Assassin 59

[mhparker]

Hi guys,

I assume that everyone's noticed by now that Jean-Christophe appears to have left this forum for good, judging by the fact that he has deleted all his recent posts on both the Assassin and Texas Jigsaw forums.

So here it is again, raised from the ashes as it were...


Assassin 59 V1.5



3x3::k:332835863076:4613:46132567:4361:4361:43613076:4613:4623:4623:462325783586:4374:4613386571953348:4374:4374:43743865:6947:7195:7195:719538796947:6947:6947:7195:41425936:5936:59362868:6947:4142:41424409:5936:517928684415:4415:4415:44095179:4933:4933:49331608:4409:44095179:5179:2639

[CathyW]

I noticed the deleted posts but I hope JC hasn't left for good. He often came up with some very clever moves that we could all learn from - including on the A59 V1.5 which I went through last night before it was deleted. I wasn't too concerned that JC had posted his own walkthrough rather than following on from my start. I'm sure it was just a case of not fully appreciating the implicit 'netiquette' of a tag solution - perhaps this should be added to Andrew's "Advice for newbies".

That said, if anyone can follow on from where I got stuck, please do!

[mhparker]

Hi Cathy,

Cheer up! Think of that lovely Assassin 60 waiting just round the corner!

That said, if anyone can follow on from where I got stuck, please do!

Assuming I managed to follow your walkthrough correctly, here's the marks pic after your step 39, so that others can join in more easily:

Assassin 59 V1.5 (tag)

Code: Select all

.-----------------------.-----------------------.-----------.-----------------------.-----------------------.
| 456789      456789    | 123         12345678  | 5789      | 12346789    123567    | 12346789    12346789  |
:-----------------------'-----------.           |           |           .-----------'-----------------------:
| 123456789   123456789   123456789 | 1234567   | 3457      | 123467    | 3456789     123456789   123456789 |
:-----------------------.-----------:           :-----------:           :-----------.-----------------------:
| 1234567     1234567   | 456789    | 3458      | 4689      | 4789      | 12346789  | 123456789   123456789 |
:-----------.           |           :-----------'           '-----------:           |           .-----------:
| 456789    | 1234567   | 456789    | 2346789     123468      123456789 | 12346789  | 12345679  | 456789    |
|           '-----------'-----------+-----------------------------------+-----------'-----------'           |
| 123456789   123456789   123456789 | 2346789     123468      123456789 | 3456789     123456789   123456789 |
|           .-----------.-----------+-----------------------------------+-----------.-----------.           |
| 345789    | 123456    | 123467    | 46789       123468      456789    | 123456    | 134567    | 345789    |
:-----------'           |           :-----------.           .-----------:           |           '-----------:
| 23456789    23456789  | 346789    | 24579     | 1246      | 24679     | 123456    | 12345678    12345678  |
:-----------------------'-----------:           :-----------:           :-----------'-----------------------:
| 1246789     1246789     1246789   | 12345     | 345       | 1234      | 456789      2456789     2456789   |
:-----------------------.-----------'           |           |           '-----------.-----------------------:
| 1245        1245      | 234         12456789  | 789       | 1246789     123567    | 134679      134679    |
'-----------------------'-----------------------'-----------'-----------------------'-----------------------'

[mhparker]

I noticed the deleted posts but I hope JC hasn't left for good. . He often came up with some very clever moves ...

Yes, but respect should not become a one-way street.

...including on the A59 V1.5

Sure, the A59 V1.5 is of course more difficult than your average Assassin, but nothing like some of the other puzzles we've been working on in the past. In fact, I already did it - 4 days ago! It just requires a single strong move and it's cracked.

With the release of the A60, A60RP ("Rejected Pattern"), Para's TJK31 V1.5 and Ruud's TJK32, there are plenty of other puzzles vying for our attention at the moment. Therefore, if no-one picks up on this one in the next couple of days, I'll probably just quietly post my original walkthrough and leave it at that.

