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Assassin 66
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Para
Yokozuna
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Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

PostPosted: Mon Sep 03, 2007 10:32 pm    Post subject: Reply with quote

Hi all

That was a good puzzle somewhere between 1.50 and 1.75. Some fancy combination work and/with 45-tests did the trick.
Enjoy Laughing

Walk-Through Assassin 66V1.5

1. 11(3) at R1C7, R2C1, R7C3 and R9C7 = {128/137/146/236/245}: no 9

2. 20(3) at R2C3 and R5C4= {389/479/569/578}: no 1,2

3. 8(3) at R2C7 = {125/134}: no 6,7,8,9

4. 23(3) at R2C8 = {689} -->> locked for N3

5. R34C5 = {17/26/35}: no 4,8,9

6. 11(2) at R5C2, R5C7 and R6C5 = {29/38/47/56}: no 1

7. 8(3) at R6C2 = {125/134}: no 6,7,8,9; 1 locked for R6

8. 22(3) at R7C8 = {589/679}: no 1,2,3,4; 9 locked for N9
8a. 9 in C7 locked for N6

9. 45 on R1: 3 innies: R1C456 = 22 = {589/679}: no 1,2,3,4; 9 locked for R1 and N2
9a. 45 on R1: 3 outies: R2C456 = 9 = {126/135/234}: no 7,8

10. 45 on N1: 3 innies: R23C3 + R3C1 = 22 = {589/679}: no 1,2,3,4
10a. 45 on N1: 1 innie and 1 outie: R3C1 = R3C4 + 2: R3C4: no 8
10b. 8 in N2 locked within R1C456: R1C456 = {589} -->> locked for R1 and N2
10c. 11(3) at R1C7 = {137}(last combo) -->> locked for R1 and N3
10d. R1C123 = {246} -->> locked for N1
10e. 11(3) at R2C1 = {137}(last combo) -->> locked for N1
10f. Clean up: R2C456 = {126/234}: 2 locked for R2 and N2; R3C4: no 4

11. 45 on N3: 1 innie and 1 outie: R3C6 + 3 = R3C9 -->> R3C6 = 1; R3C9 = 4(only possible combo)
11a. R23C7 = [52]
11b. Clean up: R2C456 = {234} -->> locked for R2 and N2
11c. R3C2 = 3(hidden)
11d. Clean up: R3C1: no 5
11e. R3C3 = 5(hidden)
11f. 6 in R2 locked for N3
11g. More Clean up: R4C5 = {12}; R5C2: no 6; R5C3: no 8; R5C8: no 6

12. 45 on R1: 1 innie and 1 outie: R2C4 + 5 = R1C6 -->> R1C6: no 5; R2C4: no 2

13. 45 on R5: 2 innies: R5C19 = 3 = {12} -->> locked for R5
13a. Clean up: R5C237: no 9
13b. 9 in R5 locked for N5
13c. Clean up: R7C5: no 2

14. 19(5) at R3C9 = 4{1257/1356}: no 8; 1,5 locked for C9
14a. Killer Pair {37} in R1C9 + 19(5) at R3C9 -->> locked for C9

15. 30(5) at R3C1 = {24789/25689}: {15789} blocked by R2C1, {34689/35679/45678} blocked by R5C1 -->> no 1,3; R5C1 = 2; 8,9 locked for C1
15a. R5C9 = 1
15b. 3 in C1 locked for N7
15c. Killer Pair {46} in R1C1 + 30(5) at R3C1 -->> locked for C1

16. 45 on N7: 1 innie and 1 outie: R7C1 = R7C4 + 3 -->> R7C4: no 7,8

17. 45 on N9: 1 innie and 1 outie: R7C6 = R7C9 + 2 -->> R7C6: no 2,3,6

18. 45 on R9: 1 innie and 1 outie: R8C6 + 2 = R9C4 -->> R9C4: no 1,2,3; R8C6: no 8,9

