Assassin 59

Our weekly <a href="http://www.sudocue.net/weeklykiller.php">Killer Sudokus</a> should not be taken too lightly. Don't turn your back on them.
CathyW
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Assassin 59

Post by CathyW »

Strangely, I found this one easier than 58.

1. 10(4) r123c6 + r1c7 = {1234}
-> r1c45 see all cells of 10(4) thus min 5 -> r2c5 <> 9

2. 18(5) N4 must have 1,2; no 9 -> r3c12 of 12(3) <> 9

3. Outies N5: r37c5 = 10 (no 5)

4. Outies r12: r3c46 = 6 = [51]/{24}

5. Outies r89: r7c46 = 16 = {79} not elsewhere in N8/r7
-> r3c5 <> 1,3; r6c3 <> 3,5; r6c7 <> 1,3; r89c5 <> 1

6. Innies r34: r4c19 = 11 (no 1)

7. Innies r67: r6c19 = 9 -> r6c9 <> 9

8. Innies r2: r2c456 = 12 -> r2c4 = (1.7)

9. Innies r8: r8c456 = 15 = {168/258/348/456}. Analysis: r8c46 <> 3

10. Outies - Innies N1: r4c23 - r1c3 = 10
r1c3 max 7, r4c23 min 11, max 17 -> r4c2 <> 1

11. O-I N7: r6c23 - r9c3 = 8 -> r6c2 <> 8
r9c3 = (1.9), r6c23 min 9, max 17

12. Innies N4: r46c23 = 27 -> r6c23 min 10, max 16 -> r9c3 <> 1,9

13. O-I N3: r4c78 - r1c7 = 7 -> r4c8 <> 7
r1c7 = (1234), r4c78 min 8, max 11

14. O-I N9: r6c78 - r9c7 = 4 -> r6c8 <> 4
r9c3 = (1.9), r6c78 min 5, max 13

15. Innies N6: r46c78 = 16
Since r4c78 is min 8 -> r6c78 is max 8 (no 8,9) -> r7c7 <> 1,2

16. O-I N2: r3c5 - r1c37 = 4
-> r3c5 = (789), r1c37 = (1234)
-> r7c5 = (123)

17. O-I N8: r9c37 - r7c5 = 8

18. 1 locked to r4567c5 -> r6c46 <> 1

19. Outies c12: r258c3 = 11 (no 9)

20. Innies c123: r19c3 = 9 = [18/27/36/45]

21. Outies c89: r258c7 = 20 (no 1,2)

22. Innies c789: r19c7 = 5 = {14/23}

23. 25(4) r789c6 + r9c7 = 7{68}4 / 9{68}2 / 9{58}3
-> r9c7 <> 1 -> r1c7 <> 4
-> r89c6 = {58/68} 8 not elsewhere in N8/c6

24. Killer Pair r89c6 and 8(2) r89c5 -> r89c4 <> 5,6
-> 4 required for 10(4) thus locked to r123c6 not elsewhere in N2/c6 -> r3c6 <> 2, r1c5 <> 9
-> 4 locked to r89c4 not elsewhere in c4
-> KP 13(2) and 8(2) in c5 -> r456c5 <> 5,6

25. Split 6(2) r3c46:
If [24] -> r1c7 = 2 -> CONFLICT: No place for 2 in N1
-> r3c4 = 5, r3c6 = 1
-> r1c7 = (23) -> r9c7 = (23) not elsewhere in c7
-> r12c5 = {67} -> r1c4, r3c5 = (89)
-> r89c5 = {35} not elsewhere in N8/c5
-> 10(2) r67c7 = {46} not elsewhere in c7
-> 10(2) r34c7 = [91]
-> r3c5 = 8, r7c5 = 2, r1c4 = 9 .

Straightforward combinations and singles from here

:)
Last edited by CathyW on Thu Aug 02, 2007 8:37 am, edited 3 times in total.
Jean-Christophe
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Post by Jean-Christophe »

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mhparker
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Post by mhparker »

Aaarrrggghhhh! Just spent a few hours writing up a solution to the V1, only to find out that Cathy cracked the puzzle the same way in her walkthrough, thus making mine unnecessary! (Looking for an emoticon for "frustrated" here, but couldn't find one!).

