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Assassin 69
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CathyW
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Joined: 31 Jan 2007
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PostPosted: Fri Sep 21, 2007 12:28 pm    Post subject: Assassin 69 Reply with quote

In contrast to today's Times Deadly (18 mins), A69 is proving extremely tricky! 17 steps so far, a few of which were not at all fruitful, and lunch break is over (can't extend it today!). Will persevere later.

Edit: Started over when I got home and have made much more progress. 14 steps and several placements. Not yet finished and am out tonight but should be able to post WT tomorrow. Smile


Last edited by CathyW on Fri Sep 21, 2007 4:48 pm; edited 1 time in total
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mhparker
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PostPosted: Fri Sep 21, 2007 2:58 pm    Post subject: Assassin 69 WT Reply with quote

Hi folks,

This was a nice puzzle. Thanks very much, Ruud! Very Happy

Edit: Checked through solution & simplified some steps. Rating probably around 1.25.

Edit: Modified WT to make UR-move optional.
Edit: Incorporated feedback from Andrew (Thanks, Andrew!)

Assassin 69 Walkthrough

1. 34(5)n12 = {46789}
1a. CPE: no 4,6,7,8,9 in r2c1

2. 6(2)n12 and 6(2)n9 = {15/24} = {(4/5)..}

3. 17(2)n23 = {89}, locked for r1
3a. -> {89} of 34(5)n12 (step 1) locked in r2c234
3b. -> no 8,9 elsewhere in r2

4. 22(5)n23 = {(15/24)367} ({89} unavailable)
4a. CPE: no 3,6,7 in r2c9
4b. 34(5)n12 and 22(5)n23 form grouped X-Wing on {67} in r12
4c. -> no 6,7 elsewhere in r12 (r12c5)
4d. max. r12c5 = 9
4e. -> min. r3c5 = 3 (no 1,2)

5. 8(2)n4 and 8(2)n7 = {17/26/35} (no 4,8,9)

6. 11(2)n6 and 11(2)n8 = {29/38/47/56}: no 1

7. 20(3)n5 and 20(3)n69 = {389/479/569/578} (no 1,2)

8. 11(3)n5 = {128/137/146/236/245} (no 9)

9. 13(2)n8 and 13(2)n9 = {49/58/67} (no 1,2,3)

10. 9(2)n7 = {18/27/36/45} (no 9)

11. Innies n8: r9c46 = 8(2) = {17/26/35} (no 4,8,9)
11a. 8(2)n7 and r9c46 require 2 of {567}
11b. -> {67} combo blocked for 13(2)n9
11c. -> 13(2)n9 = {49/58} = {(4/5)..}

12. 13(2)n9 (step 11c) and 6(2)n9 (step 2) form killer pair on {45} -> no 4,5 elsewhere in n9

13. Innie/outie (I/O) diff. r89: r8c46 = r7c5 + 15
13a. -> r7c5 = {12}, r8c46 = 16 or 17 = {79/89}
13b. -> 9 locked in r8c46 for r8 and n8
13c. cleanup: no 7,8 in r7c4; no 5,6,7,8 in r7c6

14. Innies r89: r8c456+r9c5 = 28(4) = {(47/56)89} (no 1,2,3) = {(5/7)..}

15. 3 in n8 locked in r9c46 (step 11) (-> {35}) or 11(2)n8 (-> {38})
15a. -> r9c46+r78c6 require {5/8}
15b. -> {58} combo for 13(2)n8 blocked
15c. -> 13(2)n8 = {49/67} = {(4/6)..}, {(4/7)..}

16. 13(3)n8 = {148/256} (no 7) = {(1/6)..}, {(4/6)..}
(Note: {157} blocked by r89 innies (step 14), {247} blocked by 13(2)n8 (step 15c))
16a. -> 13(2)n8 and 13(3)n8 form killer pair on {46} within n8
16b. -> no 4,6 elsewhere in n8
16c. cleanup: no 7 in r8c6, no 2 in r9c46 (step 11)