19. 45 on R4: 3 innies: R4C159 = 10 = [415/523/613/712] -->> R4C1: no 8,9; R4C9: no 6,7

20. 14(3) at R7C6 = [9]{14}/[5]{18}/[7]{16}/{347}: [5]{36} blocked by step 17 -->> R7C6: no 8
20a. Clean up: R7C9: no 6

21. 45 on N78: 3 innies: R7C156 = 19 = {7[3]9/{469/478/68[5]} -->> R7C15: no 5
21a. Clean up: R7C4: no 2

22. 45 on N8: 3 innies: R7C456 = 16 = [169]/[187]/{349}/[385]/{36}[7]/{457}

23. Combining R7C456 with I/O N7 and N9
23a. R7C14569 = [41697/41875/96375/85742/63497/63942] -->> R7C1: no 7; R7C4: no 4; R7C6: no 5; R7C9: no 3

24. 45 on N7: 3 innies: R7C1 + R78C3 = 14 = [4]{28}/[6]{17}/[8]{24}/[9]{14} -->> R78C3: no 6

25. 18(3) at R9C1 = {369/378/459/567}: {189/279/468} blocked by innies N7: no 1,2

26. 13(3) at R7C2 = {139/157/238/256/346}: {148/247} blocked by innies N7

27. Combining steps 23 and 24: R7C14569 = [85742] clashes with R7C1 + R78C3 = [8]{24}
27a. R7C1 + R78C3 = [4]{28}/[6]{17}/[9]{14} = {4|6..}: R7C1: no 8; R7C14569 = [41697/41875/96375/63497/63942]: R7C4: no 5; R7C5: no 7
27b. Clean up: R6C5: no 4,6

28. 14(3) at R7C6 = [9]{14}/[7]{16}/[4]{37}/[7]{34} -->> R78C7: no 8

29. R7C9 = {257}; 22(3) at R7C8 = {5|7..}/{6|8..}
29a. 11(3) at R9C7 = {128/146/236}: no 7 = {6|8..}: {245} blocked by R7C9 + 22(3) at R7C8
: no 5
29b. Killer Pair {68} in 22(3) at R7C8 + 11(3) at R9C7 -->> locked for N9

30. 45 on R6: 3 innies: R6C159 = 19 = [982]/[9]{37}/[487]/{568} -->> R6C1: no 7; R6C5: no 2
30a. Clean up: R7C5: no 9
30b. R7C14569 = [41697/41875/96375/63497]: R7C6: no 4; R7C9: no 2

31. 14(3) at R7C6 = [9]{14}/[7]{34} -->> R78C7 = {14/34}: no 7, 4 locked for C7 and N9
31a. Clean up: R5C8: no 7

32. 11(3) at R9C7 = {128/236} = {3|8..} -->> 2 locked for R9
32a. 18(3) at R9C1 = {369/459/567} = {4|6..}: {378} blocked by 11(3) at R9C7
32b. Killer Pair {46} in 18(3) at R9C1 + Innies N7(step 27a) -->> locked for N7

33. 2 in N8 locked in R8C456 for R8
33a. 45 on R9: 3 outies: R8C456 = {238/247/256}: no 1,9

34. 30(5) at R3C1 = 2{4789/5689}: 7 only in R4C1 -->> R4C1: no 4
34a. R4C159 = 10 = [523/613/712]: R4C9: no 5

35. 19(5) at R3C9 = 14(257/356}: One of {23} goes in R4C9 -->> R6C9: no 2,3

36. R6C159 = 19 = [937/487]/{568} -->> R6C5: no 7
36a. Clean up: R7C5: no 4

37. R7C14569 = [41697/41875/96375] -->> R7C1: no 6; R7C4: no 3

38. R7C1 + R78C3 = [4]{28}/[9]{14}: no 7 -->> 4 locked for N7

39. 18(3) at R9C1 = {369/567}: no 8 -->> 6 locked for R9

40. 11(3) at R9C7 = {128}(last combo): no 3 -->> locked for R9 and N9

41. 22(3) at R7C8 = {679}(last combo): no 5 -->> locked for N9
41a. R7C9 = 5; R7C6 = 7
41b. R8C8 = 7(hidden); R7C4 = 1(hidden); R4C5 = 1(hidden); R3C45 = [67]
41c. R7C15 = [48](step 37); R6C5 = 3

