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PostPosted: Tue Dec 18, 2012 1:42 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
AN NC 2

This is Anti-kNight: cells a knights move appart cannot be equal.

This is Non-Consecutive: vertically and horizontally adjacent cells cannot be consecutive.

Additionally the value in r5c5 is also in all the four diagonal cages.
It solves without this but I cannot follow one set of JSudoku deductions.


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JS Code (set AN and NC then add twin killer cage)
3x3:4:k:3076:12:13:14:4104:15:16:17:2306:18:3076:19:20:4104:21:22:2306:23:24:25:3076:26:4104:27:2306:28:29:30:31:32:33:34:3841:35:36:37:3078:3078:3078:38:3841:39:4359:4359:4359:40:41:42:3841:43:44:45:46:5386:47:48:5123:49:3849:50:4613:51:5386:52:5123:53:54:3849:55:56:4613:5386:5123:57:58:59:3849:3851:3851:3851:4613:

Solution (note end of row = start of next row, I wonder why?)

592638471
147295836
683741592
259386147
714952683
368417259
925863714
471529368
836174925


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PostPosted: Sun Jan 22, 2017 11:55 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Another puzzle from just over 4 years ago which I tried for the first time. Quite a challenging puzzle!

Here is my walkthrough:
This is Anti-kNight (AN), non-consecutive (NC). HATMAN commented that the value in R5C5 is in all four diagonal cages in the corner nonets; I didn't need to use that.

Prelims

a) 9(3) cage at R1C9 = {126/135/234}
b) 21(3) cage at R6C9 = {489/579} (cannot be {678} because of NC)
c) 20(3) cage at R7C3 = {389/479/569/578}, no 1,2

1. 45 rule on C5 3 innies R456C5 = 14
1a. 45 rule on C5 3 innies R5C456 = 16
1b. Total of two 15(3) cages and two hidden 14(3) and 16(3) cages = 60 with R5C5 used for all four cages, i.e. three extra uses -> R5C5 = 5 -> no 4,6 in R46C5 + R5C46 (NC)
1c. R5C5 = 5 -> R46C5 = 9 = {18/27}, no 3,9
1d. R5C5 = 5 -> R5C46 = 11 = {29/38}, no 1,7
1e. Killer pair 2,8 in R46C5 + R5C46, locked for N5
1f. R5C5 = 5 -> no 5 in R37C46 + R46C37 (AN)

2. 21(3) cage at R6C9 = {489/579}, 9 locked for C9
2a. 4 of {489} must be in R7C9 because of NC -> no 4 in R68C9, no 8 in R7C9
2b. 21(3) cage = {489/579} -> no 9 in R7C7 + R8C8 (N9 + AN)
2c. 18(3) cage at R7C7 = {378/468/567}, no 1,2
2d. 18(3) cage = {468/567} (cannot be {378} which clashes with 21(3) cage N9 + AN), no 3, 6 locked for N9
2e. 21(3) cage + 18(3) cage must contain 4, locked for N9
2f. 18(3) cage = {468/567} -> no 6 in R9C6 (R9 + AN)
2g. 15(3) cage at R9C6 = {159/249/348/357} (cannot be {258} which clashes with 18(3) cage N9 + AN)
2h. 4 of {348} must be in R9C6 -> no 8 in R9C6
2i. R8C9 = {89} (cannot be 5,7 which clash with 18(3) cage + NC)
2j. R8C9 = {89} -> no 9 in R7C9, no 8 in R8C8 + R9C9 (all NC)
2k. 8 of {468} must be in R7C7 -> no 4,6 in R7C7
2l. R9C8 = {12389} (cannot be 5,7 which clash with 18(3) cage + NC)
2m. 21(3) cage = {489/579} -> no 9 in R6C8 (AN)
2n. 15(3) cage = {159/249/348} (cannot be {357} = {57}3 which clashes with 18(3) cage + NC), no 7
2o. 18(3) cage = 8{46}/{567} -> no 7 in R7C8 + R8C7 (NC)
2p. Hidden killer pair 4,7 in 18(3) cage and R7C9 for N9, 18(3) cage contains one of 4,7 -> R7C9 = {47}
2q. 21(3) cage = {489/579}, R7C9 = {47} -> no 7 in R6C9

3. 16(3) cage at R1C5 = {169/349/367} (cannot be {178/268} which clash with R46C5), no 2,8
3a. 3,9 of {349/367} must be in R2C5 -> no 4,7 in R2C5

4. 12(3) cage at R5C1 = {129/138/147/246} (cannot be {237} which clashes with R5C46)
4a. 9 of {129} must be in R5C2 -> no 9 in R5C13

5. 15(3) cage at R7C5 = {168/249/267/348}
5a. 2 of {267} must be in R8C5 -> no 7 in R8C5

