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Ed
 Post subject: ANOv2 Revisit
PostPosted: Sun Dec 13, 2020 1:07 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
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A New One v2 Revisit

Andrew suggested I post this a little early since Christmas is just around the corner. Not a problem!

These Revisits are old puzzles from the archive that get a score from 1.5-2.00 and a rating above E1.50. Skipped a few that don't fit those criteria to get to this puzzle, ANOv2, originally made by nd. He thought it was easier than the original but the posted solution doesn't agree. It has some big cages so is ripe for a revisit. SS gives it 1.70, Jsudoku uses 7 'complex intersections'. We are looking for 'nice', (and to use a word that nd used to use often) elegant solutions. Though he never posted one unfortunately.
triple-click code:
3x3::k:2816:8449:8449:8449:2818:6147:5124:5124:5124:2816:8449:4869:8449:2818:6147:6147:5124:6150:4869:4869:4869:2311:2311:2311:6147:5124:6150:2312:2312:4869:8969:8969:8969:8969:8969:6150:2312:8458:8458:8969:8458:8458:8715:8715:6150:7436:7436:8458:8458:8458:4877:8715:8715:8715:7436:7436:3086:3086:3086:4877:8715:4367:4367:5392:5392:5392:3601:3086:4877:4877:4367:4370:5392:5392:3601:3601:3603:3603:3603:4370:4370:
solution: 
+-------+-------+-------+
| 6 3 8 | 7 9 4 | 5 2 1 |
| 5 9 1 | 6 2 8 | 3 4 7 |
| 7 4 2 | 1 5 3 | 9 8 6 |
+-------+-------+-------+
| 3 2 5 | 9 4 6 | 7 1 8 |
| 4 1 9 | 8 7 2 | 6 5 3 |
| 8 7 6 | 5 3 1 | 4 9 2 |
+-------+-------+-------+
| 9 5 3 | 2 1 7 | 8 6 4 |
| 1 8 4 | 3 6 9 | 2 7 5 |
| 2 6 7 | 4 8 5 | 1 3 9 |
+-------+-------+-------+
Cheers
Ed
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Andrew
 Post subject: Re: ANOv2 Revisit
PostPosted: Wed Dec 16, 2020 11:12 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Thanks Ed for posting this latest Revisit. I enjoyed this one with its large cages. I see that I originally commented that the v2 took away a useful 45 without giving anything useful in return; not quite true, which will explain why SS gave a lower score for the v2.

Here is my walkthrough for ANOv2 Revisit:
Prelims

a) R12C1 = {29/38/47/56}, no 1
b) R12C5 = {29/38/47/56}, no 1
c) 9(3) cage at R3C4 = {126/135/234}, no 7,8,9
d) 9(3) cage at R4C1 = {126/135/234}, no 7,8,9
e) 29(4) cage at R6C1 = {5789}
f) 12(4) cage at R7C3 = {1236/1245}, no 7,8,9
g) 33(5) cage at R1C2 = {36789/45789}, no 1,2

1a. 45 rule on N9 3 innies R789C7 = 11 = {128/137/146/236/245}, no 9
1b. 34(6) cage at R5C7 must contain 9, locked for N6
1c. 35(6) cage at R4C4 must contain 9, locked for N5
1d. 45 rule on R789 3 innies R7C127 = 1 outie R6C6 + 21
1e. Max R7C127 = 24 -> max R6C6 = 3
1f. Min R7C127 = 22 -> min R7C7 = 5
1g. R7C7 = {5678} -> no 5,6,7,8 in R89C7
1h. 19(4) cage at R6C6 cannot contain more than two of 1,2,3,4 -> no 1,2,3,4 in R78C6