17. Naked pair (NP) on {89} at r18c6 -> no 8,9 elsewhere in c6

18. r7c5 blocks {129} combo for 12(3)n2
18a. -> no 9 in r3c5
18b. furthermore, 12(3)n2 must have 1 of {12} (hidden killer pair in c5 w/ 13(3)n8)
18c. -> {345} combo blocked
18d. also, {156/246} both blocked by 13(3)n8 (step 16)
18e. -> 12(3)n2 = {138/147/237} (no 5,6)
18f. max. r12c5 = 7
18g. -> no 3,4 in r3c5

19. 9 in c5 locked in n5 -> not elsewhere in n5
19a. -> 20(3)n5 (step 7) = {(38/47/56)9}

20. from step 17, {128} combo now unavailable for 11(3)n5
20a. -> cannot have both of {12}
20b. -> 11(3)n5 and 14(3)n5 form hidden killer pair on {12} in n5
20c. -> 14(3)n5 = {(1/2)..} = {158/167/248/257} (no 3)

21. 5 in r7 locked in n7 -> not elsewhere in n7
21a. -> 8(2)n7 = {17/26} (no 3) = {(1/6)..}, {(2/7)..}
21b. -> {27} combo for 9(2)n7 blocked
21c. -> 9(2)n7 = {18/36} (no 4) = {(1/6)..}
21b. -> 9(2)n7 and 8(2)n7 form killer pair on {16} -> no 1,6 elsewhere in n7

22. Hidden killer triple on {367} in n9, as follows:
22a. only places for {367} in n9 are r7c7, r7c89 and r89c7
22b. r7c89 cannot have 2 of {367} due to 20(3)n69 cage sum
22c. r89c7 cannot have 2 of {367} due to 13(3)n89 cage sum, and because 4 unavailable in r9c6
22d. -> r7c7, r7c89 and r89c7 must each contain exactly 1 of {367}
22e. -> r7c7 = {367}; no 8 in r89c7; no 3,7 in r9c6; no 3 in r9c7
(Note: possibilities for r9c6+r89c7 = [1][39]/[5]{17}/[5]{26})
22f. cleanup: no 1,5 in r9c4 (step 11)

23. 3 in r9 locked in 13(3)n78 = {3..} = {238} (only possible combo)
23a. -> r9c46 = [35] (step 11); r89c3 = {28}, locked for c3 and n7
23b. cleanup: no 4 in r1c4, no 8 in r9c89, no 1 in r8c12, no 6 in r9c12

24. r78c6 = [29]; r1c67 = [89]; r78c4 = [67]; r37c5 = [71]
24a. cleanup: no 4 in r12c5

25. Hidden single (HS) in r2 at r2c2 = 8

26. HS in r9 at r9c7 = 6
26a. -> r8c7 = 2
26b. cleanup: no 4 in r8c89

27. r89c3 = [82], r89c5 = [48]

28. NP on {49} at r9c89 -> no 9 in r7c89

29. Hidden triple (HT) in n7 at r7c123 = {459} (no 3,7)

30. NP on {23} at r12c5 -> no 2,3 elsewhere in c5 and n2
30a. cleanup: no 4 in r1c3

31. NP on {15} at r1c34 -> no 1,5 elsewhere in r1

32. Naked triple (NT) on {569} at r456c5 -> no 5,6 elsewhere in n5

33. 11(3)n5 = {137} (only remaining combo), locked for c6 and n5

34. NP on {46} at r23c6 -> no 4 elsewhere in n2

35. Naked single (NS) at r2c4 = 9

36. NT on {467} at r1c12+r2c3 -> no 4,6 elsewhere in n1

[Note: following step optional and may be skipped!]
37. Unique rectangle (UR) type I on {15} at r13c34
37a. -> no 1,5 in r3c3

38. I/O diff. n1: r4c1 = r13c3
[with step 37:]
38a. -> r3c3 = 3, r4c1 = {48}
[if step 37 skipped:]
38a. -> r3c3 = {135}, r4c1 = {468}