BTW, I thought this was an excellent V1 (thanks Ruud), requiring some quite advanced techniques. The series of moves leading to the first placement was really neat. I much prefer puzzles (like this one) that involve a strategy, rather that those that simply represent a "war of attrition" (in other words, just scraping away at candidates here and there without any real plan). It will be interesting to see how others solve this one.
Cheers,
Mike
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Post by CathyW »

I agree with Mike about enjoyability of puzzles where there's a strategy. Will be interesting to see how Para solves this one. My solution path certainly involved several outies - innies and killer pairs.

Great minds evidently do think alike! Amazing that we solved the same way. :lol:
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Post by Andrew »

mhparker wrote:Aaarrrggghhhh! Just spent a few hours writing up a solution to the V1, only to find out that Cathy cracked the puzzle the same way in her walkthrough, thus making mine unnecessary!

BTW, I thought this was an excellent V1 (thanks Ruud), requiring some quite advanced techniques. The series of moves leading to the first placement was really neat. I much prefer puzzles (like this one) that involve a strategy, rather that those that simply represent a "war of attrition" (in other words, just scraping away at candidates here and there without any real plan). It will be interesting to see how others solve this one.
Mike

After the first 12 Assassins, which were ones that I caught up on after I first saw a link to this website, I've always written a walkthrough even if it wasn't posted on this forum. I find it very useful to concentrate my thoughts. I write it as I work through the puzzle. At one time they were entirely in the order that I made the moves, apart from any necessary corrections. These days if I see a move that I should have seen earlier I usually find where it should have been and put it in there. I think that makes the walkthrough more readable.

I haven't yet finished V1 so I can't comment on it. However I agree that ones with a strategy are preferable to those that are a "war of attrition". Most Assassins have sufficient flexibility of solving paths that they can have both.

Keep up the good work Ruud, those others who produce variants and everyone who posts walkthroughs.
sudokuEd
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Post by sudokuEd »

mhparker wrote: a "war of attrition"
Pretty close to the way this puzzle went for me. Loved it. Missed the neat way that Cathy and Mike used so made it much more difficult. Still couldn't find their short-cut doing it a second time after reading
Cathy wrote:Strangely, I found this one easier than 58.
description.

Instead, had to use a contradiction move to finally unlock it: the {1469} combo in the 20(4)r7c4 forces the 25(4)r7c6 = {4678}, but this means 2 6's in r9 -> no {1469} in 20(4). Fortunately this meant staying out of the snake-pit in n5.

Haven't tried Jean-Christophe's V1.5 yet but will later this week.

Cheers
Ed
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Post by Jean-Christophe »

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Para
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Post by Para »

CathyW wrote: Will be interesting to see how Para solves this one.
Well here it is.

Walk-through Assassin 59

1. 10(4) at R1C6 = {1234}

2. R1C12, R34C7, R67C3 and R9C89 = {19/28/37/46}: no 5

3. R1C89 and R67C4 = {39/48/57}: no 1,2,6

4. R12C5, R34C3 and R9C12 = {49/56/67): no 1,2,3

5. R89C5 = {17/26/35}: no 4,8,9

6. 20(3) at R2C1 = {389/479/569/578}: no 1,2

7. 18(5) at R4C1 = {12348/12357/12456}: no 9; {12} locked for N4

8. 45 on R89: 2 outies: R7C46 = 16 = {79} -->> locked for R7 and N8
8a. Clean up: R6C3: no 3,5; R6C7: no 1,3; R89C5: no 1

9. 45 on R12: 2 outies: R3C46 = 6 = {24}/[51]: R3C4 = {245}; R3C6 = {124}

10. 45 on N2: 2 outies and 1 innie: R3C5 = R1C37 + 4: Min R3C5 = 7; Max R1C37 = 5 -->> R3C5 = {789}; R1C3 = {1234}

11. Killer Quad {1234} in R1C12 + R1C367 -->> locked for R1
11a. R1C89 = {57} -->> locked for R1 and N3
11b. Clean up: R1C12: no 3, R2C5: no 6,8,9; R4C7: no 3

12. 45 on N5: 2 outies: R37C5 = 10 = [73/82/91]: R7C7 = {123}

13. 45 on C123: 2 innies: R19C3 = 9 = [18/27/36/45]: R9C3 = {5678}

14. 45 on C789: 2 innies: R19C7 = 5 = {14/23}: R9C7 = {1234}

15. 25(4) = {2689/3589/4678}(other combo’s blocked, can only have one of {79})
15a. R9C7 = {234}; R89C6 = {58/68} -->> 8 locked for C6 and N8
15b. Killer Pair {56} in R89C5 and R89C6 -->> locked for N8
15c. 4 in N8 locked for C4
15d. Clean up: R1C7: no 4; R3C6: no 2

