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Nasenbaer
Master

Joined: 20 Jul 2006
Posts: 167
Location: Fellbach (Deutschland)

 Posted: Fri Oct 20, 2006 9:55 pm    Post subject: Assassin 21 This was a nice killer, harder than number 20 (just my opinion). It took some time to get the first number, but then it unfolded nicely. Here it is (in spoiler proofed tiny font): Walkthrough Assassin Killer 21 EDIT: There are some wrong assumptions (see below, thanks nd), sorry. I will look at them as soon as possible. 1. r12c4 = {13} -> locked in c4 and n2 2. r1c56 = {79} -> locked in r1 and n2 3. r89c6 = {12} -> locked in c6 and n8 4. r34c7 = {12} -> locked in c7 5. r4c89 = 6 = {15}/{24} 5a. 6(2) in n6 has 1 or 2 -> 1,2 locked in n6 and r4 6. n7 : 30(4) = {6789} -> locked in n7 7. n7 : 9(3) = {135}/{234} -> 3 locked in n7 8. 45 on n2 : r3c3+r4c6+r2c7 = 17 9. 45 on n4 : r3c2+r4c4+r7c3 = 19 -> no 1 possible 9a. -> r7c3 = {24} -> r6c3 = {13} 9b. -> n7 : 9(3) = {135} only possible combination 9c. -> r8c3 = {24} -> 2,4 locked in c3 10. from step 9 : 19(3) = {289}/{469}/{478} -> 2,4 locked in r7c3 -> not in r3c2 and r4c4 11. 10(2) in r3 : {[8]2}/{[6]4} 12. from step 8 : 17(3) = {368}/{458}/{467} (combinations with 1 and 2 not possible because of steps 3 and 5a) 13. 45 on n6 : r3c7+r6c6+r7c8 = 15 -> 1 or 2 in r3c7 -> 15(3) = {159}/{168}/{249}/{258}/{267} 14. 45 on n8 : r8c3+r6c4+r7c7 = 9 = {126}/{234} (has to have 2 or 4, see step 9c) 14a. -> locked 1 in c4 and c7 (steps 1 and 4) -> 9(3) = {234} 14b. -> locked 3 in c4 -> r6c4 = {24} -> r7c7 = 3 15. r7c6 = 7 -> r1c6 = 9 -> r1c5 = 7 16. r7c12 = {15} -> locked in r7 and n7 -> r8c2 = 3 17. 45 on r89 -> r7c9 = 6 18. r8 : 17(3) = {[2]69}/{[4]58} (step 9c) 19. r9 : 9(2) = {3[6]}/{45} 20. n58 : 15(3) = {249} only possible combination -> r6c4 = 2 20a. r7c45 = {49} -> locked in r7 and n8 21. -> r7c8 = 8, r7c3 = 2 -> r6c3 = 3, r8c3 = 4, r3c4 = 4 -> r3c3 = 6, r7c4 = 9 -> r7c5 = 4 22. -> r8c45 = {58} (only possible combination) -> locked in r8 and n8 23. -> r9c4 = 6 -> r9c5 = 3 24. -> r8c1 = 6 -> {789} locked in r9c13 -> locked in r9 25. c1 : 15(2) = {78} -> locked in c1 and n4 -> r9c1 = 9 26. from step 8 : 19(3) = {289} only possible combination -> r4c4 = 8 -> r3c2 = 9 27. -> r8c4 = 5 -> r8c5 = 8, r5c4 = 7 (leftover c4) 28. -> r5c1 = 8 -> r4c1 = 7 29. r6 : 10(2) = {46} only possible combination -> r6c1 = 4 -> r6c2 = 6 From here on it is straight forward. Here are some more steps: 30. -> r4c2 = 5 (only possible) -> r5c2 = 2 31. -> r4c3 = 9 (only possible) -> r5c3 = 1 32. r7c2 = 1 -> r7c1 = 5 33. n1 : {123} locked in r123c1 and 14(4) -> r1c2 = 8 34. r9c2 = 7 -> r9c3 = 8 35. r2c2 = 4, r1c3 = 5 -> r2c3 = 7 36. only possible: r4c5 = 6, r6c6 = 5 -> r3c6 = 8 -> r2c3 = 6, r5c5 = 9 -> r6c5 = 1 -> r5c6 = 4 -> r4c6 = 3 37. -> r3c5 = 5 -> r2c5 = 2, r2c6 = 6 -> r2c7 = 8 The rest is for you to solve on your own. Please feel free to comment on this walkthrough. PeterLast edited by Nasenbaer on Sat Oct 21, 2006 8:03 am; edited 1 time in total
nd
Hooked

