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 Post subject: Assassin 410
PostPosted: Tue Feb 01, 2022 5:17 am 
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Posts: 1040
Location: Sydney, Australia
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Assassin 410
A nice hard one, mainly because the moves are well hidden. But one heavy bit needed the way I did it. Right on my limit but always felt like there was somewhere to explore. This started out as a Jsudoku generated cage pattern but then joined up some cages. SudokuSolver gives it 2.00 but Jsudoku uses just 3 advanced steps which gave me the courage.
triple click code:
3x3::k:4096:4096:5377:4098:4098:1795:1795:3588:3588:4096:5377:5377:4098:3845:5382:6151:6151:3588:5377:5377:3848:3848:3845:5382:6151:6151:2569:4106:4106:3848:9739:9739:5382:5382:5382:2569:4106:3340:3340:9739:9739:9739:1549:1549:3086:2831:5392:5392:5392:9739:9739:3601:3086:3086:2831:5650:5650:5392:2323:3601:3601:5908:5908:4373:5650:5650:5392:2323:4630:5908:5908:4887:4373:4373:1816:1816:4630:4630:5908:4887:4887:
solution:
+-------+-------+-------+
| 9 2 6 | 1 7 3 | 4 8 5 |
| 5 3 7 | 8 9 4 | 6 2 1 |
| 1 4 8 | 2 6 5 | 7 9 3 |
+-------+-------+-------+
| 2 6 5 | 9 4 1 | 8 3 7 |
| 8 9 4 | 3 2 7 | 5 1 6 |
| 7 1 3 | 6 5 8 | 9 4 2 |
+-------+-------+-------+
| 4 5 1 | 7 8 2 | 3 6 9 |
| 3 7 9 | 4 1 6 | 2 5 8 |
| 6 8 2 | 5 3 9 | 1 7 4 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 410
PostPosted: Fri Feb 04, 2022 6:11 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
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Thanks Ed for your latest Assassin. Quite a hard one, although it didn't feel like 2.0.

Here's my walkthrough for Assassin 410:
Prelims

a) R1C67 = {16/25/34}
b) R23C5 = {69/78}
c) R34C9 = {19/28/37/46}, no 5
d) R5C23 = {49/58/67}, no 1,2,3
e) R5C78 = {15/24}
f) R67C1 = {29/38/47/56}, no 1
g) R78C5 = {18/27/36/45}, no 9
h) R9C34 = {16/25/34}
i) 19(3) cage at R8C9 = {289/379/469/478/568}, no 1

1a. 45 rule on N1 R3C3 = 8 -> R3C4 + R4C3 = 7 = {16/25/34}, no 7,9, clean-up: no 7 in R2C5, no 2 in R4C9, no 5 in R5C2
1b. 45 rule on C6789 4 innies R5689C6 = 30 = {6789}, locked for C6, clean-up: no 1 in R1C7
1c. Min R89C6 = 13 -> max R9C5 = 5
1d. 45 rule on N9 R7C7 = 3 -> R6C7 + R7C6 = 11 = [65/74/92], clean-up: no 4 in R1C6, no 6 in R8C5
1e. 45 rule on N3 2 innies R1C7 + R3C9 = 7 = [43/52/61], R1C6 = {123}, R4C9 = {789}
1f. 45 rule on N5 2 innies R4C6 + R6C4 = 7 = [16]/{25/34}
1g. 45 rule on N7 2 innies R7C1 + R9C3 = 6 = [24/42/51], R6C1 = {679}, R9C4 = {356}

