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 Post subject: Assassin 218
PostPosted: Thu Aug 04, 2011 11:11 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Much easier than the last monster. Almost doesn't feel hard enough to be an Assassin in 2011!

Assassin 218

Image

code: copy & paste into solver:
3x3::k:5121:5121:2562:2562:5635:5635:5635:3844:3844:5121:3077:3077:2562:6406:5635:6406:6406:3335:5121:8200:11529:11529:6406:6406:6406:4874:3335:4363:8200:1804:11529:11529:3597:3597:4874:4874:4363:8200:1804:11529:11529:3597:4874:4874:4366:4363:8200:2831:2831:11529:11529:3088:3088:4366:3345:8200:8200:3858:3858:11529:3347:3347:4366:3345:3345:3860:8981:3858:8981:8981:3347:3347:1558:1558:3860:8981:8981:8981:4375:4375:4375:
solution:
+-------+-------+-------+
| 5 6 3 | 1 9 4 | 2 8 7 |
| 2 8 4 | 6 3 7 | 1 5 9 |
| 7 9 1 | 5 2 8 | 6 3 4 |
+-------+-------+-------+
| 3 7 2 | 9 6 5 | 8 4 1 |
| 6 4 5 | 7 8 1 | 9 2 3 |
| 8 1 9 | 2 4 3 | 5 7 6 |
+-------+-------+-------+
| 9 5 6 | 3 7 2 | 4 1 8 |
| 1 3 8 | 4 5 9 | 7 6 2 |
| 4 2 7 | 8 1 6 | 3 9 5 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 218
PostPosted: Sat Aug 06, 2011 5:30 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Ed wrote:
Much easier than the last monster. Almost doesn't feel hard enough to be an Assassin in 2011!
It certainly is! I still haven't managed to finish A217 and I hate being unable to finish a V1; must find time to concentrate on it again and try to find what I'm missing.

Thanks Ed for a fun puzzle!

Dare I say:
that the way I solved it was a "one trick pony". I don't often manage to find the key ways to solve those quickly. A218 was definitely worth posting as an Assassin. I hope my breakthrough step was what Ed intended.

Rating Comment:
I'll rate A218 at 1.5 because I used a simple "clone". I also used a combined cage; that might not have been necessary but it led to step 12.

Here is my walkthrough for A218:
It's fairly late so I haven't had time to check my walkthrough yet. Edit, done that now; a couple of typos corrected and part of step 29 removed. Also thanks Ed for pointing out another typo.

Prelims

a) R1C89 = {69/78}
b) R2C23 = {39/48/57}, no 1,2,6
c) R23C9 = {49/58/67}, no 1,2,3
d) R45C3 = {16/25/34}, no 7,8,9
e) R6C34 = {29/38/47/56}, no 1
f) R6C78 = {39/48/57}, no 1,2,6
g) R89C3 = {69/78}
h) R9C12 = {15/24}
i) 10(3) cage at R1C3 = {127/136/145/235}, no 8,9
j) 13(4) cage in N9 = {1237/1246/1345}, no 8,9
k) and, of course, 45(9) cage at R3C3 = {123456789}

1. 13(4) cage in N9 = {1237/1246/1345}, 1 locked for N9

2. R23C9 = {49/58} (cannot be {67} which clashes with R1C89), no 6,7
2a. Killer pair 8,9 in R1C89 and R23C9, locked for N3

3. 45 rule on N7 2 innies R7C23 = 11 = {29/38/47/56}, no 1

4. 45 rule on N8 1 outie R8C7 = 1 innie R7C6 + 5, R7C6 = {1234}, R8C7 = {6789}

5. 45 rule on N9 2 innie R7C9 + R8C7 = 15 = {69/78}

6. 45 rule on C12 2 outies R27C3 = 10 = [37/46/64/82], no 5,9, no 4,8 in R7C3, clean-up: no 3,7 in R2C2, no 2,3,6,7 in R7C2 (step 3)