Joined: 25 Jun 2006
Posts: 45
Location: Toronto

 Posted: Sat Oct 21, 2006 4:47 am    Post subject: Not a pushover puzzle but actually you can place numbers within a couple of steps using overlapping cages -- there are several instances of 2-cell cages on which you can superimpose another 2-cell cage (via 45 rule). 1. In N2, 4(2) cage = {13}, 16(2) cage = {79}. 45 rule on N3 -> R23C7 = 10 = [91|82] -> R2C56 = {2(5|6)} with 2 locked in R2 and N2 in those cells. 2. 45 rule on N1 -> R3C23 = 15. The difference between this cage & the overlapping 10(2) cage -> R3C4 = R3C2 - 5 -> R3C4 = 4 (because {123} are blocked in N2), R3C23 = [96]. 3. 45 rule on N123 -> R4C67 = [31], R4C89 = {24}, R23C7 = [82], R2C56 = [26] (because the 3(2) cage in N8 = {12}), R3C56 = {58}. R3C189 = {137}. 4. There is a naked quad in C1 on {6789} in R4589C1 (since 15(2) and 30(4) must contain only those numbers) -> R3C1 = {13} -> R3C89 = {(1|3)7}. But a 13(3) cage cannot contain {37} (since this would make the 3rd digit a 2nd 3) -> R3C1 = 3, R3C89 = {17}, R2C8 = 5, R2C9 = 9, R1C789 = {346}, R12C4 = [13]. 5. 45 rule on N9 -> R7C78 = 11. Use cage superimposition again -> R7C6 = R7C8 - 1 -> R7C6 = 7 and R7C8 = 8 (only possibility not blocked!), R7C7 = 3, R1C56 = [79], R56C8 = {39}, R1C9 = 3. 45 rule on N6 -> R6C6 = 5, R3C56 = [58], R5C6 = 4, R56C9 = [58], R56C7 = {67}, R1C78 = [46], R89C7 = {59}, R8C8 = 2, R89C6 = [12], R4C89 = [42]. 6. Mop-up. 45 rule on N5 -> R46C4 = 10 = [82], R5C4 = 7, R56C7 = [67]. 45 rule on N7 -> R78C3 = 6 = [15|24], R8C2 = 3, R7C45 = [94], R7C12 = {15}, R78C3 = [24], R6C3 = 3, R56C8 = [39], R7C9 = 6, R8C45 = [58], R8C9 = 7, R89C7 = [95], R8C1 = 6, R9C45 = [63] & you carry on....Last edited by nd on Wed Nov 01, 2006 11:35 pm; edited 7 times in total
nd
Hooked

Joined: 25 Jun 2006
Posts: 45
Location: Toronto

Posted: Sat Oct 21, 2006 5:11 am    Post subject: Re: Assassin 21

Peter--hm.... just looking at your walkthrough. There are a few problem spots I think, mostly involving assumptions about combinations. I haven't gone through the whole thing, just the first few steps.

 Nasenbaer wrote: 9. 45 on n4 : r3c2+r4c4+r7c3 = 19 -> no 1 possible

No: the cells are not consecutive so {199} is possible.

 Quote: 11. 10(2) in r3 : {[8]2}/{[6]4}

Hmm.... not sure about what the notation means here (the square brackets around only a single number). Currently the 10(2) cage = {46} or {28} but there's no indication of the order (the preceding steps don't affect these cells). Have you left out the 45 rule on N1 = R3C23 = 15 = {69|78}? This is the missing step needed.

 Quote: 12. from step 8 : 17(3) = {368}/{458}/{467} (combinations with 1 and 2 not possible because of steps 3 and 5a)

Again, a problem with combinations involving nonconsecutive/nonbuddy numbers. {566} is valid too.
sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

Posted: Sun Oct 22, 2006 6:29 am    Post subject:

Nice walk-through nd. Totally missed that quad.

I don't really get the overlapping cages though. Doesn't a simple innie/outie difference do the same thing? Is there an advantage in thinking 'overlapping cages' rather than i/o?

BTW - one little typo:
 Quote: 5. ....R7C89 = {59}
should be r89c7 = {59}

Tried to make a V2 for Assassin 21 - but couldn't solve it. But there are a huge number of logical steps in this V2 - feels like just needs one more - eg maybe the quad etc. I don't have any time to try again at the moment, but maybe someone else can. If it is solvable - it will certainly spur me on to try again.

