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Assassin 68
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frank
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Joined: 07 Oct 2006
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Location: Victoria, B.C., Canada

PostPosted: Fri Sep 14, 2007 5:39 am    Post subject: Assassin 68 Reply with quote

Nice one. Here is my jiggery-pokery (T & E) walkthrough:

1. Outies of C12 => R19C3 = 6

2. Innies of R89 => R8C37 = 5

3. Innies of N8 => R7C456 = 8

4. Outies of N7 => R6C1+R7C4 = 4

=> R6C1+R7C4 = [13], [22], or [31]

Now comes the jiggery-pokery part Smile

[13] quickly leads to a contradiction

[22] quickly leads to a contradiction

So R6C1+R7C4 = [31]

And the rest is plain sailing


Last edited by frank on Fri Sep 14, 2007 7:27 pm; edited 2 times in total
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CathyW
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PostPosted: Fri Sep 14, 2007 7:52 am    Post subject: Reply with quote

Good to see you here Frank! I've only just printed out the puzzle to look at in lunch break later.
ttfn
Cathy x

Edit:
Here's my WT (took a couple of hours including doing the WT - Yes it was an extended lunch break!!) - a reasonable challenge if you don't use Frank's method! Estimated rating 1-1.25 (Para's trick doesn't apply either!)

Prelims

a) 13(2) N2 = {49/58/67}
b) 10(3) @r1c6 = {127/136/145/235}
c) 26(4) @r3c3: no 1
d) 14(4) @r3c7: no 9
e) 22(3) @r5c7 = {589/679} -> r6c89 <> 9
f) 19(3) @r5c8, r6c9: no 1
g) 11(3) @r6c1, r7c3: no 9
h) 6(2) r67c5 = {15/24}
i) 10(2) N7: no 5
j) 20(3) N8: no 1,2
k) 5(2) N9 = {14/23}

1. Outies c12: r19c3 = 6 = {15/24}

2. Outies c89: r19c7 = 13 = {49/58/67}

3. Outies r1234: r5c456 = 10 = {127/136/145/235}

4. Outies r789: r6c159 = 11 = {128/137/146/236/245}

5. Innies r89: r8c37 = 5 = {14/23}

6. Innies N8: r7c456 = 8 = {125/134} -> 1 n/e r7/N8

7. Outies N7: r6c1 + r7c4 = 4 = {13/22} -> max from r6c15 = 3+5=8 -> r6c9 <> 2,3

8. Outies N9: r6c9 + r7c6 = 9 = [45/54/63/72/81]

9. 9 locked to r7c789
-> 16(3) N9 does not contain 9 -> r1c7 <> 4

10. 17(3) @r8c2: r9c3 max 5 -> r89c2 min 12 -> r89c2 <> 1,2
Options: {179/269/278/458/467/539}

11. 11(3) @r7c3: r78c3 = 8; 9; 10 = [71/62]; [81/63]; [82/73/64] -> r7c3 = (678)
(combo [542] blocked by split 6(2) r19c3)

12. 14(3) @r7c6: max from r7c6+r8c7 = 5+4=9 -> r7c7 = (6789)

13. 1 locked to r4c789 n/e r4

14. Innies c1234: r389c4 = 20 = {389/479/569/578} (r3c4 <> 1,2; r9c4 <> 2)

15. Innies c6789: r389c6 = 17

16. 11(3) @r6c1 = [128/182/137/173/146/164/236/263/326/362]
[245/254] blocked by split 8(3) in r7c456 -> r7c12 <> 5
17(3) @r8c2 = 5{39/48}
-> r9c3 = (45) -> r1c3 = (12).

17. 17(3) r9c456 = {269/278/368/467}; {359/458} blocked by split 8(3)
-> r9c456 <> 5

18. Innies N7: r7c123 + r8c3 = 18 = {1278/1368/1467/2367} -> r8c3 <> 4 -> r8c7 <> 1
Mike noted that {1368} blocked by 10(2)n9 (or 17(3)n9) which I missed at the time I did the puzzle. Would have been helpful if I'd seen it!


