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 Post subject: Assassin 377
PostPosted: Sat Jun 01, 2019 11:42 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 811
Location: Sydney, Australia
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13 years since Ruud gave us Assassin 01. viewtopic.php?p=707#p707

Assassin 377
Found this surprisingly difficult but nothing like as hard as the last puzzle. It gets 1.40 and JSudoku has some trouble needing 13 chains! Quite a few areas to explore which I always like in my puzzles. Perhaps I missed an area.

code:
3x3::k:4352:4352:2049:2049:2049:10242:2819:5636:5636:4869:4869:5382:5382:10242:10242:2819:5636:5636:4869:4869:5382:5382:10242:10242:10242:8455:8455:2056:2056:4617:2058:2058:10242:10242:8455:8455:7435:7435:4617:4617:4617:3596:3596:3596:8455:7435:2829:2829:4366:4366:5903:5903:5903:8455:7435:4368:4368:4366:5903:5903:6678:6678:6678:7435:1042:2835:4366:1812:1812:4885:6678:6678:7435:1042:2835:5649:5649:5649:4885:4885:4885:
solution:
Code:
+-------+-------+-------+
| 8 9 1 | 5 2 7 | 3 6 4 |
| 4 2 3 | 9 6 1 | 8 7 5 |
| 7 6 5 | 4 3 8 | 2 1 9 |
+-------+-------+-------+
| 1 7 2 | 3 5 9 | 4 8 6 |
| 9 4 8 | 7 1 6 | 5 3 2 |
| 3 5 6 | 2 8 4 | 1 9 7 |
+-------+-------+-------+
| 5 8 9 | 1 7 2 | 6 4 3 |
| 2 1 7 | 6 4 3 | 9 5 8 |
| 6 3 4 | 8 9 5 | 7 2 1 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 377
PostPosted: Tue Jun 04, 2019 8:35 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1658
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin. It seemed tricky until I found my step 7 and particularly step 7a.

Here is my walkthrough for Assassin 377:
Prelims

a) R1C12 = {89}
b) R12C7 = {29/38/47/56}, no 1
c) R4C12 = {17/26/35}, no 4,8,9
d) R4C45 = {17/26/35}, no 4,8,9
e) R6C23 = {29/38/47/56}, no 1
f) R7C23 = {89}
g) R89C2 = {13}
h) R89C3 = {29/38/47/56}, no 1
i) R8C56 = {16/25/34}, no 7,8,9
j) 8(3) cage at R1C3 = {125/134}
k) 22(3) cage at R9C4 = {589/679}
l) 40(8) cage at R1C6 = {12346789}, no 5

Steps resulting from Prelims
1a. Naked pair {89} in R1C12, locked for R1 and N1, clean-up: no 2,3 in R2C7
1b. Naked pair {89} in R7C23, locked for R7 and N7, clean-up: no 2,3 in R89C3
1c. Naked pair {13} in R89C2, locked for C2 and N7, clean-up: no 5,7 in R4C1, no 8 in R6C3
1d. Naked pair {89} in R17C2, locked for C2, clean-up: no 2,3 in R6C3
1e. 8(3) cage at R1C3 = {125/134}, 1 locked for R1
1f. 22(3) cage at R9C4 = {589/679}, 9 locked for R9 and N8
1g. 2 in N7 only in R789C1, locked for C1 and 29(6) cage at R5C1, clean-up: no 6 in R4C2
1h. 6,7 in R1 only in R1C6789, CPE no 6,7 in R3C7

2a. 45 rule on C12 2 outies R67C3 = 15 = [69/78], R6C3 = {67} -> R6C2 = {45}
2b. Killer pair 6,7 in R6C3 and R89C3, locked for C3
2c. 45 rule on N1 3 innies R123C3 = 9 = {135/234}, 3 locked for C3 and N1
2d. Killer pair 4,5 in R123C3 and R89C3, locked for C3
2e. Max R23C3 = 8 -> min R23C4 = 13, no 1,2,3 in R23C4
2f. 45 rule on N47 2 innies R45C3 = 10 = {19/28}
2g. 45 rule on N47 2 outies R5C45 = 8 = {17/26/35}, no 4,8,9

