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Ed
 Post subject: Assassin 452
PostPosted: Thu Feb 15, 2024 6:04 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Attachment:
a452.png
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Note: numbers in the zero cells can repeat but normal sudoku rules still apply

Assassin 452
No gimmes at the start! This came out pretty quickly for me. Couldn't find a way to toughen it up but it has some interesting steps anyway so no harm done. It gets 1.70 and JSudoku has to work.

triple click code:
3x3::k:2816:2816:7937:7937:7937:7937:7937:2818:2818:8707:8707:5124:7937:1285:1285:5382:5382:2818:8707:5124:5124:2055:2055:5640:5382:5382:5129:8707:5124:2826:2826:3595:5640:5640:1292:5129:8707:2829:2826:3854:3595:3595:0000:1292:5129:8707:2829:0000:3854:3854:3345:0000:3858:3858:8707:4371:0000:0000:3345:3345:5908:5908:3858:2837:4371:4371:2070:2070:5908:5908:9743:3344:2837:2837:9743:9743:9743:9743:9743:9743:3344:
soluton: 
+-------+-------+-------+
| 4 7 1 | 8 9 2 | 5 6 3 |
| 5 3 8 | 6 4 1 | 7 9 2 |
| 9 6 2 | 3 5 7 | 1 4 8 |
+-------+-------+-------+
| 8 4 7 | 1 2 6 | 9 3 5 |
| 1 9 3 | 5 8 4 | 6 2 7 |
| 6 2 5 | 7 3 9 | 4 8 1 |
+-------+-------+-------+
| 2 5 9 | 4 1 3 | 8 7 6 |
| 7 8 4 | 2 6 5 | 3 1 9 |
| 3 1 6 | 9 7 8 | 2 5 4 |
+-------+-------+-------+
Cheers
Ed
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wellbeback
 Post subject: Re: Assassin 452
PostPosted: Fri Feb 16, 2024 4:04 am 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
Thanks Ed!
Came out pretty quickly for me as well. Here's how I did it...
Assassin 452 WT:
1. Outies r1 = r2c49 = +8(2)
5(2)n2 cannot be {23} since that would put both H+8(2)r2c49 and 8(2)n2 = {17}
-> 5(2)n2 = {14}
-> One of h+8(2)r2c49 and 8(2)n2 is {35} and the other is {26}

2. Outies c9 = r16c8 = +14(2) = {59} or {68}
-> In 11(3)n3 r2c9 is max 4.
-> H+8(2)r2c49 = [62] or [53]
But it cannot be [53] since that puts r1c89 = [62] which leaves no place for 5 in r1
-> r2c49 = [62] and 8(2)n2 = {35}

3. Innies n3 = [r1c7,r3c9] = +13(2) - No (123)
-> (for (36) in r1) either r1c89 = [63] or r1c123 = [{56}3]
Outies c1 = r129c2 = +11(2)
-> r1c123 cannot be [{56}3] since that puts min r2c2 = 7 contradicting previous line.
-> r1c89 = [63]

4. -> (Outies c9) r6c8 = 8
-> r67c9 = {16}

Also (HS 1 in r1) r1c3 = 1
-> (HS 5 in r1) r1c7 = 5
-> (Innies n3) r3c9 = 8
-> 21(4)n3 = [{79}{14}]
-> 5(2)n6 = {23}

Also 2 in n2 only in r1
-> 11(2)r1 = {47}
-> r3c6 = 7
-> r4c67 = {69}

Also 13(2)c9 = {49}
-> 20(3)c9 = [8{57}]
-> 4 in n6 in r56c7
-> r3c78 = [14]
-> (HS 1 in n6) r67c9 = [16]
-> r456c7 = {469} and r789c7 = {238}
-> r789c8 = {157}

Also r2c123 = {358} and r3c123 = {269}

5. 9 must be in 38(7)r8c8 -> 13(2)n9 = [94]
-> 38(7) = {1256789}
-> 3 in r9c12

6. 34(7) r2c1 is missing an +11(2) and one of the values missing is in r1c1.
Since r1c12 = +11(2) - both those values are the missing values from 34(7)
-> (47) in c1 in r189c1
Also value from r2c2 only in r89c1
Outies c1 = r129c2 = +11(3)
3 not in r9c2 since r12c2 cannot be [71] or [44]
-> r9c1 = 3
-> r2c2 = 3
-> r129c2 = [731]
-> r1c1 = 4
Also -> 11(3)n7 = [731]
-> r8c8 = 1