19. Innies N9: r7c789 + r8c7 = 24 = {2589/2679/3579/4569} -> r7c89 <> 2,3,4
({3489} blocked by 5(2) r89c9)
Thanks to Andrew: for completeness {3678} is blocked by r7c3. {4578} is also impossible because r8c3 = 1, r7c3 = 6 cannot give a valid combination in the 11(3) cage.


20. 19(3) @ r6c9 = {469/478/568} -> r6c9 <> 7 -> r7c6 <> 2
Must have at least one of 6,8 within r7c89 -> 16(3) @r8c8 <> {268} -> r89c8 <> 2

21. KP {68}: split 18 (4) r7c123 and r7c89 must have at least one of 6,8
-> r7c7 <> 6,8
14(3) = [194/392/572/374/473]
-> 22(3) @r5c7 can’t have both {79} within r56c7 -> r6c6 <> 6

22. Outies N8: r7c37 + r6c5 = 18 = {189/279/567} -> r6c5 <> 4 -> r7c5 <> 2

23. Split 11(3) r6c159 = [128/218/326/254} -> 2 n/e r6; r6c9 <> 5 -> r7c6 <> 4
-> r8c7 <> 3 (see options for 14(3) in step 21) -> r8c3 <> 2

24. KP {24} in N9 between 5(2) and r8c7 -> 16(3) <> 4 -> r1c7 <> 9

25. 2 locked to r789c1+r7c2 in N9 -> r6c1 <> 2 -> r7c4 <> 2 -> r7c3 = 7
-> r89c1 <> 3; -> r7c7 = 9 -> r6c6 = 9; r7c6 <> 5, r6c9 <> 4

26. NP {13} r7c46 n/e N8/r7 -> r7c5 = 4 -> r6c5 = 2

27. HS: r9c6 = 2 -> r8c9 <> 3, r8c1 <> 8

28. r7c12 = {26/28} -> r8c1 <> 2; r9c1 <> 8

29. r7c89 = {56/58} -> 16(3) N9 <> 5 -> 16(3) = 7{18/36} -> r1c7 <> 8

30. Naked Quad {5678} r1569c7 -> r234c7 = (1234)

31. KP {68} in N6 between r6c9 and r56c7 -> 6,8 n/e N6
-> 19(3) @ r5c8 = {379} n/e N6
-> r56c7 = {58} n/e c7/N6; r19c7 = {67}
-> 9 locked to r5c89 n/e r5
-> r4c789 = {124} n/e r4
-> r6c9 = 6 -> r7c6 = 3 -> r7c4 = 1 …

Fairly straightforward from here with singles and cage combos.
Very Happy


Last edited by CathyW on Mon Sep 24, 2007 7:57 am; edited 3 times in total
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Howard S
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PostPosted: Fri Sep 14, 2007 4:10 pm    Post subject: Reply with quote

Well done Frank - you made that look easy - I have been staring at this on and off all day without success.

Why is Frank's approach deemed to be T&E? To me it is a sensible elimination using the 3 options available. It is not as if you need to go through the whole killer to find the contradiction, they are very quickly spotted.

T&E is surely picking a number out of the ether and trying it until a contradiction arises and then trying another etc with no logical support.

PS Congratulations Frank on joining me at Rookie level Smile

Edit - PPS Just noticed I have made it to Regular level with this post - come on Frank keep up. Laughing


Last edited by Howard S on Fri Sep 14, 2007 4:13 pm; edited 1 time in total
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Para
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Joined: 08 Nov 2006
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PostPosted: Fri Sep 14, 2007 4:12 pm    Post subject: Reply with quote

Hi all

It was a really fun puzzle. I enjoyed solving it. I agree with Cathy about the rating. About a 1.00-1.25.

Cathy wrote:
Para's trick doesn't apply either!