7. 45 rule on C1 3 outies R189C2 = 11 = {128/137/146/236/245}, no 9

[The next step may be a bit of overkill, but it looks a nice way forward.]
8. R7C6 “sees” all cells in N5 except for R6C4 -> R6C4 and R7C6 must be “clones” -> R6C4 = R7C6 = {234}, R6C3 = {789}, clean-up: no 6 in R8C6 (step 4), no 9 in R7C9 (step 5)

9. Killer quad 6,7,8,9 in R27C3, R6C3 and R89C3, locked for C3, clean-up: no 1 in R45C3
9a. 1 in C3 only in R13C3, locked for N1
9b. 6 in C3 only in R789C3, locked for N7

10. 6 in N7 only in R789C3 -> combined cage R7C23 (step 3) + R89C3 = [47]{69}/[56]{78}/[83]{69}, no 9 in R7C2, no 2 in R7C3, clean-up: no 8 in R2C3 (step 6), no 4 in R2C2

11. 1 in C3 only in R13C3
11a. 45 rule on C123 3 innies R136C3 = 13 = {139/148/157}, no 2

12. 2 in C3 only in R45C3 = {25}, locked for C3 and N4, clean-up: no 7 in R6C3 (step 11a), no 4 in R6C4, no 4 in R7C6 (step 8), no 9 in R8C7 (step 4), no 6 in R7C9 (step 5)

13. Naked pair {78} in R7C9 + R8C7, locked for N9

14. 9 in N9 only in 17(3) cage at R9C7, locked for R9, clean-up: no 6 in R8C3
14a. 17(3) cage = {269/359}, no 4
14b. Killer pair 2,5 in R9C12 and 17(3) cage, locked for R9

15. 35(6) cage at R8C4 = {146789/236789/245789/345689}, 9 locked for R8 and N8, clean-up: no 6 in R9C3
[I didn’t spot that {236789} clashes with R7C6, but it didn’t matter after step 15a.]
15a. Naked pair {78} in R89C3, locked for C3 and N7 -> R6C3 = 9, R6C4 = 2, R7C6 = 2 (step 8), R8C7 = 7 (step 4), R7C9 = 8, R89C3 = [87], clean-up: no 7 in R1C8, no 5 in R2C2, no 5 in R23C9, no 3 in R6C7, no 3,5 in R6C8

16. Naked pair {49} in R23C9, locked for C9 and N3, clean-up: no 6 in R1C89
16a. R1C89 = [87], clean-up: no 4 in R6C7

[Forgot some clean-ups from step 15a so I’ll do them now.]
17. R7C23 (step 3) = [56], R2C3 = 4 (step 6), R2C2 = 8, clean-up: no 1 in R9C12

18. Naked pair {24} in R9C12, locked for R9 and N7, clean-up: no 6 in 17(3) cage at R9C7 (step 14a)
18a. Naked pair {13} in R8C12, locked for R8 and N7 -> R7C1 = 9
18b. Naked triple {359} in 17(3) cage at R9C7, locked for R9 and N9

19. 8 in C1 only in 17(3) cage at R4C1 = {368} (only remaining combination), locked for C1 and N4 -> R8C12 = [13]

20. R3C2 = 9 (hidden single in C2), R23C9 = [94]

21. Naked pair {14} in R7C78, locked for R7 and N9

22. R7C45 = {37} = 10 -> R8C5 = 5
22a. R8C46 = {49} (hidden pair in R8)

23. R7C9 = 8 -> R56C9 = 9 = {36}, locked for C9 and N6 -> R8C89 = [62], R9C9 = 5, R4C9 = 1

24. 45 rule on N69 1 remaining innie R4C7 = 1 outie R3C8 + 5 -> R3C8 = 3, R4C7 = 8, R3C3 = 1, R1C3 = 3, R6C7 = 5, R6C8 = 7, R9C78 = [39]

25. Naked triple {126} in R123C7, locked for C7 and N3 -> R2C8 = 5, R7C78 = [41], R5C7 = 9

26. Naked triple {257} in R123C1, locked for C1 and N1 -> R1C2 = 6, R9C12 = [42]

27. R1C3 = 3 -> R12C4 = 7 = [16], R9C4 = 8, R1C7 = 2, R1C1 = 5, R23C7 = [16]

28. R4C7 = 8 -> R45C6 = 6 = [51], R45C3 = [25], R45C8 = [42], R4C2 = 7, R56C2 = [41], R9C56 = [16]

29. R1C56 = {49} = 13, R1C7 = 2 -> R2C6 = 7 (cage total)

30. R6C6 = 3 (hidden single in C6)

and the rest is naked singles.