7. Only possibilities for 20(4)r2c3 are [5{29}4] and [8{26}4]
I.e., r4c2 = 4
-> 8 in c2 only in r78c2
-> 17(3)n7 = [{58}4]
-> 11(2)n4 = {29}-> 20(4)n1 = [8624]
-> 34(7)r2c1 = [539{168}2]
-> r456c3 = {357}
-> 11(3)r4c3 = <317> and r6c3 = 5
-> (Remaining Innies n5) r46c6 = {69}
-> 7 in r6 in r6c45
-> 15(3)n5 = [5{37}]

8. 3 in n9 in r78c7
-> For 23(4)n9 r78c7 = {38), r7c8 = 7 and r8c6 = 5
-> 8(2)n8 = [26]
etc.
It's been a crazy start of the year for me. Assassin 450 was a very interesting struggle! Will try and write something for that if I can remember how I did it and find the time.
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Andrew
 Post subject: Re: Assassin 452
PostPosted: Thu Feb 22, 2024 5:42 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin. It looks like it took me rather longer to solve it. One of the middle steps in my solving path was fairly heavy combination analysis; I don't know whether that was necessary and I'm too busy at the moment to go through again and find out what would happen if I skipped that step. Ed's and wellbeback's comments suggest it wasn't necessary. I'll be interested to see how they did it after Ed has posted his WT and I've got time to go through them. After that heavy step I started finding interesting steps again.

Here's how I solved Assassin 452:
Prelims

a) R1C12 = {29/38/47/56}, no 1
b) R2C56 = {14/23}
c) R3C45 = {17/26/35}, no 4,8,9
d) R45C8 = {14/23}
e) R56C2 = {29/38/47/56}, no 1
f) R8C45 = {17/26/35}, no 4,8,9
g) R89C9 = {49/58/67}, no 1,2,3
h) 11(3) cage at R1C8 = {128/137/146/236/245}, no 9
i) 22(3) cage at R3C6 = {589/679}
j) 20(3) cage at R3C9 = {389/479/569/578}, no 1,2
k) 11(3) cage at R4C3 = {128/137/146/236/245}, no 9
l) 11(3) cage at R8C1 = {128/137/146/236/245}, no 9

1a. 45 rule on R1 2 outies R2C49 = 8 = {17/26/35}, no 4,8,9
1b. Combined cage R2C49 + R2C56 = {17}{23}/{26}{14}/{35}{14}, 1 locked for R2
1c. 34(7) cage at R2C1 = {1234789/1235689/1245679/1345678}, 1 locked for C1
1d. 45 rule on R89 3 outies R7C278 = 20 = {389/479/569/578}, no 1,2
1e. 45 rule on C1 3 outies R129C2 = 11 = {128/137/146/236/245}, no 9, clean-up: no 2 in R1C1
1f. 45 rule on C9 4 innies R1267C9 = 12 = {1236/1245}, no 7,8,9, clean-up: no 1 in R2C4
1g. 45 rule on C9 2 outies R16C8 = 14 = [59/68/86]
1h. 11(3) cage at R1C8 can only contain one of 5,6,8 -> no 5,6 in R12C9, clean-up: no 2,3 in R2C4
1i. R67C9 must contain one of 5,6 -> 15(3) cage at R6C8 = 6{45}/8{16}/8{25}/9{15}, no 3 in R67C9
1j. 20(3) cage at R3C9 = {389/479/578} (cannot be {569} which clashes with R89C9), no 6
1k. 38(7) cage at R8C8 = {1256789/1346789/2345789}, CPE no 7,8,9 in R9C9, clean-up: no 4,5,6 in R8C9
1l. 45 rule on N2 5 innies R1C456 + R2C4 + R3C6 = 32 = {26789/35789/45689}, no 1
1m. 45 rule on N3 2 innies R1C7 + R3C9 = 13 = {49/58}/[67], no 1,2,3, no 7 in R1C7
1n. 45 rule on N23 2 innies R3C69 = 1 outie R1C3 + 14
1o. Max R3C69 = 17 -> max R1C3 = 3
1p. Min R3C69 = 15, no 4,5 in R3C69, clean-up: no 8,9 in R1C7