Yeah that would be getting a bit one-sided. Also the center nonets lose their transformers look a bit, and start looking more like Dr. Eggman from the Sonic game. (Childhood memories, ahhhh(yeah i played Sega))

Here's my Walk-Through

Walk-through Assassin 68

1. R12C5 = {49/58/67}: no 1,2,3

2. 10(3) at R1C6 = {127/136/145/235}: no 8,9

3. 26(4) at R3C3 = {2789/3689/4589/4679/5678}: no 1

4. 14(4) at R3C7 = {1238/1247/1256/1346/2345}: no 9

5. 22(3) at R5C7 = {589/679}: no 1,2,3,4; 9 locked in cage -->> R6C89: no 9

6. 19(3) at R5C8 and R6C9 = {289/379/469/478/568}: no 1

7. 11(3) at R6C1 and R7C3 = {128/137/146/236/245}: no 9

8. R67C5 = {15/24}: no 3,6,7,8,9

9. R89C1 = {19/28/37/46}: no 5

10. 20(3) at R8C4 = {389/479/569/578}: no 1,2

11. R89C9 = {14/23}: no 5,6,7,8,9

12. 45 on R89: 2 innies: R8C37 = 5 = {14/23}: no 5,6,7,8,9

13. 16(3) at R8C8 = {169/178/259/268/358/367/457} = {1|2|3|4..}: {349} blocked by R89C9
13a. Killer Quad {1234} in R8C7 + R89C9 + 16(3) at R8C8 -->> locked for N9

14. 45 on N7: 2 outies: R6C1 + R7C4 = 4 = {13/22}: no 4,5,6,7,8,9

15. 11(3) at R6C1 = {128/137/146/236/245}: needs one of {5678} in R7C12
15a. 11(3) at R7C3 = {128/137/146/236/245}: needs one of {5678} in R7C3 -->> R7C3 = {5678}
15b. Killer Quint {56789} in R7C12 + R7C3 + R7C789 -->> locked for R7
15c. Clean up: R6C5: no 1

16. 45 on N8: 3 innies: R7C456 = 8 = {134}: no 2 -->> locked for R7 and N8
16a. 2 in R7 locked for N7 and 11(3) cage at R6C1
16b. 11(3) at R6C1 = [1]{28}/[3]{26}: no 5,7
16c. Clean up: R8C7: no 3(step 12); R89C1: no 8

17. 45 on N9: 2 outies: R6C9 + R7C6 = 9 = [54/63/81] -->> R6C9 = {568}
17a. 19(3) at R6C9 = {568}(last combo): no 7,9
17b. R7C7 = 9(hidden); R6C6 = 9(last place in 22(3) cage at R5C6)
17c. R7C3 = 7(hidden)
17d. 5 in R7 locked for N9 and 19(3) at R6C9
17e. Clean up: R89C1: no 3

18. 11(3) at R7C3 = 7{13}: no 4
18a. Clean up: R8C7: no 1

19. 14(3) at R7C6 = 9[14/32]: R7C6: no 4
19a. R7C5 = 4(hidden); R6C5 = 2

20. 22(3) at R5C7 = 9{58/67} -->> R56C7 = {58/67} = {6|8..}
20a. Killer Pair {68} at R56C7 + R6C9 -->> locked for N6

21. 19(3) at R5C8 = {379}(last combo): no 2,4,5; locked for N6; 9 locked in R5C89 for R5
21a. R56C7 = {58}(last combo): no 6 -->> locked for C7 and N6
21b. R6C9 = 6

22. R7C89 = {58} -->> locked for R7 and N9
22a. 11(3) at R6C1 = {236} -->> R6C1 = 3; R7C12 = {26} -->> locked for N7
22b. R6C8 = 7
22c. R89C1 = {19}(last combo) -->> locked for C1 and N7
22d. R8C3 = 3; R7C46 = [13]; R8C7 = 2; R9C7 = 7(hidden)

23. 16(3) at R8C8 = 7[63](last combo): R89C8 = [63]
23a. R5C89 = [93]

24. 45 on C89: 1 outie: R1C7 = 6

25. 20(3) at R8C4 = {578}(last combo) -->> locked for R8 and N8
25a. R8C2 = 4; R89C9 = [14]; R89C1 = [91]; R4C9 = 2

