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Assassin 57
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Andrew
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PostPosted: Fri Jul 06, 2007 8:39 pm    Post subject: Reply with quote

A couple more moves after Richard's excellent combination crunching in step 18.

18e. Valid remaining combinations for r2c3+r3c12 and r123c3 are
{168} {359}
{238} {469}
{258} {467}
{348} {269}
{358} {169}
{568} {139}
{568} {247}
18f. 8 locked in r3c12, locked for r3, n1 and 22(4) cage, no 8 in r4c1

22. Step 19 eliminated 2,4 from r2c2 and therefore eliminated all combinations for r2c3+r3c12 which required 2 or 4 in r2c2. Valid remaining combinations for r2c3+r3c12 and r123c3 are
{168} {359}
{258} {467}
{358} {169}
{568} {139}
{568} {247}
22a. r3c12 = {2368}, no 5
22b. From the remaining combinations for r2c3+r3c12, r4c1 = {367}, no 9 [8 has now been deleted in step 18e]

23. r2c3 = {79} -> 13(3)n1 must contain 7/9
13(3)n1 = {139/247} (cannot be {157} which clashes with r2c2), no 5,6
[Alternatively this elimination of 5,6 could have come from killer pair 5/6 in the remaining combinations for r2c3+r3c12 and r123c3]
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rcbroughton
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PostPosted: Sat Jul 07, 2007 9:29 am    Post subject: Reply with quote

24. 8 in n1 now locked in r3c12 - locked for r3 realised this was covered by Andrew in 18f

25. 8 in n2 now locked in r12c6 - locked for c6
Cleanup
25a. 10(2)n56 - no 2 at r5c7
25b. 10(2)n89 - no 2 at r7c7
25c. 11(2)n89 - no 3 at r8c7

26. 22(4)n14 - can't place combo {2389} - no other combo with 9 (and this one in his 22b) only remaining combos {1678/2578/3568}.

27 45 Rule on n3 - innies r2c78 r3c789 r1c7 total 28 = {123589}/{123679}/{124579}/{124678}/{134569}/{134578] cannot be {234568} because 1 locked in 15(4)n36 in n3
27a To fit with 15(4)n36 r2c8 + r3c89 must total 10,9,8,7,6 only
10 = 1{36} - needs {279} or {459} in r123c7 - not possible
10= 1{45} not possible
9 = {135}
8 = {125}/{134}
7 = 1{24}
6 = 1{23}
27b. No 6 at r3c89
27c. Cleanup - no 5 in r4c9 because we can't make 10 on the other 3 cells


Last edited by rcbroughton on Sun Jul 08, 2007 8:09 am; edited 1 time in total
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Andrew
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PostPosted: Sun Jul 08, 2007 5:56 am    Post subject: Reply with quote

Following Richardís example

28. 45 rule on n7 6 innies r7c12+r8c1 and r789c3 = 25 and must contain 1,2 = {123469/123478/123568/124567}
28a. From r6c2 = {2789} r7c12+r8c1 must total 17, 12, 11 or 10 and have valid cage combinations with r6c2, can only be
Total 17 - cannot be {269/278} when r6c2 = 2, cannot be {179/359/458} which arenít in 25(6)
{368/467} need r789c3 = {125} - not possible
Total 12 - cannot be {147/237} when r6c2 = 7, cannot be {345} which arenít in 25(6)
{129/138/156/246} all give possible combinations for r789c3 and 20(3)
Total 11 - cannot be {128} when r6c2 = 8
{137/146/236/245} all give possible combinations for r789c3 and 20(3)
Total 10
{127/136/145/235} all give possible combinations for r789c3 and 20(3)
28b. No 2 in r6c2
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mhparker
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PostPosted: Sun Jul 08, 2007 5:57 pm    Post subject: Reply with quote

Firstly, thanks to Andrew for eliminating that one candidate in step 28 to keep the ball rolling and prevent the 24-hour timer from expiring.

