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 Post subject: Assassin 406
PostPosted: Wed Dec 01, 2021 6:09 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 947
Location: Sydney, Australia
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a406.png
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NOTE: 1-9 cannot repeat on either diagonal. A broken 17(3)r3c5.

Assassin 406
Found a really nice way to crack this one. Just enough mind-bending throughout to make an interesting puzzle. SS gives it 1.60, JSudoku uses 4 advanced steps.
triple click code:
3x3:d:k:2560:6657:6657:6657:3330:3587:3587:3587:1796:2053:2560:6657:3330:3330:3078:3078:1796:5895:3848:2053:10249:10249:4362:8203:8203:5895:5895:3848:3848:10249:4876:4876:4876:8203:2317:5895:2574:2574:10249:4623:4362:4876:8203:2317:3344:3857:3857:10249:4623:4623:4362:8203:3344:3344:2834:3857:10249:10249:4623:8203:8203:3091:3091:2834:2836:3093:5398:5398:1559:5656:2329:3091:2836:3093:3093:5398:5398:1559:5656:5656:2329:
solution:
+-------+-------+-------+
| 4 9 3 | 6 7 5 | 1 8 2 |
| 1 6 8 | 2 4 9 | 3 5 7 |
| 2 7 5 | 8 1 3 | 4 6 9 |
+-------+-------+-------+
| 8 5 4 | 3 2 6 | 9 7 1 |
| 7 3 1 | 4 9 8 | 5 2 6 |
| 9 2 6 | 1 5 7 | 8 4 3 |
+-------+-------+-------+
| 6 4 7 | 9 8 1 | 2 3 5 |
| 5 8 9 | 7 3 2 | 6 1 4 |
| 3 1 2 | 5 6 4 | 7 9 8 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 406
PostPosted: Sat Dec 04, 2021 8:39 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 251
Location: California, out of London
Another fun one! Not as tricky as the previous one and will be interested to see if Ed's way to crack it is similar to mine.
Corrections thanks to Andrew.
Assassin 406 WT:
1. Outies c123 = r137c4 = +23(3) = {689}
40(7)r3c3 = {56789} + {14} or {23}
-> r34567c3 = {57} + either {14} or {23} + one of {689}
-> r1289c3 = Two of (689) + either {23} or {14}

2. Innies n7 = r7c23 = +11(2)
-> 1 in n7 in 12(3)

3. Innies c12 = r19c2 = +10(2) (No 5)
-> 5 not in 26(4)r1c2
-> 26(4)r1c2 from {2789}, {3689}, or {4679}

4. Innies n9 = r7c7 = 2
-> 10(2)n1 not {28}

5. If 7 in 26(4)r1c2 it can only go in r1c2
This puts r9c2 = 3 and r89c3 = {18} which puts 4 in r12c3
-> 26(4)r1c2 cannot be {2789}

6. 2 in n1 only in 8(2) or r3c1
5 in n1 only in 8(2) or r3c13
IOD n1 -> r3c1 + r3c3 = r1c4 + 1
-> r3c13 = max +10(2)

7! Consider where 8 goes in n1.
8 not in r3c1 since that would put 8(2)n1 = {26} and r3c3 = 5 making r3c13 more than the max.
-> 8 in n1 in 26(4) or in r3c3 which also puts it in r1c4
-> 8 in 26(4)r1c2
-> 26(4)r1c2 = {3689} with 3 in n1

8. 10(2) from [19] or [46] -> r12c23 = {3689}
-> 8(2)n1 = {17}
-> 10(2)n1 = [46]
-> r1c4 = 6 and {389} in 26(4) in n1
Also -> r3c1 = 2
-> r3c3 = 5
Also r37c4 = {89}

9. Remaining innies D\ (in n5) = +19(3) (No 1)
-> 9(2)n9 = {18}
-> D\ (n5) = {379}
Innies c6789 = r456c6 = +21(3)
-> Min r45c6 = +12(2) -> Max r4c45 = +7(2)
-> D\ n5 = [3{79}]
-> 17(3)r3c5 = [1{79}]
-> (Outies n5) r7c5 = 8
-> 8 in n5 in r45c6
-> 19(4)n5 = [32{68}]
-> r6c6 = 7 and r5c5 = 9
-> 18(4)r5c4 = [{145}8]

