Hi Andrew
Yes it is. I had a look back as well, and arranged the three puzzles in the wrong order. This was actually the hardest. Sorry for the confusion. It goes through some really tough innie-outie differences and multiple CPE moves if i remember corrctly. Next time i'll keep a walk-through for my V2's.
greetings
Para
Assassin 44
start of walkthrough for Assassin 44V1.5
Hello,
I've noticed that there is no walkthrough posted for Assasin 44V1.5 (and saw that it was noted in the "Unsolvable" thread.
I have solved it - but afraid I only have a partial walkthrough as at the end I wasn't quite sure what I was doing. Would love if someone could help me out.
Here is my complete walkthrough:
I've redone it to fit the formatting guidelines and have borrowed some steps from Mike below so that I could finish it. I think it is still somewhat different as I occasionally seemed to take a much different route.
Assassin 44V1.5 Walkthrough (vCaida)
Prelims (as written by Mike below)
a) 20(3)n23 = {389/479/569/578} (no 1,2) = {(5/9)..}
b) 21(3)n1 = {489/579/678} (no 1..3) = {(6/9)..}, {(7/8)..}
c) 8(2)n2 = {17/26/35} (no 4,8,9)
d) 12(4)n14 = {1236/1245} (no 7..9); {12} locked -> no 1,2 in r1c2
e) 12(2)n2 = {39/48/57} (no 1,2,6)
f) 13(2)n8 = {49/58/67} (no 1..3)
g) 6(2)n8 = {15/24} (no 3,6..9)
h) 19(3)n78 = {289/379/469/478/568} (no 1)
1. Outies n1: r1c4+r4c2 = 6(2) = {15/24]} (no 3,6..9)
(Note: cannot be {33} as this would leave nowhere to place the 3 in n1)
2. Outies n9: r6c8+r9c7 = 7(2) = {16/25/34} (no 7..9)
3. Innies r1: r1c59 = 7(2) = {16/25/34} (no 7..9)
4. Innies r9: r9c15 = 6(2) = {15/24} (no 3,6..9)
5. Outies c123: r189c4 = 11(3) = {128/137/146/236/245} (no 9)
6. Outies c789: r129c6 = 19(3) = {289/379/469/478/568} (no 1)
6a. max r9c6 = 6 -> min r12c6 = 13
6b. -> no 2,3 in r12c6
6c. cleanup: no 6 in r6c8 (step 2)
7. as worded by mhparker
h6(2) at 49c15 (step 4) has one cell within n8
7a. -> h6(2) at r9c15 and 6(2)n8 cannot contain the same combo
7b. -> r78c6+r9c15 = 12(4) = {1245}
7c. -> no 2,4,5, in r9c46 (common peers)
7d. cleanup: no 2,3,5 in r6c8 (step 2)
7e. r129c6 (step 6) = {79}[3]/{49}[6]/{58}[6]
7f. -> no 6 in r12c6
10. trial and error #1: consider r4c2 = 1
10a. -> r1c4=5 (step 1)
10b. -> r1c59 (step 3) = {16/34}
10c. -> 12(4)n14 = {245}[1]/{236}[1]
10d. -> 18(4)n12 = {139}[5]/{148}[5] – all of these are in conflict with step 10b.
10e. RESULT no 1 in r4c2; no 5 in r1c4
10f. RESULT 1 locked in 12(4)n14 in r3n1
10g. cleanup: no 7 in r2c4
11. trial and error #2a: consider 12(4)n14 = {136}[2]
11a. -> r1c4=4 (step 1)
11b. -> 12(2)n2 = {57}
11c. -> 8(2)n2 = [62]
11d. -> all possibilities for r129c6 (step 6) are eliminated
11e. CONCLUSION 12(4)n14 <> {136}[2]
12. trial and error #2b: consider 12(4)n12 = {145}[2]
12a. -> r1c4=4 (step 1)
12b. ->12(2)n2 = {39}
12c. -> 12(3)n1 = {678}
12d. -> 18(4)n14 = {239}[4]
12e. -> r1c59 = [61] (step 3)
12f. -> r2c6 = 5
12g. -> 6(2)n8 = {24}
12f. -> 21(4)n28 = [6285]
12g. -> 13(2)n8 = {67}
12h. -> all possibilities for r129c6 (step 6) are eliminated
12i. CONCLUSION 12(4)n14 <> {145}[2]
12j. RESULT no 2 in r4c2 and no 4 in r1c4
12k. RESULT 2 locked in 12(4)n14 in r3n1
12l. cleanup: no 3,6 in 12(4)n14; no 6 in r2c4
12m. cleanup: 3 locked in 18(4)n12
12n. cleanup: no 4 in r1c59 (step 3)
12o. cleanup: no 4,5 in r12c2 (b/c locked in 12(4)
12p. 18(4)n12 = {368}[1]/{349}[2]
Note all other options blocked either by h7(2) from step 3 or 12(4)n14