Also, I see there's some serious combination crunching going on here Cool, so hope you don't mind if I join the party! Mind you, you may need a stiff drink before trying to follow one or two of my steps below! Smile

Assassin 57V2X Tag Walkthrough (continued)

29. r8c2+r9c1 cannot contain both of {79} due to r3c7
29a. -> no 4 in r9c2

30. Now apply a few simple AIC's based on the 2-cell cages in n23
30a. AIC: (6=8)r2c7-(8)r2c6=(8)r1c6,(4)r1c7
30b. -> 17(3)r1c8 cannot contain both of {46}
30c. -> {467} combo blocked
30d. {368} blocked by r2c7
30e. -> remaining combos are {269/278/359/458} = {(2/5)..}
30f. -> {125} combo blocked from r2c8+r3c89
30g. AIC: (4)r123c89=(4)r1c7,(8)r1c6-(8=6)r2c6-(6=8)r2c7
30h. -> if r123c89 does not contain a 4, 17(3)r1c8 cannot contain an 8
30i. -> in other words, if 17(3)r1c8 contains an 8, r123c89 must contain a 4
30j. AIC: (5)r123c89=(5)r1c7,(7)r1c6-(7=9)r3c6-(9=7)r3c7
30k. -> if 17(3)r1c8 contains a 7, r123c89 must contain a 5
30l. Now analyze pairs of combinations for r1c89+r2c9/r2c8+r3c89:
269/134
269/135
278/134 (blocked by AIC of step 30k)
278/135 (blocked by AIC of step 30i)
359/124
458/123
30m. -> {278} combo blocked for 17(3)r1c8
30n. -> 17(3) = {269/359/458} (no 7)
30o. {58} only in r1c89
30p. -> no 4 in r1c89
30q. (from step 30l) 3 locked in r123c89
30r. -> no 3 in r1c7
30s. -> no 9 in r1c6

31. Killer AIC removes 9 from r3c7:
31a. (9=4)r1c89|r2c9-(4)r2c4=(2)r2c4,(9)r2c3-(9)r1c12=(9)r1c7
31b. -> 9 in n3 either in 17(3)r1c8 or r1c7
31c. -> no 9 in r3c7
31d. -> r3c67 = [97]
31e. Cleanup: no 1 in r1c5 and r57c7, no 2 in r8c7, no 5 in r1c6, no 3 in r57c6, no 4 in r8c6

Marks pic after step 31e:

Code:
.-----------------------.-----------------------.-----------.-----------------------.-----------------------.
| 123479      123479    | 123456      123456    | 367       | 3478        4589      | 235689      235689    |
|           .-----------+-----------------------:           :-----------------------+-----------.           |
| 23479     | 15        | 79          24        | 347       | 68          68        | 15        | 2349      |
:-----------'           :-----------------------+-----------+-----------------------:           '-----------:
| 2368        2368      | 123456      123456    | 23456     | 9           7         | 12345       12345     |
|           .-----------:                       |           :-----------------------+-----------.           |
| 367       | 123456789 | 1234567     1234567   | 2345689   | 123456      123456    | 123456789 | 6789      |
:-----------'           :-----------------------:           :-----------------------:           '-----------:
| 123456789   123456789 | 456789      456789    | 234689    | 12467       34689     | 123456789   123456789 |
|           .-----------+-----------------------+-----------+-----------------------+-----------.           |
| 123456    | 789       | 5689        5689      | 1278      | 123456      123456    | 12789     | 123456    |
:-----------'           :-----------------------:           :-----------------------:           '-----------:
| 123456789   123456789 | 12345689    123456789 | 1278      | 12467       34689     | 123456789   123456789 |
|           .-----------:                       :-----------+-----------------------+-----------.           |
| 123456789 | 34689     | 123456789   123456789 | 123456    | 23567       45689     | 2346789   | 123456789 |
:-----------'           :-----------------------:           :-----------------------:           '-----------:
| 345689      356789    | 3467        3467      | 345678    | 12          12        | 3456789     3456789   |
'-----------------------'-----------------------'-----------'-----------------------'-----------------------'

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Mike


Last edited by mhparker on Mon Jul 09, 2007 9:25 pm; edited 1 time in total
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rcbroughton
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PostPosted: Sun Jul 08, 2007 7:39 pm    Post subject: Reply with quote

mhparker wrote:
so hope you don't mind if I join the party
The more the merrier.
Quote:
you may need a stiff drink before trying to follow one or two of my steps below
or something! Step 30 is a move on steroids! ... And I thought I posted convoluted moves!!

