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 Post subject: Assassin 358
PostPosted: Mon Sep 10, 2018 8:33 am 
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Grand Master
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Posts: 774
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1-9 cannot repeat on the diagonals.

Assassin 358

Had a zero version of this puzzle but changed my mind at the last minute. Hoping someone will find a better way than I did. More cages should give a better chance of that. This one gets 1.55 JSudoku uses a couple of chains.
code:
3x3:d:k:2816:3585:3585:3330:3330:3330:8451:8451:2564:2816:7173:2054:2054:3335:1544:1544:8451:2564:7173:7173:7173:4361:3335:2314:8459:8451:8451:7173:4876:7173:4361:4361:2314:8459:8459:8451:4876:4876:7181:7181:7181:7181:1806:8459:8459:4876:4623:4623:4880:7181:7181:1806:8459:5137:3602:4623:4623:4880:4880:3603:3860:3860:5137:3602:4885:5654:5654:5654:3603:3860:5137:5137:3602:4885:4885:5654:5654:3603:3863:3863:3863:
solution:
Code:
+-------+-------+-------+
| 2 8 6 | 3 1 9 | 7 5 4 |
| 9 1 3 | 5 7 4 | 2 8 6 |
| 4 5 7 | 8 6 2 | 1 3 9 |
+-------+-------+-------+
| 3 2 8 | 4 5 7 | 6 9 1 |
| 6 4 5 | 2 9 1 | 3 7 8 |
| 7 9 1 | 6 3 8 | 4 2 5 |
+-------+-------+-------+
| 8 6 2 | 9 4 3 | 5 1 7 |
| 1 3 4 | 7 8 5 | 9 6 2 |
| 5 7 9 | 1 2 6 | 8 4 3 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 358
PostPosted: Fri Sep 14, 2018 4:30 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1602
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin. Not sure that I found a better way. I got stuck for a while, then found my step 5 which isn't the sort of step that I'd normally expect to need to use in one of your puzzles; I hope there's a better way at that stage. Step 5h was very useful in making step 6 simple.

Here is my walkthrough for Assassin 358:
Prelims

a) R12C1 = {29/38/47/56}, no 1
b) R1C23 = {59/68}
c) R12C9 = {19/28/37/46}, no 5
d) R2C34 = {17/26/35}
e) R23C5 = {49/58/67}, no 1,2,3
f) R2C67 = {15/24}
g) R34C6 = {18/27/36/45}, no 9
h) R56C7 = {16/25/34}, no 7,8,9
i) 19(3) cage at R6C4 = {289/379/469/478/568}, no 1
j) 19(3) cage at R8C2 = {289/379/469/478/568}, no 1

1a. 45 rule on C789 1 outie R2C6 = 4 -> R2C7 = 2, clean-up: no 7,9 in R1C1, no 6,8 in R1C9, no 6 in R2C34, no 9 in R23C5, no 8 in R2C9, no 5 in R34C6, no 5 in R56C7
1b. 45 rule on N9 1 outie R6C9 = 5
1c. 45 rule on N3 or N6, 1 outie R4C9 = 1 innie R3C7, no 5 in R3C7, no 2 in R4C9
1d. R12C1 = [29/38/47/83] (cannot be {56} which clashes with R1C23), no 5,6
1e. 45 rule on N8 2(1+1) outies R6C4 + R8C3 = 10 = {28/37/46}/[91], no 5,9 in R8C3

2. 45 rule on N2 3 remaining innies R2C4 + R3C46 = 15 = {159/258/267} (cannot be {168/357} which clash with R23C5), no 3, clean-up: no 5 in R2C3, no 6 in R4C6
2a. 5,7 of {159/258/267} must be in R2C4 -> R2C4 = {57}, clean-up: no 7 in R2C3
2b. 9 of {159} must be in R3C4 -> no 1 in R3C3
2c. Killer pair 5,7 in R2C4 and R23C5, locked for N2, clean-up: no 2 in R4C6
2d. 3 in N2 only in 13(3) cage at R1C4, locked for R1, clean-up: no 8 in R2C1, no 7 in R2C9
2e. 13(3) cage = {139/238}, no 6
2f. Killer pair 8,9 in R1C23 and 13(3) cage, locked for R1, clean-up: no 3 in R2C1, no 1 in R2C9
2g. 7 in R1 only in R1C789, locked for N3, clean-up: no 7 in R4C9 (step 1c)

3. 45 rule on C6 3 innies R156C6 = 18 = {189/279/369} (cannot be {378} which clashes with R34C6, cannot be {567} because no 5,6,7 in R1C6), no 5, 9 locked for C6
3a. 5 in C6 only in 14(3) cage at R7C6, locked for N8
3b. 14(3) cage = {158/257/356}

