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Ed
 Post subject: Assassin 446
PostPosted: Wed Nov 15, 2023 8:44 am 
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x-sudoku so 1-9 cannot repeat on the diagonals

Assassin 446
Enjoyed this with some nice hidden steps that are not difficult once spotted. SudokuSolver gives it 1.95 but JSudoku finds it very easy which is why I tried it.

triple click code:
3x3:d:k:3328:3328:7937:7937:4610:2819:2819:3076:3076:3328:5381:4358:7937:4610:6421:6421:6421:3076:2056:5381:4358:7937:4610:4610:6420:6421:6420:2056:5381:7937:7937:2826:2826:6421:6420:6420:4619:5381:5381:5388:2061:2061:4622:4622:4622:4619:2831:5388:5388:7696:7185:3090:3090:3090:4619:2831:2831:5388:7696:7185:7185:3603:3603:11015:11015:11015:5388:7696:7696:7185:7185:2569:11015:11015:11015:11015:11015:7696:7696:7185:2569:
solution: 
+-------+-------+-------+
| 2 7 6 | 4 9 8 | 3 5 1 |
| 4 3 9 | 7 1 5 | 2 8 6 |
| 5 1 8 | 3 2 6 | 7 4 9 |
+-------+-------+-------+
| 3 5 2 | 9 7 4 | 6 1 8 |
| 1 8 4 | 6 5 3 | 9 7 2 |
| 9 6 7 | 2 8 1 | 5 3 4 |
+-------+-------+-------+
| 8 2 3 | 1 6 7 | 4 9 5 |
| 7 9 1 | 5 4 2 | 8 6 3 |
| 6 4 5 | 8 3 9 | 1 2 7 |
+-------+-------+-------+
Cheers
Ed
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wellbeback
 Post subject: Re: Assassin 446
PostPosted: Sun Nov 19, 2023 8:00 pm 
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Thanks again Ed! My solution path was a bit messy with a number of "Either a) or b) -> c)" style steps. Interested to see what others do!
Assassin 446 WT:
1. 17(2)n1 = {89}
Innies r1234 = r234c2 = +9(3)
-> r5c23 = +12(2)
Outies c1234 -> r9c5 = 3
No 2 in 43(8)r8c1
-> (23) in n7 in r7c123

2. Outies r6789 = r5c14 = +7(2)
-> Max r5c1 = 6
-> Min r7c1 = 3
-> 2 in r7c23
One of:
a) 3 in r7c23 puts 11(3)r6c2 = [6{23}] which puts 21(5)r2c2 = [{135}84]
b) 3 in r7c1 puts r5c1 = 6 which puts r5c4 = 1 and 8(2)r5c5 = [53] which puts r5c23 = [84]
Either way, r5c23 = [84]
-> 4 in n1 in 13(3)n1
Also (HS 9 in n4) r6c1 = 9
-> 9 in n7 in r89c2
Also 9 in n5 in r4c456
-> 9 in n6 in 18(3)n6
Also 8 in n7 in r789c1

3. 21(5)r2c2 and 11(3)r6c2 can only be one of:
a) [{135}84] and [623]
b) [{126}84] and [542]
But the latter case puts 18(3)c1 = [693] and (HS 3 in c2) r1c2 = 3, which leaves no solution for 13(3)n1
-> 21(5)r2c2 = {135} and 11(3)r6c2 = [623]

4. IOD c1 -> r8c1 + r9c1 = r1c2 + 6
r1c2 from (47)
a) r1c2 = 4 puts r89c1 = +10(2) = {46} puts r1c3 = 6
b) r1c2 = 7 puts r12c1 = {24} puts (HS 6 in n1) r1c3 = 6
Either way, r3c1 = 6
-> Remaining Innies n1 = r2c2 + r3c12 = +9(3) (No 7)
-> 13(3)n1 = {247} with 2 in r12c1
-> Remaining Innies n1 = {135} and r3c1 = r4c2
Since Outies r6789 = r5c14 = +7(2) and r5c3 = 4 -> 3 not in r5c1
-> 3 in n4 in r4c12
-> Either 8(2)c1 = [53] and r234c2 = [{13}5], or r234c2 = [{15}3] and 8(2)c1 = [35]

5. -> 18(3)c1 = [198]
-> r5c4 = 6
-> 8(2)r5 = [53]
-> 18(3)r5 = {279}
Also r46c3 = {27} and r89c3 = {15}
Also 14(2)n9 = {59}
Also (HS 8 in 43(8)) r9c4 = 8

