Assassin 55
Assassin 55
Ruud has certainly ramped up the difficulty level this week. 20 steps so far and very little progress. This one may well take all week - I don't think I'll be attempting Ruud's promised V2!
Hi all
What a work to fill 81 little cells. But in the end i liked the result. I keep missing obvious things but i get back to those some steps later. There's some nice steps in there. Especially one interesting 45 move.
I am fearing this monster Ruud has set up for us though.
Walkthrough Assassin 55
1. 10(3) in R1C1 = {127/136/145/235}: no 8, 9
2. 19(3) in R1C3 = {289/379/469/478/568}: no 1
3. R12C5 and R23C8 = {49/58/67}: no 1,2,3
4. R23C2 = {16/25/34}: no 7,8,9
5. 21(3) in R3C1 = {489/579/678}: no 1,2,3
6. 11(3) in R5C6 = {128/137/146/236/245}: no 9
7. R89C5 = {19/28/37/46}: no 5
8. 22(3) in R8C7 = {589/679}: no 1,2,3,4; 9 locked in 22(3) cage: R9C89: no 9
9. 45 on R5: 3 innie: R5C159 = 22 = {589/679}: no 1,2,3,4; 9 locked in R5C159 for R5
10. 45 on R89: 4 innies: R8C2468 = 13 = {1237/1246/1345}: no 8,9; 1 locked in R8C2468 for R8
10a. Clean up: R9C5: no 9
11. 45 on C89: 3 innies: R456C8 = 7 = {124} -->> locked for C8 and N6
11a. Clean up: R23C8: no 9
12. 45 on C12: 3 innies: R456C2 = 19 = {289/379/469/478/568}: no 1
13. 45 on N9: 2 outies and 1 innie: R6C9 + R9C6 = R7C7 + 14; Max R6C9 + R9C6 = 18 -->> Max R7C7 = 4; Min R7C7 = 1 -->> Min R6C9 + R9C6 = 15 (Min 6 and Max 9) Conclusion: R6C9 = {6789}, R7C7 = {1234}, R9C6 = {6789}
13a. R6C9 + R9C6 see all 9’s in N9: can’t both be 9 -->> Max R6C9 + R9C6 = 17 -->> R7C7: no 4
13b. 4 in N9 locked for C9
14. 45 on N1: 2 innies and 1 outie: R3C13 = R1C4 + 9; Max R3C13 = 17 -->> Max R1C4 = 8; Min R1C4 = 2 -->> Min R3C3 = 11; Conclusion: R3C13: no 1; R1C4: no 9
14a. 45 on N1: 4 innies: R123C3 + R3C1 = 28 = {4789/5689}: no 2,3
Hmm, missed these.
15. 45 on C1234: 2 innies: R46C4 = 17 = {89} -->> locked for C4 and N5
15a. Clean up: R5C19: no 5(no {89} in R5C5)(step 9)
16. 45 on C6789: 2 innies: R46C6 = 3 = {12} -->> locked for C6 and N5
17. 23(5) in R3C5 needs one of {89} in R4C4, one of {12} in R4C6 and one of {567} in R5C5 -->> 23(5) = {12479/12569/12578/13469/13478/13568/23459/23468}: can only have one of {89} and it goes in R4C4 -->> R3C5: no 8,9
18. 19(4) in R6C4 can only have one of {89} and it goes in R6C4 -->> R7C5: no 8,9
19. {89} in C5 locked in 13(2) in R1C5 and 10(2) in R8C5; they can’t have both so both need one of {89} -->> R12C5 = {49/58}: no 6,7; R89C5 = {28}/[91]: no 3,4,6,7
20. 23(5) in R3C5 only has {45} in R345C5. Can’t have both {45} because it clashes with R12C5 -->> 23(5): no {23459}; 23(5) = {12479/12569/1258/12469/12478/13568/23468}
20a. Killer Pair {45} in R12C5 + R345C5 -->> locked for C5
21. 45 on N9: 4 innies and 1 outie: R7C789 + R8C8 = R9C6 + 5: Max R9C6 = 9 -->> Max R7C789 + R8C8 = 14: Conclusion: R7C789 + R8C8: no 9;
21a. R1C8 = 9(hidden)
21b. Clean up: R2C5: no 4
21c. 16(3) in R1C8 = 9{16/25}: no 3,7,8
22. Killer Pair {56} in R12C9 + R23C8 -->> locked for N3
23. Follow up 45 on N9: R7C789 + R8C8= 11/12/13/14
23a. When 11: R78C8 = {35}, R7C79 = {12}
23b. When 12: R78C8 = {36}, R7C79 = {12}
23c. When 13: R78C8 = {35}, R7C79 = [14]; R7C78 = {37}, R7C79 = {12}
23d. When 14: R78C8 = {35}, R7C79 = [24]; R78C8 = {36}, R7C79 = [14]; R78C8 = [83], R7C79 = {12}(no {56} clashes with R23C8)
23e. Conclusion: R7C7: no 3; R7C9: no 3,5,6,7,8; R78C8 = {35/36/37}/[83]: 3 locked for N9
24. 3 in C9 locked in 17(3) cage in R3C9 -->> 17(3) = {359/368}: no 1,2,7
24a. Killer Pair {56} in R12C9 + 17(3) cage in R3C9 -->> locked for C9
25. 18(3) cage in R8C9 = [981/972/954]/{8[6]4}: R89C9: no 7
25a. R6C9 = 7(hidden)
25b. Clean up: R9C6: no 6,7(step 13(R6C9 + R9C6 is minimal 15)
25c. Killer Pair {89} in R89C5 + R9C6 -->> locked for N8
26. 26(5) in R6C7 needs 2 of {124} in R6C8 + R7C7 -->> 26(5) = {12689/14579/14678/23489/24569/24578}: Only 2 of {124} so no {124} anywhere else in 26(5) -->> R78C6: no 4; 26(5) needs one of {89} -->> only place for {89} is R6C7: R6C7 = {89}
26a. Naked Pair {89} in R6C47 -->> locked for R6
27. 19(4) in R6C4 needs one of {89} in R6C4, one of {36} in R6C5 and one of {12} in R6C6 -->> 19(4) = {1369/1378/2368}: Needs one of {12}, this goes in R6C6 so nowhere else in 19(4) cage: R7C5: no 1,2; 3 locked in R67C5 -->> locked for C5
28. Hidden Killer Pair {12} in C5 in R3C5 + R89C5: R3C5 = {12}
28a. 23(5) in R3C5 in needs both {12} in R3C5+ R4C6 -->> 23(5) = {12479/12569/12578}
29. 4 in N8 locked for C4
29a. 5 in R6 locked for N4
29b. 9 in R7 locked for N7
30. 11(3) in R5C6 = [731]/[461]/{36}[2]/[452]: R5C6: no 5; R5C7: no 8; R5C8: no 4
Missed something at step 26.
31. 26(5) in R6C7 can’t have both {89} -->> 26(5) = {14579/14678/24569/24578}: 4 locked in 26(5) -->> R6C8 = 4(only place in cage): R78C6 = {56/57/67}: no 3
31a. Naked Pair {12} in R4C68 -->> locked for R4
32. 45 on C9: 3 outies: R789C8 = 16; R78C8 = {35/36}/[83](step 23) -->> R9C8 = {578}
32a. 18(3) in R8C9 = [981/972/954] -->> R8C9 = 9; R9C9: no 8
32b. R9C6 = 9(hidden); R2C5 = 9; R1C5 = 4
32c. Clean up: R9C5: no 1
32d. R89C5 = {28} -->> locked for C5 and N8
32e. R3C5 = 1; R4C6 = 2; R45C8 = [12]; R6C6 = 1; R7C7 = 2(step 13)
32f. R5C3 = 1(hidden); R5C6 = 4(hidden); R5C7 = 5; R4C5 = 5(hidden)
32g. Clean up: R89C7 = {67} -->> locked for C7 and N9
33. 17(3) in R3C9 = {368} -->> locked for C9
33a. Clean up: R12C9 = {25} -->> locked for N3
33b. Clean up: R23C8 = {67}
34. 26(5) in R6C7 = 24{569/578} -->> 5 locked in R78C6 for C6 and N8
35. 12(3) in R5C2 = [813](last possible combination)
35a. R6C5 = 6; R5C5 = 7; R5C9 = 6; R5C1 = 9; R4C4 = 8; R6C4 = 9; R7C5 = 3
35b. R34C9 = [83]; R46C7 = [98]; R8C8 = 3(hidden); R3C3 = 9(hidden); R7C2 = 9(hidden)
36. 23(5) in R2C6 = {13469}(last possible combination) -->> R3C7 = 4(only place in cage); R23C6 = {36} -->> locked for C6 and N2
36a. R1C6 = 8(hidden); R2C3 = 8(hidden)
37. 19(3) in R1c3 = [658](last possible combination): R1C3= 6; R1C4 = 5
37a. R12C9 = [25]
38. 10(3) in R1C1 = {127}(last possible combination) -->> R2C1 = 2; R1C12 = {17} -->> locked for R1 and N1
38a. R12C7 = [31]; R23C4 = [72]; R23C8 = [67]; R23C6 = [36]; R23C2 = [43]; R3C1 = 5
38b. R4C1 = 7; R4C23 = [64]; R1C12 = [17]; R6C1 = 3
38c. R9C3 = 3(hidden); R9C7 = 7(hidden); R8C7 = 6;
39. R6C23 = {25} -->> locked for 19(5) cage in R6C2: 19(5) = {12457}: no 6
And now all singles.