10. Also r37c4 = [89]
-> (Innies n8) r7c6 = 1
-> 6(2)n8 = {24}
-> r89c5 = {36}
-> r89c4 = {57}
-> r6c5 = 5 and r45c4 = {14}
-> 13(3)n2 = [724]

11. Also 8(2)n1 = [17]
-> r19c2 cannot be [37] -> 3 in r12c3
-> 2 in r89c3
-> r19c2 = [91]
-> r89c3 = {29}

12. Also 9(2)n9 = [18]
-> 22(3)n9 = {679}
-> 12(3)n9 = {345}
-> Innies r89 = r8c19 = +9(2) = [54] (Since r89c5 = {36} and 4 already in c1)
-> 11(2)r8c2 = [83]
-> 21(4)n8 = [7356] and 6(2)n8 = [24]
-> 22(3)n9 = [6{79}]
-> r89c3 = [92]

13. Also 2 in D/ only in r1c9
-> r2c8 = 5
etc.


Last edited by wellbeback on Sat Dec 11, 2021 7:37 pm, edited 1 time in total.

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 Post subject: Re: Assassin 406
PostPosted: Wed Dec 08, 2021 5:26 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1820
Location: Lethbridge, Alberta, Canada
wellbeback found a better place to work so more a fun puzzle for him. I hope some of my steps are of interest.

Thanks Ed for pointing out that I'd got my mental arithmetic wrong in my original step 5d! Fortunately there was a nice step which provided a short re-work; only two steps replaced.

Here's my walkthrough for Assassin 406:
Prelims

a) 10(2) cage at R1C1 = {19/28/37/46}, no 5
b) 7(2) cage at R1C9 = {16/25/34}, no 7,8,9
c) 8(2) cage at R2C1 = {17/26/35}, no 4,8,9
d) R2C67 = {39/48/57}, no 1,2,6
e) R45C8 = {18/27/36/45}, no 9
f) R5C12 = {19/28/37/46}, no 5
g) R78C1 = {29/38/47/56}, no 1
h) 11(2) cage at R8C2 = {29/38/47/56}, no 1
i) R89C6 = {15/24}
j) 9(2) cage at R8C8 = {18/27/36/45}, no 9
k) 22(3) at R8C7 = {589/679}
l) 26(4) cage at R1C2 = {2789/3689/4589/4679/5678}, no 1

1a. 45 rule on N9 1 innie R7C7 = 2, placed for D\, clean-up: no 8 in 10(2) cage at R1C1, no 9 in R8C1, no 7 in 9(2) cage at R8C8
1b. 12(3) cage at R7C8 = {138/147/345} (cannot be {156} which clashes with 22(3) cage at R8C7), no 6,9
1c. 45 rule on C89 2 innies R19C8 = 17 = {89}, locked for C8, clean-up: no 1 in R45C8, no 1 in R9C9
1d. Min R1C8 = 8 -> max R1C67 = 6, no 6,7,8,9 in R1C67
1e. 45 rule on N7 2 innies R7C23 = 11 = {38/47/56}, no 1,9
1f. 45 rule on C12 2 innies R19C2 = 10 = {28/37/46}/[91], no 5, no 9 in R9C2
1g. 45 rule on N5 2 outies R37C5 = 9 = {18/36/45}/[27], no 9, no 7 in R3C5
1h. 45 rule on N8 3 innies R7C456 = 18 = {189/369/378/567} (cannot be {459} which clashes with R89C6, cannot be {468} which clashes with R7C23), no 4, clean-up: no 5 in R3C5
1i. 45 rule on C123 3 outies R137C4 = 23 = {689}, locked for C4
1j. 40(7) cage at R3C3 = {1456789/2356789}, 5,7 locked for C3
1k. 1 in N7 only in 12(3) cage at R8C3 = {129/138/147}, no 6, clean-up: no 4 in R1C2
1l. 7 of {147} must be in R9C2 -> no 4 in R9C2, clean-up: no 6 in R1C2
1m. 8 of {138} must be in R89C3 (R89C3 cannot be {13} which clashes with 40(7) cage), no 8 in R9C2, clean-up: no 2 in R1C2
1n. 26(4) cage at R1C2 = {2789/3689/4679}
1o. 2,3 of {2789/3689} must be in R1C23 (R1C234 cannot be {689/789} which clash with R1C8), no 2,3 in R2C3
1p. 26(4) cage = {2789/3689/4679}, CPE no 9 in R1C1, clean-up: no 1 in R2C2

