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Assassin 61
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CathyW
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Joined: 31 Jan 2007
Posts: 161
Location: Hertfordshire, UK

PostPosted: Tue Jul 31, 2007 10:26 am    Post subject: Reply with quote

Here's my WT for the A61X. Do let me know about any typos!

Preliminaries

a) 3(2) r89c1 = {12} not elsewhere in N9/c1

b) 24(3) r5c456 = {789} not elsewhere in N5/r5

c) 13(4) r345c9+r4c8 must have 1 -> r6c9 <> 1


1. Innies N7: r7c1 + r9c3 = 8 = {35} not elsewhere in N7
-> r78c3 <> 8 -> 8 locked to r789c2, not elsewhere in c2

2. Innies N9: r7c9 + r9c7 = 9 = {18/27/36/45}

3. Outies c12: r35c3 = 7 = {16/25/34}

4. Outies c89: r35c7 = 6 = {15/24}

5. Innies r123: r3c1469 = 23

6. Innies r789: r7c1469 = 15 = 3{129/147/156/246}; 5{127/136} (no 8)
-> r9c7 <> 1

7. Outies r1234: r5c239 = 6 = {123} -> r3c3 <> 1,2,3
-> r5c18 = (456), r5c7 = (45) -> r3c7 = (12)
-> NP {12} not elsewhere on diagonal /
-> r9c7 <> 2 -> r7c9 <> 7
-> 9(2) r78c7 <> {45}

8. 16(3) r9c3+r89c4 = 3{49/58/67}; 5{29/38/47} -> r89c4 <> 1

9. Innies N1: r12c3+r3c1 = 14

10. Innies N3: r12c7+r3c9 = 15

11. Innies c5: r456c5 = 18 (only one of 7,8,9) = {567/468/369/459}
-> HS r6c6 = 1 -> r6c7, r7c6 <> 2,3; no 8 in r7c6 -> r6c7 <> 5
-> HS r4c4 = 2 -> r3c4, r4c3 = {89}
Clean up r8c7 <> 7,8

12. 14(4) r6c345+r7c4: min from r6c45 is 7 -> r6c3+r7c4 is max 7
-> r6c3 <> 7,8; r7c4 <> 7
14(4) options: {1256/1346/2345} -> r7c4 <> 6

13. Outies - Innies r12: r2c28 - r3c5 = 3
Min from r2c28 is 7 -> r3c5 = (4…9)
-> r12c5 <> 8,9

14. O-I N8: r9c37 - r7c46 = 2

15. Outies c1: r1456c2+r5c3 = 18 (no 8); 1,2,3 can be repeated between r1c2 and r5c3

16. Outies c9: r1456c8+r5c7 = 24; 4,5 can be repeated between r1c8 and r5c7

17. 13(3) N1: r3c3 min 4 -> r23c2 max 9 -> r23c2 <> 9
Options: 4[72]/{36}; 5[71/62]; 6[52]/{34} -> r3c2 <> 7

18. Pointing cells: r7c9 sees all with 2 in N6 -> r7c9 <> 2 -> r9c7 <> 7
-> split 15(4) r7c1469 now 3{147/156}; 5{136} -> r7c6 <> 9 -> r6c7 <> 4
Must have 1 -> r7c58 <> 1
Must have 3 -> r7c578 <> 3 -> r8c7 <> 6
If 3{147}, r7c6 = 7 -> r7c6 <> 4 -> r6c7 <> 9

19. 26(5) N69: {14579/14678/24569/34568}
Must have 4 -> r4c9 <> 4 since sees all cells of 26(5)

20. Elimination chain:
a) r3c4 = 9 -> r12c6, r3c8 <> 9
b) r3c4 = 8 -> r4c3 = 9 -> (r4c7 <> 9) 9 locked to r12c7 -> r12c6, r3c8 <> 9
Either case r12c6, r3c8 <> 9

21. 16(3) r9c3+r89c4 = 3{49/58/67}; 5{38/47}
Must have one of 7,8,9 within r89c4 -> Killer triple with r35c4 -> r12c4 <> 7,8,9
-> 9 locked to r3c456 -> r3c1 <> 9

22. a) if r7c6 = 5, r6c7 = 8
b) if r7c6 = 6, r6c7 = 7 -> r7c7 = 8
c) if r7c6 = 7, r6c7 = 6 -> r7c7 = 8
-> 8 locked to r67c7 not elsewhere in c7 -> r7c9 <> 1
-> 8 locked to r6c789 -> r6c1 <> 8

23. 21(4) r567c1+r5c2 = {3459/3567} -> r6c2 <> 2
Must have 3 and 5, r4c1 sees all cells of 21(4) -> r4c1 <> 3,5

24. HS r7c4 = 1
-> r6c345 = {256/346} 6 not elsewhere in r6 -> r7c6 <> 7
-> split 15(4) r7c1469 = {1356} -> r8c7 <> 3, r9c7 <> 5
-> NP {12} r38c7, not elsewhere in c7
-> NP {12} r8c17, not elsewhere in r8 -> r8c3 <> 7
-> NP {78} r67c7, r124c7 <> 7