39. Innies r1234: r4c456 = 12(3) = [291/453] (no 6,7,8) = {(4/9)..}

40. Outies n1: r13c4+r4c123 = 25(5)
40a. r13c4 = {15} = 6(2) -> r4c123 = 19(3) = [8]([29]/{47})
(Note: [4]{69} (and [6]{49}, if step 37 skipped) blocked by r4c456 (step 39))
(Note: [8]{56} blocked because it would also require a 5 in r3c4 for 19(4) cage sum)
40b. -> r4c1 = 8
[with step 37:]
40c. -> r1c3 = 5 (step 38)
[if step 37 skipped:]
40c. -> r13c3 = [53] (step 38)

41. r13c4 = [15]
42. I/O diff. n7: r6c1 = r7c3 - 1
42a. -> r6c1 = 3, r7c3 = 4
42b. cleanup: no 5 in r5c12, no 7 in r4c2

43. r8c12 = [63]
43a. cleanup: no 2 in r5c2

44. HS in c1 at r1c1 = 4

45. HS in c1 at r7c1 = 5
45a. -> r7c2 = 9

46. HS in r3/c1/n1 at r3c1 = 9

47. HS in c2/n4 at r4c2 = 4
47a. -> r4c3 = 7 (cage sum)
47b. -> r4c456 = [291] (step 39)
47c. cleanup from step 47a: no 1 in r5c12

48. NS at r2c3 = 6
48a. -> r1c2 = 7

49. NT on {356} at r4c789 -> no 3,5,6 elsewhere in n6
49a. cleanup: no 8 in r5c89

50. I/O diff. n9: r6c9 = r7c7 + 2
50a. -> r6c9 = 9, r7c7 = 7

51. NS at r2c6 = 4
51a. -> split 18(4) at r1c89+r2c78 = {2367} (no 1,5) (step 4)
51b. -> r2c78 = [37]; r1c89 = {26}, locked for r1 and n3

Now it's all naked singles to end.
_________________
Cheers,
Mike


Last edited by mhparker on Thu Oct 04, 2007 2:12 pm; edited 3 times in total
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CathyW
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PostPosted: Fri Sep 21, 2007 10:22 pm    Post subject: Reply with quote

Couldn't resist finishing it off when I got home! I guess I must be addicted after all.

Quite a challenge - estimated rating 1.5. Not quite sure what Mike means by tackling it like a V2 - will have a look at his WT tomorrow.

Edit: Revised WT on page 2


Last edited by CathyW on Mon Oct 01, 2007 3:32 pm; edited 4 times in total
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gary w
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PostPosted: Sat Sep 22, 2007 1:26 pm    Post subject: assassin 69 Reply with quote

Hi all,

Very much enjoyed your wt Mike..haven't looked at Cathy's yet.
Just one step I didn't see easily...37 Why is 1/5 ruled out at r3c3 by forming a "unique rectangle"(what is that?)?That would force r2c78 to be {15} which can readily be seen leads to a contradiction but I don't see how the UR rules out this possibility per se.

Used my own hypotheticals to solve quite readily but I have to say I'm getting very unhappy myself with such routes having seen how it "should" be done.

Great puzzle.

Regards

Gary
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gary w
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PostPosted: Sat Sep 22, 2007 7:04 pm    Post subject: assassin 69 Reply with quote

Enjoyed going through Cathy's wt too.Solving path seemed very similar to Mike's..I guess this means the path is quite tightly constrained.My own method certainly concentrated on the same parts of the puzzle as well.I really appreciate the effort in putting the wts on..the compelling logic on going through them is impressive.Even though I don't understand what a UR is/implies.I think that's what stymied my own attempt at solving without resorting to hypotheticals...got to where Mike/Cathy reached but didn't see that r3c3 had to be a 3.

Regards

Gary

P.S Did the Times deadlys yesterday(quite a hard one by recent standards) ~ 45 mins,today ~ 30 mins.
Also solved the killersudokuonline no.89 in ~ 60 mins so this,too,is much easier than Ruud's dastardly concoctions.
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gary w
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PostPosted: Sat Sep 22, 2007 7:15 pm    Post subject: assassin 69 unique rectangle Reply with quote

Looked up URs on another site...of course!!..I'm really rather appalled that I didn't see the implications of the two 1/5 combos.I don't think it's come up in any other sudoku I've done but no excuses!.I'll know the next time.