3. 45 rule on N7 3 innies R789C1 = 13 = {247/256}
3a. 29(6) cage at R5C1 contains 2 = {124679/234569/234578} (cannot be {123689/124589/125678} which clash with R45C3) -> R5C12 + R6C1 = {169/349/358}, no 7
[Note. R5C12 + R6C1 = 16.]
3b. R5C2 = {456} -> no 4,5,6 in R56C1
3c. 4 in N4 only in R56C2, locked for C2

4a. 45 rule on N12 3 outies R3C7 + R4C67 = 15 without 5 so must contain one of 1,2,3 -> R1C6 + R23C56 must contain two of 1,2,3 for 40(8) cage
4b. Hidden killer triple 1,2,3 in R1C45 and R1C6 + R23C56 for N2, R1C6 + R23C56 contain two of 1,2,3 -> R1C45 must contain one of 1,2,3
4c. 8(3) cage at R1C3 must contain two of 1,2,3 -> R1C3 = {123}

5. 45 rule on N1 4 outies R1C45 + R23C4 = 20 and must contain 5 for N2 = {1568/2459/2567/3458}

6. 45 rule on R6789 5(2+3) innies R6C19 + R789C1 = 23, R789C1 = 13 (step 3) -> R6C19 = 10 = [19/37/82/91]

7. 45 rule on R5 using R5C45 = 8 (step 2g) 4 innies R5C1239 = 23
7a. R5C123 cannot total 16, which clashes with R5C12 + R6C1 (step 3a) = 16 (combo crossover clash) -> no 7 in R5C9
[Inserting this step simplifies things a lot; I originally analysed all nine combinations for R5C1239 = 23]
7b. R5C1239 = {1589/2489/3569} (cannot be {4568} because 4,5,6 only in R5C29), 9 locked for R5
7c. 4,5 of {1589/2489} must be in R5C2, 9 of {3569} must be in R5C3 -> R5C12 + R6C1 (step 3a) = {349/358} (cannot be {169}), no 1,6, 3 locked for N4, clean-up: no 9 in R6C9 (step 6)
7d. Naked pair {45} in R56C2, locked for C2
7e. 6 in C2 only in R23C2, locked for N1
7f. R5C1239 = {1589/2489} (cannot be {3569} because R4C3 + R5C1239 = [13596] clashes with R5C45), no 3,6
7g. R6C1 = 3 (hidden single in N4) -> R6C9 = 7 (step 6), R6C3 = 6 -> R6C2 = 5, R5C2 = 4 -> R5C1 = 9 (cage sum), R4C1 = 1 -> R4C2 = 7, clean-up: no 7 in R789C1 (step 3)
7h. R5C12 = [94] -> R5C39 = {28}, locked for R5, clean-up: no 6 in R5C45 (step 2g)
7i. Naked pair {28} in R45C3, locked for C3, clean-up: no 4 in R23C3 (step 2c)
7j. Naked triple {135} in R123C3, 5 locked for C3, N1 and 21(4) cage at R2C3
7k. 5 in N2 only in R1C45 -> 8(3) cage at R1C3 = {125} -> R1C3 = 1, R1C45 = {25}, locked for R1 and N2, clean-up: no 6,9 in R2C7
7l. 7 in 40(8) cage only in R1C6 + R23C56, locked for N2
7m. R23C3 = {35} = 8 -> R23C4 = 13 = {49}, locked for C4 and N2
7n. 4 in R1 only in R1C789, locked for N3, clean-up: no 7 in R1C7
7o. 40(8) cage at R1C6 = {12346789} -> R3C7 = {29}, R4C67 = {249}, 4 locked for R4
7p. 45 rule on N3 3 innies R3C789 = 12 = {129} (only remaining combination, cannot be {138/156} because R3C7 only contains 2,9), locked for R3 and N3 -> R23C2 = [26], R23C4 =
[94], R23C1 = [47]
7q. Naked pair {38} in R3C56, locked for R3 and N2 -> R23C3 = [35]
[I really hadn’t expected to get so much from that 45 for R5.]

