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 Post subject: Assassin 392
PostPosted: Sun Mar 01, 2020 7:01 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 841
Location: Sydney, Australia
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Killer X - 1-9 cannot repeat on either diagonal.

Assassin 392
Kept finding easier ways to do this, but still not easy. A lot harder than a391 for example. JSudoku found it pretty easy which is why I first tried it. Found one really interesting step which fortunately, makes a big difference.

triple click code:
3x3:d:k:5120:5120:6657:5634:5634:2563:2308:2308:6917:5120:6657:6657:5634:2563:2563:6662:6917:6917:5120:3591:6657:6657:3336:3081:6662:6662:6917:5120:3591:3591:5386:3336:3081:3083:6662:6917:9996:9996:5386:5386:5386:3081:3083:1805:4878:9996:9996:9996:5386:4623:9488:3083:1805:4878:9996:9996:2833:2833:4623:6162:9488:9488:4878:7187:7187:2833:2580:4623:6162:9488:9488:4878:7187:7187:7187:2580:4623:6162:6162:9488:9488:
solution:
Code:
+-------+-------+-------+
| 4 2 7 | 5 9 1 | 6 3 8 |
| 1 5 9 | 8 3 6 | 4 7 2 |
| 6 8 3 | 2 7 4 | 9 5 1 |
+-------+-------+-------+
| 7 4 2 | 1 6 5 | 3 8 9 |
| 8 1 5 | 9 2 3 | 7 6 4 |
| 9 3 6 | 4 8 7 | 2 1 5 |
+-------+-------+-------+
| 5 7 1 | 6 4 9 | 8 2 3 |
| 2 6 4 | 3 5 8 | 1 9 7 |
| 3 9 8 | 7 1 2 | 5 4 6 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 392
PostPosted: Sat Mar 14, 2020 7:56 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 205
Location: California, out of London
Thanks again Ed. I found this one not very difficult - but I did use a small contradiction chain in my Step 3. Maybe there's a way around that.
Here's how I did it.
Corrections thanks to Ed.
Assassin 392 WT:
1. Outies n3 = r4c89 = +17(2) = {89}
-> Remaining Innies n6 = r56c9 = +9(2)
-> r78c9 = +10(2)

2. Outies c6789 = r2c5 = 3
-> n2 is either:
(a) 22(3) = {679}, 10(3) = {235}, r3c456 = {148}
(b) 22(3) = {589}, 10(3) = {136}, r3c456 = {247}
-> 13(2)r3c5 either (a)[85] or (b)[76]. (Cannot be [49] since 9 already in r4.)
-> Whichever of (56) is in r12c6 is in r4c5 and in n8 in r789c4

3. IOD c123 -> r578c3 = r3c4 + 8
Since r3c4 is max 4 -> r578c3 = max +12(3)
Trying 2(a) puts r7c4 = 5 puts r78c3 = {24} puts r7c12 = {38}
Since in this case Max r5c3 is 6 -> no place for 8 in n4
-> 2(b) must be correct.
-> 22(3)n3 = {589}, r12c6 = {16}, 13(2)r3c5 = [76], r3c46 = {24}
Also 6 in n8 in r789c4

4. Outies c1234 = r15c5 = +11(2) = [92]
-> r12c4 = {58}

5. 12(3)r3c6 only from [4{35}] or [2{37}]
-> 3 in n8 in r789c4
-> r789c4 from [6{37}] or [3{46}]
-> 2 in n8 in r789c6
-> r3c46 = [24]
-> IOD c123 -> r7c4 = r5c3 + 1
-> Since r5c3 cannot be 2 -> r7c4 cannot be 3
-> r7c4 = 6 and 10(2)n8 = {37}
-> r456c4 = {149} and r5c3 = 5
Also 12(3)r3c6 = [453]
-> r6c56 = [87]

