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Assassin 341 X
http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1395
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Author:  HATMAN [ Thu Dec 15, 2016 4:05 pm ]
Post subject:  Assassin 341 X

Assassin 341 X

It is X.

I went for a restrained puzzle - it is unique with the four 5-cages and five 3-cages.

SS gives it 1.5 and JS uses two medium fishes.


Image


JS Code:
3x3:d:k:22:23:5121:5121:3080:5122:5122:3847:3847:24:5121:25:26:3080:27:28:5122:3847:5121:4880:2316:29:3080:30:3853:3349:5122:5121:4880:4880:2316:2316:3853:3349:3349:5122:4618:4618:4618:4111:31:3853:4617:4617:4617:5124:3857:3857:4111:4878:4878:5908:5908:5123:5124:3857:4111:4882:4619:3347:4878:5908:5123:2309:5124:4882:4882:4619:3347:3347:5123:2310:2309:2309:5124:5124:4619:5123:5123:2310:2310:

Solution:
982634157
645871923
731529468
197268345
864153279
253947681
576312894
419786532
328495716

Author:  Andrew [ Sun Dec 25, 2016 5:50 am ]
Post subject:  Re: Assassin 341 X

Thanks HATMAN for an interesting and challenging puzzle!

Here is my walkthrough for Assassin 341 X:
Prelims

a) 19(3) cage at R3C2 = {289/379/469/478/568}, no 1
b) 9(3) cage at R3C3 = {126/135/234}, no 7,8,9
c) 19(3) cage at R6C5 = {289/379/469/478/568}, no 1
d) 23(3) cage at R6C7 = {689}
e) 19(3) cage at R7C4 = {289/379/469/478/568}, no 1
f) 9(3) cage at R8C1 = {126/135/234}, no 7,8,9
g) 9(3) cage at R8C9 = {126/135/234}, no 7,8,9

1. 45 rule on R6789 1 outie R5C4 = 1 -> R6C4 + R7C3 = 15 = {69/78}
1a. 45 rule on R5 2 innies R5C56 = 8 = {26/35}
1b. Min R4C45 = 6 (cannot be {23} which clashes with R5C56) -> max R3C3 = 3
1c. 45 rule on C5 3 innies R456C5 = 15, R45C5 cannot total 8 (which would clash with R5C56 CCC) -> no 7 in R6C5
1d. R456C5 = 15 = {249/258/348/456}
1e. 8,9 of {249/258/348} must be in R6C5 -> no 2,3 in R6C5
1f. R45C6 cannot total 8 (which would clash with R5C56 CCC) -> no 7 in R3C7
1g. 23(3) cage at R6C7 = {689}, CPE no 6,8,9 in R45C8

2. 45 rule on N7 3 innies R7C23 + R8C3 = 2 outies R6C1 + R9C4 + 16
2a. Min R6C1 + R9C4 = 3 -> min R7C23 + R8C3 = 19, no 1 in R7C2
2b. Max R7C23 + R8C3 = 24 -> max R6C1 + R9C4 = 8, no 7,8,9 in R6C1, no 8,9 in R9C4

3. 45 rule on N9 3 innies R7C78 + R8C7 = 2 outies R6C9 + R9C6 + 16
3a. Min R6C9 + R9C6 = 3 -> min R7C78 + R8C7 = 19, no 1 in R8C7
3b. Max R7C78 + R8C7 = 24 -> max R6C9 + R9C6 = 8, no 8,9 in R6C9 + R9C6

4. Hidden killer pair 8,9 in 18(3) cage at R5C1 and 18(3) cage at R5C7 for R5, 18(3) cage cannot contain both of 8,9 (because no 1) -> each 18(3) cage must contain one of 8,9 = {279/369/378/459/468}
4a. 6 of 23(3) cage at R6C7 must be in R6C78 (R6C78 cannot be {89} which clashes with 18(3) cage at R5C7), 6 locked for R6, N6 and 23(3) cage, no 6 in R7C8, clean-up: no 9 in R7C3 (step 1)
4b. 18(3) cage at R5C7 = {279/378/459}
4c. Killer pair 8,9 in 18(3) cage and 23(3) cage, locked for N6
4d. Max R4C78 = 9 (cannot be {47/57} which clash with 18(3) cage, cannot be {37} because 13(3) cage at R3C8 cannot be 3{37}) -> min R3C8 = 4
4e. 19(3) cage at R6C5 = {289/379/469/478/568}
4f. 6 of {568} must be in R7C7 -> no 5 in R7C7
4g. 2 of {289} must be in R6C6 (R6C56 cannot be {89} which clashes with 23(3) cage), no 2 in R7C7
4h. 3 of {379} must be in R6C6 (R6C56 cannot be {79} which clashes with R6C4 + 23(3) cage), no 3 in R7C7
4i. 6 of {469} must be in R7C7, R6C56 cannot be {78} which clashes with R6C4 + 23(3) cage), no 4 in R7C7
4j. 19(3) cage must contain one of 2,3,4,5 in R6C56 -> must contain one of 7,8,9 in R6C56
4k. Hidden killer triple 7,8,9 in R4C6, R6C4 and R6C56 for N5 -> R4C6 = {789}
4l. Min R45C6 = 9 -> max R3C7 = 6 but R3C7 + R4C6 cannot be [67] which clashes with R6C4 + R7C3 -> max R3C7 = 5