Nice going.

32. 9 locked in 17(3) at r45c5 for c5 and n5.
32a. 17(3) = {269}/{359} - no 4, 8

33. 9 locked in r78c4 for 22(4)n78
33a cleanup - no 4 at r5c3 for 9(2)n45
33b no 5 at r6c3 for 14(2)n45

34. 8 locked in D\ in n9 - nowhere else in n9
34a. cleanup - no 3 at r8c6 for 11(2)n89

35. 9(2)n8 can't be {36} - blocked by 10(2)n2

36. 3 now locked in c4 in n8 - locked for c4
36a. cleanup no 4 at r1c3 for 7(2)n12

37. Revisiting step 22.
{568} - r123c3 5@r2c2 = {247} now needs to be [274]
37a No other combo for r123c3 with 2 - so no 2 at r3c3

Rgds
Richard


Last edited by rcbroughton on Mon Jul 09, 2007 8:24 am; edited 1 time in total
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mhparker
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PostPosted: Sun Jul 08, 2007 10:21 pm    Post subject: Reply with quote

Just a few more moves to see this one off:

Assassin 57V2X Tag Walkthrough (final part)

38. Only 2 places for 7 in n2: r1c6 or 10(2)r12c5
38a. if 7 in 10(2), 3 of n2 also locked in r12c5
38b. either way -> no 3 in r1c6
38c. -> no 9 in r1c7

39. 3 in c6 locked in r46c6 -> not elsewhere in n5
39a. -> either 7(2)r4c6 = [34] or 7(2)r6c6 = [34]
39b. -> 4 locked in r46c7 for c6 and n6
39c. Cleanup: no 6 in r57c6, no 8 in r1c6, no 7 in r8c6
39d. 17(3)r3c5: 3 now only in r3c5 -> no 5 in r3c5

40. Hidden single (HS) in c6 at r2c6 = 8
40a. -> r2c7 = 6
40b. Cleanup: no 1 in r46c6, no 4 in r57c6, no 5 in r8c6

41. Hidden pair on {35} in c6 at r46c6
41a. -> r46c6 = {35}, locked for n5
41b. -> r46c7 = {24}
41c. -> 2 locked in r46c7 for n6 -> not elsewhere in n6
41d. Cleanup: no 8 in r5c3

42. HS in c6 at r1c6 = 4
42a. -> r1c7 = 8
42b. Cleanup: no 3 in r1c3, no 2 in r57c6

43. Naked single (NS) at r2c4 = 2
43a. -> r2c3 = 9
43b. Cleanup: no 5 in r1c3, no 4 in r5c4

The rest is just hidden and naked singles.
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Andrew
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PostPosted: Mon Jul 09, 2007 2:48 am    Post subject: Reply with quote

Congratulations to Mike and Richard for finishing off this tough one and to Ed for converting the V1 into such a challenging puzzle.

mhparker wrote:
you may need a stiff drink before trying to follow one or two of my steps below!

rcbroughton wrote:
or something! Step 30 is a move on steroids! ... And I thought I posted convoluted moves!!

Having worked through Mike's step 30 I'll settle for the stiff drink after working through it; I'll have a large neat single malt while watching TV this evening. Would never had managed step 30 with the drink first. Wink

I'm gradually learning how to do Richard's combination crunching but I still needed the pointer with his first set of 6 innies. I did actually also try a similar set for n9, which looked more promising than n7 with r9c7 = {12}, but it didn't yield any eliminations so I never posted it.

Now that several people on the forum are using advanced 'vanilla sudoku' techniques such as AIC, ALS, Colouring and Nice Loops I clearly need to learn them. I bought Andrew Stuart's book when it came out earlier this year but still haven't found the time to study it and learn these techniques.