[rcbroughton]

With the release of the A60, A60RP ("Rejected Pattern"), Para's TJK31 V1.5 and Ruud's TJK32, there are plenty of other puzzles vying for our attention at the moment. Therefore, if no-one picks up on this one in the next couple of days, I'll probably just quietly post my original walkthrough and leave it at that.

I've traded a couple of moves with Cathy which might prompt a solution walkthrough from the last position.

The "A60RP" is going to take a bit of a team effort I think

Rgds
Richard

[Para]

I am still stuck on A60. I also had a quick glance at TJK 32. And i am not too hopeful so far of finishing anything soon. So A60RP has to wait for a while.

Para

[CathyW]

I am still stuck on A60.

Me too - 34 steps so far without placement, Ruud has certainly ramped up the difficulty this week (still stuck on the A59 V1.5 as well though haven't looked at it again today)

No more time now and there'll be the final Harry Potter book to start reading this weekend too! Hope it's quiet at work next week!

[Para]

You could always call in sick. A weekend burn-out

[CathyW]

I'll probably just quietly post my original walkthrough and leave it at that.

Please do post your walkthrough of the V1.5. I've already spent too much time today struggling (although eventually successfully!) on the A60 and Harry Potter is beckoning loudly!

[mhparker]

I'll probably just quietly post my original walkthrough and leave it at that.

Please do post your walkthrough of the V1.5. I've already spent too much time today struggling (although eventually successfully!) on the A60 and Harry Potter is beckoning loudly!

Hope you're having a good time pottering around, if you know what I mean!

Here's my walkthrough for the A59 V1.5 as I originally wrote it nearly a week ago:


Assassin 59 V1.5 Walkthrough

1. 13/2 at R1C1: no 1,2,3

2. 14/4 at R1C3: no 9

3. 12/2 at R1C5: no 1,2,6

4. 10/2 at R1C8: no 5

5. 10/3 at R3C1: no 8,9

6. 13/2 at R3C3: no 1,2,3

7. 10/2 at R3C7: no 5

8. 10/2 at R6C3: no 5

9. 7/2 at R6C7: no 7,8,9

10. 11/3 at R6C8: no 9

11. 12/2 at R8C5: no 1,2,6

12. 19/3 at R8C7: no 1

13. 6/2 at R9C1 = {15/24} = {(4/5)..}

14. 10/2 at R9C8: no 5

15. Innies C123 = R19C3 = 5/2 = {14/23}

16. Innies C789 = R19C7 = 8/2: no 4,8,9

17. Outies N5 = R37C5 = 10/2: no 5

18. Split C5 innie cage at R456C5 = 11/3 = {128/137/146/236/245} (no 9)

19. Hidden killer quad on {6789} in C5, as follows:
19a. Both of 12/2 must each contain exactly 1 of {6789}
19b. R37C5 (step 17) must contain exactly 1 of {6789}
19c. -> 11/3 at R456C5 must contain exactly 1 of {6789}
19d. -> {245} combo blocked
19e. -> no 5 in R456C5

20. 5 in C5 locked in R1289C5
20a. -> one of the 2 12/2 cages at R1C5 and R8C5 must be {57}
20b. -> no 7 in R34567C5
20c. Cleanup: no 3 in R37C5 (step 17)
20d. 11/3 at R456C5 now restricted to {128/146/236} = {((6/8)..}

21. Outies R12 = R3C46 = 12/2: no 1,2,6
21a. no 3 in R3C6

22. Outies R89 = R7C46 = 11/2: no 1

23. Innies C1234 = R456C4 = 19/3
23a. {568} blocked by R456C5 (step 20d)
23b. 19/3 at R456C4 = {289/379/469/478} (no 1,5)

24. 5 in N5 locked in C6 -> no 5 elsewhere in C6
24a. Cleanup: no 7 in R3C4 (step 21), no 6 in R7C4 (step 22)