2a. 11(3) cage at R1C8 = {128/146/236/245}, R1267C9 (step 1f) = {1236/1245}, R1C7 + R3C9 (step 1m) = [49/58/67]
2b. Consider permutations for R89C9 = [76/85/94]
R89C9 = [76], no 7 in R3C9 => no 6 in R1C7 = {45} => 11(3) cage at R1C8 = {128/146/236} (cannot be {245} which clashes with R1C7)
or R89C9 = [85/94] => R1267C9 = {1236} => 11(3) cage at R1C8 = {128/236}
-> 11(3) cage at R1C8 = {128/146/236}, no 5, clean-up: no 9 in R6C8 (step 1g)
2c. Naked pair {68} in R16C8, locked for C8
2d. Consider placements for R1C8 = {68}
R1C8 = 6
or R1C8 = 8 => R67C9 = {45}, locked for C9 => R89C9 = [76]
-> R1C7 + R3C9 = [49/58], no 6,7

3a. 9 in R9 only in R9C345678, locked for 38(7) cage at R8C8, no 9 in R8C8
3b. 38(7) cage = {1256789/1346789/2345789}, 8 locked for R9
3c. 38(7) cage = {1256789/1346789/2345789} sees R9C9 = {456} -> 38(7) cage + R9C9 must contain all of 4,5,6, 6 locked for R9
3d. 45 rule on R9 3 outies R8C189 = 17
3e. Consider placement for 7 in R9
7 in R9C12 => 11(3) cage at R8C1 = [371] -> R8C89 = 14 = [59]
or 7 in R9C345678, locked for 38(7) cage
-> no 7 in R8C8
3f. 38(7) cage = {1256789/1346789/2345789}, 7 locked for R9
3g. 11(3) cage = {128/137/146/236/245}
3h. 6,7 of {137/236} must be in R8C1 -> no 3 in R8C1
3i. R8C189 = 17, max R8C89 = 14 -> no 2 in R8C1
3j. R89C9 = 13 -> R8C89 cannot total 13 (CCC) -> no 4 in R8C1
3k. 5 of {245} must be in R8C1 -> no 5 in R9C12
3l. 5 in R9 only in R9C3456789, CPE no 5 in R8C8

4a. 31(6) cage at R1C3 = {125689/134689/135679/145678/234589/234679/235678} (cannot be {124789} = 1{289}[47] which clashes with R1C12 + R1C8)
4b. R1C7 + R3C9 (step 2d) = [49/58], R3C69 = R1C3 + 14 (step 1n)
4c. 31(6) cage = {125689/135679/145678/234589/234679/235678} (cannot be {134689} = 1{389}[46] because then R3C69 = [79] = 16)
4d. 31(6) cage = {125689/145678/234589/234679/235678} (cannot be {135679} = 1{369}[57] because R3C69 cannot both be 8)
4e. 31(6) cage = {125689/234589/234679/235678} (cannot be {145678} = 1{4578}6 because R3C69 = [98] = 17), 2 locked for R2, clean-up: no 9 in R1C1
4f. 9 in R1 only in 31(6) cage = {125689/234589/234679}, locked for N2
4g. 22(3) cage at R3C6 = {589/679}, 9 locked for R4
4h. 8 of {589} must be in R3C6 -> no 8 in R4C67

[I was slow in spotting]
5a. 4 in N2 only in R1C456 or R2C56 = {14}
5b. Consider permutations for R1C7 + R3C9 (step 2d) = [49/58]
R1C7 + R3C9 = [49]
or R1C7 + R3C9 = [58] => R1C8 = 6, R12C9 = 5 = [32], R2C56 = {14}
or R1C8 = 6, R12C9 = 5 = [41]
-> no 4 in R1C456
5c. 4 in N2 only in R2C56 = {14}, locked for R2, 1 locked for N2, clean-up: no 7 in R2C4 (step 1a), no 7 in R3C45
5d. Killer pair 5,6 in R2C4 and R3C45, locked for N2
5e. 11(3) cage at R1C8 = [632/812] -> R1C9 = {13}, R2C9 = 2
5f. R1C12 = {47/56} (cannot be {38} which clashes with R1C89), no 3,8
5g. 31(6) cage at R1C3 (step 4f) = {125689/234679} (cannot be {234589} which clashes with R1C12) -> R2C4 = 6, clean-up: no 2 in R3C45, no 2 in R8C5
5h. 22(3) cage at R3C6 = {589/679}
5i. 7 of {679} must be in R3C6 -> no 7 in R4C67

