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 Post subject: Assassin 353
PostPosted: Sat Jun 30, 2018 7:05 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 353
A very easy start but then.....tricky to get into the middle. Nothing really advanced but have to be thoughful. This one started out as an easier one to remember 12 years since Ruud started making Assassins but couldn't help myself and had to toughen it up. You can guess what the easier puzzle cage was if you want to make it a bit easier. The one shown here gets a score of 1.40 JSudoku has no trouble.
code:
3x3::k:3584:3584:5121:5121:3074:3074:5635:5635:5635:3584:3584:6916:5121:3074:3845:3845:3845:5635:6916:6916:6916:3334:3334:3335:1032:1032:5385:5642:5642:0000:3334:3335:3335:5385:5385:5385:5642:4619:0000:0000:0000:0000:0000:5385:5388:4619:4619:4619:5133:5133:4366:0000:5388:5388:4619:3087:3087:5133:4366:4366:4624:4624:4624:5137:3346:3346:3346:3091:4116:4624:5397:5397:5137:5137:5137:3091:3091:4116:4116:5397:5397:
solution:
Code:
+-------+-------+-------+
| 3 8 4 | 7 1 5 | 2 6 9 |
| 1 2 7 | 9 6 3 | 4 8 5 |
| 5 9 6 | 4 8 2 | 3 1 7 |
+-------+-------+-------+
| 9 6 8 | 1 4 7 | 5 3 2 |
| 7 1 3 | 5 2 6 | 9 4 8 |
| 2 4 5 | 3 9 8 | 1 7 6 |
+-------+-------+-------+
| 6 3 9 | 8 5 4 | 7 2 1 |
| 4 5 2 | 6 7 1 | 8 9 3 |
| 8 7 1 | 2 3 9 | 6 5 4 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 353
PostPosted: Sat Jul 07, 2018 3:21 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin. Mostly fairly routine, but it took me a while to spot my breakthrough step.

Here is my walkthrough for Assassin 353:
Prelims

a) R3C78 = {13}
b) R7C23 = {39/48/57}, no 1,2,6
c) 20(3) cage at R1C3 = {389/479/569/578}, no 1,2
d) 22(3) cage at R4C1 = {589/679}
e) 21(3) cage at R5C9 = {489/579/678}, no 1,2,3
f) 20(3) cage at R6C4 = {389/479/569/578}, no 1,2
g) 14(4) cage at R1C1 = {1238/1247/1256/1346/2345}, no 9
h) 27(4) cage at R2C3 = {3789/4689/5679}, no 1,2
i) 18(5) cage in R5C2 = {12348/12357/12456}, no 9

Steps resulting from Prelims
1a. R3C78 = {13}, locked for R3 and N3
1b. 22(3) cage at R4C1 = {589/679}, 9 locked for N4
1c. 27(4) cage at R2C3 = {3789/4689/5679}, 9 locked for N1

[Immediate placements.]
2a. 45 rule on N1 1 innie R1C3 = 4 -> R12C4 = 16 = {79}, locked for C4 and N2, clean-up: no 8 in R7C2
2b. 45 rule on N9 1 innie R9C7 = 6 -> R89C6 = 10 = {19/28/37}, no 4,5
2c. 45 rule on R12 1 innies R2C3 = 7, R12C4 = [79], R3C123 = 20 = {569}, locked for R3 and N1, clean-up: no 5 in R7C2
2d. Naked triple {248} in R3C456, locked for R3 and N2 -> R3C9 = 7
2e. 45 rule on N3 1 outie R2C6 = 3 -> R2C78 = 12 = {48}, locked for R2 and N3, clean-up: no 7 in R89C6
2f. Naked pair {12} in R2C12, locked for R2 and N1
2g. 45 rule on R89 1 innie R8C7 = 8, R2C78 = [48], R7C789 = 10 = {127/145/235}, no 9, clean-up: no 2 in R9C6
2h. R7C23 = {39}/[48] (cannot be [75] which clashes with R7C789), no 5,7

3a. R3C9 = 7 -> R4C789 + R5C8 = 14 = {1238/1256/1346/2345}, no 9
3b. 45 rule on N6 using R4C789 + R5C8 = 14, 2 innies R56C7 = 10 = {19/37}, no 2,5
3c. R4C789 + R5C8 = {1256/2345} (cannot be {1238/1346} which clash with R56C7, no 8, 5 locked for N6
3d. Killer pair 1,3 in R3C7 and R56C7, locked for C7

