Rejected versions for Assassin 39

[rcbroughton]

60. r5c5 = {56}, r5c9 = {56}: locked for r5

I didn't see from your picture and subsequet steps where the {89} got eliminated in r5c9 - did I miss seeing a step? I can see that 45 on r1234 means r4c9-r5c5=3 - and so following step 59 you can put r4c9={89} -> r5c9={56}. Then your step 60 follows.

Apologies if I missed this step somewhere else - there was a lot of activity on this one!

62. 45 on c1234 - r14c4 total 7 - [34] blocked by 12(2)n5 - so no 3 r1c4, no 4 r4c4

63. 45 on n7 - r6c23+r789c4 total 31
63a. possibilities for r6c23 are [65]/[75]/[85]/[95]/{69}/{79}/{76}/{86}/{89} - {78} blocked by 11(2)n4
Looks like a lot - but only totals are 11,12,13,14,15,16,17
63b. so r789c4 total 14,15,16,17,18,19,20 but
1) combos with {38}/{34}/{89} blocked by 12(2)n5
2) combos with {69}/{78}/{79}/ blocked by 22(4)n8
3) combos with {56}/{26} blocked by the h7(2) from step 62
4) total for r78c4 must be between 4 and 11
63c. 14,15,16 - 8 or 9 restricted to r9c4
63d. 17 - {35}9 or {467}
63d. 18 - {45}9 or {46}8
64e. 19 - no possibles
64f. 20 - no possibles.
so . . .
64g. no 8,9 in r78c4

(There must be a simpler way . . . !!)

65. following on from above logic. can't have a 7 in r6c3 - because only possibiles with 7 here give r6c23 total 13 or 16
65a. =13 -> r789c4 total 18 (63d) but 17(4)n478 now becomes 7+{1/4}+{45}/{46} - not possible
65b. =16 -> r789c4 total 15
1) {24}9 - would need 4 at r7c3 for 17(4) - contradiction
2) {25}8 - would need a 3 at r7c3 to complete the 17(4)
3) {35}7 - would need a 2 at r7c3 to complete the 17(4)


[Edit] Aaagh - I was too slow. I'll take in the other moves and see what it does!!

[Para]

60. r5c5 = {56}, r5c9 = {56}: locked for r5

I didn't see from your picture and subsequet steps where the {89} got eliminated in r5c9 - did I miss seeing a step? I can see that 45 on r1234 means r4c9-r5c5=3 - and so following step 59 you can put r4c9={89} -> r5c9={56}. Then your step 60 follows.

It works through the 45 test on r1234 which gives you R5C5 and R5C9 as outies equalling 11.

Para

[Para]

62. 45 on c1234 - r14c4 total 7 - [34] blocked by 12(2)n5 - so no 3 r1c4, no 4 r4c4

What about [75] in 12(2) N5?

Never mind. Killer Pair {57} in 22(4) now. ugh.

[Para]

Ok time to reorganize and see what everything does.

Para

[Para]

63. 45 on n7 - r6c23+r789c4 total 31
63a. possibilities for r6c23 are [65]/[75]/[85]/[95]/{69}/{79}/{76}/{86}/{89} - {78} blocked by 11(2)n4
Looks like a lot - but only totals are 11,12,13,14,15,16,17
63b. so r789c4 total 14,15,16,17,18,19,20 but
1) combos with {38}/{34}/{89} blocked by 12(2)n5
2) combos with {69}/{78}/{79}/ blocked by 22(4)n8
3) combos with {56}/{26} blocked by the h7(2) from step 62
4) total for r78c4 must be between 4 and 11
63c. 14,15,16 - 8 or 9 restricted to r9c4
63d. 17 - {35}9 or {467}
63d. 18 - {45}9 or {46}8
64e. 19 - no possibles
64f. 20 - no possibles.
so . . .
64g. no 8,9 in r78c4

63d. {467} clashes with 22(4)
For later on

But it seems ok. i agree lot of work for 4 eliminations.


