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Assassin 372
http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1482
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Author:  Ed [ Fri Mar 15, 2019 7:59 am ]
Post subject:  Assassin 372

Attachment:
a372.JPG
a372.JPG [ 70.5 KiB | Viewed 6833 times ]
This is an X-puzzle. 1-9 cannot repeat on either diagonal.

Assassin 372
Found this very difficult to crack first time though my optimised solution is very short. Made a zero version and the SSscore went up but didn't make any difference to how I did it. SS must use different cages than me so have gone with the vanilla. SScore=1.25 JSudoku uses a couple of chains. I used a difficult key step so will be interested to see what I missed.
code:
3x3:d:k:2048:7681:7681:7681:7681:4610:4610:4610:4099:2048:7681:2052:2052:7681:5893:5893:5893:4099:4870:3847:3847:1544:1544:3593:5893:2058:4099:4870:4870:1547:1547:6668:3593:5893:2058:1037:3854:3854:1551:1551:6668:6668:6937:6937:1037:3854:3089:3089:4114:4114:6668:6937:9232:6937:4884:3089:5909:4114:9232:9232:9232:9232:9232:4884:4884:5909:5909:9232:2582:3095:3095:3864:4884:3347:3347:5909:2582:2582:2582:3095:3864:
solution:
Code:
+-------+-------+-------+
| 3 4 8 | 9 1 5 | 6 7 2 |
| 5 2 1 | 7 6 3 | 4 8 9 |
| 7 9 6 | 2 4 8 | 1 3 5 |
+-------+-------+-------+
| 9 3 2 | 4 8 6 | 7 5 1 |
| 6 8 5 | 1 7 2 | 9 4 3 |
| 1 7 4 | 3 5 9 | 8 2 6 |
+-------+-------+-------+
| 2 1 9 | 8 3 7 | 5 6 4 |
| 8 5 3 | 6 9 4 | 2 1 7 |
| 4 6 7 | 5 2 1 | 3 9 8 |
+-------+-------+-------+
Cheers
Ed

Author:  Andrew [ Tue Mar 19, 2019 4:33 am ]
Post subject:  Re: Assassin 372

Thanks Ed for your latest Assassin. I have no idea how SudokuSolver got that score; it felt quite a lot harder to me, more like our current level of Assassins.

Here is my walkthrough for Assassin 372:
Prelims

a) R12C1 = {17/27/35}, no 4,8,9
b) R2C34 = {17/27/35}, no 4,8,9
c) R3C23 = {69/78}
d) R3C45 = {15/24}
e) R34C6 = {59/68}
f) R34C8 = {17/27/35}, no 4,8,9
g) R4C34 = {15/24}
h) R45C9 = {13}
i) R5C34 = {15/24}
j) R89C9 = {69/78}
k) R9C23 = {49/58/67}, no 1,2,3
l) 19(3) cage at R3C1 = {289/379/469/478/568}, no 1
m) 26(4) cage at R4C5 = {2789/3689/4589/4679/5678}, no 1
n) 27(4) cage at R5C7 = {3789/4689/5679}, no 1,2
o) 10(4) cage at R8C6 = {1234}

Steps resulting from Prelims
1a. Naked pair {13} in R45C9, locked for C9 and N6, clean-up: no 5,7 in R3C8
1b. 27(4) cage at R5C7 now = {4689/5679}, 6,9 locked for N6, clean-up: no 2 in R3C8
1c. 2 in N6 only in R4C78 + R6C8, CPE no 2 in R2C8
1d. Naked quad {1234} in 10(4) cage at R8C6, CPE no 1,2,3,4 in R9C4
1e. 1 in N5 only in R456C4 + R6C5, CPE no 1 in R7C4
1f. 36(7) cage at R6C8 must contain 9, not now in R6C8, CPE no 9 in R7C4

