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Assassin 33

 
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rcbroughton
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PostPosted: Fri Jan 12, 2007 10:56 am    Post subject: Assassin 33 Reply with quote

I find with these type of arrangements you get a good kick into the solution looking at "lined up" cages.

Consider the 14(2) 17(3) and 11(2) in r5. 14(2) can only be {59} or {68} so we can't have {58} or {69} in the other cages on this row. This doesn't help us looking at them individually - but if you consider the combined cage 28(5), you get some limited options {13789} {15679} {34579} {34678} - break that back out to the 17(3) and the 11(2) you find that 17(3) cannot be {278} and 11(2) cannot be {56}.

That type of logic applied to a number of places in this puzzle and it falls quite quickly.
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PostPosted: Fri Jan 12, 2007 7:28 pm    Post subject: Re: Assassin 33 Reply with quote

rcbroughton wrote:
I find with these type of arrangements you get a good kick into the solution looking at "lined up" cages.

Consider the 14(2) 17(3) and 11(2) in r5. 14(2) can only be {59} or {68} so we can't have {58} or {69} in the other cages on this row. This doesn't help us looking at them individually - but if you consider the combined cage 28(5), you get some limited options {13789} {15679} {34579} {34678} - break that back out to the 17(3) and the 11(2) you find that 17(3) cannot be {278} and 11(2) cannot be {56}.

That type of logic applied to a number of places in this puzzle and it falls quite quickly.

Hmmmm..... I can't see any advantage to treating two cages as a single larger (combined) cage, as this simply reduces rather than adds to the amount of information you have to work with. -- Your example doesn't convince me because it's doing things the hard way. 17(3) cannot be {278} because 45 rule on R5 shows that R5C19 = 3 = {12}. 11(2) cannot be {56} because this would conflict with the 14(2) cage. Neither deduction requires unwieldy combination work.

That said, I'm still working on this one, & probably won't have a chance to make a serious try until the weekend.
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PostPosted: Sat Jan 13, 2007 1:46 am    Post subject: Reply with quote

Hm, well, I just solved it but failed to discover any really elegant way of doing so--just picked away at it bit by bit until it cracked. Ruud's comment with the puzzle implies that there are some techniques that can solve it more efficiently so I'd be interested to know of any better way of cracking it. Mostly ended up doing a lot of laborious combination work & plus-minus stuff.
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PostPosted: Sat Jan 13, 2007 7:30 pm    Post subject: Reply with quote

Sorry, made a mistake and this one isn't valid now, Solution was correct, just not the logic

Hi

This is my first walkthrough. Tell me what you think of it. Any suggestions are appreciated. Especially how i can shorten this a bit Wink
This is at least how i solved it.


1 Simple counting
1a. No 1 in 13(2), 20(3),19 (3), 14(2), 11(2), 13(2), 21 (3), 20(3)
1b No 2 in 13(2), 20(3), 14(2), 13(2), 21 (3), 20(3)
1c No 3 in 13(2), 14(2), 13(2), 21 (3)
1d No 4 in 14(2)
1e No 5 in 5(2), 10 (2)
1f No 6 in 5(2)
1g No 7 in 9(3), 7(2), 14(2), 9(3), 5(2)
1h No 8 in 9(3), 13(4), 10 (3), 7(2), 9(3), 10(3), 5(2)
1i No 9 in 9(3), 13(4), 10 (3), 7(2), 9(3), 10(3), 5(2)

2. 45 test
2a. 45 on R5, 2 innies R5C19= 3 -->> {1,2} -->> 1,2 locked for R5.
2b. 45 on R1234, 2 innies R4C19 = 14 -->> {59|68}
2c. 45 on R6789, 2 innies R6C19 = 11 -->> no 1 in R6C19
2d. 45 on C2, 3 outies R159C3 = 19 -->> no 1 in R9C3; R9C3= max 4. R15C3= min 15 -->> no 4,5 in R15C3z
2e. 45 on C8, 3 outies R159C7 = 19 -->> no 1 in R9C7
2f. 45 on N1, 2 outies R2C4 + R4C2= 13 -->> no 1,2,3 in R2C4 and R4C2
2g. 45 on N3, 2 outies R2C6 + R4C8= 17 -->> no 1,2,3,4,5,6,7 in R2C6 and R4C8
2h. 45 on N7, 2 outies R6C2 + R9C4 = 8 -->> no 4,8,9 in R6C2 and R8C4
2i. 45 on N9, 2 outies R6C8 + R8C6 = 14 -->> no 1,2,3,4,7 in R6C8 and R8C6
2j. 45 on C234, 3 innies R159C4 = 21 -->> no 1,2,3 in R159C4
2k. 45 on C678, 3 innies R159C6 = 9 -->> no 7,8,9 in R159C6, R5C6 = min 3, R19C6= max 6 -->> no 6 in R15C6

