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 Post subject: CDK v3 Revisit
PostPosted: Sun Jan 03, 2021 1:07 am 
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Posts: 875
Location: Sydney, Australia
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CDK v3 Revisit

Sorry, bit late with this. Home internet crashed. Lots of small cages so hope there is a new, interesting way to solve this. Gets a score of 1.75 and JSudoku uses 2 'complex intersections'.

triple click code:
3x3::k:2048:2817:2817:3075:3075:3075:4870:4870:4870:2048:10:4619:4619:13:3854:4870:16:2833:3090:3090:4619:4619:3854:3854:2328:2833:2833:6427:6427:6427:6427:2335:2335:2328:4386:4386:3620:37:3366:3366:40:2857:2857:43:4386:3620:3620:2095:2096:2096:5426:5426:5426:5426:6198:6198:2095:5433:5433:3643:3643:1085:1085:6198:64:3905:5433:67:3643:3643:70:6727:3905:3905:3905:3915:3915:3915:6727:6727:6727:
solution:
+-------+-------+-------+
| 6 4 7 | 8 3 1 | 9 2 5 |
| 2 5 1 | 4 9 7 | 3 8 6 |
| 3 9 8 | 5 2 6 | 7 4 1 |
+-------+-------+-------+
| 7 6 9 | 3 1 8 | 2 5 4 |
| 1 2 4 | 9 7 5 | 6 3 8 |
| 5 8 3 | 2 6 4 | 1 9 7 |
+-------+-------+-------+
| 9 7 5 | 6 8 2 | 4 1 3 |
| 8 1 2 | 7 4 3 | 5 6 9 |
| 4 3 6 | 1 5 9 | 8 7 2 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: CDK v3 Revisit
PostPosted: Tue Jan 05, 2021 5:43 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1739
Location: Lethbridge, Alberta, Canada
Thanks Ed for the latest Revisit. A nice puzzle! I read the original comments in the archive but, based on them, it didn't seem that hard this time. The original version CDK contained a 45(9) cage R258C258; of course that couldn't be used for the harder variants.

Here is my walkthrough for CDKv3 Revisit:
Prelims

a) R12C1 = {17/26/35}, no 4,8,9
b) R1C23 = {29/38/47/56}, no 1
c) R3C12 = {39/48/57}, no 1,2,6
d) R34C7 = {18/27/36/45}, no 9
e) R4C56 = {18/27/36/45}, no 9
f) R5C34 = {49/58/67}, no 1,2,3
g) R5C67 = {29/38/47/56}, no 1
h) R67C3 = {17/26/35}, no 4,8,9
i) R6C45 = {17/26/35}, no 4,8,9
j) R7C89 = {13}
k) 11(3) cage at R2C9 = {128/137/146/236/245}, no 9
l) 24(3) cage at R7C1 = {789}
m) 21(3) cage at R7C4 = {489/579/678}, no 1,2,3
n) 14(4) cage at R7C6 = {1238/1247/1256/1346/2345}, no 9
o) 26(4) cage at R8C9 = {2789/3689/4589/4679/5678}, no 1

Steps resulting from Prelims
1a. Naked pair {13} in R7C89, locked for R7 and N9, clean-up: no 5,7 in R6C3
1b. Naked triple {789} in 24(3) cage at R7C1, locked for N7, clean-up: no 1 in R6C3
1c. 14(4) cage at R7C6 = {1247/1256/2345} (cannot be {1238/1346} because 1,3 only in R8C6), no 8 -> R8C6 = {13}

2a. {456} of 21(3) cage at R7C4 must be in R7C45 (R7C45 cannot be {78/79/89} which clash with R7C12, ALS block) -> no 4,5,6 in R8C4
2b. Killer triple 7,8,9 in R7C12 and R7C45, locked for R7

