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 Post subject: Variations on 111
PostPosted: Wed Dec 23, 2015 7:42 pm 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
Having a bit of time I've been going back on some old puzzles. I really liked some of the inferences of this design by Nasenbaer in July 2008, so created a couple more.
The first has a SudokuSolver rating of 1.4, the second is 1.9.
Merry Christmas all!

Image

Code: 3x3::k:5633:5633:9474:3331:3331:1028:1028:4101:4101:5633:9474:9474:7174:3331:3847:3847:4101:5128:9474:3593:9474:7174:7174:8970:3847:5128:5128:3593:3593:9474:7174:8970:8970:2571:2571:5128:4108:3593:9474:8970:2574:8970:1807:2571:8976:4108:4108:5901:5901:2574:8970:1807:8976:8976:4108:2833:5901:5901:3602:8970:8976:4371:8976:3860:2833:2833:3602:3602:7190:7190:4371:8976:3860:3093:3093:791:791:7190:7190:4371:8976:

Image

Code: 3x3::k:5377:5377:8194:3331:3331:1028:1028:2821:2821:5377:8194:8194:5894:3331:5383:5383:2821:6664:8194:5641:8194:5894:5894:8458:5383:6664:6664:5641:5641:8194:5894:8458:8458:3595:3595:6664:3852:5641:8194:8458:1806:8458:2063:3595:8208:3852:3852:5645:5645:1806:8458:2063:8208:8208:3852:3857:5645:5645:4114:8458:8208:5139:8208:2580:3857:3857:4114:4114:6166:6166:5139:8208:2580:2581:2581:1559:1559:6166:6166:5139:8208:


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 Post subject: Re: Variations on 111
PostPosted: Sun Dec 27, 2015 7:22 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks wellbeback for posting variants on Nasenbaer's neat cage pattern for Assassin 111 which included three figure 1s.

Looking at the rating table for Archive D, I see that I gave my walkthrough for the original puzzle a higher rating than others had, so I must have found it hard. Another good reason not to look at how I solved A111 back then and start from scratch for the different cage totals.

Here is my walkthrough for the first variant:
Prelims

a) R1C67 = {13}
b) R56C5 = {19/28/37/46}, no 5
c) R56C7 = {16/25/34}, no 7,8,9
d) R89C1 = {69/78}
e) R9C23 = {39/48/57}, no 1,2,6
f) R9C45 = {12}
g) 22(3) cage at R1C1 = {589/679}
h) 10(3) cage at R4C7 = {127/136/145/235}, no 8,9
i) 11(3) cage at R7C2 = {128/137/146/236/245}, no 9
j) 28(4) cage at R2C4 = {4789/5689}, no 1,2,3
k) 14(4) cage at R3C2 = {1238/1247/1256/1346/2345}, no 9
l) 28(4) cage at R8C6 = {4789/5689}, no 1,2,3

Steps resulting from Prelims
1a. R1C67 = {13}, locked for R1
1b. R9C45 = {12}, locked for R9 and N8
1c. 22(3) cage at R1C1 = {589/679}, 9 locked for N1
1d. 28(4) cage at R2C4 = {4789/5689}, CPE no 8,9 in R1C4

2. 37(7) cage at R1C3 must contain 9 in R45C3, locked for C3 and N4, clean-up: no 3 in R9C2
2a. 45 rule on N1 2 outies R45C9 = 1 innie R3C2 + 1 -> R3C2 = {123}, R45C9 = 15,16,17 = {69/79/89}

3. 45 rule on N7 2 innies R7C13 = 7 = {16/25/34}, no 7,8,9
3a. 11(3) cage at R7C2 = {128/137/146/245} (cannot be {236} which clashes with R7C13)

4. 14(3) cage at R7C5 = {347/356}, no 8,9, 3 locked for N8

5. 45 rule on C6789 2 outies R4C5 + R5C4 = 7 = {16/25/34}, no 7,8,9

6. 45 rule on C89 2 outies R47C7 = 8 = {17/26/35}, no 4,8,9

7. 45 rule on C123 2 innies R67C3 = 8 = {26/35}/[71], no 4,8, no 1 in R6C3, clean-up: no 3 in R7C1 (step 3)
7a. 45 rule on C123 2 outies R67C4 = 15 = {69/78}
7b. 28(4) cage at R2C4 = {4789/5689} -> R3C5 = {89} (R234C4 cannot contain both of 8,9 which would clash with R67C4)
7c. Killer pair 6,7 in 28(4) cage and R67C4, locked for C4, clean-up: no 1 in R4C5 (step 5)

8. Grouped X-Wing for 9 in R89C1 + R9C2 and 28(4) cage at R8C6, no other 9 in R89

9. 45 rule on N689 3(1+2) innies R4C9 + R7C46 = 21, max R7C46 = 17 -> min R4C9 = 4