2a. 45 rule on C6789 1 innie R6C6 = 2 outies R4C45 + 2
2b. Min R4C45 = 3 -> min R6C6 = 5
2c. Max R4C45 = 7, no 7,8,9 in R4C45
2d. 45 rule on C6789 3 innies R456C6 = 21 = {489/579/678}, no 1,2,3

3a. 45 rule on R12 2 innies R2C19 = 8 = {17/26/35}, no 4,8,9
3b. 45 rule on R89 2 innies R8C19 = 9 = [27/45/54/63/81], no 3,7 in R8C1, no 8 in R8C9, clean-up: no 4,8 in R7C1

4a. 45 rule on D\ 2 remaining innies R3C3 + R4C4 = 1 outie R3C5 + 7
4b. Min R3C3 + R4C4 = 8, max R4C4 = 5 -> min R3C3 = 3
4c. Max R3C3 + R4C4 = 14 -> no 8 in R3C5, clean-up: no 1 in R7C5 (step 1g)

5a. R7C456 (step 1h) = {189/369/378/567}
5b. Consider combinations for R456C6 (step 2d) = {489/579/678}
R456C6 = {489}, 4 locked for C6 => R89C6 = {15}, 5 locked for N8
or R456C5 = {579/678}, 7 locked for C6 => R7C456 = {189/369}/[675/873], no 5 in R7C5
-> no 5 in R7C5, clean-up: no 4 in R3C5 (step 1g)
5c. Consider placement for 6 in N5
6 in R456C5, locked for C5
or 6 in R456C6 = {678} => R5C5 + R6C6 cannot be 14 = {59/68} (the latter CCC)
-> no 3 in R3C5, clean-up: no 6 in R7C5 (step 1g)
5d. 17(3) cage at R3C5 = 1{79}/2{69}/2{78}/[647] (cannot be [638] which clashes with R37C5 = [63], step 1g) -> R5C5 = {46789}, R6C6 = {6789}
First time I’d got my mental arithmetic wrong! :oops:
5e. 17(3) cage = 1{79}/2{69}/2{78} (cannot be [647] which gives R7C5 = 3 (step 1g), R4C4 = 3 (hidden single in N5) when R4C4 + R5C5 = [34] clash with 1,3,4 in 10(2) cage at R1C1 + 9(2) cage at R8C8, killer ALS block)
-> R3C5 = {12}, R5C5 + R6C6 = {69/78/79}, no 4, clean-up: no 3 in R7C5
5f. R3C5 = {12} -> R3C3 + R4C4 (step 4a) = 8,9, no 9 in R3C3
5g. R7C456 = {189/378/567} (cannot be {369} because R7C5 only contains 7,8)
5h. 1,3,5 only in R7C6 -> R7C6 = {135}
5i. Min R7C5 = 7 -> max R5C4 + R6C45 = 11, no 9 in R6C5
5j. Consider permutations for R37C5 = [18/27]
R37C5 = [18] => R7C456 = [981]
or R37C5 = [27] => 2 in C6 only in R89C6
-> R89C6 = {24}, locked for C6 and N8, clean-up: no 8 in R2C7
[Cracked. The rest is fairly straightforward.]
5k. R456C6 = {579/678}, 7 locked for C6 and N5, clean-up: no 5 in R2C7, no 8 in R6C6 (step 5e)
5l. 4 in N2 only in 13(3) cage at R1C5 = {148/247/346}, no 5,9
5m. 5 in N2 only in R123C6, locked for C6, clean-up: no 9 in R456C6
5n. Naked triple {678} in R456C6, 6,8 locked for C6 and N5 -> R5C5 = 9, placed for both diagonals, clean-up: no 1 in R1C1, no 4 in R2C7, no 1 in R5C12, no 2 in 11(2) cage at R8C2
5o. 9 in C6 only in R23C6, locked for N2
5p. Naked pair {68} in R13C4, locked for C4 and N2
5q. 13(3) cage = {247} (only remaining combination), 2 locked for N2
5r. R3C5 = 1, R7C5 = 8 (step 1g) -> R7C456 = [981], clean-up: no 7 in R2C1, no 1 in R2C9 (step 3a), no 3 in R7C23 (step 1e), no 2 in R8C1
5s. R3C5 = 1, R5C5 = 9 -> R6C6 = 7 (cage sum), placed for D\, clean-up: no 3 in 10(2) cage at R1C1
5t. Naked pair {46} in 10(2) cage, locked for N1 and D\, clean-up: no 2 in 8(2) cage at R2C1, no 2,6 in R2C9 (step 3a), no 3,5 in 9(2) cage at R8C8
5u. R8C8 = 1, R9C9 = 8, both placed for D\ -> R9C8 = 9, R1C8 = 8, R1C67 = 6 = [51], R89C7 = 13 = {67}, locked for C7, 7 locked for N9, clean-up: no 6 in R1C9, no 2,6 in R2C8, no 3 in R8C2, no 2 in R9C2 (step 1f)

