assassin 64

[Glyn]

This might get confusing

Paras step 48 replicates Richards step 45a and his step 49 is the same as Richards step 48 as well.
We all went for the area Cathy was looking at, here is the same move as I explained it.

48. If r3c2=3|4 forms quadruplet r3c1289={1234}, requires r3c5=9. therefore r3c2<>3|4.

Now we move on

49. a) If r3c5=7 innies of n2=17 require r1c4=[82]|[64].
Taken together they block {17} and {26} combos for 8(2) cage. Only remaining alternative is {35}.
b) Alternative is r3c5=5.
Combining a) and b) 20(4) cage cannot contain 5. r2c45<>5.

50. Examining 49 further we either have r3c5=5 or r23c6={35} common peers r46c6<>5.

51. If 12(3) cage in n6={147} => 12(2) cage in n3= {39} together block all combos for 11(2) cage in n9.
r456c9<>7.

I am itching to find some chain off the bivalue cell r3c5 that forces a placement, but apart from finding that r3c2<>5 from a forcing chain through n6 and back into r3, which I won't give here but leave for you to find for own amuusement I'll have to call it a day.

Latest candidates below

All the best

Glyn

Code: Select all

.-------------------------------.-------------------------------.-------------------------------.
| 23456789  123456789 123456789 | 689       1234678   234       | 23456789  23456789  345789    |
| 23456789  123456789 123456789 | 12346789  12346789  12356     | 123456789 123456789 345789    |
| 1234      5678      2346789   | 2346789   57        23567     | 23456789  1234      1234      |
:-------------------------------+-------------------------------+-------------------------------:
| 3456789   123456    123456    | 123456789 123456789 1236789   | 345678    345678    1234568   |
| 345678    123456    123456    | 2345      12345678  567       | 6789      6789      1234568   |
| 3456789   56789     56789     | 123456789 123456789 1236789   | 12345     12345     1234568   |
:-------------------------------+-------------------------------+-------------------------------:
| 1234      56789     12345678  | 1245      23456     2356789   | 123456789 1234      6789      |
| 345678    12345678  12345678  | 1245      23456789  2356789   | 123456789 123456789 789       |
| 123456    789       789       | 789       23456     1234      | 123456    123456    234       |
'-------------------------------.-------------------------------.-------------------------------'

[rcbroughton]

revisting step 44 we have:

[Edit - oops! Too slow - steps are still valid, I think - so just renumber them from Glyn's]


Need to ignore this step for now - 52a needs some further thought!

52. 45 on n9 - outies r6c78+r9c6 total 8. = {152}/{134}/{224}/{233} (we already removed {116})
52a. {224} - means r7c89 must total 9=[36] and r9c78 totals 8={53} or {62} - so not possible
52b. {233} - means r7c89 must total 10=[19] and r9c78 total 7={25} - but [19] would force 12(3)n6=1{47}/{56} and r3c9=1 contradiction - so not possible
52c. remaining is {152}/{134} - must use a 1, so no 1 at r6c6

53. 9(2)n7 cannot be {45} - blocks 11(2)n1 and 20(3)n3 - 11(2) would be forced to {29} or {38} and 20(3) to {389} - contradiction

Rgds
Richard

[Andrew]

Hope people don't mind me interrupting the "tag" solution for A64V2 to post my walkthrough for the original Assassin.

Only finished it yesterday and then worked through Para's and Cathy's walkthroughs today before deciding to post it. I agree with Para that it's rated between 1.0 and 1.25

I seem to have done more with the combinations for sub-cages, which is unusual for me, but they were helpful this time. I would have finished earlier but I found I'd deleted one combination too many from a sub-cage, which happened to be part of the solution. Still that's better than deleting a different combination and having a flawed solution path. Lots of killers, which I enjoyed.

Here is my walkthrough for A64.

1. R12C1 = {49/58/67}, no ,1,2,3

2. R12C9 = {15/24}

3. R23C6 = {16/25/34}, no 7,8,9

4. R78C4 = {18/27/36/45}, no 9

5. R89C1 = {17/26/35}, no 4,8,9

6. R89C9 = {19/28/37/46}, no 5

7. 8(3) cage in N1 = 1{25/34}, 1 locked for N1

8. R456C1 = {128/137/146/236/245}, no 9

9. R456C9 = {289/379/469/478/568}, no 1

10. R5C678 = {128/137/146/236/245}, no 9

11. 20(3) cage in N7 = {389/479/569/578}, no 1,2

12. 23(3) cage in N9 = {689}, locked for N9, clean-up: no 1,2,4 in R89C9
12a. Naked pair {37} in R89C9, locked for C9 and N9