10. Killer triple 1,2,3 in R1C7, R47C7 and R56C7, locked for C7
10a. 2 in C7 only in R47C7 (step 6) = {26} or R56C7 = {25} -> R47C7 = {17/26} (cannot be {35}, locking-out cages), R56C7 = {25/34} (cannot be {16}, locking-out cages)
10b. 10(3) cage at R4C7 = {127/136} (cannot be {145/235} which clash with R56C7), no 4,5, 1 locked for N6
10c. Killer pair 2,3 in 10(3) cage at R56C7, locked for N6

11. 45 rule on N3 3(2+1) outies R12C6 + R4C9 = 10
11a. Min R12C6 = 3 -> max R4C9 = 7
11b. Min R4C9 = 4 -> max R12C6 = 6, no 6,7,8,9 in R4C9
11c. R4C9 + R7C46 = 21 (step 9), max R4C9 = 7 -> min R7C46 = 14, no 4 in R7C6

12. 8,9 in N6 locked for 35(7) cage at R5C9, no 8,9 in R789C9
12a. 35(7) cage contains both of 8,9 = {1235789/1245689}, 1,2 locked for N9

13. 13(3) cage at R1C4 = {148/157/238/247/256/346} (cannot be {139} which clashes with R1C6), no 9
13a. 13a. Min R1C45 = 6 -> max R2C5 = 7
13b. 9 in C5 only in R356C5, CPE no 9 in R4C4
13c. 28(4) cage at R2C4 = {4789/5689}, 9 locked for N2

14. 15(3) cage at R2C6 = {159/168/249/258/348/357/456} (cannot be {267} which clashes with R47C7)
14a. Hidden killer quad 1,3,5,7 in R1C7, R47C7 and R56C7 for C7, R1C7 = {13}, R47C7 must contain one of 1,7, R56C7 must contain one of 3,5 -> R2389C7 can only contain one of 1,3,5,7
14b. 15(3) cage at R2C6 = {159/168/249/258/348/456} (cannot be {357} = 3{57} which contains two of 5,7 in R23C7), no 7

15. 1,2 in N9 only in 35(7) cage at R5C9 = {1235789/1245689}, CPE no 5 in R4C9
15a. R12C6 + R4C9 = 10 (step 11), R4C9 = {467} -> R12C6 = 3,4,6 = [12/13/31/15] (cannot be {24} because 2,4 only in R2C6), no 4, 1 locked for C6 and N2

16. 13(3) cage at R1C4 (step 13) = {238/247/256/346}, 14(3) cage at R7C5 (step 4) = {347/356}
16a. Hidden killer triple 3,6,7 in 13(3) cage, 14(3) cage and rest of C5 for C5, 13(3) cage must have at least one of 3,6,7 in C5, 14(3) cage must have one of 6,7 in C5 -> rest of C5 cannot have more than one of 3,7 -> R56C5 = {19/28/46} (cannot be {37} which contains both of 3,7), no 3,7
16b. 3 in N5 only in R4C5 + R5C4 + R456C6, locked for 35(7) cage at R3C6 -> no 3 in R3C6
16c. 35(7) cage contains 3 = {1235789/1345679/2345678}, 7 locked for C6
16d. R4C5 + R5C4 (step 5) = {25/34}/[61]
16e. 35(7) cage = {1345679/2345678} (cannot be {1235789} because 1 only possible in R4C5 + R5C4 = [61])
16f. Hidden killer pair 1,2 in 35(7) cage and R56C5 for N5, 35(7) cage contains one of 1,2 -> R56C5 (step 16a) = {19/28} (cannot be {46} which doesn’t contain 1 or 2), no 4,6
16g. Killer pair 8,9 in R3C5 and R56C5, locked for C5
16h. 8 in C5 only in R356C5, CPE no 8 in R4C4
16i. 28(4) cage at R2C4 = {4789/5689}, 8 locked for N2

17. Hidden killer pair 8,9 in 35(7) cage at R3C6 and 28(4) cage at R8C6 for C6, 35(7) cage (step 16e) contains one of 8,9 -> 28(4) cage must contain one of 8,9 in C6 -> 28(4) cage must contain one of 8,9 in C7
17a. Hidden killer pair 8,9 in 15(3) cage at R2C6 and 28(4) cage at R8C6 for C7, 28(4) cage contains one of 8,9 in C7 -> 15(3) cage must contain one of 8,9 -> 15(3) cage (step 14b) = {159/168/249/258/348} (cannot be {456} which doesn’t contain 8 or 9)
17b. 1,2,3 only in R2C6 -> R2C6 = {123}
17c. R12C6 + R4C9 = 10 (step 11), R12C6 = [12/13/31] = 3,4 -> R4C9 = {67}
17d. 13(3) cage at R1C4 (step 16) = {247/256/346}
17e. Killer pair 2,3 in 13(3) cage and R12C6, locked for N2
17f. 28(4) cage at R8C6 = {4789/5689}
17g. 4 of {4789} must be in R89C6 (because R89C6 only contain one of 8,9) -> no 4 in R89C7