[I saw this after the next step, but it simplifies things to do this step first.]
5. R5C56 (step 1a) = {26/35}
5a. 19(3) cage at R6C5 = {469/478/568} (cannot be {289/379} because R5C56 + R6C6 = {26}3/{35}2 clash with 9(3) cage at R3C3), no 2,3
5b. 19(3) cage = {478/568} (cannot be {469} = {49}6 because R4C6 + R6C4 = {78} clashes with 16(3) cage at R5C4 = 1{78}), no 9
5c. 9 in N5 only in R4C6 + R6C4, locked for D/
[With hindsight I ought to have done step 7b next.]

[I was slow in spotting the design feature of this puzzle; the previous step also seems to be an important design feature.]
6. R37C37 + R5C5 form a 5-cell cage because of the diagonals
6a. 45 rule on N5 4 outies R37C37 = 1 innie R5C5 + 14
6b. R5C5 = {2356} -> R37C37 = 16,17,19,20 -> R37C37 + R5C5 = 18,20,24,26
6c. Min R7C37 = 13, min R3C37 + R5C5 = 6 -> min R37C37 + R5C5 = 20 -> no 2 in R5C5, clean-up: no 6 in R5C6 (step 1a)
[I continued with step 6d, but feel free to jump forward to step 7.]
6d. Possible combinations for R37C37 + R5C5, taking account that 20(5) must contain 3 for R5C5, 24(5) must contain 5 for R5C5 and 26(5) must contain 6 for R5C5 and must contain two of 6,7,8 in R7C37
20(5) = {12368}
24(5) = {13578/14568/23568/24567}
26(5) = {14678/23678}
-> R3C37 = {12/13/14/23/24}, no 5 in R3C7, R7C37 = {67/68/78}

7. R5C56 (step 6c) = [35/53/62]
7a. 19(3) cage at R6C5 (step 5b) = {478} (only remaining combination, cannot be {568} = {58}6 which clashes with R5C56 using D\), 4 locked for R6 and N5, 7 locked for D\, CPE no 8 in R6C7, no 7 in R7C6
7b. Caged X-Wing for 8 in 19(3) cage and 23(3) cage at R6C7, no other 8 in R67 -> 16(3) cage at R5C4 = [196], 6 placed for D/, clean-up: no 2 in R5C6 (step 1a)
7c. R6C78 = [68] -> 19(3) cage = [478], 8 placed for D\, R7C8 = 9
7d. Naked pair {35} in R5C56, locked for R5 and N5
7e. R4C6 = 8, placed for D/ -> R3C7 + R5C6 = 7 = [25/43]
7f. Naked pair {26} in R4C45, locked for R4, R3C3 = 1 (cage sum), placed for D\
7g. 18(3) cage at R5C7 = {279} (only remaining combination), locked for R5 and N6
7h. Naked triple {468} in 18(3) cage at R5C1, locked for N4

8. 15(3) cage at R6C2 = {357} (only remaining combination) -> R6C23 = {35}, locked for R6 and N4, R7C2 = 7, R4C23 = [97], R3C2 = 3 (cage sum), R46C1 = [12], R6C23 = [53], R6C9 = 1
8a. 13(3) cage at R3C8 = {346} (only possible combination) -> R3C8 = 6, R4C78 = {34}, R4C9 = 5
8b. R9C8 = 1 (hidden single in N9) -> R89C9 = 8 = {26}, locked for C9 and N9
8c. 9(3) cage at R8C1 = {234} (only remaining combination) -> R9C2 = 2, R89C1 = {34}, locked for C1 and N7 -> R7C1 = 5, R8C2 = 1 (hidden single in N7), R9C9 = 6, placed for D\, R8C9 = 2, R4C4 = 2, R4C5 = 6, R1C1 = 9, R2C2 = 4, placed for D\
8d. Naked pair {89} in R89C3, locked for C3 -> R5C3 = 4
8e. 20(5) cage at R6C9 = {13457} -> R9C7 = 7
8f. Naked pair {34} in R9C14, locked for R9 -> R9C6 = 5, R5C56 = [53], R3C7 = 4 (cage sum), placed for D/, R8C8 = 3, R7C9 = 4, R9C1 = 3, placed for D/
8g. R8C7 = 5 -> R78C6 = 8 = [26]
8h. R7C4 = 3 -> R8C34 = 16 = [97]
8i. R1C9 + R2C8 = [72] -> R12C3 = [25], R1C8 = 5
8j. R1C3 + R2C2 + R4C1 = [241] = 7 -> R1C4 + R3C1 = 13 = [67]

and the rest is naked singles, without using the diagonals.