42. 8(3) at R6C2 = {125}(last combo): no 4

And the rest is all naked and hidden singles

Andrew a welcome to the Masters from the Grandmaster Wink Laughing

greetings

Para


Last edited by Para on Thu Nov 01, 2007 11:29 pm; edited 5 times in total
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Andrew
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PostPosted: Mon Sep 03, 2007 11:59 pm    Post subject: Reply with quote

Thanks Para!

Your comment "some fancy combination work" gives me the incentive to keep trying. I can see some potentially promising areas such as outies from N4 and outies from N6 but haven't been able to use them yet.


Last edited by Andrew on Wed Sep 12, 2007 4:49 am; edited 1 time in total
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CathyW
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PostPosted: Tue Sep 04, 2007 8:24 am    Post subject: Reply with quote

Well done Para. Won't look at your WT yet but will keep trying too.

And congratulations Andrew on your "Master" status. I'll have to keep posting to catch you up! Smile
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mhparker
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PostPosted: Wed Sep 05, 2007 5:01 pm    Post subject: Reply with quote

Hi all,

Re: Andrew's post on the Assassin 53 thread:

Andrew wrote:
Some interesting thoughts from Ed. If there are too many variants each week then people will limit the number of puzzles that they try to solve. For Assassin 53 I've solved Ruud's original and am part way through his V2. I probably won't try the other ones as they are clearly much harder and because we now have Assassin 54. I'm pleased to see that Ed then posted his walkthrough for J-C's V3. I'll try to find time to look at that.

Since we don't want to discourage any forum member from posting variants, I would suggest that anyone except Ruud should limit themselves to one variant each week.

I'm beginning to wonder whether one variant is always enough. I get the impression that, whilst some forum members would like a really tough puzzle (say, rating 1.75 or 2.0) that they can work on for a couple of days (possibly opening it up as a tag solution), others just want a kind of "mid-week V1" (rating around 1.25) that is less frustrating to solve. Therefore, assuming that both versions are available, I would argue that it's not a bad idea to also post a "Lite" version whenever a real "toughie" is published. However, if the variant is around the 1.5 rating, then this may not be necessary, as the difficulty may be just hard/easy enough to satisfy both demands.

Also, I want to stress that, whilst I'm happy to publish these mid-week variants on a (semi-)regular basis, I'm also happy if someone else wants to publish one instead. BTW, I won't publish any variants of my own without prior consultation if either Ruud himself has indicated that a variant is planned for the forum, or if I am notified via a forum post or PM that another forum member is working on one. I also won't normally post a variant until at least a couple of days after the V1 has appeared, so that ample time has been provided for the V1 to be fully "digested", as it were. If anybody else has already published a variant during this time period, I won't normally publish any myself, even if I have been working on one.

Would be interested in other people's opinions on either of these topics.
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Mike
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CathyW
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PostPosted: Thu Sep 06, 2007 8:52 pm    Post subject: Reply with quote

I gave up on the A66 V1.5 and just checked through Para's WT. Nothing short of brilliant imho! Especially combining steps to create new restrictions on combinations.

In response to Mike's post, and speaking as one who could stop any time but chooses not to (!), I certainly welcome the variants, especially when the original Assassin has been at an easier level. Even if I don't manage to solve a puzzle myself - see above! - going through someone else's WT all adds to the learning curve.
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Andrew
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PostPosted: Fri Sep 07, 2007 9:59 pm    Post subject: Reply with quote

I'm still "nibbling away" at A66V1.5 but haven't looked at it for a couple of days. I decided I'd better have a go at Mike's Vortex Lite, which I started yesterday evening, then I'll have to see what A67 is like.

