in N1 only in R12C3, locked for C3, clean-up: no 7 in R9C4

7. 45 rule on N3 2 innies R1C7 + R3C9 = 6 = [15/42], R1C7 = {14}, R3C9 = {25}, R1C6 = {36}, R4C9 = {36}
7a. R67C9 = {16/34} (cannot be {25} which clashes with R3C9), no 2,5
7b. Killer pair 3,6 in R4C9 + R67C9, locked for C9
7c. 45 rule on N3 2(1+1) outies R1C6 + R4C9 = 9 = {36}, CPE no 3,6 in R4C6
7d. 1 in R3 only in R3C46, locked for N2, clean-up: no 5 in R1C3, no 2 in R3C1 (step 6), no 6 in R4C1

8. 45 rule on N7 2 innies R7C1 + R9C3 = 6 = [15/42], R7C1 = {14}, R9C3 = {25}, R6C1 = {36}, R9C4 = {36}
8a. Killer pair 2,5 in R9C3 and R9C67, locked for R9
8b. 45 rule on N7 2(1+1) outies R6C1 + R9C4 = 9 = {36}, CPE no 3,6 in R6C4

9. 45 rule on N9 2 innies R7C9 + R9C7 = 8 = [35/62], R7C9 = {36}, R9C7 = {25}, R6C9 = {14}, R9C6 = {14}
[I originally made a typo in this step, giving R7C9 + R9C7 = [35/61] which inadvertently led to the correct solution very quickly. I’ve re-worked from here.]
9a. 45 rule on N9 2(1+1) outies R6C9 + R9C6 = 5 = {14}, CPE no 1,4 in R6C6 + R9C9
9b. 20(3) cage at R8C9 = {389/479/578} (cannot be {569} which clashes with R7C9 + R9C7), no 6

10. 45 rule on N8 2 innies R7C46 = 2 outies R9C37
10a. R9C37 = {25} = 7 -> R7C46 = 7 = {16/25/34}, no 7,8

11. 45 rule on N6 4 innies R46C79 = 14 = {1256/1346/2345} (cannot be {1247} because R4C9 only contains 3,6), no 7
11a. R1469C7 (step 5) = {1246/1345}, 4 locked for C7
11b. R9C7 = {25} -> no 2,5 in R46C7
11c. Naked quad {1346} in R46C79, locked for N6

12. R3C1469 (step 2) = {1237/1246/1345}
12a. R3C9 = {25} -> no 2,5 in R3C146, clean-up: no 2 in R1C3 (step 6), no 4 in R1C4, no 3 in R4C1
12b. R3C1 = {36} -> no 3,6 in R3C46
12c. 45 rule on N1 2(1+1) outies R1C4 + R4C1 = 7 = {25}, CPE no 2,5 in R1C1 + R4C4

13. 45 rule on N78, using R7C46 = 7 (step 10a), 2(1+1) remaining innies R7C1 + R9C6 = 5 = {14}, CPE no 1,4 in R7C456 + R9C12, clean-up: no 3,6 in R7C46 (step 10a)
13a. Naked pair {25} in R7C46, locked for R7 and N8
13b. Naked pair {25} in R17C4, locked for C4
13c. 1 in R9 only in R9C56, locked for N8

14. 45 rule on N89, using R7C46 = 7 (step 10a), 2(1+1) remaining innies R7C9 + R9C4 = 9 = {36}, CPE no 3,6 in R7C5 + R9C8
14a. 20(3) cage at R8C9 (step 9b) = {479/578}, 7 locked for N9
14b. 5 of {578} must be in R8C9 -> no 8 in R8C9

15. 45 rule on N4 4 innies R46C13 = 16 = {2347/2356}, no 8, 2,3 locked for N4
15a. R1469C3 = 14 (step 4) = {1247/1256/2345} (cannot be {1346} because R9C3 only contains 2,5), 2 locked for C3
15b. 2 in N1 only in R13C2, locked for C2
15c. 18(4) cage at R1C2 contains 2 = {1269/1278/2349/2358/2457} (cannot be {2367} which clashes with R3C1)
15d. 1,4 of {1269/1278/2349} must be in R2C3 -> no 6,7,9 in R2C3

16. 45 rule on N2 2 outies R1C37 = 2 innies R3C46
16a. R1C37 = {14} = 5 -> R3C46 = 5 = {14}, locked for R3 and N2
17b. Naked pair {14} in R39C6, locked for C6

18. 16(3) cage at R6C3 = {259/268/358/457} (cannot be {169/178/349/367} because R7C4 only contains 2,5), no 1

19. 18(3) cage at R8C2 = {189/369/378/468/567} (cannot be {459} which clashes with R7C1 + R9C3)
19a. Min R9C12 = 10 -> max R8C2 = 8
19b. Min R12C1 = 12 -> max R2C2 = 8

20. By chaining through R1C3 + R3C1, R36C1 and R67C1, R1C3 + R7C1 = naked pair {14}, CPE no 4 in R2C1 + R78C3

21. 17(4) cage at R7C7 contains 1 = {1268/1349/1358} (cannot be {1259} which clashes with R9C7)
21a. 1,4 of {1349} must be in R78C8 -> no 9 in R78C8

22. 45 rule on N12, using R3C46 = 5 (step 16a), 2(1+1) remaining innies R1C6 + R3C1 = 9 = {36}, CPE no 3,6 in R1C2 + R3C5

23. 45 rule on N23, using R3C46 = 5 (step 16a), 2(1+1) remaining innies R1C4 + R3C9 = 7 = {25}, CPE no 2,5 in R1C89 + R3C5

24. 20(3) cage at R1C8 = {389/479/569/578}
24a. 5 of {569/578} must be in R2C8 -> no 6,7 in R2C8

[I ought to have spotted the next step immediately after step 20.]
25. By chaining through R34C9, R47C9 and R7C9 + R9C7, R3C9 + R9C7 = naked pair {25}, CPE no 2,5 in R23C7 + R8C9

26. 20(3) cage at R8C9 (step 14a) = {479} (only remaining combination), locked for N9
26a. 4 in R7 only in R7C12, locked for N7

27. Caged X-Wing for 4 in R6C9 + R9C6 (step 9a) and 20(3) cage at R8C9, locked for R9 and C9, no 4 in R2C9 + R9C5

28. Hidden killer pair 7,9 in R23C7 and R5C7 for C7, R23C7 can only contain one of 7,9 (19(4) cage at R2C7 cannot be {1279} which clashes with 20(3) cage at R1C8) -> R23C7 must contain one of 7,9 and R5C7 = {79}
28a. 19(4) cage at R2C7 contains one of 7,9 in R23C7 -> no 7,9 in R2C9 + R3C8

29. 20(3) cage at R1C8 = {389/569/578} (cannot be {479} which clashes with 19(4) cage at R2C7), no 4

[I think the puzzle is, at last, cracked. Now I can place naked singles around much of the outer ring, which appears to be the key area of this puzzle.]

30. R1C7 = 4 (hidden single in N3), R1C6 = 3, R1C3 = 1, R1C4 = 5, R7C46 = [25], R3C1 = 6 (step 6), R4C1 = 2, R6C1 = 3, R7C1 = 4, R3C9 = 2 (step 7), R4C9 = 6, R46C7 = [31], R67C9 = [43], R9C7 = 5 (step 9), R9C6 = 1, R9C3 = 2, R9C4 = 6, R3C46 = [14]
30a. R1C2 = 2 (hidden single in N1)

31. R7C4 = 2 -> R6C34 = 14 = [59/68]
31a. R46C13 (step 15) = {2356} (only remaining combination, cannot be {2347} because 4,7 only in R4C3) -> R46C3 = [56], R6C4 = 8

32. 45 rule on N5 3 remaining innies R4C46 + R6C6 = 15 = {249} (only remaining combination) -> R6C6 = 2, R4C46 = [49]

33. R8C5 = 4 (hidden single in C5), R9C8 = 4 (hidden single in R9)
33a. Naked pair {79} in R89C9, locked for C9 -> R1C9 = 8, R5C9 = 5, R2C9 = 1

34. R1C9 = 8 -> R12C8 = 12 = [75/93]

35. R2C7 = 6 (hidden single in N3), R78C7 = [82]
35a. R5C6 = 6 (hidden single in C6), R125C5 = [625] (hidden singles in C5), R56C8 = [82] (hidden singles in C8)
35b. R7C28 = {16} (hidden pair in R7)

36. 18(3) cage at R8C2 (step 19) = {189/369/378} (cannot be {567} because 5,6 only in R8C2), no 5

37. R8C1 = 5 (hidden single in N7)
37a. R5C1 = 1 (hidden single in C1), R46C2 = [79], R6C8 = 7, R1C8 = 9, R2C8 = 3 (step 34), R1C1 = 7
37b. R2C23 = [54] (hidden pair in R2), R1C2 = 8 (cage sum)

38. R3C2 = 3, R9C12 = [98], R8C2 = 1 (step 36)

and the rest is naked singles.

It was interesting how many cells quickly reduced to {14}, {25} or {36} and how many of them led to naked pairs (some harder to prove than others) and CPEs.

It seems to be a fairly common feature of Maurice's puzzles that cages around the outside ring are critical. In this case the problem was how to fix them after quickly getting the 2-cell cages down to pairs of options.

One step I particularly enjoyed using, even though it didn't seem to be a particularly important one, was the X-Wing in step 27; one doesn't often come across that type of X-Wing making eliminations at right-angles, I think I've only used it once before.

Rating Comment. It's hard to rate my walkthrough for Human Solvable 9. My best guess is 1.5 Some steps may be difficult for software solvers but I think they are no higher than the 1.5 range.
Walkthrough by Ed:
HATMAN wrote:
it has taken me three weeks+ to actually get a workable puzzle
Thanks for your efforts to make this very nice puzzle. I did it differently to the others. Managed to get it out quicker than most Assassins and some fun moves on the way. I started the same way so have pinched parts of Andrew's walk-through to save a bit of time. Hope he doesn't mind. My cracking step 12 is the only one that's any different. [edit: thanks to Andrew for noticing a typo]

HS9 12 steps

Prelims

a) R1C34 = {15/24}
b) R1C67 = {16/25/34}, no 7,8,9
c) R34C1 = {17/26/35}, no 4,8,9
d) R34C9 = {17/26/35}, no 4,8,9
e) R67C1 = {16/25/34}, no 7,8,9
f) R67C9 = {16/25/34}, no 7,8,9
g) R9C34 = {17/26/35}, no 4,8,9
h) R9C67 = {15/24}
i) 20(3) cage at R1C1 = {389/479/569/578}, no 1,2
j) 20(3) cage at R1C8 = {389/479/569/578}, no 1,2
k) 20(3) cage at R8C9 = {389/479/569/578}, no 1,2
l) 32(4)n2: no 1

1. R1C67 = {16/34} (cannot be {25} which clashes with R1C34), no 2,5

2. 45 rule on N1 2 innies R1C3 + R3C1 = 7 = [16/25/43/52], no 1,7 in R3C1, clean-up: no 1,7 in R4C1
2a. R67C1 = {16/34} (cannot be {25} which clashes with R34C1), no 2,5

3. 45 rule on N3 2 innies R1C7 + R3C9 = 6 = [15/42], R1C7 = {14}, R3C9 = {25}, R1C6 = {36}, R4C9 = {36}
3a. R67C9 = {16/34} (cannot be {25} which clashes with R3C9), no 2,5
3b. Killer pair 3,6 in R4C9 + R67C9, locked for C9

4. 45 rule on N7 2 innies R7C1 + R9C3 = 6 = [15/42], R7C1 = {14}, R9C3 = {25}, R6C1 = {36}, R9C4 = {36}

5. 45 rule on N9 2 innies R7C9 + R9C7 = 8 = [35/62], R7C9 = {36}, R9C7 = {25}, R6C9 = {14}, R9C6 = {14}
5a. 45 rule on N9 2(1+1) outies R6C9 + R9C6 = 5 = {14}, CPE no 1,4 in R6C6 + R9C9

6. naked pair {25} in r9c37: both locked for r9

7. 45 rule on N8 2 innies R7C46 = 2 outies R9C37
7a. R9C37 = {25} = 7 -> R7C46 = 7 no 7,8,9

8. 45 rule on N78, using R7C46 = 7 (step 7a), 2(1+1) remaining innies R7C1 + R9C6 = 5 = {14}, CPE no 1,4 in R7C456 + R9C12
8a. -> h7(2)r7c46 = {25} only, both locked for R7 and N8
8b. 1 in R9 only in R9C56, locked for N8

9. 45 rule on N89, using R7C46 = 7 (step 7a), 2(1+1) remaining innies R7C9 + R9C4 = 9 = {36}, CPE no 3,6 in R7C5 + R9C8

10. if 2 in r9c7 -> 6 in r7c9 (h8(2)r7c9+r9c7) -> 8(2)r3c9 = [53] (ie, 5 in r3c9)
10a. or 5 in r9c7
10c. CPE: no 5 in r23c7 nor r8c9

11. 20(3)n9 = {479} only: all locked for n9

12. 1 in r9 only in r9c56:
12a. "45" on r9: 1 innie r9c5 + 7 = 2 outies r8c29
12b. if 1 in r9c5 -> r8c29 = 8 = [17] only (ie, no 4 in r8c9)
12c. if 1 in r9c6 -> r6c9 = 4 (outies n9 = 5) (ie no 4 in r8c9)
12d. -> no 4 in r8c9
12e. r9c8 = 4 (hsingle n9)

cracked


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PostPosted: Thu Oct 30, 2014 7:39 pm 
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Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Assassin 222 by Ed (October 2011) here
Puzzle Diagram:
Image
Code: Select, Copy & Paste into solver:
3x3::k:6656:1025:1025:2562:2562:5379:6660:6660:6660:6656:6656:5637:5637:5637:5379:5379:5379:6660:6656:3590:3590:2567:5637:3080:3080:5379:5379:6656:3590:5385:2567:3850:3850:5387:3340:3340:3597:3597:5385:5385:5902:5902:5387:5903:5903:5648:4113:5385:2578:5902:5387:5387:5387:5903:5648:4113:4113:2578:5902:5902:2835:5903:5903:5648:5648:4372:4372:2837:4118:2835:2071:2328:5648:2585:2585:4372:2837:4118:4118:2071:2328:
Solution:
+-------+-------+-------+
| 5 1 3 | 6 4 2 | 8 9 7 |
| 4 8 9 | 1 7 3 | 5 6 2 |
| 6 7 2 | 8 5 9 | 3 1 4 |
+-------+-------+-------+
| 3 5 1 | 2 8 7 | 6 4 9 |
| 8 6 4 | 9 1 5 | 2 7 3 |
| 9 2 7 | 3 6 4 | 1 8 5 |
+-------+-------+-------+
| 1 9 5 | 7 3 8 | 4 2 6 |
| 2 3 8 | 4 9 6 | 7 5 1 |
| 7 4 6 | 5 2 1 | 9 3 8 |
+-------+-------+-------+
Quote:
Ed: This was originally going to 221 but found a (minor) flaw in my solution at the last minute. Many thanks to manu for posting his instead.
I found this A222 a really enjoyable puzzle to solve. Found some advanced moves very early which got me hooked. Worked pretty much in the one area with the rest of the solution (18 steps) . However, many other spots are tempting so there must be other ways.

Andrew: Many thanks Ed for another challenging Assassin!
Your introduction suggests that you found more interesting steps than I did, although I found some; I particularly enjoyed my step 14a.
I'll be interested to see how Ed solved it and whether any other forum members found a better solving path than I did.

Ed: Thanks for your WT Andrew. Step 14a is really nice. We worked in the same key areas but differently because of my steps 5&6. I've included the link for some more examples of that technique in the techniques forum. Andrew found something that would have made my solution shorter by half as I comment on after step 8. I often miss blocks.

Walkthrough by Andrew:
Many thanks Ed for another challenging Assassin!

Your introduction suggests that you found more interesting steps than I did, although I found some; I particularly enjoyed my step 14a.

I'll be interested to see how Ed solved it and whether any other forum members found a better solving path than I did.

Here is my walkthrough for A222
Thanks Ed for your comments. I've corrected a typo and added explanation to step 14f.