25. Innies R1234 = R4C19 = 13/2: no 1,2,3

26. Innies R6789 = R6C19 = 12/2: no 1,2,6

27. Outies C12 = R258C3 = 17/3 = {179/269/359/278/368/458/467}
27a. -> must contain exactly 1 of {1234}
27b. 10/2 at R6C3 must also contain exactly 1 of {1234}
27c. -> 17/3 at R258C3, 10/2 at R6C3 and R19C3 form naked killer quad on {1234} in C3
27d. -> no {49} combo in 13/2 at R34C3 = {58/67} = {(6/8)..}, {(7/8)..}
27e. -> {278} and {368} combos for 17/3 at R258C3 blocked by 13/2 at R34C3 (step 29d)
27f. -> 17/3 at R258C3 = {179/269/359/458/467}

28. Outies C89 = R258C7 = 20/3 = {389/479/569/578} (no 1,2)

29. Innies R2 = R2C456 = 10/3 = {127/136/145/235} (no 8,9)
29a. {12} only in R2C46
29b. -> no 7 in R2C46
29c. Cleanup: no 3,4 in R1C5

30. Innies R8 = R8C456 = 9/3 = {135/234} (no 6,7,8,9)
30a. ({126} blocked because none of these digits are present in R8C5)
30b. 3 locked in R8C456 for R8 and N8
30c. Cleanup: no 4,5 in R9C5; no 8 in R7C46 (step 22)

31. 3 in N7 now locked in 22/4 innies at R7C123+R9C3 = {2389/3469/3478/3568} (no 1)
31a. Cleanup: no 9 in R6C3, no 4 in R1C3 (step 15)

32. {458} combo for 17/3 at R8C1 blocked by 6/2 at R9C1
32a. -> 17/3 at R8C1 = {179/269/278/467} (no 5)

33. Either 9/3 at R8C456 = {135}, or...
33a. ...9/3 at R8C456 = {234} -> R7C46 (step 22) = [56]
33b. -> 5 in N8 locked in R7C4+R8C45
33c. -> no 5 in R9C4

34. I/O difference N2: R3C5 = R1C3 + R1C7 + 1
34a. min. sum of R1C3 + R1C7 = 3
34b. -> min. of R3C5 = 4 (no 1,2)
34c. Cleanup: no 8,9 in R7C5 (step 17)

35. 8 in N8 locked in R9 -> not elsewhere in R9
35a. Cleanup: no 2 in 10/2 at R9C8 = {19/37/46}

36. Common Peer Elimination (CPE): R7C46 see all candidate positions for {12} in C5
36a. -> no 1,2 in R7C46

37. Discontinuous complex Nice Loop with 2 strong links at discontinuity:
37a. (will use "=>" notation for strong links)
37b. Either R8C456 contains a 1, OR...
37c. ...contains a 2 (internal strong link)
37d. -> R7C46 <> {29}
37e. => R7C123 must contain a 9
37f. -> 17/3 at R8C1 cannot contain a 9, and thus also cannot contain a 1
37g. => R8C456 must contain a 1
37h. Conclusion: 9/3 at R8C456 = {135} (step 30); 1,5 locked for R8 and N8
37i. Cleanup: no 8 in R9C5, no 9 in R3C5 (step 17), no 6 in R7C6 (step 22)

38. 1 in C5 locked in N5 -> not elsewhere in N5
38a. -> 11/3 at R456C5 = {128/146} (no 3)

39. 3 in C5 now locked in R28C5
39a. -> one of the 2 12/2 cages at R1C5 and R8C5 must be {39}
39b. The other 12/2 cage is {57} (step 20a)
39c. -> no {48} combo for 12/2 at R1C5

40. {179} combo not available for 17/3 at R8C1 = {269/278/467} = {(2/4)..}
40a. -> {24} combo blocked for 6/2 at R9C1
40b. -> 6/2 at R9C1 = {15}, locked for R9 and N7
40c. no 9 in 10/2 at R9C8 = {37/46}; no 3,7 in R1C7 (step 16)

41. 9 in R9 locked in N8 -> not elsewhere in N8
41a. Cleanup: no 2 in R7C46 (step 22)