8. 1 in R3 only in 33(6) cage at R3C8 = {126789/135789}, 8 locked for N6

9. 14(3) cage at R5C6 = {167/356}
9a. 7 of {167} must be in R5C6 -> no 1 in R5C6

10. 17(4) cage at R6C4 = {1268/1358/2348} (cannot be {1259/2456} which clash with R1C4, cannot be {1349} because 4,9 only in R6C5, cannot be {1457} which clashes with 22(3) cage at R9C4, cannot be {1367/2357} because 3,5,6,7 only in R78C4), no 7,9
10a. R9C5 = 9 (hidden single in C5)
10b. 45 rule on N8 4 innies R7C456 + R8C4 = 16 = {1258/1267/1348/2347} (cannot be {1357/1456/2356} which clash with 22(3) cage at R9C4)
10c. 17(4) cage = {1268/2348} (cannot be {1358} = {18}{35} because R7C456 + R8C4 cannot contain both of 3,5), no 5
[Alternatively {1358} prevents 8 being placed in N8.]
10d. Consider placements for R1C45 = {25}
R1C4 = 2 => 17(4) cage = {1268}
or R1C4 = 5, R1C5 = 2, no 6 in R4C4 => R4C4 = {23} => 17(4) cage = {1268} (cannot be {2348} which clashes with R4C4)
-> 17(4) cage = {1268}, no 3,4, 6 locked for C4 and N8, clean-up: no 2 in R4C5, no 1 in R8C56, no 7 in 22(3) cage at R9C4
[Alternatively 17(4) cage = {2348} forces R1C45 = [52], R4C45 = [62].]
10e. Naked triple {589} in 22(3) cage at R9C4, 5,8 locked for R9 and N8, clean-up: no 2 in R8C56
10f. Naked pair {34} in R8C56, locked for R8 and N8 -> R89C2 = [13], R89C3 = [74]
10g. R5C4 = 7 (hidden single in C4) -> R5C5 = 1 (step 2g)
10h. R4C4 = 3 (hidden single in C4) -> R4C5 = 5, R5C6 = 6, R1C6 = 7, R2C56 = [61], R7C56 = [72], R78C4 = [16]

11. 1,7 in R9 only in R9C789 -> 19(4) cage at R8C7 = {1279} (cannot be {1567} which clashes with R7C789, ALS block) -> R8C7 = 9, R9C789 = {127}, locked for R9 and N9
11a. R2C7 = 8 (hidden single in C7) -> R1C7 = 3

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A377 at Easy 1.5. After combination analysis, helped by a CCC, I used to short forcing chain to finally crack it.


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 Post subject: Re: Assassin 377
PostPosted: Mon Jun 10, 2019 6:45 am 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 811
Location: Sydney, Australia
7a is a beauty Andrew! It makes the rest of step 7 very powerful. I looked in that area a lot but just couldn't reduce it. I was glad you still had to work a bit at the end of the puzzle to finally get it. I had to go different ways. Thanks to Andrew for some corrections.
a377 WT:
Preliminaries from SudokuSolver
Cage 4(2) n7 - cells ={13}
Cage 17(2) n1 - cells ={89}
Cage 17(2) n7 - cells ={89}
Cage 7(2) n8 - cells do not use 789
Cage 8(2) n5 - cells do not use 489
Cage 8(2) n4 - cells do not use 489
Cage 11(2) n7 - cells do not use 1
Cage 11(2) n4 - cells do not use 1
Cage 11(2) n3 - cells do not use 1
Cage 8(3) n12 - cells do not use 6789
Cage 22(3) n8 - cells do not use 1234
Cage 40(8) n2356 - cells ={12346789}

No routine clean-up done unless stated.