39a. R7C3 = 7; R78C6 = [57]; R7C89 = [81]; R9C89 = [54]; R78C4 = [41]
39b. R7C1 = 6; R8C2 = 2; R6C23 = [52]; R9C12 = [81]; R8C1 = 4; R8C3 = 5
39c. R89C5 = [82]; R9C4 = 6
greetings
Para
What a work to fill 81 little cells. But in the end i liked the result. I keep missing obvious things but i get back to those some steps later. There's some nice steps in there. Especially one interesting 45 move.
I am fearing this monster Ruud has set up for us though.
Walkthrough Assassin 55
1. 10(3) in R1C1 = {127/136/145/235}: no 8, 9
2. 19(3) in R1C3 = {289/379/469/478/568}: no 1
3. R12C5 and R23C8 = {49/58/67}: no 1,2,3
4. R23C2 = {16/25/34}: no 7,8,9
5. 21(3) in R3C1 = {489/579/678}: no 1,2,3
6. 11(3) in R5C6 = {128/137/146/236/245}: no 9
7. R89C5 = {19/28/37/46}: no 5
8. 22(3) in R8C7 = {589/679}: no 1,2,3,4; 9 locked in 22(3) cage: R9C89: no 9
9. 45 on R5: 3 innie: R5C159 = 22 = {589/679}: no 1,2,3,4; 9 locked in R5C159 for R5
10. 45 on R89: 4 innies: R8C2468 = 13 = {1237/1246/1345}: no 8,9; 1 locked in R8C2468 for R8
10a. Clean up: R9C5: no 9
11. 45 on C89: 3 innies: R456C8 = 7 = {124} -->> locked for C8 and N6
11a. Clean up: R23C8: no 9
12. 45 on C12: 3 innies: R456C2 = 19 = {289/379/469/478/568}: no 1
13. 45 on N9: 2 outies and 1 innie: R6C9 + R9C6 = R7C7 + 14; Max R6C9 + R9C6 = 18 -->> Max R7C7 = 4; Min R7C7 = 1 -->> Min R6C9 + R9C6 = 15 (Min 6 and Max 9) Conclusion: R6C9 = {6789}, R7C7 = {1234}, R9C6 = {6789}
13a. R6C9 + R9C6 see all 9’s in N9: can’t both be 9 -->> Max R6C9 + R9C6 = 17 -->> R7C7: no 4
13b. 4 in N9 locked for C9
14. 45 on N1: 2 innies and 1 outie: R3C13 = R1C4 + 9; Max R3C13 = 17 -->> Max R1C4 = 8; Min R1C4 = 2 -->> Min R3C3 = 11; Conclusion: R3C13: no 1; R1C4: no 9
14a. 45 on N1: 4 innies: R123C3 + R3C1 = 28 = {4789/5689}: no 2,3
Hmm, missed these.
15. 45 on C1234: 2 innies: R46C4 = 17 = {89} -->> locked for C4 and N5
15a. Clean up: R5C19: no 5(no {89} in R5C5)(step 9)
16. 45 on C6789: 2 innies: R46C6 = 3 = {12} -->> locked for C6 and N5
17. 23(5) in R3C5 needs one of {89} in R4C4, one of {12} in R4C6 and one of {567} in R5C5 -->> 23(5) = {12479/12569/12578/13469/13478/13568/23459/23468}: can only have one of {89} and it goes in R4C4 -->> R3C5: no 8,9
18. 19(4) in R6C4 can only have one of {89} and it goes in R6C4 -->> R7C5: no 8,9
19. {89} in C5 locked in 13(2) in R1C5 and 10(2) in R8C5; they can’t have both so both need one of {89} -->> R12C5 = {49/58}: no 6,7; R89C5 = {28}/[91]: no 3,4,6,7
20. 23(5) in R3C5 only has {45} in R345C5. Can’t have both {45} because it clashes with R12C5 -->> 23(5): no {23459}; 23(5) = {12479/12569/1258/12469/12478/13568/23468}
20a. Killer Pair {45} in R12C5 + R345C5 -->> locked for C5
21. 45 on N9: 4 innies and 1 outie: R7C789 + R8C8 = R9C6 + 5: Max R9C6 = 9 -->> Max R7C789 + R8C8 = 14: Conclusion: R7C789 + R8C8: no 9;
21a. R1C8 = 9(hidden)
21b. Clean up: R2C5: no 4
21c. 16(3) in R1C8 = 9{16/25}: no 3,7,8
22. Killer Pair {56} in R12C9 + R23C8 -->> locked for N3
23. Follow up 45 on N9: R7C789 + R8C8= 11/12/13/14
23a. When 11: R78C8 = {35}, R7C79 = {12}
23b. When 12: R78C8 = {36}, R7C79 = {12}
23c. When 13: R78C8 = {35}, R7C79 = [14]; R7C78 = {37}, R7C79 = {12}
23d. When 14: R78C8 = {35}, R7C79 = [24]; R78C8 = {36}, R7C79 = [14]; R78C8 = [83], R7C79 = {12}(no {56} clashes with R23C8)
23e. Conclusion: R7C7: no 3; R7C9: no 3,5,6,7,8; R78C8 = {35/36/37}/[83]: 3 locked for N9
24. 3 in C9 locked in 17(3) cage in R3C9 -->> 17(3) = {359/368}: no 1,2,7
24a. Killer Pair {56} in R12C9 + 17(3) cage in R3C9 -->> locked for C9
25. 18(3) cage in R8C9 = [981/972/954]/{8[6]4}: R89C9: no 7
25a. R6C9 = 7(hidden)
25b. Clean up: R9C6: no 6,7(step 13(R6C9 + R9C6 is minimal 15)
25c. Killer Pair {89} in R89C5 + R9C6 -->> locked for N8
26. 26(5) in R6C7 needs 2 of {124} in R6C8 + R7C7 -->> 26(5) = {12689/14579/14678/23489/24569/24578}: Only 2 of {124} so no {124} anywhere else in 26(5) -->> R78C6: no 4; 26(5) needs one of {89} -->> only place for {89} is R6C7: R6C7 = {89}
26a. Naked Pair {89} in R6C47 -->> locked for R6
27. 19(4) in R6C4 needs one of {89} in R6C4, one of {36} in R6C5 and one of {12} in R6C6 -->> 19(4) = {1369/1378/2368}: Needs one of {12}, this goes in R6C6 so nowhere else in 19(4) cage: R7C5: no 1,2; 3 locked in R67C5 -->> locked for C5
28. Hidden Killer Pair {12} in C5 in R3C5 + R89C5: R3C5 = {12}
28a. 23(5) in R3C5 in needs both {12} in R3C5+ R4C6 -->> 23(5) = {12479/12569/12578}
29. 4 in N8 locked for C4
29a. 5 in R6 locked for N4
29b. 9 in R7 locked for N7
30. 11(3) in R5C6 = [731]/[461]/{36}[2]/[452]: R5C6: no 5; R5C7: no 8; R5C8: no 4
Missed something at step 26.
31. 26(5) in R6C7 can’t have both {89} -->> 26(5) = {14579/14678/24569/24578}: 4 locked in 26(5) -->> R6C8 = 4(only place in cage): R78C6 = {56/57/67}: no 3
31a. Naked Pair {12} in R4C68 -->> locked for R4
32. 45 on C9: 3 outies: R789C8 = 16; R78C8 = {35/36}/[83](step 23) -->> R9C8 = {578}
32a. 18(3) in R8C9 = [981/972/954] -->> R8C9 = 9; R9C9: no 8
32b. R9C6 = 9(hidden); R2C5 = 9; R1C5 = 4
32c. Clean up: R9C5: no 1
32d. R89C5 = {28} -->> locked for C5 and N8
32e. R3C5 = 1; R4C6 = 2; R45C8 = [12]; R6C6 = 1; R7C7 = 2(step 13)
32f. R5C3 = 1(hidden); R5C6 = 4(hidden); R5C7 = 5; R4C5 = 5(hidden)
32g. Clean up: R89C7 = {67} -->> locked for C7 and N9
33. 17(3) in R3C9 = {368} -->> locked for C9
33a. Clean up: R12C9 = {25} -->> locked for N3
33b. Clean up: R23C8 = {67}
34. 26(5) in R6C7 = 24{569/578} -->> 5 locked in R78C6 for C6 and N8
35. 12(3) in R5C2 = [813](last possible combination)
35a. R6C5 = 6; R5C5 = 7; R5C9 = 6; R5C1 = 9; R4C4 = 8; R6C4 = 9; R7C5 = 3
35b. R34C9 = [83]; R46C7 = [98]; R8C8 = 3(hidden); R3C3 = 9(hidden); R7C2 = 9(hidden)
36. 23(5) in R2C6 = {13469}(last possible combination) -->> R3C7 = 4(only place in cage); R23C6 = {36} -->> locked for C6 and N2
36a. R1C6 = 8(hidden); R2C3 = 8(hidden)
37. 19(3) in R1c3 = [658](last possible combination): R1C3= 6; R1C4 = 5
37a. R12C9 = [25]
38. 10(3) in R1C1 = {127}(last possible combination) -->> R2C1 = 2; R1C12 = {17} -->> locked for R1 and N1
38a. R12C7 = [31]; R23C4 = [72]; R23C8 = [67]; R23C6 = [36]; R23C2 = [43]; R3C1 = 5
38b. R4C1 = 7; R4C23 = [64]; R1C12 = [17]; R6C1 = 3
38c. R9C3 = 3(hidden); R9C7 = 7(hidden); R8C7 = 6;
39. R6C23 = {25} -->> locked for 19(5) cage in R6C2: 19(5) = {12457}: no 6
And now all singles.