25. 1 locked to 13(4) within N6 -> r3c9 <> 1

26. 4 locked to 26(5) within N6 -> r4c78 <> 4

27. 26(5) N69 = {24569/34568} Must have 4,5,6
6 in either r5c8 or r7c9 -> r4c9, r89c8 see both therefore <> 6
split 9(2) N9 = {36} -> r89c89 <> 3 -> r89c9 <> 9
-> 15(3) r789c8 = {159/249}, 9 not elsewhere in c8
-> r9c8 <> 5
(Andrew commented that perhaps additional explanation required for this step: Elimination chain r7c9 = 5 -> r5c7 = 4 clashes with r9c7 = 4.)

28. 3 locked to r8c456 -> r9c456 <> 3

29. 26(5) must have 4,5,6; 5 now within 26(5) within N6 -> r4c789 <> 5

30. 13(4) r345c9+r4c8 = {1237/1246} -> r3c9 <> 5
-> 2 locked to r35c9 -> r26c9 <> 2

31. 3 locked to r1c1, r2c2 on diagonal \, not elsewhere in N1

32. 6 locked to r1c1, r2c2, r3c3 on diagonal \, not elsewhere in N1

33. Pointing cells: 7 locked to r1234c8 -> r3c9 <> 7

34. Split 15(3) r12c7+r3c9 = {249/456} 4 not elsewhere in N3

35. 13(4) r345c9+r4c8:
a) if {1237} -> r3c9 = 2
b) if {1246} -> r3c9 = 4
-> r3c9 <> 6

36. 12(3) N3 = {138/156/237} -> r3c8 <> 1,2

37. 13(3) r23c2+r3c3 = [724/715/625/526/346]
-> r2c2 <> 4, r3c2 <> 5
-> NT {124} r3c279, not elsewhere in r3 -> r5c3 <> 3; r12c5 <> 7
-> 13(3) N1 = [715/625/526/346]

38. 12(3) r123c5 = {12}9/{13}8/{14/23}7/15/24}6/{16}5 ({345} blocked by r46c5}

39. Split 14(3) N1: {158/248/257}
-> r12c3 <> 9
-> 18(3) N1 = 9{18/27}; {936} not possible as 3,6 only in r1c1, {945} blocked by split 14(3)
-> r12c1 = (789); r1c2 = (12)
-> 9 locked to r12c1, not elsewhere in c1
-> {248} blocked for split 14(3) -> r12c3 <> 4
-> HS r3c2 = 4 -> r3c9 = 2, r3c7 = 1
-> r8c7 = 2, r7c7 = 7, r6c7 = 8, r9c1 = 2, r8c1 = 1, r7c6 = 5, r7c1 = 3, r7c9 = 6 …

Straightforward combos and singles from here

Very Happy


Last edited by CathyW on Sun Aug 19, 2007 8:01 pm; edited 3 times in total
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Andrew
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PostPosted: Wed Aug 01, 2007 1:57 am    Post subject: Reply with quote

Having worked through Cathy's A61 walkthrough before I posted my one, and then worked through Para's WT this evening I had another look at my WT to find the main difference.

I think my key move was step 25 which eliminated 1,2 from R5C3 and therefore 8,9 from R3C3; this gave a naked quad in C3.


Last edited by Andrew on Fri Aug 03, 2007 12:40 am; edited 1 time in total
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CathyW
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PostPosted: Wed Aug 01, 2007 8:16 pm    Post subject: Reply with quote

Did anyone else try Mike's A61X?
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Andrew
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PostPosted: Wed Aug 01, 2007 8:36 pm    Post subject: Reply with quote

CathyW wrote:
Did anyone else try Mike's A61X?

I started it last night but it was nearly bedtime so I didn't get much further than the preliminary steps and some easy 45s.

Must find time to work through the posted A60 walkthroughs and decide whether to post my one. Then I'll have a proper try at A61X.
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Para
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PostPosted: Wed Aug 01, 2007 9:09 pm    Post subject: Reply with quote

Hi Cathy

Yes i did it as well. I just didn't keep a walk-through. But your path seems very similar to mine. Especially step 22 was really formiliar. I made the exact same elimination to get past that point of the puzzle.

Para
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CathyW
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PostPosted: Wed Aug 01, 2007 9:14 pm    Post subject: Reply with quote

I was rather pleased with that move - good to know a "Grandmaster" thought the same thing. I must be improving! Very Happy
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Para
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PostPosted: Wed Aug 01, 2007 10:02 pm    Post subject: Reply with quote

"Grandmaster" just means i talk a lot Wink
I was rather pleased with it myself too. Didn't see anything special i did that didn't appear in your walk-through so didn't think of doing it again to write a walk-through. I guess i am just lazy, but it also like to solve puzzles without walk-throughs some times to see how fast i can solve a puzzle.