26. 45 on C12: 2outies: R19C3 = 6 = [15](only combo)
26a. R9C2 = 8

27. 14(3) at R5C3 = {248}(last combo) -->> R5C3 = 2; R6C34 = {48} -->> locked for R6
27a. R56C7 = [85]; R6C2 = 1

28. Naked Pair {14} in R4C78 -->> locked for R4

29. 14(3) at R5C1 = 1{67}(last combo): R5C12 = {67} -->> locked for R5 and N4
29a. R6C3 = 4(hidden); R6C4 = 8

30. 6 in C3 locked for N1

31. 14(4) at R3C7 = {1346}(last possible combo): no 5,7; R4C6 = 6; R3C7 = 3
31a. R4C5 = 3(hidden); R4C4 = 7(hidden); R9C6 = 2; R8C4 = 5
31b. R5C456 = [451]; R4C78 = [41]; R2C7 = 1; R3C5 = 1(hidden)

32. 13(3) at R1C5 = [76](last combo)

33. 15(4) at R2C9 = 12{57}(last combo): R23C9 = {57} -->> locked for C9 and N3

And the rest is all naked and hidden singles

greetings

Para


Last edited by Para on Thu Nov 01, 2007 11:38 pm; edited 1 time in total
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Para
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PostPosted: Fri Sep 14, 2007 4:28 pm    Post subject: Reply with quote

Howard S wrote:

T&E is surely picking a number out of the ether and trying it until a contradiction arises and then trying another etc with no logical support.


Hi

It all depends on how you constitute T&E. Under your definition it is more like a wildly inappropriate hypothetical, because there's a lot of things to work with that Frank hadn't looked at. It is mostly deemed as T&E because you have no way of knowing if you are trying the valid or invalid combination first, so you could just as well have started with the valid one and finished the puzzle and then go back to disprove the invalid ones, just to show your first choice was the only correct one.

But as i said before, it all depends on the choice you make. If your only purpose is to solve the puzzle, then you're free to use those moves. If you also want to find a nice path to the solution, than there's a lot of other moves to work with to help you to your solution.

greetings

Para
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Andrew
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PostPosted: Fri Sep 14, 2007 4:57 pm    Post subject: Reply with quote

Cathy wrote:
Para's trick doesn't apply either!

Para wrote:
Yeah that would be getting a bit one-sided. Also the center nonets lose their transformers look a bit, and start looking more like Dr. Eggman from the Sonic game. (Childhood memories, ahhhh(yeah i played Sega))

I set up the diagrams for A68 and the two Transformers Killer-Xs late last night and did notice that Ruud had changed the diagram a bit in the lower half. Not surprised therefore that Para's trick no longer applies although I haven't tried any of these puzzle yet so I've no idea what it is except that it's about the cage pattern; I assume it must be in the lower half because that's what Ruud changed.

It will be a few days before I try any of these puzzles. I must get some other things finished first. If I post anything about these puzzles it won't be until next week.
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gary w
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PostPosted: Sat Sep 15, 2007 11:44 pm    Post subject: a68 Reply with quote

I wonder if the moves outlined below constitute an AIC or a hypothetical?
In any event,having seen this route I was disinclined to look for others!

prelims...

r6c9+r7c6=9
r6c1+r7c4=4 > r6c1,r7c4=13/22/31
r8c3=r8c7=5 > 1/4 2/3
r6c159=11
r7c456=8 > (1/3/4) (1/2/5)

r6c9 <> 8 otherwise in N6 R4C789 would be 5 maximum..impossible

thus r7c6<> 1

If R7C5=1 >R6C5=5 >R6C9=4 >R6C1=2 >R7C456=215
thus r7c12=(3/6)..X

Now,both r78c3 and r78c7 =9 and the 5 at r7c6 > r8c3+r8c7<> (1/4)>r8c3=3
Contradiction wth X.

Thus r7c4=1.This breaks puzzle.