Last edited by Andrew on Tue Aug 23, 2011 4:59 am, edited 1 time in total.

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 Post subject: Re: Assassin 218
PostPosted: Fri Aug 12, 2011 10:54 pm 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Thanks Andrew for your WT! Indeed, I did originally have your "one-trick" in mind for the cage structure so thanks for pointing it out and using it so effectively. I used a different way for this puzzle and am always very pleased that there are multiple paths. Steps 4 & 5 are the key for my way.

A218
13 steps:
This is an optimised solution so some possible eliminations are not included. However, I do try and do all clean-up as I go. Please let me know of any mistakes or clarifications needed. Thanks Andrew!

Prelims
i. 10(3)r1c3: no 8,9
ii. 15(2)n3 = {69/78}
iii. 12(2)n1: no 1,2,6
iv. 13(2)n3: no 1,2,3
v. 7(2)n4: no 7,8,9
vi. 11(2)r6c3: no 1
vii. 12(2)n6: no 1,2,6
viii. 13(4)n9: no 8,9
ix. 15(2)n7 = {69/78}
x. 6(2)n7 = {15/24}

1. "45" on n7: 2 innies r7c23 = 11 (no 1)

2. "45" on c12: 2 outies r27c3 = 10
2a. = {37}/[46/82](no 5,9) = [one of 6/7/8..]
2a. no 4,8 in r7c3
2b. no 2,3,6,7 in r7c2 (h11(2)r7c23)
2c. no 3,7 in r2c2

3. "45" on c123: 3 innies r136c3 = 13
3a. = {139/148/157/238/247/256/346} = [one of 6/7/8/9..] (no eliminations yet)

4. Killer quad 6,7,8,9 in c3 in h13(3) (step 3a), h10(2) (step 2a) and 15(2)n7; 6 locked for c3
4a. no 1 in 7(2)n4

5. 9 in c3 in 15(2)n7 = {69} or h13(3)r136c3
5a. -> in h13(3)(step 3a), combo's with 6 must also have 9 or there would be no 9 for c3
5b. -> {256/346} blocked (Locking-out cages)
5c. -> h13(3) = {139/148/157/238/247}(no 6)
5d. no 5 in r6c4

6. 6 in c3 only in n7: locked for n7

7. 13(3)n7 = {139/148/157/238/247} = [1/2..]
7a. 6(2)n7 = {15/24} = [1/2..]
7b. Killer pair 1,2: 2 locked for n7
7c. no 9 in r7c2 (h11(2)r7c23)
7d. no 8 in r2c3 (h10(2)r27c3)
7e. no 4 in r2c2

8. h10(2)r27c3 = {37}/[46] = [3/4..]
8a. -> {34} blocked from 7(2)n4
8b. 7(2)n4 = {25}: both locked for c3 & n4
8c. no 6,9 in r6c4

9. "45" on n47: 1 outie r3c2 = 1 innie r6c3
9a. no 1,2,5,6 in r3c2

10. 32(6)r3c2 = {134789/135689/145679}
10a. must use 1 & 9: both locked for c2
10b. 1 locked for n4
10c. no 5 in r9c1
10d. no 3 in r2c3
10e. no 7 in r7c3 (h10(2)r27c3)
10f. no 4 in r7c2 (h11(2)r7c23)

11. naked pair {58} in r27c2: both locked for c2

12. 6(2)n7 = {24} only: both locked for r9 and n7

13. "45" on c1: 3 outies r189c2 = 11: must have 2 for c2 and 3/7 for r8c3 = {236} only
13a. = [632] only

cracked
Cheers
Ed


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