12q. no 5,7 in 18(4)n12
13. r189c4 (step 5) = 11(3) = [128]/[1]{37}/[146]/[218]
13a. -> r8c4 no 5,6,8
13b. -> 1 locked in r189 for c4
13c. cleanup: no 7 in r3c4
13d. 3 locked in 8(2)n2 and h11(3) (step 5) for c4 (no 3 in r456c4)
14. h7(2)n23 (step 3) blocks {569} combo for 20(3)n23 (Prelims a) (no 6)
15.{12} placement in c56 (taken almost verbatim from mhparker)
15a. 6(2)n9 must contain exactly 1 of {12}
15b. -> r456c6 must contain the other {12} pairing (contains exactly 1 of {12})
15c. 21(4)n28 cannot contain 2 of {12} -> {1299} not possible
15d. because of step 15b. r456c5 cannot contain more than 1 of {12}
15e. 21(4)n28 and r456c5 must each contain exactly 1 of {12}
15f. no 1,2 in r456c4
15g. cleanup: 2 now locked for c4 in 8(2)n2 and h11(3)n23 (step 3) blocks combo [146] from h11(3) (step 13)
15h. no 4,6 in r89c4
16. 21(3)n1 = {579/678} (no 4)
16a. 21(4)n28 = {1479/1569/1578/2469/2478/2568} (no 3)
16b. cleanup: h19(3)(step 6) = {79}[3]/{49}[6]/{58}[6]
16c. -> r12c6 no 3,6
16d. 3 in n2 locked in 8(2) and 12(2)
16e. no {57} in 12(2)n2
17. r1c6 = 7 (only 7 in n2 as others blocked by n1)
17a. cleanup: h19(3)(step 6) = [793]
17b. r6c4 = 4 (step 2)
17c. 12(2)n2 = {48}
17d. r4c2 = 4 (required for 12(4)n14)
17e. r1c4 = 2 (step 1)
17f. 18(4)n12 = {349}[2]
18. r8c4 = 1 (only 1 in c4)
18a. r9c4 = 8 (step 5)
18b. {24} locked in 6(2)n8
18c. -> r9c15 = [15] (step 4)
18d. 21(4)n28 = [6195]
18e. 8(2)n2 = [53]
18f. 13(2)n8 = [67]
19. r5c4 = 4 (only 4 in c)
19a. r1c9 = 1 (step 3)
19b. 12(2)n2 = [48]
19c. cleanup: no 6 in r9c23; no 7 in r9c3
20. 17(4)n3 = [1349]
20a. r2c7 = 2
21. 19(4)n69 = [4]{159}
21a. 20(3)n78 = {29}[8]
21b. 21(4)n89 = [34}{67}
22. r7c6 = 2
22a. r8c6 = 4
24. 24(4)n236 = [92]{67}
24a. {67} locked in r39 for c9 and in r9c89 for r9n9 and in r34 for c7
25. 22(4)n6 = [8]{67}[1]
25a. r8c3 = 5
26. 15(4)n478 = [6351]
26a. r7c1 = 4
26b. {67} locked in r28c2
26c. r1c3 = 4
27. 16(4)n47 = {237}4
27a. r1c12 = [93]
27b. r2c123 = [867]
27c. r3c1 = [5]
27d. r8c12 = [67]
28. 13(3)n5 = [4][36/81] (no 2,5)
29. 20(4)n6 = {1379/1568} (no 2)
29a. r8c8 = 2
30. 16(3)n5r1 = [781/925] can’t be [736] as this is blocked by r4c7 (no 3,6)
30a. 13(3)n5 = [436]
31. 20(4)n6 = {1379/1568}
31a. -> r4c8 (no 8) (it contains only 6)
31b. -> r6c7 (no 9) (in contains only 3)
32. Innies r4: r4c13789 = 25 = {13678} (all other options blocked by 16(3) = [781/925]
(Note option {23569} blocked by 20(4)n6
32a. 16(3)n5r4 = [925]
32b. 16(3)n5r6 = [781]
32c. Revisit of 20(4)n6 (step 31); r5c78 no 5
32d. {38} locked in r48c9
33. 21(4)n36 = [73]{29}/[68]{25}
33a. but if r34c9 = [73] then r4c1 = 7 then r4c7 = 6 then blocks r3c7
33b RESULT 21(4)n36 = [68]{25}
34. Now everything just falls into place (yay!)
Any suggestions/pointers/corrections would be most appreciated!
Caida
edited to complete the walkthrough and fix the formating
if anyone has any suggestions please let me know!
I've noticed that there is no walkthrough posted for Assasin 44V1.5 (and saw that it was noted in the "Unsolvable" thread.
I have solved it - but afraid I only have a partial walkthrough as at the end I wasn't quite sure what I was doing. Would love if someone could help me out.
Here is my complete walkthrough:
I've redone it to fit the formatting guidelines and have borrowed some steps from Mike below so that I could finish it. I think it is still somewhat different as I occasionally seemed to take a much different route.