42. 11/2 at R7C46 (step 22) = {47}, locked for R7 and N8
42a. Cleanup: no 5 in R8C5; no 3,6 in R6C3; no 3 in R6C7; no 6 in R3C5 (step 17)

43. 12/2 at R8C5 = [39]
43a. -> 12/2 at R1C5 = {57}, locked for N2

44. Naked single (NS) at R8C6 = 1
44a. -> R8C4 = 5

45. 1 in C4 locked in R12C4
45a. -> no 1 in R1C3 (same cage)
45b. -> 5/2 at R19C3 (step 15) = {23}, locked for C3
45c. Cleanup: no 7,8 in 10/2 at R6C3 = [19/46]

46. 4 in R9 locked in R9C89
46a. -> 10/2 at R9C8 = {46}, locked for R9 and N9
46b. Cleanup: no 1 in R6C7, no 2 in R1C7 (step 16)

47. Hidden single (HS) in R9 at R9C7 = 7
47a. -> R1C7 = 1 (step 16)
47b. Cleanup: no 6 in R6C7; no 9 in 10/2 at R1C8; no 3,9 in 10/2 at R3C7 = {28/46}

48. Naked single (NS) at R7C6 = 4
48a. -> R9C6 = 8 (cage sum)
48b. Cleanup: no 4,8 in R3C4 (step 21)

49. R79C4 = [72] (naked singles)
49a. -> R7C5 = 6, R9C3 = 3
49b. -> R1C3 = 2, R3C5 = 4 (step 17)
49c. Cleanup: no 4 in R6C3, no 6 in R4C7, no 8 in 10/2 at R1C8 = {37/46}

50. 10/2 at R6C3 = [19]

51. 12/2 at R3C46 (step 21) = [39]

52. R12C6 = [62]
52a. Cleanup: no 7 in 13/2 at R1C1, no 4 in 10/2 at R1C8

53. R12C4 = [81]
53a. Cleanup: no 5 in 13/2 at R1C1

54. HS in R1 at R1C5 = 5
54a. -> R2C5 = 7

55. Naked pair (NP) on {28} in R7/N7 at R7C12
55a. -> no 2,8 elsewhere in R7 and N7
55b. R6C2 = 6 (cage split)
55c. Cleanup: no 5 in R6C7, no 7 in R3C3

56. Hidden pair (HP) on {17} in N1 at R3C12
56a. -> R3C12 = {17} (no 5,6)
56b. R4C2 = 2 (cage split)
56c. Cleanup: no 8 in R3C7

57. NS at R7C2 = 8
57a. -> R7C1 = 2

58. 10/2 at R1C8 = {37}, locked for N3

59. 13/2 at R1C1 = {49}, locked for N1

60. 17/3 at R2C1 = {368} (no 5), locked for R2 and N1
60a. Cleanup: no 5,7 in R4C3

Now it's all naked and hidden singles to the end.

[Andrew]

I only finished Assassin 59 yesterday evening. I wasted a lot of time by making a totally illogical move and couldn't find where I'd gone wrong. It can be so difficult to find one's own mistakes! Thanks Ed for finding it for me.

Cathy's steps 11 and 12 were neat with steps 13 and 14 using the same principle as step 11. That's a technique that I don't think I've seen before. It works as follows

If there are 2 cells in one row RaCxy which total n more than another cell RbCx in a different row, then cell RaCy cannot contain n. Of course the same principle applies for CaRxy and CbRx. Maybe someone can post this technique as diagrams for the two cases? It should be worth putting into Ruud's killer guide.

The one surprising thing was that Cathy didn't use the killer quad in R1. Mike, since you used the same solution path, did you miss it?

My solution path started much more like Para's walkthrough which included the killer quad. Para's step 17 was a neat one.

Here is my walkthrough. In addition to Cathy's neat technique, which I'll try to remember for future use, I also missed the innies in rows 2 and 8.

