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 Post subject: Assassin 365
PostPosted: Mon Dec 10, 2018 7:26 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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a365.JPG
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This is X-puzzle. 1-9 cannot repeat on either diagonal
code:
3x3:d:k:5376:5376:5121:5121:5121:5121:1282:3331:3331:5376:5376:3844:3589:2310:5127:1282:2568:2568:2569:3844:3844:3589:2310:5127:5127:2568:4618:2569:1035:1035:3589:2828:4877:4877:4877:4618:11278:11278:11278:11278:2828:2319:2319:4368:4618:5649:11278:0000:0000:0000:0000:0000:4368:4368:5649:11278:3092:0000:2837:4630:4630:2071:2071:5649:11278:3092:3092:2837:4630:4632:4114:2071:5649:11278:3091:3091:3091:4632:4632:4114:4114:
solution:
Code:
+-------+-------+-------+
| 3 8 5 | 4 9 2 | 1 7 6 |
| 1 9 7 | 6 8 5 | 4 3 2 |
| 4 2 6 | 3 1 7 | 8 5 9 |
+-------+-------+-------+
| 6 1 3 | 5 7 9 | 2 8 4 |
| 9 7 2 | 8 4 6 | 3 1 5 |
| 5 4 8 | 2 3 1 | 6 9 7 |
+-------+-------+-------+
| 2 6 1 | 9 5 8 | 7 4 3 |
| 8 5 4 | 7 6 3 | 9 2 1 |
| 7 3 9 | 1 2 4 | 5 6 8 |
+-------+-------+-------+


Two key steps for me but pretty easy ones compared to many other recent puzzles. Hope its not too easy for you guys! I made logical errors the first couple of solves so felt harder than it is. Some interesting features to this pattern but I couldn't find a puzzle which used them. SudokuSolver gives it 1.90 and JSudoku has a very tough time too.

Cheers
Ed


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 Post subject: Re: Assassin 365
PostPosted: Sun Dec 16, 2018 10:52 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Thanks Ed. I too had to start over on this one. But went OK the second time through. My WT has two short elimination chains which made it a lot easier for me. (Look for the word "Trying"). Will be interested to see how you avoid them! :)
Assassin 365 WT:
1. Innies n3 = r3c79 = +17(2) = {89}
-> 13(2)n3 = {67}
-> 5 in n3 in 10(3)

2. Innies c89 = r4c8 = 8

3. 4(2)r4 = {13}
Innies r1234 = r3c9 + r4c59 = +20(3)
-> Trying r3c79 = [98]
puts r4c59 = +12(2) = {57}
and puts r4c14 = +10(2) = {46}
which leaves no place for 9 in r4

-> r3c79 = [89]
-> r9c9 = 8
-> Innies n9 = r789c7 = +21(3) = {579}
-> r89c8 = {26}
-> 8(3)n9 = {134}
Also 13(2)n1 = [76]

4. Innies n1 = r1c3 + r3c1 = +9(2)
-> Since 6 already in r1 -> r3c1 cannot be 3.
-> 10(2)c1 only from [19] or {46}

5. r456c7 from {236} or {146}
-> r4c67 only from [92], [74], [56]. I.e., r4c6 from (579).
r23c6 from {57} or [93]
-> 9 in c6 in r24c6.

6! 9 in r4 in one of:
a) r4c1 -> r4c14 = [92]
b) r4c5 -> r45c5 = [92]
c) r4c6 -> r4c67 = [92]
Trying r23c6 = [93]
-> c) is eliminated
It puts 6 in n2 in r23c4 -> a) is eliminated since 14(3) cannot contain both 6 and 2
Also puts 7 in n2 in 9(2) = {27} -> b) is eliminated
-> r23c6 cannot be [93]
-> r23c6 = {57}
-> r4c67 = [92]
-> 10(2)c1 = {46}