2. 45 rule on C9 2 innies R67C9 = 10 = [46/64/82], no 5,7,9, no 8 in R7C9
2a. 5 in C9 only in 16(3) cage at R1C9, locked for N3
2b. 16(3) cage = {259/457}, no 6,8

3. 45 rule on R1234 2 innies R4C59 = 9 = [63/81]
3a. Variable hidden killer pair 3,7 in 26(4) cage at R4C5 and 16(3) cage at R6C4 for N5, 26(4) cage cannot contain more than one of 3,7 -> 16(3) cage must contain at least one of 3,7 = {178/349/358/367/457} (cannot be {169/259/268} which don’t contain 3 or 7), no 2

4. 45 rule on N7 2 outies R89C4 = 1 innie R7C2 + 10
4a. Min R89C4 = 11, no 1 in R8C4
4b. Max R89C4 = 17 -> max R7C2 = 7

5. 45 rule on N5 3 innies R45C4 + R4C6 = 1 outie R7C4 + 3
5a. Max R7C4 = 8 -> max R45C4 + R4C6 = 11, no 9 in R4C6, clean-up: no 5 in R3C6
5b. Min R45C4 + R4C6 = 8 -> min R7C4 = 5
5c. 26(4) cage at R4C5 = {2789/3689/4589/4679} (cannot be {5678} which clashes with R4C6), 9 locked for N5
5d. 9 in R4 only in R4C12, locked for N4 and 19(3) cage at R3C1, no 9 in R3C1
5e. 19(3) cage must contain 9 in R4C12 = {289/379/469}, no 5

6. 45 rule on R12 3(2+1) outies R34C7 + R3C9 = 13
6a. R34C7 cannot total 4 -> no 9 in R3C9

7. R3C23 = {69/78}, R3C6 = {689} -> combined cage R3C236 = {69}8/{78}6/{78}9, 8 locked for R3
7a. 19(3) cage at R3C1 (step 5e) = {289/379/469}
7b. 2 of {289} must be in R3C1 -> no 2 in R4C12
7c. Consider combinations for R4C34 = {15/24}
R4C34 = {15}, locked for R4 => R4C56 = {68}, locked for R4 => 19(3) cage = {379/469}
or R4C34 = {24}, R4C9 = 1 (hidden single in R4) => 3 in R4 only in R4C12 => 19(3) cage = {379}
-> 19(3) cage = {379/469}, no 2,8
[Maybe SudokuSolver sees this as a huge block 19(3) cage cannot be 2{89} because R4C123456 = {89}{24}[65] clashes with R4C78.]