Assassin 44V1.5 Walkthrough (vCaida)
Prelims (as written by Mike below)
a) 20(3)n23 = {389/479/569/578} (no 1,2) = {(5/9)..}
b) 21(3)n1 = {489/579/678} (no 1..3) = {(6/9)..}, {(7/8)..}
c) 8(2)n2 = {17/26/35} (no 4,8,9)
d) 12(4)n14 = {1236/1245} (no 7..9); {12} locked -> no 1,2 in r1c2
e) 12(2)n2 = {39/48/57} (no 1,2,6)
f) 13(2)n8 = {49/58/67} (no 1..3)
g) 6(2)n8 = {15/24} (no 3,6..9)
h) 19(3)n78 = {289/379/469/478/568} (no 1)
1. Outies n1: r1c4+r4c2 = 6(2) = {15/24]} (no 3,6..9)
(Note: cannot be {33} as this would leave nowhere to place the 3 in n1)
2. Outies n9: r6c8+r9c7 = 7(2) = {16/25/34} (no 7..9)
3. Innies r1: r1c59 = 7(2) = {16/25/34} (no 7..9)
4. Innies r9: r9c15 = 6(2) = {15/24} (no 3,6..9)
5. Outies c123: r189c4 = 11(3) = {128/137/146/236/245} (no 9)
6. Outies c789: r129c6 = 19(3) = {289/379/469/478/568} (no 1)
6a. max r9c6 = 6 -> min r12c6 = 13
6b. -> no 2,3 in r12c6
6c. cleanup: no 6 in r6c8 (step 2)
7. as worded by mhparker
h6(2) at 49c15 (step 4) has one cell within n8
7a. -> h6(2) at r9c15 and 6(2)n8 cannot contain the same combo
7b. -> r78c6+r9c15 = 12(4) = {1245}
7c. -> no 2,4,5, in r9c46 (common peers)
7d. cleanup: no 2,3,5 in r6c8 (step 2)
7e. r129c6 (step 6) = {79}[3]/{49}[6]/{58}[6]
7f. -> no 6 in r12c6
10. trial and error #1: consider r4c2 = 1
10a. -> r1c4=5 (step 1)
10b. -> r1c59 (step 3) = {16/34}
10c. -> 12(4)n14 = {245}[1]/{236}[1]
10d. -> 18(4)n12 = {139}[5]/{148}[5] – all of these are in conflict with step 10b.
10e. RESULT no 1 in r4c2; no 5 in r1c4
10f. RESULT 1 locked in 12(4)n14 in r3n1
10g. cleanup: no 7 in r2c4
11. trial and error #2a: consider 12(4)n14 = {136}[2]
11a. -> r1c4=4 (step 1)
11b. -> 12(2)n2 = {57}
11c. -> 8(2)n2 = [62]
11d. -> all possibilities for r129c6 (step 6) are eliminated
11e. CONCLUSION 12(4)n14 <> {136}[2]
12. trial and error #2b: consider 12(4)n12 = {145}[2]
12a. -> r1c4=4 (step 1)
12b. ->12(2)n2 = {39}
12c. -> 12(3)n1 = {678}
12d. -> 18(4)n14 = {239}[4]
12e. -> r1c59 = [61] (step 3)
12f. -> r2c6 = 5
12g. -> 6(2)n8 = {24}
12f. -> 21(4)n28 = [6285]
12g. -> 13(2)n8 = {67}
12h. -> all possibilities for r129c6 (step 6) are eliminated
12i. CONCLUSION 12(4)n14 <> {145}[2]
12j. RESULT no 2 in r4c2 and no 4 in r1c4
12k. RESULT 2 locked in 12(4)n14 in r3n1
12l. cleanup: no 3,6 in 12(4)n14; no 6 in r2c4
12m. cleanup: 3 locked in 18(4)n12
12n. cleanup: no 4 in r1c59 (step 3)
12o. cleanup: no 4,5 in r12c2 (b/c locked in 12(4)
12p. 18(4)n12 = {368}[1]/{349}[2]
Note all other options blocked either by h7(2) from step 3 or 12(4)n14
12q. no 5,7 in 18(4)n12
13. r189c4 (step 5) = 11(3) = [128]/[1]{37}/[146]/[218]
13a. -> r8c4 no 5,6,8
13b. -> 1 locked in r189 for c4
13c. cleanup: no 7 in r3c4
13d. 3 locked in 8(2)n2 and h11(3) (step 5) for c4 (no 3 in r456c4)
14. h7(2)n23 (step 3) blocks {569} combo for 20(3)n23 (Prelims a) (no 6)
15.{12} placement in c56 (taken almost verbatim from mhparker)
15a. 6(2)n9 must contain exactly 1 of {12}
15b. -> r456c6 must contain the other {12} pairing (contains exactly 1 of {12})
15c. 21(4)n28 cannot contain 2 of {12} -> {1299} not possible
15d. because of step 15b. r456c5 cannot contain more than 1 of {12}
15e. 21(4)n28 and r456c5 must each contain exactly 1 of {12}
15f. no 1,2 in r456c4
15g. cleanup: 2 now locked for c4 in 8(2)n2 and h11(3)n23 (step 3) blocks combo [146] from h11(3) (step 13)
15h. no 4,6 in r89c4
16. 21(3)n1 = {579/678} (no 4)
16a. 21(4)n28 = {1479/1569/1578/2469/2478/2568} (no 3)
16b. cleanup: h19(3)(step 6) = {79}[3]/{49}[6]/{58}[6]
16c. -> r12c6 no 3,6
16d. 3 in n2 locked in 8(2) and 12(2)
16e. no {57} in 12(2)n2
17. r1c6 = 7 (only 7 in n2 as others blocked by n1)
17a. cleanup: h19(3)(step 6) = [793]
17b. r6c4 = 4 (step 2)
17c. 12(2)n2 = {48}
17d. r4c2 = 4 (required for 12(4)n14)
17e. r1c4 = 2 (step 1)