[A couple of IOUs.]
4a. 45 rule on N1 2 outies R4C13 = 1 innie R2C3 + 8, no 8 in R4C1
4b. 45 rule on N7 2 outies R6C23 = 1 outie R8C3 + 6, no 6 in R6C2

[I’ve analysed 28(6) cage at R2C2 by looking at the values missing from combinations.]
5. 28(6) cage = {123589/124579/124678/134578/234568} (cannot be {123679} because cannot place both of 5,8 in R1C23, cannot be {134569} because cannot place both of 2,7 in R12C1)
5a. 2 in R1 only in R1C1 and 13(3) cage at R1C4 (step 2e) = {238}
5b. 28(6) cage = {124579/124678/134578/234568} (cannot be {123589} because R1C1 = 4 + R1C23 = {68} clash with 13(3) cage = {238})
5c. 28(6) cage = {124579/124678/134578} (cannot be {234568} because R2C1 = 7 clashes with R2C34 = [17])
[And then interactions between the innies and outies for N1.]
5d. 28(6) cage = {124678/134578} (cannot be {124579} because R12C1 = [29] clashes with R4C13 = {29} and R12C1 = [47] clashes with R4C13 = {47}), no 9
5e. 28(6) cage = {134578} (cannot be {124678} because R1C23 = {59} -> R12C1 = [47] clashes with R4C13 = {47}), no 2,6
5f. R1C1 = 2 (hidden single in N1), placed for D\, R2C1 = 9, clean-up: no 5 in R1C23, no 8 in 13(3) cage at R1C4 (step 2e), no 1 in R1C9
5g. 13(3) cage = {139}, 1 locked for R1 and N2, clean-up: no 8 in R4C6
5h. Naked pair {68} in R1C23, locked for R1 and N1
5i. 28(6) cage = {134578} -> R4C3 = 8, R1C23 = [86], clean-up: no 8 in R3C7 (step 1c), no 2,4 in R6C4 (step 1e)
5j. Naked triple {457} in R1C789, locked for N3, clean-up: no 4 in R4C9 (step 1c)
5k. 45 rule on N1 1 innie R2C3 = 1 remaining outie R4C1 -> R4C1 = {13}
5l. 19(3) cage at R8C2 = {379/469}, no 2,5, 9 locked for N7
5m. 8 in C1 only in 14(3) cage at R7C1 = {158} (only remaining combination), locked for C1 and N7 -> R4C1 = 3, R2C3 = 3 -> R2C4 = 5, R2C9 = 6 -> R1C9 = 4, placed for D/, clean-up: no 8 in R2C5, no 7,8 in R3C5, no 6 in R3C6, no 3 in R3C7 (step 1c), no 7,9 in R6C4 (step 1e)
5n. R23C5 = [76], R2C2 = 1, placed for D\, R2C8 = 8, placed for D/, clean-up: no 2 in R8C3 (step 1e)

6. 45 rule on N2 3 remaining outies R4C456 = 16 = {457} (only possible combination, cannot be {169} which clashes with R4C9, cannot be {259/268} because R4C6 only contains 1,7) = [457], 4 placed for D\, 7 placed for D/, R3C6 = 2, R3C4 = 8
[Cracked. The rest is straightforward.]
6a. 1 in R4 only in R4C789, locked for N6, clean-up: no 6 in R56C7
6b. Naked pair {34} in R56C7, locked for C7 and N6

7. R6C9 = 5 -> 20(4) cage at R6C9 = {1568/2567}, no 3,9
7a. 6 in 20(4) cage only in R8C8 -> R8C8 = 6, placed for D\
7b. R4C7 + R5C9 = [68] (hidden pair in N6), clean-up: no 1 in 20(4) cage
7c. Naked pair {27} in R78C9, locked for N9
7d. R3C3 + R7C7 = [75] (hidden pair on D\), R7C3 = 2, R78C9 = [72], R8C3 = 4 -> R6C4 = 6 (step 1e), R9C3 = 9, R8C2 = 3, placed for D/, R9C2 = 7 (cage sum), R3C12 = [45]
7e. R5C5 = 9, placed for D/, R3C7 = 1
7f. R56C1 = [67] = 13 -> R45C2 = 6 = [24]
7g. R6C4 = 6 -> R7C45 = 13 = [94]

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

Rating Comment:
I'll rate my WT for A358 at Hard 1.25 for the combo analysis in step 5; must admit I thought about rating that step higher but it probably isn't higher, it didn't quite use chains.


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 Post subject: Re: Assassin 358
PostPosted: Mon Sep 17, 2018 5:26 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 774
Location: Sydney, Australia
Andrew wrote:
my step 5 which isn't the sort of step that I'd normally expect to need to use in one of your puzzles
Great! Love to have lots of variety. That's why I keep enjoying killers.

