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 Post subject: Maverick 2 Revisit
PostPosted: Tue Nov 15, 2022 6:19 am 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Maverick 2 (aka M2) by mhparker (Nov 07)
Puzzle pic:
Image
M2 JSudoku has a terribly hard time with this so don't like my chances. It gets 1.70 so there should be a decent way through.
Code: Select, Copy & Paste into solver:
3x3::k:4864:4864:4864:4611:4611:3077:3077:6151:6151:4617:4864:4611:4611:2061:4366:4366:4366:6151:4617:2835:2835:5141:2061:1303:4366:4121:6151:4617:4380:5141:5141:2061:1303:8481:4121:6151:4380:4380:8481:8481:8481:8481:8481:3627:3627:6957:1326:8481:2864:4401:3378:3378:3627:4661:6957:1326:5432:2864:4401:3378:2876:2876:4661:6957:5432:5432:5432:4401:5700:5700:4678:4661:6957:6957:2634:2634:5700:5700:4678:4678:4678:
Solution:
+-------+-------+-------+
| 2 9 7 | 1 6 8 | 4 5 3 |
| 8 1 4 | 7 3 5 | 9 2 6 |
| 3 6 5 | 9 4 2 | 1 7 8 |
+-------+-------+-------+
| 7 4 6 | 5 1 3 | 8 9 2 |
| 5 8 2 | 6 9 4 | 3 1 7 |
| 9 3 1 | 8 2 7 | 5 6 4 |
+-------+-------+-------+
| 6 2 9 | 3 8 1 | 7 4 5 |
| 1 5 3 | 4 7 6 | 2 8 9 |
| 4 7 8 | 2 5 9 | 6 3 1 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Maverick 2 Revisit
PostPosted: Tue Nov 22, 2022 3:36 am 
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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 Ed for this latest revisit. An enjoyable on once I'd found my key steps.

Here's how I solved Maverick 2 Revisited:
Prelims

a) R1C67 = {39/48/57}, no 1,2,6
b) R3C23 = {29/38/47/56}, no 1
c) R34C6 = {14/23}
d) R34C8 = {79}
e) R67C2 = {14/23}
f) R67C4 = {29/38/47/56}, no 1
g) R7C78 = {29/38/47/56}, no 1
h) R9C34 = {19/28/37/46}, no 5
i) 8(3) cage at R2C5 = {125/134}
j) 20(3) cage at R3C4 = {389/479/569/578}, no 1,2

1a. Naked pair {79} in R34C8, locked for C8, clean-up: no 2,4 in R7C7
1b. 8(3) cage at R2C5 = {125/134}, 1 locked for C5
1c. 45 rule on C5 3 innies R159C5 = 20 = {389/479/569/578}, no 2
1d. 17(3) cage at R6C5 = {269/278/368/467} (cannot be {359/458} which clash with 8(3) cage), no 5
1e. 45 rule on R1234 2 innies R4C27 = 12 = {39/48/57}, no 1,2,6
1f. 45 rule on R6789 2 innies R6C38 = 7 = {16/25/34}, no 7,8,9
1g. 33(7) cage at R4C7 must contain 1,2,6 in R5C34567 + R6C3, CPE no 1,2,6 in R5C12
1h. 17(3) cage at R4C2 = {359/458}, no 7, 5 locked for N4, clean-up: no 5 in R4C7, no 2 in R6C8
1h. 7 in R5 only in R5C345679, CPE no 7 in R4C7, clean-up: no 5 in R4C2
1i. 5 in N4 only in R5C12, locked for R5
1j. 33(7) cage = {1234689} (only remaining combination), no 7
1k. R5C9 = 7 (hidden single in R5) -> R34C8 = [79], clean-up: no 5 in R1C6, no 4 in R3C23, no 3 in R4C27
1l. Naked pair {48} in R4C27, locked for R4, clean-up: no 1 in R3C6
1m. 17(3) cage = {458} (cannot be {359} because R4C2 only contains 4,8), 4,8 locked for N4, clean-up: no 3 in R6C8, no 1 in R7C2
1n. 20(3) cage at R3C4 = {569/578} (cannot be {389/479} because 4,8,9 only in R3C4) -> R3C4 = {89}, R4C4 = 5, R4C3 = {67}, clean-up: no 6 in R67C4
1o. R5C9 = 7 -> R56C8 = [16/25/34/61], no 4,8 in R5C8
1p. 45 rule on N1 1 outie R4C1 = 1 innie R2C3 + 3 -> R2C3 = {34}, R4C1 = {67}
1q. Naked pair {67} in R4C13, locked for N4, 6 locked for R4, clean-up: no 1 in R6C8, no 6 in R5C8
1r. 45 rule on N9 1 outie R6C9 = 1 innie R8C7 + 2, no 1,2 in R6C9, no 5,7,8,9 in R8C7