13. 27(4) cage in N2 = {3789/4689/5679} = 9{378/468/567}, no 1,2, 9 locked for N2

14. 30(4) cage at R3C1 = {6789}

15. 14(4) cage at R6C2 = {1238/1247/1256/1346/2345}, no 9

16. 13(4) cage at R6C7 = {1237/1246/1345} = 1{237/246/345}, no 8,9

17. 45 rule on R5 3 innies R5C159 = 19 = {289/379/469/478/568}, no 1

18. 45 rule on N5 2 outies R37C5 = 2 innies R5C46, specifically that R37C5 and R5C46 must contain the same two numbers, in other words Law of Leftovers.

19. 45 rule on C1 2 innies R37C1 = 13 = {67}/[85/94], R7C1 = {4567}

20. 9 in C1 locked in R123C1, locked for N1
20a. 9 in C1 locked in R123C1 -> 4 in C1 must be in R12C1 or R7C1 (R12C3 and R37C1 are both 13(2) cages), locked for C1
20b. R456C1 (step 8) = {128/137/236}, no 5

21. 45 rule on C9 2 innies R37C9 = 10 = [64/82/91]

22. R9C678 = {149/158/248/257} -> R9C6 = {789}
22a. R9C78 = {14/15/24/25} -> R7C89 = {14/15/24/25} (cannot be {12/45} which clash with R9C78)
22b. 13(4) cage at R6C7 = {1246/1345} (cannot be {1237} because R7C89 cannot be {12} from step 22a), no 7
22c. {1246} = {16/26} in R6C78, {14/24} in R7C89
22d. {1345} = {34/35} in R6C78, {14}/[51] in R7C89
22e. R6C78 = {16/26/34/35}, R7C89 = {14/24}/[51] -> R9C78 = {15/24/25}
22f. R456C9 (step 9) = {289/469/568}
22g. R6C78 = {16/34/35} (cannot be {26} which clashes with R456C9), no 2

23. 45 rule on C789 3 outies R159C6 = 13 = {139/148/157/238/247} (cannot be {256/346} because only 7,8,9 in R9C6) -> R15C6 = {13/14/15/23/24}, no 6,7,8
23a. If R15C6 = {23/24} -> R23C6 = {16}
23b. Killer single 1 in R15C6 and R23C6, locked for C6
[Alternatively 45 rule on C6 4 innies R4678C6 = 25 but cannot be {1789} which clashes with R9C6 -> no 1]

24. 45 rule on C123 3 outies R159C4 = 11 = {128/137/146/236/245}, no 9

25. 45 rule on R89 4 outies R7C3467 = 30 = {6789}, locked for R7, clean-up: no 6,7 in R3C1 (step 19), R8C4 = {123}

26. Killer pair 4,5 in R7C1 and R7C89 (step 22e), locked for R7

27. 45 rule on N1 2 innies R3C12 – 10 = 1 outie R1C4, min R3C12 = 14, max R3C12 = 17 -> R1C4 = {4567}

28. 45 rule on N7 2 innies R7C12 – 4 = 1 outie R9C4
28a. R7C12 = [42/43/51/52/53] (cannot be [41] which clash with R7C89) -> R9C4 = {234}

29. 45 rule on N8 3 innies R7C5 + R9C46 = 13 = {139/148/247} (cannot be {238} which clashes with R78C4)
29a. 1 of {139} in R7C5 -> no 3 in R7C5
29b. 2 of {247} in R7C5 -> no 2 in R9C4

30. R7C2 = 3 (hidden single in R7), clean-up: no 5 in R89C1
30a. 14(4) cage at R6C2 = {1346/2345} (cannot be {1238 because R7C1 only contains 4,5} = 34{16/25}, no 7,8

31. 3 in C1 locked in R456C1 (step 20b) = {137/236}, no 8, 3 locked for N4
31a. Killer pair 1,2 in R6C23 (step 30a) and R456C1, locked for N4
31b. Min R5C23 = 9 -> max R5C4 = 6, clean-up: no 7,8 in R3C5 (step 18)

32. Hidden killer pair 1,2 in R89C1 and R9C23 -> R9C23 must contain 1/2
32a. R9C234 = {139/148/238/247} (cannot be {157/256} because R9C4 only contains 3,4), no 5,6
32b. 20(3) cage in N7 = {479/569/578}
32c. Killer pair 4,5 in R7C1 and 20(3) cage, locked for N7
32d. Killer pair 6,7 in R89C1 and 20(3) cage, locked for N7
32e. R9C234 = {139/148/238}