2a. 12(3) cage at R5C9 = {138/156/237/246} (cannot be {129/147/345} which clash with R5C78), no 9
2b. Killer pair 1,2 in R5C78 and 12(3) cage, locked for N6
2c. Consider combinations for 12(3) cage
12(3) cage = {138} => R5C78 = {24}, 5 in N6 only in R4C78, locked for 21(4) cage at R2C6 => R7C6 = 5 (hidden single in C6) => R6C7 = 6 (cage sum)
or 12(3) cage = {156/246} => R4C8 = 3 (hidden single in N6) => R1C6 = 3 (hidden single in C6), R1C7 = 4, R3C9 = 3 (step 1e), R4C9 = 7
or 12(3) cage = {237}, 7 locked for N6
-> no 7 in R6C7, clean-up: no 4 in R7C6 (step 1d)
2d. 4 in C6 only in R234C6, locked for 21(4) cage at R2C6, no 4 in R4C78
2e. Consider permutations for R6C7 + R7C6 = [65/92]
R6C7 + R7C6 = [65]
or R6C7 + R7C6 = [92], no 9 in R4C9 => no 1 in R3C9, no 6 in R1C7, no 1 in R1C6 => R1C6 = 3, R234C6 = {145} = 10 => R4C78 = 11 = [83]
-> no 6 in R4C78
2f. 4 in N6 only in R5C78 = {24} or 12(3) cage = {246} -> 12(3) cage = {138/156/246} (cannot be {237}, locking-out cages), no 7
2g. 7 in N6 only in R4C789, locked for R4
2h. Consider combinations for 12(3) cage
12(3) cage = {138}, 8 locked for N6
or 12(3) cage = {156/246} => R4C8 = 3 (hidden single in N6) … => R4C9 = 7 (as in step 2c)
-> no 8 in R4C9, no 2 in R3C9, no 5 in R1C7 (step 1e), no 2 in R1C6
2i. 21(5) cage at R2C6 contains 4 for C6 but no 6 = {12459/13458/23457}

[With hindsight step 5 could have been found at this stage]

3a. 45 rule on N7 1 outie R9C4 = 1 innie R7C1 + 1
3b. R7C1 + R9C4 + R7C6 = [235/452/562], 2 locked for R7, clean-up: no 7 in R8C5
3c. R7C1 + R9C4 + R7C6 = [235/452/562] -> no 5 in R7C45, clean-up: no 4 in R8C5
3d. 2 in R7 only in R7C16, 2 in R7C1 => R6C1 = 9 or 2 in R7C6 => R6C7 = 9 (cage sum) -> 9 in R6C17 (locking cages), locked for R6
3e. 18(3) cage at R8C6 = {189/279/369/378/468} (cannot be {459} because 4,5 only in R9C5, cannot be {567} which clashes with R7C1 + R9C4 + R7C6), no 5

4a. 12(3) cage at R5C9 (step 2f) = {138/156/246}
4b. 45 rule on R6789 4 innies R6C5689 = 19 = {1378/1468/2368/2458} (cannot be {1567/2467} which clash with R6C16, ALS block, cannot be {3457} because 12(3) cage cannot contain two of 3,4,5), 8 locked for R6

[I was a bit slow to spot this, the first key step]
5a. R1C7 + R3C9 = 7 (step 1e)
5b. Consider placement for R6C7 = {69}
R6C7 = 6 => R1C7 = 4, R3C9 = 3 => R4C9 = 7
or R6C7 = 9 => R4C9 = 7
-> R4C9 = 7, R3C9 = 3, R1C7 = 4, R1C6 = 3, clean-up: no 4 in R4C3 (step 1a), no 2 in R5C8, no 4 in R6C4 (step 1f)
5c. 14(3) cage at R1C8 = {158/167/257}, no 9
5d. 9 in C9 only in R789C9, locked for N9
5e. 45 rule on N2 3 remaining innies R23C6 + R3C4 = 11 = {146/245}, 4 locked for N2
5f. 6 of {146} must be in R3C4 -> no 1 in R3C4, clean-up: no 6 in R4C3 (step 1a)

6a. 21(5) cage at R2C6 (step 2i) = {12459/13458} -> R4C78 = {59}/[83]
6b. 45 rule on R123 4 remaining outies R4C3678 = 17 = {12}{59}/{15}[83]/[24][83], no 3 in R4C3, clean-up: no 4 in R3C4 (step 1a)
6c. 4 in N2 only in R23C6, locked for C6, clean-up: no 3 in R6C4 (step 1f)
6d. R4C3678 = {12}{59}/{15}[83], 1,5 locked in R4
6e. 1 in R4 only in R4C36, R4C3 = 1 => R3C4 = 6 (cage sum) or R4C6 = 1 => R6C4 = 6 (step 1f) -> 6 in R36C4 (locking cages), locked for C4, clean-up: no 1 in R9C3, no 5 in R7C1 (step 1g), no 6 in R6C1
6f. Naked pair {24} in R7C1 + R9C3, locked for N7
6g. R7C1 + R9C4 + R7C6 (step 3c) = [235/452], 5 locked for N8, clean-up: no 4 in R7C5