Here is the text code for Assassin 21V2. It has the same solution, but some cages have been changed around.
3x3::k:3584:3584:4098:5123:5123:5123:5638:5638:5638:3584:4098:4098:5123:4109:4109:4109:3344:5638:3584:6675:6675:6675:4118:4118:792:3344:3344:3867:6675:4637:4637:6943:4118:792:1570:1570:3867:6675:4637:6943:6943:6943:4650:5163:3372:2605:2605:1327:3888:6943:4650:4650:5163:3372:2358:2358:1327:3888:3888:2619:2619:5163:4670:7743:2358:5185:5185:5185:5185:4165:4165:4670:7743:7743:7743:2379:2379:5185:4165:4670:4670:

Last edited by sudokuEd on Mon Aug 06, 2007 9:08 pm; edited 3 times in total
nd
Hooked

Joined: 25 Jun 2006
Posts: 45
Location: Toronto

 Posted: Sun Oct 22, 2006 7:21 am    Post subject: Ed--thanks, I've fixed the typo. There's no difference between doing things as innie/outie difference & superimposing cages; the reason why I do it the other way is that I do as much of the math in my head as possible, & it's quicker to subtract 10 from 15 than to calculate 45 - (14 + 16 + 10).
sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

 Posted: Sun Oct 22, 2006 1:07 pm    Post subject: Thanks nd - now I get it. That's a great shortcut.
Andrew
Grandmaster

Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

 Posted: Tue Oct 24, 2006 3:24 am    Post subject: sudokuEd has persuaded me to post my walkthrough since it differs in some ways from those already posted, mainly I think because of things that nd had spotted but I hadn't. Thanks sudokuEd for the changes that you suggested to me off-forum in private messages. I've also taken the liberty of including a couple of his improvements which would make the solution a lot quicker. Step 1 R12C4 = {13}, R1C56 = {79}, R34C7 = {12}, 30(4) cage in N7 = {6789}, R89C6 = {12} -> R9C45 = {36/45} Step 2 45 rule on N3 2 innies = 10 -> R2C7 = {89}, 45 rule on N7 2 innies = 6 = {15/24}, R67C3 = {14/23} -> R7C3 = {124}, R6C3 = {134}, R8C3 = {245}, 3 in N7 locked in 9(3) cage = 3{15/24}, 45 rule on R89 1 innie + 3 = 1 outie -> R8C2 + 3 = R7C9, R8C2 = {12345} -> R7C9 = {45678} Step 3 45 rule on N1 2 innies = 15 = {69/78}, R3C23 -> [78]/[96] because R3C4 cannot be 1 or 3, R3C4 = {24} Step 4 45 rule on C123 2 outies – 8 = 1 innie, min. R8C3 = 2 -> min. R34C4 = 10 -> R4C4 = min.6, 5 is only odd digit in R8C3 -> R4C4 cannot be 7 -> R4C4 = {689} Step 5 45 rule on C12 1 innie + 4 = 1 outie -> R2C2 + 4 = R9C3 -> R2C2 = {2345} -> R12C3 = min.11 -> 1 in N1 locked in 14(4) cage Step 6 R7C67 = {37/46} (because 1 and 2 locked in C6 and also in C7) 45 rule on N9 2 innies = 11, R7C67 = 10 -> R7C6 + 1 = R7C8 -> R7C8 = {4578} 1 and 2 in N9 must be in R89 (one in each row), killer pair with R89C6 = {12} so no other 1/2 in R89 -> R8C2 = {345}, R8C3 = {45}, R7C3 = {12}, R6C3 = {34} Step 7 9 in R7 locked in R7C45 -> 15(3) cage in N58 = {249} (cannot be {159} because 1 blocked in C4 and N8) -> R6C4 = 2 -> R7C45 = {49}, R7C67 = {37}, R7C8 = 8 (from second line of step 6, 4 blocked from R7) -> R7C67 = [73], R8C2 = 3, R7C12 = {15}, R7C9 = 6, R7C3 = 2, R6C3 = 3, R8C3 = 4, R3C34 = [64], R3C2 = 9, R9C45 = [63], R8C45 = {58}, R8C1 = 6 Step 8 R45C1 = {78}, R9C1 = 9, 9 in N4 must be in C3 (the 10(2) cage in N4 cannot be [91] because the 1 in C2 is either in 14(4) cage in N1 or in 9(3) cage in N7) -> R4C4 = 8, R45C3 = {19}, R45C2 = {25}, R45C1 = [78], R6C12 = [46], R7C12 = [51], R123C1 = {123}, R1C2 = 8, R9C23 = [78], R2C2 = 4, R12C3 = [57] Step 9 4 and 6 in N3 locked in R1, rest of 22(4) cage in N3 must be {39} -> R2C9 = 9, R1C789 = {346}, R2C7 = 8, R34C7 = [21], 6 in N2 locked in R2 -> R2C56 = [26], R1C1 = 2, R12C4 = [13], R23C1 = [13], R2C8 = 5, R3C89 = {17}, R3C56 = {58}, R4C6 = 3, R45C3 = [91], R4C89 = {24}, R45C2 = [52], R4C5 = 6 Step 10 45 rule on C89 1 innie + 2 = 1 outie -> R8C8 + 2 = R1C7 -> R8C8 = 2 (4 blocked in R8) -> R1C7 = 4, R1C89 = [63], R89C7 = [95], R89C6 = [12], R8C9 = 7, R9C89 = {14}, R56C9 = [58], R56C8 = [39], R56C7 = [67], R6C6 = 5, 27(5) cage in N5 = {14679} and carry on …. the rest is simple elimination sudokuEd has also pointed out to me that after my step 3, innies/outies on R123 give direct placements for R3C2 and R4C67 after which R3C34 follow immediately from step 3. He has also pointed out to me that after my step 6, innies/outies on R789 give direct placements for R6C34 and R7C8. Alternatively, as I noticed while correcting my walkthrough, the innies/outies from my step 4 can be used after my step 6 to fix R4C4.
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Mon Aug 06, 2007 4:52 pm    Post subject: Hi all One down, 9 to go. Walk-through Assassin 21V2 1. 14(4) at R1C1 = {1238/1247/1256/1346/2345}: no 9 2. R34C7 = {12} -->> locked for C7 3. R45C1 = {69/78}: no 1,2,3,4,5 4. R4C89 = {15/24}: no 3,6,7,8,9 4a. Killer Pair {12} at R4C7 + R4C89 locked for R4 and N6 5. 20(3) at R5C8 = {389/479/569/578}: no 1,2 6. R56C9 = {49/58/67}: no 3 7. R6C12 = {19/28/37/46}: no 5 8. R67C3 = {14/23}: no 5,6,7,8,9 9. 9(3) at R7C1 = {125/135/234}: no 7,8,9 10. R7C67 = [19/28]/{37/{46}: no 5; R7C6: no 8,9 11. 30(4) at R8C1 = {6789} -->> locked for N7 11a. 9(3) at R7C1 = {135/234} -->> 3 locked for N7 11b. Clean up: R6C3: no 2 12. R9C45 = {18/27/36/45}: no 9 13. 45 on N1: 2 innies: R3C23 = 15 = {69/78}: no 2,3,4,5 14. 45 on N3: 2 innies: R23C7 = 10 = [82/91]: R2C7 = {89} 15. 45 on N6: 2 outies and 1 innie: R4C7 + 12 = R6C6 + R7C8: R4C7 = {12} -->> R6C6+R7C8 = 13/14 = {49/58/67/59/68}: no 1,2,3 16. 45 on N7: 2 innies: R78C3 = [15/24/42]: R8C3: no 1 17. 