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 Post subject: Assassin 354
PostPosted: Sat Jul 14, 2018 7:57 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
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Assassin 354
Two key steps for me. If you miss either one you might find this as hard as what SudokuSolver (1.90) and JSudoku do (chains). Hopefully, as usual, you guys find different key steps.
code:
3x3::k:6912:6912:6912:4865:4354:5123:5123:5123:5124:0000:0000:6912:4865:4354:5123:3847:5124:5124:0000:0000:4865:4865:4354:8712:3847:5124:3849:0000:2826:2826:2826:8712:8712:5643:5124:3849:2572:5645:5645:5645:8712:5643:5643:5643:3849:2572:8206:5645:8712:8712:3599:3599:3599:3600:2572:8206:2321:8712:4370:1797:1797:2836:3600:8206:8206:2321:2837:4370:2822:6163:2836:3600:8206:2837:2837:2837:4370:2822:6163:6163:6163:
solution:
Code:
+-------+-------+-------+
| 6 5 9 | 4 1 3 | 8 7 2 |
| 3 4 7 | 8 9 2 | 6 1 5 |
| 8 1 2 | 5 7 6 | 9 3 4 |
+-------+-------+-------+
| 7 2 6 | 3 4 1 | 5 9 8 |
| 1 9 4 | 6 5 8 | 7 2 3 |
| 5 8 3 | 7 2 9 | 1 4 6 |
+-------+-------+-------+
| 4 7 8 | 9 3 5 | 2 6 1 |
| 9 6 1 | 2 8 4 | 3 5 7 |
| 2 3 5 | 1 6 7 | 4 8 9 |
+-------+-------+-------+

Cheers
Ed


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 Post subject: Re: Assassin 354
PostPosted: Sun Jul 15, 2018 6:43 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
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I enjoyed this puzzle very much. Congratulations on a gem Ed!
Some edits to more fully explain the first key step.
Second Edit. Added a simpler equivalent way to do my first key step. (Step 2).

Assassin 354 WT:
1. Innies c6789 = r34c6 = +7(2)
Innies c5 = r456c5 = +11(3)
Innies c1234 = r67c4 = +16(2) = {79}
-> 34(7) = {12479(38|56)}
Also whichever of (79) is in r7c4 goes in r56c6 in n5 and in r123c5 in n2

2! First important move
Consider 11(2) in n8.
Since 34(7) is missing two values that add to 11 -> The values in 11(2)n8 are either both in or both not in 34(7).
If it was {38} or {65} the values cannot both go in r456c5 (since r456c5 = +11(3))
-> at least one of them would have to go in n5 in r45c4 (i.e., not anywhere in the 34(7))
-> both not in 34(7)
-> both in n5 in r45c4
(But this leaves no place in n5 for whatever is in r3c6!)
-> both would have to go in n2/c5 in r123c5 which leaves the last remaining value in r123c5 as 6.
which contradicts r123c5 containing at least one of (79) (last line of Step 1).

-> 11(2)n8 from {29} or {47)
-> whichever of (79) is in r6c4 goes in r89c6
-> it also goes in n2/c5 in r123c5
and since whatever is in r7c4 (the other one from (79)) is also in n2 in r123c5 (last line of step 1)
-> 17(3)n2 = {179}

Since writing the above I've realized a more straightforward way of seeing the above step as follows:
Whatever goes in r7c4 (7 or 9) must go in n5 in r56c6.
Similarly whatever goes in r3c6 must go in n5 in r45c5.
-> Of the two numbers not in 34(7) (which are +11(2)), one also must go in r56c6 and the other in r45c4.
-> the values in 11(2)n8 cannot be the same as the 11(2) missing from 34(7). These values must both be in the 34(7) somewhere.
However they cannot both go in r456c5 since that is +11(3).
-> One of them must go in r6c4 - which is from (79).
-> 11(2)n8 must be {47} or {29}.