17f. 18(4)n12 = {349}[2]
18. r8c4 = 1 (only 1 in c4)
18a. r9c4 = 8 (step 5)
18b. {24} locked in 6(2)n8
18c. -> r9c15 = [15] (step 4)
18d. 21(4)n28 = [6195]
18e. 8(2)n2 = [53]
18f. 13(2)n8 = [67]
19. r5c4 = 4 (only 4 in c)
19a. r1c9 = 1 (step 3)
19b. 12(2)n2 = [48]
19c. cleanup: no 6 in r9c23; no 7 in r9c3
20. 17(4)n3 = [1349]
20a. r2c7 = 2
21. 19(4)n69 = [4]{159}
21a. 20(3)n78 = {29}[8]
21b. 21(4)n89 = [34}{67}
22. r7c6 = 2
22a. r8c6 = 4
24. 24(4)n236 = [92]{67}
24a. {67} locked in r39 for c9 and in r9c89 for r9n9 and in r34 for c7
25. 22(4)n6 = [8]{67}[1]
25a. r8c3 = 5
26. 15(4)n478 = [6351]
26a. r7c1 = 4
26b. {67} locked in r28c2
26c. r1c3 = 4
27. 16(4)n47 = {237}4
27a. r1c12 = [93]
27b. r2c123 = [867]
27c. r3c1 = [5]
27d. r8c12 = [67]
28. 13(3)n5 = [4][36/81] (no 2,5)
29. 20(4)n6 = {1379/1568} (no 2)
29a. r8c8 = 2
30. 16(3)n5r1 = [781/925] can’t be [736] as this is blocked by r4c7 (no 3,6)
30a. 13(3)n5 = [436]
31. 20(4)n6 = {1379/1568}
31a. -> r4c8 (no 8) (it contains only 6)
31b. -> r6c7 (no 9) (in contains only 3)
32. Innies r4: r4c13789 = 25 = {13678} (all other options blocked by 16(3) = [781/925]
(Note option {23569} blocked by 20(4)n6
32a. 16(3)n5r4 = [925]
32b. 16(3)n5r6 = [781]
32c. Revisit of 20(4)n6 (step 31); r5c78 no 5
32d. {38} locked in r48c9
33. 21(4)n36 = [73]{29}/[68]{25}
33a. but if r34c9 = [73] then r4c1 = 7 then r4c7 = 6 then blocks r3c7
33b RESULT 21(4)n36 = [68]{25}
34. Now everything just falls into place (yay!)
Any suggestions/pointers/corrections would be most appreciated!
Caida
edited to complete the walkthrough and fix the formating
if anyone has any suggestions please let me know!
Last edited by Caida on Tue Nov 06, 2007 11:53 pm, edited 2 times in total.
Hi Caida,
I hope you don't mind, but I've completed the puzzle based on your first 8 steps only, since I managed to find a non-tryfurcation solving path from this position. I also re-wrote your steps to conform more to the established quasi-standard on this forum, involving (principally):
Look forward to seeing you on the A76 thread!
P.S. Belated thanks for the very enjoyable puzzle, Para! The puzzle consisted of three phases: an interesting and challenging start, a long straightforward bit in the middle, and difficult phase (where the puzzle seized up again) at the end.
Assassin 44 V1.5 Walkthrough
Prelims
a) 20(3)n23 = {389/479/569/578} (no 1,2) = {(5/9)..}
b) 21(3)n1 = {489/579/678} (no 1..3) = {(6/9)..}, {(7/8)..}
c) 8(2)n2 = {17/26/35} (no 4,8,9)
d) 12(4)n14 = {1236/1245} (no 7..9); {12} locked -> no 1,2 in r1c2
e) 12(2)n2 = {39/48/57} (no 1,2,6)
f) 13(2)n8 = {49/58/67} (no 1..3)
g) 6(2)n8 = {15/24} (no 3,6..9)
h) 19(3)n78 = {289/379/469/478/568} (no 1)
First 7 steps from Caida:
1. Outies n1: r1c4+r4c2 = 6(2) = {15/24} (no 3,6..9)
(Note: cannot be {33}, as this would leave nowhere to place the 3 in n1)
2. Outies n9: r6c8+r9c7 = 7(2) = {16/25/34} (no 7..9)
3. Innies r1: r1c59 = 7(2) = {16/25/34} (no 7..9)
4. Innies r9: r9c15 = 6(2) = {15/24} (no 3,6..9)
5. Outies c123: r189c4 = 11(3) = {128/137/146/236/245} (no 9)
6. Outies c789: r129c6 = 19(3) = {289/379/469/478/568} (no 1)
6a. max. r9c6 = 6 -> min. r12c6 = 13
6b. -> no 2,3 in r12c6
6c. cleanup: no 6 in r6c8 (step 2)
7. h6(2) at r9c15 (step 4) has one cell within n8
7a. -> h6(2) at r9c15 and 6(2)n8 cannot contain the same combo
7b. -> r78c6+r9c15 = 12(4) = {1245}
7c. -> no 2,4,5 in r9c46 (common peers)
7d. cleanup: no 2,3,5 in r6c8 (step 2)
7e. r129c6 (step 6) = {79}[3]/{49}[6]/{58}[6]
7f. -> no 6 in r12c6
Remaining steps from mhparker:
8. r1c59 (step 3) cannot contain both of {12}
8a. only other place in r1 for {12} is within 18(4) cage
8a. -> 18(4)n12 must contain at least 1 of {12} ({3456} blocked)
8b. furthermore 18(4) cannot be {1359} (blocked by 20(3)n23 (Prelims a))...