6. Innies n6 = r4c789 = +15(3)
-> Remaining Outies n3 = r12c6 = +13(2)
-> Remaining Innies n2 = r123c4 = +14(3)
-> Remaining Cells 31(6)r3c1 = r4c34 = +11(2)
(Since r4c12 = {35}) -> r4c3456 = [29{47}] or [74{29}]
-> r4c789 = {168}
-> 12(3)n6 = {345}

7. 3 in c4/n2 in r123c4
-> 31(6)r1c3 from [6{347}29] or [6{239}74]
-> H+13(2)r12c6 not {49}
Since 11(2)r1c6 not {56} -> H+13(2)r12c6 from [85] or [76]
But the latter puts 6 in n3 in r3c79 which leaves no place for 6 in r4c789
-> H+13(2)r12c6 = [85]

8. Continuing...
-> r1c7 = 3
Also 5 in n1 in r3c12
-> 5 in r1 in r1c8 (5 already in D/)
-> 12(3)n3 = [516]
Also (HS 8 in c5) r6c5 = 8
Also (HS 1 in r6) r6c6 = 1 (1 already in D/)
-> (HS 1 in n9) r9c7 = 1
Also (HS 5 in c4) r8c4 = 5
-> r89c3 = [15]
-> (HS 1 in c4) r7c4 = 1
-> (Remaining Innie n8) r7c6 = 7
-> (HS 7 in n9) 10(2)n9 = [37]
Also since r3c7 'sees' all of r4c789 -> r3c7 not 8
-> (HS 8 in D/) r2c8 = 8
-> (HS 8 in r8) r8c7 = 8
-> (HS 8 in r4) r4c9 = 8
-> r4c78 = [61]
-> r3c79 = [79]
-> (HS 7 in n5) r4c5 = 7
etc.
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Andrew
 Post subject: Re: Assassin 446
PostPosted: Tue Nov 21, 2023 10:55 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
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Thanks Ed; I enjoyed your latest Assassin.

I haven't yet had time to look at wellbeback's WT apart from a quick glance, I'll wait to see whether Ed posts anything; I did notice that the middle of his step 3 is basically the same forcing chain that's in my step 2b.

I found that my step 2g was wrong, similar to seeing an Innie-Outie difference the wrong way round; I ought to have highlighted those candidates on my Excel worksheet when I'd have immediately realised that it was wrong. Fortunately the rework wasn't difficult, changes in blue. Even better would have been spotting Ed's step 5.

Here's how I solved Assassin 446:
Prelims

a) R1C67 = {29/38/47/56}, no 1
b) R23C3 = {89}
c) R34C1 = {17/26/35}, no 4,8,9
d) R4C56 = {29/38/47/56}, no 1
e) R5C56 = {17/26/35}, no 4,8,9
f) R7C89 = {59/68}
g) R89C9 = {19/28/37/46}, no 5
h) 11(3) cage at R6C2 = {128/137/146/236/245}, no 9
i) 43(8) cage at R8C1 = {13456789}, no 2

1a. Naked pair {89} in R23C3, locked for C3 and N1
1b. 45 rule on C1234 1 outie R9C5 = 3, clean-up: no 8 in R4C6, no 5 in R5C6, no 7 in R8C9
1c. 45 rule on R1234 2 outies R5C23 = 12 = [57/75/84/93]
1d. 45 rule on R1234 3 innies R234C2 = 9 = {126/135/234}, no 7,8,9
1e. 45 rule on R6789 2 outies R5C14 = 7 = {16/25/34}, no 7,8,9
1f. 45 rule on N36 2 outies R12C6 = 13 = {49/58/67}, no 1,2,3, clean-up: no 8,9 in R1C7
1g. 2,3 in N7 only in R7C123, locked for R7
1h. 45 rule on N6 2 outies R3C79 = 1 innie R4C7 + 10
1i. Min R3C79 = 11, no 1 in R3C79
1j. Max R3C79 = 17 -> max R4C7 = 7
1k. 18(3) cage at R5C1 = {189/279/369/378/459/468} (cannot be {567} which clashes with R34C1)
1l. Max R56C1 = 15 -> min R7C1 = 3
1m. Max R57C1 = 15 -> min R6C1 = 3
1n. 2 in N7 only in R7C23 -> 11(3) cage at R6C2 = {128/236/245}, no 7, no 2 in R6C2