1. 17(2)n1 = {89}: both locked for r1 and n1

2. 17(2)n7 = {89}: both locked for r7 and n7

3. naked pair {89} in r17c2: both locked for c2

4. 4(2)n7 = {13}: both locked for c2 and n7

5. "45" on c12: 2 outies r67c3 = 15 = [69/78]
5a. r6c2 = (45)

6. 11(2)n7 = {47/56}(no 2)
6a. killer pair 6,7 with r6c3: both locked for c3

7. "45" on n47: 2 innies r45c3 = 10 = {19/28}(no 3,4,5)

8. "45" on n7: 3 innies r789c1 = 13 and must have 2 for n7 = {247/256}
8a. 2 locked for c1 and 29(6)r5c1

9. "45" on n7: 3 outies r5c2 + r56c1 = 16
9a. but {178} blocked by h10(2)r45c3 = 1 or 8
9b. {367} blocked by r6c3 = (67)
9c. {457} blocked by r6c2 = (45)
9d. = {169/349/358}(no 7)
9e. = one of 4,5,6 which must go in r5c2 -> no 4,5,6 in r56c1

10."45" on n1: 3 innies r123c3 = 9 = {135/234}
10a. 3 locked for n1

11. 8(3)r1c3 = {125/134}: must have 1: locked for r1

12. "45" on n1: 4 outies r1c45 + r23c4 = 20 and must have 5 for n2
12a. = {1568/2459/2567/3458}
12b. -> with combos 8(3) (step 11), r1c45 = {15}/{25}/{34}
12c. -> r1c45 = 6 or 7 -> r1c3 = (12)

Important one.
13. h9(3)r123c3 = {135/234}
13a. 1 or 2 must be in r1c3 -> no 1,2 in r23c3
13b. r23c3 = {35/34}: 3 locked for 21(4)
13c. note: 1 in r1c3 must have 5 in 23c3
13d. -> 1 in r1c3 must have 5 in r1c45 or there would be no 5 for n2 (Locking-out cages)
13e. -> 8(3)r1c3: [1]{34} blocked since leaves no 5 for n2
13f. = {125} only: all locked for r1, 5 locked for n2

14. h20(4)n2 = {1568/2459/2567}
14a. -> r23c4 = {68/49/67}(no 1,2)

15. 3 in n2 only in 40(8): locked for 40(8)

16. "45" on n3: 3 innies r3c789 = 12
16a. but {237/246/345} all blocked by r1c789 = three of {3467}
16b. = {129/138/147/156}
16c. must have 1: locked for r3 and n3

17. 4,6 & 7 in r1 only in r1c6789: r3c7 sees all those -> no 4,6,7 in r3c7

18. "45" on n12: 3 outies r3c7 + r4c67 = 15 and form a hidden cage since they are all in the 40(8)
18a. but {267} as [2]{67} only, blocked by one of the two 8(2) in r4 must have at least one of {26/17}
18b. = {168/249}(no 7)
18c. {16} blocked from r4c67 by two 8(2) in r4 must have at least one of {17/26}
18d. no 8 in r3c7, no 1 in r4c67

Took a long time to find this
19. if 1 in 40(8) is in r3c7 -> in outies n12 = 15 -> r4c67 = {68} only (step 18b) -> no 2 in 8(2)r4c1 -> 2 in n4 only in c3, locked for c3 -> r1c3 = 1
OR 1 which must be in 40(8) in r2c56 -> 1 in n1 only in r1c3
19a. -> r1c3 = 1
19b. -> h9(3)r123c3 = [1]{35} only: 5 locked for c3 and n1
19c. -> 11(2)n7 = {47} only: both locked for n7, 7 for c3
19d. r6c23 = [56], r7c3 = 9 (h15(2)r67c3)
19e. r157c2 = [948], r1c1 = 8
19f. r4c12 = [17] only permutation