18. 13(3) cage at R1C4 (step 17d) = {247/256/346}, 14(3) cage at R7C5 (step 4) = {347/356}
18a. Killer pair 6,7 in 13(3) cage and 14(3) cage, locked for C5, clean-up: no 1 in R5C4 (step 5)
18b. 1 in C5 only in R56C5 = {19}, locked for C5 and N5 -> R3C5 = 8, R9C45 = [12]
18c. 8 in C4 only in R67C4 (step 7a) = {78}, locked for C4
[I forgot to lock 7 for 23(4) cage at R6C3, but it’s eliminated in step 19e.]
18d. 28(4) cage at R2C4 = {5689} (only remaining combination), 5,6 locked for C4

19. 35(7) cage at R3C6 = {2345678} (only remaining combination), no 9, 8 locked for C6
19a. R7C8 = 9 (hidden single in R7) -> R89C8 = 8 = {35}, locked for C8 and N9
19b. 10(3) cage at R4C7 (step 10b) = {127} (only remaining combination), locked for N6 -> R4C9 = 6, R4C4 = 5, clean-up: no 2 in R5C4 (step 5), no 5 in R56C7 (step 10a), no 2 in R7C7 (step 10a)
19c. Naked pair {34} in R4C5 + R5C4, locked for 35(7) cage at R3C6
19d. Naked pair {34} in R56C7, locked for C7 and N6 -> R1C67 = [31], R6C8 = 8, clean-up: no 7 in R47C7 (step 10a)
19e. R47C7 = [26], clean-up: no 1 in R7C13 (step 3), no 2,7 in R6C3 (step 7)
19f. R2C6 = 1 (hidden single in C6) -> R23C7 = 14 = {59}, locked for N3

20. R89C6 = {49} (hidden pair in C6), locked for N8 -> R8C4 = 3, R89C8 = [53], clean-up: no 9 in R9C2
20a. R8C4 = 3 -> R78C5 = 11 = [56], clean-up: no 2 in R7C13 (step 3)
20b. R7C13 = [43], R6C3 = 5 (step 7), R6C9 = 9, R56C5 = [91], clean-up: no 8,9 in R9C23
20c. R9C23 = [57], locked for R9 and N7 -> R78C7 = [78], R9C169 = [694], R8C1 = 9 (cage sum)
20c. 22(3) cage at R1C1 = {589} (only remaining combination, cannot be {679} because 6,9 only in R1C2) -> R1C2 = 9, R12C1 = {58}, locked for C1 and N1

21. R1C4 = 2 (hidden single in N2)
21a. 6 in N3 only in 16(3) cage at R1C8 = {268} (only remaining combination) -> R1C9 = 8, R12C3 = [62], R1C3 = 4

22. R38C7 = {12} (hidden pair in C3)
22a. Naked pair {12} in R8C39, locked for R8 -> R8C2 = 8

23. R2C3 = 6, R45C3 = [98] -> R3C2 = 3 (step 2a)
23a. R3C2 = 3 -> R4C12 + R5C2 = 11 = {146} (only possible combination) -> R4C12 = [14], R5C2 = 6

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough at Easy 1.5; I used locking-out cages and several hidden killers including a hidden killer quad. I'm not sure whether I needed all those steps, but they were what I saw at the time.

Solution:
+-------+-------+-------+
| 5 9 4 | 2 7 3 | 1 6 8 |
| 8 7 6 | 9 4 1 | 5 2 3 |
| 2 3 1 | 6 8 5 | 9 4 7 |
+-------+-------+-------+
| 1 4 9 | 5 3 8 | 2 7 6 |
| 7 6 8 | 4 9 2 | 3 1 5 |
| 3 2 5 | 7 1 6 | 4 8 9 |
+-------+-------+-------+
| 4 1 3 | 8 5 7 | 6 9 2 |
| 9 8 2 | 3 6 4 | 7 5 1 |
| 6 5 7 | 1 2 9 | 8 3 4 |
+-------+-------+-------+


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 Post subject: Re: Variations on 111
PostPosted: Wed Dec 30, 2015 10:01 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
And then I tried the second variant. It's not as hard as the SSscore suggests because SudokuSolver isn't programmed to spot my step 5d (see also my Outies post in the Killer Techniques forum).