4. no 9 in 11(2) R5 (see 2a)

5 killer pair in R5 for {56} in 14(2) and 17(2). No 5 and 6 anywhere else in R5

6. 9(3) N1 canít be 126-->> No 6 in 9(3) N1 -->> 3 locked in 9(3) N1, No 3 anywhere else in N1 or C1

7. killer pair in C1 for 1,2 in 9 (3) N1 and R5C1. No 1 and 2 anywhere else in C1.

8. killer pair in R4 for 8,9 in R4C189. No 8 and 9 anywhere else in R4
8a. no 4,5 in R2C4 (see 2f)
8b. no 2 in R3C6 in 16(3) N256

9. No 2,8,9 in R6C9, not compatible with 13(3) N6 -->> no 9 in R6C1(see 2c)

10. No 4 in R9C2 in 5(2) N9 (see 2h)

11. No 8,9 in R1C2 in 13(2) N1 (see 2d)

12. No 8,9 in R23C7 in 16(3) N23 (see 2g)

13. No 7,8,9 in R23C8 in 15(3) N36 (see 2g)

14. No 9 in R5C2 in 14(2) in N4 (see 2d)

15. No 9 in R9C8 in 10(2) in N9 (see 2e)

16. 1 locked in 13(4) N2 no 1 anywhere else in N2. -->> no 6 in R4C5 in 7(2) N25
16a. No 7 in R12C5 in 13(4) N2: 13 canít be made

17. 25(4) N8: R9C6 = max 5. R8C5+R9C45 = min 20 -->> no 1,2 in R89C5

18. 45 on C78: 2innies, 2outies: outies Ė innies = 10 -->> outies max 17 -->> innies max 7 -->> no 7 in R46C7 no 7.
18b. outies canít be 16 (no 7 in outies). So innies canít be 6.

19. 12(3) N9 : no {129} because of R5C9. -->> no 9 in 12(3) N9

20. No 4 in 13(3) N6
20a. 13(3) N6 needs a 1 or 2(R5C9), so no {346}
20b. 13(3) N6 needs a 5,6,8 or 9(R4C9), so no {247}
20c. 11(2) N6 needs a 4 or 8, so 13(3) N6 no {148}
20d. No 4 in R6C9.

21. No 7 in R6C1 (see 2c + 20d)
21a. 7 locked in 21(3) N7 for C1. -->> no 4 in 21(3) N7

22. No 4 in R46C7.
22a. 4 in R4C67 -->> 11(2)N6 = 83 -->> 1 in R4C67(see 18b) -->> 13(3) N6 {256} no room for 7 in N6 so no 4 in R4C67.
22b. 4 locked in 11(2) N6 -->> 11(2) N6 {47}-->> 7 locked in R5 and N6 not anywhere else in R5 and N6

23. 3 locked in 17(3) N5 in R5.
23a. no 4 in R3C5 in 7(2) N25

24. No 4 in R1C9 (see 2c)
24a. 4 locked in 9(3) N1 for C1
24b. 9(3) N1 = {2,3,4) -->> no 2 and 4 in C1 and N1
24c. naked single 1 in R5C1 -->> naked single 2 in R5C9

25. naked quad 5689 in R46C1 + R5C23 for N4 -->> no 5689 anywhere else in N4

26. 1 locked in R46C7 in N6 for C7. no 1 anywhere else in C7
26a. no 7 in 16(3) N23
26b. no 5 in R46C7 (see 18b)

27 1 locked in 12(3) N9 for C9. No 1 anywhere else in N9 -->> no 9
27a. no 9 in 10(2) N9

28. 1 locked in 15(3) N36 for N3
28a. 15(3) N36 =1{59|68}

29. 13(3) N6= 2{38|56} so no 9 in R4C9
29a 9 locked in 20(3) N3 in C9
29b. no 2 in 11(2) N3

30. 2 locked for C7 in N3
30a. no 8 in R9C8 in 10(2) N9
30b. 2 locked in 16(3) N23.
30c. 16(3) N23 = 2{59|68}

31. killer pair 56 in R23C78 for N3.

