Rejected versions for Assassin 39

[mhparker]

Hi Ed,

Nice work! Here are a few comments:

49c. 2 or 9 in r4c45 -> from combos in 22(4) = {2479} -> 7 in r4c6 -> [29] in r56c6

I agree as far as the 9 is concerned. However, in the case of a 2 in r4c45, a 9 could go in r4c6 instead of a 7. Thus, only the 9 can be eliminated from r4c45, invalidating the subsequent steps. New "step 49" (renumbered as skipped step 48) would be:

48. From step 36. h26(4)r4569c6 = [7928/6839/9836]
48a. in summary, 7 in r4c6 -> [92] in r56c6
48b. -> no 9 in r4c45. Here's how.
48c. 9 in r4c45 -> from combos in 22(4) = {2479} -> 7 in r4c6 -> [29] in r56c6
48d. but this means 2 9s in n5
48e. -> no 9 r4c45
48f. no 2 r1c4

However, we don't really need this step at all, because of your great idea:

The two h26(4) cages in r4c4569 and in r4569c6 have 1 cell shared and 2 cells in the same house - so perhaps the 1 leftover cell have to equal each other = {89}? Or is that just luck that it worked out this way?

This is brilliant! You're absolutely right. Here's the proof:

49. h26(4) at r4569c6 = {2789/3689}; h26(4) at r4c4569 = {2789/4679/5678}
49a. -> both h26(4) cages each have 3 of {6..9}
49b. -> (geometical considerations aside) both h26(4) cages must share at least 2 digits from {6..9}
49c. only places for these common digits are the intersection r4c6 and the "leftover" cells (r4c9 and r9c6)
49d. -> r4c9 and r9c6 must be identical
49e. -> no 6 in r9c6

50. Naked pair on {89} at r59c6 -> no 9 in r4c6

Now continue with slightly modified versions of your steps above (renumbered accordingly):

51. 22(4) = {4567}(no 2): all locked for n5
51a. 6 locked for r4
51b. cleanup: no 9 in r1c4

52. {89} naked pair in r5c46: both locked for r5
52a. r56c1 = [65]

53. 15(3)n1 = {249/348}(no 7) = 4{29/38}
53a. 4 locked for c1

54a. r23c6 must have just 1 of 8/9 for c4 -> combo's in 22(4)n2 must have exactly 1 of 8/9
54a. 22(4) = {2479/2569/2578/3469/3478/3568}

Now for the rest:

55. Naked pair on {45} at r4c7 and r5c9 -> no 4 in r6c789
55a. cleanup: no 9 in r78c9

56. 4 in r6 locked in n4 -> not elsewhere in n4

57. 4 in c1 locked in n1 -> not elsewhere in n1

58. 26(5)n47 cannot have both of {56}, as {56} only available in r7c2
58a. -> {24569/34568} both blocked
58b. {23579/23678} blocked by 15(3)n14 (need {23} in c1), as is {24578} (needs {28} in c1)
58c. {23489} blocked by 15(3)n14 (needs 1 of {23}) and 14(2)n7 (needs 1 of {89})
58d. remaining combos are: {12689/13589/13679/14579/14678} = {(5/6)..}
58e. 1 locked in r789c1 for c1 and n7
58f. -> r7c2 = {56} (only place for {56} in 26(5) cage)

59. r7c2 and 14(2)n7 form killer pair on {56} -> no 5,6 elsewhere in n7

60. 9(3)n78 = {234} (only combo possible), locked for r9

61. Hidden single(r9) at r9c1 = 1

62. Hidden single(c8) at r5c8 = 2
62a. -> r5c7 = 3

63. Hidden single(c7/n9) at r8c7 = 2

64. 3 in c8/n9 locked in r78c8
64a. -> 11(2)n9 = {38}, 8 locked for c8 and n9

65. 4 in c8 locked in n3 -> not elsewhere in n3

66. Hidden single(c7) at r4c7 = 4

67. r45c9 = [85], r5c5 = 4
67a. cleanup: no 7 in r1c4

68. 5 in r9 locked in n9 -> not elsewhere in n9

69. 27(4)n689 = {5679} (only combo possible)
69a. -> r7c6 = 5

70. Naked single at r7c2 = 6

71. 14(2)n7 = {59}, locked for r8 and n7

72. Outie n69: r8c6 = 4

73. Hidden single(c6) at r4c6 = 7
73a. -> r569c6 = [928] (step 48)
73b. cleanup: no 7 in r1c4

74. r56c4 = [83]

75. Naked single at r9c4 = 2

76. Hidden single(c4) at r3c4 = 9

77. Hidden single(c4) at r2c4 = 4

78. Naked pair on {23} at r4c23 -> no 2,3 elsewhere in n4
78a. -> r3c3 = 6 (20(4) cage sum)

Now just singles and cage sums to end.