6a. R8C3 = 9 (hidden single in N7) -> R9C23 = 3 = [12], R89C6 = [24], R1C2 = 9 (step 1f), R12C3 = [38], R1C4 = 6, R3C3 = 5, placed for D\, R1C1 = 4, R2C2 = 6, R1C9 = 2 -> R2C8 = 5, both placed for D/, R2C1 = 1, R2C9 = 7 (step 3a), R3C2 = 7, clean-up: no 4 in R45C8, no 3 in R5C1, no 7 in R7C1, no 4 in R7C3 (step 1e), no 6 in R9C1
6b. R4C4 = 3 -> R89C4 = {57}, locked for N8, 5 locked for C4, clean-up: no 6 in R5C8
6c. R6C4 = 1 (hidden single on D/), R7C5 = 8 -> R5C4 + R6C5 = 9 = [45], clean-up: no 6 in R5C1
6d. R7C3 = 7 (hidden single in R7), R7C2 = 4 (step 1e), R8C2 = 8, R9C1 = 3, both placed for D/
6e. R7C1 = 6 (hidden single in R7) -> R8C1 = 5
6f. R45C6 = [68], clean-up: no 2 in R5C12, no 3 in R5C8
6g. R3C1 = 2 -> R4C12 = 13 = [85]
6h. R234C7 = [349], R2C9 = 7, R3C89 = [69] -> R4C9 = 1 (cage sum)

and the rest is naked singles, without using the diagonals.


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 Post subject: Re: Assassin 406
PostPosted: Fri Dec 10, 2021 8:12 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 947
Location: Sydney, Australia
A 3rd way to do this puzzle. Mine is probably closer to wellbeback's but as so often happens, I use a more circuitous route to get the same key elimination. An interesting detour non-the-less! [Thanks to wellbeback and Andrew for checking my WT noticing some things]
a406 WT:
Preliminaries from SudokuSolver
Cage 6(2) n8 - cells only uses 1245
Cage 7(2) n3 - cells do not use 789
Cage 8(2) n1 - cells do not use 489
Cage 12(2) n23 - cells do not use 126
Cage 9(2) n6 - cells do not use 9
Cage 9(2) n9 - cells do not use 9
Cage 10(2) n4 - cells do not use 5
Cage 10(2) n1 - cells do not use 5
Cage 11(2) n7 - cells do not use 1
Cage 11(2) n7 - cells do not use 1
Cage 22(3) n9 - cells do not use 1234
Cage 26(4) n12 - cells do not use 1

This is a highly optimised solution so any clean-up is stated.
1. "45" on c123: 3 outies r137c4 = 23 = {689} all locked for c4

2. 40(7)r3c3 = 56789{14/23}
2a. must have 5 & 7 which are only in c3: both locked for c3

3. "45" on c12: 2 innies r19c2 = 10 (no 5)

4. "45" on n7: 2 innies r7c23 = 11 (no 1)