45 on N8: 3 outies: R6C4 + R7C7 + R8C3 = 9 = [432/342/234/162/144/135] -->> R6C4 = {1234}; R7C7 = {346} 17a. Clean up: R7C6 = {467} 18. 45 on N9: 2 innies: R7C78 = 11 = [38/47/65] : R7C8 = {578} 19. 45 on C789: 2 outies and 1 innie: R78C6 = R2C7: R2C7 = {89} -->> R67C6 = 12/13 = [84/57/94/76/67]: R6C6: no 4 20. 45 on R789: 2 outies and 1 innie: R6C34 + 3 = R7C8: Min R6C34 = 3 -->>Min R7C8 = 6: R7C8 = {78} -->>R6C34 = 4/5 = {13}/{14}/[32] 20a. Clean up: R7C7: no 6; R7C6: no 4 20b. R67C6 = [57/67/76]: no 8,9; 7 locked for C6 21. 45 on R123: 4 outies: R5C2 + R4C267 = 11 = [1]{36}[1]/[1]{45}[1]/[1]{35}[2]/[2]{35}[1]/[2]{34}[2]/[3][431] -->> R5C2 = {123}; R4C26 = {3456} 22. 45 on N23: 1 innie and 2 outies: R3C4 = R4C67: min R4C67 = 4 -->> R3C4: no 2,3; Max R4C67 = 8 -->> R3C4: no 9 23. 45 on N1: 3 outies: R3C4 + R45C2 = 11(no duplicates, all in same cage) = [4][61]/[4][52]/[5][42]/[6][41]/[6][32]/[7][31] -->>R3C4 : no 8; R5C2: no 3 24. 16(3) at R2C5 needs one of {89} in R2C7 -->> R2C56 = 7/8 -->> R2C56 no 8,9 25. 20(3) at R5C8 = {39}[8]/{49}[7]/{58][7]/{57}[8]: no 6 26. 45 on R12: R3C189 = 11 = {137/146/236/245}= {1|2..}({128} blocked by R3C7): no 8,9 26a. Killer Pair {1|2..} in R3C189 + R3C7 -->> locked for R3 27. Hidden Killer Pair {89} in R3; R3C23 needs one of {89}; R3C56 needs one of {89} 27a. 16(3) at R3C5 = {349/358}: no 6,7; 3 locked 3 within cage -->> R12C6: no 3 28. R7C678 = [738/647] -->> 7 locked for R7 29. 9(3) at R7C1: [43][2] blocked by R7C7 -->> R8C2: no 2 30. 45 on R89: 1 innie and 1 outie: R8C2 + 3 = R7C9 -->> R8C2 = {135}; R7C9 = {468} 31. R7C6789 = [738]{46}/[6478] -->> 8 locked within R7C89 for R7 and N9 32. 9 in R7 lock within 15(3) at R6C4 -->> 15(3) = [1]{59}/[2]{49}/[4]{29} -->> R6C4 = {124}; R7C45 = {29/49/59}: no 1,3,6; 9 locked for N8 32a. 1 in R7 locked for N7 32b. Clean up: R7C9: no 4(step 30) 33. Only place for {12} in N2 is R2C56 within 16(3) cage at R2C5 or 20(4) at R1C4 33a. 20(4) at R1C4: {1289} blocked by R3C56, so neither cage can contain both {12}, so both need one of {12} 33b. 16(3) at R2C5 = {16}[9]/{25}[9]/[718]/{26}[8]: no 3,4 33c. 20(4) = {1379/1469/1478/1568/2369/2378/2459/2468} 34. 16(3) at R8C7 = {69}[1]/{59}[2]/{457}:{349} blocked by R7C7; {367} blocked by R7C89: no 3; R8C8: no 6,9 35. R7C789 = [478/386] 35a. 18(4) at R7C9 = [6]{129/147}/[8]{136/235}(others blocked by R7C78) 36. 45 on N4: 3 innies and 1 outie: R4C4 + 2 = R45C2 + R6C3; Min. R45C2 + R6C2 = 6 -->> Min. R4C4 = 4: no 3 37. 45 on N2356: 3 innies and 1 outie: R7C8 + 6 = R346C4: R7C8 = {78} -->> R346C4 = 13/14 -->> R346C4 = {47}[2]/{56}[2]/[48][1/2]/{57}[1/2]/[49][1]/[58][1]/{67}[1] -->> R6C4: no 4 37a. Clean up: R7C45: no 2 37b. 2 in R7 locked for N7 37c. Clean up: R7C3: no 4; R6C3: no 1 38. 1 and 8 in N8 contain with 20(5) at R8C3 and 9(2) at R9C4; 9(2) needs either both {18} or neither, so the same goes for 20(5) at R8C3 38a. 20(5) at R8C3 needs one of {45} in R8C3 -->> 20(5) = {12458/23456}: no 7; 2 locked in 20(5) cage for N8 38b. Clean up: R9C45: no 7 38c. R7C6 = 7(hidden); R7C789 = [386] 38d. Clean up: R56C9 = {49/58}= {5|9..}: no 7; R56C8 = {39/57} = {5|9..}: no 4 39. Killer Pair {59} in N6 within R56C8 + R56C9 -->> locked for N6 39a. Clean up: R4C89 = {24} -->> locked for R4 and N6 39b. R34C7 = [21] 39c. Clean up: R56C9 = {58} -->> locked for C9 and N6 39d. R56C7 = {67} -->> locked for C7, N6 and 18(3) cage at R5C7 39e. R6C6 = 5; R56C9 = [58]; R4C6 = 3 39f. R3C4 = 4(step 22); R4C56 = [58](last remaining combo within 16(3) at R3C5) 39g. R8C2 = 3(hidden) 39h. R56C8 = {39} -->> locked for C8 40. R3C23 = {69}(step 13) -->> locked for N1 and R3 and 26(5) at R3C2 40a. R45C2 = [52] 41. 45 on C789: 3 innies: R256C7 = 21 = [8]{67} -->> R2C7 = 8 42. 13(3) at R2C8 = {157} (last remaining combo) -->> R2C8 = ; R3C89 = {17} -->> locked for R3 and N3 42a. R3C1 = 3; R1C8 = 6(hidden) 43. 20(5) at R8C3 = {12458}: no 6 43a. R8C1 = 6(hidden) 43b. Naked Triple {789} in R9C123 -->> locked for R9 43c. Naked Triple {124} in R9C689 -->> locked for R9 43d. R9C7 = 5 44. R45C1 = {78}(last combo) -->> locked for C1 and N4 44a. R9C1 = 9 44b. Clean up: R6C2: no 1,4 44c. Naked Pair {69} in R4C3 + R6C2 -->> locked for N4 45. 16(3) at R1C3 = {178/457}: no 2 45a. R7C3 = 2(hidden); R6C3 = 3; R8C3 = 4(step 16) And naked singles till the end. greetings Para
Andrew
Grandmaster

Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

 Posted: Tue Feb 23, 2010 4:34 am    Post subject: I've been working on some of my unfinished puzzles on Richard's site and decided to have a go at some of the ones on this site too. I only realised later that A21 V2 was one of the ones in the Unsolvables thread. I found that I hadn't even started this one; not sure why as I managed to solve it in less than a day after I started, one of my quickest variants. I think there must be a fairly narrow solving path; Para and I used similar key steps. I'll rate my walkthrough at least Hard 1.5 because of step 24; if I'd found Para's step 37 then it would have simplified my step 24. Here is my walkthrough for A21 V2. Prelims a) R34C7 = {12} b) R45C1 = {69/78} c) R4C89 = {15/24} d) R56C9 = {49/58/67}, no 1,2,3 e) R6C12 = {19/28/37/46}, no 5 f) R67C3 = {14/23} g) R7C67 = {19/28/37/46}, no 5 h) R9C45 = {18/27/36/45}, no 9 i) 20(3) cage at R5C8 = {389/479/569/578}, no 1,2 j) 9(3) cage in N7 = {126/135/234}, no 7,8,9 l) 14(4) cage in N1 = {1238/1247/1256/1346/2345}, no 9 m) 30(4) cage in N7 = {6789} Steps resulting from Prelims 1a. Naked pair {12} in R34C7, locked for C7, clean-up: no 8,9 in R7C6 1b. Naked quad {6789} in 30(4) cage, locked for N7 1c. Naked quad {6789} in R4589C1, locked for C1, clean-up: no 1,2,3,4 in R6C2 1d. Killer pair 1,2 in R4C7 and R4C89, locked for R4 2. 45 rule on N1 2 innies R3C23 = 15 = {69/78} 2a. R3C23 = 15 -> R3C4 + R45C2 = 11 = {128/137/146/236/245}, no 9 2b. 8 of {128} must be in R4C2 -> no 8 in R3C4 + R5C2 3. 45 rule on N3 2 innies R23C7 = 10 = [82/91] 3a. 16(3) at R2C5 cannot have both of 8,9 -> no 8,9 in R2C56 4. 45 rule on N7 2 innies R78C3 = 6 = [15/24/42], no 3, no 1 in R8C3, clean-up: no 2 in R6C3 5. 45 rule on N9 2 innies R7C78 = 11 = {38/47}/[65], no 9, no 6 in R7C8, clean-up: no 1 in R7C6 6. 45 rule on R789 1 innie R7C8 = 2 outies R6C34 + 3 6a. Min R6C34 = 3 -> no 3,4,5 in R7C8, clean-up: no 6,7,8 in R7C7 (step 5), no 2,3,4 in R7C6 6b. R7C8 = {78} -> R6C34 = 4,5 = {13/14}/[32], no 5,6,7,8,9 6c. Max R6C4 = 4 -> min R7C45 = 11, no 1 6d. 20(3) cage at R5C8 = {389/479/578} (cannot be {569} because R7C8 only contains 7,8), no 6 7. 9(3) cage in N7 = {135/234} 7a. 2 of {234} must be in R7C12 (R7C12 cannot be {34} which clashes with R7C7), no 2 in R8C2 8. 45 rule on R12 1 innie R2C8 = 1 outie R3C1 +2 -> R2C8 = {34567} 9. 45 rule on R89 1 outie R7C9 = 1 innie R8C2 + 3 -> R7C9 = {4678} 9a. 1 in R7 only in R7C123, locked for N7, clean-up: no 4 in R7C9 9b. Naked triple {678} in R7C689, locked for R7 9c. 8 in R7 only in R7C89, locked for N9 10. 9 in R7 only in R7C45, locked for N8 10a. 15(3) cage at R6C4 = {159/249}, no 3 11. 45 rule on C789 2 outies R67C6 = 1 innie R2C7 + 4 11a. R2C7 = {89} -> R67C6 = 12,13 = [57/67/76], 7 locked for C6 12. 45 rule on N23 2 outies R4C67 = 1 innie R3C4 12a. Min R4C67 = 4 -> min R3C4 = 4 12b. Max R3C4 = 7 -> max R4C67 = 7, max R4C6 = 6 13. R3C4 + R45C2 (step 2a) = {137/146/236/245} (cannot be {128} because 1,2 only in R5C2), no 8 13a. 1,2 only in R5C2 -> R5C2 = {12} 13b. Min R3C4 + R5C2 = 5 -> max R4C2 = 6 14. 45 rule on C89 1 outie R1C7 = 1 innie R8C8 + 2, no 9 in R8C8 15. 45 rule on C12 3(2+1) outies R39C3 + R3C4 = 1 innie R2C2 + 14 15a. Min R39C3 + R3C4 = {67} + 4 = 17 -> min R2C2 = 2 16. 16(3) cage in N9 = {169/259/457} (cannot be {349} which clashes with R7C7, cannot be {367} which clashes with R7C89, Killer ALS block), no 3, clean-up: no 5 in R1C7 (step 14) 16a. 1 of {169} must be in R8C8 -> no 6 in R8C8, clean-up: no 8 in R1C7 (step 14) 17. R67C3 = [14/32/41], R78C3 = [15/24/42] -> R678C3 = [142/324/415], 4 locked for C3 18. Max R4C4 = 9 -> min R45C3 = 9 must contain at least one of 6,7,8,9 18a. Killer quad 6,7,8,9 in R45C1, R45C3 and R6C2, locked for N4 18b. Max R45C3 = {59} = 14 (cannot contain two of 6,7,8,9 which would clash with R45C1 + R6C2, ALS block) -> min R4C4 = 4 18c. 18(3) at R4C3 = {189/279/369/378/459/468/567} 18d. 9 of {189/279/369/459} must be in R4C34 (R4C34 cannot be {45} which clashes with R4C89, cannot be {36} which clashes with R4C2689, Killer ALS block), no 9 in R5C3 [This step was interesting but probably not essential.] 