Regards

Gary
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Andrew
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PostPosted: Sat Sep 22, 2007 7:44 pm    Post subject: Reply with quote

Gary

The use of URs is something that isn't universally acceptable.

It can lead to quicker solutions, which I suppose is why some people use them. Alternatively it could be because some people like using "advanced" techniques.

Others don't like the philosophy of relying on a puzzle having a unique solution, even though all Assassins and other puzzles posted on this forum will have a unique solution.

I think the regulars on this forum are probably split about 50/50 on this.

I try to avoid them whenever I can. For that reason I don't look for them and don't often spot them.
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CathyW
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PostPosted: Sat Sep 22, 2007 10:20 pm    Post subject: Reply with quote

You do have to be careful with use of URs in Killer Sudokus. They can only apply in situations involving two rows, two columns, two nonets AND two cages in non-diagonal puzzles.

Obviously I'm in the camp that feels use of URs is perfectly acceptable!

I took 20 mins on today's Times Deadly - I think dealing with Ruud's puzzles has helped me speed up.
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Afmob
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PostPosted: Sun Sep 23, 2007 9:04 am    Post subject: Reply with quote

Hi!

I think the best way to introduce myself is a walkthrough Smile . My first walkthrough had about 60 steps, so I cut out all the moves which led to
nothing. So some clean-up might be missing but it isn't essential for solving this assassin. The important moves are marked with a ! . I'm not sure if step 9 (previous step 15) is too much T&E, so I'm trying to solve it again without this step. But first, here is my walkthrough:

1. R12
a) 17(2) = {89} locked for R1
b) 34(5) = {46789} locked -> R2C1 <> 4,6,7,8,9
c) 8,9 locked in R2C234 for R2
d) 22(5): 3,6,7 locked (no 8,9 because of 1a, 1b) -> R2C9 <> 3,6,7
e) R12C5 <> 6,7 since 6,7 is locked in 34(5) and 22(5)

2. N8
a) Innies = 8(2) -> R9C46 = {17/26/35}

3. R89
a) Innies = 28(4) -> no 1,2,3 in R8C456+R9C5
b) 8,9 locked for N8 (in Innies)
c) R8C4 <> 4,5
d) Outies = 9(3) -> no 7 in R7C456
-> R8C4 <> 6
-> R8C6 <> 4
e) R7C56 <> 4,5,6 since outies of R89 would be
larger than 9 because of R7C7 (456)
f) R8C6 <> 5,6,7

4. C6
a) Naked pair in R18C6 (8,9) locked

5. R8
a) Killer pair (23,39) in 11(2) blocks following combinations
of 13(3): {139,238} -> 13(3) <> 9
b) 9 locked in R8C46

6. R9
a) 8(2) + Innies of N8 = 8(2) -> 13(2) <> 6,7 since both 8(2) would be {35}

7. N9
a) Killer pair (45) in 6(2) + 13(2) -> no 4,5 elsewhere
b) Killer pair (45) in 6(2) blocks (45) in 9(2) in N7 -> R8C12 <> 4,5
c) 20(3): R6C9 <> 6,7 since they would lead to a conflicting
cage sum
d) R6C9 <> 3 since R7C89 would be 89 and that's not possible
because of killer pair (89) of 13(2)
e) 13(3): R9C6 <> 6,7 since they need a 5 which is not possible in R89C7

8. N8
a) Innies = 8(2): R9C4 <> 1,2

9. N7 !
a) 13(3) <> 9 because of following conflict:
13(3) = {139} -> 8(2) = {26} -> 9(2) has no possible combination

10. R7
a) 9 locked in R7C123

11. N9
a) 20(3): R6C9 <> 4,8 would lead to conflicting cage sum
b) 20(3) <> 6 because R6C9 is the only position where
5 and 9 is possible
c) 8 locked in R7C89 for R7 and N9
d) 13(2) = {49} -> locked for N9 and R9
e) 6(2) = {15} -> locked for N9 and R8
f) 5 locked in 13(3) -> R9C6 = 5

12. N89
a) Innies = 8(2): R9C4 = 3
b) R89C7 = {26} locked for C7
c) R7C6 = 2, R8C6 = 9, R7C5 = 1
d) 13(2) = {67} -> R7C4 = 6, R8C4 = 7, R9C5 = 8, R8C5 = 4