Para

[Para]

72. R23C2 = {14} locked for N1 and C2
72a. R5C3: no 7

73. Clean up: R8C8 : no 5

74. 17(4) in N478 = 1 + [952]/[8]{35}/[6]{37}/: [9]{34}/[8]{62} clashes with C4 (like Richard pointed out), [6]{47} clashes with 22(4) N8.
74a. R6C3 = {689}, R7C4 = {357}, {2357}

75. 2 locked in R9 for N9 and 24(5).
75a. Hidden single 2 in R8C4 for N8

76. R14C4 = [61] (45 on r1234)
76a. R6C3 = 9, R7C4 = 5 (step 74)

77. 45 on N3. R3C7 - 3 = R1C6 + R4C8
77a. R1C6 + R4C8 = 3 or 4. Min R4C8 =2 -->> max R1C6 = 2: no 3

[Para]

78. Hidden single 1 in R5C1: R56C1 = [12]
78a. Hidden single 2 in R5C6 -->> R1C6 = 1
78b. R6C56 = {36/45}
78c. Killer Pair {56} in R6C56 + R5C5 -->> locked for N5
78d. Killer Pair {34} in R6C56 + 12(2) in R5C4 -->> locked for N5
78e. Naked triple {789} in R4C569 -->> locked for R4

79. 3 locked in N5 for R6
79a. 6 locked in R4 for N4.
79b. Naked triple {789} in R234C6 -->> locked for C6
79c. R6C6 = 5(hidden), R6C5 = 4, R5C5 = 6, R45C9= [95]
79d. R56C4 = [93]

80. 22(4) N8 = {1489}: {1678} clashes with R4C5.
80a. R9C6 = 4, R789C5 = {189} locked for C5 and N8.
80b. R4C56 = [78], R9C4 = 7, R9C23 = [93]

[Para]

81. R8C23 = {68} -->> locked for R8 and N7
81a. R9C1 = 5, R8C7 = 5(hidden)
81b. R78C1 = {47} -->> locked for C1 and 26(5) N47
81c. R6C2 = 8, R5C23 = [74], R8C23 = [68], R78C6 = [63], R78C8 = [34]
81d. RC78 = [38], R4C78 = [42]
81e. 14(3) in N69 = [671], R9C789 = [862], R7C7 = 9
81f. R78C1 = [47], R789C5 = [891]
81g. R2C78 = [75], R1C78 = [29], R3C78 = [61], R6C78 = [17]
81h.R23C6 = [97], R23C2 = [14], R23C4 = [48], R123C9 = [483]
81i. R13C1 = [89], R1C3 = 7(hidden)
81j. 30(5) = 874{56} -->> R1C2 = 5, R2C3 = 6
81k. R123C5 = [325], R24C1 = [36], R4C2 = 3, R34C3 = [25]

OK this is it. But i think there must be a nicer way to get past Peters long chain. Don't really mind those chains, are nice short cuts. But think there is always a nicer alternative to them.

Para

[Nasenbaer]

Back from lunch. And nothing left to do. Oh well...

I really like it when we have to work together to crack the puzzle.

Peter

[Para]

Hi

Just made this observation. Always find it nice to give certain eliminations names. Here is one for one of Peters eliminations.

49. no 4 in r123c3
49a. r7c3 = 4 -> no 4 in r123c3
49b. r7c3 = 1 -> r7c2 = 2 -> r23c2 = {14} -> no 4 in r123c3

I really like this step btw as you can see it as a Killer XY-wing

R7C2 = {12}; R7C3 = {14}; R23C2 needs 2 or 4 (killer pair)
Creating an XY-wing. Eliminating the 4 from all cells that see both R7C3 and R23C2.

greetings

Para

[rcbroughton]

OK this is it. But i think there must be a nicer way to get past Peters long chain. Don't really mind those chains, are nice short cuts. But think there is always a nicer alternative to them.

In response to Para asking if there was a way to avoid the long contradiction chain, I spent a bit of time looking at this again.