7. Also -> 5(2)n3 = {14}
-> r2c9 = 2 and r23c8 = {35}
Also r56c7 = {36}
-> 9(2)r5 = {36}
Also 17(3)n6 = [197]
-> r45c9 = {45}
-> 8(3)n9 = [4{13}]
-> Outies r1234 = r5c59 = +9(2) = {45}

8. -> r5c1234 = {2789}
-> r6789c2 = {3456}
-> 4(2)n4 = [13]
-> 3 in n1 in r12c1
-> Innies n1 -> r3c1 cannot be 6
-> 10(2)c1 = [46]
-> r1c3 = 5
-> Remaining Outie n2 -> r4c4 = 5
-> r45c9 = [45]
-> 11(2)n5 = [74]

9. Also r23c4 = +9(2)
-> r1c456 = {249}
-> r1c127 = [381]
-> r2c12 = [19]
-> Remaining Innie c1 -> r5c1 = 9

10. HS 1 on D\ -> r6c6 = 1
-> Remaining Innies c6789 = r1c6 + r6c7 = +8(2)
Can only be r1c6 = 2, r6c7 = 6
-> 9(2)r5 = [63]
-> r789c6 = {348}
Given r789c7 = {579}
-> r789c6 - [{38}4] and r789c7 = [7{59}]

11. r1c45 = [49]
-> 11(2)n8 = {56}
-> 9(2)n2 = [81]
-> r23c4 = {36}
-> r6c5 = 3, r9c5 = 2
-> r89c8 = [26]
15(3)n1 = [726]
etc.

Will be traveling over the holidays so Merry Christmas to all and see you in the New Year!


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 Post subject: Re: Assassin 365
PostPosted: Mon Dec 17, 2018 9:45 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Merry Christmas to you too! I'll still post my next A on the 20th and have it waiting as a nice NY pressie for you.

I used to look for numbers that were in just two places in a row or column. But after wellbeback's a350 WT step 6 I realised I was limiting myself too much (see also Andrew's step 13 in his a361 WT). Made a big difference for me with a365 (step 6). Thanks to Andrew for some typos
a365 WT:
Preliminaries
Cage 4(2) n4 - cells ={13}
Cage 5(2) n3 - cells only uses 1234
Cage 13(2) n3 - cells do not use 123
Cage 9(2) n2 - cells do not use 9
Cage 9(2) n56 - cells do not use 9
Cage 10(2) n14 - cells do not use 5
Cage 11(2) n8 - cells do not use 1
Cage 11(2) n5 - cells do not use 1
Cage 8(3) n9 - cells do not use 6789
Cage 10(3) n3 - cells do not use 89
Cage 20(3) n23 - cells do not use 12
Cage 19(3) n56 - cells do not use 1
Cage 44(8) n457 - cells ={23456789}

No routine clean-up unless stated.
1. 4(2)n4 = {13} only: both locked for r4 and n4

2. "45" on c89: 1 innie r4c8 = 8
2a. no 3 in r5c5

3. "45" on n3: 2 innies r3c79 = 17 = {89} only: both locked for r3 and n3
3a. no 2 in r4c1

4. 13(2)n3 = {67} only: both locked for r1 and n3

5. "45" on r1234: 2 outies r5c59 = 9 (no 9)
5a. no 2 in r4c5

The ! step. Consider 2 in r4
6. "45" on n12: 3 outies r3c7 + r4c14 = 19
6a.-> 2 in r4c4 must have 8 in r3c7 -> no 9 in r3c7
6b. or 2 in r4 in sp11(2)r4c67 = {29} -> no 9 in r3c7 through D/
6c. or 2 in r4 in r4c9 in 18(3) = [927] only
6d. -> no 9 in r3c7
6e. -> r3c7 = 8 (Placed for D/), r3c9 = 9
6f. r3c7 = 8 -> r23c6 = 12 = {57}/[93](no 4,6, no 3 in r2c6)

7. 8(3)n9 = {125/134} = 3 or 5
7a. must have 1: 1 locked for n9

8. r9c9 = 8 (hsingle n9) (Placed for D\)
8a. -> r89c8 = 8 but {35} blocked by 8(3) (step 7)
8b. r89c8 = {26} only: both locked for n9 and c8
8c. r1c89 = [76] (6 placed for D/)