6a. Naked pair {35} in R3C56, locked for R3, 3 locked for N2
6b. R1C39 = {13} (hidden pair in R1)
6c. 34(7) cage at R2C1 is missing two numbers totalling 11, R1C12 = 11 and R1C1 ‘sees’ all of the 34(7) cage -> the value in R1C2 must be in R89C1
6d. R1C2 = {4567} -> no 8 in R8C1
6e. 8 in C1 only in R234567C1, locked for 34(7) cage, no 8 in R2C2
6f. R129C2 (step 1e) = {137/236/245} (cannot be {146} because R2C2 only contains 3,5,7) = [731/632/452] -> R9C2 = {12}, no 5 in R1C2, no 7 in R2C2, clean-up: no 6 in R1C1
6g. No 5 in R1C2 -> no 5 in R8C1
6h. 11(3) cage at R8C1 = {137/146/236}, R9C2 = {12} -> no 2 in R9C1

7a. 7 in N3 only in 21(4) cage at R2C7 = {1479/1578} (cannot be {3567} which clashes with R2C2), 1 locked for R3 and N3
[Cracked, the rest is fairly straightforward.]
7b. R12C9 = [32] -> R1C8 = 6 (cage sum), R6C8 = 8 -> R67C9 = 7 = {16}, 6 locked for C9, clean-up: no 3 in R5C2, no 7 in R8C9
7c. R1C3 = 1 -> 31(6) cage at R1C3 (step 5g) = {125689} -> R1C7 = 5, R3C9 = 8, R8C9 = 9 -> R9C9 = 4, R9C1 = 3, clean-up: no 6 in R1C2
7d. Naked pair {79} in R2C78, locked for R2 and N3
7e. Naked pair {14} in R3C78, 4 locked for R3
7f. R3C6 = 7 -> R4C67 = 15 = {69}, 6 locked for R4
7g. R1C2 = R8C1 (step 6c) -> R1C2 = R8C1 = 7, R1C1 = 4, R9C2 = 1 (cage sum), clean-up: no 4 in R56C2
7h. R8C189 = 17 (step 3d), R8C19 = [79] -> R8C8 = 1, R3C78 = [14], R67C9 = [16]
[Alternatively 38(7) cage at R8C8 contains 6 so must contain 1.]
7i. Naked pair {23} in R45C8, locked for C8 and N6
7j. Naked pair {57} in R79C8, 7 locked for C8 and N9 -> R2C78 = [79]
7k. R7C278 (step 1d) = {578} (only possible combination because R7C7 only contained 3,8 and R7C8 only contained 5,7) = [587], R9C78 = [25], R8C7 = 3 -> R8C6 = 5 (cage sum), R8C45 = [26], R8C23 = {48}, locked for N7, clean-up: no 6 in R56C2
7l. Naked pair {29} in R7C13, 9 locked for R7 and N7 -> R9C3 = 6
7m. Naked pair {29} in R37C3, locked for C3
7n. 11(3) cage at R4C3 = {137} (only remaining combination) -> R4C4 = 1, R45C3 = {37}, locked for N4, 3 locked for C3, clean-up: no 8 in R5C2
7o. 1 in N8 only in R7C56 -> 13(3) cage at R6C6 = {139} (only remaining combination) -> R6C6 = 9, R7C56 = {13}, 3 locked for R7, R7C4 = 4, R4C67 = [69], R19C6 = [28], clean-up: no 2 in R5C2
7p. 7 in R6 only in R6C45, locked for N5
7q. 15(3) cage at R5C4 = {357} (only possible combination), 3,5 locked for N5 -> R5C6 = 4
7r. 45 rule on the whole grid R56C6 + R67C3 + R7C4 = 28, R56C6 = [64], R6C3 = 5, R7C4 = 4 -> R7C3 = 9
7s. R23C3 = [82], R4C2 = 4 -> R3C2 = 6 (cage sum)

and the rest is naked singles.
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Ed
 Post subject: Re: Assassin 452
PostPosted: Sat Feb 24, 2024 6:55 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Loved wellbeback's step 1! So direct. I circled around it.
Andrew wrote:
find out what would happen if I skipped that step (4)
Can't skip it sorry. Things fall apart on 5b and 5e.

We all ended up following the same path so 'narrow solution' comes to mind. A disadvantage of making a "zero" puzzle. If I'd left those other cages in it may have opened up a completely different path.