On Mike's question of how many variants should be posted, the number will be right if walkthroughs get posted for all of them, individually or as "tag" solutions, except possibly ones that are so hard that they qualify as Unsolvables. If there are too many variants posted, then they won't all get walkthroughs.

I try to solve variants when I can but can't keep up with all of them. Still I wouldn't want to deprive others of them so keep it up Mike, Ruud and others!
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gary w
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PostPosted: Sun Sep 09, 2007 12:40 am    Post subject: Reply with quote

i haven't got time to post a complete wt but my solution to a66 v1.5 was along the following line...hope you can follow it.

various easy prelims include showing r2c12=1/7 therefore r5c1=2 and r5c9=1 because of combos in 30/5 cage

r7c1 is not a 9 otherwise both r3c4 and r7c4 would be a 6

r67c1 + r67c9 = 25 r7c159 = 17

in c9 2 is either in the 19/5 cage (at r4c9) or at r9c9 but if the latter then r67c9 = 65 (only combo that works for the 19/5 cage) which then >> r67c1 = 86

but then there are two 6s in r7c159

thus the 19/5 cage in c9 is (r3-r7) 421 5/7... this breaks the puzzle



couldn't solve 65v3 without a hypothetical but it wasn't TOO hard to deduce the number that has to go in r7c9 after which no more hypos were needed.Has anyone solved it logically yet?

both a66 and a67 took me about 1.5-2 hrs to do.A "deadly" killer in The Times normally takes me about 0.5 hrs so i'ld rate both of these as significantly harder
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sudokuEd
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PostPosted: Sun Sep 09, 2007 12:43 pm    Post subject: Reply with quote

mhparker wrote:
Assassin 66 V1.5.. I would be interested to know what rating Richard's software would give to this:
The most recent method would get a score of 1.52.

The same scoring method gets a 1.25 for A66.

Some of the scores sound OK - throw in A67 as 0.89 - but the problem is the number of forum ratings we/I have given that are way off the Sudoku Solver scores.

One thing I'm trying is to get numerical scores based on forum walk-throughs to give some substance to your gut ratings. Here are the first attempts.
BTW - would really love some help doing this for a range of past walk-throughs (Andrew's probably - see below. Smile ).


Assassin 66
CathyW wrote:
I only used Innies, Outies, Outies-Innies, cage combinations, killer combos and combination analysis
and some very clever combo conflicts that are very hard to give a simple score to. Nice work!

Code:
A66 CathyWT   Score: 0.98
4   Naked2   
3   Naked3   
    Naked4   
    NakedTriangles   
   
    Hidden2   
    Hidden3   
    Hidden4   
   CageCleanup   
   CageSumSimple   
1   CageSumExtended   
7   LockedCages   
2   LockedCagesExtended   
9   CageCombinationsSimple   
3   CageCombinationsExtended   
   LockedCandidates   
6   CagePlacementSimple   
7   CagePlacementExtended   
   CagePlacementInsane   
3   KillerPairs   
   Rule45SingleCell   
9   Rule45DoubleCell   2
   Rule45SingleHouse   2
5   Rule45SingleHouse   3
   Rule45SingleHouse   4
   Rule45SingleHouse   5
   Rule45SingleHouse   6
   Rule45SingleHouse   7
   Rule45MultiHouses   2
2   Rule45MultiHouses   3
   Rule45MultiHouses   4
   Rule45MultiHouses   5
   Rule45MultiHouses   6
   Rule45MultiHouses   7
4   CageBlockersSimple   
3   CageBlockersExtended   