(size corrected - Ruud)
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mhparker
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PostPosted: Mon Sep 17, 2007 6:57 pm    Post subject: Reply with quote

Hi all,

This post is aimed at answering one or two of the questions above.

frank wrote:
Here is my jiggery-pokery (T & E) walkthrough

Howard S wrote:
Why is Frank's approach deemed to be T&E?

Unfortunately, I'm not able to answer that question in this specific case, because Frank did not provide any detail as to why or how each of the two combinations he mentioned "quickly leads to a contradiction". This is a pity, because this is the most interesting bit! Furthermore, the logic leading to the contradiction is not obvious enough to render an explanation superfluous. Although Frank could no doubt provide more information here, it's my experience that, on having to explain the logic to others, the logic used suddenly appears to be nowhere near as simple as one originally thought!

gary w wrote:
I wonder if the moves outlined below constitute an AIC or a hypothetical?

Thanks very much for listing the steps you took in some detail, Gary. Much appreciated. To answer your question: hypothetical - yes, AIC - no.

In order to qualify as an AIC, three key criteria need to be met:

  1. The implications must be expressible as a linear chain or loop. That is, there should be no branching.
  2. Each link must not depend on any previous links. Note that, since each implication is in itself bidirectional, this requirement makes the entire chain bidirectional.
  3. The chain should start on a false premise and end on a true premise. The idea is to show that at least one of the premises at either end of the chain must be true.

Note that, although your logic is great as a hypothetical, it fails to meet any of the above requirements for being an AIC. In particular:

Select text in box (e.g., by triple-clicking it) to see what I wrote:
The rejection of [72] for R78C3 depends on remembering that we placed a 2 in R7C4 in an earlier link in the chain. Similarly, the rejection of [54] for the same two cells relies on remembering that the starting premise placed a 5 in R7C6 (as part of the [215] permutation for the N8 innies). Last but not least, the rejection of [81] here requires creating a branch to visit the 14(3) cage at R7C6.

For comparison, to see an example of an AIC in practice, consider the following position reached on my solving path for the A68:

Code:
.-----------------------------------.-----------.-----------.-----------.-----------------------------------.
| 123456789   123456789   12        | 123456789 | 456789    | 1234567   | 56789       123456789   123456789 |
:-----------.           .-----------'           |           |           '-----------.           .-----------:
| 123456789 | 123456789 | 12345689    123456789 | 456789    | 1234567     1234567   | 123456789 | 123456789 |
|           |           :-----------.-----------'-----------'-----------.-----------:           |           |
| 123456789 | 123456789 | 2345689   | 3456789     123456789   123456789 | 12345678  | 123456789 | 123456789 |
|           '-----------:           '-----------.           .-----------'           :-----------'           |
| 23456789    23456789  | 2345689     23456789  | 23456789  | 2345678     12345678  | 123456789   123456789 |
:-----------------------+-----------.           |           |           .-----------+-----------------------:
| 123456789   123456789 | 12345689  | 234567    | 1234567   | 1234567   | 56789     | 23456789    23456789  |
:-----------.           |           '-----------+-----------+-----------'           |           .-----------:
| 13        | 123456789 | 12345689    2456789   | 1245      | 56789       56789     | 2345678   | 45678     |
|           '-----------+-----------------------:           :-----------------------+-----------'           |
| 2468        2468      | 7           13        | 1245      | 12345       2345689   | 2345689     2345689   |
:-----------.-----------:           .-----------'-----------'-----------.           :-----------.-----------:
| 124689    | 34589     | 13        | 3456789     3456789     3456789   | 24        | 12345678  | 1234      |
|           |           '-----------+-----------------------------------+-----------'           |           |
| 124689    | 34589       45        | 3456789     23456789    23456789  | 45678       12345678  | 1234      |
'-----------'-----------------------'-----------------------------------'-----------------------'-----------'


Select text in box (e.g., by triple-clicking it) to see what I wrote:
From this position, I found the following grouped AIC (Eureka notation):

(3)r8c3=(3)r89c2,(5)r9c3,(1)r1c3-(1=3)r8c3

This AIC forms a discontinuous loop, with two strong links on the same candidate at the discontinuity (r8c3).