2a. R234C2 (step 1d) = {126/135/234}, R5C23 (step 1c) = [57/75/84/93]
[Here I first found a forcing chain based on the placement for 8 in N4 followed by one for 3 in N7 but better is going directly to …
2b. Consider placement for 3 in N7
R7C1 = 3 => R56C1 = 15 = [69] => R5C23 = [84] (cannot be {57} which clashes with R5C56 = [17/53/71] when R5C1 = 6)
or 3 in R7C23 = {23} => R6C2 = 6 (cage sum), R234C2 = {135} (cannot be {234} which clashes with R7C2), 3,5 locked for 21(5) cage at R2C2 => R5C23 = [84]
-> R5C23 = [84], also 6 in R5C1 + R6C2, locked for N4, clean-up: no 2 in R3C1, no 3 in R5C14 (step 1e)
2c. R6C1 = 9 (hidden single in N4) -> R57C1 = 9 = [18/27/54/63], no 5,6 in R7C1
2d. 9 in R5 only in 18(3) cage at R5C7 = {279/369}, no 1,5, 9 locked for N6
2e. Killer pair 2,6 in R5C14 and 18(3) cage, locked for R5
2f. 11(3) cage at R6C2 (step 1n) = {236/245}, no 1
2g. 5 of {245} must be in R6C2 -> no 5 in R7C23
[At this stage I ought to have spotted Ed’s step 5, eliminating 3 from R6C2.]
2h. R234C2 = {126/135}, 1 locked for C2
2i. 1,5,9 in N7 only in R89C123, locked for 43(8) cage at R7C1, no 1,5,9 in R9C4
2j. 12(3) cage at R6C7 = {138/147/156/345} (cannot be {237/246} which clash with 18(3) cage at R5C7), no 2
2k. 45 rule on N6 3 innies R4C789 = 15 = {168/258/348/456} (cannot be {267} which clashes with 18(3) cage, cannot be {357} which clashes with 12(3) cage), no 7
2l. Combined half cage 9(3) cage + R6C2 = {126}3/{126}5/{135}6, 6 locked for C2
2m. 4 in N1 only in 13(3) cage at R1C1 = {247/346}, no 1,5

3a. R234C2 (step 2h) = {126/135}, 13(3) cage at R1C1 = {247/346}
3b. {126} must be {16}2 (cannot be {26}1 which clashes with 13(3) cage), no 2 in R23C2
3c. 45 rule on N1 4 innies R1C3 + R2C2 + R3C12 = 15 = {1257/1356}
3d. 2 of {1257} must be in R1C3 -> no 7 in R1C3
3e. Again consider placement for 3 in R7
R7C1 = 3 => R34C1 = [17/62/71] then with R1C3 + R2C2 + R3C12 = {1356} R23C2 must contain at least one of 3,5 while with R1C3 + R2C2 + R3C12 = {1257} R23C2 must be {15}
or 3 in R7C23 = {23} => R6C2 = 6 (cage sum) => R234C2 = {135}
-> R234C2 = {135}, 3,5 locked for C2
-> R6C2 = 6, R7C23 = [23], 3 placed for D/, clean-up: no 8 in R4C5, no 1 in R5C4 (step 1e)
3f. Consider placement for 2 in C1
2 in 13(3) locked for N1
or 2 in R4C1 => R3C1 = 6 => R1C3 + R2C2 + R3C12 = {1356}
or 2 in R5C1 => R67C1 = 16 = [97], no 7 in R3C1 => R1C3 + R2C2 + R3C12 = {1356}
-> R1C3 + R2C2 + R3C12 = {1356}, no 2,7, clean-up: no 1 in R4C1
3g. 2 in N1 only in 13(3) cage at R1C1 = {247}, 2 locked for C1, clean-up: no 6 in R3C1, no 5 in R5C4, no 7 in R7C1
3h. Killer pair 1,5 in R34C1 and R5C1, locked for C1
3i. R89C3 = {15} (hidden pair in N7), locked for C3 -> R1C3 = 6, clean-up: no 5 in R1C67
3j. Naked pair {27} in R46C3, locked for N4, clean-up: no 1 in R3C1
3k. R5C1 = 1 (hidden single in C1) -> R5C4 = 6, R6C1 = 9 -> R7C1 = 8 (cage sum), clean-up: no 7 in R5C56, no 6 in R7C89
3l. R5C56 = [53], 5 placed for both diagonals
3m. Naked pair {35} in R4C12, locked for R4
3n. Killer pair {27} in R4C3 and R4C56, locked for R4
3o. R4C789 (step 2k) = {168} (only remaining combination), 1,8 locked for R4 and N6, 8 locked for 25(4) cage at R3C7
3p. Naked triple {168} in R4C789, CPE no 6 in R3C7
3q. Naked triple {279} in 18(3) cage at R5C7, 7 locked for N6
3r. Naked triple {345} in 12(3) cage at R6C7, 4 locked for R6
3s. 43(8) cage at R8C1 = {13456789} -> R9C4 = 8, clean-up: no 2 in R8C9
3t. Naked pair {59} in R7C89, locked for R7 and N9, clean-up: no 1 in R9C89
3u. 30(6) cage at R6C5 = {124689/125679}, 9 locked for N8
3v. 21(5) cage at R5C4 = {12567} (only remaining combination), no 4 -> R8C4 = 5, 1 locked for C4, 2 locked for R6, R89C3 = [15]
3w. R1C67 = [83/92] (cannot be {47} which clashes with R1C2) -> R2C6 = {45} (step 1f)
3x. 31(6) cage at R1C3 = {234679} must have one of 2,7 and one of 4,9 in R123C4
3y. R12C6 = [85] (cannot be [94] which clashes with 31(6) cage) -> R1C7 = 3