Continuing from Ed's posted position at Step 50, we can get to a concluding position in a dozen or so steps without the real "ugly" contradiction:


51. 45 on n8 - r789c4+r78c6= 23 with no 1 (1 is locked in the 22(4)n8)
51a possibilities are {23459}, {23468}, {23567} - only {23459} uses a 9 and this must have 9 in r9c4 - so no 9 in r78c4

52. 45 rule on n78 - r6c23 minus r78c6 totals 8 - max of outies is 17, so max of innies is 9 - no 7 in r8c6 and no 8 in r7c6

53. 45 rule on c6789 r49c6 minus r6c5 totals 8: no possibilities with 8 at r9c6 (see table below)

Code: Select all

         r4c6=4 5 6 7 8 9
r9c6-> 1 r6c5=X X X X X 2
       4 r6c5=X 1 2 3 X 5
       5 r6c5=1 X 3 4 X 6
       6 r6c5=2 3 X 5 X 7
       7 r6c5=3 4 5 X X 8
       8 r6c5=4 5 6 7 X 9
       9 r6c5=5 6 7 8 X X

54. Can't have an 8 at r3c7 as it removes all possible 8s in row 1
54a. r3c7=8 -> r1c789<>8
54b. r3c7=8 -> r3c46<>8, r2c6<>8 -> r2c4=8 -> r1c123<>8

55. Can't have a 9 at r3c7 as it removes all possible 9s in row 1
55a. r3c7=9 -> r1c8<>9
55b. r3c7=9 -> r3c46<>9, r2c6<>9 -> r2c4=9 -> r1c123<>9

56. 26(4)n236={786}5/{769}4/{689}3/{789}2 - no 4 or 5 in r23c6

57. 45 rule on n3 - innie minus outies=3
57a. when r3c7=6, r1c6+r4c8=3={12}
57b. when r3c7=7, r1c6+r4c8=4={13}/{22}
57c. no 4,5 in r1c6 and no 4 in r4c8

58. 45 rule on n2 - innies total 29
1) 16(4) blocks {34679}, {25679}, {35678}
58a. only options with 1 are {14789} and {15689} with the 1 in r1c6
58b. no 1 in r23c4

59. Hidden single 1 at r4c4 in column 4
59a. 22(4)n5 can only be 1{78}6 or 1{79}5

60. 11(3)n5={236}/{245}

61. Hidden single 1 at r5c1 for row 5
61a 3(2)n4=[12]

62. 11(3)n5=2{36}/{45}

63. 11(2)n6={38}/{74}

64. 3 locked in n5 for row 6

65. 2 locked in r78c4 in n8 - locked for c4 and 17(4)n478
65a. no 7 in 17(4)n478
65b. 17(4)n478={2[4]56}/{2[1]68}/{2[4]38}/{2[1]59]) - no 4 in r78c4

66. 45 on r6789 - r6c4=3
66a. 12(2)n5=[93]

. . . and the numbers fall into place quite quickly.

[sudokuEd]

Finally got a chance to look at Richard's alternate way - a very fine way it is too. Makes a very satisfying way of unlocking the puzzle. But rather than wait until step 50, why not step 16?

So, here's an alternative solution to Assassin 39V2 using Richard's key moves as quickly as possible. It's not really a condensed walk-through - perhaps "sweetened condensed" . I add many new clean-up-at-the-end steps to keep the solution simple. It makes you wonder why we thought it was difficult .

Assassin 39V2

note: L = "locked for"
1. 3(2)n4 = {12}:L c1, n4
1a. 11(2)n4: no 9

2. 14(2)n6 = {59/68} = [5/6..]
2a. {56} blocked from 11(2)n6 = {29/38/47}

3. "45"r1234 -> r5c59 = 11 = [29/38/56/65]
3a. r5c5 = {2356}

4. "45" c1234 -> r14c4 = 7 (no 789)

5. 22(4) n5 - no 1 in r4c56 - here's how
5a. 2 combos with 1 - {1579} & {1678} - ({1489} blocked by r5c5)
5b. {1579} - 7,9 must be in r4c56
5c. {1678} - 7, 8 must be in r4c56
5d. -> no 1 r4c56

6. 1 in n7 only in r7: L r7
6a. no 6 r8c8

7. "45" c9: r34c8 - r9c9 = 1
7a. -> min r9c9 = 2

8. no 1 in r8c4. Here's how
8a. r8c4 = 1 -> r4c8 = 1 -> r1c7 = 1 -> no place for 1 in n9
8b. -> no 1 in r8c4