3a. 45 rule on N3 2 innies R2C8 + R3C7 = 15 = [78/87/96] -> R4C7 = {123}
3b. 45 rule on N7 2 innies R7C3 + R8C2 = 6 = [24/51] -> R6C3 = {36}
3c. 14(4) cage at R7C6 (step 1c) = {1256/2345} (cannot be {1247} = {24}[17] which clashes with R7C3 + R8C2), no 7
3d. 45 rule on N12 2 innies R2C25 = 14 = {59/68}
3e. 45 rule on N1 3 innies R2C23 + R3C3 = 14, min R2C2 = 5 -> max R23C3 = 9, no 9 in R23C3
3f. 45 rule on N1 2 outies R23C4 = 1 innie R2C2 + 4, IOU no 4 in R3C4
3g. 45 rule on N89 2 innies R8C58 = 10 = [19/37]/{28/46}, no 5, no 7,9 in R8C5
3h. 45 rule on R1 2 outies R2C17 = 5 = [14/23/32] -> R1C1 = {567}
3i. 1 in N1 only in R2C1 + R23C3, CPE no 1 in R2C4
3j. 45 rule on R9 2 outies R8C39 = 11 = [29/38/47]/{56}, no 1 in R8C3, no 2,4 in R8C9
3k. 45 rule on N47 2 outies R45C4 = 2 innies R58C2 + 9
3l. Min R58C2 = 3 -> min R45C4 = 12, no 1,2 in R4C4
3m. Max R45C4 = 17 -> max R58C2 = 8, no 8,9 in R5C2
3n. 45 rule on N8 3 innies R7C6 + R8C56 = 9 = {126/135/234}, no 8, clean-up: no 2 in R8C8

4a. 45 rule on R4 3 innies R4C789 = 11 = {128/137/146/236/245}, no 9
4b. 45 rule on R4 1 outie R5C9 = 1 innie R4C7 + 6, R4C7 = {123} -> R5C9 = {789}
4c. 45 rule on R6 1 innie R6C3 = 1 outie R5C1 + 2, R6C3 = {36} -> R5C1 = {14}
4d. 21(4) cage at R6C6 = {1389/1479/1578/2469/2478/2568/3459} (cannot be {1569/2379/3567} which clash with R6C45, cannot be {3468} which clashes with R6C3)
4e. 21(4) cage = {1479/1578/2469/2478/2568/3459} (cannot be {1389} because 21(4) cage combined with R6C45 = {26} clashes with R6C3)

5a. 14(4) cage at R7C6 (step 3c) = {1256/2345}, R7C6 + R8C56 (step 3n) = {126/135/234}
5b. Consider permutations for R7C3 + R8C2 (step 3b) = [24/51]
R7C3 + R8C3 = [24] => R8C7 = 2 => R7C6 + R8C56 = {135}
or R7C3 + R8C3 = [51] => R8C6 = 3 => R7C6 + R8C56 = {135/234}
-> R7C6 + R8C56 = {135/234}, no 6, 3 locked for R8 and N8, clean-up: no 4 in R8C8 (step 3g), no 8 in R8C9 (step 3j)
5c. Hidden killer pair 4,6 in 21(3) cage at R7C4 and 14(4) cage for R7, neither can contain both of 4,6 -> each must contain one of 4,6 -> 21(3) cage = {489/678}, no 5, 8 locked for N8
5d. 4,6 of 21(4) cage must be in R7C67 -> no 4,6 in R8C7
5e. 8 in R9 only in R9C789, locked for N9, clean-up: no 2 in R8C5 (step 3g)
5f. 45 rule on N9 3 innies R7C7 + R8C78 = 15 = {249/267/456}
5g. R8C7 = {25} -> no 2,5 in R7C7
5h. R7C67 contains one of 4,6, R7C7 = {46} -> no 4 in R7C6
[Alternatively R7C36 = {25} (hidden pair in R7).]
5i. R7C6 + R8C56 = {135/234}
5j. Naked triple {134} in R8C256, 4 locked for R8

6a. 45 rule on R6 3 innies R6C123 = 16 = {169/268/349/358} (cannot be {178/259/457} because R6C3 only contains 3,6, cannot be {367} which clashes with R6C45), no 7
6b. R6C3 = {36} -> no 3,6 in R6C12

7a. R2C8 + R34C7 (step 3a) = [781/872/963]
7b. R7C7 + R8C78 (step 5f) = {249/456} (cannot be {267} = [627] which clashes with R2C8 + R34C7) -> R7C7 = 4, R8C8 = {69}, R8C6 = 3 (cage sum), clean-up: no 1 in R2C1 (step 3h), no 7 in R1C1, no 6 in R4C5, no 7 in R5C6, no 8 in R5C7
7c. Naked pair {23} in R2C17, locked for R2
7d. R1C23 = {29/38/47} (cannot be {56} which clashes with R1C1), no 5,6
7e. 21(3) cage at R7C4 = {678} (only remaining combination), 6,7 locked for N8
7f. 9 in R7 only in R7C12, locked for N7
7g. Naked pair {78} in R8C14, 7 locked for R8
7h. 9 in R8 only in R8C89, locked for N9
7i. 1 in N1 only in R23C3, locked for C3 and 18(4) cage at R2C3