7a. 18(3) cage at R8C6 (step 3e) = {189/279/369/468} (cannot be {378} which clashes with R78C5)
7b. Consider permutations for R67C1 = [74/92]
R67C1 = [74] => R6C6 = {68} => 18(3) cage = {189/279/369/378} (cannot be {468} which clashes with R6C6
or R67C1 = [92] => R9C3 = 4
-> 18(3) cage = {189/279/369}, no 4, 9 locked for C6 and N8
7c. 4 in N8 only in R78C4, locked for C4 and 21(5) cage at R6C2, no 4 in R6C23
7d. 45 rule on N8 4 innies R789C4 + R7C6 = 18 with 4 in R78C4 and 5 in R7C6 + R9C4 = {2457} (only possible combination) -> R7C6 = 2, R9C4 = 5, R7C1 = 4, R6C1 = 7, R9C3 = 2, R78C4 = [74], clean-up: no 6 in R5C23
[A lot easier from here]
7e. R78C4 = [74] = 11 -> R6C234 = 10 = {13}6/{35}2
7f. Naked triple {135} in R4C3 + R6C23, locked for N4, 3 locked for R6, clean-up: no 8 in R5C2
7g. Naked pair {49} in R5C23, locked for R5 and N4, clean-up: no 2 in R5C7
7h. Naked pair {15} in R5C78, locked for R5 and N6
7i. 12(3) cage at R5C9 = {246} (only remaining combination), 6 locked for N6 -> R6C9 = 9 (or cage total from step 7e), R4C78 = [83]

8a. Naked pair {26} in R4C12, locked for R4 and N4 -> R4C45 = [94], R5C1 = 8
8b. Naked pair {26} in R36C4, 2 locked for C4
8c. Naked pair {18} in R12C4, locked for N2, R1C5 = 7 (cage sum)
8d. Naked pair {69} in R23C5, 6 locked for C5 and N2, clean-up: no 3 in R8C5
8e. Naked pair {18} in R78C5, locked for C5 and N8, R9C5 = 3
8f. R3C4 = 2 -> R4C3 = 5 (cage sum)

[Now back to the corners]
9a. 14(3) cage at R1C8 = {158} (only remaining combination), locked for N3
9b. Naked triple {158} in R1C489, 1,5 locked for R1, 5 locked for N3
9c. 1 in R3 only in R3C12, locked for N1
9d. 2 in R1 only in 16(3) cage at R1C1 = {259} -> R1C12 = {29}, locked for N1, R2C1 = 5
9e. R1C3 = 6, R2C6 = 4, R2C23 = {37}, 7 locked for R2 and N1
9f. 17(3) cage at R8C1 = {368} (cannot be {179} which clashes with R7C3) -> R89C1 = [36], R9C2 = 8
9g. R9C6 = 9, R9C89 = [74] -> R8C9 = 8 (cage sum)

and the rest is naked singles.


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 Post subject: Re: Assassin 410
PostPosted: Thu Feb 10, 2022 8:44 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
I followed a very similiar path to Andrew but as often happens, saw things very differently. Andrew is much purer than I am with his lovely forcing chains.
a410 WT:
Preliminaries by SudokuSolver
Cage 6(2) n6 - cells only uses 1245
Cage 15(2) n2 - cells only uses 6789
Cage 7(2) n78 - cells do not use 789
Cage 7(2) n23 - cells do not use 789
Cage 13(2) n4 - cells do not use 123
Cage 9(2) n8 - cells do not use 9
Cage 10(2) n36 - cells do not use 5
Cage 11(2) n47 - cells do not use 1
Cage 19(3) n9 - cells do not use 1

This is a highly optimised solution. Any clean-up needed will be stated.

1. "45" in n1: 1 innie r3c3 = 8
1a. r3c4 + r4c3 = 7 (no 7,9)

2. "45" on c6789: 4 innies r5689c6 = 30 = {6789}: all locked for c6
2a. min. r89c6 = 13 -> max. r9c5 = 5

3. "45" on n9: 1 innie r7c7 = 3
3a. r6c7 + r7c6 = 11 = [92/74/65]
3b. no 4 in r1c6

4. "45" on n3: 2 outies r1c6 + r4c9 = 10 = [19/28/37]

5. 4 in c6 in sp11(2)r6c7 + r7c6 or in 21(4)r2c6
5a. -> 4 in r4c78 must have 7 in r6c7 (Locking out cages)
5b. "45" on n6: 4 innies r4c789 + r6c7 = 27
5c. -> {4689} blocked
5d. = {3789/5679}(no 1,2,4) = 3 or 5 in r4c78
5e. 7 & 9 locked for n6