32. 8 in C1 locked in R123C1, locked for N1
32a. Killer pair 6,7 in R456C1 (step 31) and R89C1, locked for C1
32b. Killer pair 4,5 in R12C1 and 8(3) cage, locked for N1

33. R1C234 = {257/347/356}
33a. 4,5 only in R1C4 -> R1C4 = {45}
33b. 3 of {356} in R1C3 -> no 6 in R1C3

34. R3C12 – 10 = R1C4 (step 27)
34a. R1C4 = {45} -> R3C12 = 14,15 = [86/87/96], R4C23 = {69/78/79}
34b. 45 rule on R123 5 innies R3C12589 = 28, min R3C1259 = 21 -> max R3C8 = 7

35. 45 rule on N2 3 innies R1C46 + R3C5 = 11, max R1C46 = 9 -> min R3C5 = 2
35a. R1C46 + R3C5 = {146/245} (cannot be {236} because only 4,5 in R1C4) = 4{16/25}, no 3, 4 locked for N2, clean-up: no 3 in R23C6, no 3 in R5C46 (step 18)
35b. Killer pair 5,6 in R1C46 + R3C5 and R23C6, locked for N2
35c. 1 in N2 locked in R123C6, locked for C6

36. R5C678 = {128/146/236/245} (cannot be {137} because R5C6 only contains 2,4,5), no 7

37. 7 in N6 locked in R4C78, locked for R4 and 20(4) cage at R3C8
[This makes R3C2 a hidden single in the 30(4) cage but step 37a does more!]
37a. R4C23 (step 34a) = {69} (only remaining combination), locked for R4, N4 and 30(4) cage at R3C1 -> R3C12 = [87], R7C1 = 5 (step 19), clean-up: no 2 in R456C1 (step 31), no 2 in R7C9 (step 21)

38. Naked triple {137} in R456C1, locked for C1 and N4

39. Naked pair {24} in R6C23, locked for R6 and N4
39a. Naked pair {58} in R5C23, locked for R5 -> R5C4 = 2, R5C6 = 4 -> R37C5 = [42] (step 18), R9C46 = [47] (step 29), R1C4 = 5, R89C9 = [73], clean-up: no 2 in R23C6, no 1 in R2C9

40. R7C3 = 7 (hidden single in C3), R1C6 = 2 (hidden single in N2), R1C23 = [63], clean-up: no 4 in R2C9

41. Naked pair {26} in R89C1, locked for N7
41a. 20(3) cage in N7 = {479} (only remaining combination) -> R8C23 = {49}, locked for R8 and N7

41. Naked pair {14} in R7C89, locked for 13(4) cage at R6C7
41a. 13(4) cage at R6C7 = {1345} (last remaining combination) -> R6C78 = {35}, locked for R6 and N6

42. 9 in N6 locked in R456C9, locked for C9 -> R3C9 = 6, R5C9 = 9, R6C9 = 8, R4C9 = 2, R7C9 = 4 (step 21), R7C8 = 1, R12C9 = [15], R23C6 = [61]

43. Naked pair {68} in R8C78, locked for R8 and N9 -> R7C7 = 9, R89C1 = [26]

44. R5C78 = [16]

45. R1C6 = 2 -> R1C78 = 16 = [79]

46. R3C9 = 6, R4C78 = {47} -> R3C8 = 3 (cage sum)

and the rest is naked singles

[mhparker]

Hi Richard,

52a. {224} - means r7c89 must total 9=[36] and r9c78 totals 8={53} or {62} - so not possible
52b. {233} - means r7c89 must total 10=[19] and r9c78 total 7={25} - but [19] would force 12(3)n6=1{47}/{56} and r3c9=1 contradiction - so not possible

Could you briefly re-check this, please?

52a.: What about r7c89 = [18]? (seems to work with outies = {224})
52b.: What about r7c89 = [46]? (seems to work with outies = {233})

TIA.

[CathyW]

I agree with Mike on the possibilities for r7c89!

Meanwhile not sure what number to give this minor step but ...

Revisiting step 16 - Innies N8: r7c5 + r9c46 = 14 = {167/257/347/158/248/149/239}
Options {257/149} excluded as would eliminate all options for the 6(2).

That's all for now folks

[rcbroughton]

Hi Richard,

52a. {224} - means r7c89 must total 9=[36] and r9c78 totals 8={53} or {62} - so not possible
52b. {233} - means r7c89 must total 10=[19] and r9c78 total 7={25} - but [19] would force 12(3)n6=1{47}/{56} and r3c9=1 contradiction - so not possible

Could you briefly re-check this, please?