39a. R7C3 = 7; R78C6 = [57]; R7C89 = [81]; R9C89 = [54]; R78C4 = [41]
39b. R7C1 = 6; R8C2 = 2; R6C23 = [52]; R9C12 = [81]; R8C1 = 4; R8C3 = 5
39c. R89C5 = [82]; R9C4 = 6
greetings
Para
Last edited by Para on Sat Jun 23, 2007 11:56 am, edited 2 times in total.
Here is the monster, disguised as a piece of candy
3x3::k:5120:5120281833335895:5895:51202818:615669261040:58952058:6156:6156:8982:6926:692643786156:6156:8982:8982:8982:6926:6926:437823412341:898233694378:4397:7982:7982:4400:4400:4400:6963:6963:5429:4397:4397:7982:7982:4400:6963:6963:5429:5429:4159:4397:4161:7982696354294159:4159:4161:416131413143
Enjoy the ride!
Ruud
3x3::k:5120:5120281833335895:5895:51202818:615669261040:58952058:6156:6156:8982:6926:692643786156:6156:8982:8982:8982:6926:6926:437823412341:898233694378:4397:7982:7982:4400:4400:4400:6963:6963:5429:4397:4397:7982:7982:4400:6963:6963:5429:5429:4159:4397:4161:7982696354294159:4159:4161:416131413143
Enjoy the ride!
Ruud
“If the human brain were so simple that we could understand it, we would be so simple that we couldn't.” - Emerson M Pugh
Hi guys,
In the meantime, thought I'd join Para in posting a V1 walkthrough, because I unfortunately won't be able to work on the V2 this week.
This V1 wasn't too bad. There was quite a narrow solving path at the beginning, which I suspect nearly everyone will take. There weren't any really difficult techniques required. Some of the earlier Assassins (e.g. 33,34) were worse IMO. I suspect Andrew's approach will be quite good on this one.
Having said all that, I thought the difficulty of this puzzle was about optimal for a V1. Very well chosen. Thanks, Ruud.
So, enough pre-amble, here's my A55V1 Walkthrough.
Assassin 55 Walkthrough
1. 10/3 at R1C1: no 8,9
2. 19/3 at R1C3: no 1
3. 13/2 at R1C5: no 1,2,3
4. 7/2 at R2C2: no 7,8,9
5. 13/2 at R2C8: no 1,2,3
6. 21/3 at R3C1: no 1,2,3
7. 11/3 at R5C6: no 9
8. 10/2 at R8C5: no 5
9. 22/3 at R8C7 = {(58|67)9}
9a. CPE: no 9 in R9C89
10. Innies C1234: R46C4 = 17/2 = {89}, locked for C4 and N5
11. Innies C6789: R46C6 = 3/2 = {12}, locked for C6 and N5
11a. {128} combo now blocked for 11/3 at R5C6, because none of these digits in R5C6
11b. -> no 8 in R5C67
12. Innies R5: R5C159 = 22/3 = {(58|67)9} = {(5|6)..}
12a. 9 locked for R5
12b. {89} only in R5C19 -> no 5 in R5C19
13. Innies C89: R456C8 = 7/3 = {124}, locked for C8 and N6
13a. Cleanup: no 9 in 13/2 at R2C8 = {58|67}
13b. no 7 in R3C9 (due to no {124} in R45C9)
13c. min. R45C9 = 9 -> max. R3C9 = 8
13d. -> no 9 in R3C9
14. 11/3 at R5C6 = {137|146|236|245) = {(3|4|5)..}
14a. {12} only in R5C8
14b. -> R5C8 = {12}
14c. 4 only in R5C6
14d. -> no 5 in R5C6
15. 12/3 at R5C2 = {138|147|237|156|246|345}
15a. {156} blocked by h22/3 R5 innies (step 12)
15b. {345} blocked by 11/3 (step 14)
15c. -> no 5 in 12/3 at R5C2
16. 5 in N5 locked in C5 -> not elsewhere in C5
16a. Cleanup: no 8 in 13/2 at R1C5 = {49|67} = {(4|6)..}
17. {46} for 10/2 at R8C5 blocked by 13/2 at R1C5 (step 16a)
17a. -> no 4,6 in R89C5
18. 19/4 at R6C4 can only contain 1 of {89} due to 19/4 cage sum, which must go in R6C4
18a. -> no {89} elsewhere in 19/4
18b. -> no 8,9 in R7C5
19. 23/5 at R3C5 cannot contain {12389}, because none of these digits are in R5C5
19a. -> 23/5 at R3C5 must contain exactly one of {89} (in R4C4)
19b. -> no {89} elsewhere in 23/5
19c. -> no 8,9 in R3C5
20. 8 in C5 now locked in 10/2 at R8C5
20a. -> 10/2 at R8C5 = {28}, locked for C5 and N8
21. 9 in C5 now locked in 13/2 at R1C5
21a. -> 13/2 at R1C5 = {49}, locked for C5 and N2
22. 4 in N5 locked in R5 -> not elsewhere in R5
22a. Cleanup: 4 in 12/3 at R5C2 only in R5C4 -> no 6 in R5C4 (see steps 15,15a)
23. 23/5 at R2C6: min. R23C6+R4C78 = 3+5+6+1 = 15
23a. -> no 9 in R3C7
24. 9 in R3 locked in N1 -> not elsewhere in N1
24a. -> 19/3 at R1C3 = {(47|56)8} (no 2,3)
24b. 8 locked in R12C3 for C3 and N1
25. 2 in C4 locked in 28/5 at R2C4 = {2..}
25a. -> cannot also contain a 1 (otherwise 28/5 cage sum unreachable)
25b. -> no 1 in 28/5 at R2C4 (otherwise 28/5 cage sum unreachable)
25c. 2 in 28/5 at R2C4 locked in R23C4
25d. -> no 2 in R3C3+R4C23
26. HS in N2 at R3C5 = 1
26a. -> R4C6 = 2
26b. -> R6C6 = 1
26c. Cleanup: no 6 in R2C2
27. HS in R4 at R4C8 = 1
27a. -> R56C8 = [24]
28. HS in C6 at R5C6 = 4
28a. -> R5C7 = 5 (last digit in cage)
29. 12/3 at R5C2 = {138} (only 12/3 combo that doesn't conflict w/ {245} at R5C678)
29a. -> R5C234 = [813] (only possible permutation)
30. HS in C5 at R7C5 = 3
30a. Split 15/2 at R6C45 = [87|96]
30b. -> no 5 in R6C5
31. HS in C5 at R4C5 = 5
32. 28/5 at R2C4 contains a 2 (step 25), but 8 now unavailable
32a. -> 28/5 at R2C4 = {24679}
32b. 9 locked in R3C3+R4C23
32c. -> no 9 in R6C3 (CPE)
33. Hidden triple (HT) on {235} in N4 at R6C123
33a. -> R6C123 = {235}, locked for R6
34. 21/3 at R3C1 = {579}
35a. R3C1 = 5 (HS in cage)
35b. -> {79} locked in R45C1 for C1 and N4
35c. Cleanup: no 2 in R23C2, no 8 in R2C8
36. HS in R3 at R3C3 = 9
37. Outie N1: R1C4 = 5
37a. -> Split 14/2 at R12C3 = {68}
37b. -> 6 locked in R12C3 for C3 and N1
38. NS at R4C3 = 4
38a. -> R4C2 = 6
38b. Cleanup: R23C4 = {27}, 7 locked for C4 and N2
39. HS in N1 at R1C2 = 7
39a. -> R12C1 = {12}, locked for C1 and N1
39b. -> 7/2 at R23C2 = {34}, locked for C2
40. NS at R6C1 = 3
41. NP on {25} in 19/5 at R6C23
41a. -> no 2,5 in R7C3
42. NS at R7C3 = 7
43. Outie N7: R9C4 = 6
43a. -> Split 8/2 at R89C3 = {35}, locked for C3 and N7
44. NS at R6C3 = 2
44a. -> R6C2 = 5
45. 13/3 at R8C1 = {148} (only remaining combo, due to {237} being unavailable)
45a. -> R9C2 = 1 (HS)
45b. -> R89C1 = {48}, locked for C1/N7
46. NS at R7C1 = 6
Now switch attention to right-hand side of grid...
47. deleted
48. I/O diff. N9: R6C9 + R9C6 = R7C7 + 14
48a. -> max. R6C9 + R9C6 = 18
48b. -> max. R7C7 = 4
48c. -> no 8,9 in R7C7
48d. min. R7C7 = 1
48e. -> min. R6C9 + R9C6 = 15
48f. -> no 5 in R9C6
49. 8 in R7 locked in R7C89 -> not elsewhere in N9
49a. no 8 in R6C9
50. 22/3 at R8C7 = {679}
50a. -> R8C7 = 6 (HS)
50b. -> R9C67 = {79}, locked for R9
51. 18/3 at R8C9 must have 1 of {35} (due to R9C8), w/ {68} unavailable
51a. -> 18/3 at R8C9 = {459}
51b. -> R9C8 = 5, R89C9 = [94]
Next 2 moves just to get down to all naked singles...