1. R1C12 = {19/28/37/46}, no 5

2. R12C5 = {49/58/67}, no 1,2 3

3. R1C89 = {39/48/57}, no 1,2,6

4. R34C3 = {49/58/67}, no 1,2,3

5. R34C7 = {19/28/37/46}, no 5

6. R67C3 = {39/48/57}, no 1,2,6

7. R67C7 = {19/28/37/46}, no 5

8. R89C5 = {17/26/35}, no 4,8,9

9. R9C12 = {49/58/67}, no 1,2,3

10. R9C89 = {19/28/37/46}, no 5

11. R2C123 = {389/479/569/578}, no 1,2

12. 10(4) cage at R1C6 = {1234}, no 1,2,3,4 in R1C45, no 9 in R2C5

13. 18(5) cage in N4 = 12{348/357/456}, no 9, 1,2 locked for N4

14. 45 rule on R12 2 outies R3C46 = 6 = {24}/[51]

15. 45 rule on R89 2 outies R7C46 = 16 = {79}, locked for R7 and N8, clean-up: no 3,5 in R6C3, no 1,3 in R6C7, no 1 in R89C5

16. 45 rule on R1234 2 innies R4C19 = 11 = [29]/{38/47/56}, no 1, no 2 in R4C9

17. 45 rule on R6789 2 innies R6C19 = 9 = {18/27/36/45}, no 9 in R6C9

18. 45 rule on C12 3 outies R258C3 = 11 = {128/137/146/236/245}, no 9
18a. 1,2 only in R58C3 -> 8 can only be in R2C3 -> no 8 in R58C3

19. 45 rule on C89 3 outies R258C7 = 20 = {389/479/569/578}, no 1,2

20. 45 rule on C123 2 innies R19C3 = 9 = {18/27/36/45}, no 9

21. 45 rule on C789 2 innies R19C7 = 5 = {14/23}

22. 25(4) cage at R7C6, max R9C7 = 4 -> min R789C6 = 21, no 1,2,3
22a. 25(4) = {2689/3589/4678} (cannot be {1789/3679/4579} because 7,9 only in R7C6), no 1, clean-up: no 4 in R1C7 (step 21)
22b. 4 in {4678} must be in R9C7 -> no 4 in R89C6
22c. 8 locked in 25(4) cage in R89C6, locked for C6 and N8

23. 10(4) cage at R1C6, 4 locked in R123C6, locked for C6 and N2, clean-up: no 9 in R1C5, no 2 in R3C6 (step 14)

24. Killer pair 5/6 in R89C6 and R89C5, locked for N8

25. Killer pair 5/6 in R12C5 and R89C5, locked for C5

26. 45 rule on N9 2 outies R6C78 – 4 = 1 innie R9C7, max R9C7 = 4 -> max R6C78 = 8, no 8,9 in R6C7, no 7,8,9 in R6C8, clean-up: no 1,2 in R7C7

27. 45 rule on C1234 3 innies R456C4 = 16
27a. 45 rule on C6789 3 innies R456C6 = 15
27b. R456C5 = 14
27c. 45 rule on C5 5 innies R34567C5 = 24 -> R37C5 = 10 = [73/82/91] ->R3C5 = {789}, no 4 in R7C5
[Alternatively 45 rule on N5 2 outies R37C5 = 10 but I only saw that later]

28. 45 rule on N2 1 innie R3C5 – 4 = 2 outies R1C37, max R3C5 = 9 -> max R1C37 = 5 -> R1C3 = {1234}, clean-up: no 1,2,3,4 in R9C3 (step 20)
[At this stage I ought to have seen step 33.]

29. 4 in N8 locked in R89C4, locked for C4

30. 20(4) cage at R7C4 must contain 4 (step 29) = 4{169/178/259/367} (cannot be {2468/3458} because 5,6,8 only in R9C3)
30a. {3467} must use 7 in R7C4 -> no 7 in R9C3, clean-up: no 2 in R1C3 (step 20)

31. 45 rule on N7 2 outies R6C23 – 8 = 1 innie R9C3, min R9C3 = 5 -> min R6C23 = 13, no 3
31a. R9C3 = {568} -> R6C23 = 13,14,16 no 5,6 in R6C3 -> no 8 in R6C2
[Step 31a is doing it the hard way. With Cathy's technique I would have just said no 8 in R6C2 because R6C23 8 more than R9C3.]