A few comments on today's posted moves.

Step 29 was simple but a nice use of the diagonal and in step 34 8 locked on the diagonal in n9 probably wasn't easy to spot.

After looking at the AICs in steps 30a, 30g, 30j and 31a I realised that they are really a formalised way of expressing contradiction moves that I've been using for a while having learned them from Ed's walkthroughs. There may well be more to AICs than that; I'll find out when I come to study the book. The only one I had any problem understanding was step 31a until I realised from step 30n that the 17(3) must contain 4 or 9 and then it was a simple contradiction move.

After step 30n, 7 in n3 locked in r13c7, locked for c7. No problem, step 31d covers this.

Step 37 was a nice example of remembering an earlier step.

After step 38c I was then expecting 9 locked in 17(3)n3 (step 31b) = 9{26/35}, no 4,8. Mike clearly saw "no 3 in r1c6" leading to the locked 3 in c6.


Last edited by Andrew on Mon Jul 09, 2007 10:13 pm; edited 1 time in total
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PostPosted: Mon Jul 09, 2007 11:28 am    Post subject: Reply with quote

Well done team. Am very relieved that soft hints were not needed. Some really interesting moves you found..and 1 you didn't.. Cool

Here's a simplified walk-through for Assassin 57V2X. It includes a couple of short-cuts and a really nice (fish?) move. However, the basic solution is very similiar. An alternative path would have been much longer.- though there are some good leads that could show another way. Incidentally, this puzzle has a unique solution without the diagonals (X) if you want some torture.

BTW - I found the way Andrew did the innies hypothetical in n7 (original WT step 28) very easy to follow. I like easy - so have used his format for the 1 innies move I use in n3 rather than Mikes AIC's. Not that I have anything against AIC's! But this way seemed a fraction simpler.

Please let me know of any corrections or simplifications. Thanks. Ed


Simplified walk-through for Assassin 57V2X (1-9 cannot repeat on the diagonals)

0. Cage 7(2) n12 - no 789
0a. Cage 10(2) n2 - no 5
0b. Cage 12(2) n23 -no 126
0c. Cage 11(2) n12 - no 1
0d. Cage 14(2) n23 = {59}/{68}
0e. Cage 13(4) n1245 - no 89
0f. Cage 16(2) n23 = {79}
0g. Cage 7(2) n56 @r4c6 - no 789
0h. Cage 13(2) n45 - no 123
0i. Cage 10(2) n56 - no 5
0j. Cage 14(2) n45 - = {59}/{68}
0k. Cage 9(2) n58 - no 9
0l. Cage 7(2) n56 @r6c6 - no 789
0m. Cage 10(2) n89 - no 5
0n. Cage 20(3) n7 - no 12
0o. Cage 9(2) n8 - no 9
0p. Cage 11(2) n89 - no 1
0q. Cage 10(2) n78 - no 5
0r. Cage 3(2) n89 ={12}

1. 3(2)n8 = {12}: both locked for r9
1a. -> no 7 or 8 in r8c5
1b. and 10(2)n7 = {37/46}(no 8,9)

2. 16(2)n2 = {79}: both locked for r3

3. {179} combo. in 17(3)n3 blocked by {79} at r3c7
3a. no 1 in 17(3)n3

4. 1 in n3 now only in 15(4)n3
4a. no 1 in r4c9
4b. 15(4) can't be {2346}

5. 14(2)n2 = [5/6..]
5a. -> 11(2)n1 = {29/38/47}(no 5,6) ({56}combo blocked by 14(2)n2)

6. "45" n3: outies r123c6 + r4c9 = 29
6a. max. r123c6 = {789} = 24 -> min. r4c9 = 5

7. "45" n7: outies r789c4 + r6c2 = 26
7a. max r789c4 = {789} = -> min r6c2 = 2

8. "45" r6789: innies r6c19 = 7 = h7(2)r6
8a. = {16/25/34}(no 789)

9. "45" r789: outies r6c258 = 17 = h17(3)r6
9a. must have 7 for r6
9b. = 7{19/28}(no 3456) ({467} blocked: forces both 7(2)n5 and h7(2)r6 to {25})
9c. -> no 3,4,5,6 r7c5

10. 9(2)n5 = {18/27} = [1/7,2/8..]
10a. -> 10(2) n2 = {19/37/46/}(no 2,8) ({28} blocked by 9(2)n5)

11. 17(3)n2 = {269/359/368/458/467}(no 1) ({179}/{278} blocked by 9(2)n5 step 10.)