Thanks Ed for detail corrections to my walkthrough; I've also added a few more of my own corrections, spotted while checking through Ed's changes.

Rating Comment:
After checking my walkthrough and making minor changes, I'll rate it not more than Hard 1.25; even that may be a bit high.


Happy Christmas, everybody! :santa:

Author:  Ed [ Tue Dec 19, 2017 7:08 pm ]
Post subject:  Re: Assassin 341 X

And Merry Christmas to you Andrew! :D

As Andrew said, a very interesting puzzle. My solution is quite different. The way Andrew does those myriad blocking moves reminds of Para.

A341 X
start only:
Prelims courtesy of SudokuSolver
Cage 23(3) n69 - cells ={689}
Cage 9(3) n15 - cells do not use 789
Cage 9(3) n9 - cells do not use 789
Cage 9(3) n7 - cells do not use 789
Cage 19(3) n59 - cells do not use 1
Cage 19(3) n78 - cells do not use 1
Cage 19(3) n14 - cells do not use 1

1. "45" on r6789: 1 outie r5c4 = 1
1a. split 15(2)r6c4+r7c3 = 15 = {69/78}

2. Hidden killer triple 7,8,9 in r9 since the two 20(5) cages at r6c19 can have at most one of 7,8,9 each
2a. r9c5 = (789)
2b. the two 20(5) cages must have at least one of 7,8,9 = {12359/12368/12458/12467/13457}
2c. the one 7,8,9 for each of the 20(5) must be in r9 -> no 7,8,9 in r67c19 nor r8c28
2d. the 20(5) cages must both have 1 -> no 1 in r9c19 since each sees all one 20(5)

3. 8,9 in n9 only in r789c7 or r7c8 -> no 8,9 in r6c7 (CPE)
3a. r6c7 = 6
3b. r67c8 = {89}: both locked for c8
3c. 7 in n9 only in c7: locked for c7
3d. no 9 in r7c3 (sp15(2)r6c4+r7c3))

4. "45" on r789: 2 outies r6c19 - 27 = 4 innies r7c2378
4a. ->r6c19 = 3 = {12} only: both locked for r6
4b. r7c2378 = 30 = {6789} only: all locked for r7
4c. 6 in r7 only in n7: 6 locked for n7

5. Hidden quad 6,7,8,9 in n7 -> r7c23+r89c3 from {6789} only

6. 1 in n4 only in c1: locked for c1

7. 9(3)n7 = {135/234}
7a. must have 3 -> 3 locked for n7

8. r6c1 = {12} and sees all 1 & 2 in n7 apart from r9c2 -> r9c2 = {12}
(Alternatively: r67c1+r8c2 = three of {1245} -> 9(3)n7: {24}[3]/{23}[4] blocked -> r9c2 = {12} only)(personally I find the clone way much easy to find so chuffed I found both!)
8a. 3 in n7 only in c1: locked for c1

9. 9(3)n9 = {126/135/234}
9a. 6 blocked from r8c9 since {12} in r9c89 clashes with r9c2
9b. 6 blocked from r9c8 since {12} in r89c9 clashes with r6c9

10. 6 in n9 only in r8c8, r9c9 only on D\: 6 locked for D\ and no 6 in r9c6 (CPE)

Have a step in common with Andrew so perhaps is the key one for this puzzle. However, I checked SS and it doesn't use it.
11. "45" on n5: 4 outies r3c37+r7c37 - 14 = 1 innie r5c5
11a. Min. 4 outies = {12}[67] = 16 -> min. r5c5 = 3 (can't be 2 since r3c37 has a 2 and it sees r5c5 through the two diagonals

12. "45" on r5: 2 innies r5c56 = 8 = {35} only valid combination: both locked for r5 and n5, and not in r3c7 (CPE)

13. 18(3)n6 = {279} only valid combination: all locked for r5 and n6

14. r6c89 = [81]

15. "45" on c5: 3 innies r456c5 = 15 = [654]: only valid permutation. 5 placed for both D.

On from there. Lots of naked subsets, hidden singles, last combos or cage sums right till the end.
Cheers
Ed

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