18e. 5 in N4 only in R4C23 + R5C3, CPE no 5 in R4C4 19. 14(4) cage in N1 = {1238/1247/1256/1346/2345} 19a. 6,7,8 of {1238/1247/1256/1346} must be in R1C2 -> no 1 in R1C2 20. 45 rule on N6 2(1+1) outies R6C6 + R7C8 = 1 innie R4C7 + 12 20a. R4C7 = {12} -> R6C6 + R7C8 = 13,14 = [58/68/77] (cannot be [67] which clashes with R67C6, CCC) 20b. R6C6 + R7C8 = [58/68/77], R67C6 = [57/67/76] (step 11a) -> R7C68 = [78/78/67], 7 locked for R7, clean-up: no 4 in R8C2 (step 9) [Para got this a bit more directly by using the overlap of R7C67 and R7C78. Ed pointed out that step 20a can be simplified to R4C7 = {12} -> R6C6 + R7C8 = 13,14 = [58/68/77] (cannot be [67] which clashes with R7C6). Thanks Ed!] 21. 45 rule on R12 3 outies R3C189 = 11 = {137/146/236/245} (cannot be {128} which clashes with R3C7), no 8,9 21a. Killer pair 1,2 in R3C189 and R3C7, locked for R3 [I ought to have spotted this step much earlier.] 22. Hidden killer pair 8,9 in R3C23 and R3C56 for R3, R3C23 contains one of 8,9 -> R3C56 must contain one of 8,9 22a. 16(3) cage at R3C5 = {349/358} (cannot be {367/457} which don’t contain 8 or 9), no 6,7, CPE no 3 in R12C6 23. R4C67 = R3C4 (step 12) 23a. Max R4C67 = 6 (cannot be [52] which clashes with R4C89) -> max R3C4 = 6 23b. Killer triple {345} in R4C2, R4C6 and R4C89, locked for R4 23c. Naked quad {6789} in R45C1, R4C3 and R6C2, locked for N4 24. Hidden killer pair 1,8 in 20(5) cage at R8C3 and R9C45 for N8, R9C45 must contain both of 1,8 or 20(5) cage must contain both of 1,8 24a. 20(5) cage at R8C3 = {12368/12458/23456} (cannot be {12467/13467} which only contain one of 1,8), no 7 24b. {12368/23456}, 6 locked for N8 => R7C6 = 7 {12458} => 3 in N8 only in R9C45 = {36}, locked for N8 => R7C6 = 7 24c. -> R7C6 = 7, R7C7 = 3, R7C89 = [86], clean-up: no 7 in R56C9, no 1 in R8C8 (step 14), no 2 in R9C45 25. 6 in N6 only in R56C7, locked for 18(3) cage at R5C7 -> R6C6 = 5, R56C7 = 13 = {67}, locked for C7 and N6, clean-up: no 8 in R5C9 26. R8C2 = 3 (hidden single in N7) 26a. R2C7 = 8 (hidden single in C7), R2C56 = 8 = [26/62/71], no 3,4, no 1,5 in R2C5 27. 5 in C7 only in R89C7, locked for N9 27a. 16(3) cage in N9 = {259/457} 27b. 2,7 only in R8C8 -> R8C8 = {27} 28. R7C8 = 8 -> R56C8 = 12 = {39} (only remaining combination), locked for C8 and N6, clean-up: no 4 in R56C9 28a. R56C9 = [58], clean-up: no 1 in R4C89, no 2 in R6C1 28b. Naked pair {24} in R4C89, locked for R4 -> R4C2 = 5, R4C6 = 3, R34C7 = [21] 29. R3C4 + R45C2 (step 2a) = {245} (only remaining combination) -> R3C4 = 4, R5C2 = 2, clean-up: no 5 in R9C5 30. 16(3) cage at R3C5 (step 22a) = {358} (only remaining combination) -> R3C56 = [58] 31. R3C23 (step 2) = {69} (only remaining combination), locked for R3 and N1 32. 7 in R3 only in R3C89, locked for N3 32a. 13(3) cage in N3 = {157} (only remaining combination) -> R2C8 = 5, R3C89 = {17}, locked for R3 and N3 -> R3C1 = 3, clean-up: no 7 in R6C2 32b. R1C8 = 6 (hidden single in N3) 32c. Naked pair {69} in R36C2, locked for C2 33. R7C2 = 1 (hidden single in C2), R7C1 = 5 (cage sum) 33a. Naked pair {24} in R78C3, locked for C3 34. 15(3) cage at R6C4 (step 10a) = {249} (only remaining combination) -> R6C4 = 2, R7C45 = [94], R7C3 = 2, R6C3 = 3, R8C3 = 4, R5C3 = 1, R6C1 = 4, R6C2 = 6, R3C23 = [96], R2C23 = [47], R1C23 = [85], R9C2 = 7, clean-up: no 1 in R2C6 (step 26a), no 9 in R45C1, no 5 in R9C4 35. Naked pair {78} in R45C1, locked for C1 and N4 -> R4C3 = 9, R4C4 = 8 (cage sum), clean-up: no 1 in R9C5 36. Naked pair {26} in R2C56, locked for R2 and N2 -> R2C1 = 1, R2C4 = 3, R2C9 = 9, R1C7 = 4 36a. R89C7 = {59} -> R8C8 = 2 (step 27a) and the rest is naked singles.Last edited by Andrew on Sun Mar 07, 2010 7:19 am; edited 1 time in total
sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