8. 45 rule on N6 3 innies R4C78 + R6C8 = 14 must contain 2 for N6 = {248/257}
8a. Consider combinations for R4C34 = {15/24}
R4C34 = {15}, locked for R4 => R4C56 = {68}, locked for R4
or R4C34 = {24}, locked for R4 => R4C8 = {57} => R4C78 + R6C8 = {257}, no 8
-> no 8 in R4C7
8b. 8 in R4 only in R4C56, locked for N5
[Extending that forcing chain.]
8c. R4C34 = {15}, locked for R4 => R4C56 = {68}, locked for R4
or R4C34 = {24}, locked for R4 => R4C78 = {57}, locked for R4 => R4C56 = {68}
-> R4C56 = {68}, clean-up: no 9 in R3C6
8d. Naked pair {68} in R4C56, locked for R4 and N5
8e. Naked pair {68} in R34C6, locked for C6
8f. 26(4) cage at R4C5 (step 5c) = {2789/4589/4679} (cannot be {3689} because 6,8 only in R4C5), no 3
8g. 3 in N5 only in R6C45, locked for R6
8h. 16(3) cage at R6C4 (step 3a) must contain 3 = {358/367}, no 1,4 -> R6C45 = {35/37}, R7C4 = {68}
[At this stage I saw
1 in N5 only in R45C4, locked for C4, clean-up: no 7 in R2C3, no 5 in R3C5
Either R4C34 = [51] or R5C34 = [51] (locking cages) -> 5 in R45C3, locked for C3 and N4
but this quickly becomes unnecessary.]
8i. Killer pair 6,8 in R3C23 and R3C6, locked for R3, clean-up: no 2 in R4C8
8j. 19(3) cage at R3C1 = {379} (only remaining combination), no 4
8k. 19(3) cage = 7{39} (cannot be 3[79] which clashes with R34C8 = [17/35]) -> R3C1 = 7, R4C12 = {39}, 3 locked for R4 and N4 -> R45C9 = [13], clean-up: no 1 in R12C1, no 8 in R3C23, no 1 in R2C4, no 5 in R4C34
8l. R5C4 = 1 (hidden single in N5) -> R5C3 = 5, clean-up: no 3 in R2C4, no 5 in R3C5, no 8 in R9C2
8n. Naked pair {24} in R4C34, locked for R4 -> R4C78 = {57}, locked for N6
8o. R6C8 = 2 (hidden single in N6), clean-up: no 8 in R6C9 (step 2)
8p. Naked pair {46} in R67C9, locked for C9, clean-up: no 9 in R89C9
8q. Naked pair {78} in R89C9, locked for C9 and N9
8r. Naked triple {259} in 16(3) cage at R1C9, locked for N3
8s. Naked pair {69} in R3C23, locked for R3 and N1 -> R3C6 = 8, R4C6 = 6, placed for D/, R4C5 = 8, clean-up: no 2 in R12C1, no 2 in R2C4
8t. Naked pair {35} in R12C1, locked for C1 and N1 -> R4C12 = [93], clean-up: no 5 in R2C4

9. 36(7) cage at R6C8 = {2345679} (only remaining combination), no 1, 7 locked for N8
9a. 1 in N8 only in R8C6 + R9C56, locked for 10(4) cage at R8C6, no 1 in R9C7
9b. 1 in N9 only in 12(3) cage at R8C7 = {129/156}, no 3,4
9c. 2 of {129} must be in R8C7 -> no 9 in R8C7
9d. 36(7) cage = {2345679}, CPE no 6 in R7C4
9e. R7C4 = 8 -> R6C45 = 8 = {35}, 5 locked for N5
9f. R45C4 + R4C6 = R7C4 + 3 (step 5) -> R45C4 + R4C6 = 11, R4C6 = 6, R5C4 = 1 -> R4C4 = 4, placed for D\, R4C3 = 2, R2C3 = 1 -> R2C4 = 7, clean-up: no 2 in R3C5

10. R8C8 = 1 (hidden single on D\), R3C8 = 3 -> R4C8 = 5, R4C7 = 7
10a. 1 in N8 only in R9C56, locked for R9
10b. R3C7 = 1 (hidden single on D/), R3C5 = 4, R3C4 = 2, R3C9 = 5

11. 45 rule on N47 5(1+4) outies R3C1 + R4589C4 = 23, R3C1 = 7, R45C4 = [41] -> R89C4 = 11 = {56}, locked for C4, N8 and 23(4) cage at R7C3
11a. R6C4 = 3, placed for D/, R6C5 = 5
11b. 36(7) cage at R6C8 = {2345679} -> R7C7 = 5, placed for D\, 6 locked for R7 and N9 -> R8C7 = 2, R9C8 = 9, clean-up: no 4 in R9C23
11c. Naked pair {46} in R7C89, locked for R7 and N9 -> R9C7 = 3, R8C6 = 4
11d. R89C4 = 11 -> R78C3 = 12 = [93], 9 placed for D/ -> R1C9 = 2, placed for D/, R5C5 = 7, placed for both diagonals
11e. R9C1 = 4 (hidden single in R9), placed for D/
11f. R2C8 = 8, R34C7 = [17] -> R2C67 = 7 = [34]
11g. R8C2 = 5 (hidden single on D/), R9C1 = 4 -> R78C1 = 10 = [28]
11h. R56C1 = [61] -> R5C2 = 8 (cage sum)

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

Rating Comment:
I'll rate my walkthrough for A372 at Easy 1.5. I used the same forcing chain three times in slightly different ways.