I did the end a little differently to the other two. [Many thanks to Andrew for checking my WT]
a452 WT:
Preliminaries by SudokuSolver
Cage 5(2) n2 - cells only uses 1234
Cage 5(2) n6 - cells only uses 1234
Cage 8(2) n8 - cells do not use 489
Cage 8(2) n2 - cells do not use 489
Cage 13(2) n9 - cells do not use 123
Cage 11(2) n1 - cells do not use 1
Cage 11(2) n4 - cells do not use 1
Cage 22(3) n256 - cells do not use 1234
Cage 20(3) n36 - cells do not use 12
Cage 11(3) n3 - cells do not use 9
Cage 11(3) n7 - cells do not use 9
Cage 11(3) n45 - cells do not use 9

1. "45" on c9: 2 outies r16c8 = 14 = [59]/{68}
1a. min. r1c8 = 5 -> max. r12c9 = 6 (no 5,6,7,8)
1b. 11(3)n3 = {128/146/236/245}
1c.note: if it has 3 must also have 6 (and no 5)

2. "45" on r1: 2 outies r2c49 = 8 = [71/62/53]
2a. [53] forces 5 in r1 into 11(2)n1 = {56} but this clashes with 6 also in r1c8 (step 1c)
2b. h8(2) = [71/62]

3. killer pair 1,2 in 5(2)n2 and r2c9: both locked for r2

4.i. if 1 in r2 in r1c9 -> 7 in r2c4 (h8(2))
ii. or 1 in r2 in 5(2)n2
4a. -> {17} blocked from 8(2)n2
4b. = {26/35}(no 1,7) = 2 or 3

5. -> {23} blocked from 5(2)n2
5a. = {14} only: both locked for r2 and n2
5b. -> r2c49 = [62]

6. 8(2)n2 = {35}: both locked for r3 and n2

7. sp25(5)r1c3 must have 2 for n2 and two of 7,8,9 for r1c456
7a. = {125689/234679/235678}
7b. has two of 7,8,9 -> no 7,8,9 in r1c37
7c. 2 locked for r1
7d. no 9 in 11(2)n1

8. "45" on n3: 2 innies r1c7 + r3c9 = 13 = [49/58]

9. r1c89 = 9 (cage sum): but [54] blocked by r1c7
9a. = [81/63]
9b. note: if it has 3 must also have 6

10. [63] in r1c89 or 6 in r1 in 11(2)n1
10a. -> {38} blocked from 11(2) (Locking out cages)
10b. = {56/47}(no 3,8)

11. "45" on c1: 1 innie r1c1 = 2 outies r29c2
11a. -> max. r29c2 = 7
11b. -> r2c2 = (35)
11c. r9c2 = (1234)

12. "45" on c1: 3 outies r129c2 = 11 and must have 3 or 5 for r2c2
12a. = {137/236/245}
12b. but {236} can only be [632] but this leaves no 3 for n3! (step 9b)
12c. = {137/245} = [731/452]

13. 11(2)n1 = {47} only: both locked for r1 and n1
13a. r1c7 = 5
13b. -> r3c9 = 8 (h13(2)n3)
13c. -> r16c8 = [68]
13d. r1c9 = 3 (cage sum)
13e. r1c3 = 1

14. r2c78 = {79}: both locked for n3, 9 for r2

15. 5(2)n6: {14} blocked by r3c8 = (14)
15a. = {23}: both locked for c8 and n6

16. r45c9 = 12 (cage sum) = {57} only: both locked for c9, 7 for n6

17. 13(2)n9 = {49} only: both locked for n9, c9

18. r789c8 = {157} only: 1 & 7 locked for c8 and n9
18a. r67c9 = [16]
18b. r23c78 = [7914]

19. 38(7)n9 can't have both 4 & 9 because r9c9 = (49)
19a. = {1256789}(no 3,4)
19b. must have 2,6,8,9: all locked for r9
19c. r89c9 = [94]
19d. r9c2 = 1
19e. -> r12c2 = [73](h11(3)c2)
19f. r8c8 = 1 (38(7) must have 1)

20. r9c1 = 3 (hsingle r9)
20a. r8c1 = 7 (cage sum)

21. 23(4)n9 must have 3 for n9
21a. only option is {38} in r78c7: 8 locked for n9 and 23(4), and r8c6 = 5, r7c8 = 7

22. 8(2)n8 = [26] only

23. 20(3)n1 = {2459/2468}
23a. -> r4c2 = 4

24. 17(3)n7 = [584]

25. 11(2)n4 = {29} only: both locked for c2 and n4

26. 20(4)n1 = [8624]

On from there
Cheers
Ed
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