52. 13/2 at R2C8 = {67}, locked for C8 and N3
53. HS in C9 at R2C9 = 5
As promised, rest is naked singles now.
I hope not! How are you getting on?Cathy wrote:This one may well take all week
In the meantime, thought I'd join Para in posting a V1 walkthrough, because I unfortunately won't be able to work on the V2 this week.
This V1 wasn't too bad. There was quite a narrow solving path at the beginning, which I suspect nearly everyone will take. There weren't any really difficult techniques required. Some of the earlier Assassins (e.g. 33,34) were worse IMO. I suspect Andrew's approach will be quite good on this one.
Having said all that, I thought the difficulty of this puzzle was about optimal for a V1. Very well chosen. Thanks, Ruud.
So, enough pre-amble, here's my A55V1 Walkthrough.
Assassin 55 Walkthrough
1. 10/3 at R1C1: no 8,9
2. 19/3 at R1C3: no 1
3. 13/2 at R1C5: no 1,2,3
4. 7/2 at R2C2: no 7,8,9
5. 13/2 at R2C8: no 1,2,3
6. 21/3 at R3C1: no 1,2,3
7. 11/3 at R5C6: no 9
8. 10/2 at R8C5: no 5
9. 22/3 at R8C7 = {(58|67)9}
9a. CPE: no 9 in R9C89
10. Innies C1234: R46C4 = 17/2 = {89}, locked for C4 and N5
11. Innies C6789: R46C6 = 3/2 = {12}, locked for C6 and N5
11a. {128} combo now blocked for 11/3 at R5C6, because none of these digits in R5C6
11b. -> no 8 in R5C67
12. Innies R5: R5C159 = 22/3 = {(58|67)9} = {(5|6)..}
12a. 9 locked for R5
12b. {89} only in R5C19 -> no 5 in R5C19
13. Innies C89: R456C8 = 7/3 = {124}, locked for C8 and N6
13a. Cleanup: no 9 in 13/2 at R2C8 = {58|67}
13b. no 7 in R3C9 (due to no {124} in R45C9)
13c. min. R45C9 = 9 -> max. R3C9 = 8
13d. -> no 9 in R3C9
14. 11/3 at R5C6 = {137|146|236|245) = {(3|4|5)..}
14a. {12} only in R5C8
14b. -> R5C8 = {12}
14c. 4 only in R5C6
14d. -> no 5 in R5C6
15. 12/3 at R5C2 = {138|147|237|156|246|345}
15a. {156} blocked by h22/3 R5 innies (step 12)
15b. {345} blocked by 11/3 (step 14)
15c. -> no 5 in 12/3 at R5C2
16. 5 in N5 locked in C5 -> not elsewhere in C5
16a. Cleanup: no 8 in 13/2 at R1C5 = {49|67} = {(4|6)..}
17. {46} for 10/2 at R8C5 blocked by 13/2 at R1C5 (step 16a)
17a. -> no 4,6 in R89C5
18. 19/4 at R6C4 can only contain 1 of {89} due to 19/4 cage sum, which must go in R6C4
18a. -> no {89} elsewhere in 19/4
18b. -> no 8,9 in R7C5
19. 23/5 at R3C5 cannot contain {12389}, because none of these digits are in R5C5
19a. -> 23/5 at R3C5 must contain exactly one of {89} (in R4C4)
19b. -> no {89} elsewhere in 23/5
19c. -> no 8,9 in R3C5
20. 8 in C5 now locked in 10/2 at R8C5
20a. -> 10/2 at R8C5 = {28}, locked for C5 and N8
21. 9 in C5 now locked in 13/2 at R1C5
21a. -> 13/2 at R1C5 = {49}, locked for C5 and N2
22. 4 in N5 locked in R5 -> not elsewhere in R5
22a. Cleanup: 4 in 12/3 at R5C2 only in R5C4 -> no 6 in R5C4 (see steps 15,15a)
23. 23/5 at R2C6: min. R23C6+R4C78 = 3+5+6+1 = 15
23a. -> no 9 in R3C7
24. 9 in R3 locked in N1 -> not elsewhere in N1
24a. -> 19/3 at R1C3 = {(47|56)8} (no 2,3)
24b. 8 locked in R12C3 for C3 and N1
25. 2 in C4 locked in 28/5 at R2C4 = {2..}
25a. -> cannot also contain a 1 (otherwise 28/5 cage sum unreachable)
25b. -> no 1 in 28/5 at R2C4 (otherwise 28/5 cage sum unreachable)
25c. 2 in 28/5 at R2C4 locked in R23C4
25d. -> no 2 in R3C3+R4C23
26. HS in N2 at R3C5 = 1
26a. -> R4C6 = 2
26b. -> R6C6 = 1
26c. Cleanup: no 6 in R2C2
27. HS in R4 at R4C8 = 1
27a. -> R56C8 = [24]
28. HS in C6 at R5C6 = 4
28a. -> R5C7 = 5 (last digit in cage)
29. 12/3 at R5C2 = {138} (only 12/3 combo that doesn't conflict w/ {245} at R5C678)
29a. -> R5C234 = [813] (only possible permutation)
30. HS in C5 at R7C5 = 3
30a. Split 15/2 at R6C45 = [87|96]
30b. -> no 5 in R6C5
31. HS in C5 at R4C5 = 5
32. 28/5 at R2C4 contains a 2 (step 25), but 8 now unavailable
32a. -> 28/5 at R2C4 = {24679}
32b. 9 locked in R3C3+R4C23
32c. -> no 9 in R6C3 (CPE)
33. Hidden triple (HT) on {235} in N4 at R6C123
33a. -> R6C123 = {235}, locked for R6
34. 21/3 at R3C1 = {579}
35a. R3C1 = 5 (HS in cage)
35b. -> {79} locked in R45C1 for C1 and N4
35c. Cleanup: no 2 in R23C2, no 8 in R2C8
36. HS in R3 at R3C3 = 9
37. Outie N1: R1C4 = 5
37a. -> Split 14/2 at R12C3 = {68}
37b. -> 6 locked in R12C3 for C3 and N1
38. NS at R4C3 = 4
38a. -> R4C2 = 6
38b. Cleanup: R23C4 = {27}, 7 locked for C4 and N2
39. HS in N1 at R1C2 = 7
39a. -> R12C1 = {12}, locked for C1 and N1
39b. -> 7/2 at R23C2 = {34}, locked for C2
40. NS at R6C1 = 3
41. NP on {25} in 19/5 at R6C23
41a. -> no 2,5 in R7C3
42. NS at R7C3 = 7
43. Outie N7: R9C4 = 6
43a. -> Split 8/2 at R89C3 = {35}, locked for C3 and N7
44. NS at R6C3 = 2
44a. -> R6C2 = 5
45. 13/3 at R8C1 = {148} (only remaining combo, due to {237} being unavailable)
45a. -> R9C2 = 1 (HS)
45b. -> R89C1 = {48}, locked for C1/N7
46. NS at R7C1 = 6
Now switch attention to right-hand side of grid...
47. deleted
48. I/O diff. N9: R6C9 + R9C6 = R7C7 + 14
48a. -> max. R6C9 + R9C6 = 18
48b. -> max. R7C7 = 4
48c. -> no 8,9 in R7C7
48d. min. R7C7 = 1
48e. -> min. R6C9 + R9C6 = 15
48f. -> no 5 in R9C6
49. 8 in R7 locked in R7C89 -> not elsewhere in N9
49a. no 8 in R6C9
50. 22/3 at R8C7 = {679}
50a. -> R8C7 = 6 (HS)
50b. -> R9C67 = {79}, locked for R9
51. 18/3 at R8C9 must have 1 of {35} (due to R9C8), w/ {68} unavailable
51a. -> 18/3 at R8C9 = {459}
51b. -> R9C8 = 5, R89C9 = [94]
Next 2 moves just to get down to all naked singles...
52. 13/2 at R2C8 = {67}, locked for C8 and N3
53. HS in C9 at R2C9 = 5
As promised, rest is naked singles now.
Last edited by mhparker on Sat Jun 30, 2007 9:09 am, edited 1 time in total.
Cheers,
Mike
Mike
27 "Cathy" steps are normally equivalent to 50 of anyone else's, so it sounds like you've almost made it!CathyW wrote:27 steps now but not much further on.
As far as difficult V1 Assassins go, I should also add A50V1 to the list, which you sailed through in no time at all, whilst the rest of us were really struggling. In terms of techniques, the A50V1 was harder than this one. Hope I've motivated you a bit there!
However, it's not just the techniques used that make a puzzle easy or difficult, but also how narrow the solution paths are. I suspect that's what makes this one more difficult than most, because it appears to be quite dependent in places on chipping off that 1 extra candidate that turns out to be crucial later. In my case, it was the small step 12b that just about made step 13d possible, combining with the equally small step 23a later, in turn making steps 24 and 25 possible that cracked open the puzzle. That's what I call narrow!
Cheers,
Mike
Mike
Some of the earlier ones were also difficult. A18 was the first one I found hard and then A24 was the first really difficult one. I had to use a long hypothetical but Ed and Richard had neat solutions. I've also still got three, A26, A31 and A34, in my unfinished backlog. Must have another look at them and see if I can finish them now with what I've learnt since then. If I remember correctly A26 was a tag solution on the forum. Ruud introduced A34 as "be careful what you ask for" after some people, not including me, thought A33 was easier than expected.