9. 8(3)n9 = {134} only: 3,4 locked for n9

10. naked triple {579} in r789c7: all locked for c7
10a. no 1,2,4 in r5c6

11. sp11(2)r4c6 = [92/74/56]
11a. r23c6 = [39]/{57} and r4c6 = (579) -> 9 locked for c6

12. "45" on n9: 1 innie r7c7 - 3 = 1 outie r9c6 = [52/74/96]
12a. -> r9c6 = (246)

13. "45" on n9: 3 outies r789c6 = 15 and must have 2,4,6 for r9c6
13a. but {258} as {58}[2] only blocked by 5 in r7c7 (step 12.)
13b. = {168/267/348/456} = one of 1/2/4

Next key step
14. hidden killer triple 1,2,4 in c6 since r789c6 can only have one of them
14a. -> r16c6 from {124} only

15. 3 in D\ only in n1: 3 locked for n1
15a. 10(2)r3c1 = [19]/{46}(no 2,7)

16. "45" on n1: 2 innies r1c3 + r3c1 = 9 = [81/54]

17. 20(4)r1c3 = {1289/2459/3458} = 3 or 9 (no eliminations)

18. sp12(2)r23c6: [39] blocked by 20(3)r1c3
18a. = {57} only: both locked for n2 and c6

cracked
19. r4c6 = 9, Placed for D/, r4c7 = 2 (cage sum), 5(2)n3 = {14} only, both locked for n3 and c7

20. r45c9 = 9 = {45} only: both locked for c9 and n6, r56c7 = {36}: 3 locked for n6, r78c9 = {13}: both locked for c9 and r7c8 = 4

21. 9(2)r5c6 = {36} only: both locked for r5

22. h9(2)r5c59 = {45} only: both locked for r5

23. r5c1234 = {2789} only -> none of those in r6789c2
23a. 9 locked for r5 and n4
23b. r56c8 = [19], r26c9 = [27]

24. r6789c2 = {3456} only: all locked for c2, 3 locked for n7
24a. r4c23 = [13]

25. hidden single 3 in n1 -> r1c1 = 3

26. 20(4)r1c3 = {1289/2459}: must have both 2 & 9: both locked for r1 and n2
26a. r1c23 = [85], r3c1 = 4 (h9(2)), r4c1 = 6
26b. r1c12 = 11 -> r2c12 = 10 = [19] only, 9 placed for D\
26c. 20(4)r1c3 = [5]{249}, 4 locked for r1 and n2

27. hidden single 1 in D\ -> r6c6 = 1

28. "45" on c6789: 2 innies r1c6 + r6c7 = 8 = [26] only

on from there.


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 Post subject: Re: Assassin 365
PostPosted: Tue Dec 25, 2018 2:41 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Merry Christmas :santa: and a Happy New Year to all Assassins!

I'm still out of practice at Assassin level puzzles after those two trips, plus being busy on other things in recent weeks.

My solving path was fairly similar to Ed's until I failed to spot his final key step and had to do a lot of hard work to get the same result. Loved Ed's step 6; my step 8 for the same result was one of my better steps.

Thanks Ed for pointing out my omissions in step 13d.
Here is my walkthrough for Assassin 365:
Prelims

a) R12C7 = {14/23}
b) R1C89 = {49/58/67}, no 1,2,3
c) R23C5 = {18/27/36/45}, no 9
d) R34C1 = {19/28/37/46}, no 5
e) R4C23 = {13}
f) R45C5 = {29/38/47/56}, no 1
g) R5C67 = {18/27/36/45}, no 9
h) R78C5 = {29/38/47/56}, no 1
i) 20(3) cage at R2C6 = {389/479/569/578}, no 1,2
j) 10(3) cage at R2C8 = {127/136/145/235}, no 8,9
k) 19(3) cage at R4C6 = {289/379/469/478/568}, no 1
l) 8(3) cage at R7C8 = {125/134}
m) 44(8) cage at R5C1 = {23456789}, no 1