8. 45 rule on R45 5(1+4) innies R4C7 + R5C1258 = 15 -> max R5C1258 = 14, no 9 in R5C58

9a. R2C8 + R34C7 (step 3a) = [781/872/963], R5C9 = R4C7 + 6 (step 4b)
9b. Consider placement of 9 in R8
R8C8 = 9 => R2C8 + R3C7 = [781/872]
or R8C9 = 9 => R5C9 = {78}, R4C7 = {12}
-> R2C8 + R3C7 = [781/872], 7,8 locked for N3, R4C7 = {12}, R5C9 = {78}
With hindsight, Ed might have seen this as
45 rule on N3 1 innie R2C8 = 1 outie R4C7 + 6, 45 rule on R4 1 outie R5C9 = 1 innie R4C7 + 6 -> R2C8 = R5C9 = [77/88/99]
9 in R8 only in R8C89 -> R2C8 = R5C9 = [77/88] (cannot be [99]) -> R4C7 = {12}

Alternatively there’s the unusual but direct 45 rule on N3 + R4 (valid since they’re adjacent but don’t overlap) 1 outie R5C9 = 1 innie R2C8 = [77/88/99] …
9c. R34C7 + R5C9 = [728/817] -> naked pair {78} in R3C7 + R5C9, CPE no 7 in R5C7, clean-up: no 4 in R5C7
9d. 11(3) cage at R2C9 = {146/245} (cannot be {236} which clashes with R2C7), no 3, 4 locked for N3
9e. 9 in N3 only in R1C789, locked for R1, clean-up: no 2 in R1C23
9f. 19(4) cage at R1C7 contains 3,9 for N3 = {1369/2359}
9g. Killer pair 5,6 in R1C1 and 19(4) cage, locked for R1
9h. 12(3) cage at R1C4 = {138/147} (cannot be {237} much clashes with R1C23), no 2, 1 locked for R1 and N2
9i. 19(4) cage = {2359} -> 11(3) cage = {146}
9j. R1C1 = 6 (hidden single in R1) -> R2C17 = [23], clean-up: no 8 in R2C5 (step 3d), no 8 in R5C6
9k. R3C12 = {39/57} (cannot be {48} which clashes with R1C23), no 4,8

10a. 1 in N1 only in R23C3
10b. R2C23 + R3C3 (step 3e) = 14 = {149/158}, no 3,7
10c. R2C25 (step 3d) = [59/86/95]
10d. 45 rule on N2 3 innies R2C45 + R3C4 = 18 = {279/369/378/459/468/567}
10e. {459} cannot be [459] which clashes with R2C2 + R23C3 = 9{14} using R2C25, {567} cannot be [657/756] which clashes with R78C4, ALS block) -> no 5 in R2C5, clean-up: no 9 in R2C2
[Cracked. Straightforward from here.]
10f. 9 in N1 only in R3C12 = {39}, locked for R3, 3 locked for N1, clean-up: no 8 in R1C23
10g. Naked pair {47} in R1C23, locked for R1, 4 locked for N1
10h. Naked triple {138} in 12(3) cage at R1C4, 8 locked for N1
10i. R2C23 + R3C3 = {158} -> R23C3 = {18} (cannot be {15} because R23C4 cannot total 12 without 5), 8 locked for C3 and N1, clean-up: no 5 in R5C4
10j. R23C3 = {18} = 9 -> R23C4 = 9 = [45/72]
10k. R2C2 = 5 -> R2C5 = 9

11a. 15(3) cage at R2C6 = {267/456}
11b. 7 of {267} must be in R2C6 -> no 7 in R3C56
11c. R3C7 = 7 (hidden single in R3) -> R4C7 = 2, R5C9 = 8 (step 4b), R8C7 = 5, R1C7 = 9, R5C7 = 6 -> R5C6 = 5, R7C36 = [52] -> R6C3 = 3, R8C2 = 1 (step 3b), R6C7 = 1, R5C1 = 1 (step 4c), clean-up: no 7 in R6C45
11d. Naked pair {26} in R6C45, locked for R6 and N5
11e. 1 in R4 only in R4C56 = {18}, 8 locked for R4 and N5
11f. R5C9 = 8 -> R4C89 = 9 = {45}, locked for R4 and N6
11g. Naked pair {79} in R6C89, locked for R6, 7 locked for N6 -> R6C6 = 4
11h. R5C58 = [73]
11i. R23C6 = [76] -> R3C5 = 2 (cage sum)

and the rest is naked singles.