Thanks Mike! Time to enter all candidates on my grid then so I can chip away at them.mhparker wrote:Hope I've motivated you a bit there!
I do have the two naked pairs in N5, naked triple in N6 and a couple of locked candidates in r5 and r8 which I assume everybody has got but still no placements at all . However, I'm not going to cheat by looking at your walkthrough or Para's (yet).
At last! It fell fairly quickly after I'd eventually made the first placements in step 35 of 38. Step 28 was a real marathon of combination analysis but it proved important.
Now I'll go see how Para and Mike did it.
1. Only initial candidate entry!
22(3) r8c7+r9c67 must have 9 -> r9c89 can see all cells of this 22(3) therefore <> 9
2. Innies c12: r456c2 = 19 (no 1)
3. Innies c89: r456c8 = 7 = {124} -> 13(2) r23c8 = {58/67}
4. Innies r5: r5c159 = 22 = {589/679}, 9 not elsewhere in r5.
5. Innies c1234: r46c4 = 17 = {89}, not elsewhere in c4/N5.
-> split 22(3) in r5c159: r5c19 <> 5
6. Innies c6789: r46c6 = 3 = {12}, not elsewhere in c6/N5 -> since r1c6 <> 1 or 2, r12c7 of 12(3) <> 9.
7. Outies – Innies N7: r6c1 + r9c4 – r7c3 = 2.
8. O-I N9: r6c9 + r9c6 – r7c7 = 14 (15-1, 16-2, 17-3, 18-4)
-> r7c7 = (1234); r6c9 + r9c6 = 15 {69/78} or 16 {79/88} or 17 {89} or 18 {99}
-> r6c9, r9c6 = (6789)
9. 11(3) r5c678: r5c6 (34567), r5c7 (35678), can’t make 7 from these candidates without repetition -> r5c8 <> 4
-> combination options: {137/146/236/245} -> r5c6 <> 5, r5c7 <> 8.
10. Innies r12: r2c2468 = 20.
11. Innies r89: r8c2468 = 13 (must have 1 which is thus locked to r8c24) not elsewhere in r8 -> r9c5 <> 9. Options: {1237/1246/1345}
a) If {1237} r8c24 = {12}, r8c68 = {37}
b) If {1246} r8c24 = {12}, r8c68 = [46]
c) If {1345} r8c2 = (1345), r8c4 = (1345), r8c6 = (345), r8c8 = (35)
-> r8c24 = (12345), r8c6 = (3457), r8c8 = (3567)
12. Innies N1: r3c1 + r123c3 = 28 = {4789/5689} -> 10(3) in N1 cannot have 4 since if split 28(4) = {5689}, 7(2) would be {34} -> options for 10(3): {127/136/235}
13. Innies N3: r3c9 + r123c7 = 16.
14. O-I N5: r5c46 – r37c5 = 3 -> r5c46 is max 13, min 7; r37c5 is max 10, min 4.
15. 17(3) r345c9: min from r45c9 = 9 -> r3c9 <> 9.
16. 16(3) in N3 can’t be {178/268/457} else no options for 13(2). Remaining options: {169/259/349/358/367}
17. Outies c1: r1789c2 = 19.
18. Outies c9: r1789c8 = 25 = {3589/3679}
19. 12(3) r5c234: options {138/147/156/237/246/345}. Analysis -> r5c3 <> 8
20. 19(4) r67c9 + r78c8: minimum from r6c9 + r78c8 = 14 -> r7c9 = max 5.
Options: {1369/1378/1567/2359/2368/3457} Analysis -> r7c9 <> 3,5
21. Pointing pair: r8c24 must have 1 -> r7c3 <> 1.
22. 17(3) r345c9: can’t make 10 from candidates in r45c9 without repetition -> r3c9 <> 7.
23. 18(3) in N9: max from r9c89 = 15 -> r8c9 <> 2.
24. 19(4) r6c456 + r7c5 must have at least one of (12) and (89) -> r6c46 = 9, 10 or 11
a) if r6c46 = [81] -> r67c5 = 10 = {37/46}
b) if r6c46 = [82/91] -> r67c5 = 9 = {27/36/45}
c) if r6c46 = [92] -> r67c5 = 8 = {17/35}
-> r7c5 <> 8, 9.
25. 23(5) r345c5 + r4c46 must have at least one of (12) and (89) in r4c46 -> r345c5 = 12, 13 or 14
a) if r4c46 = [81] -> r345c5 = 14 = {257/347/356}
b) if r4c46 = [82/91] -> r345c5 = 13 = {157/247/346}
c) if r4c46 = [92] -> r345c5 = 12 = {147/156/345}
-> r3c5 <> 7, 8, 9
26. Innies c5: r34567c5 = 22 = {13567/23467}-> 3, 6, 7 not elsewhere in c5 -> r89c5 <> 4
-> 7 locked to r456c5 -> r5c46 <> 7
27. 26(5) r6c78 + r7c67 + r8c6:
Can’t have both {12} as can’t make 23 {689} from remaining candidates
Max from r6c8 + r7c7 + r8c6 = 14 [437] leaving r6c7 + r7c6 = 12 -> Conflict as no options remaining
-> Max from r6c8 + r7c7 + r8c6 = 12 {435} leaving r6c7 + r7c6 = min 14 {59/68}
-> r6c7 <> 3, r7c6 <> 3,4
28. Analysis of options for 23(5) r345c5 + r4c46:
a) if r4c46 = [81], r345c5 = {257} -> r12c5 = {49}, r89c5 = {28}, Conflict – no options for r67c5 of 19(4)
b) if r4c46 = [81], r345c5 = {347} -> r12c5 = {58}, r89c5 = [91], r67c5 = [62] Conflict in 19(4)
c) if r4c46 = [81], r345c5 = {356} -> r12c5 = {49}, r89c5 = {28}, r67c5 = [71] OK
d) if r4c46 = [82], r345c5 = {157}-> r12c5 = {49}, r89c5 = {28}, r67c5 = {36} OK
e) if r4c46 = [82], r345c5 = {346} -> r12c5 = {58}, r89c5 = [91], r67c5 = [72] OK
f) if r4c46 = [91], r345c5 = {247} -> r12c5 = {58}, r89c5 = [91], r67c5 = {36} OK
g) if r4c46 = [91], r345c5 = {346} -> r12c5 = {58}, r89c5 = [91], r67c5 = [72] Conflict in 19(4)
h) if r4c46 = [92], r345c5 = {147} -> r12c5 = {58}, r89c5 = [91] -> Conflict as no options left for remaining 10(2) in r67c5.
i) if r4c46 = [92], r345c5 = {156} -> r12c5 = {49}, r89c5 = {28}-> r67c5 = [73] OK
j) if r4c46 = [92], r345c5 = 345 -> Conflict as no options left for 13(2) in r12c5.
Summary:
a) if r4c46 = [81], r345c5 = {356}, r67c5 = [71]
b) if r4c46 = [82], r345c5 = {157/346}, r67c5 = {36}/[72]
c) if r4c46 = [91], r345c5 = {247}, r67c5 = {36}
d) if r4c46 = [92], r345c5 = {156}, r67c5 = [73]
Conclusion: 13(2) in c5 must have at least one of (45), r345c5 of 23(5) must have at least one of (45) (but can’t have both) -> 45 not elsewhere in c5: r6c5 = (367), r7c5 = (1236)
29. 12(3) r5c234 = {138|147|237|246|345} – can’t be {156} else no options for split 22(3) in r5.
30. 11(3) r5c678 = {137|146|236|245}. Must have 3 or 4 -> 12(3) r5c234 can’t be {345}
-> eliminate 5 from r5c234
-> 5 locked to r45c5 in N5 -> eliminate 5 from r123c5 -> 13(2) = {49} not elsewhere in N2/c5
-> 10(2) in c5 = {28} not elsewhere in N8/c5
-> 4 locked to r5 in N5 -> r5c23 <> 4
31. 12(3) r5c234 = {138/147/237/246} Analysis -> r5c23 <> 3, r5c4 <> 6
32. Multi-colouring 1s: r3c5 <=> r7c5, r8c2 <=> r8c4
r7c5 and r8c4 both in N8 -> buddy of r3c5 and r8c2 <> 1
-> r3c2 <> 1 -> r2c2 <> 6
(I hope this is correct for nice loop notation:
[r3c5]=1=[r7c5]-1-[r8c4]=1=[r8c2]-1-[r3c2])
33. Multi-colouring 1s: r3c5 <=> r4c6 <=> r6c6 <=> r7c5; r8c2 <=> r8c4
r7c5 and r8c4 both in N8 -> r6c1 (buddy of r6c6 and r8c2) <> 1
-> 1 locked to r456c3 -> r9c3 <> 1
[r6c6]=1=[r7c5]-1-[r8c4]=1=[r8c2]-1-[r6c1]
34. Cage 23/5 r23c6 + r34c7 + r4c8 = {12389/12569/12578/13469/13478/13568/14567/23459/23468/23567}
For any option with 9, must be in r4c7 -> r3c7 <> 9 -> 16(3) in N3 must have 9
-> 16(3) <> 7,8
-> 9 locked to r3c13 -> r12c3 <> 9
35. 19(3) r1c34 + r2c3 = {478/568} (Must have 8)
-> 8 not elsewhere in c3
-> r1c4 <> 2, 3
-> 2 locked to r23c4 in N2/cage 28(5) -> r4c23 <> 2
-> Cage 28(5) must have 2 -> can’t also have 1 thus r23c4 <> 1, r4c3 <> 1
-> First placement!! HS r3c5 = 1
-> r4c6 = 2, r6c6 = 1, r4c8 = 1, r5c8 = 2, r6c8 = 4
-> r5c3 = 1
36. HS r5c6 = 4 -> r5c7 = 5, r5c4 = 3, r5c2 = 8, r4c5 = 5, r7c5 = 3
37. 21(3) r345c1 = {579} not elsewhere in c1.