3. -> HS 1 in 34(7) -> r4c6 = 1
-> Innies c6789 -> r3c6 = 6
-> r456c5 = {245}
-> 17(3)n8 = {368}

Since 11(4)r8c4 = {1235}
-> r89c4,r9c23 = {12},{35} or {15},{23}

4. Since r4c6 = 1 and r456c5 = {245}
-> 11(3)r4c2 can only be {236} with 2 in r4c23

5! Second important move
r45c4 are either {68} or {36}
r89c4 are either {12} or {15}
Innies - Outies n47 -> r4c1 + 5 = r4589c4
-> r4589c4 are max +14(4)
-> r45c4 = {36}, r89c4 = {12}, r4c1 = 7

6. Continuing
-> r123c4 = {458}
-> r3c3 = 2

Also r9c23 = {35}
-> 9(2)n7 = {18}
-> 32(5) = [8{2679}]
-> r7c1 = 4
-> 10(3)r5c1 = [{15}4]

Also 11(2)n8 = {47}
-> r67c4 = [79]
-> r56c6 = [89]

Also 7(2)r7c6 = [52]
-> Remaining outies n9 -> r6c9 = 6

7. 7 in n7 in r78c2
-> 7 in n1 in r12c3
-> 27(4)n1 = {79(38|56)}
-> r23c2 = {14}
-> 5 in n1 in 27(4)
-> 27(4)n1 = {5679}
-> r23c1 = {38}

8. r12c6 = {23}
-> r1c78 = +15(2)
Since at least one of (69) in r1 in n1
-> r1c78 = {78}
-> 27(4)n1 = [{569}7]
Also 15(2)n3 = [69]
-> 17(3)n2 = [197]

9. Innies - Outies n3 -> [r3c9,r4c8] from [38] or [49]
(Cannot be [16] or [27] by previous placements)
But [38] leaves no place for 9 in r4/n6
-> 20(5)n2 = [{1235}9]
and -> 15(3)r3c9 = [483]

-> r45c4 = [36]
-> r4c23 = [26]

Also 22(4)n4 = [9463]
Also r6c78 = {14}
-> 22(4)r4c7 = [5872]
-> r456c5 = [452]

Also r1c78 = [87]
Also r78c9 = {17}
-> HS 5 in n9 -> 11(2)n9 = [65]
-> 24(4)n9 = [3489]
-> 17(3)n8 = [386]
etc.


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 Post subject: Re: Assassin 354
PostPosted: Tue Jul 24, 2018 4:19 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks Ed for another fun Assassin. Looks like I found a very different way to solve it.

Here is my walkthrough for Assassin 354:
Prelims

a) R23C7 = {69/78}
b) R78C3 = {18/27/36/45}, no 9
c) R7C67 = {16/25/34}, no 7,8,9
d) R78C8 = {29/38/47/56}, no 1
e) R89C6 = {29/38/47/56}, no 1
f) 11(3) cage at R4C2 = {128/137/146/236/245}, no 9
g) 10(3) cage at R5C1 = {127/136/145/235}, no 8,9
h) 27(4) cage at R1C1 = {3789/4689/5679}, no 1,2
i) 11(4) cage at R8C4 = {1235}
j) 32(5) cage at R6C2 = {26789/35789/45689}, no 1

Steps resulting from Prelims
1a. 27(4) cage at R1C1 = {3789/4689/5679}, 9 locked for N1
1b. 11(4) cage at R8C4 = {1235}, CPE no 1,2,3,5 in R9C56, clean-up: no 6,8,9 in R8C6

2a. 45 rule on C56789 2 outies R67C4 = 16 = {79}, locked for C4 and 34(7) cage at R3C6
2b. 45 rule on C6789 2 innies R34C6 = 7 = {16/25/34}, no 8
2c. 45 rule on C5 3 innies R456C5 = 11 = {128/146/245} (cannot be {236} which clashes with R34C6 in 34(7) cage), no 3
2d. 17(3) cage at R7C5 = {278/359/368/467} (cannot be {179} which clashes with R7C4, cannot be {269/458} which clashes with R456C5), no 1
2e. 45 rule on N9 2(1+1) outies R6C9 + R7C6 = 11 = [56/65/74/83/92] -> R6C9 = {56789}, R7C6 = {23456}, clean-up: no 6 in R7C7
2f. R6C9 + R7C6 = 11 -> R67C9 cannot total 11 (CCC) -> no 3 in R8C9
2g. 45 rule on N9 3 innies R7C7 + R78C9 = 10 = {127/136/145/235}, no 8,9 in R78C9
2h. 1 in N8 only in R89C4, locked for C4 and 11(4) cage at R8C4