16. 4 in 10(4) at R1C6 locked for C6 and N2
16a. Clean up: R1C5: no 9
16b. Killer Pair {56} in R12C5 and R89C5 -->> locked for C5

Addition suggested by Mike:
17x. 13(2) at R9C1 = {49/58/67} = {4|5|6..}
17y. 15(3) at R8C1 = {159/168/249/258/267/348/357} = {1|2|3..): {456} blocked by 13(2) at R9C1

17. Distribution {123} in N7: 15(3) at R8C1 can only have one so R7C123 needs 2 of {123} -->> Killer Triple {123} in R7C123 + R7C5 -->> locked for R7
17a. Clean up: R6C7: no 7,8,9
17b. Killer Pair {24} in R19C7 + R67C7 -->> locked for C7
17c. Clean up: R34C7: no 6,8

18. 14(3) at R6C8 needs 2 of {4568} in R7C89: 14(3) = [1]{58}/[2]{48}/[3]{56} -->> R6C8 = {123}

19. 45 on N36: 3 innies: R1C7 + R6C78 = 9 = [162/243/261/342]: R6C7: no 2

20. R67C7 = {46} -->> locked for C7
20a. R1C19 = {23} -->> locked for C7
20b. R34C7 = {19} -->> locked for C7
20c. R2C7 = 8
20d. 13(3) at R2C7 = 8{14}(8{23} blocked by R1C7): R2C89 = {14} -->> locked for R2 and N3
20e. R34C7 = [91]

21. 45 on N2: R3C5 = R1C37 + 4 -->> R3C5 = {78} -->> R1C37 = 3 or 4 = [12/13] -->> R1C3 = 1
21a. R9C3 = 8; R3C6 = 1(hidden); R3C4 = 5; R12C5 = [67]; R3C5 = 8; R7C5 = 2
21b. R12C4 = [93]; R12C6 = [42]; R1C7 = 3; R7C46 = [79]; R9C7 = 2
21c. R8C6 = 8(hidden); R9C6 = 6; R6C78 = [42](step 19); R7C7 = 6; R3C89 = [62]
21d. R4C8 = 9; R34C3 = [76](last possible combo); R67C3 = [93]; R2C3 = 5
21e. R4C4 = 2; R4C56 = [47](last possible combo); R5C5 = 9(hidden); R6C5 = 1(hidden)

22. 45 on N1: 1 outie: R4C2 = 5

23. 45 on N4: 1 innie: R6C2 = 7
23a. R6C12 = [51](last remaining combo in 13(3) cage); R5C1 = 1(hidden)
23b. R6C46 = [85](last remaining combo in 16(4) cage); R5C46 = [63]; R6C19 = [36]
23c. R4C19 = [83]; R3C12 = [43]; R1C12 = [28]; R9C12 = [94](last possible combo)
23d. R2C12 = [69]; R5C23 = [24]; R8C123 = [762]; R89C4 = [41]; R9C89 = [37]
23e. R1C89 = [75]; R89C5 = [35]; R8C789 = [519]; R2C89 = [41]; R7C89 = [84]; R5C789 = [758]

And we are done.

greetings

Para
Last edited by Para on Wed Jul 25, 2007 11:17 pm, edited 3 times in total.
mhparker
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Post by mhparker »

Andrew,
Andrew wrote:I've always written a walkthrough even if it wasn't posted on this forum. I find it very useful to concentrate my thoughts. I write it as I work through the puzzle. At one time they were entirely in the order that I made the moves, apart from any necessary corrections. These days if I see a move that I should have seen earlier I usually find where it should have been and put it in there. I think that makes the walkthrough more readable.
Unfortunately, my walkthroughs involve a lot of work these days. Maybe I'm becoming too much of a perfectionist. I also type in my steps as I solve the puzzle, and - like you do now - I don't simply publish the steps as I originally made them. Instead, I try to take the ideas and moves used and piece them together into a more coherent whole, maybe changing their order. At the same time, I also look out for anything I might have missed the first time through.

I use software for saving the state of the game after each step so that, if someone refers to a step in the walkthrough later, I don't have to start from scratch to confirm his or her observations. I also use software to check for both missed cleanups and naked subsets, the latter often indicating a missed "locked for..." clause earlier on. If I find anything I missed, I have to unwind a few steps and make the correction, making sure that these new eliminations don't invalidate the subsequent steps (e.g., by producing hidden singles, etc.).

