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Ed
 Post subject: Assassin 428
PostPosted: Wed Mar 01, 2023 6:26 am 
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Attachment:
a428.png
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Note: an x-puzzle so 1-9 cannot repeat on either diagonal.
Also, a broken 15(3) on D\ at r6c6

Assassin 428

Loved this one! Very interesting. JSudoku uses one unusual technique but otherwise, didn't have much trouble. That gave me the courage to try it with a score of 2.00.
triple click code:
3x3:d:k:3072:5121:5121:5121:3330:3330:8451:8451:8451:3072:3072:2820:2820:3330:8451:8451:8451:6917:3846:3846:3846:3846:6663:3592:3592:3592:6917:2569:2569:8970:8970:6663:6663:6663:6917:6917:8971:8971:8971:8970:8970:6663:6917:6917:4620:8971:8971:8971:8970:8970:3853:4622:4620:4620:8971:5903:5903:7696:7696:4622:4622:4622:2833:5903:5903:7696:7696:3090:3859:4622:3853:2833:5903:7696:7696:7696:3090:3859:3092:3092:3853:
solution: 
+-------+-------+-------+
| 7 5 9 | 6 4 1 | 2 3 8 |
| 2 3 6 | 5 8 7 | 9 4 1 |
| 1 8 4 | 2 9 3 | 6 5 7 |
+-------+-------+-------+
| 6 4 8 | 9 1 5 | 7 2 3 |
| 9 1 7 | 3 2 4 | 8 6 5 |
| 5 2 3 | 7 6 8 | 1 9 4 |
+-------+-------+-------+
| 8 6 1 | 4 3 2 | 5 7 9 |
| 4 9 5 | 8 7 6 | 3 1 2 |
| 3 7 2 | 1 5 9 | 4 8 6 |
+-------+-------+-------+
Cheers
Ed
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Ed
 Post subject: Re: Assassin 428
PostPosted: Fri Mar 10, 2023 8:33 am 
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Here's how I got started. [Thanks to Andrew for some corrections]

a428 start:
Preliminaries by SudokuSolver
Cage 15(2) n8 - cells only uses 6789
Cage 12(2) n8 - cells do not use 126
Cage 12(2) n9 - cells do not use 126
Cage 10(2) n4 - cells do not use 5
Cage 11(2) n12 - cells do not use 1
Cage 11(2) n9 - cells do not use 1
Cage 20(3) n12 - cells do not use 12
Cage 18(5) n689 - cells do not use 9


1. "45" on r12: one innie r2c9 = 1

2. "45" on r3: 2 innies r3c59 = 16 = {79} only: both locked for r3

3. the (79) in r3c9 sees all of n6 apart from r46c7 or r6c8 so must repeat in one of those
3a. and "45" on n6: 1 remaining outie r3c9 + 1 = 2 innies r46c7
3b. -> if r3c9 repeats in one of those innies the other innie must be 1
3c. or 7,9 are in r6c8
3d. either way, no 1 in r6c8
3e. -> 1 in n6 only in innies n6 -> the other one of r46c7 = r3c9 = (79)
3f. ie, r46c7 = {17}/[91]
3g. 1 locked for c7

4. "45" on n9: 3 outies r6c6 + r6c7 + r7c6 = 11
4a. if r6c7 = 7 -> r67c6 = 4 = {13}
4b. or r6c7 = 1
4c. ie, outies n9 must have 1
4d. the only two cages with 1 in n9 also have 1 in outies n9 -> both cages must have 1
4e. -> 15(3)r6c6 = {159/168}(no 2347)
4f. 1 locked for d\
4g. -> 3 outies n9 = 11 = [173/614/812] (note: can't have repeats since all 3 cells see each other through 18(4) cage)
4h. ie, r6c6 from (168), r7c6 from (234)
4i. 1 locked for r6

5. "45" on n78: 2 innies r7c16 = 10 = [82/73/64], r7c1 = (678)