9. no 1 in r8c6
9a. r8c6 = 1 -> no place for 1 in n9
9b. -> no 1 in r8c6

10. n8 : 1 locked in 22(4) -> 22(4) = 1{489|579|678} -> no 2,3

11. no 1 in r1c4: here's how.
11a. r1c4 = 1 -> r4c8 = 1 -> no place for 1 in n3
11b. -> no 6 in r4c4 (step 4)

12. 16(4)n2: any combination with 1 also needs a 6: here's why
12a. 16(4) has 1 -->> R123C5 = 1 -->> R9C6 = 1 -->> R4C4 = 1 -->> R1C4 = 6 (step 4)

13a. 16(4) = {1267/1456/2347/2356}(no 8 or 9)

14. "45" n6789 -> r6c23 - 11 = r4c78
14a. -> min r6c23 = 14 -> no 3,4 r6c23
14b. max r6c23 = 17 -> max r4c78 = 6
14c. max r4c7 = 5
14d. min r4c7 = 2 -->> max r4c8 = 4

15. "45" n3 -> 2 outies + 3 = 1 innie
15a. max r3c7 = 9 -> max r1c6 + r4c8 = 6
15b. max r1c6 = 5

16. Can't have 8 at r3c7 as it removes all possible 8s in row 1. Here's why
16a. r3c7=8 -> r1c789<>8
16b. r3c7=8 -> r3c46<>8, r2c6<>8 -> r2c4=8 -> r1c123<>8
16c. -> no 8 r3c7

17. Can't have 9 at r3c7 as it removes all possible 9s in row 1. Here's why
17a. r3c7 = 9 -> r1c78<>9
17b. r3c7=9 -> r3c46<>9, r2c6<>9 -> r2c4=9 -> r1c123<>9
17c. -> no 9 r3c7

18. "45" n3 -> 2 outies + 3 = 1 innie
18a. min r1c6 + r4c8 = {12} = 3 (can't have {11}:leaves no 1 for n3)
18b. min r3c7 = 6
18c. max r3c7 = 7 -> max. 2 outies = 4
-> max value r1c6 & r4c8 = 3

19. 26(4)n236: must have 6/7 = {2789/3689/4679/5678}
19a. 2,3,4,5 are all required in r4c7 -> no 2,3,4,5 r23c6

20. "45"n2 : 5 innies = 29 = 29(5)
20a. 16(4)n2 blocks {34679}, {25679}, {35678}
20b. only combinations for 29(5) with a 1 are {14789} and {15689} with the 1 only in r1c6
20c. ->no 1 in r23c4

21. r4c4 = 1 (h single c4)
21a. 22(4)n5 now 21(3) = {579/678}(no 234) ({489} blocked by r5c5)
21b. = 7{59/68}: 7 L n5, r4
21c. -> r5c5 = {56}
21d. r4c56 = {789}

22. 11(3)n5= 2{36/45}(no 8)

23. r5c1 = 1 (hsingle r5)
23a r6c1 = 2

24. r5c6 = 2 (hsingle n5)

25. r5c9 = {56} (step 3)
25a. {56} pair r5c59: L r5
25b. r4c9 = {89}
25c. Killer Pair {89} with r4c56:L r4

26. 11(2)n4 & 6={38/74}(no 9):all L r5

27. r5c4 = 9 (single r5), r6c4 = 3

28. r6c56 = {45}: L n5, r6
28a. r5c59 = [65], r4c9 = 9
28b. r4c56 = {78}:L r4

29. r1c4 = 6 (step 4)
29a. r123c5 = {127/235}(no 4) ({145} blocked by r6c5)
29b. = 2{17/35}: 2 L n2

30. 6 r4 in n4: L n4

31. 9 n2 in c6: L c6

32. 9 n8 in 22(4) = 19{48/57} (no 6)

33. "45"n3 -> r1c6 + r4c8 + 3 = r3c7 = 6/7
33a. ->r1c6 + r4c8 = 3/4 = [12/13]
33b. -> r1c6 = 1

34. 16(4)n2 now 10(3) = {235}: L n2,c5
34a. r6c56 = [45]