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 Post subject: Re: CDK v3 Revisit
PostPosted: Wed Jan 13, 2021 4:28 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 875
Location: Sydney, Australia
Yeah, I solved another Revisit! I found this one fun since I always felt I was getting somewhere, but it is quite long. Mostly different to Andrew's WT but as so often happens, we picked on the same areas. I found some really nice clones but they were not essential to this optimized solution.
CDK v3 WT:
Preliminaries from SudokuSolver
Cage 4(2) n9 - cells ={13}
Cage 8(2) n1 - cells do not use 489
Cage 8(2) n5 - cells do not use 489
Cage 8(2) n47 - cells do not use 489
Cage 12(2) n1 - cells do not use 126
Cage 13(2) n45 - cells do not use 123
Cage 9(2) n36 - cells do not use 9
Cage 9(2) n5 - cells do not use 9
Cage 11(2) n1 - cells do not use 1
Cage 11(2) n56 - cells do not use 1
Cage 24(3) n7 - cells ={789}
Cage 21(3) n8 - cells do not use 123
Cage 11(3) n3 - cells do not use 9
Cage 14(4) n89 - cells do not use 9
Cage 26(4) n9 - cells do not use 1

1. 4(2n7 = {13}: both locked for r7 and n9
1a. 24(3)n7 = {789}: all locked for n7

2. 21(3)n8 = {489/579/678}: has two of 7,8,9 only one of which can be in r7c45 to avoid clashing with r7c12.
2a. -> r8c4 from (789)
2b. and killer triple, 7,8,9 in r7c1245, 7 & 8 locked for r7

3. "45" on n7: 2 innies r7c3+r8c2 = 6 = [24/51]
3a. r6c3 = (36)

4. 14(4)r7c6 can only have one of 1 and 3 since they are only in r8c6
4a. = {1247/1256/2345}(no 8)
4b. must have 1 or 3 -> r8c6 = (13)
4c. with 1 in r8c6 must have 2 in r8c7 to avoid clash with h6(2)n7
4d. with 3 in r8c6 must have 2 or 5 in r8c7 to avoid clash with r7c3 = (25)
4e. -> r8c7 = (25)
4f. -> 14(4) = {1256/2345}
4g. must have both 2 & 5, one of which must be in r7
4e. -> killer pair 2,5 with r7c3: 5 locked for r7

5. 21(3)n8 = {489/678}
5. must have 8: locked for n8

6. 8 in r9 only in r9c789 in 26(4): 8 locked for n9 & 26(4)
6a. -> 26(4) = {2789/4589/5678}
6b. = 2 or 5

7. Killer pair 2,5 between 26(4)n9 and r8c7: both locked for n9
7a. 14(4)r7c6 = {1256/2345}
7b. must have both 2 & 5 -> r7c6 = (25)
7c. -> no 2,5 in r8c5 since it sees both 2 & 5 in 14(4) (Common Peer Elimination, CPE)

8. "45" on n3: 2 innies r2c8 + r2c7 = 15 = [96]/{78}
8a. -> r4c7 = (123)
8b. -> combined half cage r2c8 + r2c7 + r4c7 = [963/872/781] = 2 or 6 or 7

Andrew found this really nice block step also. Loved it.
9. "45" on n9: 3 innies r78c7 + r8c8 = 15
9a. but {267} as [627] only: blocked by combined half cage n3 (step 8b)
9b. = {249/456}(no 7)
9c. 4 locked for n9

10. "45" on n89: 2 innies r8c58 = 10 = [19]/{46} = 1 or 4

11. killer pair 1,4 in r8c258: both locked for r8
11a. -> r8c6 = 3
11b. -> 14(4) = {2345} only
11c. -> r7c8 = 4
11d. -> no 6 in r8c5 (h10(2))

12. 21(3)n8 = {678} only: 6 & 7 locked for n8

13. "45" on r2: 2 outies r2c17 = 5 = {23} only: both locked for r2
13a. r1c1 = (56)

14. "45" on r1: 1 innie r1c1 - 3 = 1 outie r2c7 = [52/63]
14a. -> 19(4)n3: {359}[2]/{458}[2] blocked
14b. also {1279/2368}/2467} blocked by h15(2)n3 (step 8)
14c. -> no 2 in r2c7
14d. -> r2c17 = [23], r1c1 = 6
14e. no 6 in r3c7