6. "45" on n6: 1 outie r7c1 = 1 innie r4c3 = (678)
6a. note: with those two places, the same number on d/ can only be in n3
6b. if 7 in r4c3 -> 7 in n5 only in r5c6 -> r3c5 = 9 (same cage) -> r3c9 = 7: which contradicts 6a.
6c. -> no 7 in r4c3, no 7 in r7c1, no 3 in r7c6 (h10(2))
6d. -> 3 outies n9 = 11 = [614/812]
6e. ie, r6c7 = 1
6f. r8c8 = 1 (Hsingle n9)
6g. r6c6 + r9c9 = {68} only: both locked for d\ and no 6,8 in r6c9 nor r9c6
6h. -> naked pair {68} in r68c6: both locked for c6
6i. naked pair {79} in r3c5 + r4c7: both locked for 26(5)r3c5

7. "45" on n12: 3 innies r2c6 + r3c56 = 19 = {79}[3] only
7a. 7 & 9 locked for n2
7b. and r3c78 = 11 = {56} only: both locked for n3 and r3

8. 1 in n1 only in r3: locked for r3

9. "45" on n1: 3 outies r123c4 = 13 = {56}[2] only: 5 & 6 locked for n2 and c4

10. "45" on r789: 1 outie r6c6 = 1 innie r7c1
10a. -> naked pair 6,8 between r7c1 and r8c6: no 6,8 in r7c45 nor r8c123
10b. and naked pair 6,8 between r7c1 and r9c9: no 6,8 in r7c89 nor r9c123

11. r8c6 = 6 (hsingle n8), r9c6 = 9

12. 12(2)r8c5: {48} blocked by r2c5 = (48) -> = {57} only: both locked for c5 and n8

13. 1,3 in n8 only in 30(7): locked for that cage
13a. 30(7)r7c4 = {1234578}(no 9)
13b. 5,7 locked for n7

Much easier now.
Cheers
Ed
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wellbeback
 Post subject: Re: Assassin 428
PostPosted: Sat Mar 11, 2023 7:03 pm 
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I delayed a bit posting my WT since I was trying to find a 'nicer' way of doing the placement in my eventual Step 3. Ed and I used the same initial conclusions but quite different paths to do that placement. His was probably the better. Here's how I did it...
Assassin 428 WT:
1. Innies r12 -> r2c9 = 1
Innies r3 = r3c59 = +16(2) = {79}
IOD n6 -> r2c9 + r3c9 = r4c7 + r6c7
Since the only three places in n6 for the values from r23c9 are in r4c7, r6c7, and r6c8 - at least one of the values from r23c9 must go in n6 in one of r46c7.
-> Both values from r23c9 must go in r46c7.
Since 18(5) cannot contain a 9 -> r46c7 = [91], [71], or [17]
Also both (19) or (17) in r789c8
-> 1 in r78c8

2. Outies n9 = r6c6, r6c7, r7c6 = +11(3). Since they all 'see' each other - they must all be different.

3. Can 1 be in r7c8?
That would put r6c7 (and r3c9) = 7
Which puts (outies n9) r67c6 = [13] and (18(5)) r78c7 = {25}
But this leaves no place for 7 in n9 since 15(3)D\ cannot be [177], nor 12(2) = [57]
-> r8c8 = 1
(An alternate contradiction is that 3 in r7c6 puts (Innies n78) r7c1 = 7 and (IOD n4) r4c3 = 7 which leaves no place for 7 in D/).

4. -> (15(3)D\) r6c6 is min 5
-> (Outies n9) r6c7 is max 4
-> (Since r6c7 only from (17)) r6c7 = 1
-> 15(3)D\ = <519> or <618>
Since (outies n9) r67c6 = +10(2) cannot be [55] or [91] -> 15(3)D\ = <618>

5. -> 15(2)n8 cannot be [96]
Combined cage 12(2)n8 and 15(2)n8 = 27(4) - must contain a 9
-> Either 12(2)n8 = {39} or 15(2) = [69]
-> 12(2)n9 cannot be {39}
-> (HS 9 in n9) 11(2)n9 = {29}
-> r3c59 = [97]
-> 12(2)n8 cannot be {39}
-> 15(2)n8 = [69]
-> 15(3)D\ = [816]
-> (Outies n9) r7c6 = 2
-> 11(2)n9 = [92]