2a. 45 rule on C1 2 innies R15C1 = 1 outie R9C2
2b. Min R15C1 = 5 -> min R9C2 = 5
2c. Max R9C2 = 9 -> max R15C1 = 9 -> max R1C1 = 5
2d. 17(3) cage at R4C2 = {458}, R15C1 = R9C2 -> no 5,8 in R9C2
2e. 45 rule on C89 2 outies R79C7 = 1 innie R2C8 + 11
2f. Max R79C7 = 17 -> max R2C8 = 6
2g. Min R79C7 = 12, no 1,2 in R9C7
2h. 13(3) cage at R6C6 = {148/157/238/247/256/346} (cannot be {139} which clashes with R6C123, ALS block), no 9
2i. 18(3) cage at R2C1 = {279/378/468/567} (cannot be {189/459} because R4C1 only has 6,7, cannot be {369} = {39}6 which clashes with R2C3 + R4C1 = [36], step 1p), no 1
2j. 18(3) cage at R6C9 = {189/369/459/468}, no 2
2k. 45 rule on C9 1 outie R1C8 = 1 remaining innie R9C9 + 4 -> R1C8 = {568}, R9C9 = {124}
2l. 45 rule on N3 3(2+1) remaining outies R12C6 + R4C9 = 15
2m. Max R4C9 = 3 -> min R12C6 = 12, no 1,2 in R2C6
2n. 7 in C7 only in R79C7, min R79C7 = 12 -> no 3,4 in R79C7, clean-up: no 8 in R7C8
2o. R79C7 = {57/67/78/79} = 12,13,15,16 -> R2C8 = {1245}

3a. 18(3) cage at R6C9 (step 2j) = {189/369/459/468}
3b. Consider placement for 8 in C8
R1C8 = 8 => R9C9 = 4 (step 2k), 8 in C9 only in 18(3) cage = {189} = 8{19}
or 8 in R89C8, locked for N9
-> no 8 in R89C9
3c. 18(3) cage = {189/369/459} (cannot be {468} = 8{46} which classes with R6C9 + R8C7 = [86], step 1r), 9 locked for C9 and N9, clean-up: no 2 in R7C8
3d. 24(5) cage at R1C8 = {14568/23568}, 5,6,8 locked for N3, clean-up: no 4,7 in R1C6
3e. 17(4) cage at R2C6 = {1259/1349} (cannot be {1268/1358/1367/1457/2357/2456} because 5,6,7,8 only in R2C6, cannot be {2348} which clashes with 24(5) cage), no 6,7,8, 1 locked for N3

4a. 6,7 in N2 only in 18(4) cage at R1C4 = {1467/2367}, no 5,8,9
4b. R2C3 = {34} -> no 3,4 in R1C45 + R2C4
4c. R159C5 (step 1c) = {479/569/578} (cannot be {389} because R1C5 only contains 6,7), no 3
4d. R1C5 = {67} -> no 6,7 in R59C5
4e. 8 of {578} must be in R5C5 -> no 8 in R9C5
4f. 17(3) cage at R6C5 (step 1d) = {269/278/368} (cannot be {467} which clashes with R1C5), no 4

[I’d spotted this 45 earlier but it’s only useful now; very useful.]
5a. 18(4) cage at R1C4 (step 4a) = {1467/2367}
5b. 45 rule on C1234 3 innies R5C34 + R6C3 = 1 outie R1C5 + 3
5c. R1C5 = {67} -> R5C34 + R6C3 = 9,10 = {126/234/136}, no 8,9 -> R4C6 = {46}
5d. R1C5 + R5C34 + R6C3 = 6{12}6/6{23}4 (cannot be 7{13}6 which clashes with 18(4) cage -> R1C5 = 6, R56C3 = {12/23}, 2 locked for C3, N4 and 33(7) cage at R4C7, clean-up: no 9 in R3C2, no 3 in R7C2, no 8 in R9C4
5e. R1C5 = 6 -> R59C5 (step 4c) = 14 = [95], clean-up: no 2 in R7C4
5f. 7 in N2 only in R12C4, locked for C4, clean-up: no 4 in R67C4, no 3 in R9C3
5g. 8(3) cage at R2C5 = {134} (only remaining combination), 3 locked for C5, 4 locked for N2, clean-up: no 1 in R4C6
5h. R2C3 + R4C1 (step 1p) = [36/47], R3C4 + R4C3 = [87/96] -> R2C3 + R3C4 = [38/49]
5i. R2C3 + R123C4 = 4{17}9 (cannot be 3{27}8} which clashes with R67C4) -> R2C3 = 4, R3C4 = 9, R12C4 = {17}, 1 locked for C4 and N2, R4C3 = 6 (cage sum), R4C1 = 7, clean-up: no 3 in R1C7, no 2,5 in R3C2, no 2 in R6C4, no 1,9 in R9C3, no 4,6 in R9C4
[Cracked. Clean-ups omitted from here.]