It is a painstaking, laborious (but sometimes rewarding and satisfying) process that takes me several hours to complete.

However, I'm not by any means saying that this is the way everyone else should do it. On the contrary, such an approach would probably put others off writing walkthroughs for good, which would be counter-productive!

BTW, this is one reason (apart from the team factor) why I like tag solutions, namely because I can avoid most of these more mundane aspects and just concentrate on being creative.
Andrew wrote:Keep up the good work Ruud, those others who produce variants and everyone who posts walkthroughs.
I agree with this wholeheartedly on all counts. Firstly, every serious walkthrough is potentially valuable, even if written by a relative novice, containing several mistakes and/or overlooked moves. Everyone has to start somewhere. If any newbie is reading this, feel free to post the way you solved a puzzle, and don't feel you have to be a masochist like I obviously am! :)

Secondly, the contribution made by Ruud to the Sudoku community is absolutely enormous (indeed, almost immeasurable). Sure, SumoCue has become a bit long in the tooth now, but what would be the benefit of a perfect piece of software if it meant (due to time constraints) sacrificing the superb quality of this site?
Cheers,
Mike
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Post by CathyW »

I think Para had basically the same strategy making use of the Outies-Innies and killer combinations on the V1. Solving path must be relatively narrow.

In contrast to Mike, my walkthroughs will probably continue to be as I do the puzzles - rarely have the extra time (or to be honest the inclination) to do a puzzle two or three times to modify order of steps.

I'm stuck on the V1.5 so hopefully someone can continue - I currently have 39 steps and no placement. Looks like lots of combo crunching required!

1. Outies N5: r37c5 = 10 = {19/28/37/46}-> r456c5 = 11 (no 9)

2. Outies r12: r3c46 = 12 = [39]/{48/57}

3. Outies r89: r7c46 = 11 = {29/38/47/56}

4. Innies r2: r2c456 = 10 (no 8,9) -> r1c5 <> 3,4

5. Innies r8: r8c456 = 9 (no 7,8,9)
-> r9c5 <> 3,4,5
r8c456 = {135/234} 3 not elsewhere in r8/N8 -> r3c5 <> 7, r7c46 <> 8

6. Innies r34: r4c19 = 13 = {49/58/67}

7. Innies r67: r6c19 = 12 = {39/48/57}

8. Outies – Innies N1: r4c23 – r1c3 = 8
-> r4c23 min 9, max 16

9. O – I N7: r6c23 – r9c3 = 4
-> r6c23 min 5, max 13

10. Innies N4: r46c23 = 17
-> r4c23 max 12 -> r1c3 = (1234)
-> r6c23 max 8 -> r9c3 = (1234)

11. Outies c12: r258c3 = 17

12. Innies c3: r19c3 = 5 = {14/23}

13. O-I N3: r4c78 – r1c7 = 8 -> r4c8 <> 8
-> r4c78 min 9, max 16

14. O-I N9: r6c78 – r9c7 = 2 -> r6c8 <> 2
-> r6c78 min 3, max 11

15. Innies N6: r46c78 = 18
-> r4c78 max 15, r6c78 max 9

16. Outies c89: r258c7 = 20 (no 1,2)

17. Innies c7: r19c7 = 8 = {17/26/35}

18. O-I N2: r3c5 – r1c37 = 1
r1c37 min 3 -> r3c5 = (4689) -> r7c5 = (1246)

19. O-I N8: r9c37 – r7c5 = 4

20. Innies c1234: r456c4 = 19 (no 1)

21. Innies c6789: r456c6 = 15

22. In c5 there are two 12(2)s one of which must be {57} since if one is {39} and the other {48} there would be no options for the split 10(2) in r37c5.
-> split 11(3) r456c5 <> 5,7 -> options: {128/146/236}(Must have at least one of 24/26/68)
-> blocks {568}for split 19(3) r456c4
-> blocks {168/249/267} for split 15(3) r456c6