52a.: What about r7c89 = [18]? (seems to work with outies = {224})
52b.: What about r7c89 = [46]? (seems to work with outies = {233})

TIA.

Hmm - good point on 52a. I was convinced when I looked at this last night that [18] didn't work for some reason. . .

{233} doesn't work with r7c89 must total 10=[46] either - r9c78 total 7={25} - but that would eliminate all candidates in r9c9
- so that's still OK - but I need to look more closely at the {224}

Rgds

[Glyn]

For the moment I've ignored Richards step 52 which I'll have a look at myself after I've rewetted the towel.

I'll call this one step 54.

54) Forcing Chain.
If r3c5=7=>r5c678=7{69}=>r7c9=6 (last vacant place in c9) => 25(4) cage in n8 must contain 6 whilst blocked for 7 in c56. Only combo is {2689} forcing 6(2) cage r78c4={15}.
Alternative r3c5=5.
Common peers of r3c5 and r78c4 => r46c4+r789c5<>5.

[CathyW]

Actually, I think if the outies of N9 is 224, it would force r7c89 to [18]

r9c6 would be 2 -> r9c78 = {35} -> r9c9 = 4 -> r8c9 = 7
so only option left for r7c89 would be [18] leaving {269} for the 17(3)

Will check again when I get home - meantime, end of lunch break and back to work!

[goooders]

maybe there is a different way to start

column1 1 can only go in the 5(row3 and7) or the 9 in the bottom nonet

1 will not go in the 9 because that requires542 for the 11(neither 641 nor 623 works) but if that is so columns 1 and 2 of row 7 are 63 which works with neither 68 nor 59 required by the bottom left outies of 22 (note 59 cant work with the 20 in middle left nonet)

so that means rows 3 and 7 of column 1 are 14

that means the 9 in bottom left nonet can only be 72 or 63

but 72 will not work because the 11 cant be made at all

so the 9 in the bottom left nonet is 36

that makes the 11 cage 542 and columns 1 and 2 of row 7 are 17

that makes the other part of the 23 cage 69 ( given the 22 outie of the bottom left nonet)

so in column 1 the 11 is 29 and the 20 is 875

the 14 cage in top left nonet has a 4 in it which means the 9 cage in middle left nonet must be 423 and therefore the 14 cage above is 4631`

that means column 4 row 5 is 3

that means column 4 row 1 is 9 ( outies of leftside 3 nonets are 19)

easy now because columns 4 8 and 9 of row 3 have to be 135

the only other info you need is that row 1 column 6 which looks like it is either 3 or 5 or 7 can only be 3 or 5 because the outies of the 3 rightside nonets are12

should also say 16 cage in top left nonet has to be 358 because of either 3 or 5 in row 1 column 6 and row 3 columns 8 and 9 being two of 135

[mhparker]

54) Forcing Chain.
If r3c5=7=>r5c678=7{69}=>r7c9=6 (last vacant place in c9) => 25(4) cage in n8 must contain 6...

Thanks, Glyn! Just for info, this move can be simplified a little. No need to go round the houses (n6/c9/r7) to show that r7c5 can't be a 6 if r3c5 is a 7. Instead, we can just use LoL on n5 (step 12) here. In other words, because the innies (r5c46) cannot be {67}, the outies (r37c5) cannot be {67} either.

Moving on:

55. 5 in r9 locked in r9c78 -> not elsewhere in n9
55a. 10(3)n89 = {145/235} (no 6)

56. no 6 in r7c5. Here's how.
56a. recall outies c123 (step 17): r159c4 = 19(3)
56b. only options with 7 in r4c9 are [847/937]
56c. -> r7c5 = {34} (LoL(n5), step 12)
56d. -> innies n8 = {347}
56e. -> {167} blocked for n8 innies (step 53(?))
56f. -> no 6 in r7c5

57. LoL(n5): no 6 in r5c6

58. 22(3)n56 = [5]{89}/[7]{69}
58a. -> no 7 in r5c78

59. 7 in n6 locked in r4c78 -> not elsewhere in r4
59a. 16(4)n36 = {(126/135/234)7} (no 8)

60. 6 in n8 locked in 25(4) = {2689/3679/4678} (no 5)

61. 5 in n8 locked in r78c4
61a. -> 6(2)n8 = {15}, locked for c4 and n8

62. both of {16} now unavailable to c789 outies
62a. -> r159c6 = {345} (only remaining combo, see step 29a)
62b. -> r5c6 = 5
62c. r19c6 = {34}, locked for c6