6a. 8(3) cage at R2C5 = [341] -> R1C6 = 8 -> R1C7 = 4, R1C8 = 5 -> R9C9 = 1 (step 2k), R234C6 = [523], R14C9 = [32], R4C27 = [48], R67C4 = [83], R9C4 = 2 -> R9C3 = 8
6b. Naked pair {78} in R78C5, 7 locked for C5 and N8 -> R6C5 = 2
6c. R7C2 = 2 -> R6C2 = 3, R56C3 = [21], R6C8 = 6 (step 1f) -> R5C8 = 1 (cage sum)
6d. R7C8 = 4 -> R7C7 = 7
6e. R6C7 = 5 -> R67C6 = 8 = [71]
6f. R589C7 = [326], R9C5 = 5 -> R89C6 = 15 = [69]
6g. R9C2 = 7 -> R15C1 = 7 (step 2a) = [25], R5C2 = 8, R3C2 = 6 -> R3C3 = 5

and the rest is naked singles.


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 Post subject: Re: Maverick 2 Revisit
PostPosted: Tue Nov 29, 2022 7:35 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 Ed. This one needed quite a few steps and probably has several solving paths. MY WT has one or two small contradiction chains.
Assassin Maverick 2 WT:
1. Innies r6789 = r6c38 = +7(2)
Whatever goes in r6c3 goes in r5 in r5c89
-> The other value in 14(3)n6 must be a 7
-> 16(2)c8 = [79]
-> r5c9 = 7

Similarly Innies r1234 = r4c27 = +12(2)
Whatever is in r4c7 is in r5 in r5c12
-> The other value in 17(3)n4 must be a 5
-> (Since 9 already in r4) r4c27 = {48} and 17(3)n4 = {458} with 5 in r5c12

2. Possibilities for 29(3)r3c4 are [965] or [875]
I.e., r4c4 = 5

3. IOD n1 -> r4c1 = r2c3 + 3
-> r4c1 from (67) (Since 4589 already in r4)
-> One of:
(A) r2c3 = 3, r3c4 = 8, r4c13 = [67]
(B) r2c3 = 4, r3c4 = 9, r4c13 = [76]
-> r4c569 = {123}

4! r1c5 cannot be 9 since that puts r3c4 = 8, r2c3 = 3, r12c4 = {24} which leaves no solution for 11(2)c4
Also r1c5 cannot be 8 since that puts r3c4 = 9 and r2c3 = 4 which leaves no solution for 18(4)r1c4
IOD c1234 -> r56c3 + r5c4 = r1c5 + 3
Since r1c5 is Max 7 -> r56c3 + r5c4 is Max +10(3) (I.e., no (89))
-> (HS 9 in n4) r6c1 = 9
Remaining cells in n4 = r5c3,r6c23 = {123}

5. [r2c3,r3c4] are [38] or [49]
Remaining outies n7 = r6c2,r89c4 are +9(3)
Either 3 in r56c3 -> [r2c3,r3c4] = [49]
Or 3 in r6c2 -> r89c4 = {24} and 11(2)c4 = {38} which again -> [r2c3,r3c4] = [49]
-> [r2c3,r3c4] = [49]
-> r4c13 = [76]
-> r23c1 = +11(2)
-> None of (179) can go in either of the 11(2)s in n1
-> 19(4)n1 = {1279}

6. IOD c1 -> r1c1 + r5c1 = r9c2
-> r9c2 cannot be the same as r5c1. (I.e., not 458).
-> Possibilities for [r1c1,r5c1,r9c2] are [156], [246], [257]
But the first of those leaves no place for 6 in n1
-> r1c1 = 2 and 8 in r45c2

7! -> Remaining Outies n3 = r12c6 and r4c9 = +15(3)
Since r4c9 from (123) -> r12c6 one of (+14, +13, +12)
-> 3 not in r12c6 (Since 9 already in n2)
Also, for the numbers (78) in n2 - one must go in 18(4) and the other in r12c6
-> 18(4)n2 (in n2) from {167}, {257}, or {158}
Remaining IOD c9 -> r1c8 is Min 5
-> (HS 3 in r1) r1c9 = 3

8. -> 3 not in r4c9
-> 3 in r4 in r4c56. r4c9 from (12)
-> r12c6 = +14 or +13
-> 4 in n2 in r3c56
-> (HS 4 in r1) -> r1c7 = 4
-> 12(2)r1 = [84]
Also r4c27 = [48]
Also [r2c6,r4c9] from [52] or [61]

9. 3 in n2 in 8(3) or 5(2)
Also 3 in r4 in r4c56
-> 3 in both 8(3)n2 and 5(2)n2
-> 5(2)n2 = {23}
-> 8(3)n2 = <143>

10. (From Step 6) since r4c2 = 4 -> r5c1 = 5 and r9c2 = 7
-> r1c3 = 7 and r12c2 = {19}
-> r2c4 = 7
Also -> (HS 9 in r1) r12c2 = [91]
-> r1c45 = [16]
-> 8(3)c5 = [341]
-> 5(2)c6 = [23]
Also r2c6 = 5, r4c9 = 2, and r9c9 = 1
Also 11(2)c4 = [83]
Also 27(5)r6c1 can only be [9{146}7]
etc.


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