Para
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mhparker
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PostPosted: Wed Aug 01, 2007 11:26 pm    Post subject: Reply with quote

Para wrote:
Especially step 22 was really formiliar. I made the exact same elimination to get past that point of the puzzle.

Cathy wrote:
I was rather pleased with that move - good to know a "Grandmaster" thought the same thing...

I'm glad you mentioned that move, because after Cathy's earlier remark about the puzzle being "easier than A61", I was naturally inquisitive to find out what had not gone according to plan, so dived into Cathy's WT at the first opportunity!

It turned out to be exactly thνs move that shortcut the puzzle. The problem is that current solvers (even the latest JSudoku!) don't detect such moves, no doubt considering them to be hypotheticals (which, in turn, are assumed to be synonymous with limited trial and error (T&E)).

However, it's clear in this case that whilst this neat move is indeed strictly speaking a hypothetical, it is definitely not T&E, but (instead) perfectly sound human logic.

Clearly hypotheticals therefore need to be formalized and the boundaries of T&E precisely defined. This has opened up a whole new avenue of research I had not been considering up to now. For that reason alone, this puzzle has already proved to be more than worthwhile.

Many thanks for providing this insight. Very Happy
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CathyW
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PostPosted: Thu Aug 02, 2007 11:28 am    Post subject: Reply with quote

You're welcome Smile

imho:

Hypothetical = logical "if ... then ... so ..." statements leading to elimination of candidate(s) and/or placement
Bifurcation/Trifurcation included in this provided each option is pursued to logical conclusion to prove conflict or validity.

So where does one draw the line for 'unacceptable' T&E?
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mhparker
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PostPosted: Thu Aug 02, 2007 2:27 pm    Post subject: Reply with quote

Hi Cathy,

I'll take your points in reverse order.

Cathy wrote:
So where does one draw the line for 'unacceptable' T&E?

Speaking for myself, T&E is only acceptable if there is no other obvious way to proceed with a puzzle. That happens a lot around here! Smile

However, what I was trying to say is not that T&E is acceptable, but rather that some very simple hypotheticals like this one should not be considered to be T&E at all. My reasoning is that humans are very quickly able to spot a promising-looking pattern. If the chains are very short and the options at this position are so few that they can be enumerated on the fingers of one hand, then I don't see how this has anything more to do with T&E than, say, Aligned Pair Exclusion (APE) does.

Cathy wrote:
Hypothetical = logical "if ... then ... so ..." statements leading to elimination of candidate(s) and/or placement
Bifurcation/Trifurcation included in this provided each option is pursued to logical conclusion to prove conflict or validity.

Yes, you're spot on there! Many people immediately raise the question as to why XY-Chains, Nice Loops, and so on, aren't also considered to be hypotheticals. After all, they're also a case of "if ... then ... so ...", aren't they? However, it's exactly the bifurcative aspect of hypotheticals you mention that is their major distinguishing feature.

The basic rule for a single chain or loop is: if you can identify the various (strong, weak, etc.) links in advance, construct the chain or loop from them, and come to some productive conclusion (candidate/combo elimination, etc.) without touching (i.e., altering) the state of the grid, then it's not a hypothetical.

If, on the other hand, one or more of the links in the chain depends on some side-effect (e.g., candidate elimination) of an earlier link in the chain, then one is dealing with a hypothetical. For example, consider the following x-cycle on the digit 'x' (in verbose notation, where "=>" and "->" imply strong and weak links, respectively):

A<>x => B=x -> ... -> E<>x => F=x

If (in this example) the strong link between E and F didn't exist at the start of the chain, but only arose dynamically due to the eliminations on x caused by the premise "B=x" at the end of the first link, then it's a hypothetical. The implication here is that one is building up the chain as one goes along, re-evaluating the situation at each step, creating new links dynamically as they become available. This "try first, evaluate later" approach explains why hypotheticals are generally considered to be a form of trial and error. The bifurcation you mention comes into play because a computer program has to bifurcate (i.e., fork a copy of the grid) in order to maintain the side-effects, such that they can be undone (e.g., by restoring the original grid) at the end of the sequence.

Bifurcation is also required in order to record and compare the effects of assuming each of a set of initial "possibilities" (candidates in a cell, combos in a cage, etc.) in turn. This is the case that Andrew mentioned above, which corresponds closely to forcing chains.

The "rule of thumb" I use is:

If bifurcation is required, then it's a hypothetical.