6. Remaining Innies r1234 = r4c47 = +4(2) = [13]
-> r4c123 = {247}

7. Remaining cells c6 -> r789c6 = {289}
-> r9c7 = 5
Also r789c5 = {145}

8. 7 in r6c6 -> 7 in n9 in r78c8
-> -> r78c9 = {37}
-> 19(4)c9 = [45{37}]
-> 7(2)n6 = {16}
-> 12(3)n6 = [372]

9. (35) in D\ in n1
-> 14(3)r3c2 = [8{24}]
-> r4c1 = 7
-> 39(7)r5c1 must be [8{1369}57]
-> (HS 5 on D\) r2c2 = 5
Also HS 8 in n7 -> r9c3 = 8
Also HS 7 on D/ -> r2c8 = 7
-> HS 7 on r1 -> r1c3 = 7
-> 26(5)r1c3 = [75932]
Also (HS 3 in c8) -> 9(2)n3 = [63]
etc.


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 Post subject: Re: Assassin 392
PostPosted: Fri Mar 20, 2020 10:51 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1710
Location: Lethbridge, Alberta, Canada
Thanks Ed for this Assassin. After starting it, I had a week's break when I was busy on other things, then found my breakthrough step, parts of which I'd seen earlier but not how to put it all together.

Just gone through wellbeback's walkthrough. Our breakthroughs were both based on 8 in N4 and 11(3) cage = {24}5, but in very different ways.

Here is my walkthrough for Assassin 392:
Prelims

a) R1C78 = {18/27/36/45}, no 9
b) R34C5 = {49/58/67}, no 1,2,3
c) R56C8 = {16/25/34}, no 7,8,9
d) R89C4 = {19/28/37/46}, no 5
e) 22(3) cage at R1C4 = {589/679}
f) 10(3) cage at R1C6 = {127/136/145/235}, no 8,9
g) 11(3) cage at R7C3 = {128/137/146/236/245}, no 9
h) 26(4) cage at R2C7 = {2789/3689/4589/4679/5678}, no 1

1a. 45 rule on C6789 1 outie R2C5 = 3 -> R12C6 = 7 = {16/25}
1b. 22(3) cage at R1C4 = {589/679}, 9 locked for N2, clean-up: no 4 in R4C5
1c. Killer pair 5,6 in 22(3) cage at R12C6, locked for N2, clean-up: no 7,8 in R4C5
1d. 4 in N2 only in R3C456, locked for R3
1e. 45 rule on N3 2 outies R4C89 = 17 = {89}, locked for R4 and N6, clean-up: no 4 in R3C5
1f. Naked pair {89} in R4C89, CPE no 8,9 in R2C8
1g. Killer pair 7,8 in 22(3) cage and R3C5, locked for N2
1h. 45 rule on N36 2 innies R56C9 = 9 = {27/36/45}, no 1
1i. 45 rule on N36 2 outies R78C9 = 10 = {19/28/37/46}, no 5

2a. 45 rule on C1234 2 outies R15C5 = 11 = [74/92] (cannot be {56} which clashes with R4C5)
2b. 45 rule on C789 4 outies R6789C6 = 26 = {2789/3689/4589/4679} (cannot be {5678} which clashes with R12C6), no 1, 9 locked for C6
2c. 45 rule on C789 1 outie R6C6 = 1 innie R9C7 + 2, no 2 in R6C6, no 8,9 in R9C7
2d. 45 rule on C9 1 outie R2C8 = 1 innie R9C9 + 1, no 1 in R2C8, no 2,7,8,9 in R9C9
2e. 12(3) cage at R3C6 = {138/147/237/345} (cannot be {156/246} which clash with R12C6), no 6
2f. 12(3) cage at R4C7 = {147/156/237/345} (cannot be {246} which clashes with R56C8)

3. 37(7) cage at R6C6 = {1246789/1345789/2345689}, R6C6 = R9C7 + 2 (step 2c)
3a. Consider placement for 4 in 37(7) cage
R6C6 = 4
or 4 in R78C78 + R9C89, locked for N9, no 4 in R9C7 => no 6 in R6C6
-> no 6 in R6C6, clean-up: no 4 in R9C7
3b. Consider placement for 5 in N9
5 in R78C78 + R9C89 => no 5 in R6C6
or R9C7 = 5 => R6C6 = 7
-> no 5 in R6C6, clean-up: no 3 in R9C7
[Alternatively 37(7) cage must contain 4 -> R6C6 + R9C7 cannot be [64], 37(7) cage must contain both or neither of 3,5 -> R6C6 + R9C7 cannot be [53].]