[sudokuEd]

Here's the proof

Great proof Mike - just the clear headed explanation I couldn't get too. Thanks. Great that a mistake has turned out positive for once . Really good finish too - the combo work in 58 is real easy to follow. Great to have another unsolvable ticked off.

I think this neat technique we've stumbled on could be quite useful. For one, it will work from the beginning of this puzzle since all the combinations in a 26(4) share 2 digits with every other combo. I'll do a condensed walk-through to see how much difference it makes early on. [edit: doesn't - but is a fantastic dead-lock-breaker. No condensed walk-through needed.]

It works like this.
26(4) = {2789/3689/4589/4679/5678}
{2789} shares 8 & 9 with {3689/4589}, 7 & 9 with {4679}, 7 & 8 with {5678}
{3689} shares 8 & 9 with {4589}, 6 & 9 with {4679}, 6 & 8 with {5678} (already did with {2789} - its 8 & 9)
{4589} shares 4 & 9 with {4679}, 5 & 9 with {5678}
{4679} shares 6 & 7 with {5678}.
Each combination shares two numbers with each of the others. The shared numbers for shared 26(4) cages are {456789}. Whichever two combinations are in the 2 overlapping 26(4) cages, must have 1 of {4..9} iin the shared cell, and 1 of {4..9} in both cells that are outside the house that the other 5 cells are in. This is the only way to get the 8 candidates (for two 4-cell cages) into 7 cells.

Another extension of this is that heaps of cages share at least 1 number from each combination with each other permutation. For example, a 19(3) = {127/136/145/235}. The
{127} shares a 1 with {136}, a 1 with {145} and 3 with {235}.
{136} shares 1 with {145}, a 3 with {235} (already done the sharing with {127} - it's a 1)
{145} shares a 5 with {235}
The shared numbers for a 19(3) are 1,3,5.

This could be very handy for when two 19(3) cages (at least 1 hidden) overlap one cell and that are both completely in the same house (c, r or n). The overlap cell must contain one of the common candidates - how else can two 3-cell cages fit into just 5 cells?

With two 19(3)s, the shared cell can only be {135}. The 2,4,6 & 7 can all be eliminated.

The more complex application of this shared number principle is what we have stumbled across - where it works for 2 cells that have escaped from the same house. This can only work for cages where each permutation has two shared numbers with each other combination.

Other cage sizes (besides 26(4) from this puzzle) that have the same inherent property of two shared numbers are - 12(4), 13(4), 14(4), 26(4), 27(4), 28(4), 20(5), 21(5), 22(5). Maybe others as well, haven't gone through em all yet.

For example, a 14(4) = {1238/1247/1256/1346/2345}
{1238} shares 1 & 2 with {1247/1256}, 1 & 3 with {1346}, 2 & 3 with {2345}
{1247} shares 1 & 2 with {1256}, 1 & 4 with {1346}, 2 & 4 with {2345}
{1256} shares 1 & 6 with {1346}, 2 & 5 with {2345}
{1346} shares 3 & 4 with {2345}
The shared numbers for a 14(4) are 1,2,3,4,5,6. Therefore, 7 can be eliminated from the overlapping cell and the two escaped cells.

Further, as we have worked out, the 2 escaped cells have to be equal to fit in the second shared number. As a candidate is eliminated from one, the same candidate can be eliminated from the second escaped cell as well.

Of course, all cage sizes may have this property as certain combinations are eliminated. It may be that this principle still works with cages of different sizes that happen to share numbers from each permutation in each cage. Haven't got that far yet.

Hope this is all valid. I keep thinking of prison and escapees to help reminding me of the importance of the shared house to make this powerful. Perhaps its all really a CON (Cage Overlap Number )

Cheers
Ed