63. HS in n2 at r3c5 = 5

That should do for now.

marks pic after step 63:

Code: Select all

.-----------.-----------------------------------.-----------.-----------------------------------.-----------.
| 23456789  | 123456789   123456789   689       | 1234678   | 34          23456789    23456789  | 345789    |
|           :-----------------------.-----------'           :-----------.-----------------------:           |
| 23456789  | 123456789   123456789 | 2346789     12346789  | 126       | 123456789   123456789 | 345789    |
:-----------'-----------.           |           .-----------:           |           .-----------'-----------:
| 1234        678       | 2346789   | 2346789   | 5         | 267       | 2346789   | 1234        1234      |
:-----------.           '-----------+-----------'           '-----------+-----------'           .-----------:
| 345689    | 123456      123456    | 234689      1234689     12689     | 34567       34567     | 1234568   |
|           :-----------------------'-----------.           .-----------'-----------------------:           |
| 34678     | 12346       12346       234       | 1234678   | 5           689         689       | 123468    |
|           :-----------------------.-----------'           '-----------.-----------------------:           |
| 3456789   | 56789       56789     | 2346789     12346789    126789    | 12345       12345     | 1234568   |
:-----------'           .-----------+-----------.           .-----------+-----------.           '-----------:
| 1234        56789     | 12345678  | 15        | 234       | 26789     | 12346789  | 1234        6789      |
:-----------.-----------'           |           :-----------'           |           '-----------.-----------:
| 3678      | 12345678    12345678  | 15        | 2346789     26789     | 12346789    12346789  | 789       |
|           :-----------------------'-----------:           .-----------'-----------------------:           |
| 1236      | 789         789         789       | 2346      | 34          12345       12345     | 234       |
'-----------'-----------------------------------'-----------'-----------------------------------'-----------'

[rcbroughton]

64. From 62b 22(3)n56=5{89} {89} locked for r5/n6

65. 12(3)n6 - {138} not now possible because no 8, so no 3 either
65a. now = 6{15}/{24} - 6 locked for n6 and c9
65b. no 4 at r3c9 (innies of c9=10)

[CathyW]

Moving on:

55. 5 in r9 locked in r9c78 -> not elsewhere in n9
55a. 10(3)n89 = {145/235} (no 6)

At this point I still have a 5 in r9c1! When did that get eliminated or have you already used Goooders information?

Thanks for your input Goooders. That's a fair implication chain which I found a bit difficult to follow but have rephrased the first part thus:

Step No ?!.

1 locked to split 5(2) r37c1 and r9c1

a) r9c1 = 1 -> r8c1 = 8 -> r9c23 = {79}
-> split 5(2) r37c1 = {23}
-> 11(3) N9 = {245}
-> r7c1 = 3, r7c2 = 6 -> r6c23 = {59}
BUT 20(3) r456c1 requires at least one of 8,9 CONFLICT

b) split 5(2) r37c1 = {14} -> 9(2) r89c1 = {27/36}
If 9(2) = {27} no options for 11(3) -> 9(2) = {36} not elsewhere in N7
-> 11(2) r12c1 = {29} not elsewhere in N1
-> 20(3) r456c1 = {578} not elsewhere in N4

now follow through newly revealed naked pair and then carry on with Mike's steps above ...!

[mhparker]

At this point I still have a 5 in r9c1! When did that get eliminated...

Richard eliminated the {45} combo for 9(2)n7 in step 53.

[CathyW]

Thanks - missed that after querying step 52!!

Goooders implication chain is still valid and I have now reached solution!

Well done everyone a great team effort. If anyone is inclined, may be worth consolidating the steps for a complete WT.

[Andrew]

Well done everyone a great team effort. If anyone is inclined, may be worth consolidating the steps for a complete WT.

Yes, a great team effort. Looks like it was all the regulars except for Ed, who is currently busy working on ratings and doing a great job on them. I'll be interested to see what rating Ed eventually comes up with for this puzzle; it's clearly much, much harder than A64. Sorry I wasn't able to take part but I did start working through the posted moves yesterday evening and had got as far as step 32. I'd hoped that I might be able to catch up with the moves today and possibly even contribute something but that final spurt beat me to it. Well done everyone!

If Cathy is just asking for a consolidated WT, rather than a condensed WT, then I'll volunteer to do it. If someone has already started, or one of the solving team wants to do it, please contact me by PM and that will save some wasted effort. If as Mike suggested early on, this one might have a very narrow solving path, then a consolidated WT is probably almost the same as a condensed WT.