Prelims

a) R1C23 = {13}
b) R1C45 = {19/28/37/46}, no 5
d) R34C4 = {19/28/37/46}, no 5
e) R3C67 = {39/48/57}, no 1,2,6
f) R4C56 = {69/78}
g) R4C89 = {49/58/67}, no 1,2,3
h) R5C12 = {59/68}
i) R67C4 = {19/28/37/46}, no 5
j) R78C7 = {29/38/47/56}, no 1
k) R89C5 = {29/38/47/56}, no 1
l) R89C8 = {17/26/35}, no 4,8,9
m) R89C9 = {18/27/36/45}, no 9
n) R9C23 = {19/28/37/46}, no 5
o) 26(4) cage at R1C7 = {2789/3689/4589/4679/5678}, no 1
p) 21(6) cage at R1C6 = {123456}, no 7,8,9

Steps resulting from Prelims
1a. Naked pair {13} in R1C23, locked for R1 and N1, clean-up: no 7,9 in R1C45
1b. R4C89 = {49/58} (cannot be {67} which clashes with R4C56), no 6,7
1c. Killer pair 8,9 in R4C56 and R4C89, locked for R4, clean-up: no 1,2 in R3C4
1d. 1 in N3 only in R2C78 + R3C89, locked for 21(6) cage at R1C6, no 1 in R2C6

2. 45 rule on N3 3 outies R123C6 = 14 = {239/257/347/356} (cannot be {248} which clashes with R1C45), no 8, clean-up: no 4 in R3C7
2a. 7 of {347} must be in R3C6 -> no 4 in R3C6, clean-up: no 8 in R3C7

3. 45 rule on R1234 2 innies R4C37 = 7 = {16/25/34}, no 7

4. 45 rule on R123 3 outies R4C124 = 10 = {127/136/235} (cannot be {145} which clashes with R4C89), no 4, clean-up: no 6 in R3C4

5. 45 rule on C6789 3 innies R457C6 = 20 = {389/479/569/578}, no 1,2

6. 45 rule on N23 2(1+1) outies R2C3 + R4C4 = 11 = [47/56/83/92], no 2,6,7 in R2C3, no 1 in R4C4, clean-up: no 9 in R3C4

7. 1 in C6 only in R689C6
7a. 45 rule on N69 3 outies R689C6 = 11 = {128/146} (cannot be {137} which clashes with R123C6), no 3,5,7,9

8. 45 rule on N69 1 innie R9C7 = 1 outie R6C6 + 5, R6C6 = {124}, R9C7 = {679}
8a. 16(3) cage at R8C6 = {169/178/268}, no 4

9. 1 in N2 must be in 22(4) cage at R2C3 = {1489/1579/1678}, no 2,3

10. 8 in N3 must be in 26(4) cage at R1C7 = {2789/4589/5678} (cannot be {3689} which clashes with R1C45), no 3
10a. 5 of {4589/5678} must be in R1C789 (R1C789 cannot be {489/568} which clash with R1C45), no 5 in R2C9
10b. 3 in R2 only in R2C678, locked for 21(6) cage at R1C6, no 3 in R3C89

11. 1 in N2 only in R2C45 + R3C5
11a. 45 rule on N23 4 innies R23C45 = 21 = {1389/1479/1578} (cannot be {1569} because R3C4 only contains 3,4,7,8), no 6
11b. R123C6 (step 2) = {239/257/356} (cannot be {347} which clashes with R23C45), no 4
11c. Killer pair 2,6 in R123C6 and R689C6, locked for C6, clean-up: no 9 in R4C5

12. 4 in 21(6) cage at R1C6 only in R2C78 + R3C89, locked for N3
12a. 26(4) cage at R1C7 (step 10) = {2789/5678}, 7 locked for N3, clean-up: no 5 in R3C6

13. R457C6 = 20 (step 5)
13a. R45C6 cannot total 15 (because of CCC clash with R4C56) -> no 5 in R7C6

14. 45 rule on C1234 3(1+2) innies R1C4 + R2C34 (or R2C3 + R12C4) = 16
14a. R1C4 + R2C3 cannot total 11 (because of CCC clash with R2C3 + R4C4, step 6) -> no 5 in R2C4
14b. R2C34 cannot total 8 -> no 8 in R1C4, clean-up: no 2 in R1C5
14c. R12C4 cannot total 8 -> no 8 in R2C3, clean-up: no 3 in R4C4 (step 6), no 7 in R3C4
14d. 22(4) cage at R2C3 (step 9) = {1489/1579}
14e. R2C3 + R12C4 = 4[48]/5[29/47]/9[61], no 4 in R2C4
14f. 4,8 of 22(4) cage = {1489} must be in R2C34 (cannot be 9{148} which clashes with R1C45 and 4{189} can only be 48{19} because of step 14e), no 4,8 in R23C5

15. R4C124 (step 4) = {127/136/235}
15a. 6 of {136} must be in R4C4 -> no 6 in R4C12

16. 45 rule on N23 1 outie R2C3 = 1 innie R3C4 + 1
16a. R2C3 + R12C4 (step 14e) = 4[48]/5[29]/9[61] (cannot be 5[47] which clashes with R2C3 + R3C4 = [54]) -> no 7 in R2C4

17. Consider combinations for R23C45 (step 11a) = {1389/1479/1578}
17a. R23C45 = {1389} => R3C4 = 3, R4C4 = 7
or R23C45 = {1479/1578}, 7 locked for C5
-> no 7 in R4C5, clean-up: no 8 in R4C6

18. 45 rule on C1234 3 outies R123C5 = 16, R123C6 = 14 (step 2) -> R123C4 = 15
18a. R12C4 (step 16a) = [29/48/61] -> R123C4 = [294/483/618]
18b. Consider permutations for R123C4
R123C4 = [294/618] => R67C4 = {37}
or R123C4 = [483] => R34C4 = [37]
-> Killer pair 3,7 in R34C4 and R67C4, locked for C4
18c. R67C4 = {19/37} (cannot be {28/46} which clash with R123C4), no 2,4,6,8

19. R123C5 = 16 (step 18) = {169/178/457}
19a. 45 rule on C6789 4 outies R4567C5 = 18 = {1269/1368/2358} (cannot be {1359/2349/2457} because R4C5 only contains 6,8, cannot be {1278/1458/1467/2367} which clash with R123C5, cannot be {3456} = 6{345} which clashes with R457C6), no 4,7
19b. 9 of {1269} must be in R7C5 (R45C5/R46C5 cannot be [69] which clashes with R4C56, CCC) -> no 9 in R56C5
19c. R123C5 = {169/457} (cannot be {178} which clashes with R4567C5), no 8, clean-up: no 2 in R1C4
19d. 11(2) cage at R8C5 = {29/38/47} (cannot be {56} which clashes with R4567C5), no 5,6

20. Naked pair {46} in R1C45, locked for R1 and N2, clean-up: no 6 in R4C4, no 5 in R2C3 (step 6)
20a. R2C3 + R12C4 (step 16a) = 4[48]/9[61], no 9 in R2C4

21. 2 in N2 only in R12C6, locked for C6 and 21(6) cage at R1C6, no 2 in R2C78 + R3C89
21a. 2 in N3 only in 26(4) cage at R1C7 (step 12a) = {2789} (only remaining combination), locked for N3, clean-up: no 3 in R3C6
21b. Naked pair {79} in R34C6, locked for C6

22. R689C6 (step 7a) = {146} (only remaining combination) -> R6C6 = 4, R89C6 = {16}, locked for N8, R9C7 = 9 (step 8), clean-up: no 9 in R6C4, no 2 in R78C7, no 2 in R8C5, no 1 in R9C23

23. 45 rule on N9 2 remaining innies R7C89 = 8 = {17/26/35}, no 4,8

24. 11(2) cage at R8C9 (step 19d) = {47}/[92] (cannot be {38} which clashes with R7C6), no 3,8
24a. 3 in N8 only in R7C456, locked for R7, clean-up: no 5 in R7C89 (step 23), no 8 in R8C7
24b. 9(2) cage at R8C9 = {18/36/45} (cannot be {27} which clashes with R7C89), no 2,7

25. 5 in C4 only in R589C4
25a. 45 rule on C123 3 outies R589C4 = 1 innie R2C3 + 9
25b. R2C3 = {49} -> R589C4 = 13,18 = {256/459}, no 1,8
25c. 6 of {256} must be in R5C4 -> no 2 in R5C4
25d. 5 of {459} must be in R89C4 (R89C4 cannot be {49} which clashes with 11(2) cage at R8C9), no 5 in R5C4
25e. 5 in C4 only in R89C4, locked for N8 and 17(3) cage at R8C3, no 5 in R8C3
25f. Killer pair 2,4 in R89C4 and 11(2) cage at R8C5, locked for N8

26. 8 in C6 only in 23(5) cage at R5C5, no 8 in R567C5
26a. R4C5 = 8 (hidden single in C5), R4C6 = 7, R3C6 = 9, R3C7 = 3, R34C4 = [82], R2C4 = 1, R6C4 = 3, R7C4 = 7, R12C6 = [23] (hidden pair in N2), R57C6 = [58], R1C1 = 5 (hidden single in R1), R2C9 = 2 (hidden single in N3), clean-up: no 5 in R4C89, no 9 in R5C12, no 1 in R7C89 (step 23), no 4 in R89C5, no 4 in R8C7

27. R89C5 = [92], R7C5 = 3, R5C4 = 9, R1C4 = 6 (hidden singles in C4), R1C5 = 4
27a. R7C89 = [26], clean-up: no 5 in R78C7, no 3 in R89C9
27b. R78C7 = [47], R1C7 = 8, clean-up: no 1 in R89C8, no 5 in R89C9
27c. R89C4 = {45} -> R8C3 = 8 (cage sum), R89C9 = [18], R89C6 = [61]
27d. Naked pair {49} in R4C89, locked for R4 and N6
27e. Naked pair {68} in R5C12, locked for R5 and N4 -> R56C5 = [16], R5C7 = 2, R4C7 = 6 (hidden single in R4), R2C7 = 5, R3C9 = 4, R2C8 = 6, R3C8 = 1, R23C5 = [75], R6C7 = 1, R4C89 = [49], R1C89 = [97], R5C89 = [73], R6C89 = [85], R5C3 = 4, R2C3 = 9, clean-up: no 6 in R9C2

28. R5C34 = [49] = 13 -> R46C3 = 8 = [17], R1C23 = [13], R9C3 = 6, R9C2 = 4

and the rest is naked singles.

Rating Comment. I'll rate my walkthrough for A222 at Hard 1.5. Although none of the individual steps feel to me higher than 1.5, some of them were hard to find so I've gone for the slightly higher rating.
Walkthrough by Ed:
Thanks for your WT Andrew. Step 14a is really nice. We worked in the same key areas but differently because of my steps 5&6. I've included the link for some more examples of that technique in the techniques forum. Andrew found something that would have made my solution shorter by half as I comment on after step 8. I often miss blocks.

A222 17 steps

Prelims
i. 4(2)n1 = {13}
ii. 10(2)cages at r1c4 & r3c4: no 5
iii. 21(6)r1c6: no 7,8,9
iv. 26(4)n3: no 1
v. 1292)r3c6: no 1,2,6
vi. 15(2)n5 = {69/78}
vii. 13(2)n6: no 1,2,3
viii. 14(2)n4 = {59/68}
ix. 10(2)r6c4: no 5
x. 11(2) cages in n8 & 9: no 1
xi. 8(2)n9: no 4,8,9
xii. 9(2)n9: no 9
xiii. 10(2)n7: no 5

1. 4(2)n1 = {13}: both locked for n1 and r1
1a. no 7,9 in 10(2)n2

2. 1 in n3 only in 21(6) -> no 1 in r2c6

3. "45" on n3: 3 outies r123c6 = 14
3a. {248} blocked by 10(2)n2 = [2/4..]
3b. = {239/257/347/356}(no 8)
3c. no 4 in r3c7
3d. 4 in {347} must be in r1c6 -> no 4 in r23c6
3e. no 8 in r3c7

4. 21(6)r1c6 = {123456}: r2c9 sees all 3s -> no 3 in r2c9

Straight into the first advanced step
5. "45" on r1: 1 outie r2c9 + 5 = 2 innies r1c16
5a. since the 1 outie sees all 5s (also the IOD) in r1 apart from the 2 innies -> no 5 in r2c9 (IOE - ie, innies can't be [55])(Andrew did this differently in his step 10a)
5b. and note that if r1c6 = r2c9 -> the IOD of 5 must be in r1c1 (IOE, no eliminations yet)

6. r1c6 sees all n3 except r2c9 or r3c7 -> must be cloned in one of those two cells
6a. from step 5b, if r1c6 = r2c9 -> r1c1 = 5
6b. if r1c6 = r3c7, the only common digit is 5
6c. -> 5 locked in r1c1 or r3c7
6d. no 5 in common peers in r3c123 nor r1c789

7. 26(4)n3 must have 8 for n3 = {2789} only: locked for n3
7a. no 3,5 in r3c6

8. 2 in 21(6)r1c6 only in r12c6: 2 locked for n2 and c6
8a. -> 10(2)n2 = {46} only: both locked for r1 and n2
8b. no 4,6,8 in r4c4

[NOTE: after reading Andrew's WT (his step 7a), the puzzle is actually pretty cracked from here. From my step 3b. r123c6 = {239/257} = [3/7..] -> {137} blocked from r689c6 -> my step 17a should now work!]

9. "45" on r123: 3 outies r4c124 = 10 (no 8,9)
9a. no 1 in r3c4

10. "45" on n23: 1 innie r3c4 + 1 = 1 outie r2c3
10a. no 9 in r3c4
10b. r2c3 from (489)
10c. no 1 in r4c4

11. 1 in n2 must be in 22(4)r2c3
11a. = {1489/1579}(no 3)
11b. must have 4 in r2c3 or 9 in r2c3
11c. -> no 8 in r2c3
11d. no 7 in r3c4 (IOD n23 = +1)
11e. no 3 in r4c4

12. 5 in n1 only in 26(5): no 5 in r4c1

13. h10(3)r4c124 must have 2 or 7 for r4c4
13a. = {127/235}(no 4,6)
13b. 5 in {235} must be in 4c2 -> no 3 in r4c2

14. "45" on n23: 2 outies r2c3 + r4c4 = 11 = [47/92] = [4/2..] (no eliminations yet)

15. 14(3)r3c2
15a. {49}[1] blocked by r2c3 = (49)
15b. {158} blocked by 1 & 5 only in r4c2
15c. {48}[2] blocked by outies n23 (step 14)
15d. = {167/257}(no 4,8,9)
15e. must have 1/5 only in r4c2 -> r4c2 = (15)
15f. must have 7: 7 locked for n1 and r3

16. r3c67 = [93]
16a. 3 in 21(6)r1c6 only at r2c6 -> r2c6 = 3
16b. r1c6 = 2

17. "45" on n69: 3 outies r689c6 = 11
17a. = {146} only: all locked for c6


cracked


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PostPosted: Thu Oct 30, 2014 8:17 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Assassin 223 by manu (October 2011) here
Puzzle Diagram:
Image
Ed's Coloured Diagram:
Image
Code: Select, Copy & Paste into solver:
3x3::k:4352:6913:6913:6913:5124:5124:2822:2822:4104:4352:6913:3339:3339:3339:5124:2822:2832:4104:4352:6913:2068:6933:6933:6933:6933:2832:1306:4352:4380:2068:6933:2847:2847:4385:4385:1306:4380:4380:2068:2343:3112:3112:3626:4385:4385:4653:4653:4653:2343:2865:3112:3626:3636:3636:3638:2871:2871:1849:2865:3387:3387:3636:3902:3638:2880:2880:1849:3651:3908:3387:4678:3902:3638:2377:2377:3651:3651:3908:4678:4678:3902:
Solution:
+-------+-------+-------+
| 5 6 3 | 1 9 4 | 2 8 7 |
| 2 8 4 | 6 3 7 | 1 5 9 |
| 7 9 1 | 5 2 8 | 3 6 4 |
+-------+-------+-------+
| 3 7 2 | 9 6 5 | 8 4 1 |
| 6 4 5 | 7 8 1 | 9 2 3 |
| 8 1 9 | 2 4 3 | 5 7 6 |
+-------+-------+-------+
| 9 5 6 | 3 7 2 | 4 1 8 |
| 1 3 8 | 4 5 6 | 7 9 2 |
| 4 2 7 | 8 1 9 | 6 3 5 |
+-------+-------+-------+
Quote:
manu: A soft(er) one for this week
Enjoy !