35. 4 n2 in c4: L c4

36. 22(4)n8 = {1489}
36a. r9c6 = 4
36b. r789c5 = {189} L n8, c5
36c. r4c56 = [78]

37. r23c6 = {79}:L n2,c6, 26(4)
37a. -> r34c7 = [64]

38. 11(2)n6 = {38}: L n6, r5
38a. r4c8 = 2, r1c78 = [29] (hsingle 2)

39. 11(2)n4 = {47} L n4

40. 23(4)n6 must have 8/9 -> r7c7 = {89}

41. "45"n9 -> r6c9 + r8c6 = r7c7 = 8/9
41a. -> r6c9 + r8c6 = [63], r7c7 = 9, r7c6 = 6, r9c8 = 6 (hsingle n9), 7(2)n9 = [34], r78c9 = [71]
etc
[edit step 20 for clarity]

[sudokuEd]

Managed to get out V3

Hmm. Don't know how. So, this is one of Para & Glyn's Unsolvables. Thanks Ruud.
Managed to find some nice tricks, but stuck now. Thought I'd had it.... Pretty sure this far is correct. Anyone want to lend a hand? Over to A44 V1.5 for the interim.

Cheers
Ed

A39 V3
Prelims
i. 17(5)n3: no 8,9
ii. 15(2)n1 = {69/78}
iii. 10(2)n3: no 5
iv. 13(2)n6: no 1,2,3
v. 11(2)n4, 5 & 9: no 1
vi. 8(2)n4: no 4,8,9
vii. 5(2)n6 = {14/23}
viii. 27(4)n6: no 1,2
ix. 26(4)n8: no 1
x. 14(2)n7 = {59/68}
xi. 9(3)n7: no 7,8,9

1. "45" r6789: r5c146 = h23(3) = {689}
1a. all locked for r5
1b. r4c9: no 4,5,7
1c. r6c14 = {235}
1d. no 2 r5c23

2. "45" r1234: r5c59 = h9(2) = [27/45/54](no 1,3)
2a. r5c5 = {245}

3. "45"r1234: r4c4569 = h26(4)

4. "45" r6789: r6c1456 = h11(4) = {1235}: all locked for r6
4a. 1 only in r6c56 in 12(3)
4b. 1 locked for n5 & 12(3) must have 1 = {129/138/156}

5. "45" c6789: r4569c6 = h26(4) = {2789/3689/5678}(no 1, 4) ({4589/4679} are blocked by combo's in 12(3)n5)
5a. = 8{279/369/567}
5b. 8 locked for c6

6. 12(3)n5 = {129/138/156} = 1{..}
6a. r6c5 = 1
6b. r56c6 = [92/83/65]
6c. -> from step 5, r49c6 = {78/69} = {6789}

7. 22(4)n5 must have 4 & 7 for n5 = 47{29/38/56}.
7a. r4c456+r4c9 = h26(4) must have 3 cells overlapping with the 22(4) = {2789/4679/5678}
7b. = {279}[8]/{467}[9]/{479}[6]/{567}[8]
7c. no 3 or 8 r4c456
7d. 22(4) = {2479/4567}
7e. 7 locked for r4
7f. r4c456 = [6/9..]

8. 3 in n5 only in r6: 3 locked for r6
8a. no 8 r5c1

9. deleted

10.from step 5, r4569c6 = h26(4) = {2789/3689/5678} =
10a. = [7928]/{6[83]9}/[7658]
10a. no 7 r9c6

11. "45" c1234: r14c4 = h11(2) = {29/47/56}(no 1,3,8)

12. 11(2)n4 = [92/65]
12a. 15(3)n1 = {168/249/348/357/456} ({159/258/267} all blocked by 11(2))

13. "45" n3:2 outies + 1 = 1 innie r3c7
13a. min r3c7 = 3
13b. max 2 outies = 8 -> no 9 r1c6

14. "45" n236: r23c4 + r6c78 = 35
14a. max r6c789 = {679} = 22 ({789/689} blocked by 13(2)n6)
14b. -> min r23c4 = 13 (no 1..3)

