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Ed
 Post subject: Assassin 235
PostPosted: Thu Mar 29, 2012 10:21 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
An enjoyable killer. I used one of my favourite techniques. Still enjoy killers after 6 years!

A235

NOTE: This is a Killer-X ie 1-9 cannot repeat on the diagonals

Image

code: paste into solver:
3x3:d:k:4608:1537:1537:4610:4610:4610:2307:2307:3076:4608:9221:9221:9221:9221:9221:9221:9221:3076:4608:3846:4359:2824:2824:2824:2313:3338:3076:3851:3846:3846:4359:3852:2313:3338:3338:4621:3851:5902:3343:3343:3852:1808:1808:3338:4621:3851:5902:5902:785:3852:2322:4883:4883:4621:3092:5902:785:5397:5397:5397:2322:4883:3862:3092:8727:8727:8727:8727:8727:8727:8727:3862:3092:3096:3096:3609:3609:3609:2842:2842:3862:
solution: 
+-------+-------+-------+
| 4 1 5 | 3 6 9 | 7 2 8 |
| 8 2 3 | 7 4 5 | 9 6 1 |
| 6 7 9 | 1 8 2 | 5 4 3 |
+-------+-------+-------+
| 9 6 2 | 8 3 4 | 1 5 7 |
| 5 8 4 | 9 7 1 | 6 3 2 |
| 1 3 7 | 2 5 6 | 4 8 9 |
+-------+-------+-------+
| 2 5 1 | 4 9 8 | 3 7 6 |
| 7 9 6 | 5 2 3 | 8 1 4 |
| 3 4 8 | 6 1 7 | 2 9 5 |
+-------+-------+-------+
Cheerio
Ed
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Andrew
 Post subject: Re: Assassin 235
PostPosted: Sat Mar 31, 2012 5:12 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks Ed for an enjoyable killer; once I'd spotted my breakthrough step. ;)

Ed wrote:
Still enjoy killers after 6 years!
Where did you first come across killers? I've just checked and Ruud didn't post his first Assassin until June 2006. I first came across killers on 1st April 2006 on http://www.sudoku.org.uk, so a few days short of 6 years, but didn't manage to solve one until the third one appeared two days later. I wondered at first whether they were an April Fool's Day joke; I'm happy that they weren't! :D I'm still "hooked" on killers and still trying to solve harder ones which I gave up on at the time or didn't attempt then.

Ed wrote ""I used one of my favourite techniques":
Not sure which of my steps that might apply to. Step 18 seems like it might be a sort of IOE. Step 19 is a block using a column and crossovers; I'm guessing it's that one, since it's my breakthrough step. The note before step 27 is also a candidate, since I know that Ed likes locking, blocking and locking-out cages.

Rating Comment:
I'll rate my walkthrough for A235 at least Hard 1.5. The SS score is 1.55 but I think my block using a column and crossovers is more like the top end of the 1.5 range.

Here is my walkthrough for A235:
Prelims

a) R1C23 = {15/24}
b) R1C78 = {18/27/36/45}, no 9
c) 17(2) cage at R3C3 = {89}
d) 9(2) cage at R3C7 = {18/27/36/45}
e) R5C34 = {49/58/67}, no 1,2,3
f) R5C67 = {16/25/34}, no 7,8,9
g) 3(2) cage at R6C4 = {12}
h) 9(2) cage at R6C6 = {18/27/36/45}, no 9
i) R9C23 = {39/48/57}, no 1,2,6
j) R9C78 = {29/38/47/56}, no 1
k) 11(3) cage at R3C4 = {128/137/146/236/245}, no 9
l) 19(3) cage at R6C7 = {289/379/469/478/568}, no 1
m) 21(3) cage at R7C4 = {489/579/678}, no 1,2,3
n) 13(4) cage at R3C8 = {1237/1246/1345}, no 8,9

Steps resulting from Prelims
1. R1C78 = {18/27/36} (cannot be {45} which clash with R1C23), no 4,5
1a. Naked pair {89} in 17(2) cage at R3C3, locked for D\, CPE no 8,9 in R3C4 + R4C3, clean-up: no 1 in 9(2) cage at R6C6
1b. Naked pair {12} in 3(2) cage at R6C4, locked for D/, CPE no 1,2 in R6C3, clean-up: no 7,8 in 9(2) cage at R3C7

2. 45 rule on R1 2 innies R1C19 = 12 = [39/48/57/75], no 1,2,6 in R1C1, no 3,4,6 in R1C9