23. 1 locked to r4567c5 -> r6c6 <> 1

24. 2 locked to r4567c5 -> r6c46 <> 2

25. (Should have seen this earlier) 8 locked to r9c456 -> r9c89 <> 2,8

26. 23(4) r6c456 + r7c5: max from r67c5 = 8+6 = 14 -> r6c46 <> 3

27. Innies N14: r1c3 + r6c23 = 9 = {117/126/135/234} -> r6c2 <> 7

28. Innies N69: r4c78 + r9c7 = 16

29. Innies N36: r1c7 + r6c78 = 10 = {127/136/145/235/334} -> r6c8 <> 8

30. Innies N47: r4c23 + r9c3 = 13

31. Outies N25: r1c37 + r7c5 = 9 = {117/126/135/144/225/234}

32. Outies N58: r3c5 + r9c37 = 14

33. 17(3) r8c123 = {179/269/278/467} ({458} blocked by 6(2) r9c12) -> no 5 in 17(3)

34. Split 19(3) = {289/379/469/478} (no 5)
-> 5 locked to r456c6, not elsewhere in c6
-> r3c4 <> 7
-> r7c4 <> 6
-> split 15(3) = 5{19/28/37/46}

35. Innies N1: r1c3 + r3c123 = 15

36. Innies N3: r1c7 + r3c789 = 17

37. Innies N7: r7c123 + r9c3 = 22 (must have 3 -> no 1) -> r1c3 <> 4

38. Innies N9: r7c789 + r9c7 = 16

39. 19(3) r8c789 = {289/469/478/568} (at least one of 4,8)
-> split 16(4) in N9 not {1348}

tbc …

Incidentally, I ran this through JSudoku (latest version) and used the Deduce one move feature after I got stuck. Many cells are shaded with lots of colours to indicate various split combinations. It does eventually reach a solution using complex xy chains! :?

Beyond me so over to you guys! :)
Last edited by CathyW on Thu Aug 02, 2007 8:39 am, edited 2 times in total.
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Post by Jean-Christophe »

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mhparker
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Post by mhparker »

Jean-Christophe wrote:Here is my walkthrough for V1.5
Couldn't resist the temptation of posting a solution to your own puzzle, then?! :wink:

Good job you're not the anchor man in the Belgian 4 x 100m relay team, otherwise you'd be running from the starting line each time instead of taking over from the previous guy! :-)

Just joking, of course, but there's a somewhat more serious side to all this. Unlike what I've seen so far on other forums, many of the puzzles on this site are played as a team. One therefore has to think twice before posting individual walkthroughs when such a team ("tag") solution is in progress, even more so when one is the creator of the puzzle!

Basically, the unwritten "rule" is that as soon as one forum member publishes a partially-completed puzzle, asking for assistance, he or she is implicitly starting off the tag. The idea then is for someone else to pick up the tag and "run" with it, where possible, rather than interrupting the tag by "going it alone".

In this case, the impact was fairly small, but on the Texas Jigsaw forum (TJK31) a tag solution was interrupted (by another one!) much later on. I hope that those involved on the original tag didn't take offence at this in any way.
Last edited by mhparker on Wed Jul 18, 2007 7:27 pm, edited 1 time in total.
Cheers,
Mike
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Post by sudokuEd »

mhparker wrote:In this case, the impact was fairly small
A fair summary for this (Assassin) forum Mike. I like the way you put into words the unwritten 'rules'. Though would hate for them to become rules.

I don't feel real comfortable when you talk about the other forums though - (including TJK).

Hope I can add to Cathy's start to V1.5 solution on the weekend.

Cheers
Ed
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Post by mhparker »

sudokuEd wrote:I don't feel real comfortable when you talk about the other forums though - (including TJK).
In retrospect, the situation is apparently not as bad as I feared. Therefore, I've re-worded my original post accordingly.
sudokuEd wrote:Hope I can add to Cathy's start to V1.5 solution on the weekend.
I did this one a few days ago (around the time I was complaining about how much effort it was writing the walkthroughs!). My walkthrough is very similar to JC's. I've already seen a second way to crack it though.

By the way, going back to Cathy's assertion regarding the V1:
Cathy wrote:I think Para had basically the same strategy...Solving path must be relatively narrow.
Well, I've now checked all walkthroughs I have to hand. Mine (unpublished) and Cathy's were - as I said before - similar, but both Para's and JC's walkthroughs were totally different (although very effective and recommended reading!). In the meantime, I've discovered yet another way in, so there are at least 4 different ways to "peel this onion". In other words, it's now clear that the solving path for this one isn't narrow at all.
Cheers,
Mike
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Post by CathyW »

I stand corrected :oops:

Didn't really have time to go through Para's walkthrough properly - just noticed he had used a few outies-innies which were crucial to the way I did it.
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