6. 21(5)r2c8....
6a. {123}{69/78} blocked by r1c6 = (123) (or doesn't have 3 or 5 in r4c78)
6b. {12468} blocked by no 3 or 5 for r4c78 (step 5d)
6c. {125}{67} blocked by outies n3 = 10 (or boring way, doesn't have 3 or 5 for r4c78)
6d. {23457} as {234}{57} only, blocked by [16]+[65] in r1c67 & r6c7+r7c6
6e. = {12459/13458/13467}
6f. 1 & 4 locked for c6
6g. -> no 9 in r4c9 (outiesn3=10)
6h. r6c7 = (69)

7. 5 in c6 in r7c6 -> 6 in r6c7; or 5 in c6 in 21(5)r2c6
7a. -> 6 in r4c78, 21(5) must also have 5 for c6 (locking out cages)
7b. -> 21(5)r2c6: {13467} blocked
7c. = {12459/13458}(no 6,7)

8. r4c9 = 7 (hsingle n6), r3c9 = 3
8a. -> r1c6 = 3 (outies n3=10), r1c7 = 4

more key steps
9. "45" on n7: 2 innies r7c1 + r9c4 = 6 = [24/42/51]
9a. r6c1 = (679)
9b. r9c4 = (356)
9c. note: two outies n7 = 12 = [66/75/93]
9d. note2 from r7c1 = (679)
i. {69} in r6c17 -> r6c6 = (78)
ii. or [75] in outies n7 -> 2 in r7c6

10. 18(3)n8: {78}[3] blocked since r6c6 = (69) -> [25] in r7c6 + r9c4 (step 9dii) leaves nothing for 9(2)n8
10a."45" on n7: 1 innie r7c1 + 1 = 1 outie r9c4
-> {567} blocked from 18(3) by [22] in r7c16
10b. {459} blocked by 4 & 5 only in r9c5
10c. = {189/279/369/468}(no 5) = 1 or 2 or 6
10d. -> [516] in r7c1 + r9c34 forces 2 into both r7c6 and 18(3)n8
10e. -> no 5 in r7c1, no 1 in r9c3, no 6 in r9c4, no 6 in r6c1

11. r6c17 from {679} -> must have 6 or 7 -> 5 in r7c6 (h11(2) or r9c4 (outiesn7=12)
11a. 5 locked for n8
11b. no 4 in 9(2)n8
11c. also 9 locked in r6c17 since can't have [55] in r7c6 + r9c4 -> 9 locked for r6
11d. -> 2 locked in r7c16 for r7 (locking cages)

12. "45" on n5: 2 innies r4c6 + r6c4 = 7 = [16]/{25}/[43] (r6c4 = (2356))

13. combined cage 24(4)r2378c5 = {69}{18/27}/{78}{36}
13a. 6 locked for c5
13b. note: has 6 in r23c5 or 3 in r78c5

Took a long time to find this "45". Final crack
14. "45" on n2369: 1 innie r3c4 + 1 = 2 outies r47c6
14a. but [515] blocked by 6 in both r6c4 (h7(2)n5) and r6c7(sp11(2))
14b. and [625] -> r39c4 = [63], blocked by combined 24(4)r2378c5 (step 13b)
14c. -> no 5 in r7c6
14d. -> r7c6 = 2, r6c7 = 9, r67c1 = [74], r9c34 = [25]
14e. no 7 in 9(2)n8
14f. 18(3)n8 = {189/369/468}(no 7)

15. "45" on n8: 2 remaining innies r78c4 = 11 and must have 7 for n8 = [74] only

16. killer pair 1,3 in 9(2)n8 and r9c5: both locked for c5

17. "45" on n1236: 2 remaining outies r4c36 = 6 = {15} only, both locked for r4
17a. no 3 in r6c4 (h7(2)n5)
17b. r4c78 = [83]
17c. no 1 in r3c4

18. naked pair {26} in r36c4: both locked for c4
18a. r4c4 = 9

19. naked pair {18} in r12c4: both locked for n2 & c4
19a. -> r1c5 = 7 (cage sum)

20. r4c6 = 1 (hsingle c6)
20a. r4c3 = 5, r3c4 = 2 (cage sum)
20b. r56c4 = [36]
20c. r56c6 = [78]