Hope that helps.
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mhparker
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PostPosted: Wed Aug 15, 2007 12:37 pm    Post subject: Reply with quote

Hi folks,

Here is an alternative partial WT for the A61X starting from Cathy's step 21:


After cathy's step 21:

22. Innies n9 cannot be {45} due to r5c7 (common peer)
22a. -> no 4,5 in r7c9+r9c7

23. Hidden killer pair on {45} in n9 ({45} locked within 15(3)n9 and 12(2)n9)
23a. 15(3) cannot contain both of {45} (i.e., cannot be {456}) due to r5c8
23b. -> 15(3)n9 and 12(2)n9 must each contain exactly 1 of {45}
23c. -> 12(2)n9 = {48/57} (no 3,9)

24. 9 in n9 locked in 15(3) = {159/249} (no 3,6,7,8)
24a. 1 only in r9c8 -> no 5 in r9c8
24b. 9 locked for c8
24c. Cleanup: no 1 in r3c8

25. 3 in d\ locked in n1 -> not elsewhere in n1

26. Innie/outie diff. n1: r12c4 = r3c1 + 4
26a. min. r3c1 = 4 -> min. r12c4 = 8
26b. -> no 1 in r12c4
26c. max. r12c4 = 11 -> max. r3c1 = 7
26d. -> no 8 in r3c1

27. Hidden single (HS) in c4 at r7c4 = 1
27a. Cleanup: no 8 in r9c7 (innies n9)

28. Naked pair (NP) on {36} at r7c9+r9c7 -> no 3,6 elsewhere in n9

29. NP on {12} at r38c7 -> no 1,2 elsewhere in c7

30. NP on {12} at r8c17 -> no 2 elsewhere in r8

31. 3 in r8 locked in n8 -> not elsewhere in n8

32. 6 in d\ locked in n1 -> not elsewhere in n1

33. Split 13(3) at r6c345 = {256/346}
33a. 2 only in r6c3 -> no 5 in r6c3
33b. 6 locked in r6c345 for r6
33c. Cleanup: no 7 in r7c6

34. NP on {78} at r67c7 -> no 7,8 elsewhere in c7

35. Naked triple (NT) on {356} at r7c169 -> no 5,6 elsewhere in r7
35a. Cleanup: no 7 in r8c3

36. Hidden killer triple on {789} in c3 at r12478c3
36a. -> r12c3 must contain 1 of {789}
36b. -> min. r12c3 = 8 -> max. r12c4 = 10 -> max. r3c1 = 6 (step 26)
36c. -> no 7 in r3c1

37. 13(3)n1 = {157/247/256/346} = {(4/5)..}
37a. -> r3c1 and 13(3)n1 form killer pair on {45} -> no 4,5 elsewhere in n1

38. 18(3)n1 = {189/279/378} (no 6)
38a. (Note: {369} unplaceable, since {36} only in r1c1)
38b. {12} only in r1c2
38c. -> no 9 in r1c2

39. 6 in n1 and d\ locked in 13(3)n1 = {256/346) (no 1,7)
39a. 2 only in r3c2
39b. -> no 5 in r3c2
39c. 3 only in r2c2
39d. -> no 4 in r2c2

40. 4 in n1 locked in r3 -> not elsewhere in r3

41. 1 in n6 locked in 13(4)n36 -> no 1 in r3c9

42. HS in r3 at r3c7 = 1

This is the first of a run of naked/hidden singles that continues until the end of the puzzle.

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Mike
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Andrew
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PostPosted: Thu Aug 16, 2007 2:37 am    Post subject: Reply with quote

I found Mike's A61X a real struggle. I was stuck and left this one for a few days while I worked on other puzzles and walkthroughs. When I returned my next move, step 57, proved to be my breakthrough. Now that I've finished it and worked through Cathy's walkthough I can see where I missed key moves.

Cathy and I reached almost similar positions before her step 22 (my step 41) but using slightly different moves to get there. I also had a couple of extra 45s which locked 3 to two cells on D/ but that doesn't seem to be significant for the solving path.

Soon after that I started struggling. I missed some points about the 26(5) cage, see comments below, which probably weren't significant at the time. I think the key things I missed were the eliminations in N9 that allowed the locked 3 and 6 on D\ in N1. Cathy also used the split cage in N1 which I didn't use; I don't know why because I did use the similar split cage in N3. N9 and N1 were clearly the key areas for the final stages of this puzzle.

Thanks Mike for posting your partial walkthrough showing the intended solving path if Cathy's hypothetical hadn't provided a shortcut.

Mike's steps 22 and 23 appear to be the key ones; the equivalent of step 22 was also in Cathy's walkthrough although expressed differently. I must admit I'm still finding it hard to spot common peer eliminations although I did find a couple in this puzzle. The one in Mike's step 22 is definitely a harder one to find. Then there's the hidden killer pair in step 23. We've been getting those moves more recently but this one is harder to see because it also depends on one combination in R789C8 not being available.

Mike estimated this puzzle as 1.75 when he posted the puzzle and Ed has it currently rated as 1.5. Having seen both Cathy's and Mike's walkthroughs those ratings look right.

In case anyone is interested, here is how I solved A61X.