6. -> (Innies n78) r7c1 = 8
-> (IOD n4) r4c3 = 8

7. 30(7)n78 must contain 2 in n7 and (13) in n8
-> That part of 30(7) in n7 = +14(3) = {257}
-> 12(2)n8 = {57}
-> 12(2)n9 = {48}
-> r7c8 = 7 and r78c7 = {35}

8. Remaining Innies n12 = r23c6 = +10(2) can only be [73]
-> 14(3)r3 = [365]

9. Remaining cells c6 = r145c6 = {145}
-> Remaining cells 26(5)r3c5 = +10(3) = {145} (I.e., r1c6 = r4c5)
-> (Since 5 already in c5) 5 in r45c6 and r1c6 = r4c5 from (14)

10. Since 1 already on D\ and r2 and not in 20(3) -> 1 in n1 in r3c12
-> 1 in n2/r1 in r1c56
-> 13(3)n8 = {148}
-> r124c5 = {148} (with 8 in r12c5)
-> r567c5 = [263]
etc.
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Andrew
 Post subject: Re: Assassin 428
PostPosted: Sun Mar 12, 2023 6:06 am 
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I continue to be busy but managed to find time for this challenging puzzle; after doing the easy steps I came back to it yesterday and finished today. Haven't yet had time to see how Ed and wellbeback solved it, so I've no idea whether mine is similar or different.

My observation steps have been further clarified after reading Ed's comments below. My observation 3b, as used in step 5c, now appears to be correct. See further post with diagram.

With hindsight it's fortunate that I didn't spot step 6a earlier; if I had we might have missed all the interesting discussions in this thread.
Here's how I solved Assassin 428:
Prelims

a) R2C34 = {29/38/47/56}, no 1
b) R4C12 = {19/28/37/46}, no 5
c) R78C1 = {29/38/47/56}, no 1
d) R89C5 = {39/48/57}, no 1,2,6
e) R89C6 = {69/78}
f) R9C78 = {39/48/57}, no 1,2,6
g) 20(3) cage at R1C2 = {389/479/569/578}, no 1,2
h) 18(5) cage at R6C7 = {12348/12357/12456}, no 9

1a. 45 rule on R12 1 innie R2C9 = 1
1b. 45 rule on R3 2 innies R3C59 = 16 = {79}, locked for R3
1c. 45 rule on N12 3 innies R2C6 + R3C56 = 19 = {289/379/469/478} (cannot be {568} because R3C5 only contains 7,9), no 1,5
1d. 3 of {379} must be in R3C6 -> no 3 in R2C6
1e. 45 rule on N8 5 innies R7C456 + R89C4 = 18 = {12348/12357/12456}, no 9
1f. 45 rule on N9 3 outies R6C67 + R7C6 which ‘see’ each other = 11 = {128/137/146/236/245}, no 9
1g. 45 rule on N78 2 innies R7C16 = 10 = {28/37/46}/[91], no 5, no 1 in R7C1
1h. 45 rule on N4 1 innie R4C3 = 1 outie R7C1 -> no 1,5 in R4C3
1i. 1 in R9 only in R9C1234, CPE no 1 in R8C3
1j. 45 rule on C1234 3 outies R567C5 = 11 = {128/137/146/236/245}, no 9
1k. 35(6) cage at R4C3 = {146789/236789/245789/345689}, CPE no 8,9 in R4C56
1l. 45 rule on C6789 2 outies R34C5 = 1 innie R1C6 + 9
1m. Max R34C5 = 16 -> max R1C6 = 7