Ed: Nice puzzle manu thankyou. Hope you don't mind me doing a "Børge" and posting an extra image. The original one is a bit blurry to me.
I raced through this first time since I saw the key step to my solution (step 5) very quickly. Tried the puzzle again without using that and came to a halt. As always, this is an optimised start so some obvious eliminations are not included. Please let me know of any errors or things that could be clearer.

manu: Nice work Ed :applause: :applause: !
(sorry for the pics ...)

Andrew: Thanks manu for another good Assassin!
Nice walkthrough Ed, particularly your breakthrough step which broke the puzzle open quickly.
I found A223 much harder, partly because I didn't spot Ed's step 4; I only spotted that when I did the V2.
As often happens:
Ed found one step using an IOU in his step 11; I made the same eliminations using a CCC in my step 14.

Walkthrough by Ed:
Nice puzzle manu thankyou. Hope you don't mind me doing a "Børge" and posting an extra image. The original one is a bit blurry to me.

I raced through this first time since I saw the key step to my solution (step 5) very quickly. Tried the puzzle again without using that and came to a halt. As always, this is an optimised start so some obvious eliminations are not included. Please let me know of any errors or things that could be clearer.

A223 15 steps

Prelims
i. 20(3)n2: no 1,2
ii. 11(3)n3: no 9
iii. 16(2)n3 = {79}
iv. 11(2)n3: no 1
v. 8(3)r3c3: no 6..9
vi. 5(2)r3c9 = {14/23}
vii. 11(2)r4c5 and r6c5: no 1
viii. 9(2)n5: no 9
ix. 14(2)n6 = {59/68}
x. two 11(2) cages n7: no 1
xi. 7(2)n8: no 7,8,9
xii. 15(2)n8 = {69/78}
xiii. 9(2)n7: no 9

1. "45" on n89: 2 innies r7c58 = 8 (no 4,8,9)
1a. no 7 in r7c8
1b. no 2,3,7 in r6c5

2. "45" on n6: 2 outies r3c9+r7c8 = 5 (no 5,6)
2a. no 1 in r3c9
2b. no 2,3 in r7c5 (h8(2)r7c58)
2c. no 8,9 in r6c5

3. "45" on n3: 2 innies r3c79 = 7 = {34}/[52]
3a. r3c7 = (345)

4. "45" on r6789: 3 innies r6c467 = 10
4a. {127} blocked by no 1,2,7 in r6c7
4b. = {136/145/235}(no 7,8,9)
4c. must have 5/6 but not both -> no 5,6 in r6c46
4d. no 1,2,3,4 in r5c4
4e. no 5,6 in r5c7

The key move
5. "45" on n3689: 1 innie r3c7 + 1 = 1 outie r6c5
5a. but [56] blocked by r6c7 = (56)
5b. = [34/45] only
5c. r7c5: no 5
5d. r7c8: no 3 (h8(2)r7c58)

6. h7(2)r3c79 = {34} only: both locked for n3 and r3
6a. no 3,4 in r4c9
6b. no 7,8 in 11(2)n3

7. h10(3)r6c467: {145} blocked by r6c5 = (45)
7a. = {136/235}(no 4) = [1/2 but not both..]
7b. must have 3 -> 3 locked for n5 and r6
7c. no 5 in r5c4
7d. no 8 in 11(2)n5

8. 14(3)r6c8; max. r7c8 = 2 -> min. r6c89 = 12 (no 1,2)

9. Hidden killer pair 1,2 in r6: r6c467 can only have one of 1/2 (step 7a) -> 18(3)n4 must have 1/2
9a. 18(3) = {189/279}(no 4,5,6)
9b. must have 9 -> 9 locked for r6 and n4

10. "45" on c1: 2 innies r56c1 = 14 = [59/68] only

11. "45" on c1: 1 innie r6c1 + 3 = 2 outies r45c2
11a. -> no 3 in r45c2 (IOU)

12. 17(3)n4 must have 5/6 for r5c1 = {458/467}(no 1,2)
12a. must have 4 -> 4 locked for c2 & n4
12b. can't have both 5&6 -> no 5,6 in r45c2

13. "45" on n4: 1 outie r3c3 + 2 = 1 innie r4c1 = [13/57] only

14. r5c1 = 6 (hsingle n4)
14a. -> r6c1 = 8 (h14(2)r56c1)
14b. r6c23 = 10 = {19} only: 1 locked for r6 & n4

15. naked pair {23} in r6c46 : 2 locked for n5
15a. r6c46 = 5 -> r6c7 = 5 (h10(3)r6c467)
15a. no 9 in 11(2)n5

cracked
Walkthrough by Andrew:
Thanks manu for another good Assassin!

Nice walkthrough Ed, particularly your breakthrough step which broke the puzzle open quickly.

I found A223 much harder, partly because I didn't spot Ed's step 4; I only spotted that when I did the V2.

Here is my walkthrough for A223

Prelims

a) R12C9 = {79}
b) R23C8 = {29/38/47/56}, no 1
d) R34C9 = {14/23}
e) R4C56 = {29/38/47/56}, no 1
f) R56C4 = {18/27/36/45}, no 9
g) R56C7 = {59/68}
h) R67C5 = {29/38/47/56}, no 1
i) R7C23 = {29/38/47/56}, no 1
j) R78C4 = {16/25/34}, no 7,8,9
k) R8C23 = {29/38/47/56}, no 1
l) R89C6 = {69/78}
m) R9C23 = {18/27/36/45}, no 9
n) 20(3) cage at R1C5 = {389/479/569/578}, no 1,2
o) 11(3) cage at R1C7 = {128/137/146/236/245}, no 9
p) 8(3) cage at R3C3 = {125/134}

Steps resulting from Prelims
1a. Naked pair {79} in R12C9, locked for C9 and N3, clean-up: no 2,4 in R23C8
1b. 8(3) cage at R3C3 = {125/134}, 1 locked for C3, clean-up: no 8 in R9C2

2. 45 rule on N3 2 innies R3C79 = [61/52/43/34], R3C7 = {3456}

3. 45 rule on N6 1 outie R7C8 = 1 innie R4C9, R7C8 = {1234}

4. 45 rule on N9 1 outie R7C6 = 1 innie R7C8 + 1, R7C6 = {2345}

5. 45 rule on N8 2 innies R7C56 = 9 = [72/63/54/45], R7C5 = {4567}, clean-up: R6C5 = {4567}

6. 45 rule on N14 3 outies R1C4 + R2C45 = 10 = {127/136/145/235}, no 8,9

7. 45 rule on N124 2 outies R3C7 + R4C4 = 12 = [39/48/57], no 6 in R3C7, R4C4 = {789}, clean-up: no 1 in R3C9 (step 2), no 4 in R4C9, no 4 in R7C8 (step 3), no 5 in R7C6 (step 4), no 4 in R7C5 (step 5), no 7 in R6C5
7a. 1 in N3 only in 11(3) cage at R1C7 = {128/146}, no 3,5

8. 45 rule on N4 1 innie R4C1 = 1 outie R3C3 + 2, no 1,2,8,9 in R4C1

9. 45 rule on C1 2 innies R56C1 = 14 = {59/68}

10. 45 rule on C9 2 innies R56C9 = 9 = {18/36/45}, no 2

11. 45 rule on C123 1 innie R2C3 = 1 outie R1C4 + 3, no 7 in R1C4, no 2,3 in R2C3

12. 15(3) cage at R7C9 = {168/258/456} (cannot be {348} which clashes with R34C9), no 3

13. 45 rule on N9 3 innies R7C78 + R8C7 = 12 = {129/138/147/237} (cannot be {156/246} which clash with 15(3) cage at R7C9, cannot be {345} which clashes with R7C68 = [43], step 4), no 5,6

14. R56C1 (step 9) = 14 -> 17(3) cage at R4C2 cannot contain 3 (CCC with R56C1)
14a. R56C1 = 14 -> 18(3) cage at R6C1 cannot contain 4 (CCC with R56C1)

15. 20(3) cage at R1C5 = {389/479/569/578}
15a. 3,4 of {389/479} must be in R12C6 (R12C6 cannot be {79/89} which clash with R89C6), no 3,4 in R1C5
15a. 4 of {479} must be in R1C6 (R1C56 cannot be {79} which clashes with R1C9), no 4 in R2C6

16. 45 rule on N8 4 innies R7C456 + R8C4 = 16 = {1456/2347/2356} (cannot be {1267} which clashes with R89C6, cannot be {1357} because R78C4 must contain one even number)
16a. Killer pair 6,7 in R7C456 + R8C4 and R89C6, locked for N8

17. 45 rule on N124 3 innies R3C456 = 15 = {168/258/267/348} (cannot be {159} which clashes with 20(3) cage at R1C5, cannot be {249/357/456} which clash with R3C79), no 9
17a. 9 in N2 only in 20(3) cage at R1C5 = {389/479/569}
17b. 9 in R3 only in R3C12, locked for N1, clean-up: no 6 in R1C4 (step 11)

18. Hidden killer pair 5,6 in 11(3) cage at R1C7, R23C8 and R3C7 for N3, R23C8 contains both or neither of 5,6 -> 11(3) cage at R1C7 + R3C7 must contain both or neither of 5,6
18a. R123C7 cannot contain both of 5,6 (which would clash with R56C7) -> no 6 in R12C7
18b. 6 in N3 only in R123C8, locked for C8
18c. 11(3) cage at R1C7 (step 7a) = {128/146}
18d. 6 of {146} must be in R1C8 -> no 4 in R1C8

19. R56C9 (step 10) = {18/36/45}
19a. Consider combinations for R56C9
R56C9 = {18} => R56C7 = {59}
or R56C9 = {36/45}, killer pair 5,6 in R56C7 and R56C9, locked for N6
-> 5 must be in R56C79, locked for N6

20. 14(3) cage at R6C8 = {149/167/239/248/257/347} (cannot be {158/356} which clash with R56C7)
20a. {149} can only be [941], 1 of {167} must be in R7C8 -> no 1 in R6C89, clean-up: no 8 in R5C9 (step 10)
20b. 9 of {239} must be in R6C8, 2,3 of {248/257/347} must be in R7C8 -> no 2,3 in R6C8

21. 45 rule on C789 3 innies R378C7 = 14 = {149/158/248/257/347} (cannot be {239} = 3{29} because 13(3) cage at R7C6 cannot be 2{29})
21a. 3 of {347} must be in R3C7 (R78C7 cannot be {37} because 13(3) cage at R7C6 cannot be 3{37}), no 3 in R78C7

22. 45 rule on R789 3 outies R6C589 = 17 = {359/458/467} (cannot be {368} because 14(3) cage at R6C8 cannot be [833])
22a. 5 of {458} must be in R6C5 (R6C89 cannot be [85] which clashes with R56C7), no 5 in R6C9, clean-up: no 4 in R5C9 (step 10)

23. 18(3) cage at R6C1 = {189/279/369} (cannot be {378/567} which clashes with R6C589), no 5, 9 locked for R6 and N4, clean-up: no 5 in R5C7
23a. 1 of {189} must be in R6C2 -> no 8 in R6C2

24. 17(3) cage at R4C2 = {278/458/467}, no 1
24a. 6 of {467} must be in R5C1 -> no 6 in R45C2

25. R6C589 (step 22) = {458/467}, no 3, 4 locked for R6, clean-up: no 5 in R5C4, no 6 in R5C9 (step 10)

26. R6C589 (step 25) = {467} (only remaining combination, cannot be {458} = 5{48} which clashes with R56C7) -> R6C8 = 7, R6C58 = {46}, locked for R6, clean-up: no 8 in R5C1 (step 9), no 2,3 in R5C4, no 8 in R5C7, no 1 in R5C9 (step 10), no 6 in R7C5

27. 18(3) cage at R6C1 (step 23) = {189} (only remaining combination) -> R6C2 = 1, R6C13 = {89}, locked for R6 and N4 -> R6C7 = 5, R5C7 = 9, R5C9 = 3, R6C9 = 6 (step 10), R6C5 = 4, R7C5 = 7, R7C6 = 2 (step 5), R7C8 = 1 (step 4), R6C6 = 3, R6C4 = 2, R5C4 = 7, clean-up: no 2 in R34C9, no 8,9 in R4C56, no 4,9 in R7C23, no 5 in R7C4, no 5,6 in R8C4, no 8 in R89C6, no 8 in R9C3
27a. R3C3 = 1 (hidden single in C3)

28. R34C9 = [41], R3C7 = 3, clean-up: no 8 in R23C8
28a. R4C4 = 9 (hidden single in N5)

29. Naked pair {56} in R4C56, locked for R4 and N5

30. R5C1 = 6 (hidden single in N4), R6C1 = 8 (step 9), R6C3 = 9, clean-up: no 2 in R8C2

31. Naked triple {258} in 15(3) cage at R7C9, locked for N9 -> R78C7 = [47], R9C7 = 6, R89C6 = [69], R89C8 = [93], R7C4 = 3, R8C4 = 4, clean-up: no 8 in R7C23, no 5 in R8C2, no 2,5 in R8C3

32. Naked pair {38} in R8C23, locked for R8 and N7

33. Naked pair {56} in R7C23, locked for R7 and N7 -> R7C1 = 9, R7C9 = 8
33a. R7C1 = 9 -> R89C1 = 5 = [14]

34. R5C1 = 6 -> R45C2 = 11 = [74], R4C1 = 3, R45C3 = [25], R4C78 = [84], R5C8 = 2, R8C5 = 5, R4C56 = [65]

35. R1C8 = 8 (hidden single in C8)

36. R1C5 = 9, R12C6 = 11 = [47], R3C6 = 8, R7C23 = [56], R9C23 = [27]

37. R2C3 = 4, R2C5 = 3 (hidden single in N2) -> R2C4 = 6 (cage sum)

and the rest is naked singles.

As often happens Ed found one step using an IOU in his step 11; I made the same eliminations using a CCC in my step 14.

Rating Comment. I'll rate my walkthrough for A223 at Easy 1.5. I used a very short forcing chain in step 19; also my breakthrough step 26, the move I most enjoyed finding, might deserve this rating.


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PostPosted: Thu Oct 30, 2014 8:30 pm 
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Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Assassin 223 V2 by manu (October 2011) here
Puzzle Diagram:
Image
Ed's Coloured Diagram:
Image
Code: Select, Copy & Paste into solver:
3x3::k:4352:6913:6913:6913:5124:5124:2822:2822:4104:4352:6913:3339:3339:3339:5124:2822:2832:4104:4352:6913:2068:6933:6933:6933:6933:2832:1306:4352:4380:2068:6933:2847:2847:3105:3105:1306:4380:4380:2068:2343:3112:3112:4906:4906:4906:4653:4653:4653:2343:2865:3112:4906:3636:3636:3638:2871:2871:1849:2865:3387:3387:3636:3902:3638:2880:2880:1849:3651:3908:3387:4678:3902:3638:2377:2377:3651:3651:3908:4678:4678:3902:
Solution:
+-------+-------+-------+
| 5 6 3 | 1 9 4 | 2 8 7 |
| 2 8 4 | 6 3 7 | 1 5 9 |
| 7 9 1 | 5 2 8 | 3 6 4 |
+-------+-------+-------+
| 3 7 2 | 9 6 5 | 8 4 1 |
| 6 4 5 | 7 8 1 | 9 2 3 |
| 8 1 9 | 2 4 3 | 5 7 6 |
+-------+-------+-------+
| 9 5 6 | 3 7 2 | 4 1 8 |
| 1 3 8 | 4 5 6 | 7 9 2 |
| 4 2 7 | 8 1 9 | 6 3 5 |
+-------+-------+-------+
Quote:
manu: Nice work Ed :applause: :applause: !
I had prepared a V2 I did not plan to post , but it could be interesting for you since your key step does not work for this one :twisted:
(sorry for the pics ...)