-> r3c1 = 5 -> {79} not elsewhere in N4; r23c2 <> 2
38. O-I N1: r3c3 – r1c4 = 4 -> r3c3 = 9, r1c4 = 5
-> r12c3 = {68} -> eliminate 6 from elsewhere in N1/c3
-> 7(2) in N1 = {34} not elsewhere in c2, 10(3) in N1 = {127}
-> r4c2 = 6 …
Fairly straightforward from here.
Edit: I've added what I hope is correct nice loop notation for steps 32 and 33 where I used multi-colouring. Please let me know if it's not right.
Now I'll go see how Para and Mike did it.
1. Only initial candidate entry!
22(3) r8c7+r9c67 must have 9 -> r9c89 can see all cells of this 22(3) therefore <> 9
2. Innies c12: r456c2 = 19 (no 1)
3. Innies c89: r456c8 = 7 = {124} -> 13(2) r23c8 = {58/67}
4. Innies r5: r5c159 = 22 = {589/679}, 9 not elsewhere in r5.
5. Innies c1234: r46c4 = 17 = {89}, not elsewhere in c4/N5.
-> split 22(3) in r5c159: r5c19 <> 5
6. Innies c6789: r46c6 = 3 = {12}, not elsewhere in c6/N5 -> since r1c6 <> 1 or 2, r12c7 of 12(3) <> 9.
7. Outies – Innies N7: r6c1 + r9c4 – r7c3 = 2.
8. O-I N9: r6c9 + r9c6 – r7c7 = 14 (15-1, 16-2, 17-3, 18-4)
-> r7c7 = (1234); r6c9 + r9c6 = 15 {69/78} or 16 {79/88} or 17 {89} or 18 {99}
-> r6c9, r9c6 = (6789)
9. 11(3) r5c678: r5c6 (34567), r5c7 (35678), can’t make 7 from these candidates without repetition -> r5c8 <> 4
-> combination options: {137/146/236/245} -> r5c6 <> 5, r5c7 <> 8.
10. Innies r12: r2c2468 = 20.
11. Innies r89: r8c2468 = 13 (must have 1 which is thus locked to r8c24) not elsewhere in r8 -> r9c5 <> 9. Options: {1237/1246/1345}
a) If {1237} r8c24 = {12}, r8c68 = {37}
b) If {1246} r8c24 = {12}, r8c68 = [46]
c) If {1345} r8c2 = (1345), r8c4 = (1345), r8c6 = (345), r8c8 = (35)
-> r8c24 = (12345), r8c6 = (3457), r8c8 = (3567)
12. Innies N1: r3c1 + r123c3 = 28 = {4789/5689} -> 10(3) in N1 cannot have 4 since if split 28(4) = {5689}, 7(2) would be {34} -> options for 10(3): {127/136/235}
13. Innies N3: r3c9 + r123c7 = 16.
14. O-I N5: r5c46 – r37c5 = 3 -> r5c46 is max 13, min 7; r37c5 is max 10, min 4.
15. 17(3) r345c9: min from r45c9 = 9 -> r3c9 <> 9.
16. 16(3) in N3 can’t be {178/268/457} else no options for 13(2). Remaining options: {169/259/349/358/367}
17. Outies c1: r1789c2 = 19.
18. Outies c9: r1789c8 = 25 = {3589/3679}
19. 12(3) r5c234: options {138/147/156/237/246/345}. Analysis -> r5c3 <> 8
20. 19(4) r67c9 + r78c8: minimum from r6c9 + r78c8 = 14 -> r7c9 = max 5.
Options: {1369/1378/1567/2359/2368/3457} Analysis -> r7c9 <> 3,5
21. Pointing pair: r8c24 must have 1 -> r7c3 <> 1.
22. 17(3) r345c9: can’t make 10 from candidates in r45c9 without repetition -> r3c9 <> 7.
23. 18(3) in N9: max from r9c89 = 15 -> r8c9 <> 2.
24. 19(4) r6c456 + r7c5 must have at least one of (12) and (89) -> r6c46 = 9, 10 or 11
a) if r6c46 = [81] -> r67c5 = 10 = {37/46}
b) if r6c46 = [82/91] -> r67c5 = 9 = {27/36/45}
c) if r6c46 = [92] -> r67c5 = 8 = {17/35}
-> r7c5 <> 8, 9.
25. 23(5) r345c5 + r4c46 must have at least one of (12) and (89) in r4c46 -> r345c5 = 12, 13 or 14
a) if r4c46 = [81] -> r345c5 = 14 = {257/347/356}
b) if r4c46 = [82/91] -> r345c5 = 13 = {157/247/346}
c) if r4c46 = [92] -> r345c5 = 12 = {147/156/345}
-> r3c5 <> 7, 8, 9
26. Innies c5: r34567c5 = 22 = {13567/23467}-> 3, 6, 7 not elsewhere in c5 -> r89c5 <> 4
-> 7 locked to r456c5 -> r5c46 <> 7
27. 26(5) r6c78 + r7c67 + r8c6:
Can’t have both {12} as can’t make 23 {689} from remaining candidates
Max from r6c8 + r7c7 + r8c6 = 14 [437] leaving r6c7 + r7c6 = 12 -> Conflict as no options remaining
-> Max from r6c8 + r7c7 + r8c6 = 12 {435} leaving r6c7 + r7c6 = min 14 {59/68}
-> r6c7 <> 3, r7c6 <> 3,4
28. Analysis of options for 23(5) r345c5 + r4c46:
a) if r4c46 = [81], r345c5 = {257} -> r12c5 = {49}, r89c5 = {28}, Conflict – no options for r67c5 of 19(4)
b) if r4c46 = [81], r345c5 = {347} -> r12c5 = {58}, r89c5 = [91], r67c5 = [62] Conflict in 19(4)
c) if r4c46 = [81], r345c5 = {356} -> r12c5 = {49}, r89c5 = {28}, r67c5 = [71] OK
d) if r4c46 = [82], r345c5 = {157}-> r12c5 = {49}, r89c5 = {28}, r67c5 = {36} OK
e) if r4c46 = [82], r345c5 = {346} -> r12c5 = {58}, r89c5 = [91], r67c5 = [72] OK
f) if r4c46 = [91], r345c5 = {247} -> r12c5 = {58}, r89c5 = [91], r67c5 = {36} OK
g) if r4c46 = [91], r345c5 = {346} -> r12c5 = {58}, r89c5 = [91], r67c5 = [72] Conflict in 19(4)
h) if r4c46 = [92], r345c5 = {147} -> r12c5 = {58}, r89c5 = [91] -> Conflict as no options left for remaining 10(2) in r67c5.
i) if r4c46 = [92], r345c5 = {156} -> r12c5 = {49}, r89c5 = {28}-> r67c5 = [73] OK
j) if r4c46 = [92], r345c5 = 345 -> Conflict as no options left for 13(2) in r12c5.
Summary:
a) if r4c46 = [81], r345c5 = {356}, r67c5 = [71]
b) if r4c46 = [82], r345c5 = {157/346}, r67c5 = {36}/[72]
c) if r4c46 = [91], r345c5 = {247}, r67c5 = {36}
d) if r4c46 = [92], r345c5 = {156}, r67c5 = [73]
Conclusion: 13(2) in c5 must have at least one of (45), r345c5 of 23(5) must have at least one of (45) (but can’t have both) -> 45 not elsewhere in c5: r6c5 = (367), r7c5 = (1236)
29. 12(3) r5c234 = {138|147|237|246|345} – can’t be {156} else no options for split 22(3) in r5.
30. 11(3) r5c678 = {137|146|236|245}. Must have 3 or 4 -> 12(3) r5c234 can’t be {345}
-> eliminate 5 from r5c234
-> 5 locked to r45c5 in N5 -> eliminate 5 from r123c5 -> 13(2) = {49} not elsewhere in N2/c5
-> 10(2) in c5 = {28} not elsewhere in N8/c5
-> 4 locked to r5 in N5 -> r5c23 <> 4
31. 12(3) r5c234 = {138/147/237/246} Analysis -> r5c23 <> 3, r5c4 <> 6
32. Multi-colouring 1s: r3c5 <=> r7c5, r8c2 <=> r8c4
r7c5 and r8c4 both in N8 -> buddy of r3c5 and r8c2 <> 1
-> r3c2 <> 1 -> r2c2 <> 6
(I hope this is correct for nice loop notation:
[r3c5]=1=[r7c5]-1-[r8c4]=1=[r8c2]-1-[r3c2])
33. Multi-colouring 1s: r3c5 <=> r4c6 <=> r6c6 <=> r7c5; r8c2 <=> r8c4
r7c5 and r8c4 both in N8 -> r6c1 (buddy of r6c6 and r8c2) <> 1
-> 1 locked to r456c3 -> r9c3 <> 1
[r6c6]=1=[r7c5]-1-[r8c4]=1=[r8c2]-1-[r6c1]
34. Cage 23/5 r23c6 + r34c7 + r4c8 = {12389/12569/12578/13469/13478/13568/14567/23459/23468/23567}
For any option with 9, must be in r4c7 -> r3c7 <> 9 -> 16(3) in N3 must have 9
-> 16(3) <> 7,8
-> 9 locked to r3c13 -> r12c3 <> 9
35. 19(3) r1c34 + r2c3 = {478/568} (Must have 8)
-> 8 not elsewhere in c3
-> r1c4 <> 2, 3
-> 2 locked to r23c4 in N2/cage 28(5) -> r4c23 <> 2
-> Cage 28(5) must have 2 -> can’t also have 1 thus r23c4 <> 1, r4c3 <> 1
-> First placement!! HS r3c5 = 1
-> r4c6 = 2, r6c6 = 1, r4c8 = 1, r5c8 = 2, r6c8 = 4
-> r5c3 = 1
36. HS r5c6 = 4 -> r5c7 = 5, r5c4 = 3, r5c2 = 8, r4c5 = 5, r7c5 = 3
37. 21(3) r345c1 = {579} not elsewhere in c1.