A66 Andrew WT  Score: 1.32
7   Naked2   
3   Naked3   
2   Naked4   
   NakedTriangles   
   Hidden1   
   Hidden2   
2   Hidden3   
   Hidden4   
   CageCleanup   
4   CageSumSimple   
   CageSumExtended   
5   LockedCages   
1   LockedCagesExtended   
11   CageCombinationsSimple   
2   CageCombinationsExtended   
1   LockedCandidates   
10   CagePlacementSimple   
4   CagePlacementExtended   
   CagePlacementInsane   
3   KillerPairs   
   Rule45SingleCell   
14   Rule45DoubleCell   2
   Rule45SingleHouse   2
6   Rule45SingleHouse   3
2   Rule45SingleHouse   4
   Rule45SingleHouse   5
   Rule45SingleHouse   6
   Rule45SingleHouse   7
   Rule45MultiHouses   2
1   Rule45MultiHouses   3
2   Rule45MultiHouses   4
   Rule45MultiHouses   5
   Rule45MultiHouses   6
   Rule45MultiHouses   7
8   CageBlockersSimple   
1   CageBlockersExtended   


Cheers
Ed
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mhparker
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PostPosted: Sun Sep 09, 2007 6:05 pm    Post subject: Reply with quote

Hi Gary + welcome to the forum!

gary w wrote:
hope you can follow it.

No, unfortunately I can't, in particular the bit in bold below:

gary w wrote:
in c9 2 is either in the 19/5 cage (at r4c9) or at r9c9 but if the latter then r67c9 = 65 (only combo that works for the 19/5 cage) which then >> r67c1 = 86


Other permutations of these digits within 19(5) work for me at the position where the puzzle starts to get difficult. Without a WT, we would need at least a marks pic (i.e., candidates diagram, as produced by SumoCue or other software), in order to at least know the grid state at the time you made the move. Although, once we had this, this could give rise to other questions, like how you got rid of some of the other candidates.

However, your move to eliminate the 9 in r7c1 was very interesting and is a bona fide chain (not a hypothetical). As mentioned in my post on the Vortex Killer thread yesterday, I normally re-work them so that they can be expressed as a loop. VIZ:

Code:
r7c1=9 -> r3c1<>9 => r3c1=8, r3c4 = 6 -> r7c4<>6, r7c1<>9 (contradiction)


The links between r3c1 and r3c4, and between r7c1 and r7c4 (which I've represented by the comma above) are an example of direct links, as discussed in my recent post here. In that post, I explicitly mentioned the possibility of using direct links based on innie/outie difference groups with 1 innie and 1 outie, as you're doing here. So it's nice to see a practical example of this in action so quickly.

Many thanks and keep those thoughts and ideas coming!
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Mike
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gary w
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PostPosted: Sun Sep 09, 2007 9:47 pm    Post subject: Reply with quote

Thanks for your welcome Mike..I appreciate my brief notes are difficult to follow.Essentially if the 19/2 cage doesn't contain a 2 it must be (r3-r7) 4 (356) 1 (356)(356).But all these can be eliminated using very short chains based on

r7c1 is not 9
r67c1+ r67c9=25
r7c159=17
r7c6=r7c9+2
and the numbers in the 11/2 cage at r67c5 which fix r3c5 as either a 6 or 7 and thus r3c1 aseither an 8 or a 9.

No guesswork or pencil marks needed.Does this,then,constitute an AIC ???
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mhparker
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PostPosted: Mon Sep 10, 2007 9:14 am    Post subject: Reply with quote

Hi Gary,

I've been doing some more thinking about your moves in the light of your last post, which I'd like to clarify a bit more here so that I (and other forum members) can better understand them. To make things easier, I'll refer to the following grid state, which corresponds to how far SumoCue v1.3 gets if one manually sets r5c19 to [21] (due to logic you mentioned in your first post, corresponding to step 15 in Para's WT):