4a. R6C5 = 8 (hidden single in R6) -> 30(6) cage at R6C5 (step 3u) = {124689}, no 7
4b. 7 in N8 only in R7C46, locked for R7
4c. 28(6) cage at R6C6 = {124678}, no 3, 8 locked for N9, clean-up: no 2 in R9C9
4d. R8C9 = 3 (hidden single in N9) -> R9C9 = 7, placed for D\ -> R6C6 = 1, placed for D\
4e. R2C2 = 3 -> R34C2 = [15], R34C1 = [53]
4f. Naked pair {27} in R6C34, locked for 21(5) cage at R5C4 -> R7C4 = 1
4g. R7C6 = 7 (hidden single in R7), clean-up: no 4 in R4C5
4h. R9C7 = 1 (hidden single in N9) -> R4C7 = 6, R7C7 = 4, placed for D\ -> R1C1 + R3C3 + R4C4 + R8C8 = [2896], R2C3 = 9, R9C8 = 2, R8C7 = 8, R6C789 = [534], clean-up: no 2 in R4C56
4i. R4C56 = [74], 4 placed for D/
4j. R4C89 = {18} = 9 -> R3C79 = 16 = [79], 7 placed for D/
4k. R1C9 = 1 -> R1C8 + R2C9 = 11 = [56]

and the rest is naked singles, without using the diagonals.
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Ed
 Post subject: Re: Assassin 446
PostPosted: Fri Nov 24, 2023 10:28 pm 
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Joined: Wed Apr 16, 2008 1:16 am
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Interesting that we all followed a similiar path. However, I had different reasoning at critical spots. I generally can't find the chains that wellbeback and Andrew find yet I can see my step 16! [ I made a mistake at step 17 and step 31. Both fixed now. Thanks Andrew!]
a446 my start:
This is how I got started with A446

Preliminaries by SudokuSolver
Cage 17(2) n1 - cells ={89}
Cage 14(2) n9 - cells only uses 5689
Cage 8(2) n14 - cells do not use 489
Cage 8(2) n5 - cells do not use 489
Cage 10(2) n9 - cells do not use 5
Cage 11(2) n23 - cells do not use 1
Cage 11(2) n5 - cells do not use 1
Cage 11(3) n47 - cells do not use 9
Cage 43(8) n78 - cells ={13456789}

1. "45" on c1234: 1 outie r9c5 = 3

2. 17(2)n1 = {89}: both locked for n1 and c3

3. "45" on r1234: 3 innies r234c2 = 9 = {126/135/234}(no 7,8,9) = 2 or 3, 3 or 6
3a. -> r5c23 = 12 = [93/84]/{57}(no 6, no 3,4 in r5c2)

4. "45" on r6789: 2 outies r5c14 = 7 (no 7,8,9)
4a. -> min. r67c1 = 12 (no 1,2)

5. 2 in n7 only in 11(3)r6c2 = {128/236/245}(no 7)
5a. no 2 in r6c2
5b. {236} can only be [623] since [326/362/632] all clash with h9(3)c2 (step 3.)
5c. -> no 3 in r67c2, no 6 in r7c23

6. if 3 in n7 is in r7c1 -> r56c1 = 15 = [69]
6a. or 3 is in r7c3 -> 6 in r6c2 (step 5b)
6b. 6 locked for n4 (Locking cages)