3. 45 rule on R2 2 innies R2C19 = 9 = {18/27/36/45}, no 9

4. 45 rule on R12 2 outies R3C19 = 9 = {18/27/36/45}, no 9
4a. 9 in R3 only in R3C23, locked for N1

5. Min R1C9 = 5 -> max R23C9 = 7, no 7,8 in R23C9, clean-up: no 1,2 in R23C1 (steps 3 and 4)

6. 45 rule on R89 2 outies R7C19 = 8 = {17/26/35}, no 4,8,9

7. 45 rule on R8 2 innies R8C19 = 11 = {29/38/47/56}, no 1

8. 45 rule on R9 2 innies R9C19 = 8 = [35/53/62/71], no 4,8,9 in R9C1, no 4,6,7 in R9C9

9. Min R79C1 = 4 -> max R8C1 = 8, clean-up: no 2 in R8C9 (step 7)

10. 9 in C1 only in 15(3) cage at R4C1, locked for N4, clean-up: no 4 in R5C4
10a. Hidden killer pair 1,2 in 15(3) cage and 12(3) cage at R7C1 for C1, neither cage can contain both of 1,2 -> each must contain one of 1,2 -> 15(3) cage = {159/249}, no 3,6,7,8
10b. 12(3) cage = {138/156/237/246} (cannot be {147} which clashes with 15(4) cage)
10c. 1 of {156} must be in R7C1, 6 of {246} must be in R9C1 -> no 5,6 in R7C1, clean-up: no 2,3 in R7C9 (step 6)
10d. Killer pair 1,2 in 12(3) cage and R7C3, locked for N7
10e. 1 in N7 only in R7C13, locked for R7, clean-up: no 7 in R7C1 (step 6)

11. 45 rule on N8 3 innies R8C456 = 10 = {127/136/145/235}, no 8,9

12. 9 in R3 only in R3C23
12a. 45 rule on R123 4 innies R3C2378 = 25 = {2689/3589/3679/4579} (cannot be {1789} because R3C7 only contains 3,4,5,6), no 1
12b. Max R3C78 = 13 -> min R3C23 = 12, no 2 in R3C2

13. 13(4) cage at R3C8 = {1237/1246/1345}, 1 locked for N6, clean-up: no 6 in R5C6

14. 18(3) cage at R1C1 = {378/468/567}
14a. 4 of {468} must be in R1C1 -> no 4 in R23C1, clean-up: no 5 in R23C9 (steps 3 and 4)

15. 3(2) cage at R6C4 and 9(2) cage at R6C6 cannot both contain 2 in R6C46 -> R7C37 cannot be [17]
15a. 45 rule on R789 4 innies R7C2378 = 16 = {1249/1267/1348/1357/2356} (cannot be {1258/2347} which clash with R7C19, cannot be {1456} which clashes with 21(3) cage at R7C4)
15b. 1 of {1267/1357} must be in R7C3 -> no 7 in R7C7, clean-up: no 2 in R6C6

16. 18(3) cage at R4C9 = {279/369/378/459/468} (cannot be {567} which clashes with R7C9)
16a. Hidden killer pair 8,9 in 18(3) cage and R6C78 for N6, 18(3) cage contains one of 8,9 -> R6C78 must contain one of 8,9
16b. Max R6C78 = 16 -> min R7C8 = 3

17. 15(3) cage at R7C9 = {159/168/258/267/357/456} (cannot be {249/348} because R7C9 only contains 5,6,7)
17a. 12(3) cage at R1C9 = {129/138/147/237/345} (cannot be {246} because R1C9 only contains 5,7,8,9, cannot be {156} which clashes with 15(3) cage), no 6, clean-up: no 3 in R23C1 (steps 3 and 4)

18. 8 in N4 only in 15(3) cage at R3C2, 23(4) cage at R5C2 or R5C34, 8 in R7C2 “see” all 8s in N4 except for R5C3
18a. 45 rule on N4 3(2+1) outies R37C2 + R5C4 = 21
18b. 8 in 15(3) cage or 23(4) cage within N5 => no 8 in R7C2
or R5C34 = [85] => R37C2 = 16 = {79}
-> no 8 in R7C2
[This step doesn’t seem to work for 8 in R3C2 or for 3,6,7 because R4C3 contains 3,6,7 and R6C3 contains 3,6,7,8.]