8c. ...or either of {1269/1278} (blocked by 21(3)n1 (Prelims b))
8d. ...or either of {1458/2367} (blocked by h7(2)n23 (step 3))
8d. -> possible combos for 18(4)n12 = {1368/1467/2349/2358/2457} = {(3/4)..} (no eliminations yet)
9. 18(4)n12 blocks {34} combo for h7(2)n23 (step 3) = {16/25} (no 3,4) = {(5/6)..}
10. h7(2)n23 (step 9) blocks {569} combo for 20(3)n23 (Prelims a) = {389/479/578} (no 6)
11. Distribution of {12} in c56:
11a. r78c6 must contain exactly 1 of {12}
11b. only other place for {12} in c6 is r456c6
11c. -> r456c6 must contain exactly 1 of {12}
11d. r1289c5 cannot contain both of {12} due to 21(4) cage sum
11e. only other place for {12} in c5 is r456c5, which cannot contain BOTH of {12} due to r456c6 (step 11c)
11f. -> r456c5 must contain exactly 1 of {12}
11g. -> r1289c5 must also contain exactly 1 of {12} = {1479/1569/1578/2379/2469/2478/2568} (no 3)
(Note: {1389} unplaceable because because r19c5 only have 1 of {1389} between them (=1))
11h. r456c6 (step 11c) and r456c5 (step 11f) form killer pair on {12} within n5
11i. -> no 1,2 in r456c4
12. Distribution of 3 in c56:
12a. from step 7e: if r9c6 = 3, then r12c6 = {79} -> 12(2)n2 <> {39}
12b. -> r9c6 and 12(2)n2 cannot both contain a 3
12c. -> the second 3 of c56 must be contained within r456c56
12d. -> no 3 in r456c4
13. Innies c4: r4567c4 = 26(4), 1..3 unavailable, 9 of c4 locked = {4(58/67)9}
13a. -> 4 locked for c4
13b. cleanup: no 2 in r4c2 (step 1)
14. 4 now unavailable to r189c4 (step 5) = {1(28/37)} (no 5,6)
(Note: {236} combo blocked by r9c6)
14a. {12} only in r18c4
14b. -> no 8 in r8c4
14c. 1 locked for c4
14c. cleanup: no 7 in r23c4; no 1 in r4c2 (step 1)
Should have seen this next move right near the start:
15. r7c45, r89c4, and r8c5 form a hidden killer triple on {789} in n8
15a. -> r8c5 = {789}
Now back on track:
16. 12(4)n14 (Prelims d) must contain 1 of {45} due to r4c2 = {1245} (no 3,6) (last combo)
16a. {12} locked in r3c123 for r3 and n1
16b. no 4,5 in r12c2 (CPE)
16c. cleanup: no 6 in r2c4
17. 3 in n1 locked in 18(4)n12 (step 8d) = {1368/2349/2358} (no 7)
17a. 3 locked for r1
18. 7 in n1 locked in 21(3)n1 (Prelims b) = {579/678} (no 4)
18a. 7 locked for r2
19. Consider positions for 3 in n2:
19a. Either 3 is in 12(2)n2 = {39}, or...