Andrew: Thanks manu for the variant!
At first sight it looked as if the changes to the cage pattern in N6 would make it harder to solve; the changes removed some of the steps which I'd used for A223.
Although I found steps harder to find this time, my actual steps were technically simpler. This may have been partly because I found a few steps which I'd missed when solving A223.

Ed: Even better puzzle! Found a neat way (step 5) to get into it early which works for the V1 as well. V3? :) ;)
From there, followed the same general path as Andrew. The early key step didn't crack it comprehensively like in the original puzzle.

Walkthrough by Andrew:
Thanks manu for the variant!

At first sight it looked as if the changes to the cage pattern in N6 would make it harder to solve; the changes removed some of the steps which I'd used for A223.

Although I found steps harder to find this time, my actual steps were technically simpler. This may have been partly because I found a few steps which I'd missed when solving A223.

Here is my walkthrough for A223 V2

Prelims

a) R12C9 = {79}
b) R23C8 = {29/38/47/56}, no 1
d) R34C9 = {14/23}
e) R4C56 = {29/38/47/56}, no 1
f) R4C78 = {39/48/57}, no 1,2,6
g) R56C4 = {18/27/36/45}, no 9
h) R67C5 = {29/38/47/56}, no 1
i) R7C23 = {29/38/47/56}, no 1
j) R78C4 = {16/25/34}, no 7,8,9
k) R8C23 = {29/38/47/56}, no 1
l) R89C6 = {69/78}
m) R9C23 = {18/27/36/45}, no 9
n) 20(3) cage at R1C5 = {389/479/569/578}, no 1,2
o) 11(3) cage at R1C7 = {128/137/146/236/245}, no 9
p) 8(3) cage at R3C3 = {125/134}

Steps resulting from Prelims
1a. Naked pair {79} in R12C9, locked for C9 and N3, clean-up: no 2,4 in R23C8
1b. 8(3) cage at R3C3 = {125/134}, 1 locked for C3, clean-up: no 8 in R9C2

2. 45 rule on N3 2 innies R3C79 = [61/52/43/34], R3C7 = {3456}

3. 45 rule on N6 1 outie R7C8 = 1 innie R4C9, R7C8 = {1234}

4. 45 rule on N9 1 outie R7C6 = 1 innie R7C8 + 1, R7C6 = {2345}

5. 45 rule on N8 2 innies R7C56 = 9 = [72/63/54/45], R7C5 = {4567}, clean-up: R6C5 = {4567}

6. 45 rule on N14 3 outies R1C4 + R2C45 = 10 = {127/136/145/235}, no 8,9

7. 45 rule on N124 2 outies R3C7 + R4C4 = 12 = [39/48/57], no 6 in R3C7, R4C4 = {789}, clean-up: no 1 in R3C9 (step 2), no 4 in R4C9, no 4 in R7C8 (step 3), no 5 in R7C6 (step 4), no 4 in R7C5 (step 5), no 7 in R6C5
7a. 1 in N3 only in 11(3) cage at R1C7 = {128/146}, no 3,5

8. 45 rule on N4 1 innie R4C1 = 1 outie R3C3 + 2, no 1,2,8,9 in R4C1

9. 45 rule on C1 2 innies R56C1 = 14 = {59/68}

10. 45 rule on C9 2 innies R56C9 = 9 = {18/36/45}, no 2

11. 45 rule on C123 1 innie R2C3 = 1 outie R1C4 + 3, no 7 in R1C4, no 2,3 in R2C3

12. 15(3) cage at R7C9 = {168/258/456} (cannot be {348} which clashes with R34C9), no 3

13. 45 rule on N9 3 innies R7C78 + R8C7 = 12 = {129/138/147/237} (cannot be {156/246} which clash with 15(3) cage at R7C9, cannot be {345} which clashes with R7C68 = [43], step 4), no 5,6

14. R56C1 (step 9) = 14 -> 17(3) cage at R4C2 cannot contain 3 (CCC with R56C1)
14a. R56C1 = 14 -> 18(3) cage at R6C1 cannot contain 4 (CCC with R56C1)

15. 20(3) cage at R1C5 = {389/479/569/578}
15a. 3,4 of {389/479} must be in R12C6 (R12C6 cannot be {79/89} which clash with R89C6), no 3,4 in R1C5
15a. 4 of {479} must be in R1C6 (R1C56 cannot be {79} which clashes with R1C9), no 4 in R2C6

16. 45 rule on N8 4 innies R7C456 + R8C4 = 16 = {1456/2347/2356} (cannot be {1267} which clashes with R89C6, cannot be {1357} because R78C4 must contain one even number)
16a. Killer pair 6,7 in R7C456 + R8C4 and R89C6, locked for N8

17. 45 rule on N124 3 innies R3C456 = 15 = {168/258/267/348} (cannot be {159} which clashes with 20(3) cage at R1C5, cannot be {249/357/456} which clash with R3C79), no 9
17a. 9 in N2 only in 20(3) cage at R1C5 = {389/479/569}
17b. 9 in R3 only in R3C12, locked for N1, clean-up: no 6 in R1C4 (step 11)

[I’ve used the steps from my walkthrough for A222 so far, apart from one change in the Prelims. My original step 18 doesn’t work for the V2, because of the changes to the cage pattern in N6.]

18. 45 rule on C789 3 innies R378C7 = 14 = {149/158/248/257/347} (cannot be {239} = 3{29} because 13(3) cage at R7C6 cannot be 2{29})
18a. 3 of {347} must be in R3C7 (R78C7 cannot be {37} because 13(3) cage at R7C6 cannot be 3{37}), no 3 in R78C7

19. 45 rule on R789 3 outies R6C589 = 17 = {359/368/458/467} cannot be {179/278} because R6C5 only contains 4,5,6, cannot be {269} because 2,9 only in R6C8), no 1,2, clean-up: no 8 in R5C9 (step 10)
19a. 6 of {368} must be in R6C5, 7 of {467} must be in R6C8 -> no 6 in R6C8

20. 45 rule on N6 2 innies R6C89 = 1 outie R3C9 + 9, IOU no 9 in R6C8
20a. 45 rule on N6 3 innies R4C9 + R6C89 = 14 = {158/167/248/257/356} (cannot be {347} which clashes with R4C78)
20b. 3 of {356} must be in R4C9 -> no 3 in R6C89, clean-up: no 6 in R5C9 (step 10)
20c. R6C589 (step 19) = {458/467}, 4 locked for R6, clean-up: no 5 in R5C4

21. 45 rule on R6789 3 innies R6C467 = 10 = {127/136/235}, no 8,9, clean-up: no 1 in R5C4

22. 9 in R6 only in 18(3) cage at R6C1, locked for N4
22a. 18(3) cage = {189/279/369}, no 5
22b. 1 of {189} must be in R6C2 -> no 8 in R6C2

23. 17(3) cage at R4C2 = {278/458/467}, no 1
23a. 6 of {467} must be in R5C1 -> no 6 in R45C2

24. 45 rule on N5689 2 innies R4C49 = 10 = [73/82/91]
24a. 45 rule on R4 remaining innies R4C123 = 12 = {147/156/237/246} (cannot be {138} which clashes with R4C49, cannot be {345} which clashes with R4C78), no 8
24b. {156} must be [651] -> no 5 in R4C13, clean-up: no 3 in R3C3 (step 8)

25. R4C9 + R6C89 (step 20a) = {158/167/248/257} (cannot be {356} because R6C589 only contains one of 5,6), no 3, clean-up: no 2 in R3C9, no 5 in R3C7 (step 2), no 7 in R4C4 (step 7), no 3 in R7C8 (step 3), no 4 in R7C6 (step 4), no 5 in R7C5 (step 5), no 6 in R6C5

26. Naked pair {34} in R3C79, locked for R3 and N3, clean-up: no 6 in R4C1 (step 8)
26a. R23C8 = {56} (only remaining combination), locked for C8 and N3, clean-up: no 7 in R4C7

27. R3C456 (step 17) = {168/258/267}
27a. Killer pair 5,6 in R3C456 and R3C8, locked for R3, clean-up: no 7 in R4C1 (step 8)
27b. Killer pair 1,2 in R3C3 and R3C456, locked for R3

28. R7C78 + R8C7 (step 13) = {129/147}, no 8, 1 locked for N9
28a. 15(3) cage at R7C9 (step 12) = {258/456}, 5 locked for C9 and N9, clean-up: no 4 in R56C9 (step 10)
28b. Killer pair 2,4 in R7C78 + R8C7 and 15(3) cage at R7C9, locked for N9
28c. 1 in C9 only in R45C9, locked for N6
28d. 4 in C8 only in R456C8, locked for N6, clean-up: no 8 in R4C8

29. R6C589 (step 20c) = {458/467}
29a. R6C9 = {68} -> no 8 in R6C8

30. R7C6 = R7C8 + 1 (step 4)
30a. R7C68 = [21/32], 2 locked for R7 and 13(3) cage at R7C6, no 2 in R8C7, clean-up: no 9 in R7C23, no 5 in R8C4

31. R7C456 + R8C4 (step 16) = {2347/2356} (cannot be {1456} because R7C6 only contains 2,3), no 1, 2,3 locked for N8, clean-up: no 6 in R78C4

32. 45 rule on N4 3 innies R4C13 + R5C3 = 10 = {145/235}, 5 locked for N4, clean-up: no 9 in R6C1 (step 9)
32a. Naked pair {68} in R56C1, locked for C1 and N4

33. 18(3) cage at R6C1 (step 22a) = {189/369} (cannot be {279} because R6C1 only contains 6,8), no 2,7
33a. Naked pair {68} in R6C19, locked for R6, clean-up: no 3 in R5C4

34. 17(3) cage at R4C2 (step 23) = {278/467}, 7 locked for C2, clean-up: no 4 in R7C3, no 4 in R8C3, no 2 in R9C3

[The next step has been available for some time but I’ve only just spotted it.]
35. 6 in R4 only in R4C56 = {56}, locked for N5 -> R6C5 = 4, R6C8 = 7, R6C9 = 6 (step 20c), R5C9 = 3 (step 10), R7C8 = 1 (cage sum), R3C9 = 4, R4C9 = 1, R3C7 = 3, R56C1 = [68], R7C5 = 7, R7C6 = 2 (step 5), clean-up: no 9 in R4C78, no 2,4 in R5C4, no 3 in R6C4, no 5 in R7C4, no 8 in R89C6

36. R4C78 = [84], R4C4 = 9, R4C1 = 3, R4C3 = 2, R4C2 = 7, R5C2 = 4, R3C3 = 1, R5C3 = 5, R6C23 = [19], R6C46 = [23], R6C7 = 5, R5C4 = 7, clean-up: no 6 in R7C2, no 2,6,9 in R8C2, no 7 in R8C3, no 8 in R9C3

37. Naked pair {12} in R12C7, locked for C7 and N3 -> R1C8 = 8, R5C78 = [92], R78C7 = [47], R9C7 = 6, R78C4 = [34], R89C6 = [69], R89C8 = [93], clean-up: no 8 in R7C23, no 5 in R8C2

38. R7C23 = [56], R9C2 = 2, R9C3 = 7, R4C56 = [65], R789C1 = [914], R3C1 = 7, R3C6 = 8, R5C56 = [81], R89C5 = [51], R123C5 = [932], R3C4 = 5 (cage sum)

and the rest is naked singles.

The steps which I'd missed when solving A223 were 3 innies for R6789 and the IOU for N6. I also missed Ed's 2 innies for N89 although that didn't matter after my step 7.

Rating Comment. I'll rate my walkthrough for A223 V2 at Hard 1.25. I added the Hard because I found some steps difficult to spot. Also step 30a probably deserves this rating.

It was interesting that SudokuSolver v3.3.1 rated A223V2 slightly lower than A223. I'm still using that version of SS to get ratings for the puzzles which will be included in Archive Part H; scores using the more recent version of SS will be added before the outline for Archive Part H is posted.
Walkthrough by Ed:
Even better puzzle! Found a neat way (step 5) to get into it early which works for the V1 as well. V3? :) ;)

From there, followed the same general path as Andrew. The early key step didn't crack it comprehensively like in the original puzzle.

16 steps

Prelims
i. 20(3)n2: no 1,2
ii. 11(3)n3: no 9
iii. 16(2)n3 = {79}
iv. 11(2)n3: no 1
v. 8(3)r3c3: no 6..9
vi. 5(2)r3c9 = {14/23}
vii. 11(2)r4c5 and r6c5: no 1
viii. 9(2)n5: no 9
ix. 12(2)n6: no 1,2,6
x. two 11(2) cages n7: no 1
xi. 7(2)n8: no 7,8,9
xii. 15(2)n8 = {69/78}
xiii. 9(2)n7: no 9


1. "45" on n89: 2 innies r7c58 = 8 (no 4,8,9)
1a. no 7 in r7c8
1b. no 2,3,7 in r6c5

2. "45" on n6: 2 outies r3c9+r7c8 = 5 (no 5,6)
2a. no 1 in r3c9
2b. no 4 in r4c9
2c. no 2,3 in r7c5 (h8(2)r7c58)
2d. no 8,9 in r6c5

3. 16(2)n3 = {79}: both locked for c1 and n3
3a. no 2,4 in 11(2)n3

4. 11(2)n3 = {38/56} = [3/5..] (no eliminations yet)

5. "45" on n36: 1 innie r3c7 - 2 = 1 outie r7c8
5a. but [53] blocked by 11(2)n3 (step 4)
5b. = [42/31] only
5c. no 2 in r3c9 (outies n6 = 5)
5d. no 3 in r4c9
5e. no 5 in r7c5 (h8(2)r7c58)
5f. no 6 in r6c5

6. Naked pair {34} in r3c79: both locked for r3 and n3
6a. -> 11(2)n3 = {56}: both locked for c8 and n3
6b. no 7 in r4c7

7. "45" on n6: 1 outie r3c9 + 9 = 2 innies r6c89
7a. -> no 9 in r6c8 (IOU)

8. "45" on r6789: 3 innies r6c467 = 10 (no 8,9)

9. 9 in r6 only in n4: 9 locked for n4

10. r6c5 = (45) -> {459} blocked from 18(3)n4
10a. -> 18(3)n4 (must have 9 for n4) = {189/279/369}(no 4,5)

11. "45" on c1: 2 innies r56c1 = 14 = [59]/{68}

12. "45" on c1: 1 innie r6c1 + 3 = 2 outies r45c2
12a. -> no 3 in r45c2 (IOU)

13. 17(3)n4 = {278/458/467}(no 1)
13a. 6 of {467} must be in r5c1 -> no 6 in r45c2

14. "45" on n4: 1 outie r3c3 + 2 = 1 innie r4c1
14a. r4c1 = (347)

15. "45" on n5: 2 innies r4c4 + r6c5 = 13 = [94/85]

16. 6 in r4 only in 11(2)n5 = {56} only: both locked for n5 and 5 for r4

cracked


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PostPosted: Thu Oct 30, 2014 8:47 pm 
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Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Assassin 224 by Ed (October 2011) here
Puzzle Diagram:
Attachment:
a224.png
a224.png [ 65.07 KiB | Viewed 7320 times ]
Code: Select, Copy & Paste into solver:
3x3::k:2304:3585:1282:1282:3075:3588:3588:6405:6405:2304:3585:4870:3075:3075:4359:4359:4359:6405:4104:4104:4870:4870:10249:10249:2826:2826:6405:4363:4104:4870:2060:10249:2573:2573:2826:782:4363:4363:2060:2060:10249:4879:4879:2064:782:4113:2322:2322:10249:10249:10249:4879:2064:5395:4113:7188:7188:7188:2581:4630:4630:4630:5395:4113:4113:4119:7188:2581:4630:5144:5395:5395:3097:3097:4119:4119:5144:5144:5144:2074:2074:
Solution:
+-------+-------+-------+
| 7 8 3 | 2 1 9 | 5 6 4 |
| 2 6 1 | 4 7 5 | 3 9 8 |
| 4 9 5 | 6 8 3 | 2 1 7 |
+-------+-------+-------+
| 9 3 7 | 1 5 4 | 6 8 2 |
| 6 2 4 | 3 9 8 | 7 5 1 |
| 5 1 8 | 7 2 6 | 4 3 9 |
+-------+-------+-------+
| 1 5 6 | 9 4 2 | 8 7 3 |
| 3 7 2 | 8 6 1 | 9 4 5 |
| 8 4 9 | 5 3 7 | 1 2 6 |
+-------+-------+-------+
Quote:
Ed: I found this is a really, really tough puzzle. It may very well be too hard as a V1 Assassin. But put so much work into it and assume I missed something important since the v3.5.7 SSscore is 1.55. My solution has about 6 advanced steps! But some of them are so nice they are worth promoting in the V1. No chains used. I hope you still enjoy the challenge and find a better way.

manu: What a monster ! Many hard steps before crsacking it. Here is the most straightforward way I've found, but I need two forcing chains !! I am really impressive you did not use any one, Ed. ;clapclap;
My first move used a step by contradiction, but I managed to find an other way.