Code:
.-----------------------------------.-----------------------.-----------.-----------------------------------.
| 46          246         246       | 589         589       | 89        | 137         137         37        |
:-----------------------.-----------:           .-----------'           :-----------.-----------------------:
| 17          17        | 89        | 34        | 234         234       | 5         | 689         689       |
:-----------.           |           '-----------+-----------.-----------'           |           .-----------:
| 89        | 3         | 5           67        | 67        | 1           2         | 89        | 4         |
|           :-----------'-----------------------:           :-----------------------'-----------:           |
| 4567      | 1456789     1346789     12345678  | 12        | 2345678     346789      2345678   | 235       |
|           :-----------------------.-----------'-----------'-----------.-----------------------:           |
| 2         | 4578        3467      | 3456789     3456789     3456789   | 34678       34578     | 1         |
|           :-----------------------'-----------.-----------.-----------'-----------------------:           |
| 456789    | 145         134         1234      | 234578    | 2345678     346789      2345678   | 2567      |
|           :-----------.-----------------------:           :-----------------------.-----------:           |
| 46789     | 12456789  | 124678      13456     | 346789    | 4579        134678    | 56789     | 2357      |
:-----------'           |           .-----------'-----------+-----------.           |           '-----------:
| 1357        12456789  | 124678    | 123456789   12345678  | 234567    | 134678    | 56789       5689      |
:-----------------------'-----------:           .-----------'           :-----------'-----------------------:
| 1357        2456789     246789    | 456789    | 123456789   23456789  | 13468       1234568     2568      |
'-----------------------------------'-----------'-----------------------'-----------------------------------'


This corresponds roughly (but not exactly) to the position reached by Para after his step 21.

Note that {67} have been eliminated from r4c9 due to permutations on r4 innies (r4c159 = 10(3)) (Para's step 19). The 6 in r7c9 has been eliminated due to fact that r7c6 cannot be 8, otherwise cannot make up 14(3) cage total. Hence, via innie/outie difference n9, r7c9 can't be 6 (Para's step 20). Lastly, the 3 has gone from r6c9, because otherwise there would be no way of reaching the 19(5) cage total.

As you say, the next step is to remove the 9 from r7c1. Since my last post, I've realized that there is another way of doing this, namely by using innie/outie difference on n245689 (yes, 6 nonets!!), as follows:

Code:
2 outies (r37c1) = 2 innies (r37c4) + 5

Using this equation, r7c4 can't be a 6, because this would force r3c4 to be 7, resulting in the innies summing to 13. This would require that the outies sum to 18, which is clearly impossible, because they are peers of each other. Thus we can eliminate 6 from r7c4, which in turn eliminates 9 from n7 due to innie/outie difference on n7.

This results in the following grid state:

Code:
.-----------------------------------.-----------------------.-----------.-----------------------------------.
| 46          246         246       | 589         589       | 89        | 137         137         37        |
:-----------------------.-----------:           .-----------'           :-----------.-----------------------:
| 17          17        | 89        | 34        | 234         234       | 5         | 689         689       |
:-----------.           |           '-----------+-----------.-----------'           |           .-----------:
| 89        | 3         | 5           67        | 67        | 1           2         | 89        | 4         |
|           :-----------'-----------------------:           :-----------------------'-----------:           |
| 4567      | 1456789     1346789     12345678  | 12        | 2345678     346789      2345678   | 235       |
|           :-----------------------.-----------'-----------'-----------.-----------------------:           |
| 2         | 4578        3467      | 3456789     3456789     3456789   | 34678       34578     | 1         |
|           :-----------------------'-----------.-----------.-----------'-----------------------:           |
| 456789    | 145         134         1234      | 234578    | 2345678     346789      2345678   | 2567      |
|           :-----------.-----------------------:           :-----------------------.-----------:           |
| 4678      | 12456789  | 124678      1345      | 346789    | 4579        134678    | 56789     | 2357      |
:-----------'           |           .-----------'-----------+-----------.           |           '-----------:
| 1357        12456789  | 124678    | 123456789   12345678  | 234567    | 134678    | 56789       5689      |
:-----------------------'-----------:           .-----------'           :-----------'-----------------------:
| 1357        2456789     246789    | 456789    | 123456789   23456789  | 13468       1234568     2568      |
'-----------------------------------'-----------'-----------------------'-----------------------------------'


Now to apply your key move:

If 19(5) at r3c9 is {13456}, then r67c9 must be [63/65]. Since r67c19 = 25 (innies r6789), possible permutations for r67c19 (where "x" = r7c5) are:

Code:
9       6
7---x---3

8       6
6---x---5


However, both of these possibilities are blocked due to innies of r789 = r7c159 = 17(3), which would require x (i.e., r7c5) = 7 in the first case and x = 6 in the second. In both cases, r7c5 would clash with r7c1. Thus, we can reject the {13456} combo for 19(5) at r3c9, which must now be {12457}.