-> r3c1 = 5 -> {79} not elsewhere in N4; r23c2 <> 2
38. O-I N1: r3c3 – r1c4 = 4 -> r3c3 = 9, r1c4 = 5
-> r12c3 = {68} -> eliminate 6 from elsewhere in N1/c3
-> 7(2) in N1 = {34} not elsewhere in c2, 10(3) in N1 = {127}
-> r4c2 = 6 …
Fairly straightforward from here.
Edit: I've added what I hope is correct nice loop notation for steps 32 and 33 where I used multi-colouring. Please let me know if it's not right.
Last edited by CathyW on Tue Jul 10, 2007 3:31 pm, edited 6 times in total.
Mike forgot to add it was so narrow that only he and Cathy could find it. Me and Para had to claw up a rock cliff around N9. Was a really tough, fun challenge. Thanks Ruud.mhparker wrote:This V1 wasn't too bad. There was quite a narrow solving path at the beginning, .
Don't need a V2 after that - but can't resist trying the lollie. But will need lots of help. Any other suckers (:roll:)? No tiny text this time. I've been caught so many times with missed steps etc when copying. (Let me know if this is not OK by others.)
Found a nearly generalized X-wing that is interesting: doesn't lead to much yet.
Assassin 55V2
1. 23(3)n3 = {689}: all locked for n3
2. 4(2)n3 = {13}: both locked for n3 and c8
3. 13(3)n2 must have 2 of 2,4,5,7 = [1{57}]/{247}/[6{25}]
3a. r1c6 = {12467}
3b. r3c79 = {24/57/45/25/47}
4. 35(5)n2 = {56789} -> no {56789} in r6c5
5. "45"c1234: r46c4 = 13 = h13(2)n5 = {58/67}/[94]
5a. r6c4 = {4..8}
6. "45" c6789: r46c6 = 11 = h11(2)n5 = {56}/[74/83/92]
6a. r6c6 = {2..6}
7. "45" r5: r5c159 = 23 = h23(3)r5 = {689}
7a. 6,8,9 locked for r5
8. 9(3)n4 = {135/234} = 3{..}
8a. 3 locked for r5
9. 13(3)n5 = {157/247} = 7{..}
10. naked triple {689} r125c9: all locked for c9
11. "45" c89: r456c8 = 13 = h13(3)n6 = {247/256}(no 8,9)
11a. = 2{..}: 2 locked for n6 and c8
12. 17(3)n3 = {179/278/359/458} (no 6)({467} blocked: 6 in r5c9 forces h13(3)n6 = {247} step 11: but {47} in r4c9 clashes with {47} in h13(3))
13. 6 in c9 only in n1: no 6 r1c8
14. 20(3)n1 = {389/479/569/578}(no 12)
14a. 3 blocked from r2c1 by r1c8
14b. 21(4)n6: {1389} combo. blocked by r1c8
Now things start to get interesting.
15. No 2 in r5c6, no 4 in r5c8. Here's how.
15a. 13(3)n6 = {157/247} and h13(3)n6 = {247/256}
15b. With {247} combo. in 13(3)n6, 4 and 7 cannot be in r5c8 as the h13(3) can only be {247} which will clash with [4/7] in r5c7
15c. -> 13(3)n6 = {157}/{47}[2]
15d. no 2 r5c6, no 4 r5c8
16. no 7 in r46c8. Here's how.
16a. 7 in r46c8 -> h13(3) = {247} -> r5c8 = 2 -> 13(3)n6 = {247} -> r5c7 = [4/7] which clashes with {47} in r46c8
17. r5c7 + r456c8 = [1]{256}/[7]{256}/[1]{247}/[5]{247}/[4]{256}/[7]{256}
18. 17(3)n3 = {179/278/359/458}
18a. = [719/278/539/458/548] = [5/7..]
19. from step 18a. r34c9 = [5/7]
19a. since 35(5)n2 must have 5 & 7 in r34 -> 5/7 (but not both) have a generalized X-Wing for r34
19b. -> no other cage can have both 5 & 7 in r34. There is only room for 1 more of 1 of them somewhere else in r34.
20. For now, step 19a. only means no 5 in r4c7 & r6c9: Here's how.
20a. r5c7 + r456c8 all have 5 except [1]{247} (step 17)
20b. With this combo, r4c9 = {3/5}.
20c. when r4c9 = 3 r3c9 = 5
20d. from step 19a. r3c9 = 5 -> no 5 in r6c9 and 5 in 35(5) in r4 -> no 5 in r4c7
21. {2568} combo. blocked from 21(4)n6 by r6c9
21a. {2469} combo. blocked from 21(4) by 4 in r6c9 and 6 in c8: {46} clashes with h13(3) = [4/6..]
22. 12(3)n9: {138} combo blocked: forces r1c8 = 9 -> r5c9 = 9 -> 17(3) = [1/3](step 12): but this clashes with {13} in r89c9
22a. no 8 r9c8
OK. That will do for now.
Cheers
Ed
Marks pic for here: copy-paste into Sudocue.
Code: Select all
.-----------------------.-----------------------.-----------.-----------------------.-----------------------.
| 3456789 3456789 | 12345678 12345678 | 1245 | 12467 2457 | 89 689 |
| .-----------: .-----------: :-----------. :-----------. |
| 456789 | 123567 | 12345678 | 123456789 | 1245 | 123456789 | 2457 | 13 | 689 |
:-----------: :-----------' :-----------: '-----------: :-----------:
| 123456789 | 123567 | 123456789 123456789 | 56789 | 123456789 2457 | 13 | 2457 |
| :-----------' .-----------' '-----------. '-----------: |
| 123456789 | 123456789 123456789 | 56789 56789 56789 | 1346789 2456 | 13457 |
| :-----------------------'-----------. .-----------'-----------------------: |
| 689 | 12345 12345 12345 | 689 | 1457 1457 257 | 89 |
:-----------+-----------------------.-----------'-----------'-----------.-----------------------+-----------:
| 123456789 | 123456789 123456789 | 45678 1234 23456 | 13456789 2456 | 1347 |
| '-----------. '-----------. .-----------' .-----------' |
| 123456789 123456789 | 123456789 123456789 | 123456789 | 123456789 123456789 | 456789 123457 |
:-----------. :-----------. :-----------: .-----------: .-----------:
| 123456789 | 123456789 | 123456789 | 123456789 | 23456789 | 123456789 | 123456789 | 456789 | 123457 |
| '-----------: '-----------: :-----------' :-----------' |
| 123456789 123456789 | 123456789 123456789 | 23456789 | 123456789 123456789 | 45679 123457 |
'-----------------------'-----------------------'-----------'-----------------------'-----------------------'
Taking the female prerogative and changing my mind about the V2, especially as Ed has started things off . These few don't get us much further but may be useful later:
23. 15(3) r345c1: r34c1 is max 9 -> no 9 in r34c1.
24. Outies - Innies N5: r37c5 - r5c46 = 7
-> r37c5 is min 10, max 17
-> r5c46 is min 3, max 10
25. O-I N3: r3c79 - r1c6 = 5
Options: (24) - 1, (25) - 2, (27) - 4, (45) - 4, (47) - 6, (57) - 7.
26. O-I N1: r3c13 - r1c4 = 6
r1c4 = (12345678), r3c13 is min 7, max 14
27. O-I N7: r6c1 + r9c4 - r7c3 = 4
-> r6c1 + r9c4 is min 5, max 13
28. O-I N9: r6c9 + r9c6 - r7c7 = 0
-> r6c9 + r9c6 = r7c7 -> min 2, max 9
-> r6c9 = (1347), r9c6 = (1..8), r7c7 = (2..9)
29. 9 locked in r4c456 + r5c5 of 35(5) -> r3c5 <> 9
30. 3 locked to r1c1234 -> r2c3 <> 3
31. 9 locked to r3c346 -> r2c4 <> 9
32. See Ed's step 19/20: we can also eliminate 7 from r4c7:
If r3c9 = 7 -> 7 locked to r4c456 in 35(5), thus not elsewhere in r4/N5
If one of r123c7 = 7, r4c7 <> 7.
33. Progressing step 32 to eliminate 7 from r6c7:
If r3c9 = 7 -> 7 locked to r4c456 in 35(5), not elsewhere in r4/N5 -> 7 locked to r5c78, not elsewhere in N6.