Author:  wellbeback [ Sat Mar 23, 2019 8:43 pm ]
Post subject:  Re: Assassin 372

Thanks again Ed. Interesting puzzle! I found a couple of key steps which made the rest straightforward. Here is how I started - not using the diagonals at all. Andrew and I used the same basic 45s, IODs etc. - but in quite different ways.
Assassin 372 WT:
1. 4(2)n6 = {13}
-> Innies r1234 = r4c59 = +9(2) from [81] or [63]

2. 27(4)n6 = {69(48|57)}
Innies n6 = r4c78 + r6c8 = {2{48|57)}

3. Innies c9 = r67c9 = +10(2) = [82] or {46}

4! 36(7)r6c8 contains three of (1234)
If r89c8 were both from (1234) those values would be in 10(4) in n8 and neither in 36(7)
-> at most one of r89c8 is from (1234)
-> 12(3)n9 cannot be from [9{12}], [8{13}], [7{14}], [7{23}], [6{24}], [5{34}] (Important later)

5! r6c8
r6c8 from (24578)
IOD n89 -> r789c4 = r6c8 + 17
-> r6c8 is max 7
Trying r6c8 = 7 puts r4c78 = {25} contradicting 6(2)r4
Trying r6c8 = 4 puts 27(4)n6 = {5679} which leaves no solution for H10(2)r67c9
-> r6c8 from (25)
Trying r6c8 = 5 puts 5 in n9 in r8c7 ...
... puts 12(3)n9 = [5{16}] ...
... which leaves no solution for 8(2)c8
-> r6c8 = 2

Easy from here.

6. -> Remaining Innies n6 = r4c78 = +12(2) which can only be {57} (neither of (48) in r4c8)
-> 6(2)r4 = {24}
HS 1 in r4 -> 4(3)n6 = [13]
Innies r1234 -> r4c5 = 8
HS 3 in r4 -> 19(3)r3c1 = [7{39}]
-> NS 6 in r4 -> 14(2)r3c6 = [86]
-> 15(2)n1 = {69}
-> 8(2)n1 = {35}
-> r4c12 = [93]

7. 3 in n5 cannot be in 26(4) (since r4c6 = 6) - must be in r6c45
-> HS 1 in n5 -> 6(2)r5c3 = [51]
-> Remaining outies n4 = r4c4 + r7c2 = +5. Can only be r4c4 = 4 and r7c2 = 1.
-> r4c3 = 2
-> Remaining outie n5 -> r7c4 = 8
-> 16(3)n5 = [{35}8]
-> 12(3)n4 = [{47}1]

8. Remaining outies n3 = r12c6 = +8(2)
Only from {17} or {35}
-> 6(2)n2 = [24]
-> r3c789 = [{13}5]

9. Innies c9 r67c9 can only be [64]
-> 15(2)n9 = {78}
-> 16(3)c9 = [{29}5]
etc.

Author:  Ed [ Sun Mar 24, 2019 10:35 pm ]
Post subject:  Re: Assassin 372

Interesting! We all had to think carefully. As usual, wellbeback gets the prize for the most creative. A good read. My way is very close to Andrew, though as often happens, he used several easier steps while I make one complicated one (step 9).

SS does it very differently. Machines are good at some things so its score makes some sense! I have included its way after Andrew's step 6a. I started the continuation in Andrew's WT style but couldn't keep it up. Had to concentrate too hard! Thanks Andrew for finding a mistake.