4. 13(3) cage at R3C4 = {28}3/{48}1, 8 locked for R3, R4C4 = {13}
4a. 13(3) cage at R3C6 = {148/238/247/256/346} (cannot be {139/157} because R3C6 only contains 2,4), no 9
4b. 9 in R4 only in R4C12, locked for N4

5. 4 in N4 only in 18(5) cage at R5C2 -> no 4 in R7C1
5a. 18(5) cage = {12348/12456}, no 7
5b. 7 in N4 only in 22(3) cage at R4C1 = {679}, 6 locked for N4

6. 6 in R7 only in R7C1456
6a. 45 rule on R789 4 innies R7C1456 = 23 = {1679/2678/3569/4568}
6b. 45 rule on N7 1 innie R7C1 = 1 outie R8C4 = {12356}
6c. R7C1 = R8C4 -> R7C1456 = R7C456 + R8C4 = {1679/2678/3569/4568}, 6 locked for N8
6d. Killer pair 8,9 in R7C456 + R8C4 and R89C6, locked for N8
[Ed pointed out that I’d missed 9 in N8 only in R7C56 + R89C6, CPE no 9 in R6C6]

7. 20(3) cage at R6C4 = {389/569/578} (cannot be {479} because 7,9 only in R6C5), no 4
7a. 7,9 only in R6C5 = {79}

8. 45 rule on R1234 2 outies R5C18 = 1 innie R4C3 + 3
8a. Min R5C18 = 7 -> no 1,2,3 in R4C3
8b. R4C3 = {58} -> R5C18 = 8,11 = [62/71/65/74], no 3,6 in R5C8

9. 45 rule on R6789 2 outies R5C29 = 1 innie R5C7 + 8, IOU no 8 in R5C2
9a. Max R5C29 = 14 -> no 7,9 in R6C7, clean-up: no 1,3 in R5C7 (step 3b)
9b. R6C7 = {13} -> R5C29 = 9,11 = [18/36/54/29/38/56], no 4 in R5C2
9c. 4 in N4 only in R6C12, locked for R6
9d. 21(3) cage at R5C9 = {489/678} = [498/678/876] -> R5C9 = {468}, R6C8 = {79}, R6C9 = {68}
9e. R5C29 = [18/36/54/38/56], no 2
9f. Naked pair {79} in R6C58, locked for R6
9g. R4C789 + R5C8 (step 3c) = {1256/2345}
9h. R5C29 + R6C7 = [181/361/383] (cannot be [541/563] because R5C9 + R6C7 = [41/63] clash with R4C789 + R5C8), no 5 in R5C2, no 4 in R5C9
[Cracked; the rest is fairly straightforward.]
9i. Naked pair {68} in R56C9, locked for C9 and N6, R6C8 = 7 (cage sum), R5C7 = 9 -> R6C7 = 1 (step 3b), R2C9 = 5, R1C789 = [269], R3C78 = [31], R4C7 = 5
9j. R78C7 = [78] = 15 -> R7C89 = 3 = [21], R5C8 = 4, R4C89 = [32], R4C4 = 1 -> R3C45 = 12 = {48}, clean-up: no 2 in R8C4 (step 6b)
9k. R3C6 = 2 (hidden single in N2) -> R4C56 = 11 = {47}, 7 locked for R4 and N5, clean-up: no 8 in R9C6 (step 2b)
9l. Naked pair {19} in R89C6, locked for C6 and N8
9m. R4C3 = 8, clean-up: no 4 in R7C2
9n. Naked pair {39} in R7C23, locked for R7 and N7, clean-up: no 3 in R8C4 (step 6b)

10. 18(5) cage at R5C2 (step 5a) = {12456} -> R5C2 = 1, R6C123 = {245}, locked for R6 and N4, R7C1 = 6, R8C4 = 6 (step 6b), R2C12 = [12]
10a. R6C4 = 3 (hidden single in R6), R6C5 = 9 -> R7C4 = 8 (cage sum)
10b. R47C6 = [74] (hidden pair in C6), R7C5 = 5 -> R6C6 = 8 (cage sum)
10c. R8C4 = 6 -> R8C23 = 7 = [52]

and the rest is naked singles.
Thanks Ed for correcting my typos and pointing out one sub-step, fortunately not a critical one, which I'd missed.