15. r23c4 = 13..17 -> r6c789 = 22..18
15a. r6c789 =
15b. = 18 Blocked: {468} clashes with 13(2)n6
15c. = 19 = {469} ({478} clashes with 13(2)
15d. = 20 = {479}
15e. = 21 = {489/678}
15f. = 22 = {679}
15f. [89] must be in r6c789 or 13(2) for n6: no 89 in r4c7


16. Not sure if this is strictly logical. No 4 in r5c78. Here's how.
16a. from step 15c..f: 4 must be in r6c789
16b. or it is forced into 13(2) by {678}
16c. or it is forced into r5c5 by {679} through h9(2)r5c59 = [45]
16d. -> no 4 r5c78

17. 5(2)n6 = {23}: both locked for n6 & r5

18. 8(2)n4 = {17}: both locked for n4 & r5
18a. no 6 r4c9
18b. 13(2)n6 = [94/85]

19. from step 15d..f. r6c789 must have 7
19a. = 20 = {479}
19b. = 21 = {678}
19c. = 22 = {679}
19d. -> innies n236 = 35 -> r23c4 = 13-15

20. 1 in c4 only in n8: 1 locked for n8

21. from step 19a.r6c789 = 20{479}/21{678}/22{679}
21a. "45" n9: 5 outies = 31
21b. min r6c789 = 20 -> max r78c6 = 11
21c. min r7c6 = 3 -> max r8c6 = 8

22. Now a nice trick to get of 9 from r7c6
22a. 9 in r7c6 -> since max r78c6 = 11 (step 21b) and since min r8c6 = 2 -> r78c6 can only be [92] = 11
22b. from outies n9 = 31, when r78c6 = 11 -> r6c789 = 20 = {479} only (step 22)
22c. since r6c78 is in the same cage as r7c6 -> 9 can only go in r6c9
22d. However, cannot have {47} in r6c78 in a 27(4) cage {4+7+9+7} clash
22e. -> no 9 r7c6

23. Generalized X-wing on 9 in c78: must be in 27(4)n6 or c78 in n3: 9 locked for c78
23a. no 2 r78c8

24. 27(4)n6
24a. {3789} cannot have {89} in r6c78 because of r4c9 -> must have 7 in r6c78 -> 3 must be in r7c6 -> no 3 in r7c7
24b. {4689} same logic -> no 4 r7c7

25. 17(5)n3 = 123{47/56} = [123]
25a. 23 locked for n3
25b. CPE: no 1 r12c8

26. 10(2)n3 = {46}/[19]
26a. since 17(4) = 123{47/56} = [14/16] -> whichever combination is in 19(2), the leftover 1/4/6 has to be in r4c8
26b. -> r2c7 = r4c8 (no 5)
26c. -> r24c8 = [46/64/91]

27. "45" c9: r34c8 + 1 = r9c9
27a. -> min r9c9 = 4
27b. -> r4c8 != r9c9
27c. -> from step 26b, the 14(3)n6 must have [1/4/6] in r78c9
27d. -> 14(3)n6 = {149/158/167/347/356}(no 2) ({248} clashes wtih 13(2)n6)
27e. 14(3)n6 = [1/3..]

28. 2 in c9 only in n3
28a. no 2 r3c8

29. because r78c9 = [1/3..] (step 27e)-> the 1 & 3 required in 17(5)n3 cannot both be in c9
29a. they also cannot both be in r34c8 since that would leave 1/3 missing from c9 (r78c9 can only have 1 of 1/3)
29b. -> 1/3 must be in r34c8 = [14/16/41/51/61/71](cannot have [36] since sum is max 8);[34] forces 8 into both r9c9 & r4c9
29c. no 3 r3c8
29d. 3 locked for c9 in r123c9
29e. 1 must be r34c8: no 1 r9c8, no 1 r123c9
29f. min r34c8 = {14} -> min r9c9 = 6
29g. r234c8 = [614/416/941/951/961/971]

30. 14(3)n6 = 1{49/58/67}
30a. 1 locked for n9
30b. 8 in {158} must be in r6c9 -> no 8 r78c9