1. R78C3 = {49/58/67}, no 1,2,3

2. R78C7 = {18/27/36/45}, no 9

3. R89C1 = {12}, locked for C1 and N7

4. R89C9 = {39/48/57}, no 1,2,6

5. 19(3) at R3C4 = {289/379/469/478/568}, no 1

6. R5C456 = {789}, locked for R5 and N5
6a. Max R4C4 = 6 -> min R3C4 + R4C3 = 13, no 2,3

7. R789C2 = {489/579/678}, no 3

8. 13(4) at R3C9 = {1237/1246/1345}, no 8,9

9. 14(4) at R6C3 = {1238/1247/1256/1346/2345}, no 9

10. 3 in N7 only in R7C1 + R9C3
10a. 45 rule on N7 2 innies R7C1 + R9C3 = 8 = {35}, locked for N7, clean-up: no 8 in R78C3
10b. R789C2 = {489/678} = 8{49/67}, 8 locked for C2
10c. Max R9C3 = 5 -> min R89C4 = 11, no 1

11. 45 rule on N9 2 innies R7C9 + R9C7 = 9 = {18/27/36/45}, no 9

12. 45 rule on C1234 1 innie R5C4 – 3 = 1 outie R6C5 -> R6C5 = {456}

13. 45 rule on C6789 1 innie R5C6 – 3 = 1 outie R4C5 -> R4C5 = {456}

14. 45 rule on C12 2 outies R35C3 = 7 = {16/25/34}, no 7,8,9

15. 45 rule on C89 2 outies R35C7 = 6 = {15/24}
15a. R78C7 (step 2) = {18/27/36} (cannot be {45} which clashes with R35C7)

16. 45 rule on R1234 3 outies R5C239 = 6 = {123}, locked for R5, clean-up: no 1,2,3 in R3C3 (step 14), no 4,5 in R3C7 (step 15)
16a. Min R3C3 = 4 -> max R2C23 = 9, no 9

17. 14(4) cage at R6C3 = {1247/1256/1346/2345} (cannot be {1238} because R6C5 only contains 4,5,6), no 8

18. 21(4) cage at R5C1 has R5C1 = {456}, R7C1 = {35}, valid combinations {1569/2568/3459/3468/3567}

19. 45 rule on R789 4 innies R7C1469 = 15 with R7C1 = {35}, valid combinations {1239/1257/1347/1356/2346}, no 8, clean-up: no 1 in R9C7 (step 11)

20. Naked pair {12} in R3C7 + R9C1, locked for D/ -> no 2 in R9C7, clean-up: no 7 in R7C9 (step 11)
[This has been there since step 16. Only just spotted it.]

21. 12(3) cage in N3 with R3C7 = {12} = {129/138/147/156/237/246}
21a. 12(3) cage at R2C8, 1,2 only in R3C78 -> no 9 in R3C8

22. R6C6 = 1 (hidden single in N5), locked for D\, clean-up: no 8 in R8C7
22a. R6C7 + R7C6 = {49}/{67}/[85], no 2,3, no 5 in R6C7

23. R4C4 = 2 (hidden single in N5), locked for D\, clean-up: no 7 in R8C7
23a. R3C4 + R4C3 = 17 = {89}

24. 3 in N5 locked in R4C6 + R6C4, locked for D/
24a. 12(3) cage in N3 (step 21) = 12(3) cage in N3 with R3C7 = {12} = {129/138/147/156/237/246}
24b. 8 of {128} must be in R2C8 -> no 8 in R3C8

25. 16(3) cage at R8C4 = {349/358/367/457} -> R89C4 must contain 7/8/9
25a. Killer triple 7,8,9 in R3C4, R5C4 and R89C4, locked for C4

26. 14(4) at R6C3 (step 17) = {1256/1346/2345} (cannot be {1247} because 2,7 only in R6C3), no 7
26a. {1256/1346} must have 1 in R7C4 -> no 6 in R7C4
26b. {1256/2345} must have 2 in R6C3 -> no 5 in R6C3

27. 45 rule on C5 3 innies R456C5 = 18 = {459/468/567}

28. R123C5 = {129/138/147/237} (cannot be {156/246/345} which clash with R456C5), no 5,6

29. R789C5 = {159/168/249/267/348/357} (cannot be {258/456} which clash with R456C5)

30. R123C5 (step 27) = {129/138/237} (cannot be {147} which clashes with all combinations for R789C5 in step 29), no 4

31. 13(3) cage at R2C2 min R2C2 + R3C3 = 7 -> max R3C2 = 6

32. R789C8 = {159/168/249/258/267/348/357} (cannot be {456} which clashes with R5C8)

33. R35C7 = 6 (step 15), 9 in C7 locked in R1246C7
33a. 45 rule on C7 5 innies R12469C7 = 30 = 9{1578/2478/3468/3567} (cannot be {25689} which clashes with R35C7)

34. CPE 2 in R7C9 can see every 2 in N6 -> no 2 in R7C9, clean-up: no 7 in R9C7 (step 11)

35. 26(5) cage at R5C7 must have R5C7 = {45}, R5C8 = {456} and R7C9 = {13456}, valid combinations {14579/14678/24569/34568}
[I missed the fact that 4 is now locked in the 26(5) cage leading to a CPE elimination.]