5. 1 in n7 only in 12(3) = {129/138/147}(no 6)

The other two WTs didn't use this
6. 6 in c3 & 4 are only in two cages so both must have 6 (forced caged x-wing)
6a. -> 26(4)r1c2 = {3689/4679}(no 2)
6b. 6 in 26(4) must be in c34 -> no 6 in r1c2

7. "45" on n9: 1 innie r7c7 = 2: placed for d\
7a. 10(2)n1 = {19/37/46}(no 8)

first key step
8. 6 & 9 must be in 26(4)n1 but both can't be in n1 since combined cages 10(2)n1 = {37} + 8(2)n1 is blocked by 6 already in n1
8a. -> 26(4) must have 6 or 9 in r1c4
8b. -> no 8 in r1c4

9. 8 in c4 only in 40(7)r3c3: 8 locked for that cage
9a. -> no 8 in r34567c3

Fun one!
10. "45" on n5: 1 outie r7c5 + 8 = 2 innies r5c5 + r6c6 (IOD=+8)
10a. r57c5 see each other so the IOD can't be 0 -> no 8 in r6c6 (IOU)

11. "45" on c89: 2 innies r19c8 = 17 = {89}: both locked for c8
11a. -> r1c67 = 5/6 (no 6,7,8,9)

Key steps
12. 32(7)r3c6 = {123}{4589/4679/5678}: if it has 6 must also have 7
12a. 22(3)r8c7 = {589}/[{67}[9]: if it has 6 must also have 7 (and both in c7)
12b. "45" on c6789: 3 innies r456c6 = 21 = {489/579/678}(no 1,2,3): if it has 6 must also have 7
12c. the two 6s in c67 are only in those 3 cages -> both 7's must be in two of those three -> 7 locked for c67
12d. -> no 5,7 in 12(2)r2c6

13. from 12a,b,c; 6 & 7 in c7 are both in r89c7 or both must be in 32(7) in c7
13a. -> 6 & 7 in 32(7) are both in c7 or both in r37c6
13b. -> 7 in h21(3)r456c6 must also have 6 or there would be no 6 for c6 (locking-out cages)
13c. -> h21(3) = {489/678}(no 5)
13d. must have 8: locked for n5 and c6

14. r9c9 = 8 (hsingle d\)
14a. -> r8c8 = 1: placed for d\
14b. r19c8 = [89]
14c. 12(2)r2c6 = {39}: both locked for r2

15. 26(4)r1c2 = {3689/4679} = 3 or 4 in n1 [alternatively: also 3 or 7 in n1 -> [37] blocked in next sub-step, also no 6 in r1c1]
15a. 10(2)n1 = {46}/[37] = 3 or 4
15b. -> killer pair 3,4: both locked for n1
15c. no 5 in 8(2)n1

Final cracker
16. "45" on n1: 1 outie r1c4 + 1 = 2 innies r3c13 which must have 5 for n1
16a. -> 2 innies can't be 10 -> no 9 in r1c4
16b. -> r1c4 = 6
16c. -> r3c13 = 7 = [25], 5 placed for d\
16d. -> 8(2)n1 = {17}, 7 locked for n1
16e. -> 10(2)n1 = [46], both placed for d\
16f. r2c3 = 8
16g. r1c23 = {39} both locked for r1

17. naked triple {379} on d\ in n5: all locked for n5

18. 17(3)r3c5 must have two of {379} in n5 = [1]{79} only
18a. -> r4c4 = 3
18b. and 8(2)n1 = [17]

19. "45" on n5: 1 remaining outie r7c5 = 8
19a. r7c4 = 9, r3c4 = 8

20. "45" on n8: 1 remaining innie r7c6 = 1
20a. -> 6(2)n8 = {24}: both locked for c6 and n8
20b. r1c67 = [51]

21. "45" on r89: 2 innies r8c19 = 9
21a. but {36} blocked by r8c457 = {3567} (Almost Locked set) [alternatively courtesy of Andrew, Step 21a r89c4 = {57}, locked for n8, then {36} blocked by r8c5 = (36)]
21b. = [54], only permutation

Much easier now with a few cage sums. Don't forget the Ds!
Cheers
Ed


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