32. 1 locked 17(3) N1; 17(3) = 179
32a R4C2 = 7
32b 19 locked for N1 in C2
32c. no 6 in R1C3 in 13(2) N1

33. killer pair 78 for R1 in R1C3 + 11(3) N3

34. 19(3) N12= {568}
34a. hidden single 7 in R1C3; naked single 6 in R1C2
34b. naked pair 58 in R23C3 in 19(3) N12 for C3
34c. R2C4 in 19(3) N12 = 6

35. 11(2) N3 = {38} locked in N3
35a. 20(3) N3 = {479} locked in C9

36. No 4 in R9C3 in 5(2) N7
36a 5(2) N7 = {23}, locked for N7 and R9

37. 13(4) N2= {1345}
37a. naked single in R3C5

And the rest is all singles.


Hope it is all clear

Para


Last edited by Para on Sat Jan 13, 2007 8:43 pm; edited 1 time in total
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PostPosted: Sat Jan 13, 2007 7:36 pm    Post subject: Reply with quote

Well one way to shorten it is to rejig the lists of exclusions as lists of inclusions. For instance in the first step you list every exclusion from 14(2) which isn't needed: just say 14(2)={59|68}.
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PostPosted: Sat Jan 13, 2007 7:51 pm    Post subject: Reply with quote

Quote:
2g. 45 on N3, 2 outies R2C6 + R4C8= 17 -->> no 1,2,3,4,5,6,7 in R2C6 and R4C8
2h. 45 on N7, 2 outies R6C2 + R9C4 = 8 -->> no 4,8,9 in R6C2 and R8C4
2i. 45 on N9, 2 outies R6C8 + R8C6 = 14 -->> no 1,2,3,4,7 in R6C8 and R8C6

The first line is an instance of what I mean: just shorten the conclusion to "R2C6 + R4C8 = {89}".
The 2nd line has a typo (R8C4 not R9C4) and a logical flaw: {44} is not excluded because the cells do not "see" each other.
The 3rd line has the same logical flaw: {77} is possible because the cells do not see each other.

I haven't continued with the walkthrough past this point.
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PostPosted: Sat Jan 13, 2007 7:55 pm    Post subject: Reply with quote

nd wrote:

The 2nd line has a typo (R8C4 not R9C4) and a logical flaw: {44} is not excluded because the cells do not "see" each other.
The 3rd line has the same logical flaw: {77} is possible because the cells do not see each other.


yes, that's true. thanks. That's kinda dumb. Now i understand where i made the mistake in an earlier puzzle as well. Smile

Para
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PostPosted: Sat Jan 13, 2007 8:16 pm    Post subject: Reply with quote

I've just been working through the rest of the solving path to see if those flaws really made a difference after all. Unfortunately they do:

Quote:

18b. outies canít be 16 (no 7 in outies). So innies canít be 6.

This step falls foul of the logical flaw mentioned earlier, as the 7 isn't excluded.

More importantly:
Quote:
20. No 4 in 13(3) N6
20a. 13(3) N6 needs a 1 or 2(R5C9), so no {346}
20b. 13(3) N6 needs a 5,6,8 or 9(R4C9), so no {247}
20c. 11(2) N6 needs a 4 or 8, so 13(3) N6 no {148}
20d. No 4 in R6C9.

I donít understand this step. I think I've followed your eliminations correctly up to this point, in which case there is still a possibility of a 4 in R6C9. And so 13(3) = [814] is thus still a possibility. & the succeeding steps in the walkthrough depend on this step, unfortunately.