2a. Hidden killer pair 3,4 in 12(4) cage at R7C3 and 17(3) cage at R7C8 for R7, 12(4) cage and 17(3) cage cannot contain both of 3,4 -> 12(4) cage must contain one of 3,4 in R7C345, 17(3) cage must contain one of 3,4 in R7C89 -> no 3,4 in R8C5, no 3,4 in R8C8
2b. 45 rule on R89 4 innies R8C5678 = 24
2c. Max R8C7 = 4 -> min R8C568 = 20, no 1,2 in R8C58
2d. 12(4) cage at R7C3 = {1236/1245}, 1,2 locked for R7
2e. R8C5 = {56} -> no 5,6 in R7C345
2f. R12C5 = {29/38/47} (cannot be {56} which clashes with R8C5), no 5,6
2g. 9(3) cage at R3C4 = {126/135} (cannot be {234} which clashes with R12C5), no 4, 1 locked for R3 and N2
2h. R2C3 = 1 (hidden single in N1)
2i. 12(4) cage = {1236/1245}, 1 locked for N8

3a. 45 rule on R789 3 outies R6C126 = 1 innie R7C7 + 8
3b. Consider placements for R7C7 = {5678}
R7C7 = {578} => that same value must be in R6C12 because of 29(4) cage at R6C1 => the rest of R6C126 must total 8 = [53/71]
or R7C7 = 6 => R6C126 = 14 = {57}2/{58}1
-> no 9 in R6C12
3c. 29(4) cage at R6C1 = {5789}, 9 locked for R7 and N7
3d. 19(5) cage at R2C3 cannot be 1{234}9/1{235}8/1{236}7/1{245}7 which clash with 9(3) cage at R3C4 -> no 7,8,9 in R4C3
3e. 9 in N4 only in 33(7) cage at R5C2 = {1234689/1235679}
3f. 1 in 33(7) cage only in R5C256 + R6C45, CPE no 1 in R5C4
3g. 7,8 in N4 only in R5C23 + R6C123, CPE no 7,8 in R6C45
3h. Variable hidden killer pair 4,5 for N5, 33(7) cage cannot contain both of 4,5 -> 35(6) cage at R4C4 must contain at least one of 4,5 = {146789/245789/345689} (cannot be {236789} which doesn’t contain either of 4,5)

4a. 9(3) cage at R4C1 = {126/135/234} contains two of 1,2,3, 33(7) cage at R5C2 (step 3e) = {1234689/1235679} contains all of 1,2,3, R6C6 = {123} -> double killer triple 1,2,3, locked for N45
4b. 19(5) cage at R2C3 = {12457/13456} (cannot be {12349/12358/12367} which clash with 9(3) cage at R3C4, no 8,9
4c. Killer pair 2,3 in 19(5) cage and 9(3) cage, locked for R3
4d. 19(5) cage = {12457/13456} = 1{247}5/1{345}6 (cannot be 1{257}4/1{356}4} which clash with 9(3) cage), no 6 in R3C123, 4 in R3C123 locked for R3, N1 and 19(5) cage, clean-up: no 7 in R12C1
4e. 19(5) cage = {12457/13456}, CPE no 5 in R1C3
4f. Combined part cage R3C123 with 9(3) cage = {247}{135}/{345}{126}, 5 locked for R3
4g. 8,9 in R3 only in R3C789, locked for N3
4h. 33(5) cage at R1C2 = {36789/45789} contains 7,8,9
4i. 45 rule on N1 3(2+1) outies R12C4 + R4C3 = 18
4j. Consider combinations for 19(5) cage = 1{247}5/1{345}6
19(5) cage = 1{247}5, 7 locked for N1, R4C3 = 5 => R12C4 = 13 must contain 7 for 33(5) cage = {67}
or 19(5) cage = 1{345}6, 3,5 locked for N1 => R12C1 = {29}, 9 locked for N1, R4C3 = 6 => R12C4 = 12 must contain 9 for 33(5) cage = {39}
-> R12C4 = {39/67}, no 4,5,8
4k. 33(5) cage = {36789}, no 5, 8 locked for N1, clean-up: no 3 in R12C1
4l. Killer pair 3,6 in R12C4 and 9(3) cage, locked for N2, clean-up: no 8 in R12C5
4m. Killer pair 7,9 in R12C4 and R12C5, locked for N2
4n. 45 rule on N12 2 innies R12C6 = 1 outie R4C3 + 7
4o. R4C3 = {56} -> R12C6 = 12,13 = {48/58}, no 2, 8 locked for C6 and 24(4) cage at R1C6
4p. 24(4) cage at R1C6 = {48}[39]/{48}[57]/{58}[29]/{58}[47] -> R2C7 = {2345}, R3C7 = {79}
4q. 45 rule on R123 2 innies R23C9 = 1 outie R4C3 + 8
4r. R4C3 = {56} -> R23C9 = 13,14, no 2,3
4s. R23C9 = 13,14 -> R45C9 = 10,11, no 1