32. 12(3) cage at R3C1, min R4C2 = 3 -> max R3C12 = 9, no 9

33. Killer quad 1/2/3/4 in R1C12, R1C3, R1C67, locked for R1, clean-up: no 8,9 in R1C89

34. Naked pair {57} in R1C89, locked for R1 and N3, clean-up: no 3 in R1C12, no 6,8 in R2C5, no 3 in R4C7
34a. R3C89 cannot be {79} -> no 1 in R4C8
34b. R3C89 cannot be {49/58/67} ({49} blocked because 17(3) cage cannot be {49}4) -> no 4 in R4C8

35. 18(4) cage at R1C3 = {1269/1359/1458/2358/2367} (cannot be {1278/3456} which clash with R12C5, cannot be {1368/1467} because R3C4 only contains 2,5, cannot be {2349} because [932] in R123C4 clashes with R123C6, cannot be {2457} because R1C4 only contains 6,8,9)
35a. {1458/2358} both require 8 in R1C4 -> no 8 in R2C4

36. 2 in C3 locked in R58C3
36a. R258C3 (step 18) = {128/236/245} = 2{18/36/45}, no 7

37. 7,9 in C3 locked in R346C3 -> R34C3 = {49/67}, no 5,8 and R67C3 = [75/93], no 4,8

38. 45 rule on N3 4 innies R1C7 +R3C789 = 20 = {1289/2369/2468} (cannot be {1469} which clashes with R3C6) = 2{189/369/468}, 2 locked for N3
38a. If {1289} R3C89 cannot be {89} -> 1,2 must be in R1C7 and R3C89 -> no 1 in R3C7 (it can still be 2 in the {2369} combination), clean-up: no 9 in R4C7
38b. R2C789 = {139/148/346}
[Mike has pointed out that after step 38 I missed 2 in R2 locked in R2C46 -> R3C4 = 5, R3C6 = 1.]

39. R2C123 (step 11) = {389/479/569} (cannot be {578} which clashes with R2C5) = 9{38/47/56}, 9 locked for R2 and N1, clean-up: no 1 in R1C12, no 4 in R4C3

40. R1C4 = 9 (hidden single in R1) -> R7C46 = [79], clean-up: no 1 in R7C5

41. 1 in N8 locked in R89C4, locked for C4
[At this stage I should have remembered 4 locked in R89C4 (step 29) -> R89C4 = {14}, locked for 20(4) cage. This was done in steps 46 and 47.]

42. 1 in N2 locked in R123C6, locked for C6 and 10(4) cage -> no 1 in R1C7, clean-up: no 4 in R9C7

43. Naked pair {23} in R19C7, locked for C7, clean-up: no 8 in R3C7, no 7,8 in R4C7, no 7 in R6C7, no 8 in R7C7

44. Naked pair {46} in R67C7, locked for C7 -> R2C7 = 8, R3C7 = 9, R4C7 = 1, clean-up: no 3 in R2C123 (step 39)

45. 9 in C3 locked in R46C3, locked for N4

46. Killer pair 2/3 in R7C5 and R89C5, locked for C5 and N8

47. R789C4 = 7{14} -> R9C3 = 8, R1C3 = 1 (step 20), clean-up: no 5 in R9C12, no 2 in R9C89
[R9C3 was also a hidden single in C3]

48. R8C6 = 8 (hidden single in N8)
48a. 8 in N9 locked in R7C89 -> 14(3) cage at R6C8 = 8{15/24}, no 3,6

49. Killer pair 7/8 in R12C5 and R3C5, locked for C5

50. 3 in R1 locked in R1C67 -> no 3 in R2C6

51. R2C7 = 8 -> R2C89 = 5 = {14}, locked for R2 and N3 -> R2C6 = 2, R1C7 = 3, R13C6 = [41], R3C4 = 5, R12C5 = [67], R2C4 = 3, R3C5 = 8, clean-up: R7C5 = 2 (step 27c), R9C7 = 2 (step 21)