36. R4C3 = {89}, R78C3 must contain 7/9 -> 18(4) cage at R1C3 must contain 7/8/9 in R12C3 with no other 7/8/9 in the cage, valid combinations {1269/1359/1368/1458/1467/2349/2358/2367/2457}

37. 45 rule on N1 2 outies R12C4 – 4 = 1 innie R3C1, max R12C4 = 11 -> max R3C1 = 7

38. 8,9 in R3 locked in R3C456, locked for N2

39. 45 rule on N89 3 innies R7C469 – 7 = 1 outie R9C3, R9C3 = {35} -> R7C469 = 10,12 = {136/145/147/156/345}, no 9, clean-up: no 4 in R6C7 (step 22a)

40. R7C1469 (step 19) = {1347/1356} = 13{47/56}, 1,3 locked for R7, clean-up: no 6 in R8C7
40a. 7 only in R7C6 -> no 4 in R7C6, clean-up: no 9 in R6C7

41. R7C6 = {567}
41a. If R7C6 = 5 -> R6C7 = 8
41b. If R7C6 = 6 -> R6C7 = 7 -> R7C7 = 8
41c. If R7C6 = 7 -> R6C7 = 6 -> R7C7 = 8
41d. 8 in C7 locked in R67C7, locked for C7, clean-up: no 1 in R7C9 (step 11)
41e. 8 in N6 locked in R6C789, locked for R6

42. R7C4 = 1 (hidden single in R7)
42a. 14(4) at R6C3 (step 26) = 1{256/346} = 16{25/34}, 6 locked for R6, clean-up: no 7 in R7C6 (step 22a)
42b. R789C5 (step 29) = {249/267/348/357}

43. 1 in C5 locked in R123C5 (step 30) = 1{29/38}, no 7
43a. 8,9 only in R3C5 -> no 1,2,3 in R3C5
43b. Naked pair {89} in R3C45, locked for R3
43c. 7 in N2 locked in R123C6, locked for C6, clean-up: no 4 in R4C5 (step 13)

44. 21(4) cage at R5C1 (step 18) = {3459/3567} = 35{49/67}, no 2
44a. CPE 3,5 in R4C1 can see every 3,5 in 21(4) cage -> no 3,5 in R4C1

45. 26(5) cage at R5C7 (step 35) = {24569/34568} = 456{29/38}, no 7
[Now we have 4,5,6 locked but I still missed the CPE elimination.]

46. R789C8 (step 32) = {159/168/249/267} (cannot be {348/357} which clash with R7C9 + R9C7, cannot be {258} -> R7C9 + R9C7 = {36} and there are no remaining combinations in R78C7 and R89C9), no 3
46a. 1 only in R9C8 -> no 5,8 in R9C8

47. 45 rule on N78 2 remaining innies R7C16 – 5 = 1 outie R9C7, R7C16 = [35/36/56] = 8,9,11 -> R9C7 = {346}, clean-up: no 4 in R7C9 (step 11)

48. Naked triple {356} in R7C169, locked for R7, clean-up: no 7 in R8C3, no 3 in R8C7

49. Naked pair {12} in R8C17, locked for R8

50. Naked quad (actually two naked pairs) {1278} in R3678C7, locked for C7
50a. R12469C7 (step 33a) = 9{3468/3567} = 369{48/57}

51. 1 in N6 locked in R4C89 + R5C9 -> no 1 in R3C9

52. 45 rule on N3 3 innies R12C7 + R3C9 = 15 = {249/357/456} (cannot be {267} because 2,7 only in R3C9)
52a. 7 only in R3C9 -> no 3 in R3C9

53. 16(3) cage at R8C6 = {268/349/358} (cannot be {259} because R9C7 only contains 3,4,6)
53a. 6 of {268} must be in R9C7 -> no 6 in R89C6
53b. 2 only in R9C6 -> no 8 in R9C6
53c. 8 only in R8C6 -> no 5 in R8C6

54. 21(4) cage at R3C6 = {2469/3459/3567} (cannot be {2379} because R4C5 only contains 5,6)
54a. Only combination with both 5,6 is {3567} when 7 must be in R3C6 -> no 5,6 in R3C6
54b. 9 in {2469/3459} must be in R4C7 -> no 4 in R4C7
54c. 21(4) cage at R3C6 must contain either 7 in R3C6 or 9 in R4C7 -> 21(4) cage at R1C6 must contain 7/9, valid combinations {2469/3459/3567}

55. 18(4) cage at R1C3 (step 36) = {1359/1368/1458/1467/2349/2358/2367/2457} (cannot be {1269} because 1,2,9 only in R12C4)
55a. R12C4 = {3456} -> R12C3 = {17/18/19/27/28/29}, no 3,4,5,6
55b. R12C3 contains 1/2 -> R13C2 must contain 1/2, R45C2 must contain 1/2 and R56C3 must contain 1/2