5a. 45 rule on N4578 (2+2/3+1) outies R4C78 + R89C7 = 1 innie R4C3 + 6
5b. Min R489C7 = 7 (cannot be 6 because R4C3 cannot be the same as R4C8), max R4C3 = 6 -> max R4C78 + R89C7 = 12 -> max R4C8 = 5
5c. Max R489C7 = 10 (cannot total 11 which clashes with R789C7 = 11, overlap clash) -> no 8 in R4C7
5d. 35(6) cage at R4C4 (step 3h) = {146789/245789/345689}, 8 locked for N5

6a. 9(3) cage at R4C1 = {126/135/234}
6b. R6C126 (step 3a) = R7C7 + 8, max R7C7 = 8 -> max R6C126 = 16
6c. Consider combinations for R6C12 = {57/58/78}
R6C12 = {57/58}, 5 locked for N4 => R4C3 = 6 => 9(3) cage = {234} => R5C2 = 1 (hidden single in N4) => R6C6 = 1 (hidden single in N5)
or R6C12 = {78} = 15 => R6C6 = 1
-> R6C6 = 1
[Getting easier now.]
6d. 33(7) cage at R5C2 (step 3e) = {1234689/1235679} -> R5C2 = 1
6e. 9(3) cage = {234}, locked for N4
6f. 1 in N6 only in R4C78 -> 35(6) cage at R4C4 (step 3h) = {146789}, no 2,3,5
6g. 5 in N5 only in R5C56 + R6C45, locked for 33(7) cage at R5C2
6h. 33(7) cage must contain 5 and 9 = {1235679}, no 4,8
6i. 4 in N5 only in R4C456 + R5C4, locked for 35(6) cage -> R5C8 = 1
6j. R6C6 = 1 -> R78C6 + R8C7 = 18 = [594/693/792] -> R8C6 = 9
6k. 8 in N4 only in R6C12, locked for R6 and 29(4) cage at R6C1
6l. 9 in N4 only in R56C3, locked for C3
6m. 8 in R7 only in R7C789, locked for N9
6n. R6C126 = {58}1/{78}1 = 14,16 -> R7C7 = {68}
6o. 17(3) cage at R7C8 = {458/467} (cannot be {368} which clashes with R7C7), no 3, 4 locked for R7 and N9
6p. R7C345 = {123} -> R8C5 = 6 (cage sum), clean-up: no 3 in R8C7
6q. 6 in R7 only in R7C789, locked for N9
6r. R68C6 = [19] = 10 -> R7C6 + R8C7 = 9 = [72]