52. Naked pair {35} in R89C5 -> R9C6 = 6, clean-up: no 7 in R9C12, no 4 in R9C89

53. Naked pair {26} in R3C89, locked for R3 -> R4C8 = 9, R4C5 = 4, clean-up: no 7 in R4C3 -> R4C3 = 6, R3C3 = 7, R2C3 = 5, R6C3 = 9, R7C3 = 3, no 1 in R9C9

54. R3C12 = {34} -> R4C2 = 5

55. Naked pair {49} in R9C12, locked for R9 and N7 -> R8C3 = 2, R5C3 = 4, R6C2 = 7, R89C4 = [41]

56. Naked pair {37} in R9C89, locked for R9 and N9 -> R89C5 = [35], R8C7 = 5, R5C7 = 7

57. R4C6 = 7 (hidden single in C6)

58. R8C7 = 5 -> R8C89 = 10 = [19] -> R8C12 = [76], R7C12 = [51], R2C12 = [69], R9C12 = [94], R3C12 = [43], R2C89 = [41], R7C89 = [84], R7C8 = 2 (cage sum), R67C7 = [46], R3C89 = [62]

59. Naked pair {35} in R5C68, locked for R5

60. R5C456 = [693] (only remaining permutation)

and the rest is naked singles

[sudokuEd]

Cathy's steps 11 and 12 were neat with steps 13 and 14 using the same principle as step 11. That's a technique that I don't think I've seen before. It works as follows

If there are 2 cells in one row RaCxy which total n more than another cell RbCx in a different row, then cell RaCy cannot contain n. Of course the same principle applies for CaRxy and CbRx.

I liked this move too Andrew and Cathy - and think it will help our walk-throughs if we give it a name. My suggestion is IOU. If an Innie & Outie are in the same r, c or n, then the 3rd cell must be Unequal to the I/O difference. Hence, IOU. For example, Cathy's step 11 says

11. O-I N7: r6c23 - r9c3 = 8

Since r6c3 (an Outie) is in the same column as r9c3 (an Innie), then the 3rd cell must be Unequal to the difference - in this case 8. From IOU -> r6c2 <> 8.

Another way of seeing this is that if r6c2 was 8 then the difference would be 0 which means that r6c3 = r9c3: which is impossible because they are in the same column.

Similarly, step 13 is

13. O-I N3: r4c78 - r1c7 = 7

Since the difference is 7: IOU -> r4c8 <> 7

Perhaps with a walk-through we could explain the logic of this move the first time and after that just refer to it as IOU. So with Cathy's step 14,

14. O-I N9: r6c78 - r9c7 = 4

then add: IOU -> r6c8 <> 4

Incidently, Assassin 60 has several of this move - including where the 1 innie (of a r or c) and 1 outie (of that r or c) are in the same nonet. This is a bit harder to see compared to the I/O of nonets.

Thanks for highlighting this trick Andrew. Good luck with Assassin 60! It's a corker!

Cheers
Ed

[mhparker]

Andrew,

The one surprising thing was that Cathy didn't use the killer quad in R1. Mike, since you used the same solution path, did you miss it?

No, I saw that one.

I also saw something else that Cathy didn't mention, namely the R1 outies at R2C456+R3C46 = 18/5 = {12..}, thus eliminating {12} from R1C6.

What I missed at the time, however, is that I could have done much, much more with this. For example, since we know that R3C46 (= 6/2) must contain one of {12}, R2C456 cannot contain both of {12}. The only other place for the other one of {12} in R2 is in the 13/3 cage at R2C7, which therefore cannot be {346} -> no 6 in 13/3 at R2C7. This then locks the 6 in N3 into R3, and the 13/3 cage now blocks the {389} combo in 20/3 at R2C1 -> no 3 in R2C123, etc.