Here is my walkthrough for the second variant:
Thanks wellbeback for spotting my typos
Prelims

a) R1C67 = {13}
b) R56C5 = {16/25/34}, no 7,8,9
c) R56C7 = {17/26/35}, no 4,8,9
d) R89C1 = {19/28/37/46}, no 5
e) R9C23 = {19/28/37/46}, no 5
f) R9C45 = {15/24}
g) 21(3) cage at R1C1 = {489/579/678}, no 1,2,3
h) 11(3) cage at R1C8 = {128/137/146/236/245}, no 9
i) 21(3) cage at R2C6 = {489/579/678}, no 1,2,3
j) 20(3) cage at R7C8 = {389/479/569/578}, no 1,2
k) 26(4) cage at R2C9 = {2789/3689/4589/4679/5678}, no 1

1. Naked pair {13} in R1C67, locked for R1
1a. 13(3) cage at R1C4 = {148/157/238/247/256/346} (cannot be {139} which clashes with R1C6), no 9
1b. Min R1C45 = 6 -> max R2C5 = 7
1c. 11(3) cage at R1C8 = {128/146/236/245} (cannot be {137} which clashes with R1C7), no 7
1d. Min R1C89 = 6 -> max R2C8 = 5
1e. 9 in R1 only in R1C123, locked for N1

2. 45 rule on N7 2 innies R7C13 = 10 = {19/28/37/46}, no 5

3. 45 rule on C123 2 innies R67C3 = 10 = {19/28/37/46}, no 5
3a. 45 rule on C123 2 outies R67C4 = 12 = {39/48/57}, no 1,2,6

4. 45 rule on C89 2 outies R47C7 = 13 = {49/58/67}, no 1,2,3

5. 45 rule on C6789 2 outies R4C5 + R5C4 = 13 = {49/58/57}, no 1,2,3
5a. 45 rule on C6789 5 innies R34567C6 = 20 = {12458/12467/23456} (cannot be {12359/12368/13457} which clash with R1C6), no 9, 2,4 locked for C6, 4 also locked for 33(7) cage at R3C6, no 4 in R4C5 + R5C4, clean-up: no 9 in R4C5 + R5C4
[With hindsight 33(7) cage at R3C6 contains 1, locked for C6 -> R1C67 = [31] would have been simpler.]
5b. R34567C6 = {12458/12467} (cannot be {23456} which clashes with R4C5 + R5C4), 1 locked for C6 -> R1C67 = [31], clean-up: no 7 in R56C7
5c. 11(3) cage at R1C8 (step 1c) = {236/245}, no 8, 2 locked for N3
5d. 45 rule on N3 2(1+1) remaining outies R2C6 + R4C9 = 14 = {59/68} (cannot be [77] because R2C6 + R4C9 ‘see’ all 7s in N3)

6. 45 rule on R1 3 outies R2C158 = 1 innie R1C3 + 4
6a. Min R2C158 = 7 -> no 2 in R1C3

7. 9 in N5 only in R46C4, locked for C4, clean-up: no 3 in R6C4 (step 3a)
7a. 9 in R4C4 or R67C4 = [93] -> no 3 in R4C4 (locking-out cages)
7b. 3 in N5 only in R56C5 = {34}, locked for C5 and N5, clean-up: no 8 in R7C4, no 2 in R9C4
7c. 9 in N2 only in R2C6 + R3C5, CPE no 9 in R3C7
7d. 13(3) cage at R1C4 (step 1a) = {148/157/247/256}
7e. 4 of {148} must be in R1C4 -> no 8 in R1C4

8. 1 in N9 only in R789C9, locked for 32(7) cage at R5C9, no 1 in R5C9 + R6C89
8a. 1 in N6 only in 14(3) cage at R4C7 = {149/158/167}, no 2,3

9. Hidden killer pair 2,3 in R56C7 and 24(4) cage at R8C6 for C7, R56C7 contains one of 2,3 -> 24(4) cage must contain one of 2,3 = {2589/2679/3489/3579/3678} (cannot be {4569/4578} which don’t contain 2 or 3)
9a. Hidden killer pair 2,3 in R56C7 and 32(7) cage at R5C9 for N6, R56C7 contains one of 2,3 -> 32(7) cage contains one of 2,3 in R5C9 + R6C78, 32(7) cage contains both of 2,3 -> it must contain one of 2,3 in R789C9
9b. Killer pair 2,3 in 24(4) cage and R789C9, locked for N9

10. 14(3) cage at R4C7 (step 8a) = {149/158/167}, 32(7) cage at R5C9 = {1234589/1234679/1235678}, R47C7 (step 4) = {49/58/67}
10a. Consider placements for 4 in N6
4 in 14(3) cage = {149}, locked for N6
or 4 in R5C9 + R6C89 => 32(7) cage = {1234589/1234679}, 14(3) cage = {158/167} => R47C7 = {58/67} => 9 in 32(7) cage must be in R5C9 + R6C89 + R789C9, CPE no 9 in R4C9
-> no 9 in R4C9, clean-up: no 5 in R2C6 (step 5d)