3. 45 rule on N23456789 1 innie R4C1 = 1 outie R3C3 + 5 -> R4C1 = {6789}, R3C3 = {1234}
3a. 19(4) cage at R1C4 = {1468/2368/2458}, 8 locked for C4 and N2

4. Consider placement for 8 in C5
8 in R456C5 (step2c) = {128}, locked for 34(7) cage at R3C6 => R34C6 (step 2b) = {34}, locked for C6
or 8 in 17(3) cage at R7C5, locked for N8
-> R89C6 = [29/47/56/74] (cannot be [38]), no 3,8
4a. 8 in N8 only in 17(3) cage at R7C5, locked for C5
4b. 17(3) cage (step 2d) = {278/368}, no 4,5,9
4c. R456C5 (step 2c) = {146/245}, 4 locked for C5, N5 and 34(7) cage at R3C6, clean-up: no 3 in R34C6 (step 2b)
4d. Deleted, replaced by step 4g
4e. R89C6 = [29/47/74] (cannot be [56] which clashes with R34C6), no 5,6
4f. Killer pair 7,9 in R7C4 and R89C6, locked for N8
4g. 7,9 in C5 only in 17(3) cage at R1C5 = {179}, locked for N2, 1 locked for C5
4h. R456C5 (step 2c) = {245} (only remaining combination), locked for C5, N5 and 34(7) cage at R3C6 -˃ R34C6 = [61], clean-up: no 9 in R2C7, no 5 in R6C9 (step 2e), no 1 in R7C7
4i. Naked triple {368} in 17(3) cage at R7C5, 3 locked for N8, clean-up: no 8 in R6C9 (step 2e), no 4 in R7C7
4j. Naked pair {36} in R45C4, 3 locked for C4 and N5
4k. 19(4) cage at R1C4 (step 3a) = {2458} (only remaining combination), no 1,3, 5 locked for C4 and N2, clean-up: no 6,8 in R4C1 (step 3)
4l. Naked pair {12} in R89C4, locked for C4, N8 and 11(4) cage at R8C4, clean-up: no 9 in R6C9 (step 2e), no 5 in R7C7, no 9 in R9C6
4m. Naked triple {458} in R123C4, 4 locked for N2 and 19(4) cage at R1C4 -˃ R3C3 = 2, R4C1 = 7 (step 3), clean-up: no 7 in R78C3
4n. 11(3) cage at R4C2 = {236} (only remaining combination, cannot be {245} which clashes with R4C5) -˃ R4C2 = 2, R4C34 = {36}, locked for R4
4o. Naked pair {47} in R89C6, locked for C6 and N8 -˃ R67C4 = [79], R6C9 = 6, R7C6 = 5 -˃ R7C7 = 2, clean-up: no 4 in R8C3, no 6,9 in R8C8
4p. R9C23 = {35}, locked for R9 and N7, clean-up: no 4,6 in R7C3, no 6 in R8C3
4q. Naked pair {18} in R78C3, locked for C3 and N7
4r. 7 in N7 only in R78C2, locked for C2
4s. 7 in C3 only in 27(4) cage at R1C1 = {3789/5679}, no 4
4t. 4 in C3 only in R56C3, locked for N4
4u. 10(3) cage at R5C1 = {136/145}, 1 locked for C1 and N4
4v. 22(4) cage at R5C2 contains 4 in R56C3 = {3469} (only remaining combination), no 5,8
4w. R6C2 = 8 (hidden single in N4)
4x. R56C1 = {15} (hidden pair in N4), 5 locked for C1, R7C1 = 4 (cage sum), clean-up: no 7 in R8C8
4y. R6C6 = 9 -˃ R6C78 = 5 = [14/32/41], R5C6 = 8
4z. 9 in R4 only in R4C789, locked for N6