Posted: Fri Mar 05, 2010 5:32 am    Post subject:

 sudokuEd wrote: If it is solvable - it will certainly spur me on to try again.
This has been bugging me for years since I like to only post variants I have solved first. Many thanks to Para and Andrew for solving A21V2. They found a really interesting both/neither move to set up their solutions (Para's step 38; Andrew's step 24). I missed that but found an equally interesting alternative way in another part of the grid (step 11). I think it is a Killer Empty Rectangle. Some extra clarification added to this step (thanks Andrew).

A21 v2
This is an optimised WT so only the essential steps have been included.
Prelims as per Andrew's WT

1. 3(2)n3 = {12}: both locked for c7
1a. no 8,9 in r7c6

2. "45" on n9: 2 innies r7c78 = 11
2a. no 9 in r7c78; no 6 in r7c8
2b. no 1 in r7c6

3. "45" on r789: 2 outies r6c34 + 3 = 1 innie r7c8
3a. -> min r7c8 = 6
3b. -> r6c34 = 4/5 (no 5..9)
3c. r7c7 = (34) (h11(2))
3d. r7c6 = (67)

4. naked quad {6789} in n7: Locked for n7

5. "45" on r89: 1 outie r7c9 - 3 = 1 innie r8c2
5a. r7c9 = (4..8)

6. 9 in r7 only in 15(3)n5: 9 locked for n8
6a. 15(3) = 9{15/24}(no 3,6,7,8)
6b. min. r7c45 = 11 (no 1)

7. hidden triple 6,7,8 in r7c689
7a. r7c9 = (678)
7b. 8 locked for n9

8. "45" on n9: 1 outie r7c6 + 1 = 1 innie r7c8
8a. = [67/78]: 7 locked for r7

9. "45" on n3: 2 innies r23c7 = 10 = [91/82]

10. "45" on c89: 3 outies r189c7 = 18
10a. = {369/378/459/468/567}
10b. no 7 in r1c7 since no 8 in r89c7 for {378} and cannot be [7]{56} in overlapping the 16(3)n9 {can't be {56}[5]}

11. r7c89 = [78/86] = [7/6] -> {67} blocked from 13(2)n6 since there is no 7 in r123c7 (Killer Empty Rectangle?)
11a. another way of seeing this:
i. {67} in r56c9 -> r7c9 = 8 -> r7c8 = 7:
ii. {67} in r56c9 -> 7 in n3 only in c8, 7 locked for c8: but this means two 7s in c8
11b. yet another way of seeing this: (this one courtesy of Andrew)
i. If r7c89 = [78] -> 7 in n3 in c9 -> no 7 in 13(2) at r56c9 -> {67} blocked
ii. If r7c89 = [86] -> {67} blocked from 13(2) at r56c9

12. 20(3)n6 must have 7/8 for r7c8 = {389/479/578}(no 6)

13. 6 in n6 only in c7: locked for c7 and not in r6c6

14. 16(3)n9 = {259/457}(no 1,3,6)({169} blocked by no 1,6 in r89c7; {349} blocked by r7c7; {367} blocked by [3/7..] in h11(2) at r7c78 step 2; or alternatively, by [6/7..] at r7c89 step 11)
14a. 5 locked for n9

15. "45" on c89: 1 outie r1c7 - 2 = 1 innie r8c8
15a. r1c7 = (49), r8c8 = (27)

16. h18(3)r189c7 = {459} only: all locked for c7

It's cracked.

Cheers
Ed
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