SS key steps:
Prelims

a) R12C1 = {17/27/35}, no 4,8,9
b) R2C34 = {17/27/35}, no 4,8,9
c) R3C23 = {69/78}
d) R3C45 = {15/24}
e) R34C6 = {59/68}
f) R34C8 = {17/27/35}, no 4,8,9
g) R4C34 = {15/24}
h) R45C9 = {13}
i) R5C34 = {15/24}
j) R89C9 = {69/78}
k) R9C23 = {49/58/67}, no 1,2,3
l) 19(3) cage at R3C1 = {289/379/469/478/568}, no 1
m) 26(4) cage at R4C5 = {2789/3689/4589/4679/5678}, no 1
n) 27(4) cage at R5C7 = {3789/4689/5679}, no 1,2
o) 10(4) cage at R8C6 = {1234}

Steps resulting from Prelims
1a. Naked pair {13} in R45C9, locked for C9 and N6, clean-up: no 5,7 in R3C8
1b. 27(4) cage at R5C7 now = {4689/5679}, 6,9 locked for N6, clean-up: no 2 in R3C8
1c. 2 in N6 only in R4C78 + R6C8, CPE no 2 in R2C8
1d. Naked quad {1234} in 10(4) cage at R8C6, CPE no 1,2,3,4 in R9C4
1e. 1 in N5 only in R456C4 + R6C5, CPE no 1 in R7C4
1f. 36(7) cage at R6C8 must contain 9, not now in R6C8, CPE no 9 in R7C4

2. 45 rule on C9 2 innies R67C9 = 10 = [46/64/82], no 5,7,9, no 8 in R7C9
2a. 5 in C9 only in 16(3) cage at R1C9, locked for N3
2b. 16(3) cage = {259/457}, no 6,8

3. 45 rule on R1234 2 innies R4C59 = 9 = [63/81]
3a. Variable hidden killer pair 3,7 in 26(4) cage at R4C5 and 16(3) cage at R6C4 for N5, 26(4) cage cannot contain more than one of 3,7 -> 16(3) cage must contain at least one of 3,7 = {178/349/358/367/457} (cannot be {169/259/268} which don’t contain 3 or 7), no 2

4. 45 rule on N7 2 outies R89C4 = 1 innie R7C2 + 10
4a. Min R89C4 = 11, no 1 in R8C4
4b. Max R89C4 = 17 -> max R7C2 = 7

5. 45 rule on N5 3 innies R45C4 + R4C6 = 1 outie R7C4 + 3
5a. Max R7C4 = 8 -> max R45C4 + R4C6 = 11, no 9 in R4C6, clean-up: no 5 in R3C6
5b. Min R45C4 + R4C6 = 8 -> min R7C4 = 5
5c. 26(4) cage at R4C5 = {2789/3689/4589/4679} (cannot be {5678} which clashes with R4C6), 9 locked for N5
5d. 9 in R4 only in R4C12, locked for N4
5e. 19(3) cage at R3C1 must contain 9 in R4C12 = {289/379/469}, no 5, no 9 in R3C1

6. 45 rule on R12 3(2+1) outies R34C7 + R3C9 = 13
6a. R34C7 cannot total 4 -> no 9 in R3C9
Optimised SS steps from here
6b. R3c9 + R4c7 cannot total 5 -> no 8 in r3c7 (Thanks to Andrew for finding a mistake I made here! - not SS!)

6c. "45" on r12: innies r12c9 - 3 = 2 outies r34c7
6d. r12c9 <> 14 -> no 9 in r3c7 (since that leaves max. r12c9 = {57} = 12


7. 45 rule on N89 1 outie r6c8 + 17 = 3 innies R789C4
7a. Max R789C4 = 24 -> max r6c8 = 7

8. 45 rule on N6 3 innies R4C78 + R6C8 = 14 must contain 2 for N6 = {248/257}
8a. 8 in {248} must be in R4C7 -> no 4 in R4C7

9. "45" on r123: 2 innies r3c16 - 3 = 2 outies r4c78
9a. r4c78 cannot total 14 or 11 -> no 8 in r3c1