[Just spotted interactions between R1C19 and R9C19 for 3,5,7 …
With hindsight it was available after step 8, but if I’d spotted it then I wouldn’t have found the interesting steps 15 and 18.]
19. 18(3) cage at R1C1 (step 14) = {378/468} (cannot be {567} which clashes with R9C19 using C1, {57} pair in R1 and both diagonals), no 5, 8 locked for C1 and N1 -> R3C3 = 9, R4C4 = 8, clean-up: no 7 in R1C9 (step 2), no 4 in R23C9 (steps 3 and 4), no 5 in R5C3, no 3 in R8C9 (step 7), no 3 in R9C2
19a. 3,4 of {378/468} only in R1C1 -> R1C1 = {34}, clean-up: no 5 in R1C9 (step 2)

20. Killer pair 8,9 in R1C9 and 18(3) cage at R4C9, locked for C9, clean-up: no 2,3 in R8C1 (step 7)

21. 12(3) cage at R1C9 (step 17a) = {129/138}, 1 locked for C9 and N3, clean-up: no 8 in R1C78, no 7 in R9C1 (step 8)

22. 18(3) cage at R1C4 = {279/369/468} (cannot be {189} which clashes with R1C9, cannot be {378/567} which clash with R1C78, cannot be {459} which clashes with R1C23), no 1,5

23. 1 in N1 only in R1C23 = {15}, locked for N1
23a. 2 in N1 only in R2C23, locked for R2, clean-up: no 7 in R2C1 (step 3)
23b. R8C8 = 1 (hidden single on D\)
23c. R4C7 = 1 (hidden single in N6)

24. R7C13 = {12} (hidden pair in N7), locked for R7, clean-up: no 7 in R6C6

25. Killer pair 3,4 in R1C1 and 9(2) cage at R6C6, locked for D\, clean-up: no 5 in R9C1 (step 8)

26. 7 on D\ only in R2C2 + R5C5, CPE no 7 in R2C58 + R58C2

[Earlier I’d spotted
1 in R9 only in R9C19 = [71] or in 14(3) cage at R9C4 -> 14(3) cage can only contain 7 if it also contains 1 -> 14(3) cage = {149/167/239/248/356} (cannot be {257/347} which contain 7 but not 1, locking-out cages, cannot be {158} which clashes with R8C456)
However this is no longer needed …]

27. R8C456 (step 11) = {235} (only remaining combination), locked for R8 and N8, clean-up: no 6 in R8C19 (step 7)
27a. Naked pair {47} in R8C19, locked for R8
27b. 21(3) cage at R7C4 = {489/678}, 8 locked for R7 and N8

28. 12(3) cage (step 10b) = {237/246} -> R7C1 = 2, R7C3 = 1, R6C4 = 2, R1C23 = [15], clean-up: no 5 in R5C7, no 7 in R9C2

29. 15(3) cage at R4C1 (step 10a) = {159} (only remaining combination, or hidden triple for C1), locked for N4

30. 9(2) cage at R3C7 = {45} (only remaining combination, cannot be {36} which clashes with R9C1 using D/), locked for D/, CPE no 4,5 in R3C6

31. R5C5 = 7 (hidden single on D/), placed for D\, clean-up: no 6 in R5C34
31a. 15(3) cage at R4C5 = {357} (only remaining combination), 3,5 locked for C5 and N5 -> R4C6 = 4, R3C7 = 5, R5C4 = 9, R5C3 = 4, R5C6 = 1, R5C7 = 6, R6C6 = 6, R7C7 = 3 (both placed for D\), R1C1 = 4, R2C2 = 2, placed for D\, R9C9 = 5, R8C19 = [74], R9C1 = 3 (step 8), R7C9 = 6 (step 6), 15(3) cage at R4C1 = [951], R9C3 = 8, R8C23 = [96], R9C2 = 4, R7C2 = 5, R8C5 = 2, R8C7 = 8
[I’ve now stopped doing routine clean-ups.]