Nice creative move! Congratulations! Very Happy

Apart from my alternative way of stripping the 9 from r7c1, does this correspond more or less to the logic you were using?

Thanks once again for providing this move sequence. Keep up the good work! BTW, do you do your puzzles by keeping close track of the remaining candidates using software, or do you do them all on paper, using only rough pencilmarks? In any case, 30 minutes for a Times Deadly is pretty good going, by any means! I used to take about 50 minutes for them on average, although it's a long time ago since I did the last one. (Cathy: if you're reading this, how long do you take to do them?).
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mhparker
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PostPosted: Mon Sep 10, 2007 10:07 am    Post subject: Reply with quote

Hi Ed,

Sounds like you and Richard are making great progress with the ratings.

sudokuEd wrote:
One thing I'm trying is to get numerical scores based on forum walk-throughs to give some substance to your gut ratings. Here are the first attempts.
BTW - would really love some help doing this for a range of past walk-throughs (Andrew's probably - see below. Smile ).

Just occurred to me that there may be a problem with doing this. Many (indeed, probably most) people on the forum don't write optimized WTs, but instead methodically work their way around the puzzle until they hit upon the key move, documenting everything exactly as they found it. The thing is, wouldn't the less productive moves (that are not on the critical solution path) that were found before stumbling across the key big move unnecessarily bump up the rating?

Expressed another way: isn't the rating of previous WTs only going to be accurate if they are optimized first?

Just a thought.
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CathyW
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PostPosted: Mon Sep 10, 2007 11:01 am    Post subject: Reply with quote

I still do the Times puzzles every day - always in less time than they suggest! Deadlys on Fri/Sat vary but, like Gary, 30 mins is probably an average - I'll have to time myself! I would be surprised if many of the Times Deadlys rated above 0.5.
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sudokuEd
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PostPosted: Mon Sep 10, 2007 11:28 am    Post subject: Reply with quote

mhparker wrote:
isn't the rating of previous WTs only going to be accurate if they are optimized first?
Could be - we'll see I guess. Sudoku Solver works like this too - not optimized at all. Would be so easy if it just did the most productive thing for each step rather than strictly in the routine order - humans are so much better at learning where to look first/next. That's why I'm using an average of different settings to get the current set of (touch-wood emoticon here) accurate ratings.

Might have to average the human WT's too - Andrew & Cathy would average out to 1.16 for A66. Or, may just use the one puzzler (read Andrew) who is the most methodical amoung us to be the reference point. Or at least to get a feel if there really is a strong link (or should I say now, direct link Very Happy ) between perceived difficulty and number/type/order of solve steps.

For example, am having a terrible time with A60 - one we generally felt was very hard and rated at 1.5 - but (current) SSscore is closer to 1.0. That is not a very good correlation. No way it was easier than A66. Was it Para Question

Anyway - still trying out things. Thanks for the thoughts Mike.

Cheers
Ed
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gary w
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PostPosted: Mon Sep 10, 2007 2:29 pm    Post subject: Reply with quote

Thanks very much Mike for your detailed reply.I have to say that your logic is virtually IDENTICAL to that I employed!!

I haven't been able to look at Para's wt because it's still in tiny text and I have to admit I don't know how to expand to normal font!

I always work with just pencil and the puzzle..no software although that's something I'ld like to get into.

Thanks once again..I'm looking forward to the next assassin/variants now I've discovered the site - it's marvellous.
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