Yes, the duplicate-cell issue is always an easy one to forget! This is why when I design a puzzle I sometimes deliberately work in a duplicate pair to trip up the unwary.....
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PostPosted: Sat Jan 13, 2007 8:26 pm    Post subject: Reply with quote

Hi

Quote:


20. No 4 in 13(3) N6
20a. 13(3) N6 needs a 1 or 2(R5C9), so no {346}
20b. 13(3) N6 needs a 5,6,8 or 9(R4C9), so no {247}
20c. 11(2) N6 needs a 4 or 8, so 13(3) N6 no {148}
20d. No 4 in R6C9.

I donít understand this step. I think I've followed your eliminations correctly up to this point, in which case there is still a possibility of a 4 in R6C9. And so 13(3) = [814] is thus still a possibility. & the succeeding steps in the walkthrough depend on this step, unfortunately.


11(2) N6 = {38|47}, so 13(3) N6 can't contain both 4 and 8.
The whole idea of this step was to eliminate that 4 in R6C9 Wink

Right?

greetings

Para

p.s. on 18b missing my logic puzzle crashes on 22. Don't think it still works out now. It is kinda important in that step.

p.p.s. glad i posted this one.

p.p.p.s. back to work ugh Sad
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PostPosted: Sat Jan 13, 2007 8:55 pm    Post subject: Reply with quote

Oh, of course--I forgot about that.

Hm, I'm still wondering what Ruud has in mind for the solving-path--I can't see any elegant way of dealing with this puzzle. It's solvable but not pretty.
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PostPosted: Sun Jan 14, 2007 5:11 am    Post subject: Reply with quote

nd wrote:
I can't see any elegant way of dealing with this puzzle
This solution is not the elegant version nd is looking for. Uses some contradictions/hypotheticals between nonets - though seemed pretty straight-forward ones. They made the puzzle quick first time through.

Please let me know if anything is not clear, accurate or valid.

Assassin 33
1. "45" r5 -> r5c19 = 3 = {12} locked for r5 -> no 9 11(2)r5

2. 14(2) r5 = {59/68} = [5/6.5/8..]
2a.-> 11(2)r5 = {38/47} ({56} blocked by 14(2))
2b.17(3) = {359/368/467} ({458} blocked by 14(2))

3. 15(3)c1 must have 1 or 2 but no combo can have both -> no 1,2 r46c1
3a. 9(3)c1 {126} blocked -> = 3{15/24} (no 6): 3 locked for c1,n1

4.21(3)c1 = {489/579/678} = [5/8] -> 15(3)c1 {258} blocked

5. 15(3)n4 = {159/168/249/267}, 14(2)n4 = {59/68}
5a. -> when 14(2) = {68}, 15(3) = 9{15/24} -> 9 locked for n4 in 14(2) or 15(3).
5b. -> when 14(2) = {59} -> 15(3) = 6{18/27} -> 6 locked for n4 in 14(2) or 15(3)

6. "45" n1 -> r2c4 + r4c2 = 13 -> r4c2 = {4578}, r2c4 = {5689}

7. "45" n3 -> r2c6 + r4c8 = 17 = {89} -> no {89} in r4c6.
7a. both the 16(3)n23 and 15(3)n36 must have [8/9] and cannot have both (8+9=17)-> no {89} in r23c78

8. "45" n9 -> r8c6 + r6c8 = 14 -> min 5 in both outies

9. "45" n7 -> r6c2 + r8c4 = 8 -> max 7 in both outies

10. "45" r1234 -> r4c19 = 14 = {59/68}

11. "45" r6789 -> r6c19 = 11 -> r6c1 = {456789}, r6c9 = {234567}

Now for the hypotheticals that break this puzzle.

12. Putting steps 1,5,10 and 11 together we can easily see what the 4 combinations for 15(3)n4 mean:

15(3)n4[r456c1] -> 14(2) = {r5c23} and 13(3)=[r456c9] -> 11(2) = {r5c78}

12a.15(3)n4 = {159} ->

-15(3)[519] -> 14(2)={68} and 13(3)=[922] Blocked (2 2's)
-> no [519] 15(3)

-15(3)[915] -> 14(2)={68} and 13(3)=[526] -> 11(2)r5={47}

12b.15(3)n4 = {168} ->

-15(3)[618] -> 14(2)={59} and 13(3)=[823] -> 11(2)r5={47}

-15(3)[816] -> 14(2)={59} and 13(3)=[625] -> 11(2)r5={38/47}

12c.15(3) = {249} -

-15(3)[924] only -> 14(2) {68} and 13(3) = [517] -> 11(2)r5={38}:Blocked, (2 8's r5)
-> no {249} 15(3) -> no 4 15(3)