32. 21(3) cage at R7C4 (step 27b) = {489} (only remaining combination) -> R7C4 = 4, R7C56 = {89}, locked for R7 and N8 -> R7C8 = 7, R9C6 = 7

33. R4C23 = [62] (hidden pair in N4), R3C2 = 7 (cage sum), R2C3 = 3, R2C9 = 1, R6C3 = 7

34. 4 in N6 only in 19(3) cage at R6C7, which also contains 7 = {478} (only remaining combination) -> R6C78 = [48]

35. R1C78 = {27} (only remaining combination, cannot be {36} because 3,6 only in R1C8) -> R1C78 = [72], R2C78 = [96], R23C1 = [86], R3C9 = 3 (step 4)

and the rest is naked singles without using diagonals.
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wellbeback
 Post subject: Re: Assassin 235
PostPosted: Sat Mar 31, 2012 8:18 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
Thanks again Ed. I went all over the place at first on this one not getting anywhere fast, but then I found a breakthrough step right near the beginning which opened it up for me.

I too would be interested to learn what is the "favorite technique" you used :).

Hidden Text:
1. Starting off

17/2@r3c3 = {89} - locked for D\
3/2@r6c4 = {12} - locked for D/

Innies r1: r1c19 = +12 (no 126)
Innies r2: r2c19 = +9 (no 9)
Outies r12: r3c19 = +9 (no 9)

Innies r9: r9c19 = +8 (no 489)
Innies r8: r8c19 = +11 (no 1)
Outies r89: r7c19 = +8 (no 489)

r9c1 from (3567) -> 12/3@r7c1 cannot contain a 9.
-> 9 in c1 in 15/3@r4c1.
-> 15/3@r4c1 = {159} | {249}


2. The Breakthrough

r19c19 all see each other.
r1c19 = +12, r9c19 = +8.

{57} in r1c19 would force r9c19 to be [62]
-> 18/3@r1c1 cannot be {567}!

(Alternatively - {567} in r123c1 leaves no solution for r9c19 = +8)


3. What follows

-> 18/3@r1c1 must contain an 8 (in r23c1). 18/3@r1c1 = {378} | {468}.
-> 17/2@r3c3 = [98]

Also since r2c19 and r3c19 both = +9 -> 1 must be in r23c9.
-> 1 in n9 must be in r8c78.
-> Innies n8 r8c456 = +10 must be {235}
-> Innies r8 r8c19 = +11 must be {47}
-> (47) locked in c1 in r1238.
-> 15/3@r4c1 = {159}
-> HS 2 in c1 r7c1 = 2
-> 3/2@r6c4 = [21]

Also Outies r89 -> r7c9 = 6
-> 14/3@r9c4 = {167}
-> Innies r9 -> r9c19 = [35]
-> 12/3@r7c1 = [273], 15/3@r7c9 = [645]

Also 12/2@r9c2 = {48}
-> r8c78 = [81]
-> r8c23 = [96]
-> r7c2 = 5
-> Outies n4 r3c2+r5c4+r7c2 = +21 -> r3c2 = 7, r5c4 = 9.

Also
18/3@r1c1 = [4{68}]
-> 12/3@r1c9 = [8{31}]
18/3@r4c9 = {279}

etc., etc.
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Frank
 Post subject: Re: Assassin 235
PostPosted: Tue Apr 03, 2012 5:31 am 
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Joined: Mon Apr 28, 2008 10:58 pm
Posts: 47
Location: Victoria, B.C., Canada
Nice puzzle Ed. I think this is the fastest I have solved an assassin.
Here is how I did it:
The top corners add to 12. The bottom corners add to 8. I worked with the bottom.
{1,7} is easy to dismiss. {2,6} is not much harder.
So the bottom corners are {3,5}.The top corners are [4,8]. And the puzzle just collapses from here.
Many thanx Ed. Great fun :)
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Ed
 Post subject: Re: Assassin 235
PostPosted: Fri Apr 06, 2012 1:45 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
I stumbled apon killers on djape's forum while looking for samurais. Can't remember exactly when. Elaine (one of our regular lurkers here) was already there as was Frank. Can't remember if HATMAN was. Can't check any dates. After that site got hacked by the Hungarian Hacker, djape has never put it back. So much lost.

Frank found a really quick way into this puzzle. Plugging in those permutations/combos does lead to quick contradictions or big progress (for the correct one). Wellbeback and Andrew found a really productive (huge!) block.

Andrew used two of my favourites in his WT that are not crucial to solving the puzzle (steps 15 & 18) Step 18 is actually a classic IOE technique by looking at it slightly differently to how Andrew expressed it.