12. {467} combo in 17(3)n2 blocked: forces the 2 9(2) cages in c5 to {18}
12a. 17(3) n2 = {269/359/368/458}(no 7)

13. "45" r12: innies r2c28 = 6 = h6(2)r2
13a. = {15/24}

14. "45" r1: outies r2c159 = 14 = h14(3)r2
14a. = {167/239/347}(no 5,8). Other combo's blocked. Here's how.
i.{149} -> 11(2)n1 = {38}: but no [8/9] left for 14(2)n2
ii. {158} clashes with [5/8] needed by 14(2)n2
iii.{248} blocked by [2/4/8] needed for 11(2)n1
iv. {257} -> by 11(2)n1 = {38}:but no [5/8] left for 14(2)n2
v. {356} blocked by [5/6] needed for 14(2)n2

A really cool move.
15. no 1 in r4c6 because of 1 required in 13(4)n1. Here's how. Maybe this is a nice fish move. Don't know which one [edit: Turbot fish! Thanks Mike].
15a. 13(4)n1 must have 1.
i.1 in r3c34 -> 1 in n3 in r2c8 -> no 1 r4c6
ii. 1 in r4c34 -> no 1 in r4c6

15b. -> no 1 r4c6, no 6 r4c7

16. 1 in D/ only in r7c7 or r2c8:
16a. cross-over -> no 1 r7c8

17. no 1 in r2c2 or r8c8 because of 1's in D/. Here's how.
17a. 1 in D/ only in r7c7 or r2c8
i. 1 in r7c7 -> 1 in n3 in r3 -> 1 required in 13(3)n1 only in r4c4 -> no 1 in r2c2 or r8c8 (on same D\ as r4c4)
ii. 1 in D/ in r2c8 -> no 1 in r2c2 or r8c8
17b. no 5 r2c8 (h6(2)r2)
17c. {189} combo blocked from 18(3)n9 (no 1 available)

[edit out redundant step]

A nice shortcut compared to the big innies move in the original WT.
Alternate step 18: better way as suggested by Mike from following post. Original step in TT below.
18. 13(3)n1 can only contain at most one of {79} due to cage sum.
18a. only other place for {79} in n1 is r2c3
18b. -> r2c3 = {79} and...
18c. ...13(3)n1 = {(7/9)..} (only other place in n1)
18d. 13(3) = {157/247/139} (no 6,8)
18e. 8 in n1 only in r3c12.
18f. 8 locked for r3 and no 8 r4c1

18. 8 locked in r3c12 because of innies n1. Here's how.
18a. "45" n1:6 innies {r2c2, r3c12} + {r123c3} = 32(6) and can only contain at most 2 of {789} from r3c12=[8] or r2c3={789}. No other place available for {789} in innies n1.
18a. -> 32(6) n1 = {135689/145679/234689/235678/245678}
18b. However, the {145679} is blocked since 7 & 9 only in r2c3.
18c. = {135689/234689/235678/245678}
18d. Each of these combinations has exactly 2 of {789} -> the 8 in r3c12 must be one of them. The [7/9] must come from r2c3.
18e. since 8 must be in r3c12: 8 locked for n1, r3 and no 8 in r4c1
18f. -> r2c3 = {79}, r2c4 = {24}


19. h6(2)r2c28 = [51] ({24} blocked by r2c4)