Steps Resulting From Prelims and Initial Placement
1a. Naked pair {13} in R4C23, locked for R4 and N4, clean-up: no 7,9 in R3C1, no 8 in R5C5
1b. 8(3) cage at R7C8 = {125/134}, 1 locked for N9
1c. 45 rule on C89 1 innie R4C8 = 8 -> R4C67 = 11 = {29/47/56}, clean-up: no 5 in R1C9, no 2 in R3C1, no 3 in R5C5, no 1 in R5C6

2a. 10(3) cage at R2C8 = {145/235} (cannot be {127/136} which clash with R12C7), no 6,7
2b. Naked quint {12345} in R12C7 + 10(3) cage, locked for N3, clean-up: no 8,9 in R1C89
2c. Naked pair {67} in R1C89, locked for R1 and N3
2d. Naked pair {89} in R3C79, locked for R3, clean-up: no 1 in R2C5, no 2 in R4C1
2e. Max R3C23 = 13 -> min R2C3 = 2
2f. Max R3C67 = 16 -> min R2C6 = 4

3a. 45 rule on R1234 2 outies R5C59 = 9 = {27/45}/[63], no 9 in R5C5, no 1,6,9 in R5C9, clean-up: no 2 in R4C5
3b. 18(3) cage at R3C9 = {279/369/378/459/468} (cannot be {567} which doesn’t contain one of 8,9 for R3C9)
3c. 8,9 must be in R3C9 -> no 9 in R4C9
3d. 1 in R5 only in R5C78, locked for N6

4. 45 rule on N9 3 innies R789C7 = 21 = {579/678} (cannot be {489} which clashes with R3C7), no 2,3,4, 7 locked for C7 and N9, clean-up: no 4 in R4C6 (step 1c), no 2 in R5C6
4a. Killer pair 8,9 in R3C7 and R789C7, locked for C7, clean-up: no 2 in R4C6 (step 1c)
4b. Min R89C7 = 11 -> max R9C6 = 7
4c. Max R89C7 = 16 (cannot be {89} which clashes with R3C7) -> min R9C6 = 2
4d. 16(3) cage at R8C8 = {259/268/349} (cannot be {358} which clashes with R789C7)
4e. 8 of {268} must be in R9C9 -> no 6 in R9C9
4f. 9 in N6 only in 17(3) cage at R5C8 = {179/269/359}, no 4
4g. 1 of {179} must be in R5C8 -> no 7 in R5C8

5. 45 rule on N1 2 innies R1C3 + R3C1 = 9 = [36/54/81], R1C3 = {358}, R3C1 = {146}, clean-up: no 7 in R4C1

6. 45 rule on C6789 3(1+2) innies R1C6 + R6C67 = 9
6a. Min R6C7 = 2 -> max R16C6 = 7, no 7,8,9 in R16C6

7. 45 rule on N2 3(1+1+1) outies R1C3 + R3C7 + R4C4 = 18
7a. Min R1C3 + R3C7 = 11 -> max R4C4 = 7

8. 45 rule on R123 2 outies R4C14 = 1 innie R3C9 + 2
8a. R3C9 = {89} -> R4C14 = 10,11 = {46}/[47/65/92]
8b. R4C67 (step 1c) = [56/65/74/92]
8c. Consider combinations for R4C14
R4C14 = {46}, locked for R4 => R4C67 = [92], 9 placed for D/ => R3C79 = [89]
or R4C14 = [47/65/92] = 11 => R3C9 = 9
-> R3C9 = 9, R4C4 = {257}
[Not the cleanest of forcing chains but the best I could see at this stage; hope Ed or wellbeback found something better to make these placements.]
8d. R3C7 = 8, placed for D/ -> R23C6 = 12 = [57/75/93], no 4,6, no 3 in R2C6
8e. R9C9 = 8 (hidden single in C9), placed for D\
8f. R789C7 (step 4) = {579} (only remaining combination), 5 locked for C7 and N9, 9 also locked for N9, clean-up: no 6 in R4C6 (step 1c), no 4 in R5C6
8g. 8(3) cage at R7C8 = {134} (only remaining combination), locked for N9
8h. Naked pair {26} in R89C8, locked for C8 -> R1C89 = [76], 6 placed for D/, clean-up: no 5 in R4C5
8i. R3C9 = 9 -> R45C9 = 9 = {27/45}, no 3
8j. 10(3) cage at R2C8 (step 2a) = {145/235}
8k. 2 of {235} must be in R2C9 -> no 3 in R2C9
8l. 17(3) cage at R5C8 (step 4f) = {179/359}, no 2
8m. R89C7 = {579} = 12,14,16 -> R9C6 = {246}
[Or 18(3) cage must contain at least one even number -> R9C6 = {246}]