10. Hidden killer pair 8,9 in r3 in 15(2)r3c2 and r3c6
10a. -> r3c6 = (89), r4c6 = (56)

11. "45" on r123: 1 innie r3c1 + 11 = 3 outies r4c678
11a. r4c678 cannot total 17 -> no 6 in r3c1

12. from step 5e. 19(3)r3c1 = {289/379/469}
12a. 4 in {469} must be in r3c1 -> no 4 in r4c12

13. 4 in r4 only in 6(2)r4c3 = {24}: both locked for r4
13a. -> hidden single 2 in n6 -> r6c8 = 2

14. naked quad {5678} in r4c5678, all locked for r4

etc.
My way:
Preliminaries
Cage 4(2) n6 - cells ={13}
Cage 6(2) n2 - cells only uses 1245
Cage 6(2) n45 - cells only uses 1245
Cage 6(2) n45 - cells only uses 1245
Cage 14(2) n25 - cells only uses 5689
Cage 15(2) n9 - cells only uses 6789
Cage 15(2) n1 - cells only uses 6789
Cage 8(2) n1 - cells do not use 489
Cage 8(2) n12 - cells do not use 489
Cage 8(2) n36 - cells do not use 489
Cage 13(2) n7 - cells do not use 123
Cage 19(3) n14 - cells do not use 1
Cage 10(4) n89 - cells ={1234}
Cage 27(4) n6 - cells do not use 12
Cage 26(4) n5 - cells do not use 1

1. 4(2)n6 = {13}: both locked for c9 and n6

2. 16(3)n3: {268} blocked by 15(2)n9 = 6 or 8
2a. = {259/457}(no 6,8)
2b. must have 5: locked for c9 and n3

3. "45" on c9: 2 innies r67c9 = 10 = {46}/[82](no 7,9; no 8 in r7c9)

4. "45" on r1234: 2 innies r4c59 = 9 = [81/63]

5. 27(4)n6 = {4689/5679}
5a. must have both 6,9: both locked for n6

6. 36(7)r6c8 must have 9 which must be in r7 or r8c5 -> no 9 in r7c4

7. "45" on n5: 4 outies r3c6 + r45c3 + r7c4 = 23.
7a. max. r45c3 + r7c4 = {458} = 17 -> min. r3c6 = 6
7b. -> no 9 in r4c6

8. 9 in r4 only in 19(3)r3c1
8a. 9 locked for n4 and no 9 in r3c1
8b. 19(3) = {289/379/469}(no 5)

The crack
9. "45" on r123: 5 outies r4c12678 = 30 and must have 7 & 9 for r4
9a. = {24789/35679}
9b. but {24789} as {29}[8]{47} blocked by 27(4)n6 = 4 or 7
9c. but {24789} as {79/49}[8](47)[2] blocked by [66] in r3c68
9d. but {24789} as {49}[827] blocked by [66] in r3c16
9e. -> all permutations for {24789} blocked
9f. r4c12678 = {35679}: 3,5 & 6 locked for r4

10. r4c59 = [81], r5c9 = 3

11. {57} in r4c78: both locked for n6 and r4
11a. r34c6 = [86](6 placed for D/)

12. r4c78 = 12 -> "45" on n6, 1 innie r6c8 = 2

13. "45" on n36: 2 outies r12c6 = 8
13a. = {17/35}(no 2,4,9) = 1 or 5

14. 6(2)r3c4: {15} blocked by r12c6
14a. = {24} only: both locked for r3 and n2

15. r4c12 = {39} = 12, 3 locked for n4
15a. r3c1 = 7

16. {69} in r3c23: both locked for n1 and r3
16a. r3c9 = 5

17. r3c78 = {13}: both locked for n3

18. r67c9 = h10(2) = {46} only: both locked for c9
18a. 15(2)n9 = {78} only: both locked for n9: 7 for c9

19. r12c9 = {29}: both locked for n3
etc
Cheers
Ed

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