6. 15(3) cage at R9C6 (step 2n) = {159/249/348}
6a. Hidden killer triple 1,2,3 in R7C8, R8C7 and 15(3) cage at R9C6 for N9, 15(3) cage only contains one of 1,2,3 -> R7C8 = {123}, R8C7 = {123}, 15(3) cage contains one of 1,2,3 in R9C78 -> no 1,2,3 in R9C6
6b. R7C8 = {123} -> no 2 in R6C8 (NC)
6c. R8C7 = {123} -> no 2 in R8C6 (NC)

7. 17(3) cage at R5C7 = {179/269/368/467} (cannot be {278} which clashes with R5C46)
7a. 4 of {467} must be in R5C8 -> no 4 in R5C79

8. 4,6 in N5 only in 15(3) cage at R4C4 = {456} or 15(3) cage at R4C6 = {456}
Because of the symmetry of candidates in these 15(3) cages and in R46C5 and R5C46 it doesn’t matter where 4,6 are place at the start of a forcing chain, so consider the forcing chain with R4C4 = 4, R6C6 = 6 -> no 3 in R5C4 (NC), no 8 in R5C6 (step 1d), no 7 in R6C5 (NC), no 2 in R4C5 (1c), R4C6 + R6C4 = {19/37}
Consider candidates for R4C6
R4C6 = {13} => R6C4 = {79} => no 8 in R5C4 (NC) no 3 in R5C6 (step 1d) => R4C6 = 3 (hidden single in N5), R6C4 = 7
or R4C6 = {79} => no 7,8 in R4C5 (NC) => R4C5 = 1, R6C5 = 8 (step 1c) => no 3 in R5C6 (NC) => R6C4 = 3 (hidden single in N5), R4C6 = 7
-> R4C6 + R6C4 = {37}
Similarly for other placements of 4,6
-> R46C46 = {3467}, locked for N5, R5C46 (step 1d) = {29}, locked for R5 and N5, R46C5 = {18}, locked for C5
8a. Naked pair {18} in R46C5 -> no 1,8 in R5C37 (AN)
8b. Naked pair {29} in R5C46 -> no 2,9 in R37C5 (AN)

9. 12(3) cage at R5C1 = {138/147}, no 6, 1 locked for R5 and N4
9a. 1 in R5C12 -> no 1 in R37C12 (AN)
9b. 6 in R5 only in 17(3) cage at R5C7, locked for N6
9c. 17(3) cage = {368/467}
9d. 4 of {467} must be in R5C8 -> no 7 in R5C8
9e. R8C9 = {47} -> {467} = [746] (AN), no 7 in R5C9
9f. 6 in R5C789 -> no 6 in R3C8 (AN)

10. 15(3) cage at R7C5 = {249/267}, no 3, 2 locked for N8
10a. 4 of {249} must be in R7C5 -> no 4 in R89C5
10b. 2 of {267} must be in R8C5 -> no 6 in R8C5
10c. 2 in R789C5 -> no 2 in R8C37 (AN)
10d. 3 in C5 only in 16(3) cage at R1C5, locked for N2
10e. 16(3) cage = {349/367}
10f. 3 of {367} must be in R2C5 -> no 6 in R2C5
10g. 9 of {349} must be in R2C5 -> no 9 in R1C5
10h. 3 in R123C5 -> no 3 in R2C37 (AN)

11. 17(3) cage at R5C7 (step 9c) = {368} (cannot be {467} = [746] which clashes with 21(3) cage at R6C9 because of AN for R5C8 + R8C9 and NC for R67C9), locked for R5 and N6
11a. Naked triple {368} in 17(3) cage -> no 3,8 in R37C8 (AN)
11b. Naked triple {147} in 12(3) cage at R5C1, locked for N4
11c. Naked triple {147} in 12(3) cage at R5C1 -> no 4,7 in R37C2 (AN)

12. 21(3) cage at R6C9 (step 2) = {489/579} = [579/948]
12a. 18(3) cage at R7C7 (step 2f) = {468/567}
12b. Killer pair 7,8 in 21(3) cage and 18(3) cage, locked for N9
12c. 15(3) cage at R9C6 (step 2n) = {159/249}, no 3, 9 locked for R9
12d. R8C7 = 3 (hidden single in N9), R5C7 = 6
12e. R8C7 = 3 -> no 4 in R8C68, no 2 in R9C7 (NC), no 3 in R6C6 (AN) -> no 7 in R4C4 (cage sum)
12f. 2 in N9 only in R79C8, locked for C8
12g. R5C7 = 6 -> no 7 in R46C7 (NC), no 6 in R3C6 (AN)
12h. 4 in N9 only in R79C9, locked for C9
12i. Naked pair {38} in R5C89 -> no 3,8 in R3C9 (AN)