13. R1
a) R1C6 = 8, R1C7 = 9

14. N2
a) R3C5 = 7 since 12(3) = {237}
b) R12C5 = {23} -> locked for N2 and C5

15. N5
a) 20(3) = {569} locked
b) 11(3) = {137} locked for N5 and C6
c) 14(3) = {248} locked for C4

16. N2
a) R2C4 = 9
b) 6(2) = {15} since R1C4 = (15) -> locked for R1

17. R7
a) R7C789 = {378} locked

18. N7
a) 3 locked in 9(2) = {36} locked for N7 and R8
b) 8(2) = {17} locked
c) R9C3 = 2, R8C3 = 8
d) 17(3) = 5{39/48} -> R6C1 = (38) because of R7C12 = 5(4/9)
e) 5 locked in R7C12

19. N9
a) R8C7 = 2, R9C7 = 6

20. R2
a) Hidden Single: R2C2 = 8

21. R1234
a) Innies = 12(3) -> R4C4 <> 8 (R4C5 is at least 5)
b) R4C6 <> 7 (same reason)
c) R4C5 <> 6 (conflict in cage sum)

22. N69 !
a) Innies = 14(3) -> R4C789 <> 8 because:
-> 248 blocked by R4C4 = (24)
-> 158 blocked by killer pair (15) of Innies 12(3)
of R1234 = {129/345}

23. R4
a) Hidden Single: R4C1 = 8
b) R6C1 = 3, R8C1 = 6, R8C2 = 3
c) Hidden Single: R3C3 = 3 @ N1

24. N7
a) 17(3) = {359} <> 4
b) Hidden Single: R7C3 = 4, R4C2 = 4 @ N4
c) R4C4 = 2

25. N1
a) 19(4) = {3457} -> no 1,6,9
b) R3C4 = 5, R4C3 = 7, R2C3 = 6, R1C2 = 7, R1C1 = 4
c) R9C2 = 1, R9C1 = 7, R1C4 = 1, R1C3 = 5, R2C6 = 4, R3C6 = 6

26. R456
a) Hidden Single: R4C9 = 6 @ R4, R1C8 = 6 @ N3
b) 8(2) = {26} -> R4C2 = 6, R4C1 = 2
c) R2C1 = 1, R3C1 = 9, R3C2 = 2, R7C1 = 5, R7C2 = 9, R6C2 = 5
d) R6C9 = 9, R6C5 = 6, R6C3 = 1, R5C3 = 9, R9C9 = 4, R9C8 = 9
e) R6C6 = 7, R5C5 = 5, R4C5 = 9
f) Hidden Single: R6C8 = 2 @ R6

27. N3
a) 22(5) = {23467} -> R1C9 = 2

28. Rest is clean-up and singles.

Edit: Had a look at Cathy's walkthrough and step 4 also eliminates 9 from R9C3 which is one of my two main steps.


Last edited by Afmob on Wed Oct 10, 2007 9:27 am; edited 7 times in total
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Para
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PostPosted: Sun Sep 23, 2007 7:26 pm    Post subject: Reply with quote

Afmob wrote:
The important moves are marked with a ! . I'm not sure if step 15 is too much T&E, so I'm trying to solve it again without this step. But first, here is my walkthrough:


Hi and welcome

I don't think step 15 is too much trial and error. I used the same step in my solving proces. But maybe other people feel differently about it. But i looked at it through Richards proposed Combined Cage move. I wrote my idea behind the step below.
Explanation in tiny text:
Combined cage of 9(2) at R8C1 + 8(2) at R9C1 needs either 1 in R89C12 or a 3 in R9C12, which eliminates the [193] combo in the 13(3) cage. But through this logic you can eliminate more options from the 13(3) cage, although the [193] is the most important one.

I'll see what the rest thinks about it. Otherwise shame on me Wink.

greetings

Para


Last edited by Para on Mon Nov 26, 2007 5:17 am; edited 1 time in total
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mhparker
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PostPosted: Sun Sep 23, 2007 9:41 pm    Post subject: Reply with quote

Hi folks,

Wow, a lot's been happening on the forum in the last couple of days, where I have been busy with other things!