I worked in the same area as Andrew's step 5 but quite differently. The next part after that is different too. Looking forward to see how wellbeback does it. Thanks to Andrew for some corrections.
walkthrough for a358:
Preliminaries courtesy of SudokuSolver
Cage 6(2) n23 - cells only uses 1245
Cage 14(2) n1 - cells only uses 5689
Cage 7(2) n6 - cells do not use 789
Cage 8(2) n12 - cells do not use 489
Cage 13(2) n2 - cells do not use 123
Cage 9(2) n25 - cells do not use 9
Cage 10(2) n3 - cells do not use 5
Cage 11(2) n1 - cells do not use 1
Cage 19(3) n58 - cells do not use 1
Cage 19(3) n7 - cells do not use 1

Note: no routine clean-up done unless stated
1. "45" on n9: 1 outie r6c9 = 5

2. "45" on c789: 1 outie r2c6 = 4, r2c7 = 2
2a. 13(2)n2: no 9
2b. 8(2)r2c3: no 6
2c. 10(2)n3: no 8
2d. 9(2)r3c6: no 5
2e. no 7,9 in r1c1

3. "45" on n2: 3 innies r2c4 + r3c35 = 15
3a. but {357} blocked by 13(2) = 5 or 7
3b. = {159/168/258/267}(no 3)
3c. no 5 in r2c3
3d. no 6 in r4c6

4. 3 in n2 only in 13(3) = {139/238}(no 5,6,7) = 1 or 8, 3 locked for r1
4a. no 7 in r2c9
4b. no 8 in r2c1

5. h15(3)n2 = {159/168/258/267}(step 3b)
5a. but {168} blocked by 13(3) = 1/8 (step 4)
5b. h15(3)n2 = {159/258/267}
5c. 1 in {159} must be in r3c6 -> no 1 in r23c4
5d. h15(3) must have 5,7 for r2c4 and can't have both 5,7 -> no 5,7 in r3c46
5e. no 2 in r4c6, no 7 in r2c3

6. 11(2)n1: {56} blocked by 14(2) = 5/6, no 5,6 in r12c1

7. r4c1 sees all 4,7 in n1 -> no 4,7 in r4c1 (CPE)

8. 7 in r1 only in n3: 7 locked for n3

9. 13(3)n2 = {139/238} = 8/9
9a. killer pair 8,9 in 13(3)n2 + 14(2)n1: both locked for r1
9b. no 3 in r2c1, no 1 in r2c9
9c. no 6 in r1c9

10. "45" on n1: 5 innies r2c23 + r3c123 = 20 and must have both 1,3 for n1 = {12359/12368/13457} (no eliminations yet)

This step is effective. It's not elegant.
11. 7 in r3 only in r3c5 or r3c123 -> [17] in r2c34 must have 7 in r3c123
11a. h20(5)n1 = {12359/12368/13457}
11b. {12368} has no 7 -> can only be 3 in r2c3 -> {1268} all in 28(6)r2c2. {1268} = 17 -> r4c13 as part of 28(5)r2c2 = 11 = {47} only: but no 4,7 in r4c1 -> {12368} blocked
11c. h20(5)n1 = {12359/13457}(no 6,8): must have 5: 5 locked for n1 and no 5 in r4c13

12. 14(2)n1 = {68}: both locked for r1

13. 13(3)n2 = {139} only: 1,9 locked for r1 and n2

14. naked triple {457} in r1c789: 4,5 locked for n3 & r1
14a. r12c1 = [29]; 2 placed for D\

15. 28(6)r2c2 must have 4,5,7 for n1: but {124579} blocked by 2,9 only in r4c3
15a.= {134578} only (no 2,6,9)
Made a mistake now. I originally eliminated 8 from r4c1 by doing IOU on n1 but forgot to put it in here. See Andrew's step 4a for the details.
15b. must have 8 -> r4c3 = 8
15c. r1c23 = [86]


16. "45" on n8: 2 outies r6c4 + r7c3 = 10
16a. no 2,4 in r6c4, no 5,9 in r7c3

17. 19(3)n7 = {379/469)(no 2,5)
17a. must have 9: locked for n7

18. hidden single 2 on D/ -> r7c3 = 2

19. "45" on n7: 2 outies r6c23 - 6 = 1 innie r8c3
19a. 1 innie and 1 outie see each other -> cannot be equal -> IOD blocked from r6c2 -> no 6 in r6c2 (IOU)

20. 2 and 6 in n4 only in 19(4) = {2467} only: 4 and 7 locked for n4

21. hidden single 5 in n4 -> r5c3 = 5

22. "45" on c123: 2 remaining innies r28c3 = 7 = [34] only
22a. r6c4 = 6 (outiesn8=10), placed for d/
22b. last innie n7, r7c2 = 6

etc
Cheers
Ed


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 Post subject: Re: Assassin 358
PostPosted: Sun Sep 23, 2018 4:33 pm 
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Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 171
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Thanks Ed! We all cracked it in essentially the same way - some less than elegant combination analysis. The specific details of those analyses were different which probably shows our different strengths in thinking patterns :).