15. h15(2)n3 = {78} only: both locked for n3

16. 19(4)n3 = {259}[3] only
16a. 2,5,9 locked for r1: 2,5 for n3

17. 12(3)n2 = {138/147}
17a. 1 locked for n2

18. 15(3)n2: only combos with 3 are {348/357} but these are both blocked by 12(3)n2 (step 17)
18a. -> no 3 in r3c5

A key step (note: highly optimised! Had to break this area down a lot more first time to see below)
19. 3 in n2 only in 12(3) = {138} or in r3c4
19a. -> no 8 in r3c4 since it would leave no 3 for n2
19b. and 8 in r2c4 must have 3 in r3c4 or there would be no 3 for n2
19c. but {38} = 11 leaves r23c3 = 7, but can't make {16/25}(can't have {34} since 3 already in r3c4)
19d. -> no 8 in r2c4 (these substeps are Locking-out cages)

20. "45" on n12, 2 innies r2c25 = 14 = [86]/{59}

21. "45" on n1: 3 innies r2c23 + r3c3 = 14 and must have 1 for n1
21a. = {149/158}(no 7)
21b. 1 locked for c3
21c. but {149} as [9]{14} only means that r23c4 = 13 = {67} only, but this is blocked by r78c4
21d. -> innies n1 = {158} only: 5,8 locked for n1: no 5 in r2c4 since it sees all those
21d. no 5 in r2c5 (h14(2)r2c25)

22. 11(2)n1 = {47} only: both locked for r1 and n1
12(2)n1 = {39} only: both locked for r3, 3 for n1

23. 18(4)r2c4 must have two of 1,5,8 for r23c3 = {1278/1458}(no 6,9)
23a. must have 8, locked for c3 and n1
23b. r23c3 = {18} = 9 -> r23c4 = 9 = [45/72]
23b. r2c25 = [59](h14(2))

24. 9 in c3 only in n4: locked for n4

25. "45" on r6: 1 outie r5c1 + 2 = 1 innie r6c3 = [13/46]
25a. 14(3)r5c1 must have 1 or 4 for r5c1 = {158/167/248/347}
25b. can't have both 1 & 4 -> no 1,4 in r6c12

26. 4 & 9 (=13) in r6 only in 21(4) -> other two cells = 8 (no 8)

27. 8 in r6 only in 14(3)n4 = {158/248}(no 3,6,7)
27a. 8 locked for n4

This was the slow one to find
28. 5 in n4 only in r6c1 -> 14(3) = [158] or 5 in r5c3 -> [58] in r5c34, or 5 in 25(5)r4c1
28a. -> {1789} blocked from 25(4)r4c1 since it can only be {179}[8] which leaves no 5 for n4
28b. -> no 1 in 25(4)

Another tricky one. This is a contradiction type step.
29. "45" on n47: 2 innies r58c2 + 9 = 2 outies r45c4
29a. and remembering 9 in r4c3, or [94] in r5c34, for c3
29b. if 1 or 4 in r5c2 -> naked pair {14} in r5c12 -> r5c34 <> 9 -> 9 in r4c3 -> no 9 in r45c4
29b and 1 or 4 in r5c2 -> innies n47 = 5 -> r45c4 = 14 = {68} only
29c. but this is blocked by r78c4
29d. -> no 1,4 in r5c2

30. r5c1 = 1 (hsingle n4)
30a. -> r6c12 = 13 = [58]
30b. and r6c3 = 3 (iodr4=-2), r7c3 = 5, r8c2 = 1 (h6(2)), r8c58 = [46](h10(2)), r8c37 = [25], r7c6 = 2

31. "45" on r4: 2 outies r3c7 + r5c9 = 15 = {78} only
31a. -> no 7 or 8 in r56c7 since they see both those cells (CPE)

32. hidden pair 7,8 in c7 -> r9c7 = (78)

33. 6 in c7 only in n6: locked for n6

34. 17(3)n6 must have 7,8 for r5c9 = {179/278/458}(no 3)

35. r5c8 = 3 (hsingle n6)
35a. r37c8 = [41]

36. 5 in n6 only in 17(3) = [548] only permutation

37. 25(4)r4c1 = [7693] only permutation
37a. r5c234 = [249]

38. r7c2 and r8c4 = 7 (hsingles)
Cheers
Ed

easier now


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