11. 45 rule on C789 4 remaining innies R2389C7 = 23 = {2489/2678/3479/3578} (cannot be {2579/3569/4568} which clash with R56C7)
10a. Min R2C6 = 6 -> max R23C7 = 15 -> 4 of {2489/3479} must be in R23C7 -> no 4 in R89C7

12. 45 rule on N689 3(1+2) innies R4C9 + R7C46 = 15, R4C9 = {568} -> R7C46 = 7,9,10 = [34/36/72/37/46] (cannot be {16/18/28} because 1,2,6,8 only in R7C6, cannot be {45}/[52] which clash with R9C45) -> R7C4 = {347}, R7C6 = {2467}, clean-up: no 7 in R6C4 (step 3a)
12a. 16(3) cage at R7C5 = {169/178/268/358} (cannot be {259/457} which clash with R9C45, cannot be {349} because 3,4 only in R8C4, cannot be {367} which clashes with R7C46), no 4
12b. Consider placement for 3 in N8
R7C4 = 3 => R7C46 = [34/36/37]
or R8C4 = 3 => 16(3) cage = {358} => R9C45 = [42] => R7C35 = [72]
-> R7C46 = [34/36/72/37], no 4 in R7C4, clean-up: no 8 in R6C4 (step 3a)
12c. Combined cage R7C46 + R9C45 = [34]{15}/[36]{15/24}/[37]{15/24} (cannot be [72]{15} which clashes with 16(3) cage -> R7C4 = 3, R7C6 = {467}, R6C4 = 9 (step 3a), clean-up: no 1,7 in R67C3 (step 3), no 1,7,9 in R7C1 (step 2)
12d. 16(3) cage = {169/178/268}, no 5

13. 45 rule on N8 3 remaining innies R789C6 = 20 = {479/569/578}
13a. 6 of {569} must be in R7C6 -> no 6 in R89C6
13b. R89C6 = {58/59/79} -> 24(4) cage at R8C6 (step 9) = {2589/2679/3579} (cannot be {3678} because R89C6 cannot contain both of 7,8)
13c. 7,9 of {2679/3579} must be in R89C6 -> no 7 in R89C7
13d. R89C7 = {26/28/29/35} -> R2389C7 (step 11) = {2489/2678/3578} (cannot be {3479} because R89C7 cannot contain two of 3,9), 8 locked for C7, clean-up: no 5 in R47C7 (step 4)
13e. 3,5 of {3578} must be in R89C7 -> no 5 in R23C7
13f. 14(3) cage at R4C7 (step 8a) = {149/167} (cannot be {158} because no 1,5,8 in R4C7), no 5,8
13g. 21(3) cage at R2C6 = {489/678}, CPE no 8 in R2C9
13h. 8 in N6 only in R45C9 + R6C89, CPE no 8 in R789C9

14. 26(4) cage at R2C9 = {3689/4589/5678} (cannot be {4679} which clashes with 11(3) cage at R1C8 + 21(3) cage at R2C6)
14a. 5 of {4589/5678} must be in R2C9 + R3C89 (R2C9 + R3C89 cannot be {489/678} which clash with 11(3) cage at R1C8 + 21(3) cage at R2C6) -> no 5 in R4C9, clean-up: no 9 in R2C6 (step 5d)
14b. R2C6 + R4C9 (step 5d) = {68}, CPE no 6,8 in R2C9 + R4C6
14c. 9 in C6 only in R89C6, locked for N8 and 24(4) cage at R8C6, no 9 in R89C7
14d. R789C6 (step 13) contains 9 = {479/569}, no 8
14e. 4,6 only in R7C6 -> R7C6 = {46}

15. 14(3) cage at R4C7 (step 13e) = {149/167}, R47C7 (step 4) = {49/67}, 32(7) cage at R5C9 = {1234589/1234679/1235678}
15a. Consider combinations for R47C7
R47C7 = {49} => 32(7) cage = {1234589/1234679}
or R47C7 = {67} => 14(3) cage = {167}, locked for N6 => R4C9 = 8 => 32(7) cage = {1234679}
-> 32(7) cage = {1234589/1234679}

16. 45 rule on N3689 1 remaining outie R2C6 = 1 remaining innie R7C6 + 2 -> R27C6 = [64/86], 6 locked for C6

17. R7C13 (step 2) = {28} (only remaining combination, cannot be {46} which clashes with R7C6), locked for R7 and N7, clean-up: no 4,6 in R6C3 (step 3)
17a. Naked pair {28} in R67C3, locked for C3
17b. 32(7) cage at R1C3 = {1234589/1234679/1235678}, 2 locked for N1