20. naked pair {25} in r1c45: 2 locked for n2

21. 40(8)r1c6 must have 2 which is only in h15(3)r3c7+r4c67 = {249} only, 4 and 9 locked for 40(8) cage
21a. hidden pair {49} in n2 -> r23c4 = {49}: both locked for c4

22. naked pair {26} in r23c2: 6 locked for n1

23. h12(3)r3c789: must have 2,9 for r3c7
23a. = {129} only: 2,9 locked for r3 and n3
23b. 1 locked for 33(6)r3c8
23c. r23c4 = [94], r23c12 = [4276]

24. naked pair {38} in r3c56: both locked for n2, 3 for r3

25. "45" on r12345: (remembering h16(3)r5c2+r56c1): 2 outies r6c19 = 10 = [37] only
25a. r5c1 = 9

26. "45" on n47: 2 outies r5c45 = 8 (no 8)

27. "45" on r5: h8(2)r5c45 -> r5c39 = 10 = {28} only: both locked for r5
27a. no 6 in r5c45

The final important steps. Took a long time to see this first one.
28. 3 & 5 in n5 only in r45c45 or in r5c6. Can't both be in r5c6 -> must both be in one of the 8(2) or h8(2) cages in n5
28a. -> no 3,5 in r5c6

29. 7 in n5 in h8(2)r5c45 = {17} or in r5c6
29a. -> no 1 in r5c6 since it would leave no 7 for n5 (Locking out cages)

30. naked pair {67} in r15c6: both locked for c6

31. 17(4)r6c4. only valid combination with 9 = {1259} but this is blocked by r1c4 = (25)
31a. no 9 in r6c5
31b. r9c5 = 9 (hsingle c5)
31c. -> r9c46 = 13 = {58} only, both locked for r9 and n8

32. 17(4)r6c4: {1367} blocked by 3,6,7 only in r78c4
32a. = {1268/2348}(no 7)
32b. must have 2 -> 2 locked for c45 with r1c45 (caged x-wing)
32c. -> 8(2)r4c4 = {35} only: both locked for r4 and n5

33. naked pair {17} in r5c45: both locked for r5, 1 for n5
33a. r125c6 = [716]

34. 7(2)n8 = {34} only: both locked for r8 and n8
34a. r89c23 = [1734]

35. 19(4)n9 = {1279/1567}(no 8)
35a. must have 5/9 which are only in r8c7 -> r8c7 = (59)

36. 8 in n5 only in r6: locked for r6

37. hsingle 8 in c7 -> r12c7 = [38]

Much easier now.
Cheers
Ed


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 Post subject: Re: Assassin 377
PostPosted: Fri Jun 14, 2019 8:24 pm 
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Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 186
Location: California, out of London
Here is mine. Just got in under the wire...
Nothing particularly fancy here - just steadily eliminating impossibilities.
Will now look at your two WTs.
Thanks for another fine puzzle Ed!
Assassin 377 WT:
1. 17(2)n7 = {89}
4(2)n7 = {13}
11(2)n7 = {47} or {56}
-> 2 in n7 in r789c1

2. 17(2)n1 = {89}
-> (89) locked in c2 in r17c2
Innies n1 = r123c3 = +9(3)
-> 7 in 19(4)n1

3. Innies c12 = r67c2 = +13(2)
-> r67c2 = [58] or [49]
-> 11(2)n4 = [56] or [47]

4. Innies n47 = r45c3 = +10(2) - must contain an 8 or 9
-> n4 is either:
(A) 11(2) = [56], r45c3 = {28}, 8(2) = [17], r56c1 = {39}, r5c2 = 4
(B) 11(2) = [47], r45c3 = {19}, 8(2) = [62], r56c1 = {38}, r5c2 = 5