Andrew: That makes three of us who found A224 a really challenging puzzle. It's certainly at the top end for a V1 Assassin. Thanks Ed for an Assassin which makes one think hard!
Congratulations to Ed for solving it without using chains! :applause: I'll be interested to see what your 6 advanced steps were.
Like manu I also used two forcing chains but not the same ones. I enjoyed manu's walkthrough; don't think I'm giving anything away when I say that he had some neat logic for the 40(7) cage.
There were times when I only managed to find two or three steps in a day, then yesterday I managed to make real progress.

Ed: Me to and agree, it is really neat. Nice to have a shortcut way to solve this puzzle. My step 9 is the way around manu's first chain. I found his chain way first but my purist's heart kept looking for a way around it.
Andrew's way to get his first placement is the best. Really neat use of killer subsets. After that, he found a longcut way but I still really appreciate the effort and have read closely the walk-through. My step 7d is the key reason why mine and Andrew's walkthrough's are different.
My other step I really enjoyed was my step 23 but Andrew got it using killer subsets again. He takes the record for number and sizes of killer subsets in his walkthrough!

wellbeback: I'm a little late to the party - and I too had a couple of short chains in my solution.
I have a lot of "Either A is true or B is true - but either way they both imply that C is true" logic.
After writing my solving path I perused the other solutions and I see a lot of commonality - but also some differences. In particular - my very first placement was the very last line in Ed's walkthrough :)
Thanks for the nice puzzle Ed ;clapclap;

Walkthrough by manu:
What a monster ! Many hard steps before crsacking it. Here is the most straightforward way I've found, but I need two forcing chains !! I am really impressive you did not use any one, Ed. ;clapclap;
My first move used a step by contradiction, but I managed to find an other way.
Opening part :

Prelim's

Cage 3(2) N6 = {12}
Cage 5(2) N12 - cells only uses 1234
Cage 14(2) N23 - cells only uses 5689
Cage 14(2) N1 - cells only uses 5689
Cage 8(2) N9 - cells do not use 489
Cage 8(2) N6 - cells do not use 489
Cage 12(2) N7 - cells do not use 126
Cage 9(2) N4 - cells do not use 9
Cage 9(2) N1 - cells do not use 9
Cage 10(2) N56 - cells do not use 5
Cage 10(2) N8 - cells do not use 5
Cage 8(3) N45 - cells do not use 6789
Cage 11(3) N36 - cells do not use 9
Cage 19(3) N56 - cells do not use 1
Cage 28(4) N78 - cells do not use 123


1. a. R45 = {12} locked for C9 and N6
b. Last combo : 8(2) at N6 = {35} locked for C8 and N6.
c. « 45 » rule for R789 : outies = h14(2) = R6C19 = [86/68/59].
d. « 45 » rule for C12 :Innies = h6(2) = R67C2= [15/24].
→ Cage sum : R6C23=[18/27]

2. a. 40(7) at N2 = {1456789/2356789} : cannot contain both {12} or {13} or {34}
b. Killer pair {12} for R6 at R6C2 + R6C456
step a → No 1,2 for the rest of cage 40(7).

3. First forcing chain
a. 1 locked for R6 at R6C2 + R6C456 :
(i) R6C2=1 → R6C3=8 → R6C19 = [59] (step 1.c.) → R6C8=3 (naked single)
(ii) 1 at R6C456 → R6C456 <> 3 (step 2. a.) → R6C8=3 (Hidden single).
The conclusion is R6C8=3.
b. 8(2) at N6 = [53]
c. R5C34 connot be {12}, blocked by R5C9={12} : R4C4 <> 5
d. No 5 at 8(2) at R4C4 : {134} locked for R5C56
e. {256789} locked for N5 at R4C56+R5C56+R6C456
→ R3C6 <> 256789 since R3C6 see R4C56+R5C56+R6C456
→ R3C6 = {34}
Since 40(7) cannot contain both of {34}, no 3 4 for the rest of cage 40(7) .

4. a. Hidden single for R6 : R6C7=4
b.Innies for C1234 : R26C4 sum to 11 : no 1.
c. « 45 » rule for C6789 : 6 + R9C5 = R3C6 + R6C6
d. Min R3C6 + R6C6 = 7 and Max R3C6 = 4 → Min R6C6 = 3 : R6C6 <> 1 , 2 .

5. a. Outies for N7 : R6C1 + R789C4 sum to 27
The last forcing chain :
b. 1 locked for R6 at R6C2 + R6C5
(i) R6C2=1 → R6C3=8 → R6C19 = [59] (step 1.c.) → R789C4 sum to 22 (step 5.a.) : R789C4 must contain 9 : no 9 at 10(2) at N8 .
(ii)R6C5=1 → 10(2) at N8 <> {19}
The conclusion is 10(2) at N8 <> {19}. 10(2) = {28/37/46}

6. a. 40(7) must contain all of {678} , and R3C6 contain none of them.
b. R345C5 cannot be {678} which would block combos of cage 10(2) at N8
→ R6C456 must contain at least one of {678}
c. R6C3 = {78} contain one of {678} and R6C456 must contain at least one of {678}, so R6C19 contain at most one of {678} : from step 1.c. , R6C9 = [59]
d. Outies for N14 : R1345C4 sum to 12 : {1236/1245}.
R1345C4 = [2613/2631/2514/2541]
→ R1C4 = 2 , R1C3 =3 (cage sum)
→ 1 locked for C4, N5, and the rest of 8(3) : R5C3=4.

Finally Cracked !

Edit : some typos have been corrected. Thanks to Andrew.
Walkthrough by Andrew:
That makes three of us who found A224 a really challenging puzzle. It's certainly at the top end for a V1 Assassin. Thanks Ed for an Assassin which makes one think hard!

Congratulations to Ed for solving it without using chains! :applause: I'll be interested to see what your 6 advanced steps were.

Like manu I also used two forcing chains but not the same ones. I enjoyed manu's walkthrough; don't think I'm giving anything away when I say that he had some neat logic for the 40(7) cage.

There were times when I only managed to find two or three steps in a day, then yesterday I managed to make real progress.

Here is my walkthrough for A224

Prelims

a) R12C1 = {18/27/36/45}, no 9
b) R12C2 = {59/68}
d) R1C34 = {14/23}
e) R1C67 = {59/68}
f) R4C67 = {19/28/37/46}, no 5
g) R45C9 = {12}
h) R56C8 = {17/26/35}, no 4,8,9
i) R6C23 = {18/27/36/45}, no 9
j) R78C5 = {19/28/37/46}, no 5
k) R9C12 = {39/48/57}, no 1,2,6
l) R9C89 = {17/26/35}, no 4,8,9
m) 11(3) cage at R3C7 = {128/137/146/236/245}, no 9
n) 8(3) cage at R4C4 = {125/134}
o) 19(3) cage at R5C6 = {289/379/469/478/568}, no 1
p) 28(4) cage at R7C2 = {4789/5689}, no 1,2,3

Steps resulting from Prelims
1a. Naked pair {12} in R45C9, locked for C9 and N6, clean-up: no 8,9 in R4C6, no 6,7 in R56C8, no 6,7 in R9C8
1b. Naked pair {35} in R56C8, locked for C8 and N6, clean-up: no 7 in R4C6, no 3,5 in R9C9
1c. Caged X-Wing for 1 in 8(3) cage at R4C4 and R45C9, no other 1 in R45, clean-up: no 9 in R4C7

2. 45 rule on C12 2 innies R67C2 = 6 = [15/24], clean-up: R6C3 = {78}
2a. 28(4) cage at R7C2 = {4789/5689}, R7C2 = {45} -> no 4,5 in R7C34 + R8C4
2b. 28(4) cage at R7C2 = {4789/5689}, CPE no 8,9 in R7C56, clean-up: no 1,2 in R8C5

3. 45 rule on R12 1 outie R3C9 = 1 innie R2C3 + 6, R2C3 = {123}, R3C9 = {789}

4. 45 rule on R9 2 outies R8C37 = 11 = {29/38/47/56}, no 1

5. 45 rule on R789 2 outies R6C19 = 14 = [59]/{68}
5a. 4 in N6 only in R4C78 + R56C7, CPE no 4 in R3C7
5b. Min R6C1 = 5 -> max R7C1 + R8C12 = 11, no 9 in R7C1 + R8C12

6. Hidden killer pair 1,2 in R6C2 and R6C456 for R6, R6C2 = {12} -> R6C456 must contain one of 1,2
6a. 40(7) cage at R3C5 = {1456789/2356789} only contains one of 1,2 -> no 1,2 in R3C56 + R45C5
6b. 8(3) cage at R4C4 = {125/134}
6c. 5 of {125} must be in R5C4 (R45C4 cannot be {12} which clashes with R6C456, R5C34 cannot be {12} which clashes with R5C9) -> no 5 in R4C4 + R5C3, no 2 in R5C4

7. 45 rule on C1234 2 innies R26C4 = 11 = {29/38/47/56}, no 1

8. Hidden killer quad 1,2,3,4 for N4, 17(3) cage at R4C1 contains one of 2,3,4, R5C3 = {1234}, R6C2 = {12} -> R4C23 must contain one of 2,3,4
8a. Killer quad 1,2,3,4 in R4C23, R4C4, R4C67 and R4C9, locked for R4
8b. Killer quint 1,2,3,4,5 in 17(3) cage at R4C1, R5C34, R5C8 and R5C9, locked for R5

9. 19(3) cage at R5C6 = {469/478} -> R6C7 = 4, clean-up: no 7 in R2C4 (step 7), no 6 in R4C6, no 7 in R8C3 (step 4)

10. 11(3) cage at R3C7 = {128/137/146/236} (cannot be {245} because R4C8 only contains 6,7,8), no 5
10a. R4C8 = {678} -> no 6,7,8 in R3C78

11. 45 rule on C89 2 innies R27C8 = 1 outie R3C7 + 14
11a. R3C7 = {123} -> R27C8 = 15,16,17 = {69/78/79/89}, no 1,2,4

12. 45 rule on C9 3 outies R189C8 = 12 = {129/147/246}, no 8
12a. R89C8 cannot total 8, which would give CCC with R9C89 -> no 4 in R1C8
12b. 4 of {147/246} must be in R8C8 -> no 6,7 in R8C8

13. 25(4) cage at R1C8 = {3589/3679/4579/4678} (cannot be {1789} which clashes with R69C9, ALS block, cannot be {2689} which clashes with R6C9), no 1,2
[Note. R1C89 cannot total 14 which clashes with R1C2 and R1C67 in R1, CCC clash, so R23C9 cannot total 11. This might be useful later.]
13a. R389C8 = {124} (hidden triple in C8)

14. Hidden killer quad 1,2,3,4 for N3, 25(4) cage at R1C8 contains one of 3,4, R3C7 = {123}, R3C8 = {124} -> R2C7 must contain one of {123} -> R2C7 = {123}

15. 17(3) cage at R2C6 = {179/269/278/359/368} (cannot be {458/467} because R2C7 only contains 1,2,3), no 4
15a. R2C7 = {123} -> no 1,2,3 in R2C6

16. 45 rule on N89 3 innies R789C4 = 1 outie R6C9 + 13
16a. Min R6C9 = 6 -> min R789C4 = 19, no 1 in R9C4

17. 5,6,7,8,9 in N3 only in R1C7, 25(4) cage at R1C8 and R2C8
17a. 25(4) cage = {3589/3679/4579/4678} (step 13) -> R1C7 + R2C8 = [58/59/67/68/86], no 9 in R1C7, clean-up: no 5 in R1C6
[I hope Ed doesn’t consider this step to be a chain.]

18. 45 rule on N4 4 innies R4C23 + R5C3 + R6C1 = 19 = {1369/1459/1468/2359/2467/3457} (cannot be {1279/1378/1567/2368/2458} which clash with R6C23
18a. 5 of {1459/2359/3457} must be in R6C1 -> no 5 in R4C23

19. 45 rule on N7 2 innies R7C23 = 2(1+1) outies R6C1 + R9C4 + 1
19a. Max R7C23 = 14 -> max R6C1 + R9C4 = 13 -> max R9C4 = 8

20. 3 in R6 only in R6C4568
20a. 45 rule on R6789 4 remaining innies R6C4568 = 18 = {1359/1368/2367} (cannot be {2358} which clashes with R6C23, other combinations don’t contain 3)
20b. 3 of {1359} must be in R6C8 (R6C456 cannot be {139} because 40(7) cage at R3C5 cannot contain both of 1,3, step 6a), 3 of {1368/2367} must be in R6C8 -> R6C8 = 3, R5C8 = 5, clean-up: no 8 in R2C4 (step 7)
20c. 5 in N4 only in R46C1, locked for C1, clean-up: no 4 in R12C1, no 7 in R9C2
20d. 5 in N5 only in R4C5 + R6C456, locked for 40(7) cage at R3C5, no 5 in R3C56
[I’d seen step 20a for some time but only just spotted step 20b.]

21. 8(3) cage at R4C4 = {134} (only remaining combination), no 2

22. Hidden killer quad 1,2,3,4 for N1, R12C1 contains one of 1,2,3, R1C3 = {1234}, R2C3 = {123} -> R3C123 must contain one of 1,2,3,4
22a. Killer quad 1,2,3,4 in R3C123, 40(7) cage at R3C5 (step 6a) must contain one of 3,4 in R3C56, R3C7 and R3C8, locked for R3

23. Hidden killer quad 1,2,3,4 for N2, R1C4 = {1234}, R3C56 contains one of 3,4 -> 12(3) cage at R1C5 must contain two of 1,2,3,4 = {129/138/147/237/246} (cannot be {156} which only contains one of 1,2,3,4, cannot be {345} which clashes with R3C56), no 5, clean-up: no 6 in R6C4 (step 7)

24. R6C4568 (step 20a) = {1359/1368/2367} -> R6C456 = {159/168/267}
24a. 8 of {168} must be in R6C4 -> no 8 in R6C56

25. R1C34 = {14/23}, 8(3) cage at R4C4 = {134} -> either 1,4 locked for C34 or 3 locked for C34
25a. 19(4) cage at R2C3 = {1279/1459/1468/1567/2359/2368/2458/2467} (cannot be {1369/1378/3457} because R234C3 cannot contain both of 1,3 or both of 3,4)
25b. 19(4) cage = {1279/1459/1468/1567} => 1 locked for C3 and 3 locked by X-Wing in R1C34 + 8(3) cage at R4C4 for C34
or 19(4) cage = {2359/2368} => 3 locked for C3 and 1 locked by X-Wing in R1C34 + 8(3) cage at R4C4 for C34
or 19(4) cage = {2458/2467}, 2,4 locked for C3 => R15C3 = {13}, locked for C3
-> no 1,3 in R89C3; clean-up: no 8 in R8C7 (step 4)
[Definitely a forcing chain this time.]