31. 16(4)n2 must have 1 because of 1's in n6. Here's how.
31a. "45" n3: 2 outies + 1 = 1 inn.
31b. 1 in r4c8 -> min r1c6 = 2 -> 1 in n2 in r23c6 in 16(4)
31c. 1 in r4c7 must be in 16(4) = 1{..}
31d. {2347/2356} blocked

32. no 1 in r1c7, r2c6 or r3c13 because of 1s in c8. Here's how.
32a. 1 in r4c8 -> 1 in r2c7 (step 26b) -> 1 in 16(4) in r3c6 -> no 1 in r1c7, r2c6 or r3c13
32b. 1 in r3c8 -> no 1 in r1c7 & r3c13 -> 1 in n6 in r4c7 -> no 1 in r2c6

33. "45" n3: 2 outies + 1 = r3c7 -> no 4 in r4c8. Here's how.
33a. 4 in r4c8 -> 4 in r2c7 (step 26b) -> 6 in r2c8
33b. 4 in r4c8 -> 1 in r4c7 -> 1 in n2 in r1c6 -> 6 in r3c7 (step 33)
33c. but this means 2 6s n3
33d. no 4 r4c8
33e. no 4 r2c7
33f. no 6 r2c8

34.when 6 in r4c8 -> 1 in r4c7 -> 1 in n2 in r1c6 -> max 2 outies n3 = 7
34a. -> max r3c7 = 8

35. {1249} combo blocked from 16(4)n2. Here's how.
35a. "45" n3: 2 outies + 1 = r3c7
35b. {1249} combo must have 4 in r3c7 -> 2 outies = 3 = [21]: but this forces 2 into both r1c6 & r23c6
35c. 16(4) = {1258/1267/1348/1357/1456} (no 9)

36. 9 in c6 only in h26(4)r4569c6
36a. = 9{278/368} (no 5)
36b. -> no 6 r5c6
36c. = [7928/6839/9836]

Code: Select all

.-------------------------------.-------------------------------.-------------------------------.
| 123456789 123456789 123456789 | 245679    23456789  1234567   | 456789    456789    234567    |
| 123456789 6789      123456789 | 456789    23456789  234567    | 16        49        234567    |
| 23456789  6789      23456789  | 456789    23456789  1234567   | 45678     14567     234567    |
:-------------------------------+-------------------------------+-------------------------------:
| 2345689   2345689   2345689   | 245679    245679    679       | 1456      16        89        |
| 69        17        17        | 689       45        89        | 23        23        45        |
| 25        4689      4689      | 235       1         23        | 46789     46789     46789     |
:-------------------------------+-------------------------------+-------------------------------:
| 123456789 123456789 123456789 | 123456789 23456789  34567     | 56789     345678    145679    |
| 123456789 5689      5689      | 123456789 23456789  234567    | 2345678   345678    145679    |
| 123456789 123456    123456    | 123456    23456789  689       | 2345678   2345678   6789      |
'-------------------------------.-------------------------------.-------------------------------'

[mhparker]

Hi Ed,

I admire your noble effort in reviving this forgotten beast. Here are a few moves to keep you going for a bit:

37. 2 in n9 locked in 24(5)n89
37a. -> no 2 in r8c6

38. "45" n689: 5 innies r4c78+r789c4 = 15
38a. min. r4c78 = {14} = 5
38b. -> max. r789c4 = 10
38c. -> no 8,9 in r78c4

39. {89} in n8 locked in 26(4) = {2789/3689} (no 4,5) = {(2/3)..}
(Note: {4589} blocked by r5c5)

40. no 6 in r4c8. Here's how.
40a. 6 in r4c8 -> 1 in r3c8 -> 5 in r123c9 (step 25)
40b. 6 in r4c8 -> 1 in r4c7 -> 5 in r5c9 (only other place for 5 in n6)
40c. but this means 2 5s c9
40d. -> no 6 in r4c8

41. Naked single at r4c8 = 1

42. Hidden single(c7/n3) at r2c7 = 1
42a. -> r2c8 = 9
42b. cleanup: no 6 in r3c2

43. Hidden single(r3) at r3c6 = 1

44. min. r34c7 = {45} = 9
44a. -> no 7 in r2c6

45. split 16(4) at r12c9+r3c89 = {23(47/56)} = {(4/6)..}

46. 17(3)n23 = [2]{78}/[3]{68}/[4]{58}/[4]{67}/[5]{48}/[6]{47}
(Note: [7]{46} blocked by split 16(4) (step 45))
46a. -> no 7 in r1c6