5. Innies r6789 = r6c19 = +10(2)
Outies n47 = r5c45 = +8(2)
Remaining outies r1234 = r5c39 + r6c9 = +17(3)

Trying Step 4 (B) puts r6c19 = [82]
puts r5c12 = [35]
also puts r5c39 = [96] and r4c3 = 1
Leaves no solution for r5c45 = +8(2)
-> Step 4 (A) must be correct

-> 11(2)n4 = [56], r45c3 = {28}, 8(2)n4 = [17], r56c1 = {39}, r5c2 = 4
Also 17(2)n7 = [89], 11(2)n7 = {47}, r789c1 = {256}
Also 17(2)n1 = [89], r123c3 = {135}, r23c1 = {47}, r23c2 = {26}

6. Given r6c19 from [37] or [91], r5c3 from (28), and r5c39 + r6c9 = +17(3)
Only possibilities are:
-> r6c19 = [37] and r5c39 = {28}
-> r5c12 = [94]

7. 40(8) has no 5
-> 5 in n2 in r1c45 or r23c4
5 also in r123c3 -> 8(3)r1 = {125}
21(4) cannot contain all of (135)
-> 8(3)r1 = [1{25}]
-> r23c3 = {35}
-> r23c4 from {67} or {49}

8. Outies n12 = r3c7 + r4c67 = +15(3)
Those must be 2 + the same numbers as in r23c4
r1c6789 = {3467}
r3c7 sees all those -> cannot be from any of those.
Since 7 already in r4 -> 7 not in outies n12 -> 7 in 40(8) in n2
-> r23c4 = {49}
-> 40(8) in n2 = {13678}
-> Outies n12 = {249}

9! Innies n3 = r3c789 = +12(3)
That has at most one of the numbers (3467) (those are in r1c6789)
and has one of (249) in r3c7
Only possibility is r3c789 = {129}
-> r23c4 = [94]
-> r23c1 = [47]
Also r23c2 = [26]

10. At least one of (35) in n5 either in 8(2)n5 or H8(2)r5c45
-> Both (35) in n5 in 8(2)n5 or H8(2)r5c45
-> 5 in c6 in n8 r789c6

11. -> 22(3)n8 either [895] or {679}
If the latter -> 8 in n8 in r8c4
-> 8 in n8 in r89c4
-> 8 in n5 in r6c56

(There are a number of different ways of doing this next step).
12! r6c45678 = {12489}
17(4)r6c4 cannot contain both (89)
Also 23(5)r6c6 cannot contain both (89) since that would put 3 in r7c56 which leaves no place for 4 in n8
-> r6c5 from (89)
-> r8c4 cannot be 8
-> 22(3)n8 = [895]
-> 9 in n5 in r46c6
-> r6c5 = 8

13. Since one of 8(2)n5 or H8(2)r5c45 = {35} and 4 already in c4 ...
... -> r678c4 = {126}
-> r1c45 = [52]
-> 8(2)n5 = [35]
-> r5c45 = [71]
-> r6c4 = 2, r78c4 = {16}
Also 14(3)r5 = [6{35}]
-> r46c6 = {49}
-> 7(2)n8 = [43]
-> r7c56 = {27}
Also 11(2)n7 = [74], 4(2)n7 = [13]
-> r78c4 = [16]

14. HS 1 in n2 -> r2c6 = 1
-> HS 8 in n2 -> r3c6 = 8
HS 5 in r3 -> r23c3 = [35]
-> NS 3 in c5 r3 -> r3c5 = 3
-> HS 6 in n2 -> r2c5 = 6
-> NS r1c6 = 7
-> r7c56 = [72]

15. r7c789 = {345} or {346}
-> 26(5)n9 = [{346}{58}]
-> 19(4)n9 = [9{127}]
-> r789c1 = [526]
Also HS 9 in 40(8) -> r4c6 = 9
-> r34c7 = [24]
-> r6c678 = [419]

16. HS 8 in c7 -> 11(2)n3 = [38]
-> r5c78 = [53]
-> r7c789 = [643]
Also r3c89 = [19]
-> r9c789 = [721]
etc.


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