18. 16(3) cage at R7C5 (step 12d) = {178/268}
18a. 6 of {268} must be in R7C5 -> no 6 in R8C45
18b. 6 in N8 only in R7C56, locked for R7, clean-up: no 7 in R4C7 (step 4)
18c. 14(3) cage at R4C7 (step 13e) = {149/167}
18d. 6 of {167} must be in R4C7 -> no 6 in R45C8

19. 20(3) cage at R7C8 = {569/578} (cannot be {479} which clashes with R7C7), 5 locked for C8 and N9
19a. 5 in C7 only in R56C7 = {35}, locked for C7 and N6
19b. 2 in C7 only in R89C7, locked for N9

20. 32(7) cage at R5C9 (step 15a) = {1234679} (only remaining combination), CPE no 6 in R4C9
20a. R4C9 = 8 -> R2C6 = 6 (step 5d), R7C6 = 4, clean-up: no 2 in R9C5
20b. R2C6 = 6 -> R23C7 = 15 = {78}, locked for C7 and N3 -> R7C7 = 9, naked pair {26} in R89C7, locked for C7 and N9 -> R4C7 = 4
20c. Naked pair {15} in R9C45, locked for R9 and N8, clean-up: no 9 in R8C1, no 9 in R9C23
20d. Naked pair {79} in R89C6, locked for C6 and N8 -> R7C5 = 6
20e. Naked pair {28} in R8C45, locked for R8 -> R89C7 = [62], clean-up: no 4 in R9C1
20f. Naked pair {57} in R78C8, locked for C8 and N9 -> R7C9 = 1, R9C8 = 8
20g. Naked pair {34} in R89C9, locked for C9

21. 33(7) cage at R3C6 = {1245678} -> R5C4 = 6, R4C5 = 7
21a. 6 in N6 only in R6C89, locked for R6
21b. 8 in N5 only in R56C6, locked for C6
21c. R3C5 = 9 (hidden single in C5) -> R2C9 = 9 (hidden single in C9)
21d. R3C5 = 9 -> R234C4 = 14 = {257} (only possible combination, cannot be {158} which clashes with R9C4, cannot be {248} which clashes with R8C4), locked for C4 -> R1C4 = 4, R12C5 = 9 = [81]

22. 15(4) cage at R5C1 = {1239/1248/1257} (cannot be {1347} because R7C1 only contains 2,8), 1 locked for N4
22a. 45 rule on N14 4 innies R5C1 + R6C123 = 15 and shares three cells with 15(4) cage so must have the same combination = {1239/1248/1257}, 2 locked for N4
22b. 32(7) cage at R1C3 contains 1, locked for N1

23. 45 rule on N1 2 outies R45C3 = 1 innie R3C2 + 8
23a. 32(7) cage at R1C3 contains 3, R45C3 cannot be {38} (because no 8 in R45C3) -> no 3 in R3C2
23b. 3 in N1 only in R2C23 + R4C13, locked for 32(7) cage, no 3 in R45C3

24. 21(3) cage at R1C1 = {579/678} (cannot be {489} because 4,8 only in R2C1), no 4, 7 locked for N1
24a. 7 in R1 only in R1C12, locked for N1

25. 3 in R4 only in R4C12, locked for N4
25a. 22(4) cage at R3C2 contains 3 = {3469/3568} (cannot be {3478} because no 4,7,8 in R4C12), no 7
25b. R5C1 + R6C123 (step 22a) = {1248/1257}, no 9

26. Consider combinations for 21(3) cage at R1C1 (step 24) = {579/678}
21(3) cage = {579}, locked for N1 => R1C3 = 6
or 21(3) cage = {678} = {67}8
-> 6 in R1C123, locked for R1 and N1
26a. R1C89 = [25] -> R2C8 = 4 (cage sum), R3C89 = [36]

27. R45C3 = R3C2 + 8 (step 23), R3C2 = {458} -> R45C3 = 12,13,16 = [57/67/94/97] -> R5C4 = {47}
27a. R5C1 + R6C123 (step 25b) = {1248/1257}
27b. Killer pair 4,7 in R5C1 + R6C123 and R5C3, locked for N4

28. 22(4) cage at R3C2 (step 25a) = {3469/3568}
28a. 3,6 only in R4C12 = R4C12 = {36}, locked for R4

[Just spotted a neat step for the final breakthrough …]
29. 32(7) cage at R1C3 = {1234589/1234679/1235678}, R45C3 (step 27) = [57/94/97]
29a. Consider placements for R1C3 = {69}
R1C3 = 6 => 32(7) cage must contain 7 => R5C3 = 7
or R1C3 = 9 => R4C3 = 5 => R5C3 = 7
-> R5C3 = 7, clean-up: no 3 in R9C2
29b. R5C3 = 7 -> 32(7) cage = {1234679/1235678} -> R1C3 = 6, clean-up: no 4 in R9C2
29c. R5C1 + R6C123 (step 25b) = {1248} (only remaining combination), locked for N5