5. R12C6 = {23} = 5 -˃ R1C78 = 15 = {69/78}
5a. Naked quad {6789} in R1C78 + R23C7, locked for N3
5b. 45 rule on N3 using R1C78 + R23C7 = 30, 1 outie R4C8 = 1 innie R3C9 + 5 –˃ R4C8 = {89}, R3C9 = {34}
5c. R6C9 = 6 -> R78C9 = 8 = [17/35/71], no 4 in R8C9
5d. 15(3) cage at R3C9 = [483] (only possible combination, cannot be [357] which clashes with R78C9, cannot be [384/492] which clash with R3C9 + R4C8 = [38/49], CCC), R4C8 = 9, clean-up: no 6 in R1C7, no 2 in R6C8 (step 4y), no 5 in R8C9
5e. R5C234 = [946], R6C3 = 3, R4C34 = [63], R9C23 = [35]
5f. Naked pair {14} in R6C78, locked for R6 and N6 -˃ R4C7 + R5C78 = [572], clean-up: no 8 in R1C8 + R23C7
5g. R23C7 = [69], R1C78 = [87], R12C3 = [97] = 16 -˃ R1C12 = 11 = [65], clean-up: no 4 in R8C8
5h. R8C8 = 5 (hidden single in N9) -> R7C8 = 6

and the rest is naked singles.
Thanks Ed for correcting some typos in step 4

Rating Comment:
I'll rate my WT for A354 at Easy 1.5; I used a short forcing chain. The high SS score is because it isn't programmed to look for 45s on large numbers of nonets.


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 Post subject: Re: Assassin 354
PostPosted: Thu Jul 26, 2018 6:16 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Another puzzle, another techniques post! Love how I'm still learning. I've managed to make a future Assassin using the techniques that wellbeback is using so might be handy to know! Glad I'm finally getting the hang of them.

Anyway, I thought a slightly different explanation of wellbeback's step 1 and 2 might be useful. This is very close to his alternate explanation but just a little embellished.
Attachment:
WellStep2a.jpg
WellStep2a.jpg [ 50.95 KiB | Viewed 6711 times ]
1. the cell at r7c4 sees all of n5, except r56c6 so must repeat there and then also repeat in c5 in n2. r7c4 is from (79) so the 17(3)n2 must have (79)
2. r3c6 sees all of n5 apart from r45c4 so must repeat there.
3. in other words, the 34(7) at r3c6 is also completely inside n5, ie, there is an (implied) 34(7) in n5
4. "45" on n5 -> the remaining two innies = 11, this is an implied 11(2) cage
5. that 11(2) must have 1 cell in one of r45c4 and one cell in one of r56c6. We don't know exactly which of those 4 cells has the 11(2), but that's not important.
6. the actual 11(2) cage at r89c6 sees both r45c6 -> the combinations in those two 11(2) cages must be different so they must have different numbers
7. since the 11(2)r89c6 also sees r3c6 (which repeats in r45c4, step 2) -> the two numbers at r45c4 must both be different to the 11(2)r89c6
8. so where are the two numbers in r89c6 going to fit in n5?
9. they can't both go in r456c5 since it is an 11(3) (hidden) cage so will go over the cage total
10. -> one of r89c6 must be in r6c4.
Attachment:
WellStep2b.jpg
WellStep2b.jpg [ 69.92 KiB | Viewed 6711 times ]

11. since r6c4 is from (79) -> the 11(2)r89c6 must have 7 or 9 = {29/47}
12. further, since the (79) in 11(2)r89c6 is repeated in r6c4, this must also repeat in n2 in the 17(3)
13. from step 1. and step 12. -> the 17(3)n2 must have both 7&9 = {179} only

Now to Andrew. Love step 3! I got the same thing but in a different way, but Andrew's is simpler. In general, I followed the same path. Thanks to Andrew for going through my WT and suggesting some improvements.

A354
WT:
Preliminaries courtesy of SudokuSolver
Cage 15(2) n3 - cells only uses 6789
Cage 7(2) n89 - cells do not use 789
Cage 9(2) n7 - cells do not use 9
Cage 11(2) n9 - cells do not use 1
Cage 11(2) n8 - cells do not use 1
Cage 10(3) n47 - cells do not use 89
Cage 11(3) n45 - cells do not use 9
Cage 11(4) n78 - cells ={1235}
Cage 27(4) n1 - cells do not use 12
Cage 32(5) n47 - cells do not use 13