19b. ...3 is in 8(2)n2 = {35} -> 12(2)n2 <> {57}
19c. Either way, 12(2)n2 cannot be {57}
19d. -> no 5,7 in r3c56
20. Hidden single (HS) in n2 at r1c6 = 7
20a. -> r29c6 = [93] (step 7e)
20b. -> r6c8 = 4 (step 2)
20c. cleanup: no 9 in r1c7, no 3 in r3c5
21. 3 in c4/n2 locked in 8(2)n2 = {35} (no 2,6)
21a. 5 locked in r23c4 for c4 and n2
21b. cleanup: no 8 in r7c5
22. 6 in c6 locked in n5 -> not elsewhere in n5
23. HS in c4 at r7c4 = 6
23a. -> r7c5 = 7
24. r9c4+r8c5 = [89]
25. 6(2)n8 and r8c4 form killer pair on {12} within n8
25a. -> no 1,2 in r9c5
25b. cleanup: no 4,5 in r9c1 (step 4)
26. Naked pair (NP) at r3c56 -> no 4,8 elsewhere in r3 and n2
27. HS in 12(4)n14 (step 16) at r4c2 = 4
27a. -> r1c4 = 2 (step 1)
27b. 5 in 12(4)n14 locked in r3c123 for r3 and n1
28. r2358c4 = [5341] (singles and cage sums)
28a. cleanup: no 5 in r78c6
29. HS in n8 at r9c5 = 5
29a. -> r9c1 = 1 (step 4)
30. HS in c5 at r3c5 = 4
30a. -> r3c6 = 8
31. Naked triple (NT) at r2c123 -> no 6,7,8 elsewhere in r2 and n1
32. r12c5 = [61]
33. HS in r1/n3 at r1c9 = 1
34. Hidden triple (HT) in c6 at r456c6 = {156} (no 2)
35. Split 16(3) at r2c89+r3c8 = [349] (last combo/permutation)
36. Naked single (NS) at r2c7 = 2
36a. split 13(3) at r34c7 = {67}, locked for c7
37. 1 unavailable to 16(4)n47 = {23(47/56)} (no 8,9)
37a. {23} locked for c1
38. HS in c1 at r1c1 = 9
38a. -> r1c23 = [34]
39. HS in r9 at r9c7 = 4
39a. r9c89 = {67} (no 2,9) (last combo), locked for r9 and n9
40. Split 15(3) at r7c789 = {159/258} (no 3)
40a. 5 locked for r7 and n9
41. NS at r7c1 = 4
42. r78c6 = [24]
43. HS in r7 at r7c3 = 3
44. Split 15(3) at r7c789 = {159} (no 8) (last combo)
45. HS in r7 at r7c2 = 8
46. HS in c1 at r2c1 = 8
47. Split 12(3) at r456c1 = {237} (no 5,6) (last combo), locked for c1 and n4
48. NS at r3c1 = 5
49. HS in c1 at r8c1 = 6
49a. -> r8c2 = 7 (cage sum)
50. HS in r8/c7 at r8c3 = 5
50a. -> r6c3 = 6 (cage sum)
51. r2c23 = [67]
52. Naked pair (NP) at r39c9 -> no 6,7 elsewhere in c9
53. 21(4)n36 must have (exactly) 1 of {67} due to r3c9, {14} unavailable = {2(379/568)} = {(3/5)..}
53a. 2 locked in r456c9 for c9 and n6
54. HS in c8/n9 at r8c8 = 2
55. Split 9(2) at r5c56 = [36/81] (no 2,5)
56. 16(3)n5 = [781/925] (no 3)
57. 16(3)n5 = [781/925] (no 3,6)
{Note: [736] blocked by r4c7)
58. Naked pair (NP) at r46c5 -> no 2,8 elsewhere in c5
58a. -> r5c5 = 3
58b. -> r5c6 = 6 (cage sum)
59. {24} unavailable to 20(4)n6, 1 of n6 locked = {1379/1568}
59a. 6 only in r4c8
59b. -> no {58} in r4c8
59c. 3 only in r6c7
59d. -> no 9 in r6c7
60. Innies r4: r4c13789 = 25(5) w/ 4 unavailable, {36} of r4 locked = {13678} (no 2,5,9), locked for r4
(Note: {23569} unplaceable because because r4c78 only have 1 of {23569} between them (=6))
61. r4c456 = [925]
62. r6c456 = [781]
63. Hidden pair (HP) in c7 at r57c7 = {19}
64. NP at r48c9 -> no 3,8 elsewhere in c9
65. 21(4)n36 and r6c7 form killer pair on {35} -> no 5 in r5c8
66. Innies n6: r4c7+r456c9 = {2568}
(Note: {2379} blocked by r4c1)
66a. -> r4c79 = [68]
66b. r56c9 = {25}, locked for c9 and n6
Now (finally!) all naked singles to end.
Thanks very much for your partial WT. This puzzle has been on several forum members' "to do" list for a long time. Andrew mentioned it again to me only a few days ago, and Ed has also been talking about starting it. You managed to beat us all to it!Caida wrote:Any suggestions/pointers/corrections/completion would be most appreciated!
I hope you don't mind, but I've completed the puzzle based on your first 8 steps only, since I managed to find a non-tryfurcation solving path from this position. I also re-wrote your steps to conform more to the established quasi-standard on this forum, involving (principally):
- Use of curly brackets {} for unordered sets (combinations), and square brackets [] for ordered sets (permutations).
- Cleanups as sub-steps, rather than steps in their own right.
- Blank line between each step.
Look forward to seeing you on the A76 thread!
P.S. Belated thanks for the very enjoyable puzzle, Para! The puzzle consisted of three phases: an interesting and challenging start, a long straightforward bit in the middle, and difficult phase (where the puzzle seized up again) at the end.