In plain English, this AIC can be expressed as follows:

"If R8C3 is not 3, then R89C2 must contain a 3 (only other place for 3 in N7), implying that r9c3 must be a 5 (because only combination for 17(3) cage containing a 3 is {359}, which would force a 5 into R9C3). This in turn forces a 1 into R1C3 (outies C12 = R19C3 = 6(2)), which prevents a 1 from existing in R8C3, implying that R8C3 must be a 3 (only other candidate in bivalue cell)."

Thus, this logic shows that if R8C3 is not a 3 (starting premise), then it must be a 3, which is a contradiction. The result is that we can immediately place a 3 in this cell!

The above AIC consists of two strong links, 1 weak link and 2 direct links, as follows:

1. R8C3 <> 3 => R89C2 contain a 3 (strong link, N7)
2. R89C2 = {3..} => R9C3 = 5 (direct link based on 17(3) cage combinations)
3. R9C3 = 5 => R1C3 = 1 (direct link, C12 outies)
4. R1C3 = 1 => R8C3 <> 1 (weak link, C3)
5. R8C3 <> 1 => R8C3 = 3 (strong link, bivalue cell R8C3)

The important thing to note is that all of the above implications work independently of each other, regardless of what has gone before. Furthermore, the logic is representable as a linear implication chain (forming a closed loop in this case). In addition, the chain begins on a false premise (R8C3 is not 3) and ends on a true premise (R8C3 is 3). Thus all key requirements for an AIC listed above are fulfilled.


Note that AICs represent an advanced solving technique. The difficulties come from getting used to the cryptic Eureka notation and in distinguishing bona fide AICs from other intuitive contradiction moves that look deceptively similar. Like many formalisms, they require a deep understanding of the basic principles.
_________________
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Mike
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Ruud
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PostPosted: Mon Sep 17, 2007 10:26 pm    Post subject: Reply with quote

This Transformer pattern was very fruitful.

Here are a few variants:

If anyone could post the pictures salut

A68V2 (rated 2.0-3.0?)
3x3::k:5376:5376:5376:6147:2308:5637:4870:4870:4870:3593:5376:6147:6147:2308:5637:5637:4870:5137:3593:5376:5396:5397:5397:5397:6424:4870:5137:3593:3593:5396:5396:5397:6424:6424:5137:5137:4644:4644:2598:5396:5397:6424:3882:3883:3883:4653:4644:2598:2598:3121:3882:3882:3883:3893:4653:4653:3384:3384:3121:3899:3899:3893:3893:3647:3136:3384:3650:3650:3650:3899:2886:3911:3647:3136:3136:3147:3147:3147:2886:2886:3911:

A68V3 (rated 3.0-4.0?)
3x3::k:4608:4608:4608:4099:3588:4357:5638:5638:5638:6665:4608:4099:4099:3588:4357:4357:5638:6161:6665:4608:5908:4629:4629:4629:4632:5638:6161:6665:6665:5908:5908:4629:4632:4632:6161:6161:3364:3364:4390:5908:4629:4632:4650:2347:2347:3885:3364:4390:4390:2353:4650:4650:2347:4149:3885:3885:3896:3896:2353:3387:3387:4149:4149:2879:4416:3896:4674:4674:4674:3387:4934:2887:2879:4416:4416:2123:2123:2123:4934:4934:2887:
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gary w
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PostPosted: Mon Sep 17, 2007 11:18 pm    Post subject: Reply with quote

Thanks very much for your reply Mike.I don't pretend to understand the full implications of what constitutes an AIC but I hope to read more about it on some of your previous postings.Given though,that you start with a premise eg r8c3 <>3 which you subsequently show to lead to a contradiction ie r8c3=3 why is finding and using such an AIC considered to be a "solving technique" whereas starting with a premise eg r7c4=1 which again can be easily (without pencil marks or feats of memory) shown to lead to a contradiction rather frowned upon?