Rating Comment:
I'll rate my WT for A353 at Hard 1.25; I decided on that, rather than just 1.25, partly because I didn't immediately spot step 9h.


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 Post subject: Re: Assassin 353
PostPosted: Sun Jul 08, 2018 8:46 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
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Thanks Ed. I found my breakthrough step quite quickly after the easy stuff. A small follow on from Andrew's Step 8.
Assassin 353 WT:
1. Innies r12 -> r2c3 = 7
Innies n1 -> r1c3 = 4
-> 20(3)r1c3 = [479]
-> HS 7 in r3 -> r3c9 = 7
-> Outies n3 -> r2c6 = 3
-> 15(3)r2c6 = [3{48}]

2. 4(3)r3c7 = {13}
-> 3 in n1 in r1c12
8 in n2 cannot be in 12(3) -> only in r3c456
-> 8 in n1 in r1c2
-> 14(4)n1 = [{38}{12}]
-> r3c123 = {569}
-> r3c456 = {248}
-> 12(3)n2 = {156} with 5 or 6 in r2c5
-> 22(4)n3 = {2569} with 5 or 6 in r2c9

3. 13(3)r3c4 cannot be [{24}7]
-> 13(3)r3c4 from [{28}3] or [{48}1] and r3c6 from 4 or 2.

4. Innies n9 -> r9c7 = 6
Innies r89 -> r8c7 = 8
-> r2c78 = [48]

5. Remaining Innies - Outies n6 -> r56c7 = +10(2)
This must be from {19} or {37}
-> 21(3)n6 cannot be {579} - must include an 8 in r56c9
-> 2 and 5 in 21(5) in n6 (Important for step 7!)

6. 4 in n4 only in 18(5)
-> 18(5) contains (124) and either (38) or (56)
-> 7 in n4 only in 22(3)
-> 22(3)n4 = {679}

7! The breakthrough step
Remaining outies n2 -> r4c456 = +12(3)
Innies - Outies r1234 -> r4c3 + 3 = r5c1 + r5c8

Since r5c1 cannot be 3 -> r4c3 cannot be the same as r5c8
-> r4c3 cannot be the same as any of the values in the 21(5) in n6
-> r4c3 cannot be 2 or 5
Also r4c3 cannot be from 4 or 7 (already in c3)
Also r4c3 cannot be from 6 or 9 (already in n4)
-> r4c3 from (138)
-> r4c456 = +12(3) cannot be {138}
-> HS 8 in r4 -> r4c3 = 8!

8. Innies - Outies n7 -> r7c1 = r8c4
Since 8 already in r8 -> r7c1 cannot be 8
-> 8 not in 18(5)n4
-> 18(5)n4 = [{1245}6]
-> r8c4 = 6
Also r4c2 = 6 and r45c1 = {79}
-> r3c123 = [596]
Also r5c3 = 3

9. Also r8c23 = {25}
-> 12(2)n7 = [39]
-> 20(4)n7 = [{48}71]

10. Only possibilities for 13(3)r3c4 and 13(3)r3c6 are
a) [{28}3] and [4{27}]
b) [{48}1] and [2{47}]
-> 7 in r4 in r4c56
-> r45c1 = [97]
-> Remaining Innies - Outies r1234 -> r5c8 = 4
-> 21(3)n6 = {678} with 7 in r6c8
-> Innies n6 -> r56c7 = {19}
-> r4c789 = {235}
-> 13(2)r3c4 = [{48}1] and 13(3)r3c6 = [2{47}]

11. Remaining outies n8 -> r6c456 = +20(3)
-> r6c6 = r7c4
Since (679) already in c4 and
... 7 already in n5 and
... 3 already in c6
-> 20(3)r6c4 = [398] and r6c6 = 8
Also HS 7 in n9 -> r7c7 = 7
-> r7c89 = {12}
etc.


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 Post subject: Re: Assassin 353
PostPosted: Mon Jul 09, 2018 9:05 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
As often happens, sounds like you guys found this puzzle easier than I did. I worked in similar areas to Andrew but saw it differently (as usual!). But first, loved wellbeback's step 7.
Attachment:
a353Wstep7.JPG
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Quote:
5. ....-> 2 and 5 in 21(5) in n6 (Important for step 7!)