Assassin 44 V1.5 Walkthrough
Prelims
a) 20(3)n23 = {389/479/569/578} (no 1,2) = {(5/9)..}
b) 21(3)n1 = {489/579/678} (no 1..3) = {(6/9)..}, {(7/8)..}
c) 8(2)n2 = {17/26/35} (no 4,8,9)
d) 12(4)n14 = {1236/1245} (no 7..9); {12} locked -> no 1,2 in r1c2
e) 12(2)n2 = {39/48/57} (no 1,2,6)
f) 13(2)n8 = {49/58/67} (no 1..3)
g) 6(2)n8 = {15/24} (no 3,6..9)
h) 19(3)n78 = {289/379/469/478/568} (no 1)
First 7 steps from Caida:
1. Outies n1: r1c4+r4c2 = 6(2) = {15/24} (no 3,6..9)
(Note: cannot be {33}, as this would leave nowhere to place the 3 in n1)
2. Outies n9: r6c8+r9c7 = 7(2) = {16/25/34} (no 7..9)
3. Innies r1: r1c59 = 7(2) = {16/25/34} (no 7..9)
4. Innies r9: r9c15 = 6(2) = {15/24} (no 3,6..9)
5. Outies c123: r189c4 = 11(3) = {128/137/146/236/245} (no 9)
6. Outies c789: r129c6 = 19(3) = {289/379/469/478/568} (no 1)
6a. max. r9c6 = 6 -> min. r12c6 = 13
6b. -> no 2,3 in r12c6
6c. cleanup: no 6 in r6c8 (step 2)
7. h6(2) at r9c15 (step 4) has one cell within n8
7a. -> h6(2) at r9c15 and 6(2)n8 cannot contain the same combo
7b. -> r78c6+r9c15 = 12(4) = {1245}
7c. -> no 2,4,5 in r9c46 (common peers)
7d. cleanup: no 2,3,5 in r6c8 (step 2)
7e. r129c6 (step 6) = {79}[3]/{49}[6]/{58}[6]
7f. -> no 6 in r12c6
Remaining steps from mhparker:
8. r1c59 (step 3) cannot contain both of {12}
8a. only other place in r1 for {12} is within 18(4) cage
8a. -> 18(4)n12 must contain at least 1 of {12} ({3456} blocked)
8b. furthermore 18(4) cannot be {1359} (blocked by 20(3)n23 (Prelims a))...
8c. ...or either of {1269/1278} (blocked by 21(3)n1 (Prelims b))
8d. ...or either of {1458/2367} (blocked by h7(2)n23 (step 3))
8d. -> possible combos for 18(4)n12 = {1368/1467/2349/2358/2457} = {(3/4)..} (no eliminations yet)
9. 18(4)n12 blocks {34} combo for h7(2)n23 (step 3) = {16/25} (no 3,4) = {(5/6)..}
10. h7(2)n23 (step 9) blocks {569} combo for 20(3)n23 (Prelims a) = {389/479/578} (no 6)
11. Distribution of {12} in c56:
11a. r78c6 must contain exactly 1 of {12}
11b. only other place for {12} in c6 is r456c6
11c. -> r456c6 must contain exactly 1 of {12}
11d. r1289c5 cannot contain both of {12} due to 21(4) cage sum
11e. only other place for {12} in c5 is r456c5, which cannot contain BOTH of {12} due to r456c6 (step 11c)
11f. -> r456c5 must contain exactly 1 of {12}
11g. -> r1289c5 must also contain exactly 1 of {12} = {1479/1569/1578/2379/2469/2478/2568} (no 3)
(Note: {1389} unplaceable because because r19c5 only have 1 of {1389} between them (=1))
11h. r456c6 (step 11c) and r456c5 (step 11f) form killer pair on {12} within n5
11i. -> no 1,2 in r456c4
12. Distribution of 3 in c56:
12a. from step 7e: if r9c6 = 3, then r12c6 = {79} -> 12(2)n2 <> {39}
12b. -> r9c6 and 12(2)n2 cannot both contain a 3
12c. -> the second 3 of c56 must be contained within r456c56
12d. -> no 3 in r456c4
13. Innies c4: r4567c4 = 26(4), 1..3 unavailable, 9 of c4 locked = {4(58/67)9}
13a. -> 4 locked for c4
13b. cleanup: no 2 in r4c2 (step 1)
14. 4 now unavailable to r189c4 (step 5) = {1(28/37)} (no 5,6)
(Note: {236} combo blocked by r9c6)
14a. {12} only in r18c4
14b. -> no 8 in r8c4
14c. 1 locked for c4
14c. cleanup: no 7 in r23c4; no 1 in r4c2 (step 1)
Should have seen this next move right near the start:
15. r7c45, r89c4, and r8c5 form a hidden killer triple on {789} in n8
15a. -> r8c5 = {789}
Now back on track:
16. 12(4)n14 (Prelims d) must contain 1 of {45} due to r4c2 = {1245} (no 3,6) (last combo)
16a. {12} locked in r3c123 for r3 and n1
16b. no 4,5 in r12c2 (CPE)
16c. cleanup: no 6 in r2c4
17. 3 in n1 locked in 18(4)n12 (step 8d) = {1368/2349/2358} (no 7)
17a. 3 locked for r1
18. 7 in n1 locked in 21(3)n1 (Prelims b) = {579/678} (no 4)
18a. 7 locked for r2
19. Consider positions for 3 in n2:
19a. Either 3 is in 12(2)n2 = {39}, or...