12d. 15(3) = {267} ->

-15(3)[627]only -> 14(2)={59} and 13(3)=[814] -> 11(2)r5 Blocked, (no 4 or 8 available)
-> no {267} 15(3)

13. In summary: 15(3)n4 = [915/618/816]= 1{59/68} (no 4,7)
13a.-> {59/68} locked for n4 in 15(3) and 14(2) -> no 5,8 r2c4 (step 6)
13a. r5c1 = 1, r4c1 = {689}, r6c1 = {568}

14. In summary:13(3)n6 = [526/823/625] = 2{38/56}
14a. r5c9 = 2, r4c9 = {568}, r6c9 = {356}

15. 9(3)c1 = {234}:locked for c1,n1
15a.-> 1 in n1 only in 17(3) = {179}
15b.-> r4c2 = 7, r2c4 = 6 (step 6)

16.r23c2 = {19}:locked for n1,c2

17.r23c3 = {58}:locked for n1,c3

18.13(2)n1 = [67]

19. 1 in r1 locked in r1c456->1 locked for n2 -> 7(2)n25 = {25/34} (no 1,6)

20.15(3)c2 = 8{25/34}:8 locked for n7,c2

21.21(3)n7 = {579}:locked n7c1

22.15(3)n47 = {348}:locked for c2 -> r9c23 = [23] -> r6c2= 3,r8c4 = 5 (step 9)

23.r78c2 = {48}:locked for n7, r78c3 = {16}

24.9(3)n458 = {234} -> r7c4 = 3, r6c34 = {24}:locked for r6

25.10(3)n254 = {127} -> r4c34 = [21]

the rest is straightforward


Last edited by sudokuEd on Fri Jan 19, 2007 12:16 pm; edited 1 time in total
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PostPosted: Sun Jan 14, 2007 11:40 am    Post subject: Reply with quote

Well solved it correctly. But used same logic in the end as Ed did. He just came to it quicker.

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PostPosted: Thu Feb 08, 2007 2:09 am    Post subject: Reply with quote

I got stuck on this puzzle when I first tried it and came back to it yesterday after getting a hint from Peter.

ďGot stuck? Here is a vague hint in tiny font.
Hint: Forget about innies/outies (for now). The key is in another approach.
Not enough? Here's a more specific hint.
Hint: Look at N4 and N6 and there at c1 and c9. Look at the possible combinations and possible killer pairs.Ē
That solved it. The extra work prompted by these hints is mainly in steps 24 and 31. I donít think Iíd been making full use of R4C19 and R6C19 before. Thanks Peter!

I think the key step that I used to break it open is sufficiently different from Ed's method so here is my walkthrough.

Clean-up is used in various steps, using the combinations in steps 1 to 10 for further eliminations from these two cell cages; it is also used for the two cell split sub-cages that are produced by applying the 45 rule to R4, R6, N1, N3, N7 and N9. In some of the later steps, clean-up is followed by further moves and sometimes more clean-up.

1. R1C23 = {49/58/67}, no 1,2,3

2. R1C78 = {29/38/47/56}, no 1

3. R34C5 = {16/25/34}, no 7,8,9

4. R5C23 = {59/68}

5. R5C78 = {29/38/47}(not {56} which would clash with R5C23), no 1,5,6

6. R67C5 = {49/58/67}, no 1,2,3

7. R9C23 = {14/23}

8. R9C78 = 10(2), no 5

9. 9(3) cage in N1 = {126/135/234}, no 7,8,9

10. 19(3) cage in N12, no 1

11. 20(3) cage in N3 = {389/479/569/578}, no 1,2

12. 10(3) cage in N254 = {127/136/145/235}, no 8,9

13. 9(3) cage in N458 {126/135/234}, no 7,8,9

14. 10(3) cage in N658 = {127/136/145/235}, no 8,9

15. 21(3) cage in N7 = {489/579/678}, no 1,2,3

16. 20(3) cage in N89 = {389/479/569/578}, no 1,2

17. 13(4) cage in N2 = 1{237/246/345}, no 8,9, 1 locked for N2, clean-up no 6 in R4C5