2a. 27(6) cage at R2C9 and 18(3) cage at R5C9 total 45
2b. R56C9 ‘see’ all the cells of the 27(6) cage -> 27(6) cage + 18(3) cage must contain at least 8 different numbers so form a 45(9) cage = {123456789}, no 1 in R6C8
2c. 1 in N6 only in R46C7, locked for C7
2d. 45 rule on N6 2 outies R23C9 = 2 innies R46C7
2e. R23C9 = 1{79} -> R46C7 = 1{79}
2f. 18(3) cage = {369/378/459/468/567} (cannot be {279} which clashes with R46C7), no 2
2g. R6C67 + R7C6 (step 1f) = {128/137/146} (cannot be {236/245} because R6C7 only contains 1,7), no 5
2g. Consider placement for 1 in R6C67 + R7C6
2h. R6C6 = 1
or 1 in R6C7 + R7C6 => R8C8 = 1 (hidden single in N9)
-> 1 in R6C6 + R8C8, locked for D\
2i. Disjoint 15(3) cage at R6C6 contains 1 = {159/168}
2j. R6C67 + R7C6 = {128/137/146}
2k. 2,3,4 only in R7C6 -> R7C6 = {234}, R7C1 = {678} (step 1g), R4C3 = {678} (step 1h)
2l. 35(6) cage at R4C3 = {146789/236789/245789/345689}, 9 locked for C4 and N5, clean-up: no 2 in R2C3
2m. 1 in R1 only in R1C56, locked for N2
2n. 13(3) cage at R1C5 = {139/148/157}, no 2,6
2o. 12(3) cage at R1C1 = {237/246/345}, no 8,9

Observations.
3a. 30(7) cage at R7C4 and R89C5 contain 9 cells but only total 42 -> R8C5 must be the same as one of R9C12 and/or R9C5 must be the same as R8C3 -> R7C789 must contain one or both of the numbers in R89C5
The original observation 3a seems still to be valid. However for clarity I’ll now look at things the other way round
New 3a. R8C3 ‘sees’ all the cells in N8 except for R7C6 + R9C56, R9C23 ‘see’ all the cells in N8 except for R7C6 + R8C56
New 3b. BECAUSE 30(7) cage at R7C4 must contain all of 1,2,3,4,5 and R7C6 only contains 2,3,4, the value in R7C6 must be repeated in R8C3 + R9C23.
The pair in R89C6 = 15 cannot be repeated in the 30(7) cage because that would leave the other possible pair totalling 15 which cannot both fit into R89C5 = 12.
Therefore, since R89C6 cannot repeat in the 30(7) cage -> R89C5 must repeat in R8C3 + R9C23 with R9C5 in R8C3 and R7C6 + R8C5 in R9C23.

4a. 45 rule on N3 2 outies R23C6 = 1 remaining innie R3C9 + 3
[Only used later]
4b. Consider placements for 9 in C6
R2C6 = 9 => R3C5 = 7, R3C6 = 3 (step 1c)
or R89C6 = {69}, locked for C6
-> no 6 in R23C6
4c. 6 in N2 only in R123C4, locked for C4
4d. 45 rule on N1 3 outies R123C4 = 13 contains 6 = {256/346}, no 7,8, clean-up: no 3,4 in R2C3
4e. 1 in N8 only in R7C45 + R89C4, locked for 30(7) cage at R7C4
4f. 1 in R9 only in R9C14, CPE no 1 in R6C4 using D/
4g. R6C67 + R7C6 (step 1f) = {128/137/146}, 1 locked for R6
[I ought to have spotted this sooner]
4h. Disjoint 15(3) cage at R6C6 (step 2i) = {159/168}
4i. Consider combinations for 18(5) cage at R6C7 = {12348/12357/12456}
18(5) cage = {12348}, 8 locked for N9 => R9C78 = {39/57} => 15(3) cage = {168} (cannot be {159} = 1{59} which clashes with R9C78
or 18(5) cage = {12357/12456}, 5 locked for N9
-> 15(3) cage = {168}, 6,8 locked for D\
4j. 9 in N9 only in R78C9 = {29} or R9C78 = {39} -> no 3 in R78C9 (locking-out cages), clean-up: no 8 in R78C9