20. 14(2)n2 = {68}:both locked for r2
20a. no 4 or 9 in r1c5

21. split-cage 17(3)r2c2 = 8{27/36}(no 1,4,9)
21a. 7 only in r4c1 -> no 2 r4c1

22. 13(3)n1 = {139/247} = [2/3..] [edit out invalid combination]

23. deleted: redundant from new step 18

24. 8 in n2 only in c6: 8 locked for c6
24a. -> no 2 in r57c7
24b. and no 3 r8c7

25. 1 in r3 only in r3c34 in 13(4)
25a. -> no 1 r4c34
note: there are lots of neat eliminations now that there are only 2 1s in r4: but can't use it to solve the puzzle.
25c. 1 in n4 only in 16(4)
25d. {2347/2356} combo's blocked [edit step numbers]

26. {368} combo blocked from 17(3)n3 by r2c7
26a. {467} combo blocked since r12c7 = [4/6] (8 in n2 is only in r12c6)
26b. = {269/278/359/458} = [2/5..]
[edit:26a added]

Now an innies move for n3 rather than using AIC's - simpler?[edit: step 27a. because of added step 26a]
27. "45" n3: 5 innies = 27 = 27(5)
27a. = {23679/24678/34569/34578} ({23589/24579} blocked by 17(3) step 26b)
27b. r3c89 = 5..9 -> r3c89 = 9..5

i. r4c9 = 5 -> r3c89 = 9 = {36} only: blocked. Here's how.
r3c89 = {36} -> r123c7 = blocked: from combinations in 27(5) step 27a. can't make {279} from candidates in r2c7
-> no 5 r4c9

ii. r4c9 = 6 -> r3c89 = 8 = {35} -> from combinations in 27(5) step 27a.
-> r123c7 = {469}: blocked: can only be [469] which forces r123c6 = [887]:but 2 8's c6
................= {478} = [487]
-> innies n3 = {34578}

iii. r4c9 = 7 -> r3c89 = 7 = {25}: blocked. NO combinations with {25} in 27a.

iv. r4c9 = 7 -> r3c89 = 7 = {34}:Blocked. Here's how.
r3c89 = {34} -> from combinations in 27(5) step 27a-> r123c7 = {569} = [569]: but this forces 2 7's into c6;
...............................................................................................= {578} = [587]-> r123c6 = [769]:but this means no 8 for r12c6
-> no 7 r4c9

v. r4c9 = 8 -> r3c89 = 6 = {24}
r3c89 = {24}-> from combinations in 27(5) step 27a.-> r123c7 = {678} = [867]
-> innies n3 = {24678}

vi. r4c9 = 9 -> r3c89 = 5 = {23}
r3c89 = {23}-> from combinations in 27(5) step 27a.r123c7 = {679} = [769/967]
-> innies n3 = {23679}

28. In summary from steps 27ii,v,vi: innies n3 = {23679/24678/34578} = 7{..}
28a. 7 locked for n3 and c7
28b. no 3 r57c6, no 4 r8c6

29. in Summary from steps 27ii,v,vi:: r3c89 = {35/24/23}(no 6)
29a. split-cage 14(3)r2c8 = {239/248/356}
29b. r4c9 = {689}

30. In summary from steps 27ii,v,vi:: r123c7 = [487/867/769/967]
30a. r1c7 = {4789}(no 3,5)
30b. r1c6 = {3458}(no 7,9)

31. 17(3)n3 = {269/359/458} = [4/9..]
31a. {58} only in r1c89 -> no 4 r1c89

32. no 9 in r3c7. Here's how.
32a. 17(3)n3 = [4/9..](step 31)
i. if [4] -> r1c89 = {58} -> 12(2)n2 = [39]-> no 9 r3c7
ii. if [9] -> no 9 r3c7

33. r3c67 = [97]
33a. Cleanup: no 1 in r1c5 or r57c7, no 2 in r8c7, no 5 in r1c6,

34. 9 in c5 only in r45c5 in 17(3): 9 locked for n5
34a. no 4 r5c3, no 5 r6c3
34a. 17(3)n2 = 9{26/35} - no 4, 8