7! The breakthrough step
Innies - Outies r1234 -> r4c3 + 3 = r5c1 + r5c8

Since r5c1 cannot be 3 -> r4c3 cannot be the same as r5c8
-> r4c3 cannot be the same as any of the values in the 21(5) in n6
-> r4c3 cannot be 2 or 5
Often see the first part of this in puzzles but never thought of combining this with locked numbers. Very obvious once you see it!

Better post my WT. Thanks to Andrew for some typos.
A353
25 steps:
Preliminaries courtesy of SudokuSolver
Cage 4(2) n3 - cells ={13}
Cage 12(2) n7 - cells do not use 126
Cage 22(3) n4 - cells do not use 1234
Cage 21(3) n6 - cells do not use 123
Cage 20(3) n12 - cells do not use 12
Cage 20(3) n58 - cells do not use 12
Cage 27(4) n1 - cells do not use 12
Cage 14(4) n1 - cells do not use 9
Cage 18(5) n47 - cells do not use 9

No routine clean-up done unless stated.
1. 4(2)n3 = {13} only: both locked for r3 and n3

2. "45" on r12: 1 innie r2c3 = 7
2a. -> r3c123 = 20 = {569} only: all locked for r3 and n1

3. "45" on n1: 1 innie r1c3 = 4
3a. -> r12c4 = 16 = [79] only permutation

4. naked triple {248} in r3c456: all locked for n2 and r3
4a. -> r3c9 = 7
4b. -> r4c789+r5c8 = 14 (no 9)

5. "45" on n3: 1 outie r2c6 = 3
5a. -> r2c78 = 12 = {48} only: both locked for n3 and 8 for r2

6. "45" on n9: 1 innie r9c7 = 6
6a. -> r89c6 = 10 = {19/28}(no 4,5,7)

7. "45" on r89: 1 innie r8c7 = 8
7a. ->r7c789 = 10 (no 9)
7b. r2c78 = [48]

8. 13(3)r3c4 must have two of 2,4,8 for r3c45 = {148/238}
8a. must have 1,3 -> r4c4 = (13)
8b. must have 8: locked for r3

9. 13(3)r3c6 must have 2,4 for r3c6 = {148/238/247/256/346}(no 9)

10. 4 in n4 only in 18(5) = {12348/12456}(no 7)
10a. no 4 in r7c1

11. 7 in n4 only in 22(3) = {679} only: 6,9 locked for n4

12. 9 in r4 only in r4c12 -> no 9 in r5c1

still some easy stuff left but time to get serious
13. "45" on r6789: 1 innie r6c7 + 8 = 2 outies r5c29
13a. one outie and 1 innie are in the same nonet -> r5c2 cannot equal the IOD of 8 -> no 8 (IOU)
13b. max. r5c29 = 14 -> max r6c7 = 6 (no 7,9)

14. "45" on c789: 2 innies r56c7 = 10 = [91/73]

15. 21(3)n6: but {579} blocked by r5c7
15a. = {489/678}(no 5) = [4->9..]
15b. must have 8: locked for n6

16. "45" on r6789: 3 outies (remembering h10(2)r56c7, r5c279 = 18
16a. but {567} blocked by r5c1
16b. but {459} as [594] only, blocked by 9 also in 21(3) (step 15a)
16c. but {468} blocked by no 7,9 for r5c7
16d. = {189/279/369/378}(no 4,5)

17. 4 in n4 only in r6, locked for r6

18. 21(3)n6 = {678} only -> r6c8 = 7, 6 locked for c9 and n6

19. r56c7 = [91] (h10(2)r56c7)
19a. -> r5c29 = 9 (step 15.) = [18/36](no 2)

20. "45" on 1234: 1 innie r4c3 + 3 = 2 outies r5c18
20a. -> no 3 in r5c8 (IOU)
20b. min. r5c18 = 8 -> min. r4c3 = 5

21. 3 in n6 only in r4: locked for r4
21a. r4c4 = 1 -> r3c45 = {48}: 4 locked for n2
21b. r3c6 = 2 -> r4c56 = 11 = {47/56}(no 8)

22. r4c3 = 8 (hsingle r4)

23. "45" on n7: 1 innie r7c1 = 1 outie r8c4 = {2356}(no 1,8,4)

24. 18(5)r5c2 = {12456} only (no 3) -> r5c2 = 1, r7c1 = 6, -> r5c9 = 8 (h9(2)r5c29) and r8c4 = 6 (IODn7=0)

25. r89c6 = 10 = {19} only: both locked for c6 and n8

pretty straightforward from here.
Cheers
Ed


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