Afmob wrote:
I think the best way to introduce myself is a walkthrough

Welcome to the forum! Nothing like jumping straight in at the deep end with a full WT on the first post!

gary w wrote:
Just one step I didn't see easily...37 Why is 1/5 ruled out at r3c3 by forming a "unique rectangle"(what is that?)?That would force r2c78 to be {15} which can readily be seen leads to a contradiction but I don't see how the UR rules out this possibility per se.

gary w wrote:
Looked up URs on another site...

Hi Gary, sorry I wasn't there to answer your question, but glad to see that you've managed to answer it yourself! BTW, if you have any questions like this in the future, you could always send me a personal message (PM). Also, don't forget that other people may be sending you PMs, too, so don't forget to check for this frequently (the status is displayed in the header area of any forum page). I've also got my profile set up to send me an e-mail when a new PM arrives, which further reduces the chance of me accidentally missing any.

Andrew wrote:
The use of URs is something that isn't universally acceptable.

It can lead to quicker solutions, which I suppose is why some people use them. Alternatively it could be because some people like using "advanced" techniques.

Sure, it's always nice to be able to use a good mix of techniques, which is why I used the UR move in this case. However, it wasn't necessary, and I've reworked my WT slightly to also show how to proceed without it. Indeed, as shown in the modified WT, the alternative without the UR move would have actually saved a step! However, because it was very easy to spot, I thought it would be interesting to mention (and use) it.
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Mike
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mhparker
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PostPosted: Tue Sep 25, 2007 8:50 am    Post subject: Reply with quote

Afmob wrote:
I'm not sure if step 15 is too much T&E, so I'm trying to solve it again without this step.

Para wrote:
I don't think step 15 is too much trial and error...I'll see what the rest thinks about it.

I can't speak for the others, but I certainly have no problems with this move:

I've used similar moves myself before. I don't regard it as T&E for two reasons, which I've stated before, but will repeat here for convenience:

Firstly, none of the links in the implication chain rely on side-effects (such as candidate elimination in peer cells) of any earlier links, even though some of the links are quite complicated in this case (because they are not based on a single candidate digit).

Secondly, the implication chain does not contain any branches. Therefore, the contradiction can be expressed as a loop. I refer to them as "complex Nice Loops". As Para said, it's not just the {139} combo that's being eliminated. For example, the same logic also eliminates the 1 from the 13(3) cage, by virtue of blocking both the {139} and {157} combos ({148} not possible because none of these digits are available in r9c4):

13(3)n78 = {1..} -> 9(2)n7 <> {18} => 9(2)n7 = {27/36} -> 8(2)n7 <> {26} => 8(2)n7 = {17/35} -> 13(3)n78 <> {1..} (contradiction)

Notes: In the above loop, I've used "=>" for strong links and "->" for weak links. There may be other contradiction chains that block other combos, too.
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Cheers,
Mike


Last edited by mhparker on Sat Oct 20, 2007 8:16 pm; edited 1 time in total
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Afmob
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PostPosted: Tue Sep 25, 2007 9:58 am    Post subject: Reply with quote

Thank you for the explanation. This step was probably the key move to solve this assassin and it was a bit different from my other steps so I wasn't sure whether it's just T&E or not.
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PostPosted: Tue Sep 25, 2007 10:36 pm    Post subject: Reply with quote

Hi all,

Looks like we're in need of more material:

Assassin 69 V1.5 (Est. rating: 1.5)



3x3::k:5888:5888:3074:3074:3332:2821:2821:6919:6919:5385:5888:5888:5888:3332:6919:6919:6919:3857:5385:5385:5908:5908:3332:4887:4887:3857:3857:5385:5908:5908:3614:4383:3616:4887:4887:3857:2852:2852:6438:3614:4383:3616:4650:3371:3371:2861:6438:6438:3614:4383:3616:4650:4650:5429:2861:2861:6438:1337:3898:2875:4650:5429:5429:2623:2623:4161:1337:3898:2875:4165:2630:2630:2376:2376:4161:4161:3898:4165:4165:1359:1359:

Enjoy!
_________________
Cheers,
Mike
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Para
Yokozuna
Yokozuna


Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

PostPosted: Wed Sep 26, 2007 9:40 am    Post subject: Reply with quote

Hi all

Here's the walk-through for Assassin 69V1.5. Again a typical Mike Killer.
The rating seems to fit. One really tricky move, which probably isn't necessary if i had seen the things i did after this step before this step.