47. 11(2) at r56c4, h11(2) at r14c4 and split 11(2) at r56c6 must be different combos (i.e., 6 different digits) to avoid clash
47a. r6c6 must have 1 of {23}
47b. -> r1456c4 cannot have both of {23}
47c. only other place for {23} in c4 is r789c4 = {(2/3)..}
47d. -> r789c4 and 26(4)n8 form killer pair on {23} within n8
47e. -> no 3 in r78c6

Code: Select all

.-----------------------------------.-----------------------.-----------------------------------.-----------.
| 123456789   123456789   123456789 | 245679      23456789  | 23456       45678       45678     | 234567    |
:-----------.-----------.           '-----------.           :-----------.-----------------------:           |
| 2345678   | 678       | 2345678     45678     | 2345678   | 23456     | 1           9         | 234567    |
|           |           :-----------------------:           |           '-----------.-----------'           |
| 23456789  | 789       | 23456789    456789    | 23456789  | 1           45678     | 4567        234567    |
|           :-----------'           .-----------'-----------'-----------.           |           .-----------:
| 2345689   | 2345689     2345689   | 245679      245679      679       | 456       | 1         | 89        |
:-----------+-----------------------+-----------.           .-----------+-----------'-----------:           |
| 69        | 17          17        | 689       | 45        | 89        | 23          23        | 45        |
|           :-----------.-----------:           :-----------'           :-----------------------+-----------:
| 25        | 4689      | 4689      | 235       | 1           23        | 46789       4678      | 46789     |
:-----------'           |           '-----------+-----------.-----------'           .-----------:           |
| 123456789   123456789 | 123456789   1234567   | 236789    | 4567        56789     | 345678    | 145679    |
|           .-----------'-----------.           |           :-----------------------:           |           |
| 123456789 | 5689        5689      | 1234567   | 236789    | 4567        2345678   | 345678    | 145679    |
|           :-----------------------'-----------:           '-----------.           '-----------'-----------:
| 123456789 | 123456      123456      123456    | 236789      689       | 2345678     2345678     6789      |
'-----------'-----------------------------------'-----------------------'-----------------------------------'

[sudokuEd]

I admire your noble effort in reviving this forgotten beast.

Thanks Mike. In fact, this puzzle has never been far from my mind. The V2 has a lot of nostalgia about it - Peter's stubbornness and Para's massaging combo's are still vivid...and their speed .

Anyway, here's a few more after Mike's great work. I also have a technique query we might be able to gestate over.

edit: a flaw in these steps: see Mike's next post for corrections
49. From step 36. h26(4)r4569c6 = [7928/6839/9836]
49a. in summary, 7 in r4c6 -> [92] in r56c6
49b. -> no 2 or 9 in r4c45. Here's how.
49c. 2 or 9 in r4c45 -> from combos in 22(4) = {2479} -> 7 in r4c6 -> [29] in r56c6
49d. but this means 2 2s and 2 9s in n5
49e. -> no 2 or 9 r4c45
49f. no 2 or 9 r1c4

Just wondering if this move above could have been simpler because of some sort of Killer LoL move to remove the 6 from r9c6?

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?

Further, the fact that the two hidden cages have the same cage sum may not be important as there could be a + or - relationship between the 2 left-over cells. Since the 2 hidden cages exactly share the same house except for 1 cell each, this may be the important bit. Need some help to think this out.

50. 22(4) = {4567}(no 9): all locked for n5
50a. 6 locked for r6

51. {89} naked pair in r5c46: both locked for r5
51a. r56c1 = [65]

52. 15(3)n1 = {249/348}(no 7) = 4{29/48}
52a. 4 locked for c9

53. h26(4)r4569c6 = [7928/6839]
53a. no 6 r9c6

53a. r23c6 must have just 1 of 8/9 for c4 -> combo's in 22(4)n2 must have exactly 1 of 8/9
53a. 22(4) = {2479/2569/2578/3469/3478/3568}