I suppose the difference lies in the logic of the two procedures? and in the formalism of the AIC you described? I feel a little like a blind man who's stumbled into a minefield here! Can/should I be "happy" at the way I solved Assassin 68 or not?It seems as if the answer should be not.

I really do look forward to seeing much more in the forum about advanced solving techniques...it's absolutely fascinating.
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mhparker
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PostPosted: Tue Sep 18, 2007 5:24 am    Post subject: Reply with quote

Ruud wrote:
If anyone could post the pictures salut

Yes, I'd like to be the one to do this (you'll see why later Wink). Will be posting the piccies within the next couple of hours. Stay tuned everybody!
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mhparker
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PostPosted: Tue Sep 18, 2007 7:13 am    Post subject: Reply with quote

Hi all,
Ruud wrote:
This Transformer pattern was very fruitful.

Here are a few variants:

Thanks, Ruud. Not sure I'd refer to a 3.0+-rated puzzle as particularly fruitful though! Smile

Ruud wrote:
If anyone could post the pictures salut

Gladly! However, in return, I trust nobody (incl. Ruud) minds too much if I substitute my V1.5 for Ruud's V3. I was just about to publish it last night anyway, before Ruud pipped me to the post. I think going with the combination of V1.5 and V2 will provide the best mix to keep us happy for the rest of the week.

Of course, the V3 is not lost. If anyone wishes to tackle it, we can always publish the image sometime later, if and when we need it. However, if the 3.0 - 4.0 rating is anything to go by, it looks it's going to be a strong candidate for the Unsolvables list!


Assassin 68 V1.5 (mhparker) (Est. rating: 1.5)



3x3::k:5120:5120:5120:4099:4356:3077:7686:7686:7686:6921:5120:4099:4099:4356:3077:3077:7686:3857:6921:5120:5140:5397:5397:5397:4120:7686:3857:6921:6921:5140:5140:5397:4120:4120:3857:3857:3364:3364:3622:5140:5397:4120:3114:5675:5675:4653:3364:3622:3622:1585:3114:3114:5675:3637:4653:4653:4152:4152:1585:4923:4923:3637:3637:3391:2112:4152:4930:4930:4930:4923:4166:2631:3391:2112:2112:2891:2891:2891:4166:4166:2631:


Assassin 68 V2 (Ruud) (Est. rating: 2.0)



3x3::k:5376:5376:5376:6147:2308:5637:4870:4870:4870:3593:5376:6147:6147:2308:5637:5637:4870:5137:3593:5376:5396:5397:5397:5397:6424:4870:5137:3593:3593:5396:5396:5397:6424:6424:5137:5137:4644:4644:2598:5396:5397:6424:3882:3883:3883:4653:4644:2598:2598:3121:3882:3882:3883:3893:4653:4653:3384:3384:3121:3899:3899:3893:3893:3647:3136:3384:3650:3650:3650:3899:2886:3911:3647:3136:3136:3147:3147:3147:2886:2886:3911:


Have fun!
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Last edited by mhparker on Tue Sep 25, 2007 10:34 pm; edited 3 times in total
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CathyW
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PostPosted: Tue Sep 18, 2007 12:52 pm    Post subject: Reply with quote

Thanks for these Mike. I think the V1.5 was hard enough for me. I haven't kept a WT first time through - nothing majorly advanced required but it was quite a slog, despite a raft of singles part way. (Or maybe I missed a trick!) Might have another go at it later/tomorrow and keep track of steps taken.

Cathy x
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CathyW
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PostPosted: Tue Sep 18, 2007 9:36 pm    Post subject: Reply with quote

Seemed a bit easier 2nd time round. Here's the WT.

Deleted! An oversight in the prelims had major effect in the rest of the WT. Reworking!!
Rolling Eyes


Last edited by CathyW on Wed Sep 19, 2007 9:47 am; edited 1 time in total
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Para
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Joined: 08 Nov 2006
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PostPosted: Tue Sep 18, 2007 10:28 pm    Post subject: Reply with quote

Just to make it complete. Here is the picture for Assassin 68V3.

Assassin 68 V3


greetings

Para[/u]
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