7. 13(3)n1 = {157/247/256/346}
7a. {346} can't have 3 in r1c2 since h9(3)c2 has 3 in c2 or 6 in n1 (step 3)
7b. -> no 3 in r1c2

8. 3 in c2 only in h9(3)c2 = {135/234}(no 6) = 4 or 5
8a. no 3 in r5c3
8b. -> no 9 in r5c2

9. r6c1 = 9 (HSingle n4)

10. killer pair 4,5 in h9(3)c2 and h12(2)r5c23
10a. no 4,5 in r6c2
[alternatively, 21(5)r2c2 = {135}[84]/{234}{57}, must have 4 & 5 -> no 4,5 in r6c2]

11. 8 in n4 only in r56c2: locked for c2

12. 11(3)r6c2: can't be {245} since none are in r6c2 = {128/236}(no 4,5)
12a. 6,8 only in r6c2 -> r6c2 = (68)

13. "45" on n7: 3 remaining outies r5c1 + r6c2 + r9c4 = 15
13a. but r5c1 + r9c4 cannot sum to 7 since this would clash with the h7(2)r5c14 (combo crossover clash CCC)
13b. -> no 8 in r6c2
13c. -> r6c2 = 6
13d. -> r7c23 = 5 = [23]: 3 placed for d/

14. h9(3)c2 = {135} only: 1,5 locked for c2, 5 for 21(5) cage
14a. -> h12(2)r5c23 = [84] only
14b. no 3 in r5c14 (h7(2))

15. "45" on n6: 3 innies r4c789 = 15
15a. -> 4 innies r4: r4c1234 = 19
15b. must have 3 for n4
15c. but {1369/2368} blocked by 6,8,9 only in r4c4
15d. = {1378/2359/3457}(no 6)
15e. 3 locked for r4
15f. must have 4/8/9 -> r4c4 = (489)

16. h7(2)r5c14 = [16]/{25}
16a. -> 1 in r5c1 or r6c3 = (17)
16b. -> {137}[8] blocked from h19(4)r4 (step 15d)
16c. = {2359/3457}(no 1,8)
16d. 5 locked for r4 and n4

17. "45" on n236: 3 outies r1c3 + r4c34 = 17 and can't have repeats since they are in the same bigger cage = {179/269/467}(no 5)

18. h19(4)r4 = {2359/3457}
18a. r4c3 = (27) -> no 2,7 in r4c1

19. 8(2)r3c1 = {35} only: both locked for c1

20. 13(3)n1 = {247} only: 2,7 locked for n1, 2 locked for c1

21. r5c1 = 1
21a. -> r5c4 = 6 (h7(2))
21b. r7c1 = 8 (cage sum)
21c. r5c56 = [53] (only permutation):5 placed for both diagonals
21d. 43(8)r8c1 must have 8 -> r9c4 = 8
21e. r1c3 = 6 (hsingle n1)

22. 11(2)n5 = {29/47}(no 8)
22a. -> hidden triple {168} in r4c789
22b. -> no 1,6,8 in r3c7 since it sees all those (Common Peer Elimination CPE)
22c. 1,8 locked for n6
22d. -> r6c789 = {345}: 4 locked for r6

23. 14(2)n9 = {59} only: both locked for n9 and r7

24. "45" on n236: 3 innies r123c4 = 14 and must have 3 for c4 = {239/347}(no 1,5)
24a. killer pair 4,9 with r4c4: both locked for c4

25. deleted

26. r8c4 = 5 (hsingle c4)

27. 11(2)r1c6: {47} blocked by r1c2 = (47)
27a. = {29}/[83]

28. "45" on n36: 2 outies r12c6 = 13
28a. but {49} blocked by r123c4 which must have one of 4/9 (step 24)
28b. = [85] only permutation, -> r1c7 = 3

29. "45" on n6: 2 outies r3c79 - 10 = 1 innie r4c7
29a. -> max. r4c7 = 7

30. 8 in r4 only in 25(4)r3c7:
30a. = {1789/2689/4678}(no 5, no 8 in r3c9)
30b. has two of 1,6,8 which must go in r4c89 -> no 1,6 in r3c9

31. r1c8 = 5 (hsingle r1),

32. 8 on d\ only in r3c3 & r8c8 -> no 8 in r3c8 (CPE)
32a. r3c3 = 8 (hsingle r3), 8 placed for d\
32a. r12c9 = 7 = [16] only permutation, 1 placed for d/

on from there
Cheers
Ed
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