4. 22(3) cage at R1C4 = {589/679}, R15C5 (step 2a) = [74/92]
4a. Consider combinations for R12C6 (step 1a) = {16/25}
R12C6 = {16} => R12C4 = {58}, 5 locked for C4, 8 locked for N2, R15C5= [92], R34C5 = [76]
or R12C6 = {25}, locked for C6 and N1, 22(3) cage = {679}, locked for N1, 6 locked for C4 => R34C5 = [85], R6C5 = 6 (hidden single in N5)
-> 6 in R46C5, locked for C5 and N5, no 5 in R456C4, no 2 in R45C6
4b. 1 in C5 only in 18(4) cage at R6C5 = {1269/1458/1467} (cannot be {1278} which clashes with R3C5)
4c. 6 of {1269/1467} must be in R6C5 -> no 2,7,9 in R6C5

5a. 45 rule on N7 2 innies R7C12 = 1 outie R7C4 + 6, IOU no 6 in R7C12
5b. 39(7) cage at R5C1 = {1356789/2346789}, 6 locked for N4
5c. 2 in N5 only in R456C4 + R5C5, locked for 21(5) cage at R4C4
5d. 21(5) cage = {12378/12459/23457}

6. 45 rule on R1234 2 innies R4C47 = 1 outie R5C6 + 1, IOU no 1 in R4C7

7. 45 rule on N1 2(1+1) outies R3C4 + R4C1 = 1 innie R3C2 + 1, IOU no 1 in R4C1
7a. Min R3C4 + R4C1 = 3 -> min R3C2 = 2
7b. R3C4 + R4C1 cannot total 10 -> no 9 in R3C2

8. 45 rule on C123 2 outies R37C4 = 1 innie R5C3 + 3
8a. R7C12 = R7C4 + 6 (step 5a)
8b. Consider placement of 8 in N5
8 in R5C12 + R6C123, locked for 39(7) cage at R5C1 => R7C12 cannot total 11 = {29/47/56} which clash with R78C3 + R7C4 = {24}5 => no 5 in R7C4
or R5C3 = 8 => R37C4 = 11 = [47]
-> no 5 in R7C4
8c. 5 in C4 only in R12C4 -> R12C4 = {58}, locked for N2, 8 locked for C4 -> R13C5 = [97], R4C5 = 6, clean-up: no 2 in R12C6 (step 1a), no 2 in R89C4
8d. Naked pair {16} in R12C6, locked for C6, 1 locked for N2
8e. Naked pair {24} in R3C46, 2 locked for R3
8f. R1C5 = 9 -> R5C5 = 2 (step 2a), placed for both diagonals, clean-up: no 5 in R6C8, no 7 in R6C9 (step 1h), no 1 in R9C9 (step 2d)
8g. 12(3) cage at R3C6 (step 2e) = {237/345}, no 8, 3 locked for C6 and N5, clean-up: no 1 in R9C7 (step 2c)
8h. 4 of {345} must be in R3C6 -> no 4 in R45C6
8i. 8 in N5 only in R6C56, locked for R6