18. 25(4) cage in N8 = {1789/2689/3589/3679/4579/4678}, must contain at least two of 7,8,9 and contains three of 5,6,7,8,9

19. 45 rule on R5 2 innies R5C19 = 3 = {12}, locked for R5, clean-up: no 9 in R5C78

20. 45 rule on R1234 2 innies R4C19 = 14 = {59/68}

21. Valid combinations for R456C1 with R5C1 = {12} are {159/168/249/267} (cannot be {258} which would clash with R5C23), no 1,2,3 in R6C1, R456C1 must contain one of 4,5,6 and one of 7,8,9

22. Valid combinations for R456C9 with R5C9 = {12} are {139/157/238/247/256} (cannot be {148} which would clash with R5C78), no 1,2 in R6C9, R456C9 must contain one of 3,4,5 and one of 6,7,8,9

23. 12(3) in R789C9 must have 1 or 2, valid combinations are {138/147/156/237/246} ({129} not valid because it has both 1 and 2), no 9, must contain one of 3,4,5 and one of 6,7,8

24. 45 rule on R6789 2 innies R6C19 = 11 = {38/47/56}, clean-up: no 9 in R6C1, no 8,9 in R6C9
24a. Only combination in R456C1 with 5 is {159} -> no 5 in R4C1, clean-up: no 9 in R4C9 -> no {139} in R456C9
24b. From step 22, no {247} in R456C9 because 4,7 both in same cell -> no 4 in R6C9, clean-up: no 7 in R6C1
24c. R456C1 = {159/168/249} [8/9], R5C23 = {59/68} [8/9], killer pair 8,9 for N4
24d. 45 rule on N4 4 innies R4C23 + R6C23 = 16, must contain 1 or 2 and must contain 3,7 -> no 6 in R4C23 + R6C23 [Edited. 3 added]

25. 45 rule on N1 2 outies R2C4 + R4C2 = 13 = {49/58/67}, no 1,2,3
25a. Clean-up: no 4,5,7 in R2C4

26. 45 rule on N3 2 outies R2C6 + R4C8 = 17 = {89} -> no 8,9 in R2C8 and R4C6
26a. Killer pair 8,9 in R4C189 for R4

27. 45 rule on N7 2 outies R6C2 + R8C4 = 8 = {17/26/35/44}, no 8,9, clean-up: no 2 in R8C4

28. 45 rule on N9 2 outies R6C8 + R8C6 = 14 = {59/68/77}, no 1,2,3,4

29. 9 in C9 locked in R123C9 = 9{38/47/56}, clean-up: no 2 in R1C78

30. 9 in N6 locked in R46C8, locked for C8, clean-up: no 1 in R9C7

31. Remaining combinations for R456C9 are {157/238/256}. It looks like the 3 cages in C9 are meant to be effectively 9-11, 2-11 and 1-11 so letís look at a contradiction move to see if {157} can be eliminated.
If R456C9 = {157}, R4C9 = 5, R5C9 = 1, R6C9 = 7 => R4C1 = 9, R5C1 = 2, R6C1 = 4 => R5C23 = {68} which clashes with R5C78 = {38} so R456C9 cannot be {157}
31a. R5C9 = 2, R5C1 = 1
31b. R46C1 = {68}/[95]
31c. R46C9 = {56}/[83]
31d. R123C9 = 9{38/47/56}
31e. R789C9 = 1{38/47/56}, 1 locked for N9, clean-up: no 9 in R9C7

32. R123C1 = {234}, locked for C1 and N1, clean-up: no 9 in R1C23

33. 7 in C1 locked in R789C1, locked for N7, R789C1 = 7{59/68}

34. 1 in N1 locked in R23C2, locked for C2, clean-up: no 4 in R9C3
34a. Only valid combination for R234C2 = {179} -> R4C2 = 7, R23C2 = {19}, locked for C2 and N1, clean-up: no 6 in R1C3, no 5 in R5C3