26. 16(3) cage at R8C3 = {259/268/367/457} (cannot be {358} which clashes with R9C12, cannot be {349} = {49}3 which clashes with R7C2 + R9C12)
26a. 3 of {367} must be in R9C4, 7 of {457} must be in R9C3 (R89C3 cannot be {45} which clashes with R7C2) -> no 4 in R9C3, no 7 in R9C4

27. Hidden killer triple 3,4,5 in R7C2, R9C12 and the rest of N7, R7C2 = {45}, R9C12 contains one of 3,4,5 -> the rest of N7 must contain one of 3,4,5
27a. 1 in N7 only in 16(4) cage at R6C1 = {1258/1267/1357/1456} (cannot be {1348} = 8{134} because 16(4) cage cannot contain both of 3,4 in N7)
27b. Consider the combinations for R12C2
R12C2 = {59}, locked for C2, no 5 in R8C2
or R12C2 = {68} => R12C1 = {27}, locked for C1 => 2 of 16(4) cage = {1258} must be in R8C2
-> no 5 in R8C2
27c. 5 of 16(4) cage = {1258} must be in R6C1 -> no 8 in R6C1, clean-up: no 6 in R6C9 (step 5)

28. 17(3) cage at R4C1 = {269/359/368/458/467} (cannot be {278} which clashes with R6C3)
28a. Killer pair 5,6 in 17(3) cage and R6C1, locked for N4

29. Hidden killer quad 6,7,8,9 in 16(4) cage at R6C1, R7C3, R89C3 and R9C12 for N7, 16(4) cage contains one of 6,7,8 in N7, R7C3 = {6789}, R9C12 contains one of 7,8,9 -> R89C3 must contain one of 6,7,8,9
29a. 16(3) cage at R8C3 (step 26) = {259/268/457} (cannot be {367} which contains both of 6,7 in R89C3), no 3

30. 4 in N9 only in 21(4) cage at R6C9 = {1479/2469/3459/3468} (cannot be {2478} which clashes with R9C89)
30a. R6C9 = {89} -> no 8,9 in R78C9
30b. Hidden killer pair 3,4,5 in 25(4) cage at R1C8 and 21(4) cage at R6C9 for C9
30c. 25(4) cage at R1C8 (step 13) = {3589/4579/4678} (cannot be {3679} because 21(4) cage cannot contain both of 4,5 in C9)

31. R789C4 = R6C9 + 13 (step 16)
31a. R6C9 = {89} -> R789C4 = 21,22, no 2 in R9C4
31b. 16(3) cage at R8C3 (step 29a) = {259/268/457}
31c. 5 of {259} must be in R9C4, 7 of {457} must be in R9C3 -> no 5 in R9C3
[With hindsight the breakthrough in step 37 has been available from here, or even earlier if I’d seen this step next after step 27. I could have finished this puzzle a day earlier if I’d spotted it immediately.]

32. 45 rule on N9 5(1+4) outies R6C9 + R78C6 + R9C56 = 22
32a. R6C9 = {89} -> R78C6 + R9C56 = 13,14, no 9 in R8C6 + R9C56

33. 45 rule on N14 3 outies R1345C4 = 1 innie R6C1 + 7
33a. R6C1 = {56} -> R1345C4 = 12,13 = {1236/1245/1237/1246/1345} -> R3C4 = {567}
33b. R145C4 cannot be 2{14}, because R15C3 cannot be [33] -> R1345C4 = {1236/1237/1345}, 3 locked for C4, clean-up: no 8 in R6C4 (step 7)
[I tried to find a better way to explain step 33b, but couldn’t think of one.]

34. 8 in C4 only in R789C4, locked for N8, clean-up: no 2 in R7C5
34a. R789C4 = 21,22 (step 31a) = {489/589/678}
34b. 4,5 of {489/589} must be in R9C4, 8 of {678} must be in R78C4 (R78C4 cannot be {67} because 28(4) cage at R7C2 only contains one of 6,7) -> no 8 in R9C4
34c. 8 in C4 only in R78C4, locked for 28(4) cage at R7C2, no 8 in R7C3

35. 16(3) cage at R8C3 (step 31b) = {259/268/457}
35a. 6 of {268} must be in R9C4 -> no 6 in R89C3

36. 45 rule on C12 1 outie R6C3 = 1 innie R7C2 + 3
36a. 28(4) cage at R7C2 = {4789/5689}
36b. 7 of {4789} must be in R78C4 (R7C23 cannot be [47] which clashes with R7C2 + R6C3), no 7 in R7C3

37. 16(3) cage at R8C3 (step 31b) = {259/268/457}
37a. 16(4) cage at R6C1 (step 27a) = {1258/1357/1456} (cannot be {1267} which clashes with 16(3) cage at R8C3) -> R6C1 = 5, R6C9 = 9 (step 5), clean-up: no 2,6 in R2C4 (step 7)
37b. R4C5 = 5 (hidden single in R4)
37c. 6 in R6 only in R6C56, locked for N5 and 40(7) cage at R3C5, no 6 in R3C56

38. R6C3 = 8 (hidden single in R6), R6C2 = 1, R7C2 = 5 (step 2), clean-up: no 7 in R9C1
38a. 28(4) cage at R7C2 = {5689} (only remaining combination), no 7

39. R789C4 (step 34a) = {489/589}, no 6, 9 locked for C4, N8 and 28(4) cage at R7C2 -> R7C3 = 6, R2C4 = 4, R6C4 = 7 (step 7), R9C4 = 5, R3C4 = 6, clean-up: no 8 in R1C7, no 1 in R7C5, no 4 in R8C5

40. Naked pair in R45C4, locked for C4, N5 and 8(3) cage at R4C4 -> R5C3 = 4, R1C4 = 2, R1C3 = 3

41. R4C6 = 4 (hidden single in N5), R4C7 = 6, R1C7 = 5, R1C6 = 9, R5C67 = [87], R5C5 = 9, R4C8 = 8, clean-up: no 6 in R1C1, no 6,7 in R2C1, no 9 in R2C2
41a. Naked pair {68} in R12C2, locked for C2 and N1, clean-up: no 1 in R12C1, no 4 in R9C1

42. R12C1 = [72], R2C3 = 1, R2C7 = 3, R4C1 = 9, clean-up: no 3 in R9C2

43. 19(4) cage at R2C3 (step 25a) = {1567} (only remaining combination) -> R34C3 = [57]

44. R3C56 = [83], R3C12 = [49], R4C2 = 3 (cage sum), R9C2 = 4, R9C1 = 8, R3C9 = 7, R9C9 = 6, R9C8 = 2

and the rest is naked singles.

I don't think I've ever used so many hidden killer quads before. Maybe after finding the first one I got into the habit of looking for more of them.

Rating Comment. I'll rate my walkthrough for A224 at very, very Hard 1.5 but I'm reluctant to go any higher, even though I found it hard to spot a lot of my steps.
Walkthrough by Ed:
Me to and agree, it is really neat. Nice to have a shortcut way to solve this puzzle. My step 9 is the way around manu's first chain. I found his chain way first but my purist's heart kept looking for a way around it.

Andrew's way to get his first placement is the best. Really neat use of killer subsets. After that, he found a longcut way but I still really appreciate the effort and have read closely the walk-through. My step 7d is the key reason why mine and Andrew's walkthrough's are different.

My other step I really enjoyed was my step 23 but Andrew got it using killer subsets again. He takes the record for number and sizes of killer subsets in his walkthrough!

A224 26 steps
This is an optimised solution to get it cracked ASAP. However, I try and do all the clean-up as I go.

Forgot the prelims again. Please use Andrew's.

1. 3(2)n6 = {12}: both locked for c9 and n6
1a. -> 8(2)n6 = {35}: both locked for c8 & n6
1b. 8(2)n9 = [17/26]
1c. no 7,8,9 in r4c6

2. "45" on r789: 2 outies r6c19 = 14 = [59]/{68}(no 1,2,3,4,7)
2a. no 9 in r6c1

3. "45" on c12: 2 innies r67c2 = 6 = [15/24]
3a. r6c3 = (78)

4. 28(4)r7c2 = {4789/5689}
4a. = one of 4/5 but not both -> no 4 or 5 in r7c34+r8c4

5. "45" on n7: 2 outies r6c1+r9c4 + 1 = 2 innies r7c23
5a. max. r7c23 = 14 -> max. two outies = 13
5b. -> no 9 in r9c4

6. "45" on c12: 1 innie r7c2 + 3 = 1 outie r6c3 [47/58] = [5/7..] (no eliminations yet)
6a. 12(2)n7 = {39/48/57} = [4/5/9;3/5/8..](no eliminations yet)

This step is to show no 5 in r89c3. It's the most complicated step in this solution. 7d is the key.
7. 16(3)cage at r8c3: combos with 5 = {259/358/457}
7a. but {59}[2] clashes with r7c2 + 12(2)n7 (step 6a) needing two of 4,5,9
7b. {358} clashes with 12(2)n7 (step 6a)
7c. {45}[7] clashes with r7c2 = (45)
7d. {57}[4] clashes with IOD c12 = +3 (step 6.)
7e. -> no 5 in r89c3

8. Hidden killer pair 1,2 in r6 -> r6c456 must have 1 or 2
8a. 40(7)r3c5 = {14/23}{56789} = one of 1/2 but not both
8b. -> no 1 or 2 elsewhere in that cage apart from r6c456
8c. note: if it has 3 in r6c456 it must also have 2 (no eliminations yet)

9. "45" on r6789: 5 innies r6c45678 = 22 and must have 3&4 for r6
9a. but {23458} blocked by 9(2)n4 = [2/8..]
9b. = {13459/13468/23467} = [3/5 and if both, no 2..]
9c. must have 3/5 for r6c8 and if 3 in r6c456 must also have 2 (step 8c)
9d. -> no 3 in r6c456

10. r6c8 = 3 (hsingle r6)
10a. -> r5c8 = 5

11. 8(3)r4c4: but {125} = [5]{12} clashes with r5c9 = (12)
11a. = {134} only

12. r3c6 sees all 5,6,7,8,9 in n5 -> r3c6 = (34) (CPE)
12a. 40(7) can only have one of 3/4 (step 8a) -> no 3,4 elsewhere in the 40(7)r3c5

13. r6c7 = 4 (hsingle in r6)
13a. no 6 in r4c6
13b. r5c56 = 15 (no 2,3)

14. 5 in n5 only in 40(7)r3c5 -> no 5 in r3c5

This next step is why step 7 was so important
15. 5 in c3 only in 19(4)r2c3 -> no 5 in r3c4

16. "45" on r12: 1 innie r2c3 + 6 = 1 outie r3c9
16a. r2c3 from (123); r3c9 = (789)

17. 21(4)r6c9: all combos have 1/2 which must go in r8c8 except....
17a. {3567} blocked by r9c9 = (67)
17b. {3459} must have 9 in r6c9 -> 4 must be in r8c8
17c. {3468} -> 4/6/8 in r8c8
17d. -> no 7 nor 9 in r8c8

18. "45" on c9: 2 outies r18c8 - 4 = 1 innie r9c9
18a. -> no 4 in r1c8 (IOU)

19. "45" on c9: 3 outies r189c8 = 12
19a. = {129/147/246}(no 8)
19b. = [9]{12}/[741/642]
19c. r1c8 = (679)
19d. r8c8 = (124)

20. "45" on c89: 1 outie r3c7 + 14 = 2 innies r27c8
20a. max. 2 innies = 17 -> r3c7 = (123)
20b. r27c8 = 15-17: no 1,2,4

21. 25(4)n3 = {3589/3679/4579/4678} = [5/6..](no eliminations yet)

22. The two 14(2) cages at r1c2 & r1c6 must have different combinations since r1c2 sees both of r1c67 -> must have both 5 & 6 (and both 8&9)
22a. -> no 5,6,8 in r1c1 (technically not needed for this solution but couldn't resist!)
22b. no 1,3,4 in r2c1

Andrew got this next elimination with some really nice implied killer subsets
23. from steps 21 and 22, three of 5 & 6 taken for r12 -> {156} blocked from 12(3)n2
23a. {345} blocked from 12(3)n2 by r3c6 = (34)
23b. 12(3) = {129/138/147/237/246}(no 5)

24. 5 in n2 only in r12c6
24a. 17(3)r2c6: The only permutation possible with 5 in r2c6 is [5]{39} = [5 -> 9..]
24b. 5 in r1c6 -> 9 in r1c7
24c. -> 9 locked in r1c7 + r2c78 for n3 (Locking cages)
24d. no 3 in r2c3 (IODr12 = +6)

25. 25(4)n3 = {4678} only: all locked for n3
25a. 4 and 8 locked for c9
25b. no 6 in r6c1 (h14(2)r6c19)

26. h12(3)r189c8 = {147/246}
26a. -> r8c8 = 4
[Andrew pointed out that the hidden cage isn't necessdary: just hidden single 4 in c8]

cracked
Walkthrough by wellbeback:
I'm a little late to the party - and I too had a couple of short chains in my solution.
I have a lot of "Either A is true or B is true - but either way they both imply that C is true" logic.

After writing my solving path I perused the other solutions and I see a lot of commonality - but also some differences. In particular - my very first placement was the very last line in Ed's walkthrough :)

Thanks for the nice puzzle Ed ;clapclap;

1. c89

3/2@r4c9 = {12)
-> 8/2@r5c8 = {35}
-> 8/2@r9c8 = [17]|[26] -> r9c9 from (67)
Outies r789 -> r6c19 = +14 = [59]|[68]|[86] -> r6c9 from (689)
-> Max r123c9 = +22 (589)
-> Min r1c8 = 3.
But 3 already in c8 -> min r1c8 = 4.

Outies r9 r189c8 = +12.
r89c8 cannot = +8 (since r9c89 = +8) -> r1c8 cannot = 4.
Also 5 already in c8 -> Min r1c8 = 6 and max r8c8 = 4.

Outies n3 r12c6+r4c8 = +22.
5 already in n6 -> min r4c8 = 6.
-> Max r3c8 = 4.
-> r389c8 = {124} with 4 in r38c8.

But r3c8 = 4 puts 9 in r1c8 and 6 in r4c8 which leaves no way to put +16 (outies n3) in r12c6.

-> r8c8 = 4.
-> r19c8 = [62]|[71]. I.e., r1c8 = r9c9 = (6|7).
Also 4 in c9 in n3 and 4 in c7 in n6.
Also 3 in r78c9 in n9 and r23c7 in n3 -> 21/4@r6c9 = {34(68|59)}.
Also 7 in c9 in r123c9 or r9c9 -> 25/4@r1c8 = {47(59|68)}.
Also r2c7+r3c78 in n3 = {123}
-> r1c7+r2c8 = +14 -> r1c6 = r2c8 which cannot be 5 -> 14/2@r1c6 cannot be [59].


2. 40/7@r3c5

Whatever is in r3c6 must go in n5 in r45c4 -> must be 5 or less.
Since 4 is in c7 in n6 then either r4c6 = 6 or r5c6 = (6789). (Fixed typo - Thx Andrew)
Whichever it is must be in 40/7@r3c5 and only place for it is in r3c5.
-> r3c5 from (6789).

The two values not in 40/7@r3c5 = +5. One must go in r45c4 and the other in r45c6.
-> r45c4 cannot = +5 -> r5c3 not = 3.


3. r6

Innies c12 r67c2 = +6. Since r7c2 is in a 28/4 that must be min 4.
-> r67c2 = [15]|[24]
-> 9/2@r6c2 = [18]|[27]

Where are 1 and 2 going in r6?
One of them goes in r6c2, the other in r6c456. (In the 40/7).

Where is 3 going in r6? Must be in r6c456 or r6c8.
Either 9/2@r6c2 = [18] -> r6c19 = [59] -> 8/2@r5c8 = [53]
Or 9/2@r6c2 = [27] -> 1 in r6c456 -> 3 cannot be in 40/7 -> HS 3 in r6c8.
Either way 8/2@r5c8 = [53]

r5c34 cannot be {12} -> 5 not in 8/3@r4c4
-> 8/3@r4c4 = {134} with 3 in r45c4.

Where is 4 going in r6? Must be in r6c456 or r6c7.
Either 9/2@r6c2 = [18] -> 2 in r6c456 -> 4 cannot be in 40/7 -> r6c7 = 4.
Or 9/2@r6c2 = [27] -> 1 in r6c456 -> r45c4 = {34} -> r6c7 = 4.
Either way r6c7 = 4.