21. 13(2)n4 = {49}: both locked for n4, 4 for r5

22. r89c6 = {69} -> r9c5 = 3

23. no 3 in r2c1 since r1c12 cannot = {67/58/49} = 13

24. r8c1 = 3 (hsingle c1)
24a. -> r9c12 = 14 = {68}: both locked for n7 and r9
24b. -> r9c6789 = [9174]


much easier now
Cheers
Ed


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 Post subject: Re: Assassin 410
PostPosted: Sat Feb 12, 2022 10:51 pm 
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Posts: 280
Location: California, out of London
Functionally equivalent steps for me as well, although the specific reasonings to place or eliminate possibilities are (again) reflective of our different approaches. Thanks Ed!
Assassin 410 WT:
1. Innies c6789 = r5689c6 = +30(4) = {6789}
Innies n9 = r7c7 = 3
-> r7c6 from (245) and r6c7 from (976)
IOD n3 -> r1c6 = r3c9
Since r3c9 cannot be 5 -> r1c6 cannot be 5
Since 3 already placed in c7 -> r1c6 cannot be 4
-> r1c6 = r3c9 from (123) and r4c9 from (987)

2! Innies n6 = r4c789,r6c7 = +27(4)
(A) {3789}
(B) {4689}
(C) {5679}
Since r4c9 is Min 7 and r6c7 from (679) -> One of (345) in r4c78

(A) -> (HS 3 in c6) r1c6 = 3 -> 10(2)r3c9 = [37]
(B) Leaves no place for 4 in c6 (Since r1c6 is Max 3 and 14(3)r6c7 cannot be [743])
(C) -> (HS 5 in c6) 14(3)r6c7 = [653]
-> (Innies n3) [r1c7,r3c9] cannot be [61] -> 9 not in r4c9 -> r4c9 = 7

In all cases 10(2)r3c9 = [37] and 7(2)r1c6 = [34]

3! Consider Step 2 case (C). Innies n6 = [{59}76]
This puts (HS 3 in n6) r6c8 = 3 and r56c9 = {18}
This puts only solution for 14(3)n3 = [7{25}]
But this leaves remaining cells c9 = r789c9 = +19(3) contradicting 19(3) cage in n9.
-> Step 2 (A) must be correct.
-> (Since r6c7 cannot be 8) Innies n6 = [8379]
-> r7c6 = 2

4. Remaining cells c6 = r234c6 = {145}
Remaining Innies n2 = r2c6,r3c46 = +11(3). Cannot contain both (15)
-> Either r234c6 = [{14}5], r3c4 = 6, (Innies n5) r6c4 = 2
Or r234c6 = [{45}1], r3c4 = 2, r6c4 = 6.
Innies n1 -> r3c3 = 8
-> r4c36 = {15} and r3c6c4 = {26}

5. Innies n7 = r7c1,r9c3 = +6(2)
Given previous placements this can only be [r7c1,r9c3] = [42]
-> r6c1 = 7 and r9c4 = 5

6. Remaining Innies n8 = r78c4 = +11(2)
Since r89c6 is Min +13 -> Max r9c5 = 5
-> 9 in n8 in r89c6
-> Min r89c6 = +15 -> Max r9c5 = 3
-> (HS 4 in n8) r78c4 = [74]
-> (HS 7 in c6) r5c6 = 7

7. 3 in n8 only in c5
-> (HS 3 in n5) r5c4 = 3
-> 3 in n4 in r6c23
-> Remaining Innies n4 = r4c3,r6c23 = [5{13}] or [1{35}]
-> 13(2)n4 = {49}
-> 16(3)n4 = [{26}8]
-> r4c45 = [94]
Also 6(2)n6 = {15}
-> 12(3)n6 = {246}

8. 1 in n8 only in c5
-> (HS 1 in n5) r4c6 = 1
-> r23c6 = {45} and r3c4 = 2
-> r4c3 = 5 and r6c23 = {13}
-> r6c4 = 6
Also (HS 5 in n5) r6c5 = 5
-> r5c5 = 2
Also r6c6 = 8
-> r89c6 = {69} and r9c5 = 3
-> 9(2)n8 = {18}
Also 12(3)n6 = [6{24}]
Also 15(2)n2 = {69}
-> r1c5 = 7 and r12c4 = {18}

9. Given previous placements only solution for 14(3)n3 is {158} with 5 in r1.
-> 24(4)n3 = {2679} with 9 in c8
Also -> 2 in r1 in r1c12
-> 16(3)n1 = [{29}5]
-> r23c6 = [45]
-> 21(5)n1 = [6{37}14]
-> 13(2)n4 = [94]
-> r1c12 = [92]
-> r4c12 = [26]
-> 17(3)n7 = [368]
etc.


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