30. R1C12 = {79} -> R2C1 = 5 (cage sum), R2C3 = 3, clean-up: no 7 in R9C2
30a. R9C23 = [64] -> R9C9 = 3, clean-up: no 7 in R8C1
30b. R4C12 = [63], R8C1 = 3 (hidden single in C1) -> R9C1 = 7, R7C2 = 5, R5C2 = 9
30c. R45C3 = [57] = 12 -> R3C2 = 4 (step 23)

and the rest is naked singles.

Rating Comment:
I'll rate my walkthough at 1.5 to Hard 1.5. I used several forcing chains, plus analysis of interactions between three cages in N3.

Solution:
+-------+-------+-------+
| 9 7 6 | 4 8 3 | 1 2 5 |
| 5 2 3 | 7 1 6 | 8 4 9 |
| 8 4 1 | 5 9 2 | 7 3 6 |
+-------+-------+-------+
| 6 3 5 | 2 7 1 | 4 9 8 |
| 4 9 7 | 6 3 8 | 5 1 2 |
| 1 8 2 | 9 4 5 | 3 6 7 |
+-------+-------+-------+
| 2 5 8 | 3 6 4 | 9 7 1 |
| 3 1 9 | 8 2 7 | 6 5 4 |
| 7 6 4 | 1 5 9 | 2 8 3 |
+-------+-------+-------+


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 Post subject: Re: Variations on 111
PostPosted: Fri Jan 01, 2016 9:49 pm 
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Grand Master
Grand Master

Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
Thanks Andrew for your WTs. I agree that both puzzles are easier than the SS rating suggests. Most likely because it does not find some implications of the design. I found the key for both puzzles is where what is in r7c7 goes in n6.

Here is my WT for the first one.

111 Variation (1) WT:
1. 3(2)r9 = {12}
4(2)r1 = {13}
Outies c89 -> r47c7 = +8(2)
-> in c7 one each of the numbers (123) in r1, r47 (= +8(2)), r56 (= +7(2))
-> 2 in c7 in r4567
But r47c7 cannot be [62] (Would put r45c8 = {13} leaving no solution for 7(2)n6)
-> 2 in c7 in r456 (in n6)

2. Whatever is in r7c7 goes in n6 in r4c9 or r45c8
But since r47c7 = +8(2) the latter would also put a 2 in r45c8 which contradicts 2 in r456c7.
-> r7c7 = r4c9
-> 35(7) outies from n9 = Innies n6 = +28(4)
-> r789c9 = +7(3) = [{12}4]
-> 35(7) outies from n9 = Innies n6 = {5689}
-> 7(2)n6 = {34}
-> 3(4)r1 = [31]
-> r47c7 = [26]
-> r4c9 = 6
-> r45c8 = {17}

3. Also Outies n3 -> r2c6 = 1
-> r23c7 = {59}
-> r89c7 = {78}
-> 17(3)n9 = {359}
Also -> r89c6 = [49]
-> 12(2)n7 = {57}
-> 15(2)n7 = [96]
-> r1c2 = 9 and r12c1 = {58}
-> Since 37(7)n12 must contain both of (89) -> r45c3 = {89}
-> Innies - Outies n1 -> r3c2 = 3

4. Also remaining Innies n8 = r7c46 = +15(2) = {78}
-> Outies c123 = r67c4 = +15(2) = {78}
-> 28(4)n25 = {5689} with r3c5 = 8 and r234c4 = {569}
Also Innies n7 -> r7c13 = {34}
Since Innies c123 = r67c3 = +8(2) (no 4) -> r7c13 = [43]
-> r6c3 = 5
etc.

Rating:
Rating 1.0. Just on feel rather than rating for individual moves.


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 Post subject: Re: Variations on 111
PostPosted: Mon Jan 04, 2016 9:46 pm 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
And here's my WT for the second one. It had a nasty sting in the tail I thought. But I really liked the early moves.

All the way WT for V(2):
1. Outies c6789 = r5c4+r4c5 = +13(2) (No 123)
-> 33(7)r3c6 must contain (12) in c6
-> 4(3)r1 = [31]
-> 1 in n9 in r789c9
-> 1 in n6 in r45c8

2! Outies c89 = r47c7 = +13(2)
Since 1 in r45c8 -> Whatever is in r7c7 also goes in r45c8
32(7)r5c9 is missing two numbers which sum to 13
Since r7c7 is in that 32(7) and since r47c7 = +13(2) -> whatever is in r4c7 is also in the 32(7) in r789c9
-> The two missing numbers from the 32(7) must have one in r4c9 and one in r56c7
Since 8(2)n6 from {26} or {35} -> r4c9 from 7 or 8.