4. r6c9

10/2@r4c6 = [46]|[28] -> r4c7 from (68).
r4c8 from (678)
r6c9 from (689)

Either r1c8 = 6 -> 6 in r9c9 and 8 in r123c9 -> r6c9 = 9.
Or r1c8 = 7 -> r4c78 = {68} -> r6c9 = 9.
Either way r6c9 = 9.


5. Breakthrough!

Innies n2356 r1345c4+r6c9 = +21
Since r6c9 = 9 -> r1345c4 = +12 and since it contains a 3 in r45c4 it cannot contain a 4.
-> r1345c4 must be [26{13}]!!

Rest is cleanup.


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PostPosted: Fri Oct 31, 2014 1:03 am 
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Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Double Diamond 1 by HATMAN (November 2011) here
Puzzle Diagram:
Image
Original link failed; replaced by Andrew's Excel image.
Code: Select, Copy & Paste into solver:
(Archive Note): There isn't currently a valid code string for this puzzle.
Solution:
+-------+-------+-------+
| 5 4 3 | 1 8 7 | 9 6 2 |
| 2 9 8 | 5 3 6 | 7 1 4 |
| 1 6 7 | 2 9 4 | 5 3 8 |
+-------+-------+-------+
| 3 8 5 | 9 1 2 | 6 4 7 |
| 6 7 9 | 3 4 5 | 8 2 1 |
| 4 1 2 | 6 7 8 | 3 9 5 |
+-------+-------+-------+
| 9 5 4 | 8 6 1 | 2 7 3 |
| 8 3 1 | 7 2 9 | 4 5 6 |
| 7 2 6 | 4 5 3 | 1 8 9 |
+-------+-------+-------+
Quote:
HATMAN: One number is missing from all four diagonals. So essentially they are four cages of the same size from 36 to 44.
Even when you know the missing number the resultant killer is still quite hard.
Note there is a very interesting innie.

simon_blow_snow: I think the interesting innie is this:
Hint:
Innies N12346789: R37C5+R5C37 = 32, <> {12345}
R37C5 and R5C37 each ranging from 15 to 17
Even knowing this, this puzzle is very diabolical! :pallid:
Any extra hint? :?:

simon_blow_snow: Turned out I missed the rule about "all 4 diagonals have the identical sum", and assumed that they could have different sums (the puzzle still has a unique solution under that assumption). Therefore I found the puzzle much harder than it should be.
Once I understand the correct constraint, everything flows nicely and I have this super quick walkthrough of only 6 steps (!):
Much thanks to HATMAN for another very nice challenging puzzle (and the hint he PMed me which I just discovered after posting this walkthrough)!
I wonder if there is any way to solve this logically assuming each diagonal can have different sums?

HATMAN: Nice the way you found the diagonal number - walkthrough close to mine after that.
I'll wait until Andrew posts before making any further comments.

Andrew: Thanks HATMAN for a challenging puzzle! I like the concept of the diagonals, particularly when the diagonals are all required to have the same unknown total. I assume from the puzzle title that more are planned.
Congratulations Simon for an excellent solving path! After a really neat way to fix the total for the diagonals, you found a couple of very useful 45s which I didn't spot; I assume from HATMAN's comment that he also found those 45s.
HATMAN wrote:
I'll wait until Andrew posts before making any further comments.
Thanks! I appreciate that. :)

Joe Casey: This is how I got started (See discussion below)
I get lost reading WTs, so I can't figure out whether this amounts to the same methods as you others.

Walkthrough by simon_blow_snow:
Turned out I missed the rule about "all 4 diagonals have the identical sum", and assumed that they could have different sums (the puzzle still has a unique solution under that assumption). Therefore I found the puzzle much harder than it should be.

Once I understand the correct constraint, everything flows nicely and I have this super quick walkthrough of only 6 steps (!):

6 steps walkthrough

1:
Innies N12346789: R37C5+R5C37 = 32 <> {12345}
--> R3C5 10(2) = [64/73/82/91]
--> R5C3 12(2) = [75/84/93]
R1C2 D\ & R1C8 D/: R5C6-R5C4 = (17+15)-(14+16) = 2
--> R5C46 = [35/46] (R5C3 12(2): not [57])
R1C2 D\ & R2C9 D/: R6C5-R4C5 = (14+16)-(13+11) = 6
--> R46C5 = [17/39] (R3C5 10(2): not [28])

2:
R1C2 D\123678 = 14(3)+16(3) = 30(6) must include [3] or [6]
(proof: {1245789} cannot produce 30(6) no matter which candidate is dropped)
--> R4C5+R5C46 <> [346]
--> R4C5 <> [3]
--> R3C5 10(2) = [91], R6C5 = [7] ([1] D\/, [7] D\/)

3:
N3 8(3) = {125/134} ([1] N3)
--> R2C9 13(3) with no [1], <> [9]
R7C4 11(3) <> [9]
--> R2C9 D/ has no [9] at all, sum to 45-9 = 36
--> all 4 diagonals sum to 36, <> [9]
--> R5C6 = 36-14-1-16 = [5] (D\/)
--> R5C3 12(2) = [93] ([3] D\/)
Hidden singles N3,R2: R1C7 = R2C2 = [9]

4:
Innie-outies N124: R7C1-(R3C6+R6C3) = 3
--> R7C1 = [9], R3C6+R6C3 = {24} (D/)
R1C2 14(3) = {248/347} ([4] R3C123)
Innie-outies N2: R1C3 = R3C46-3 <> {1245}, = [3/6/7/8]
Innie-outies N12: R4C3 = R3C16 <> {24}, = [5/6/8]

5:
R2C1 13(3) = {148/256}
Innies N12: (R2C1+R3C2)+R3C16 = 13 can't be {25}+[24]
--> R2C1+R3C2 <> {25}
R2C1 13(3): R4C3 <> [6], = [5/8]
--> R3C16 = 5/8 = [14/32/62] (R3C1 = [1/3/6]) (step 4)
--> R2C1+R3C2 = 8/5 = {14/26} must include [2/4]
--> R1C2+R2C3 <> {24}
--> R1C2 14(3): R3C4 <> [8]

6:
[5] of N1 locked in R1C1+R3C3
--> N1 21(3): R1C1+R3C3 = 12 = {57} (N1)
--> [8] of R3 locked in R3C89 (N3)
--> R1C8 17(3): R1C8+R2C7 = {67}, R3C6 = [4] (D/)
--> R3C1 = [1], R4C3 = [5] (step 5)
--> N1 21(3) = [597], R3C4 = [2]
--> R2C1 13(3) = [265] (D\), R6C3 15(3) = [258]
Hidden single N3: N3 8(3) = [215]
Hidden single N3: R2C9 13(3) = [436] (D/)
--> R1C2 14(3) = [482] (D\), R1C8 17(3) = [674], R7C4 11(3) = [812]
--> R6C7 16(3) = [376], R7C6 13(3) = [148]
C9 24(5) = [87153]

Naked singles from here.

(2 minor typos corrected thanks to Andrew.)

Much thanks to HATMAN for another very nice challenging puzzle (and the hint he PMed me which I just discovered after posting this walkthrough)!

I wonder if there is any way to solve this logically assuming each diagonal can have different sums?
Walkthrough by Andrew:
Thanks HATMAN for a challenging puzzle! I like the concept of the diagonals, particularly when the diagonals are all required to have the same unknown total. I assume from the puzzle title that more are planned.

Congratulations Simon for an excellent solving path! After a really neat way to fix the total for the diagonals, you found a couple of very useful 45s which I didn't spot; I assume from HATMAN's comment that he also found those 45s.

HATMAN wrote:
I'll wait until Andrew posts before making any further comments.
Thanks! I appreciate that. :)

Here is my walkthrough for Double Diagonal 1

Prelims

a) R34C5 = {19/28/37/46}, no 5
b) R5C34 = {39/48/57}, no 1,2,6
c) 21(3) cage at R1C1 = {489/579/678}, no 1,2,3
d) 8(3) cage at R1C9 = {125/134}
e) 11(3) cage at R7C4 = {128/137/146/236/245}, no 9
f) 19(5) cage at R9C3 = {12349/12358/12367/12457/13456}

Steps resulting from Prelims
1a. 8(3) cage at R1C9 = {125/134}, 1 locked for N3
1b. 19(5) cage at R9C3 = {12349/12358/12367/12457/13456}, 1 locked for R9

2. R2C9 to R9C2 diagonal
2a. 13(3) cage at R2C9, 11(3) cage at R7C4, max R5C6 + R6C5 = 17 -> max diagonal total = 41
2b. Min diagonal total = 36 -> min R5C6 + R6C5 = 12, no 1,2 in R5C6 + R6C5
[Note. Max diagonal total = 41 must contain 1,2,3 -> each of the 13(3) cages on this diagonal must contain one of 1,2 because 13(3) doesn’t contain both of 1,2.]

3. R1C8 to R8C1 diagonal
3a. 17(3) cage at R1C8, 15(3) cage at R6C3, max diagonal total = 41 (step 2a) -> max R4C5 + R5C4 = 9, no 7,8,9 in R4C5, no 9 in R5C4, clean-up: no 1,2,3 in R3C5, no 3 in R5C3

4. 45 rule on N12346789 2 innies R5C7 + R7C5 = 2 outies R4C5 + R5C4 + 10
4a. Min R4C5 + R5C4 = 4 -> min R5C7 + R7C5 = 14, no 1,2,3,4 in R5C7 + R7C5
4b. Max R5C7 + R7C5 = 18 (doubles possible) -> max R4C5 + R5C4 = 8, no 6 in R4C5, no 8 in R5C4, clean-up: no 4 in R3C5, no 4 in R5C3
[It would have been simpler if I’d spotted 45 rule on N12346789 4(2+2) innies R37C5 + R5C37 = 32, R37C5 = 15,16,17, no 1,2,3,4,5, R5C37 = 15,16,17, no 1,2,3,4,5]

5. Cages on the 4 diagonals total 112, max innies on these diagonals = 26 (because max innies on R1C8 to R8C1 diagonal = 11) with the innies counted twice = total on 4 diagonals = 164
[Unfortunately this doesn’t change the max amount total for the diagonals.]
5a. Similarly min totals on 4 diagonals = 4*36 = 144 -> min innies counted twice = 32 -> min innies = 16
5b. Min totals of innies, counted twice = 32,36,40,44,48 or 52 -> total of innies on the diagonals = 16,18,20, 22 or 24

6. Total of given cages on R2C9 to R9C2 diagonal = 24, total of given cages on R2C1 to R9C8 diagonal = 26 -> R5C6 + R6C5 = R5C4 + R6C6 + 2 (because specified diagonals have the same total) -> R5C6 = R5C4 + 2, no 3,4,8 in R5C6
6a. R5C46 cannot be [57] (CCC with R5C34), no 5 in R5C4, no 7 in R5C6, clean-up: no 7 in R5C3

7. Similarly total of given cages on R1C8 to R8C1 diagonal = 32, total of given cages on R2C1 to R9C8 diagonal = 26 -> R5C4 + R6C5 = R4C5 + R5C4 + 6 -> R6C5 = R4C5 + 6 -> R4C5 = {123}, R6C5 = {789}, clean-up: no 6 in R3C5

8. R6C5 = R4C5 + 6 (step 7) = [17/39] (cannot be [28], CCC with R34C5), no 2 in R4C5, no 8 in R6C5, clean-up: no 8 in R3C5
8a. Naked pair {79} in R36C5, locked for C5
8b. Max R5C6 + R6C5 = 16 -> max total on specified diagonals = 40

9. R2C9 to R9C2 diagonal
9a. Max total on specified diagonals = 40 must contain all of 1,2,3,4, 11(3) cage at R7C4 contains two of 1,2,3,4 -> 13(3) cage at R2C9 must contain two of 1,2,3,4 = {139/148/238/346} (cannot be {157/256} which only contain one of 1,2,3,4, cannot be {247} which clashes with 11(3) cage at R7C4), no 5,7
9b. 11(3) cage at R7C4 = {128/146/236/245} (cannot be {137} which clashes with 13(3) cage at R2C9), no 7
9c. Combining cages on this diagonal the only remaining candidates are {579/678} but there’s no 8 in R5C6 + R6C5 -> R5C6 + R6C5 = {579}, no 6, clean-up: no 4 in R5C4 (step 6), no 8 in R5C3
9d. Naked pair {59} in R5C36, locked for R5
9e. We don’t yet know which of 5,7,9 are in R5C6 + R6C5. However since they are all odd numbers R5C6 + R6C5 must be even so the total on the diagonals must be even = 36, 38 or 40 -> the missing number is 5,7 or 9 but two of these numbers are locked in R5C6 + R6C5 -> no other 5,7,9 on the R2C9 to R9C2 diagonal -> 13(3) cage at R2C9 = {148/238/346}, 11(3) cage at R7C4 = {128/146/236}

10. R5C46 = [35/79]
10a. R46C5 = [17] (cannot be [39] which clashes with R5C46), 1,7 placed for appropriate diagonals, R3C5 = 9, R5C4 = 3, placed for the appropriate diagonals, R5C3 = 9, R5C6 = 5, placed for the appropriate diagonals

11. From the placements in step 10, total on the specified diagonals = 36 -> no 9 on the specified diagonals

12. R5C7 + R7C5 = R4C5 + R5C4 + 10 (step 4)
12a. R4C5 + R5C4 = 4 -> R5C7 + R7C5 = 14 = {68}, CPE no 6,8 in R5C5 + R7C7
[Now I’ve reached the position I would have done if I’d spotted 4(2+2) innies for N12346789.]

13. R1C2 to R8C9 diagonal
13a. 14(3) cage at R1C2 = {248/347}, 4 locked for the diagonal
13b. 16(3) cage at R6C7 = {268/367}, 6 locked for the diagonal

14. 9 in N1 only in 21(3) cage at R1C1 = {489/579}, no 6

15. R2C9 to R9C2 diagonal
15a. 13(3) cage at R2C9 = {238/346}, 3 locked for the diagonal
15b. 8 of {238} must be in R2C9 + R3C8 (R2C9 + R3C8 cannot be {23} which clashes 8(3) cage at R1C9), no 8 in R4C7
15c. 11(3) cage at R6C4 = {128/146}, 1 only in R7C4 + R8C3, CPE no 1 in R7C3 + R8C4

16. R1C7 = 9 (hidden single in N3), R2C2 = 9 (hidden single in N1)

17. 45 rule on N2 2 innies R3C46 = 1 outie R1C3 + 3
17a. Min R3C46 = 6 -> min R1C3 = 3
17b. R3C46 cannot total 7 -> no 4 in R1C3
17c. 1 in C3 only in R89C3, locked for N7

18. 45 rule on N12 1 outie R4C3 = 2 innies R3C16
18a. Max R3C16 = 8, no 7,8 in R3C1, no 8 in R3C6
18b. Min R3C16 = 3 -> no 2 in R4C3
18c. R3C16 cannot total 4 -> no 4 in R4C3
18d. R4C3 cannot be 5 more than R3C6 -> no 5 in R3C1

19. 13(3) cage at R2C1 = {148/256}
19a. 8 of {148} must be in R4C3, no 8 in R2C1 + R3C2

20. 45 rule on N4 2 outies R37C1 = 2 innies R46C3 + 3
20a. Min R46C3 = 7 -> min R37C1 = 10, no 1,2,3 in R7C1

21. 45 rule on N3 2(1+1) outies R3C6 + R4C7 = 1 innie R3C9 + 2, IOU no 2 in R4C7
21a. Min R3C6 + R4C7 = 5 -> min R3C9 = 3

22. 7 in N3 only in R1C8 + R2C7 + R3C9, CPE no 7 in R3C6
22a. 17(3) cage at R1C8 = {278/458/467}
22b. 2 of {278} must be in R3C6 -> no 2 in R1C8 + R2C7

23. 7 in N3 only in R1C8 + R2C7 + R3C9
23a. 7 in R1C8 + R2C7 => 17(3) cage at R1C8 = {278/467}
or R3C9 = 7 => R3C6 + R4C7 = 9 (step 21) = [63] => 17(3) cage at R1C8 = 6{47}
-> 17(3) cage at R1C8 = {278/467}, no 5, 7 locked for N3 and R1C8 to R8C1 diagonal
23b. 15(3) cage at R6C3 = {258/456}