Assassin 358 WT:
1. Outies n9 -> r6c9 = 5
-> Remaining Innie n36 -> 6(2)r2c6 = [42]
-> 8(2)r2c3 from {17} or {35}

2. Remaining Innies n2 = r2c4,r3c4,r3c6 = +15(3) has at most one of (123)
-> Two of (123) in 13(3)n2
-> Possibilities for 13(3)n2 are {139} or {238}
14(2)n1 from {59} or {68}
-> {389} locked in r1 in r1c23456
-> 11(2)n1 cannot be {38}
Also since one of (56) in 14(2)n1 -> 11(2)n1 cannot be {56}
-> 11(2)n1 from [29] or [47]

3! 8(2)r2c3 is one of:
a) {17} -> 11(2)n1 = [29] -> 14(2)n1 = {68}
b) [35] -> Remaining Outies n1 = r4c13 = +11(2).
Those two values cannot be the same as the 11(2) in n1 -> must be two of the values from r1c23,r2c3
-> must be {38} -> 14(2) = {68}
Either way 14(2)n1 = {68}

-> 13(3)n2 = {139}
-> HS 2 in r1 -> 11(2)n1 = [29]
Also 8(2)r2c3 from [35] or [17]

4! The three numbers not in 28(6)n1 = +17(3).
In n1 they must all be in the 11(2), the 14(2), or in r2c3.
The cells in the 11(2) 'see' all of the 28(6) cells except for r4c3 so at most one of those 11(2) values can be in the 28(6).
-> At least one of the 11(2) values must be in the H+17(3).

Given 11(2)n1 = [29], 14(2)n1 = {68}, and r2c3 from {13}
-> Only possibility for those missing numbers is {269}
-> 28(6) = {134578}
-> Whichever of (13) is in r2c3 must also go in r4c1
-> 8 in 28(6) in r4c3

5. -> 14(2)n1 = [86]
-> HS 8 in c1 -> 14(3)n7 = {158}
-> r4c1 = 3
-> 8(2)r2c3 = [35]
-> 13(2)n2 = {67}
-> r3c46 = {28}
Also r1c789 = {457}
-> 10(2)n3 = [46]
-> r1c78 = {57}
Also r2c8 = 8
-> r3c789 = {139}
-> r2c2 = 1
-> r3c123 = {457}
-> 13(2)n2 = [76]

6! r3c46 = {28}
a) If r3c46 = [28] -> r4c456 = [691]
b) If r3c46 = [82] -> r4c456 = [457]
Remaining Innies - Outies n5 -> r6c4 = r5c3 + 1
-> Whatever is in r5c3 goes in n5 in r4c456
But it cannot be in r4c6 since that would put remaining Innies n5 = +17(3) puts r3c4 = r6c4
-> r5c3 same value as in r4c45
-> r5c3 from (4569)
But remaining Innies c123 -> [r5c3,r8c3] = +9(2) can only be from [54], [27], or [72]
-> r5c3 = 5
-> 17(3)r3c4 = [845]
-> 9(2)r3c6 = [27]
-> r6c4 = 6
Also r8c3 = 4

7. Continuing...
Remaining Innies n4 = r6c23 = +10(2) can only be [91]
-> 18(4)r6c2 = [9162]
Also r3c123 can only be [457]
-> 19(4)n4 can only be [2647]
Also 19(3)n7 - [379] (Since 7 already in D/).
-> HS 5 in D/ -> r9c1 = 5
Since 1 already in D\ -> HS 1 in D/ -> r3c7 = 1
-> NS in D/ -> r5c5 = 9
-> HS 5 in D\ -> r7c7 = 5
etc.


Last edited by wellbeback on Fri Oct 05, 2018 1:21 am, edited 2 times in total.

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 Post subject: Re: Assassin 358
PostPosted: Wed Sep 26, 2018 7:28 am 
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Posts: 774
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This is what I was hoping for - something brain twisting from wellbeback. Can you explain step 4 some more? Why can't the missing +17(3) be in r1c23 (ie 14) + 3 from r2c3: total = 17? Why does one have to be in r12c1?

Thanks


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 Post subject: Re: Assassin 358
PostPosted: Wed Sep 26, 2018 11:10 pm 
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Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 171
Location: California, out of London
Thanks as always Ed for creating these puzzles. I have updated my WT to hopefully clarify the step and answer your question.
Cheers, Wellbeback.


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