5a. R6C67 + R7C6 (step 1f) = {128/137/146} = [173/614/812], 18(5) cage at R6C7 = {12348/12357/12456}
5b. Consider combinations containing 9 in N9
R78C9 = {29}, 2 locked for N9 => 2 in 18(5) cage only in R7C6 => R6C67 + R7C6 = [812] => R78C6 = {69}
or R9C78 = {39}, locked for R9, no 3,9 in R8C5 => 9 in N8 only in R89C6 = {69}
-> R89C6 = {69}, locked for C6 and N8, clean-up: no 4 in R7C6, no 6 in R7C1 (step 1g), no 6 in R4C3 (step 1h), no 3 in R89C5
5c. R89C6 = {69} -> no 6,9 in 30(7) cage at R7C4 (step 3b, as clarified above)
5d. 1,6,9 in N7 only in 23(5) cage at R7C2 = {12569/13469}, no 7,8
5e. Disjoint 15(3) cage at R6C6 (step 4i) = {168}, 6 locked for N9, clean-up: no 5 in R78C9
[There's a killer pair in C9 here but it's not needed]

6a. R4C4 = 9 (hidden single on D\), clean-up: no 1 in R4C12
6b. Naked pair {17} in R46C7, 7 locked for C7 and N6, clean-up: no 5 in R9C8
6c. 27(6) cage at R2C9 + 18(3) cage at R5C9 form a 45(9) cage (step 2b) = {123456789} -> R3C9 = 7, R3C5 = 9, clean-up: no 4 in R78C9
6d. Naked pair {29} in R78C9, locked for C9 and N9, clean-up: no 3 in R9C78
6e. 18(5) cage at R6C7 = {12348/12357/12456} -> R7C6 = 2 -> R6C67 = [81] (step 5a), R4C7 = 7, R78C9 = [92], 15(3) disjoint cage = [816], R89C6 = [69], R4C3 = 8, R7C1 = 8 (step 1h), clean-up: no 3 in R2C4, no 2,3 in R4C12
6f. Naked pair {46} in R4C12, locked for R4 and N4
6g. R3C9 = 7 -> R23C4 = 10 (step 4a) = [73] -> R3C78 = 11 = {56}, locked for R3 and N3
6h. R123C4 (step 4d) = {256} (only remaining combination) = {56}2, 5 locked for C4 and N2, clean-up: no 9 in R2C3
6i. R3C123 = [184], 4 placed for D\
6j. 26(5) cage at R3C5 contains 1 for R4, 5 for C6 and 7,9 = [91574], 5 placed for D/ -> R3C78 = [65], 6 placed for D/
6k. Naked pair {37} in R56C4, locked for C4 and N5 -> R56C6 = [26], 2 placed for both diagonals
6l. R9C4 = 1 (hidden single in R9) -> R78C4 = [48]
6m. 8 in R9 only in R9C78 = {48}, 4 locked for R9 and N9

7a. R9C1 = 3, placed for D/
7b. Naked pair {57} in R8C35, 5 locked for R8 -> R8C7 = 3, R7C7 = 5, placed for D\
7c. R1C1 = 7, R2C2 = 3 -> R2C1 = 2 (cage sum)
7d. Naked pair {59} in R56C1, locked for N4, 9 locked for C1 -> R8C12 = [49], 9 placed for D/
7e. Naked pair {48} in R1C9 + R2C8, locked for N3
7f. 5 in N6 only in 18(3) cage at R5C9 = {459} (only remaining combination) = [594]

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

Alternative middle section if you prefer to omit my use of observation 3b in step 5c:
Omit steps 5c and 5d, continue with steps 5e and 6a to 6c, then since there's no 6 in the 30(7) cage it must be {1234578}
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Ed
 Post subject: Re: Assassin 428
PostPosted: Sat Mar 18, 2023 9:01 pm 
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Still haven't finished the WTs yet, but, really appreciate what Andrew is doing in a few of his steps. Others may be interested in these thoughts so putting them here. When the Revisits started, I put a lot of time into trying the same kinds of steps but found I kept getting tricked so am very cautious now.

Thanks Ed for your comments, which I've added to below.