9. R23C6 (step 8d) = [57/75/93], R4C6 = {579} -> combined half cage R234C6 = {57}9/[935/937], 9 locked for C6

10. 12(3) cage at R9C3 = {129/147/156/237/345} (cannot be {246} which clashes with R9C6)
10a. Killer triple 2,4,6 in 12(3) cage, R9C6 and R9C8, locked for R9

11. R23C6 (step 8d) = [57/75/93], R234C6 (step 9) = {57}9/[935/937]
11a. 45 rule on N9 3 outies R789C6 = 15 = {168/267/348/456} (cannot be {357} because R9C6 only contains 2,4,6, cannot be {258} = {58}2 which clashes with 18(3) cage at R5C6 = {58}5)
11b. Consider combinations for R789C6
R789C6 = {168}, locked for C6 => R5C6 = {357} => naked quad 3,5,7,9 in R2345C6, locked for C6
or R789C6 = {267/456}, locked for C6 => R23C6 = [93]
or R789C6 = {348}, locked for C6
-> no 3 in R16C6
[Similarly]
11c. Consider combinations for R789C6
R789C6 = {168}, locked for C6 => R5C6 = {357} => naked quad 3,5,7,9 in R2345C6, locked for C6
or R789C6 = {267}, locked for C6 => R234C6 = [935]
or R789C6 = {348}, locked for C6 => R23C6 = {57}, locked for C6
or R789C6 = {456}, locked for C6
-> no 5 in R16C6
11d. Consider combinations for R789C6
R789C6 = {168/267/456}, locked for C6
or R789C6 = {348}, locked for C6 => R23C6 = {57}, locked for C6 => R5C6 = 6
-> no 6 in R6C6
11e. 3 on D\ only in R1C1 + R2C2 + R3C3, locked for N1, clean-up: no 6 in R3C1 (step 5), no 4 in R4C1, no 7 in R4C4 (step 8c)
11f. R1C6 + R6C67 = 9 (step 6), max R16C6 = 6 -> min R6C7 = 3
11g. Max R34C4 = 12 -> min R2C4 = 2
[I ought to have spotted Ed’s hidden killer triple 1,2,4 in R16C6 and R789C6 for C6, R789C6 only contains one of 1,2,4 -> R6C6 = {124}. Much simpler than my forcing chains!]

12. R5C59 = 9 (step 3a), R45C9 (step 8i) = {27/45}
12a. Consider placements for 1 in N6
1 in R5C67 = [81]
or in 17(3) cage at R5C8 = [179] => R45C9 = {45} => R5C59 = {45}, locked for R5
-> no 5 in R5C6, no 4 in R5C7