Image

18. r7c2 sees all 8s in n4 except r5c3
18a. "45" on n4: 2 outies r37c2 - 8 = 1 innie r5c3
18b. -> when 1 innie = 1 outie, the IOD (8) must be in the third cell (8 in r3c2)-> no 8 in r7c2

Actually, my WT has three techniques I really enjoy (steps 9a, 10a & 11a). Maybe because they are all a. substeps, ie need to put a couple of things together for them to work. Andrew correctly guessed two of them! ;clapclap; :)

Worked out an interesting puzzle.

Happy Easter!

A235
19 steps:
1. 3(2)r6c4 = {12}: both locked for D/
1a. no 7,8,9 in 9(2)r3c7

2. 17(2)r3c3 = {89}: both locked for D\
2a. no 1,9 in 9(2)r6c6

3. "45" on r1: 2 innies r1c19 = 12 (no 1,2,6)
3a. no 3,4, in r1c9
3b. min r1c9 = 5 -> max. r23c9 = 7 (no 7,8,9)
3c. max. r1c1 = 7 -> min. r23c1 = 11 (no 1)

4. 11(3)n2: no 9
4a. 13(4)r3c8: no 8,9
4c. 9 in r3 only in n1: locked for n1

5. "45" on r2: 2 innies r2c19 = 9
5a. no 2 in r2c1

6. "45" on r12: 2 outies r3c19 = 9 (no 9)
6a. no 2 in r3c1

7. 18(3)r1c1 = {369/378/468/567}
7a. {468} must have 4 in r1c1 -> no 4 in r23c1
7b. no 5 in r23c9 (two h9(2) cages r23c19)

8. 6(2)n1 = {15/24}(no 3,6,7,8) = [4/5..]
8a. ->{45} blocked from 9(2)r1c7 (no 4,5,9)
8b. 9(2)r1c7 = {18/27/36} = [1/6/7..]

9. "45" on r1: 1 innie r1c1 = 2 outies r23c9
9a. but [7]{16} clashes with 9(2)r1c7 (step 8b)
9b. -> no 6 in r23c9 since max. r23c1 = 7
9c. no 3 in r23c1 (two h9(2) cages r23c19)

10. 12(3)n3 (redundant bit deleted)
10a. h12(2)r1c19 and two h9(2) cages r23c19: Difference of +3 -> r1c9 <> 3 more than either of r23c9 -> [7]{14} blocked (Blocking cages)
10b. 12(3) = {129/138/237/345} note: cannot have both {14} (important in a sec)

11. "45" on r1: 1 innie r1c1 = 2 outies r23c9
11a. = [7]{34}/[5]{23} only/[4]{13}/[3]{12}
11b. ie, must have 3 -> no 3 in common peers r1c78, r9c9
(Alternative way to see this: 12(3) = [3..] or {129}. If {129}, 9 must be in r1c9 -> 3 in r1c1; Combined half cage)
11c. no 6 in 9(2)r1c7

12. {129} blocked from 12(3)n3 by 9(2)r1c7 = [1/2..]
12a. 12(3) = {138/237/345}(no 9)
12b. no 3 in r1c1 (h12(2)r1c19)
12c. 12(3)n3 must have 3: 3 locked for n3 and c9
12d. no 6 in r4c6

13. 3 locked in r23c9 -> 6 locked in r23c1 (two h9(2) cages r23c19; Locking cages)
13a. 6 locked for n1 and c1
13b. 18(3)n1 = {468/567} = [4/7,4/5..]

14. 3, 6 & 9 in r1 only in 18(3) = {369}: all locked for n2

15. 12(3)n7: {129} blocked by r7c3
15a. {147/345} blocked by 18(3)n1 (step 13b)
15b. = {138/237}(no 4,5,9)
15c. must have 3 -> 3 locked for c1 and n7
15d. no 1,2,6,9 in 12(2)n7

16. "45" on r9: 2 innies r9c19 = 8 = [35/71]

17. "45" on r8: 2 innies r8c19 = 11 (no 1)
17a. no 8 in r8c1
17b. no 2,5,6,7 in r8c9

18. "45" on r89: 2 outies r7c19 = 8 (no 4,8,9)
18a. no 2 in r7c9

19. 12(3)n7 = {237} only: all locked for c1 and n7

cracked.
Cheerio
Ed
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HATMAN
 Post subject: Re: Assassin 235
PostPosted: Sat Apr 07, 2012 12:58 pm 
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
I was there Ed - I think third member after JC and udosuk. I did not post much to begin with as I was using my own site http://www.diceboard.co.uk. I keep it going but I have not posted to it in about four years. The puzzles there are KiMos, twin killers and disjoint groups.
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