9a. 21(5) cage (step 5d) = {12459} (only remaining combination, cannot be {12378/23457} because 3,5,8 only in R5C3) -> R5C3 = 5, R456C4 = {149}, locked for C4 and N5 -> R3C4 = 2, clean-up: no 2 in R6C8, no 4 in R6C9 (step 1h), no 6 in R89C4
9b. R3C6 = 4 -> R45C6 = 8 = [53], 5 placed for D/, R6C5 = 8, R6C6 = 7, placed for D\, R9C7 = 5 (step 2c), clean-up: no 4 in R1C8, no 4 in R6C8, no 6 in R6C9 (step 1h), no 6 in R2C8, no 4 in R9C9 (both step 2d)
9c. R7C4 = 6 (hidden single in N8) -> R78C3 = 5 = [14/32/41]
9d. 7 in N9 only in R78C9 = 10 (step 1i) = {37}, locked for C9, 3 locked for N9, R9C9 = 6, placed for D\, R2C8 = 7 (step 2d), clean-up: no 2 in R1C78
9e. R56C9 (step 1h) = 9 = [45], clean-up: no 3 in R6C8
9f. Naked pair {16} in R56C8, locked for C8 and N6 -> R5C7 = 7, R46C7 = {23}, locked for C7, clean-up: no 8 in R1C7
9g. R2C9 = 2 (hidden single in N3)
9h. 8 in R5 only in R5C12, locked for 39(7) cage at R5C1

10a. R4C47 (step 6) = R5C6 + 1, R5C6 = 3 -> R4C47 = 4 = [13], 1 placed for D\, R56C4 = [94], 4 placed for D/, R6C7 = 2
10b. R8C7 = 1 (hidden single in N9), clean-up: no 8 in R1C8
10c. 39(7) cage at R5C1 = {1356789} (cannot be {2346789} because 2,4,7 only in R7C12), no 2,4, 7 locked for R7 and N7 -> R78C9 = [37], R89C4 = [37], R7C3 = 1, placed for D/, R8C3 = 4 (step 9c), R1C9 = 8, placed for D/, R12C4 = [58], R1C8 = 3 -> R1C7 = 6, R3C7 = 9, placed for D/

and the rest is naked singles, without using the diagonals (although quicker if they are used).

Rating Comment:
I'll rate my WT for A392 at 1.5; usually I rate short forcing chains at Easy 1.5 but my breakthrough step was a bit more complex.


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 Post subject: Re: Assassin 392
PostPosted: Thu Mar 26, 2020 8:18 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 841
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Well done wellbeback and Andrew for finding that neat shortcut (compared to me)! I had to work harder so at least we know there are two quite different solutions for this one.

a392 WT:
Preliminaries courtesy of SudokuSolver
Cage 7(2) n6 - cells do not use 789
Cage 13(2) n25 - cells do not use 123
Cage 9(2) n3 - cells do not use 9
Cage 10(2) n8 - cells do not use 5
Cage 22(3) n2 - cells do not use 1234
Cage 10(3) n2 - cells do not use 89
Cage 11(3) n78 - cells do not use 9
Cage 26(4) n36 - cells do not use 1

1. "45" on n3: 2 outies r4c89 = 17 = {89} only, both locked for r4 and n4
1a. r2c8 sees both those cells -> no 8,9 (Common Peer Elimination CPE)

2. "45" on c6789: 1 outie r2c5 = 3
2a. -> r12c6 = 7 = {16/25}(no 4,7) = 5 or 6

3. 22(3)n2 = {589/679} = 5 or 6
3a. killer pair 5,6 with r12c6: both locked for n2
3b. must have 9: locked for n2
3c. note: if 22(3) has 8, must have 5; has 7, must have 6
3d. r34c5 = [76/85]

4. "45" on n2: 3 innies r3c456 = 13 and must have 4 for n2 = {148/247}
4a. 4 locked for r3
4b. 7 or 8 must be in r3c5 -> neither in r3c46

5. "45" on c1234: 2 outies r15c5 = 11
5a. but {56} blocked by r4c5 = (56)
5b. = [92/74]

Loved finding this one.
6. both the 22(3)n2 and 13(2)r3c5 have both {67} or both {58} with the 5 or 6 in c4 for 22(3) or in r4c5 -> killer pair 5,6 -> no 5,6 in r456c4