Walk-through Assassin 69V1.5

1. R1C34 = {39/48/57}: no 1,2,6

2. R1C67, R5C12 and R78C6 = {29/38/47/56}: no 1

3. R5C89 = {49/58/67}: no 1,2,3

4. 11(3) at R6C1 = {128/137/146/236/245}: no 9

5. 21(3) at R6C9 = {489/579/678}: no 1,2,3

6. R78C4 and R9C89 = {14/23}: no 5,6,7,8,9

7. R8C12 and R8C89 = {19/28/37/46}: no 5

8. R9C12 = {18/27/36/45}: no 9

9. 45 on N8: 2 innies: R9C46 = 14 = {59/68} = {5|6..}: no 1,2,3,4,7
9a. R78C6 = {29/38/47}: {56} blocked by R9C46: no 5,6

10. 45 on R89: R7C456 = 7 = {124} -->> locked for R7 and N8
10a. R8C4 = 3; R7C456 = [214]; R8C6 = 7
10b. Clean up: R1C3: no 9; R1C7: no 4,7
10c. 15(3) at R7C5 = 1{59/68} = {5|8..},{6|9..},{5|6..}

11. 11(3) at R6C1 = [2]{36}/[1]{37} -->> R6C1 = {12}; R7C12 = {36/37} -->> 3 locked for R7 and N7
11a. Clean up: R9C12: no 6

12. 14(3) at R4C4 = {149/158/167}: 1 locked for C4 and N5

13. 14(3) at R4C6 = {239/356}={6|9..}: no 8; 3 locked for C6 and N5
13a. Clean up: R1C7: no 8

14. 17(3) at R4C5 = {278/467}: {269/458} blocked by 15(3) at R7C5: no 5,9; 7 locked for C5 and N5
14a. Clean up: R456C4: no 6(step 12)

15. 13(3) at R1C5 = {238/346}= {6|8..}: {256} blocked by 15(3) at R7C5: no 5,9

16. Hidden Pair: R89C5 = {59} -->> locked for N8

17. Naked Pair: R9C46 = {68} -->> locked for R9
17a. Clean up: R9C12: no 1

18. Killer Pair {24} in R9C12 + R9C89 -->> locked for R9

19. 45 on N9: 1 outie and 3 innies: R9C6 + 14 = R7C789
19a. R9C6 = 6 -->> R7C789 = {569/578}
19b. R9C6 = 8 -->> R7C789 = {589}
19c. 5 locked for R7 and N9

20. 16(3) at R8C7(R89C7 + R9C6] = {19}[6]/[17][8]: R89C7 = {19}/[17]: 1 locked for C7 and N9
20a. R9C89 = {23}(last combo) -->> locked for R9 and N9
20b. R8C89 = {46}(last combo) -->> locked for R8 and N9
20c. R9C12 = {45} (last combo) -->> locked for R9 and N7
20d. R89C5 = [59]

21. 16(3) at R8C4 = [178/916]: R89C3 = [17/91]: 1 locked for C3 and N7
21a. R8C12 = {28} (last combo) -->> locked for N7
21b. R5C4 = 1(hidden)

22. 9 in N7 locked for C3

23. 45 on R1234: 3 innies: R4C456 = [862/925]: [826/475] blocked by 17(3) at R4C5, [943/583] blocked by 14(3) at R4C4, [529] blocked by 14(3) at R4C6; 2 locked for R4 and N5
23a. Clean up: R6C4: no 8,9; R56C5: no 6; R56C6: no 5
Just noticed this, been there for a while. Kinda cool.