Love 2b! Really direct way of doing that. This was the one case which worked directly because of the geometry. I took a few steps. I'd wondered when I found that step, and later observation steps 3a and 3b, whether the cage pattern had been deliberately designed that way, but clearly from Ed's comment it hadn't.

Quote:
3a. 30(7) cage at R7C4 and R89C5 contain 9 cells but only total 42 -> R8C5 must be the same as one of R9C12 and/or R9C5 must be the same as R8C3
I think this should read, "30(7) cage at R7C4 and R89C5 contain 9 cells but only total 42 -> R9C5 must be the same as R8C3 and/or r9c23 repeats in r8c5". Only one cell needs to repeat, which is what my "and/or" meant which replaces the number +3 to get the 9 cells down to total 42. Both cells in r89c5 can repeat as in this puzzle, but not necessary.

For example, this pattern below has those two cages total 43 but only the 1 repeats in r8c3 and r9c5 (the 4 in r8c5 doesn't repeat), and no 3 anywhere in those nine cells to get the total down to 43.
Attachment:
repeats1.png
repeats1.png [ 32.72 KiB | Viewed 2784 times ]


Quote:
3b. 30(7) cage at R7C4 and R89C6 contain 9 cells and total 45, with R89C6 each ‘seeing’ eight of the cells of the 30(7) cage, so must effectively be a 45(9) cage
I think this should read, "30(7) cage at R7C4 and R89C6 contain 9 cells and total 45, and r8c3 repeats in r9c5 or r9c23 repeats in r8c5 -> one of r89c6 'sees' at least eight of the 30(7) -> can be no repeats". In other words it needs both the other cages to be sure that one of r89c6 don't repeat in the 30(7), so it really is a 45(9). I hope I have that right. Agreed, I've clarified my step 3b to add that it depended on step 3a.

r9c6 could repeat in r8c3 for example, with a different cage total at r89c5. For example, this pattern below has those two cages total 45 but both r89c6 repeating with no repeats at r89c5. This example is different because R89C5 and R89C6 both have the same cage total.
Attachment:
repeats2.png
repeats2.png [ 32.84 KiB | Viewed 2784 times ]


Mind bending. Thanks again Andrew!
Cheers
Ed
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Ed
 Post subject: Re: Assassin 428
PostPosted: Sun Mar 26, 2023 5:34 am 
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I think we've both been tricked about 3b Andrew. See below. The first two cages sum to more than 45 but exactly the same logic applies, ie there must be at least one repeat. But r89c6 both repeat even though the cages sum to 45.
Attachment:
repeats3.png
repeats3.png [ 32.46 KiB | Viewed 2739 times ]


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Andrew
 Post subject: Re: Assassin 428
PostPosted: Thu Apr 13, 2023 8:32 pm 
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I've now further modified my observation steps after studying Ed's examples. While my assumptions appear to be correct, I realise that I hadn't looked at them carefully enough, particularly the role of R7C6.

Here's the position after my step 5b, showing N78. What is explained would work as early as the end of step 2, when the observations were made, but don't contribute usefully until after R89C6 has been reduced to one combination.

Attachment:
Assassin 428 after step 5b.jpg
Assassin 428 after step 5b.jpg [ 30.75 KiB | Viewed 2609 times ]

New 3a. R8C3 ‘sees’ all the cells in N8 except for R7C6 + R9C56, R9C23 ‘see’ all the cells in N8 except for R7C6 + R8C56
New 3b. BECAUSE 30(7) cage at R7C4 must contain all of 1,2,3,4,5 and R7C6 only contains 2,3,4, the value in R7C6 must be repeated in R8C3 + R9C23.
The pair in R89C6 = 15 cannot be repeated in the 30(7) cage because that would leave the other possible pair totalling 15 which cannot both fit into R89C5 = 12.
Therefore, since R89C6 cannot repeat in the 30(7) cage -> R89C5 must repeat in R8C3 + R9C23 with R9C5 in R8C3 and R7C6 + R8C5 in R9C23.
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