56. 13(3) cage at R2C2 = {157/247/256/346}
56a. 1,2 only in R3C2 -> no 5 in R3C2

57. R12C4 cannot be {56} (no valid combination for 18(4) cage at R1C3) -> R12C6 must contain 5/6 (cannot be both as that would clash with R7C6)
57a. Killer pair 5,6 in R12C6 and R7C6, locked for C6
57b. R12C6 contains 5/6 -> R12C4 must contain 5/6 -> 18(4) cage at R1C3 cannot be {2349}

58. 16(3) cage at R8C6 (step 53) = {268/349}
58a. R89C6 = {28/39/49}
58b. R789C5 (step 42b) = {267/357} (cannot be {249/348} which clash with R89C6) = 7{26/35}, no 4,8,9, 7 locked for C5

59. R5C4 = 7 (hidden single in C4)

60. Killer pair 5,6 in R7C6 and R89C5, locked for N8

61. Killer pair 5,6 in R4C5 and R89C5, locked for C5 -> R6C5 = 4, R4C6 = 3
[Alternatively R6C5 = 4 (hidden single in C5)]
61a. 14(4) at R6C3 (step 42a) = {1346} (only remaining combination) -> R6C34 = [36], 6 locked for D/
61b. R4C5 = 5, R9C3 = 5, R7C1 = 3, clean-up: no 7 in R8C9, no 6 in R9C7 (step 11)

62. 16(3) cage at R8C6 (step 58) = {349} (only remaining combination) -> R9C7 = 3, R7C9 = 6 (step 11), R7C6 = 5, R6C7 = 8 (cage sum), R78C7 = [72], 7 locked for D\, R3C7 = 1, locked for D/ -> R89C1 = [12], R7C5 = 2 (hidden single in R7), clean-up: no 5 in R8C9, no 9 in R89C9

63. Naked pair {49} in R89C6, locked for C6 and N8

and the rest is straightforward, remembering to make eliminations along the diagonals
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sudokuEd
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PostPosted: Sun Aug 19, 2007 12:10 pm    Post subject: Reply with quote

mhparker wrote:
here's my suggestion for an informal numeric rating scheme.
Time to touch base on this (all-consuming-of-late Shocked ) topic.

I've been trying very hard to develop a more formal/objective numeric score for Killers using the fantastic versatility of Sudoku Solver. Just wanted to show where I'm at....and to say a very big thankyou to everyone for giving ratings to the different puzzles recently. Please keep it up.

In summary, here is a list of the Sudoku Solver scores (SScores) for recent puzzles and for the puzzles in Mikes original rating post. Hope I haven't missed any.

[EDIT: These SScores are not accurate or valid. Careful!]

Quote:
Assassin 64V2 (SScore: 1.28)

Assassin 64 (SScore: 1.29) (Rating 1.0-1.25)

Assassin 63V2 (SScore: 1.40) (Rating est 1.75)

Assassin 63 V1.5 (SScore: 1.38) (Rating 1.0)

Assassin 63 (SScore: 0.98) (Rating 0.75)

Assassin 62V2 (SScore: 6.70) (Rating 2.0?)

Assassin 62 (SScore: 1.12) (Rating 0.75)

Assassin 61X V3 (DNS 17 steps No progress) (Rating 4.0)

Assassin 61X (SScore: 1.35) (Rating 1.75) [edit: score added]

Assassin 61X V3 (DNS 17 steps No progress)

Assassin 61 (SScore: 1.08) (Rating 1.25)

Special X Killer 4 V2 (DNS 8 No progress) (Rating 4.0)

Assassin 60 RP-Lite (SScore: 2.48) (Rating 1.75)

Assassin 60 RP (DNS 28 No progress) (Rating 3.0)

Assassin 60 (SScore: 1.40) (Rating 1.5)

Assassin 59 V1.5 (SScore: 1.62) (Rating 1.5)

Assassin 59 (SScore: 1.56) (Rating 1.25)

Assassin 58 V1.5 (SScore: 1.87) (Rating 1.5)

Assassin 58 (SScore: 1.20) (Rating 1.0)

Assassin 57V2X (SScore 2.13) (Rating 2.0) [edit: score added]

Assassin 57 V1.5 (SScore: 1.01) (Rating 1.0)

Assassin 57 (SScore: 0.88) (Rating 1.0)

Assassin 56 V2 (SScore: 3.04) (Rating 1.75)

Assassin 56 (SScore: 1.12) (Rating 1.25)

Assassin 55 V2 (DNS 30 No Progress) (Rating 2.0)

Assassin 55 (SScore: 1.46) (Rating 1.25)

Assassin 50 (SScore: 1.55) (Rating 1.75)

A48-Hevvie (SScore: 4.16) (Rating 2.5)

Assassin 1 (SScore: 0.69) (Rating 0.75)


The way these scores are calculated is based on 5 different combinations of killer techniques
A) Easy:
cage sums
naked/hidden pairs
45 rule - single innies/outies
locked cages

B) Essential
45 Rule - two innies/outies
45 Rule - single houses
45 Rule - multiple houses
Locked candidates

C) Default in Sudoku Solver

D) All except Bingo & Brute

E) All except brute

Assassin 1 requires only A) to solve: could be worth making a .5 rating for such as these?