24. Killer XY-Wing: R4C4 = {89}; R9C4 = {68}; 14(3) at R4C6 = {6|9..}: eliminate 6 from Common Peers of R9C4 and 14(3): R9C6: no 6
24a. R9C46 = [68]; R89C3 = [91](last combo); R89C7 = [17]
24b. Clean up: R1C7: no 3

25. 21(3) at R6C9 = [4]{89}/[7]{59}: R6C9 = {47}; R7C89 = {59/89}: 9 locked for R7

26. 18(4) at R5C7 needs one of {58} in R7C7, canít have both {47} because of R6C9 or both {17} because R6C8 only cell with {17}
26a. 18(4) = {1359/1368/1458/2358/3456}: no 7; R6C8: no 9(only place for 1)

27. R5C89 = {4|5|7..},{4|7|8..}; R6C9 = {47} -->> 18(4) at R5C7 canít have both {45} or {48} in N6 -->> 18(4): {1458} blocked

28. 45 on N9: 1 innie and 1 outie: R6C9 + 1 = R7C7: R6C9 + R7C7 = [45/78]

29. 18(4) = {1359/1368/2358}: {3456} is blocked by step 28: no 4, R6C8: no 6(only place for 1; 3 locked for N6

30. 3 in R4 locked for N4
30a. Clean up: R5C12: no 8

31. R5C5 = {478}; R5C89 = {4|7|8..} -->> R5C12: {47} combo blocked: no 4,7
31a. R5C12 = {29/56} = {5|9..}

32. 25(4) at R5C3 = {928}[6]/{478}[6]/{468}[7]: {4579} blocked by R5C12: no 1,5; R6C2: no 2(only place for 9); 8 locked for N4
32a. {478}[6] blocked by R6C5 + R6C9(canít have 2 of {478} in R6C23) -->> 25(4) = {289}[6]/{468}[7]: R5C3 + R6C23: no 7

33. 7 in N4 locked for R4

34. 45 on N47: 3 innies: R4C123 = 14 = {347}(only option because both 3 and 7 locked within these cells) -->> locked for R4 and N4

35. 25(4) at R5C3 = {289}[6] -->> R7C3 = 6; R6C2 = 9(only place in cage); R56C3 = {28} -->> locked for C3 and N4
35a. R6C1 = 1

36. Naked Pair: R5C12 = {56} -->> locked for R5
36a. R5C89 = {49}(last combo) -->> locked for N6
36b. R6C9 = 7; R56C6 = [36]; R4C56 = [25]; R6C4 = 4; R4C4 = 9; R56C5 = [78]
36c. R56C3 = [82]; R5C7 = 2; R7C7 = 8(step 28); R4C7 = 6
36d. R1C67 = [29](last combo)
36e. Clean up: R1C3: no 3

37. 23(4) at R3C3 = {3578}(last combo): no 4; R3C4 = 8(only place in cage); R3C3 = 5(only place in cage)
37a. R4C1 = 4(hidden); R9C12 = [54]; R5C12 = [65]

38. R1C34 = [75](last combo)
38a. R12C4 = [57]; R4C23 = [73]; R7C12 = [73]; R2C3 = 4
38b. R3C7 = 4(hidden); R1C5 = 4(hidden); R3C8 = 7(hidden)

39. 19(4) at R3C6 = 46[18](last combo) -->> R3C6 = 1; R4C8 = 8
39a. R2C6 = 9; R4C9 = 1; R3C1 = 9(hidden)

40. 15(4) at R2C9 = 71[52](last combo) -->> R23C9 = [52]

And the rest is all singles.

.------.------.------.
| 3 1 7| 5 4 2| 9 6 8|
| 2 8 4| 7 6 9| 3 1 5|
| 9 6 5| 8 3 1| 4 7 2|
:------+------+------:
| 4 7 3| 9 2 5| 6 8 1|
| 6 5 8| 1 7 3| 2 9 4|
| 1 9 2| 4 8 6| 5 3 7|
:------+------+------:
| 7 3 6| 2 1 4| 8 5 9|
| 8 2 9| 3 5 7| 1 4 6|
| 5 4 1| 6 9 8| 7 2 3|
'------'------'------'

greetings

Para


Last edited by Para on Thu Nov 01, 2007 11:50 pm; edited 1 time in total
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