13. 21(3) cage at R6C9 (step 12) = [579/948]
13a. 15(3) cage at R7C5 (step 10) = {267} (cannot be {249} = [492] which clashes with 21(3) cage) -> R8C5 = 2, R79C5 = {67}, locked for C5 and N8
13b. R8C5 = 2 -> no 1 in R8C46 (NC), no 2 in R9C3 (AN)
13c. Naked pair {67} in R79C5 -> no 6,7 in R8C3 (AN)
13d. R2C5 = 9 (hidden single in C5) -> no 8 in R2C46 (NC), no 9 in R1C37 + R3C3 (AN)
13e. Naked pair {34} in R13C5, locked for N2, no 4 in R2C37 (AN)
13f. 1 in R8 only in R8C13, locked for N7
14g. 9 in N3 only in R13C8, locked for C8
14h. Naked pair {12} in R79C8, locked for C8 and N9
14i. 9(3) cage at R1C9 = {135/234} (cannot be {126} because R2C8 only contains 3,4,5), no 6, 3 locked for N3

15. 18(3) cage at R7C7 (step 2f) = {468/567} = [567/765/864]
15a. Consider combinations for 21(3) cage at R6C9 (step 2) = {489/579} = [579/948]
21(3) cage = [579] => 18(3) cage = [864] => no 8 in R5C8 (AN) => R5C89 = [38]
or 21(3) cage = [948] => R5C89 = [83] => no 7 in R46C8 (NC) => R4C9 = 7 (hidden single in N6)
-> 7 in R47C9, locked for C9, 8 in R58C9, locked for C9, 18(3) cage = [756/765/864]
15b. R23C9 = {26} (cannot be {25} which clashes with 9(3) cage at R1C9, cannot be {56} because of NC), locked for C9 and N3
[Now it gets a lot easier]
15c. 9(3) cage = {135} (only remaining combination), locked for N3, no 4 in R13C8 + R2C7 (NC)
15d. R2C7 = {78} -> no 7,8 in R1C7 + R2C6 (NC)
15e. R1C7 = 4, R13C5 = [34], no 2 in R1C4, no 2,5 in R1C6 (NC)
15f. R2C8 = 3 (hidden single in N3), R5C89 = [83]
15g. R2C7 = 8 (hidden single in N3) -> R7C7 = 7, R79C5 = [67], R79C9 = [45], R1C9 + R3C7 = [15], R9C67 = [49], R9C8 = 2 (cage sum), R4C9 = 7, R68C9 = [98], R7C8 = 1
15h. Naked pair {59} in R8C46, locked for R8 and N8
15i. Naked pair {38} in R7C46, locked for R7 and N8 -> R9C4 = 1
15j. 1 in N2 only in R23C6 -> no 2 in R23C6 (NC)
15k. R5C6 = 2 (hidden single in C6), R5C4 = 9, R8C46 = [59]
15l. R5C6 = 2 -> no 3 in R4C6 (NC) -> R4C6 = 6, R6C4 = 4 (cage sum), R6C6 = 7 -> R4C4 = 3, R7C46 = [83]
15m. R4C4 = 3 -> no 2 in R3C4 + R4C3 (NC), no 3 in R3C2 (AN)
15n. R6C4 = 4 -> no 3 in R6C3 (NC), no 4 in R5C2 (AN)
15o. R7C4 = 8 -> no 9 in R7C3 (NC), no 8 in R6C2 (AN)
15p. R8C4 = 5 -> no 4 in R8C3 (NC), no 5 in R7C2 (AN)

16. R7C3 = 5 -> R8C2 + R9C1 = 15 = [78]
16a. R8C2 = 7 -> no 6 in R9C2 (NC) -> R9C23 = [36], R8C13 = [41]
16b. 12(3) cage at R5C1 = [714]
16c. R5C1 = 7 -> no 6 in R6C1 (NC)
16d. R5C2 = 1 -> no 2 in R46C2 (NC)

17. R2C4 = 2 (hidden single in N2), R2C39 = [76], R1C36 = [28]
17a. R2C6 = 5
17b. R2C2 = 4 -> R1C1 + R3C3 = 8 = [53]
17c. R1C1 = 5 -> no 6 in R1C2 -> R1C2 = 9

and the rest is naked singles, without using anti-knight or non-consecutive.

Rating Comment:
With N5 input as R5C5 = 5 (from step 1b) and two disjoint 10(2) cages, SS gives a score of 1.65.
That seems reasonable; my key forcing chain in step 8 was a fairly heavy one. SudokuSolver used Bowman's Bingo, aka simple T&E, to get the same result as the forcing chain.

I finished this puzzle on Monday, before HATMAN's most recent puzzle was posted, but I've been busy during the week so was only able to check my WT today. I'll probably try the two Anti-kNight X puzzles, which were posted on the day before this one, before I try the new Decidoku.


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