13. 15(3) cage at R2C3 = {168/249/267/357/456} (cannot be {159/348} which clash with R1C3 + R3C1, cannot be {258} which clashes with R1C3)
13a. Consider placement for 9 in C6
R2C6 = 9
or R4C6 = 9 => R4C1 = 6 => R3C1 = 4 => 15(3) cage cannot be {249}
-> 15(3) cage = {168/267/357/456}, no 9
13b. Consider combinations for R23C6 = {57}/[93]
R23C6 = {57} => caged X-wing for 5 in R23C6 and 10(3) cage at R2C8, no other 5 in R23
or R23C6 = [93]
-> 15(3) cage = {168/267/456} (cannot be {357}), no 3, 6 locked for N1
13c. 3,9 in N1 only in 21(4) cage at R1C1 = {1389/2379/3459}
13d. 45 rule on R1 3 outies R2C127 = 14 = {149/239/347} (cannot be {158} = [851] which clashes with R1C3, cannot be {248/257} because the two digits in 21(4) cage which aren’t 3 or 9 don’t total 9), no 5,8 in R2C1, no 5 in R2C2
13e. 45 rule on C1 3 innies R125C1 = 13
13f. 45 rule on R1 3 innies R1C127 = 12 = {129/138/345}
[Ed spotted that this leaves 20(4) cage at R1C3 with one of 3,9 in N2, which cracks the puzzle immediately!]
13g. R1C127 = {129} must be [291] (cannot be because [192/912] because 8 of {1389} can only be in R1C2, cannot be [921] because 21(4) cage = [9273] would make R125C1 more than 13), no 9 in R1C1, no 2 in R1C7, clean-up: no 3 in R2C7
13h. 8,9 of {129/138} must be in R1C2 -> no 1,2 in R1C2
13i. R1C127 = {138} must be [381] (cannot be [183] because 21(4) cage = [1893] and cannot make R125C1 total 13 because no 3 in R5C1), no 1 in R1C1
13j. R1C127 = {345} must be [354] (because 21(4) cage = {45}[93] would make R125C1 total more than 13
13k. So the remaining permutations for R1C127 are [291/381/354] -> R1C1 = {23}, R1C2 = {589}, R1C7 = {14}, clean-up: no 2 in R2C7
13l. Naked pair {14} in R12C7, locked for C7 and N3, clean-up: no 7 in R4C6 (step 1c), no 8 in R5C6
13m. Naked pair {35} in R23C8, locked for C8 and N3 -> R2C9 = 2, R56C8 = [19], R6C9 = 7 (cage sum), R4C5 = 7 (hidden single in R4) -> R5C5 = 4, placed for both diagonals, R45C9 = [45], clean-up: no 5 in R2C5, no 2,5 in R3C5, no 2 in R5C7
13n. Naked pair {36} in R5C67, locked for R5
13o. R4C7 = 2 (hidden single in N6) -> R4C6 = 9 (cage sum), placed for D/, R4C1 = 6 -> R3C1 = 4, R1C3 = 5 (step 5), R4C4 = 5, placed for D\, clean-up: no 3 in R3C6 (step 8d)
[Getting easier now; routine clean-ups omitted from here.]
13p. Naked pair {57} in R23C6, locked for C6, 7 also locked for N2
13q. R4C4 = 5 -> R23C4 = 9 = {36}/[81]
13r. R1C456 = {249} (hidden triple in N2), locked for R1 -> R1C12 = [38], R12C7 = [14]
13s. R12C3 = [38] = 11 -> R2C12 = 10 = {19}, 1 locked for N1
13t. R125C1 = 13, R1C1 = 3 -> R25C1 = 10 = [19], R2C2 = 9, placed for D\ -> R7C7 = 7, placed for D\
13u. R7C7 = 7 -> R78C6 = 11 = {38}, locked for C6 and N8 -> R5C67 = [63]
13v. R89C7 = {59} -> R9C6 = 4 (cage sum)
13w. R1C456 = [492], R78C5 = {56}, locked for C5 and N8, R23C5 = [81], R9C5 = 2, R9C8 = 6, R8C8 = 2, placed for D\, R6C56 = [31], R6C4 = 2, placed for D/, R5C4 = 8
13x. R23C3 = [76], R23C6 = [57], R2C8 = 3, placed for D/
13y. R7C3 = 1 -> R8C34 = 11 = [47]

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

Rating Comment:
I'll rate my walkthrough for A365 at Hard 1.5. I used forcing chains, more than really necessary, plus two-dimensional permutation analysis for my final breakthrough.


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