If one of r123c7 = 7, r6c7 <> 7.
34. r67c5 of 17(4) cannot both be {12}, {14}, {25} or {45} as would conflict with 6(2) in r12c5 -> r6c46 not 14 [86].
35. Also, r23c4 of 24(5) and r23c6 of 27(5) cannot both be {12}, {14}, {25} or {45}.
Undoubtedly there are several similar steps involving conflicting combinations.
Edit: I've just tried putting JSudoku through its paces on this one - it gets stuck without a single placement so I think it's going to take a long hypothetical to make any real progress, although JC's as yet unreleased new version will probably do better. Over to you guys!
Cathy
PS Can someone help with AIC notation? I've used a couple of multicolouring steps in my V1 walkthrough which may be clearer if I can describe them in terms of strong and weak links. Thanks.
23. 15(3) r345c1: r34c1 is max 9 -> no 9 in r34c1.
24. Outies - Innies N5: r37c5 - r5c46 = 7
-> r37c5 is min 10, max 17
-> r5c46 is min 3, max 10
25. O-I N3: r3c79 - r1c6 = 5
Options: (24) - 1, (25) - 2, (27) - 4, (45) - 4, (47) - 6, (57) - 7.
26. O-I N1: r3c13 - r1c4 = 6
r1c4 = (12345678), r3c13 is min 7, max 14
27. O-I N7: r6c1 + r9c4 - r7c3 = 4
-> r6c1 + r9c4 is min 5, max 13
28. O-I N9: r6c9 + r9c6 - r7c7 = 0
-> r6c9 + r9c6 = r7c7 -> min 2, max 9
-> r6c9 = (1347), r9c6 = (1..8), r7c7 = (2..9)
29. 9 locked in r4c456 + r5c5 of 35(5) -> r3c5 <> 9
30. 3 locked to r1c1234 -> r2c3 <> 3
31. 9 locked to r3c346 -> r2c4 <> 9
32. See Ed's step 19/20: we can also eliminate 7 from r4c7:
If r3c9 = 7 -> 7 locked to r4c456 in 35(5), thus not elsewhere in r4/N5
If one of r123c7 = 7, r4c7 <> 7.
33. Progressing step 32 to eliminate 7 from r6c7:
If r3c9 = 7 -> 7 locked to r4c456 in 35(5), not elsewhere in r4/N5 -> 7 locked to r5c78, not elsewhere in N6.
If one of r123c7 = 7, r6c7 <> 7.
34. r67c5 of 17(4) cannot both be {12}, {14}, {25} or {45} as would conflict with 6(2) in r12c5 -> r6c46 not 14 [86].
35. Also, r23c4 of 24(5) and r23c6 of 27(5) cannot both be {12}, {14}, {25} or {45}.
Undoubtedly there are several similar steps involving conflicting combinations.
Edit: I've just tried putting JSudoku through its paces on this one - it gets stuck without a single placement so I think it's going to take a long hypothetical to make any real progress, although JC's as yet unreleased new version will probably do better. Over to you guys!
Cathy
PS Can someone help with AIC notation? I've used a couple of multicolouring steps in my V1 walkthrough which may be clearer if I can describe them in terms of strong and weak links. Thanks.
Last edited by CathyW on Tue Jun 19, 2007 4:27 pm, edited 2 times in total.
Not me I'm afraid Cathy. But this is a link about nice loops notation Mike gave me once. About time I studied it.CathyW wrote:Can someone help with AIC notation?
Now to the candy. Hmmm. First a couple of little things to add.
NOTE: no 5 in r6c9. Already went in step 20. Also, can't have repeats on outies of n9: forces that repeated digit into r7c7 since 12(3) sees both outies and the cages from two outies cover the rest of the cage except r7c7.Cathy wrote:28. O-I N9: r6c9 + r9c6 - r7c7 = 0
-> r6c9 + r9c6 = r7c7 -> min 2, max 9
-> r6c9 = (13457), r9c6 = (1..8), r7c7 = (2..9)
28a. -> min outies n9 = {12} = 3 -> min r7c7 = 3
Now: found just one useful hypothetical: but scratching my balding head about where else is worth trying. Really will have to study those nice loops.
36. "45" c5: 5 innies = 28
36a. {13789} blocked (-> r345c5 = {789} -> r4c46 = {56}: Clash: h11(2)n5 must have {56} in r6c6)
36b. {14689} blocked by 6(2)n2
36c. {15679} -> r345c5 = {57}[6] -> r4c46 = {89} and r67c5 = [19]
....................................= {569} -> r4c46 = {78} and r67c5 = [17]
....................................= {579} -> r4c46 = {68} and r67c5 = [16]
....................................= {679} Blocked (-> r4c46 = {58} -> clash with h13(2)n5 with {58} in r6c4)
36d. {23689} -> r345c5 = {689} -> r4c46 = {57} and r67c5 = {23}
36e. {24589} blocked by 11(2)n8
36f. {24679} blocked: (r345c5 = {679} -> r4c46 = {58}: clash with h13(2) and {58} in r6c4)
36g. {25678} blocked by 11(2)n8
36h. {34579} blocked by 11(2)n8
36i. {34678} -> r345c5 = {678} -> r4c46 = {59} and r67c5 = {34}
37. In summary: r345c5/r67c5 =
i. {57}[6]/[19]
ii. {569}/[17]
iii. {579}/[16]
iv. {689}/{23}
v. {678}/{34}
37a. 6 locked for c5 in r3457c5
37b. no 5 r89c5
37c. no 1,5 or 8 in r7c5 [edit:typo]
37d. r4c46 = {89}/{78}/{68}/{57}/{59} = [5/8..]
Last edited by sudokuEd on Tue Jun 19, 2007 9:25 pm, edited 1 time in total.
Have amended a typo and corrected the candidates remaining in r4c7. No further on with the V2 but will check through your next steps shortly. Took me a while but I finally got why you can't have repetition on the outies of N9! Interestingly, for the outies - innies of N7, 4,4,4 is an option.
Thanks also for link to page about nice loop notation. I'll check it out later and try and incorporate into V1 walkthrough.
Thanks also for link to page about nice loop notation. I'll check it out later and try and incorporate into V1 walkthrough.
Last edited by CathyW on Tue Jun 19, 2007 9:12 pm, edited 1 time in total.
I think this should be r7c5.sudokuEd wrote:37c. no 1,5 or 8 in r6c5
Well done though!
Not really a step, more a consolidation of your 36/37:
38. Looking at c5:
If 11(2) = {29}, 6(2) = {15}, r345c5 = {678}, r67c5 = {34}
If 11(2) = {38}, 6(2) = {15/24}, r345c5 = {5679}, r67c5 = [19/17/16] must have 1 -> 6(2) = {24}
If 11(2) = {47}, 6(2) = {15}, r345c5 = {689}, r67c5 = {23}
May be worth following these a bit further to see if leads to any conflicts elsewhere in the puzzle. Otherwise I'm completely stumped!
I followed through the first option in step 38 and it led to solution!
Just need to prove the other options lead to conflicts - will post steps later.
I've ruled out 11(2) in c5 = {47} as follows:
39a)
If 11(2) in c5 = {47}, 6(2) = {15}, r345c5 = {689}, r67c5 = {23}
-> 4,7 not elsewhere in N8, 1,5 not elsewhere in N2
-> r4c46 = {57}
-> r6c6 = 4, r6c4 = 8 (only option remaining) -> r5c6 =1, r4c6 = 7, r4c4 = 5, r3c5 = 8
-> r5c78 = {57} not elsewhere in N6/r5
-> 4 locked to r5c23 not elsewhere in N4
39b)
17(3) in c9 = {179/359/458} -> r3c9 <> 2
-> 2 locked to r123c7 -> r89c7 <> 2
39c)
From step 25 (O-I N3): only options remaining are {225/647} -> r3c7 <> 5.
39d)
Options for 27(5) in r34 = {14679/23679/24678} Must have 7
-> r3c7 = 7 -> r3c9 = 4 CONFLICT - no options for 17(3) in c9.
Thus r89c5 <> {47}
Struggling with proof of conflict if r89c5 = {38} i.e. haven't reached one yet!!
Just need to prove the other options lead to conflicts - will post steps later.
I've ruled out 11(2) in c5 = {47} as follows:
39a)
If 11(2) in c5 = {47}, 6(2) = {15}, r345c5 = {689}, r67c5 = {23}
-> 4,7 not elsewhere in N8, 1,5 not elsewhere in N2
-> r4c46 = {57}
-> r6c6 = 4, r6c4 = 8 (only option remaining) -> r5c6 =1, r4c6 = 7, r4c4 = 5, r3c5 = 8
-> r5c78 = {57} not elsewhere in N6/r5
-> 4 locked to r5c23 not elsewhere in N4
39b)
17(3) in c9 = {179/359/458} -> r3c9 <> 2
-> 2 locked to r123c7 -> r89c7 <> 2
39c)
From step 25 (O-I N3): only options remaining are {225/647} -> r3c7 <> 5.
39d)
Options for 27(5) in r34 = {14679/23679/24678} Must have 7
-> r3c7 = 7 -> r3c9 = 4 CONFLICT - no options for 17(3) in c9.
Thus r89c5 <> {47}
Struggling with proof of conflict if r89c5 = {38} i.e. haven't reached one yet!!