Nice work! Here are a few comments:
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:sudokuEd wrote:49c. 2 or 9 in r4c45 -> from combos in 22(4) = {2479} -> 7 in r4c6 -> [29] in r56c6
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:
This is brilliant! You're absolutely right. Here's the proof:sudokuEd wrote: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?
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.