Most Assassins require huge slabs of B) and C). I've used a bonus points approach to try and weight how much of B and C are needed by SS.

The hardest ones don't get far with B), need lots of C) and some of D). Once again, have tried to weight with bonus points. Assassin 62V2 is the only one with any of E).

Hopefully I've been absolutely consistent with the weightings to try and keep the whole thing fairly objective.

Anyway - feedback appreciated. I'm hoping that the SScores with replace the step counts on the Rating sticky if they seem accurate enough.

[edit: added scores for 2 X puzzles: bonus o.1 added for X.]

Cheers
Ed


Last edited by sudokuEd on Mon Aug 20, 2007 10:02 pm; edited 1 time in total
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mhparker
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PostPosted: Mon Aug 20, 2007 3:38 pm    Post subject: Reply with quote

sudokuEd wrote:
I've been trying very hard to develop a more formal/objective numeric score for Killers using the fantastic versatility of Sudoku Solver. Just wanted to show where I'm at....


Firstly, thanks for all your great work on this, Ed! Very Happy

However, I'm wondering whether you've created enough categories. Apart from Bingo (which I don't know much about, other than the fact that it's T&E) and Brute, you only have three categories (A, B and C). I am therefore concerned that there is not enough differentiation between the various techniques.

IMO, it would be better to have something like what Ruud has for the Daily Jigsaws, i.e. moves rated in 9 categories (1 - 9). This shouldn't apply to Bingo + Brute, of course, because even a single occurrence of these could in theory render the puzzle effectively unsolvable. So these should get very high weightings to reflect this.

For example, a killer triple should belong to a different (harder) category than a killer pair, as should a hidden killer pair (because these are often harder to spot than the naked ones). Similarly, conflicting combinations are much harder to spot if one or both of the cages are hidden and/or split, if not all cells are peers of each other, or if the conflict is only partial.

Also, (how) are you considering the number of steps? For example, twice as much of the same doesn't make for a twice as difficult puzzle. On the other hand, rating both puzzles identically would also be wrong.

Lastly (for now), Richard's solver would in theory have to strictly apply the moves in their difficulty order according to the new rating system, when it's ready. It would also make sense for SS to directly output the stats and/or ratings, so that anybody creating a puzzle can easily obtain and publish the SScore along with any puzzle. I assume that Richard is willing to rework some of his code accordingly to achieve this?
_________________
Cheers,
Mike
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rcbroughton
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Joined: 15 Nov 2006
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PostPosted: Mon Aug 20, 2007 4:08 pm    Post subject: Reply with quote

mhparker wrote:
Apart from Bingo (which I don't know much about, other than the fact that it's T&E)
Correct - it's a T&E approach that will try candidates in a cell to see if a contradiction is discovered in less than n-moves. Effectively, a candidate is trial-placed in a cell and a series of eliminations performed from that point, up to a defined number of steps. If the puzzle becomes invalid before this number of steps - duplicate candidates in a house, no remaining candidates in a house, all cage combinations eliminated - then the candidate is removed.

If you keep the number of look-ahead steps, "n", below around 5, then this is something that a human solver is more than capable of seeing in a puzzle. It's a short-range "What if". But a human solver is better at seeing likely places to apply this than the solver.

Quote:
Lastly (for now), Richard's solver would in theory have to strictly apply the moves in their difficulty order according to the new rating system, when it's ready. It would also make sense for SS to directly output the stats and/or ratings, so that anybody creating a puzzle can easily obtain and publish the SScore along with any puzzle. I assume that Richard is willing to rework some of his code accordingly to achieve this?
It would be relatively straightforward to assign a value for each elimination type and total these up for an automated solution. As you point out, though, you would get different totals depending on what options you had enabled and what order you had them run. (Might be an interesting exercise in it's own right to see how ratings are affected by what options you choose - but there are only so many hours in a day... )

The other difficulty (no pun intended) is what value to associate with which moves. What might seem a "difficult" move to one person may seem a "medium" move to another. (You might recall that historically, I tend to favour combination work before 45 Rules.)

We might also want to weight moves depending on other complexity factors. So, for example, a move that performs a cage elimination because of 1 external constraint may get a lower score than a similar move that required a combination of 2 external constraints. At the moment, these are done with equal importance in the solver - e.g. it doesn't look for for 1 constraint cages before looking for 2 constraint cages. To allow for this kind of subtlety would be a more significant change than merely rating solver move types and would also, potentially, make for longer processing times.

Rgds
Richard
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