35. 9 in c4 only in r78c4 in 22(4)n78: 9 locked for 22(4)
35a. no 9 r67c3

36. 8 in D\ only in n9: 9 locked for n9
36a. cleanup - no 3 at r8c6

37. 9(2)n8 can't be {36} - blocked by [3/6..] needed for 10(2)n2
37a. no 3,6 r89c5

38. 3 in n8 now only in c4: 4 locked for c4
38a. cleanup no 4 at r1c3 for 7(2)n12

39. 7 in n2 only in 10(2) = {37}: both locked for n2 & c5
39a. no 9 r1c7

40. 9(2)n5 = {18} locked for c5

41. 9(2)n8 = {45}: both locked for c5, n8
41a. no 6 r1c3, no 6 r7c7 or r7c8

42. 3 in c6 only in r46c6

42a. -> either 7(2)r4c6 = [34] or 7(2)r6c6 = [34]
42b. -> 4 locked in r46c7 for c6 and n6
42c. Cleanup: no 6 in r57c6, no 8 in r1c6, no 7 in r8c6

43. r12c7 = [86], r12c6 = [48], r4c6 = 5 (HS c6), r4c7 = 2, r9c67 = [21], r8c67 = [65],

and on you go!


Last edited by sudokuEd on Tue Jul 10, 2007 7:03 am; edited 5 times in total
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mhparker
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PostPosted: Mon Jul 09, 2007 10:15 pm    Post subject: Reply with quote

Hi guys,

Just a few odd observations:

Andrew wrote:
After looking at the AICs in steps 30a, 30g, 30j and 31a I realised that they are really a formalised way of expressing contradiction moves...

Yes, most (but not all) AIC moves can be viewed as contradiction moves. Sometimes the two ends of the chain provide complementary information. For example, if an AIC starts and ends in the same cage, an AIC can show that "if cage C doesn't contain the digit x, it must contain the digit y".

The Eureka notation for AICs is quite efficient. For example, in the future I might not be so verbose as this time:

Mike wrote:
30a. AIC: (6=8)r2c7-(8)r2c6=(8)r1c6,(4)r1c7
30b. -> 17(3)r1c8 cannot contain both of {46}
30c. -> {467} combo blocked

but instead simply abbreviate this to:

Quote:
{467} combo blocked by AIC: (6=8)r2c7-(8)r2c6=(8)r1c6,(4)r1c7

That sure saves a lot of typing!

Andrew wrote:
I bought Andrew Stuart's book when it came out earlier this year

I've got it as well now. No prizes for guessing what I was reading on the beach during my recent holiday! Some of the pages are falling out already!

sudokuEd wrote:
15. no 1 in r4c6 because of 1 required in 13(4)n1. Here's how. Maybe this is a nice fish move. Don't know which one.

Yes, it's a grouped turbot fish (again).

sudokuEd wrote:
17. no 1 in r2c2 or r8c8 because of 1's in D/. Here's how.
17a. 1 in D/ only in r7c7 or r2c8
i. 1 in r7c7 -> 1 in n3 in r3 -> 1 required in 13(3)n1 only in r4c4 -> no 1 in r2c2 or r8c8 (on same D\ as r4c4)
ii. 1 in D/ in r2c8 -> no 1 in r2c2 or r8c8

That's a neat one!

sudokuEd wrote:
A nice shortcut compared to the big innies move in the original WT.
19. 8 locked in r3c12 because of innies n1. Here's how.
19a. "45" n1:6 innies {r2c2, r3c12} + {r123c3} = 32(6) and can only contain at most 2 of {789} from r3c12=[8] or r2c3={789}. No other place available for {789} in innies n1.
18a. -> 32(6) n1 = {135689/145679/234689/235678/245678}
18b. However, the {145679} is blocked since 7 & 9 only in r2c3.
18c. = {135689/234689/235678/245678}
18d. Each of these combinations has exactly 2 of {789} -> the 8 in r3c12 must be one of them. The [7/9] must come from r2c3.
18e. since 8 must be in r3c12: 8 locked for n1, r3 and no 8 in r4c1
18f. -> r2c3 = {79}, r2c4 = {24}