7a. Naked pair {59} in R7C12, 5 locked for R7, N7 and 29(4) cage at R6C1
7b. Naked pair {78} in R6C12, 7 locked for R6 and N4
7c. R6C123 = {78}1 = 16 -> R7C7 = 8 (step 3a) -> R9C7 = 1 (step 1a)
7d. R7C89 = {46} = 10 -> R8C8 = 7 (cage sum)
7e. Naked pair {69} in R56C3, 6 locked for C3 and 33(7) cage at R5C2
7f. R4C3 = 5, R2C3 = 1 -> R3C123 = 13 = {247}, 2,7 locked for R3 and N1, clean-up: no 9 in R12C1
7g. Naked pair {56} in R12C1, locked for C1, 6 locked for N1
7h. Naked triple {389} in R1C23 + R2C2, 3,9 locked for 33(5) cage at R1C2
7i. Naked pair {67} in R12C4, locked for C4 and N2, clean-up: no 4 in R12C5
7j. R12C6 = {48} (hidden pair in N2), 4 locked for C6, R3C7 = 9 -> R2C7 = 3 (cage sum)
7k. R4C67 = [67]
7l. Naked pair {68} in R3C89, 6 locked for N3
7m. Naked triple {245} in R1C78 + R2C8, locked for N3, 2 locked for C8, R2C9 = 7, R1C9 = 1 -> R3C8 = 8 (cage sum)
7n. R23C9 = [76] = 13 -> R45C9 = 11 = {38}, 3 locked for C9 and N6

8a. R6C9 = 2 (hidden single in N6)
8b. Naked pair {35} in R6C45, locked for N5, 5 locked for R6 -> R5C56 = [72]
8c. R9C7 = 1 -> R9C56 = 13 = [85]
8d. 14(3) cage at R8C4 = {347} (only remaining combination) -> R9C3 = 7, R89C4 = {34}, locked for C4, 3 locked for N8

and the rest is naked singles.

As with the earlier Revisit puzzles, I've just been treating them as new puzzles so I haven't looked at how I originally solved them.

Also I haven't been rating my Revisit walkthroughs. However I will comment that, if I was, I'd give my Revisit walkthrough a lower rating than I did when I first solved this puzzle.
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wellbeback
 Post subject: Re: ANOv2 Revisit
PostPosted: Sun Dec 20, 2020 8:34 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
Here's my start to an optimized WT. Took me a while to see some of these moves... Quite different From Andrew's WT.
ANOv2 Revisit WT:
1. Innies n9 = r789c7 = +11(3) (No 9)
IOD r789 -> r7c127 = r6c6 + 21
-> r6c6 = Max 3 and r7c127 = Min +22(3)
-> 9 in r7c12

2. Max r3c456789 = +33(6) -> Min r3c123 = +12(3)
-> Max r24c3 = +7(2) (No 789)
-> 9 in n4 in 33(7)n45
-> 33(7)n45 = {12369(48|57)}

3! -> Whichever of (123) is in r6c6 is in n4 in r5c23
Also 9(3)n4 contains two of (123)
-> Min r4c3 = 4
-> (IOD n12) Min r12c6 = +11(2)
-> 1 in n2 in 9(3)n2
-> (HS 1 in n1) r2c3 = 1

4. r4c3 from (456) -> r3c789 = Min +22(3)
-> 4 in r3 in r3c123
-> r4c3 from (56)

5! Outies n78 = r6c126 + r89c7 = +19(5)
-> Max r6c126 = +16(3)
29(4)r6c1 = {5789} with 9 in r7c12
Trying 5 not in r6c12 -> r6c127 = [{78}1] puts r5c2 = 1 puts 9(3)n4 = {234}
-> 9(3)n4 cannot be {135}
-> Whether r4c3 is 5 or 6 both -> 9(3)n4 = {234}
-> (Step 3 Line 1) -> r6c6 = 1
-> r5c2 = 1

6! 34(6)n6 must contain a 9 which must be in n6
Outies r123 = r4c3 + r45c9 = +16(3)
-> (r4c3 from (56)) r45c9 = +11(2) or +10(2)
Since r45c9 cannot contain a 9 -> they also cannot contain a 1.
-> 1 in n6 in r4c78
-> 35(6)n56 = {146789}

7! 6 in both 35(6)n56 and 33(7)n45
-> 6 cannot go in r4c3
-> r4c3 = 5

Straightforward from here.
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