Took me a while to see this step. Should have leaped out with this "cage" pattern. It makes a huge difference.
1. "45" on whole grid -> 5 zero cells at r23c12 + r4c1 = 23
1a. "45" on n1: 5 innies r23c12 + r3c3 = 18
1b. 4 of the 5 cells overlap so must be the same -> the difference in totals (5) must be from r3c3 & r4c1
1c. -> r4c1 - 5 = r3c3
1d. r4c1 = (6789), r3c3=(1234)

2. "45" on c56789: 2 outies r67c4 = 16 = {79} only: both locked for c4 and not in r34c6 nor r456c5 (same cage)

3. 19(4)r1c4 = {1468/2368/2458}
3a. must have 8: locked for c4 and n2

4. "45" on c6789: 2 innies r34c4 = 7 = {16/25/34}(no 8) = 2 or 3 or 6

5. "45" on c5: 3 innies r456c5 = 11
5a. but {236} blocked by r34c4
5b. = {128/146/245}(no 3)

6. 34(7)r3c6 = {1234789/1245679}
6a. must have 1,2,4 -> no 1,2,4 in r56c6 since they see all 1,2,4 in 34(7) (CPE)
6b. note: if it has 8 (in c5) must also have 3 (in c6)

2nd key step
7. 8 in c5 in 34(7) -> has 3 in r34c6 (step 6b) or 8 in c5 in n8
7a. -> {38} blocked from 11(2)n8, no 3,8

8. 8 in n8 only in 17(3) = {278/368/458} (no 1,9)
8a. locked for c5

9. h11(3)r456c5 = {146/245}
9a. must have 4: 4 locked for c5, n5 and no 4 in r3c6
9b. h7(2)r34c6 = {16/25}(no 3) = 5 or 6

10. 11(2)n8: {56} blocked by r34c6
10a. = {29/47}(no 5,6) = 7 or 9 (finally got to wellbeback step 2. Andrew did it the same way I did)

11. killer pair 7,9 in n8 in 11(2) & r7c4: 7 locked for n8

12. 7 & 9 in c5 only in 17(3)n2 = {179} only: all locked for n2, 1 for c5

13. h11(3)r456c5 = {245} only: 2 & 5 locked for n5 & c5 and no 2,5 in r3c6 (same cage)
13a. r34c6 = [61]

14. 4 in c4 only in 19(4)r1c4 = {2458} only and no 4 in r3c3
14a. r3c3 = 2
14b. r4c1 = 7 (iodn1=-5)
14c. naked triple {458} in r123c4: 4 & 5 locked for n2, 5 for c4

15. naked pair {23} in r12c6: both locked for c6
15a. r12c6 = 5 -> r1c78 = 15 = {69/78}

16. 11(2)n8 = {47} only: both locked for n8 and 7 for c6
16a. r67c4 = [79], r7c67 = [52]

17. "45" on n9: 1 remaining outie r6c9 = 6

18. "45" on n23: 1 remaining innie r3c9 + 5 = 1 remaining outie r4c8 = [38/49]

19. 15(3)r3c9 must have 3 or 4 for r3c9
19a. but {249} as [4]{29} only blocked by 9 in r4c8 (step 18)
19b. = {348/357}(no 1,2,9)
19c. 3 locked in r345c9 for c9

20. r6c9 = 6 -> r78c9 = 8 = {17} only: both locked for n9 and c9
20a. -> r345c9 = {348} only, 4 & 8 locked for c9 and 8 for n6
20b. r4c8 = 9 -> r3c9 = 4 (step 18),
20c. r45c9 = {38}, 3 locked for n6

21. naked pair {25} in r12c9: both locked for c9 and n3
21a. r12c9 + r4c8 = 16 -> r23c8 = 4 = {13} only: both locked for c8

22. r9c9 = 9, 11(2)n9 = [65] only permutation

23. r1c78 = {78} only (h15(2)) both locked for r1 and n3
23a. r23c7 = [69]

24. hidden pair {12} in r89c4 -> no 1,2 in r9c23
24a. -> r9c4 = 1 (hsingle r9), r8c4 = 2

25. r9c1 = 2 (hsingle r9)
25a. ->r8c1 + r678c2 = 30 = {6789} only
25b. 6 also locked for r8, 7 locked for c2 and n7

26. 9(2)n7 = {18} only: both locked for c3 and n7

etc
Cheers
Ed


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