35. R9C2 = {234} -> R678C2 must contain two of 2,3,4 -> only possible combination = {348}, locked for C2 with 8 in R78C2, locked for N7 -> R9C2 = 2, R9C3 = 3, clean-up: no 5 in R1C3, no 6 in R5C3, no 6 in R789C1 = {579}, locked for C1 and N7, no 7,8 in R9C78 -> R9C78 = {46}, locked for R9 and N9, clean-up: no 5,7 in R789C9

36. R46C1 = {68}, locked for N4 -> R5C23 = [59], R1C23 = [67], clean-up: no 4,5 in R1C78 -> R1C78 = {38}, locked for R1 and N3

37. 1 in N2 locked in R1C456

38. R1C9 = 9 (hidden single in R1)

39. R789C9 = {138}, locked for C9 and N9 -> R46C9 = {56}, locked for C9 and N6 -> R23C9 = {47}, locked for N3

40. R78C2 = {48}, locked for C2 and N7 -> R6C2 = 3
40a. R78C3 = {16}, locked for C3 -> R8C4 = 5, clean-up: no 8 in R6C5

41. 6 in N5 locked in R456C5 = 6{38/47}, 6 locked for N5, clean-up: no 7 in R7C5
[Alternatively could have used X-wing with 6s in R46C19]

42. R23C3 = {58}, locked for C3 -> R2C4 = 6, clean-up: no 1 in R4C5 [Edited. "locked for C3" added for clarity"]

43. Only remaining combination for 9(3) cage in N458 = {234} -> R6C34 = {24}, locked for R6, R7C4 = 3, clean-up: no 9 in R7C5

44. Only remaining combination for 10(3) cage in N254 is R3C4 = 7, R4C34 = [21]
-> R6C34 = [42], R23C9 = [74], R1C4 = 4, clean-up: no 5 in R3C5, no 3 in R4C5

45. R1C1 = 2, R3C1 = 3, R2C1 = 4, R1C56 = {15}, locked for N2 -> R2C5 = 3, R3C5 = 2, R4C5 = 5, R1C56 = [15], R46C9 = [65], R46C1 = [86], R4C8 = 9, R23C6 = {89}, locked for C6 [Edited. "locked for N2" added for clarity"]

and the rest is naked and hidden singles, naked pairs, simple elimination and cage sums

Ruud wrote:
Brace for impact. Here is an Assassin that is helping me teach SumoCue some new tricks. You already know these tricks... donít you?

nd wrote:
I'm still wondering what Ruud has in mind for the solving-path--I can't see any elegant way of dealing with this puzzle. It's solvable but not pretty.

Was the intended solving path to use one or more contradiction moves or is there something else that we've all missed?
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mhparker
Grandmaster
Grandmaster


Joined: 20 Jan 2007
Posts: 345
Location: Germany

PostPosted: Thu Dec 27, 2007 11:36 pm    Post subject: Reply with quote

Hi guys,

I decided to tackle this one after Ruud's call for assistance in getting some player ratings for earlier Assassins. This puzzle was the active Assassin when I first discovered the sudocue.net site. I couldn't have imagined doing such a puzzle back then, but fortunately I've learnt a lot around here since.

nd wrote:
Hm, I'm still wondering what Ruud has in mind for the solving-path--I can't see any elegant way of dealing with this puzzle. It's solvable but not pretty.

Andrew wrote:
Was the intended solving path to use one or more contradiction moves or is there something else that we've all missed?

Yes, it can be solved elegantly, and without any hypotheticals or "contradiction moves". In fact, it's probably the shortest walkthrough I've ever written! Therefore, although this puzzle has had a fearful reputation in the past, the optimum route through the puzzle probably only deserves a rating of around 1.25.


Assassin 33 Walkthrough

Prelims:

a) 9(3) at R1C1 and R6C3 = {126/135/234} (no 7..9)
b) 13(2) at R1C2 and R6C5 = {49/58/67} (no 1..3)
c) 13(4) at R1C4 = {1237/1246/1345} (no 8,9); 1 locked for N2
d) 11(2) at R1C7 = {29/38/47/56} (no 1)
e) 20(3) at R1C9 and R7C7 = {389/479/569/578} (no 1,2)
f) 19(3) at R2C3 = {289/379/469/478/568} (no 1)
g) 10(3) at R3C4 and R6C6 = {127/136/145/235} (no 8,9)
h) 7(2) at R3C5 = {16/25/34} (no 7..9); no 6 in R4C5
i) 14(2) at R5C2 = {59/68}
j) 11(2) at R5C7 = {29/38/47} (no 1,5,6) (Note: {56} blocked by 14(2) at R5C2 (prelim i))
k) 21(3) at R7C1 = {489/579/678} (no 1..3)
l) 5(2) at R9C2 = {14/23}
m) 10(2) at R9C7 = {19/28/37/46} (no 5)