19b. ...3 is in 8(2)n2 = {35} -> 12(2)n2 <> {57}
19c. Either way, 12(2)n2 cannot be {57}
19d. -> no 5,7 in r3c56
20. Hidden single (HS) in n2 at r1c6 = 7
20a. -> r29c6 = [93] (step 7e)
20b. -> r6c8 = 4 (step 2)
20c. cleanup: no 9 in r1c7, no 3 in r3c5
21. 3 in c4/n2 locked in 8(2)n2 = {35} (no 2,6)
21a. 5 locked in r23c4 for c4 and n2
21b. cleanup: no 8 in r7c5
22. 6 in c6 locked in n5 -> not elsewhere in n5
23. HS in c4 at r7c4 = 6
23a. -> r7c5 = 7
24. r9c4+r8c5 = [89]
25. 6(2)n8 and r8c4 form killer pair on {12} within n8
25a. -> no 1,2 in r9c5
25b. cleanup: no 4,5 in r9c1 (step 4)
26. Naked pair (NP) at r3c56 -> no 4,8 elsewhere in r3 and n2
27. HS in 12(4)n14 (step 16) at r4c2 = 4
27a. -> r1c4 = 2 (step 1)
27b. 5 in 12(4)n14 locked in r3c123 for r3 and n1
28. r2358c4 = [5341] (singles and cage sums)
28a. cleanup: no 5 in r78c6
29. HS in n8 at r9c5 = 5
29a. -> r9c1 = 1 (step 4)
30. HS in c5 at r3c5 = 4
30a. -> r3c6 = 8
31. Naked triple (NT) at r2c123 -> no 6,7,8 elsewhere in r2 and n1
32. r12c5 = [61]
33. HS in r1/n3 at r1c9 = 1
34. Hidden triple (HT) in c6 at r456c6 = {156} (no 2)
35. Split 16(3) at r2c89+r3c8 = [349] (last combo/permutation)
36. Naked single (NS) at r2c7 = 2
36a. split 13(3) at r34c7 = {67}, locked for c7
37. 1 unavailable to 16(4)n47 = {23(47/56)} (no 8,9)
37a. {23} locked for c1
38. HS in c1 at r1c1 = 9
38a. -> r1c23 = [34]
39. HS in r9 at r9c7 = 4
39a. r9c89 = {67} (no 2,9) (last combo), locked for r9 and n9
40. Split 15(3) at r7c789 = {159/258} (no 3)
40a. 5 locked for r7 and n9
41. NS at r7c1 = 4
42. r78c6 = [24]
43. HS in r7 at r7c3 = 3
44. Split 15(3) at r7c789 = {159} (no 8) (last combo)
45. HS in r7 at r7c2 = 8
46. HS in c1 at r2c1 = 8
47. Split 12(3) at r456c1 = {237} (no 5,6) (last combo), locked for c1 and n4
48. NS at r3c1 = 5
49. HS in c1 at r8c1 = 6
49a. -> r8c2 = 7 (cage sum)
50. HS in r8/c7 at r8c3 = 5
50a. -> r6c3 = 6 (cage sum)
51. r2c23 = [67]
52. Naked pair (NP) at r39c9 -> no 6,7 elsewhere in c9
53. 21(4)n36 must have (exactly) 1 of {67} due to r3c9, {14} unavailable = {2(379/568)} = {(3/5)..}
53a. 2 locked in r456c9 for c9 and n6
54. HS in c8/n9 at r8c8 = 2
55. Split 9(2) at r5c56 = [36/81] (no 2,5)
56. 16(3)n5 = [781/925] (no 3)
57. 16(3)n5 = [781/925] (no 3,6)
{Note: [736] blocked by r4c7)
58. Naked pair (NP) at r46c5 -> no 2,8 elsewhere in c5
58a. -> r5c5 = 3
58b. -> r5c6 = 6 (cage sum)
59. {24} unavailable to 20(4)n6, 1 of n6 locked = {1379/1568}
59a. 6 only in r4c8
59b. -> no {58} in r4c8
59c. 3 only in r6c7
59d. -> no 9 in r6c7
60. Innies r4: r4c13789 = 25(5) w/ 4 unavailable, {36} of r4 locked = {13678} (no 2,5,9), locked for r4
(Note: {23569} unplaceable because because r4c78 only have 1 of {23569} between them (=6))
61. r4c456 = [925]
62. r6c456 = [781]
63. Hidden pair (HP) in c7 at r57c7 = {19}
64. NP at r48c9 -> no 3,8 elsewhere in c9
65. 21(4)n36 and r6c7 form killer pair on {35} -> no 5 in r5c8
66. Innies n6: r4c7+r456c9 = {2568}
(Note: {2379} blocked by r4c1)
66a. -> r4c79 = [68]
66b. r56c9 = {25}, locked for c9 and n6
Now (finally!) all naked singles to end.
Cheers,
Mike
Mike
Hi,
I will definitely take look to have my next walkthrough formatted better (and perhaps even completed!)
Cheers,
Caida
I don't mind at all! I'm so glad that some of my steps were used. Thank you for the full walk through and for the tips in formatting - i was never quite sure what the difference was in the brackets.mhparker wrote:
I hope you don't mind, but I've completed the puzzle based on your first 8 steps only, since I managed to find a non-tryfurcation solving path from this position. I also re-wrote your steps to conform more to the established quasi-standard on this forum,
I will definitely take look to have my next walkthrough formatted better (and perhaps even completed!)
Cheers,
Caida
I like that one too. Found when making the V2's. It is in all A44 puzzles. I think i used a similar step once in another puzzle.mhparker wrote:I particularly liked your step 7. Very clever, and also instrumental in solving this puzzle.
You can blame Ruud for that one. The ending is the same as in the original.mhparker wrote: Now (finally!) all naked singles to end.
greetings
Para