More like an annoying drip - before the dam finally broke.Ruud wrote:it is one long stream of innie-outie differences.
This solution is a joint effort between me and Peter. We had a terrible time with this puzzle - got tricked and totally stumped so many times. In the end, found a way through that avoided most of the innie-outie nightmares (step 20).
I'll be taking a break for a while. Going to NZ next Monday for a couple of weeks doing adventure stuff. Need a total break from anything to do with Ruud for a long while after the last month .
Please let me know if there is anything not valid, accurate or clear in this walk-through.
See you all anon. Ed
1. "45" n78 -> r6c1 - 8 = r9c6 -> r6c1 = 9, r9c6 = 1
2. n8: 17(2) = {89}, locked in N8 and r7
3. n8: 16(3) = {367/457} = 7{36/45} (no 2): 7 locked for n8,r8
4. r78c1 = 10 = [28/73/{46}] = [3/6/8] & (no 1,5)
5. c1:17(3) = {278/458/467} ({368} blocked by r78c1-step 4) (no 1,3)
6. 1 in c1 locked in 3 innies r129c1 = 9 = {126|135} (no 4,7,8)
7. 1 in c1 also locked in 20(4) n1 = {1289/1379/1469/1568} ({1478} blocked because no {478} in r12c1)
7a. ->1 locked for n1 in r12c1
8. c6:14(2) = [95/{68}] = [5/6,8/9..] -> 8 & 9 locked for c6 in r567c6
9. 45 on n5: r6c4 + r4c6 = 5 = [14/{23}]
10. n56: 18(3) must have {234}(r4c6) = {279|369|378|459|468} =[2/3/4]not two of -> r45c7 = {56789}
11.a. 8 in r7c5 -> no 8 in 14(3) in N5
b. 8 in r7c6 -> 14(2) in N5 = {59} -> 12(2) in N5 = {48}
-> no 8 in 14(3) in N5
12. n5:14(3) = {149/167/239/257} ({347} blocked by 12(2), {356} blocked by 14(2))
13. c7: 8(3) = {125|134} -> 1 locked in c7
14. r9c78 = 12 = {39/48/57} (no 2,6)
15. "45" n9 -> 3 innies: r78c7 + r7c9 = 8 = {125/134} = 1{25/34} no(67):1 locked for n9
16. n9:6 locked in 25(4) = 6{289/379/478} (no 5)
17. "45" n8 -> r9c3 - 3 = r7c4 -> r9c3 = {56789}
17a. when 6 in r9c3, 3 must be in r7c4 and rest of 14(3) in n8 must be {35}. Two 3's in n8 -> no 6 in r9c3, no 3 in r7c4
17b. In summary
[5] in r9c3 -> 2 in r7c4
-> rest of 14(3) = {36} ({27} blocked by 2 in r7c4)
[7] in r9c3 -> 4 in r7c4
-> rest of 14(3) = {25} ({34} blocked by 4 in r7c4)
[8] in r9c3 -> 5 in r7c4
-> rest of 14(3) = {24}
[9] in r9c3 -> 6 in r7c4
-> rest of 14(3) = {23}
18. "45" n14 -> r16c4 = 5 = [41/23/32]
19. "45" n147 -> 5 outies = 16
20. Combining steps 17b, 18 and 19 -> r1678c4 + r9c5 = 16 = [412{36}/41524/416{23}], ([23452/23542/32452/32542] all blocked by 12(2) c4)
-> r16c4 = [41], r7c4 = {256}, r9c3 = {589}, r9c4 = {236}, r9c5 = {2346}, 14(3) = {239/248/356} = [5{36}/824/9{23}]
21. n5: r4c6 = 4 (step 9), 12(2) = {39/57}, 14(3) must have 2 for n5 = 2{39/57}(no 6): 2 locked for c5 and {39/57} are locked in these 2 cages -> 14(2) = {68}: locked for c6 -> r7c56 = [89]
22. "45" n2 -> r1c7 - 2 = r3c6: r1c7 = {579}, r3c6 = {257}
23. r1:19(3). the only valid combination left is {379}: locked for r 1, and 3 also locked for n2. r3c6 = {57}(step 22)
24. "45" n2: 3 innies r1c56 + r3c6 = 17 and must have 3 and made up of candidates {3579} which sums to 24 -> 24 - 17 = 7.-> no 7 -> r3c6 = 5, r1c567 = [937], r8c6 = 7, r2c6 = 2, r2c45 = {67}:locked for n2, r2, r3c45 = {81}
25. n5:14(3) = {257}:locked for n5, c5 -> r2c45 = [76], r45c4 = {39}:locked for c4
25. n2:15(3) must have 2 or 7 (r2c6) = {267}
24. r1c23 = 11 = {56}:locked for r1,n1. 20(4) = 1{289/379} = 19{28/37}(no 4): 9 locked for n1, c2. r23c2 = {789},
25. "45" n1-> r4c3 + 2 = r3c1: r3c1 = {47}, r4c3 = {25}
26. c3:12(3) = {237/345} = 3{27/45} (no 8) with 3 locked in n1 -> r12c1 = [21], r23c2 = [89], r3c3 = 7 (single n1), r24c2 = [32], r3c1 = 4, r78c1 = [73], r1c89 = {18} -> r2c78 = {49}:locked for n3, r3c78 = {26}:locked for n3:....the rest is straight-forward