1. Innies R5: R5C19 = 3(2) = {12}, locked for R5
1a. cleanup: no 9 in R5C78

2. Outies N1: R2C4+R4C2 = 13(2) = {49/58/67} (no 1..3)

3. Outies N3: R2C6+R4C8 = 17(2) = {89}
3a. no 8,9 in R2C8 and R4C6 (CPE)

4. Outies N7: R6C2+R8C4 = 8(2) = {17/26/35/44} (no 8,9)

5. Outies N9: R6C8+R8C6 = 14(2) = {59/68/77} (no 1..4)

6. Innies R1234: R4C19 = 14(2) = {59/68}
6a. -> R4C19 and R4C8 form killer pair on {89} -> no 8,9 elsewhere in R4
6b. cleanup: no 4,5 in R2C4 (step 2)

7. Innies R6789: R6C19 = 11(2) = {29/38/47/56} (no 1)

8. 17(3) at R5C4 and R6C5 form hidden killer pair in N5 on {89}
8a. -> R6C5 = {89}; 17(3) at R5C4 = {359/368} (no 4,7)
(Note: {458} blocked by 14(2) at R5C2 (prelim i))
8b. 3 locked in 17(3) for R5 and N5
8c. cleanup: no 4 in R3C5; no 8 in R5C78; R7C5 = {45}

9. Naked pair (NP) at R5C78 = {47}, locked for N6
9a. cleanup: no 7 in R8C6 (step 5); no 4,7 in R6C1 (step 7)

10. 7 in C1 locked in 21(3) at R7C1 (prelim k) = {579/678} (no 4)
10a. 7 locked for N7

11. 4 in C1 locked in 9(3) at R1C1 (prelim a) = {234}, locked for C1 and N1
11a. cleanup: no 9 in R1C23; no 8,9 in R6C9 (step 7)

12. R5C19 = [12]
12a. cleanup: no 7 in R8C4 (step 4); no 9 in R6C1 (step 7)

13. 1 in N1 locked in R23C2
13a. -> 17(3) at R2C2 = {179}
13b. -> R4C2 = 7; R23C2 = {19}, locked for C2 and N1
13c. -> R2C4 = 6 (step 2)
13d. cleanup: no 6 in R1C3; no 5 in R5C3; no 4 in R9C3; no 1 in R4C5; no 2 in R6C2 (step 4); no 1 in R8C4 (step 4)

14. Hidden single (HS) in R6 at R6C6 = 7
14a. -> split 3(2) at R6C7+R7C6 = [12]
14b. cleanup: no 9 in R9C8; no 6 in R6C2 (step 4)

15. HS in N1 at R1C2 = 6
15a. -> R1C3 = 7
15b. cleanup: no 4,5 in R1C78; no 8 in R5C3

16. Innies C2: R59C2 = 7(2) = [52] (last combo/permutation)
16a. -> R59C3 = [93]
16b. cleanup: no 7,8 in R9C78

17. HS in C2 at R6C2 = 3
17a. -> R8C4 = 5 (step 4)
17b. cleanup: no 9 in R6C8 (step 5)

18. HS in C4 at R9C4 = 9
18a. cleanup: no 1 in R9C8; no 5 in R6C8 (step 5)

The rest is now just singles and cage sums.
_________________
Cheers,
Mike
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Para
Yokozuna
Yokozuna


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

PostPosted: Fri Dec 28, 2007 1:12 am    Post subject: Reply with quote

Mike, so you started solving when i attempted to write my first walk-through, which was a fiasco. Guess that wasn't my best first impression. Hope i made up for it.
Haven't had time to solve much lately. I think i have 8 or more killers to check back on. I will try to get something done before the end of the year. Otherwise i'll see you all next year.

greetings

Para

ps. i have to post something once in a while, otherwise you might catch up to me.
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