7. "45" on n7: 1 outie r7c4 + 6 = 2 innies r7c12
7a. both innies see the one outie so cannot be equal -> the Innie Outie Difference (IOD) of 6 cannot be in r7c12
7b. -> no 6 in r7c12 (IOU)

8. 39(7)r5c1 = {1356789/2346789}: must have 6: locked for n4

9. 21(5)r4c4 = {12378/12459/13458/23457}
9a. = 5 or {12378} with 2 in r5c5
9b. -> r5c3 = (13578)(no 2,4,9)
9c. note: if it has 1 in r5c3, must have 3 in c4

Got a bit lucky seeing this one.
10. "45" on c1234: 1 innie r5c3 + 3 = 2 outies r37c4
10a. but [1] + [13] blocked by 3 already in c4 (step 9c)
10b. -> 1 r5c3, no 3 in r7c4

The key
11. "45" on c789: 1 outie r6c6 - 2 = 1 innie r9c7: no 1,2 in r6c6, no 8,9 in r9c7
11a. and 3 in n8 only in 10(2) = {37} or in 24(3)r7c6
11b. -> 7 in r789c6 must also have 3 (Locking-out cages)
11c. but 24(4) as {379}[5] blocked by 7 also in r6c6 (step 11.)
11d. and {378}[6] blocked by 8 also in r6c6 (step 11.)
11e. -> no 7 in r789c6

12. 7 in c6 only in n5: locked for n5

13. 21(5)r4c4 = {12378/12459/13458/23457}
13a. 5 and 7 only in r5c3 -> = (57)
13b. can't have both -> {23457} blocked
13c. -> = {12378/12459/13458}
13d. must have 1, locked for c4 and n5
13e. note: if it has 7, must also have 8 in c4

14. from step 10, 1 innie r5c3 + 3 = 2 outies r37c4
14a. = [5][26]/[7][28/46]
14b. deleted
14c. but [7][28] blocked by 8 already in c4
14d. -> = [5][26]/[7][46]
14e. -> r7c4 = 6, r3c4 = (24)

cracked
15. 22(3)n2 = {589} only -> r1c5 = 9, 5 and 8 locked for c4
15a. r5c5 = 2 (h11(2)r15c5), 2 placed for both D
15b. r34c5 = [76]
15c. -> r3c46 = 6 (from h13(3)r3c456) = {24} only: 2 locked for r3, n2
15d. r12c6 = {16} only: both locked for c6

16. 21(5)r4c4 = {12459}
16a. r5c3 = 5, r456c4 = {149}: 4 & 9 locked for n4 and n5
16b. r3c46 = [24]
16c. -> r45c6 = 8 = [53] only: 5 placed for d/
16d. r6c56 = [87], 7 placed for d\
16e. -> r9c7 = 5 (iodc789=-2)

17. "45" on n9: 2 remaining innies r78c9 = 10 and must have 7 for n9 = {37} only: both locked for c9, 3 for n9

18. "45" on c9: 1 outie r2c8 - 1 = 1 innie r9c9 = [76] only permutation, 7 placed for d/, 6 for d\

19. 8 & 9 in c9 only in 27(5)r1c9 = [7]{{1289}: 1 locked for n3 and c9
19a. r2c9 = 2
19b. r56c9 = [45]

20. 7(2)n6 = {16}: both locked for c8 and n6
20a. r5c7 = 7
20b. r46c7 = {23}: both locked for c7

21. 8 & 9 in n4 only in 39(7) -> no 8,9 in r7c12

22. "45" on n7: 2 remaining innies r7c12 = 12 = {57} only: both locked for r7, n9
22a. r78c9 = [37], r89c4 = [37]

23. r78c3 = 5 (cage sum) = {14} only: both locked for n7 and c3

24. hsingle 3 on d/ -> r9c1 = 3, r3c3 = 3 (hsingle d\)
24a. r1c8 = 3 (hsingle n3), r1c7 = 6, r8c2 = 6 (hsingle d/), r3c1 = 6 (hsingle r3

easy now
Cheers
Ed


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