3! 11(3)n3 = {236} or {245} (No 789)
Remaining outies n2 = r2c6+r4c9 = +14(2)
Cannot be [77] since 7 not in the 11(3)n3
-> r2c6 = 6, r4c9 = 8, 8(2)n6 = {35}

4. -> r23c7 = {78}
-> r47c7 = {49} and 14(3)n6 = {149}
-> 32(7)r5c9 = {1234679} with (267) in n8, (1349) in n9 with r7c7 from (49)
Also -> r89c7 = {26}
-> 20(3)n9 = {578}
Also -> r89c6 = {79}
-> r34567c6 = {12458}
-> Outies c6789 = r5c4+r4c5 = +13(2) = {67}

5. Remaining Innies n8 = r7c46 = +7(2)
-> r7c4 is max 6
Outies c123 = r67c4 = +12(2)
-> r7c4 is Min 3.
Given 7 already in n5 and n8 and 3 already in c6
-> Outies c123 = r67c4 = +12(2) can only be [93]
-> r7c6 = 4
-> 6(2)n8 = {15}
-> 16(3)n8 = {268}
Also -> r7c7 = 9
-> r789c9 = [1{34}]
Also 7(2)n5 = {34}

6. Innies n7 = r7c13 = +10(2) can only be {28} since (349) already in r7
Innies c123 = r67c3 = +10(2)
-> r6c3 = r7c1 and r67c3 = {28}
Also r8c45 = {28} and r7c5 = 6
-> r5c4 = 6 and r4c5 = 7

7. 7 in n2 in c4
7 not in r1c4 since r23c5 cannot = +6(2)
-> 7 in r23c4
13(3)n2 cannot include a 9
-> HS r3c5 = 9
Given r12c6 = [36] -> Remaining Innies - Outies n2 -> r3c6 = r4c4
-> 23(4)n2 = {2579} with r4c4 = r3c6 from (25)
-> 13(3)n2 = [481]
-> r8c45 = [82] and r9c45 = [15]

8. 11(3)n3 from [{26}3] or [254]
-> r1c123 from {579} or {679}
But it can't be the former since that puts r1c3 = r2c1
-> r1c123 = {679}
-> 11(3)n3 = [254]
-> 26(4)n3 = [9368]

9! Innies - Outies n1 -> r45c3 = r3c2 + 8
Since 8 already in c3 -> neither of r45c3 can go in r3c2
-> Both of r45c3 go in the 21(3) in n1
-> r45c3 = Min +12 and r3c2 is Min 4
Since 21(3)n1 from [{67}8] or[{79}5]
-> r45c3 from two of {579} or [67]
But since r3c2 cannot be 6 -> r45c3 cannot be +14(2) -> r45c3 cannot be {59}
-> r45c3 from ([57] [67] [97])
-> r5c3 = 7

10. Given: 3 in n4 in r4c12
and r7c1 from (28)
and 7 in r5c3
-> 15(4)r5c1 = {1248} with r7c1 from (28)

11. -> 22(4)r3c2 contains a 3 and two of (569) in n4.
But it cannot contain all of (359)
-> 6 in n4 in r4c12
-> r4c3 from (59)
-> 21(3)n1 = [{79}5] (from Step 9 line 3)
-> r1c3 = 6

12! Given 4 in n1 in r3c123
and 1 in n1 in r3c13
and (14) in n4 in r5c1, r6c12 (-> at least one of (14) in r56c1)
and 6 in n7 in r9c12
-> 4 in n7 in r8c1 or r9c3
-> 4 in c3 in r3c3 or r9c3
But the former puts r38c1 = [14] which contradicts at least one of (14) in r56c1
-> r9c3 = 4

13. Finishing
-> r9c2 = 6
Also -> r89c9 = [43]
Also -> r4c12 = [63]
-> r2c3 = 3
-> r89c1 = [37]
-> r1c12 = [97]
Also -> r89c6 = [79]
-> r789c8 = [758]
-> r7c2 = 5
-> r4c3 = 5
-> r35c2 = [49]
-> r8c23 = [19]
Also -> r45c8 = [91]
-> r6c1 = 1
-> r5c1 = 4
-> r56c5 = [34]
-> r56c7 = [53]
-> r6c6 = 5
-> r5c6 = 8
-> r4c46 = [21]
-> r3c6 = 2
-> r23c4 = [75]
-> r23c7 = [87]
-> r3c13 = [81]
-> r2c2 = [2]
Also r7c1 = 2
-> r6c23 = [82]

Rating:
Rating 1.5 - tough all the way through the way I did it

Now to compare with Andrews!


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