24. 7 in R3 only in R3C34, CPE no 7 in R1C2 + R2C3

25. 14(3) cage at R1C2 = {248/347}
25a. 7 in R3 only in R3C34
R3C3 = 7 => 21(3) cage at R1C1 = {579}
or R3C4 = 7 => R1C2 + R2C3 = {34}, locked for N1 => 21(3) cage at R1C1 = {579}
-> 21(3) cage at R1C1 = {579}, locked for N1

26. 13(3) cage at R2C1 = {148/256}
26a. 5,8 only in R4C3 -> R4C3 = {58}

27. R4C3 = R3C16 (step 18)
27a. R4C3 = {58} -> R3C16 = 5,8 = [14/26/62], no 3,4 in R3C1

28. R3C46 = R1C3 + 3 (step 17)
28a. R1C3 = {368} -> R3C46 = 6,9,11 = [24/42/72/74], no 6,8
28b. R3C16 (step 28a) = [14/62], no 2 in R3C1

29. 17(3) cage at R1C8 = {278/467} (step 23a)
29a. 4 of {467} must be in R3C6 -> no 4 in R1C8 + R2C7

30. 14(3) cage at R1C2 = {248/347}
30a. Hidden killer pair 3,8 in 14(3) cage and R1C3 for N1, 14(3) cage contains one of 3,8 -> R1C3 = {38}

31. R1C3 = {38} -> R3C46 = 6,11 (step 28a) = [24/42/74], 4 locked for R3 and N2

32. 14(3) cage at R2C4 = {158/167/356} (cannot be {257} which clashes with R3C46), no 2

33. 8 in R3 only in R3C89, locked for N3
33a. Naked pair {67} in R1C8 + R2C7, locked for N3 and R1C8 to R8C1 diagonal, R3C6 = 4 (step 23a), placed for R1C8 to R8C1 diagonal

34. 14(3) cage at R1C2 = {248/347}, 4 locked for N1
34a. 2 of {248} must be in R3C4 -> no 2 in R1C2 + R2C3

35. 13(3) cage at R2C1 = {256} (only remaining combination) -> R2C1 + R3C2 = {26}, locked for N1 and R2C1 to R9C8 diagonal, R4C3 = 5, placed for R2C1 to R9C8 diagonal, R3C1 = 1, R3C3 = 7, R1C1 = 5, R3C4 = 2, placed for R1C2 to R8C9 diagonal

36. 14(3) cage at R1C2 = {248} (only remaining combination), 4,8 locked for N1 and R1C2 to R8C9 diagonal -> R1C3 = 3

37. 16(3) cage at R6C7 = {367}, 7 locked for N9
37a. 13(3) cage at R7C4 = {148}, 4 locked for N9, CPE no 1 in R7C79 + R8C6
37b. 15(3) cage at R6C3 = {258} -> R7C2 = 5

38. 16(3) cage at R7C7 can only contain one of 1,2,3, R7C7 = {23} -> no 1,2,3 in R8C8 + R9C9

39. 16(3) cage at R4C2 = {178/367} (cannot be {268} which clashes with R6C3), no 2,4, 7 locked for C2 and N4

40. 45 rule on N4 1 remaining outie R7C1 = 1 remaining innie R6C3 + 7 -> R6C3 = 2, R7C1 = 9, R8C1 = 8 (step 37b)

41. Naked triple {346} in R456C1, locked for C1 and N4 -> R2C1 = 2, R3C2 = 6, R9C1 = 7

42. Naked triple {178} in R456C2, locked for C2 -> R1C2 = 4, R9C2 = 2, R8C2 = 3, R7C3 = 4 (cage sum)

43. 11(3) cage at R7C4 = {128} (only remaining combination) -> R7C4 = 8, R8C3 = 1

44. R8C7 = 4
44a. 45 rule on R8 2 remaining innies R8C89 = 11 = [56]

and the rest is naked singles.

At one stage I’d noted
45 rule on N9 2(1+1) outies R6C7 + R7C6 = 2 innies R7C9 + R9C7 but I don’t think that Law of Leftovers necessarily applies since there isn’t a 45(9) cage; it’s possible that R6C7 and R7C6 can contain the same number. As it happens, the solution does have the same numbers in R6C7 + R7C6 and R7C9 + R9C7. Maybe I didn’t work hard enough on these since the cells are aligned in R7 and in C7?

Rating Comment (added later). I'll rate my walkthrough for Double Diagonal 1 at 1.5 because I used a couple of short forcing chains. Simon's solving path is better and shorter so will have a lower rating.
Joe Casey's Outline Start and discussion about it:
Joe Casey wrote:

This is how I got started:

All 4 uncaged cells from the diags are in N5. Call them North South East and West. Since the diagonals all have the same sum, S is 6 more than N, and E is 2 more than W. Very few combinations are consistent, taking into account the 10- and 12-cages. I was left with N&E being either 1,3 or 3,4, so the missing number is either 9 or 6, which eliminated the last alternative.

I get lost reading WTs, so I can't figure out whether this amounts to the same methods as you others.

simon_blow_snow wrote:

Joe Casey wrote:
I was left with N&E being either 1,3 or 3,4, so the missing number is either 9 or 6, which eliminated the last alternative.

I don't understand this bit. Perhaps you meant "I was left with N&W being either 1,3 or 3,4"? But even that doesn't make sense, because you still have the possibility of N&W being 1,4 (with E&S being 6,7). So the missing number can be 6, 8, or 9. How did you eliminate the possibility of 8 without looking at other cages?

Joe Casey wrote:

Right: I did mean N & W. And I do seem to have discounted 1,4 but I can't remember why; it seemed obvious at the time. So either it was serendipity or there was some other info that I've forgotten. Sorry to be so clueless.


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PostPosted: Fri Oct 31, 2014 2:37 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Paper Solvable 9 Elevenses by HATMAN (November 2011) here
Puzzle Diagram:
Image
Code: Select, Copy & Paste into solver:
3x3::k:2817:2817:2818:2818:2819:2819:2820:2820:2831:25:26:27:28:29:30:31:2831:2821:2830:32:33:34:35:2838:2831:36:2821:2830:2836:2836:2833:37:2833:2838:38:2822:2829:2835:2837:39:2833:2840:2840:2838:2822:2829:2835:40:2837:2837:2833:41:42:2823:2828:2835:2832:2834:2839:43:2839:44:2823:2828:2832:2834:45:46:2839:47:48:2824:49:2827:2827:2826:2826:2825:2825:50:2824:
(you need to add the diagonal)
Solution:
+-------+-------+-------+
| 2 9 6 | 5 3 8 | 4 7 1 |
| 1 8 3 | 4 6 7 | 9 2 5 |
| 7 5 4 | 2 9 1 | 8 3 6 |
+-------+-------+-------+
| 4 2 9 | 1 7 3 | 6 5 8 |
| 6 7 1 | 8 5 9 | 2 4 3 |
| 5 3 8 | 6 4 2 | 1 9 7 |
+-------+-------+-------+
| 8 1 7 | 9 2 5 | 3 6 4 |
| 3 4 2 | 7 1 6 | 5 8 9 |
| 9 6 5 | 3 8 4 | 7 1 2 |
+-------+-------+-------+
Quote:
HATMAN: Been thinking of a assassin with a single cage sum for a while - this one came out as paper solvable.
Note the single Diagonal and almost symmetry around it - clearly you cannot have full symmetry with all cage totals the same.

simon_blow_snow: Here is a complete walkthrough, solving without pencilmarks
Very nice relaxing puzzle, thanks HATMAN! :-)

Andrew: Congratulations HATMAN on achieving your target of making all the cage totals 11! An enjoyable puzzle.
My solving path is fairly similar to Simon's so maybe parts of it are fairly narrow but it's not difficult.
I also tried to make my walkthrough a paper solvable one. Maybe I overdid this by limiting the number of times I locked naked combinations, so my walkthrough is a bit longer than it needs to be.

Walkthrough by simon_blow_snow:
Here is a complete walkthrough, solving without pencilmarks:

No pencilmark walkthrough (7 steps plus prelim, mop-up)

0 (prelim):
All 11(2) cages <> [1]
All 11(3) cages <> [9]
11(4) = {1235}

1:
Hidden single R1: R1C9 = [1] (D9)
Hidden singles C1,R9: R2C1 = R9C8 = [1]
Hidden single N7: R7C2 = [1]
[1] of N5 11(4) must be in R4C4+R6C6 (N5)
Hidden single R5: R5C3 = [1]

2:
R5C3 11(3): R6C45 = 10 = {46} (N5 11(4): not {28/37})
--> R1C9 11(3): D/23 = 10 <> {46}, = {28/37}
--> N5 11(4): D/2345 include all of {235} of D/
--> D/7 11(2) <> {29/39/56}, = {47}
R6C45 = [64], D/23 = {28}
--> D/45 = {35}, R4C4+R6C6={12}

3:
Hidden single D/: D/9 = [9]
R357C1 11(2) cages <> {29}
R9C246 11(2) cages <> {29}
Hidden single C1: R1C1 11(2) = [29]
Hidden single R9: R8C9 11(2) = [92]
Hidden single N7: R7C4 11(2) = [92]

4 (breakthrough):
9 of R5 locked in R5C6 11(2) = [92] (step 2)
--> D/23 = [28] (step 2)
R3C1 11(2): R4C1 <> [3]
Innies N4: R4C1+R6C3 = 12
--> R6C3 <> [9]
Hidden single N4: R4C2 11(2) = [29]

5:
R4C4+R6C6 = [12] (step 2)
Hidden singles C4,R7,N6: R3C4 = R7C5 = [2], R6C7 = [1]
Hidden singles C7,N6: R2C7 = R6C8 = [9]
Hidden single N2: R3C5 = [9]
Hidden single N2,C5: R3C6 = R8C5 = [1]
Innies N3: R3C8 = [3]

6:
R5C2 11(3): R56C2 = 10 = {37} (R4C23+R6C45: not {28/46})
--> D/7 11(2) = [74] (step 2)
--> R4C1+R6C3 = 12 = [48] (step 4)
R3C1 11(2): R3C1 = [7]
R5C1 11(2) = [65]
Hidden single N1: R2C2 = [8]

7:
R6: R6C29 = {37}
--> R3C6 11(3): R4C7+R5C8 = 10 = [64]
--> R4C9 11(2) <> {29/47/56}, = {38}
--> R6C9 11(2) <> [38], = [74]
--> R7C5 11(3): R7C7+R8C5 = 9 <> {18/45}, = [36]

Cracked. The rest are mop-up steps:

R56C2 = [73] (step 6)
R7C1 11(2) = [83]
R7: R7C68 = [56]
--> D/45 = [35] (step 2)
R4C9 11(2) = [83]
R45: R4C58 = [75], R5C4 = [8]
R8: R8C478 = [758]
C8: R1C7 11(2) = [47]
--> R9C6 11(2) = [47]
--> R9C4 11(2) = [38]
--> R1C5 11(2) = [38]
--> R1C3 11(2) = [65]
--> R9C2 11(2) = [65]
R3: R3C239 = [546]
R2: R2C34569 = [34675]

(Typo corrected thanks to Andrew.)

Very nice relaxing puzzle, thanks HATMAN! :-)
Walkthrough by Andrew:
Congratulations HATMAN on achieving your target of making all the cage totals 11! An enjoyable puzzle.

My solving path is fairly similar to Simon's so maybe parts of it are fairly narrow but it's not difficult.

I also tried to make my walkthrough a paper solvable one. Maybe I overdid this by limiting the number of times I locked naked combinations, so my walkthrough is a bit longer than it needs to be.

Here is my complete walkthrough

Prelims

a) All 11(2) cages are {29/38/47/56}, no 1
b) All 11(3) cages are {128/137/146/236/245}, no 9
c) 11(4) cage = {1235}, locked for N5

1. R1C9 = 1 (hidden single in R1), placed for D/

2. R7C2 = 1 (hidden single in N7)

3. Min R6C45 = 10 -> R5C3 = 1, R6C45 = {46}

4. 11(3) cage at R1C9 = {128/137} (cannot be {146} which clashes with R6C4 on D/)
4a. Killer triple 2,3,5 in 11(3) cage, R4C6 and R5C5, locked for D/

5. 11(2) cage at R7C3 ={47} (only remaining combination), locked for D/ -> R6C45 = [64], 6 placed for D/, R9C1 = 9 (hidden single on D/), R8C3 = 2 (only remaining place for 2 in N7), R7C4 = 9

6. R2C1 = 1 (hidden single in C1), R1C1 = 2 (only remaining place for 2 in C1), R1C2 = 9

7. R9C8 = 1 (hidden single in R9), R9C9 = 2 (only remaining place for 2 in R9), R8C9 = 9

8. 2 innies for N3 R2C7 + R3C8 = 12 and only places for 9 in N3 -> R2C7 + R3C8 = {39}

9. 11(3) cage at R1C9 only place for 2 in N3 = {128}, locked for D/
9a. R4C6 + R5C5 = {35} -> R4C4 + R6C6 = {12}

10. R7C2 = 1 -> 11(3) cage at R5C2 = {128/137} (cannot be {146} because 4,6 only in R5C2)
10a. 2 innies for N4 R4C1 + R6C3 = 12 so cannot include 6
10b. 6 in N4 only in 11(2) cage at R4C2 or 11(2) cage at R5C1 -> one of these 11(2) cages must be {56}
10c. No 5,6 in R4C1 + R6C3 -> no 5,6 in R3C1, one of 11(2) cage at R5C1 or 11(2) cage at R7C1 must be {56} for C1

11. 9 in R5 only in 11(2) cage at R5C6 = [92], because 2 in N5 already in 11(4) cage, R3C7 = 8, R2C8 = 2

12. Innie-outie for N4 R6C3 1 more than R3C1 -> R3C1 + R6C3 = [78], R4C1 = 4

13. 11(3) cage at R5C2 (step 10) = {137} (only remaining combination) -> R56C2 = {37}, R56C1 = [65], R4C23 = [29], R4C4 = 1, R6C6 = 2
13a. R2C2 = 8 (hidden single in N1)
13b. R9C23 = {56} (hidden pair in N7)
13c. R7C3 = 7 (hidden single in C3), R8C2 = 4

14. R6C78 = [19] (hidden pair in R6), R3C8 = 3, R2C7 = 9

15. R7C5 = 2 (hidden single in R7)

16. R3C45 = [29], then R3C6 = 1 (hidden pair, then hidden single in R3)
16a. R8C5 = 1 (hidden single in C5)

17. R3C6 = 1 -> 11(3) cage at R3C6 = {146} (cannot be {137} because R6C9 can only contain 3 or 7) -> R5C8 = 4, R4C7 = 6

18. 11(2) cage at R4C9 = {38} (only remaining combination because cannot contain 2,4 or 6), R6C9 = 7, R7C9 = 4, R56C2 = [73], R5C4 = 8, R4C5 = 7, R45C9 = [83], R4C8 = 5, R4C6 = 3, R5C5 = 5

19. R1C7 = 4 (hidden single in C7), R1C8 = 7
19a. R3C3 = 4 (hidden single in R3)

20. 11(2) cage at R1C3 = {56} (only remaining combination because cannot contain 2,4,7,8,9)
20a. 11(2) cage at R1C5 = [38] (hidden pair in R1), R2C5 = 6, R1C34 = [65], R23C9 = [56], R2C7 = 3, R3C2 = 5, R9C23 = [65]

21. R9C5 = 8, R9C4 = 3, R8C4 = 7, R2C46 = [47], R9C67 = [47]

22. R7C5 = 2 -> 11(3) cage at R7C5 = {236} (only remaining combination because cannot contain 4) -> R7C7 = 3, R8C6 = 6, R78C1 = [83], R7C68 = [56], R8C78 = [58]

Rating Comment. This puzzle is no more difficult than the hardest newspaper ones so, on Mike's original rating definitions, I'll rate Paper Solvable 9 - Elevenses at 0.5.


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