This is still more complicated than necessary. Better to consider just {79} rather than {789} as follows:

Quote:
18. 13(3)n1 can only contain at most one of {79} due to cage sum.
18a. only other place for {79} in n1 is r2c3
18b. -> r2c3 = {79} and...
18c. ...13(3)n1 = {(7/9)..} = {157/247/139} (no 6,8)

This leaves 8 in n1 locked in r3c12. Also, this means that we can reject the combo {346} listed in step 22, because it contains neither of {79}.
_________________
Cheers,
Mike


Last edited by mhparker on Tue Jul 10, 2007 8:33 am; edited 1 time in total
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Andrew
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PostPosted: Tue Jul 10, 2007 12:06 am    Post subject: Reply with quote

mhparker wrote:
Just a few odd observations:

Andrew wrote:
After looking at the AICs in steps 30a, 30g, 30j and 31a I realised that they are really a formalised way of expressing contradiction moves...

Yes, most (but not all) AIC moves can be viewed as contradiction moves. Sometimes the two ends of the chain provide complementary information. For example, if an AIC starts and ends in the same cage, an AIC can show that "if cage C doesn't contain the digit x, it must contain the digit y".

The Eureka notation for AICs is quite efficient. For example, in the future I might be so verbose as this time:

Mike wrote:
30a. AIC: (6=8)r2c7-(8)r2c6=(8)r1c6,(4)r1c7
30b. -> 17(3)r1c8 cannot contain both of {46}
30c. -> {467} combo blocked

but instead simply abbreviate this to:

Quote:
{467} combo blocked by AIC: (6=8)r2c7-(8)r2c6=(8)r1c6,(4)r1c7

That sure saves a lot of typing!

Andrew wrote:
I bought Andrew Stuart's book when it came out earlier this year

I've got it as well now. No prizes for guessing what I was reading on the beach during my recent holiday! Some of the pages are falling out already!

Thanks for the comments on AICs and contradiction moves. It confirms my feeling that in some cases they are equivalent while in other cases the AICs are more powerful. I liked steps 30g, 30h and 30i where one chain produced both "positive" and "negative" results.

Yes the Eureka notation is efficient and saves some typing, I wouldn't call it saving a lot. However if you do take that route then in the short term there will be people such as myself who won't understand your moves and, after the eyes glaze over, will just have to accept the result and move on to the next step. Having said that, please go ahead and use the shorter form if you want to. In your quoted example I would then just read it as "{467} combo blocked by AIC", possibly try to work it out as a contradiction move and then go to the next step. The important thing is that you will still be stating the outcome of the AIC.

Must be nice to have time to study a book while on holiday. Our holiday last year was touring the south west USA so we were on the road each day. This year, with our move from Calgary to Lethbridge, I doubt that we'll manage more than day trips.

When I have time to work through Ed's simplified walkthrough I'll make a point of also looking at Mike's comments.
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PostPosted: Wed Jul 11, 2007 12:23 pm    Post subject: Reply with quote

Andrew wrote:
Yes the Eureka notation is efficient... However if you do take that route then in the short term there will be people such as myself who won't understand your moves and, after the eyes glaze over, will just have to accept the result and move on to the next step. Having said that, please go ahead and use the shorter form if you want to. In your quoted example I would then just read it as "{467} combo blocked by AIC", possibly try to work it out as a contradiction move and then go to the next step. The important thing is that you will still be stating the outcome of the AIC.

They are really quite easy, although more complicated for Killers than in the vanilla case.

Maybe, if we use any shorthand such as this, we should write a post explaining more about the conventions we use, and get Ruud to make it a sticky topic on the forum (like your "Advice for Newbies"). What do you think?
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Mike
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PostPosted: Wed Jul 11, 2007 1:07 pm    Post subject: Reply with quote

Hi guys,

I've just taken a look at what the latest JSudoku makes of Ed's A57V2X, and it found an interesting move combination that we all missed, which I think is worth sharing:

Here's the grid state after step 25 in Ed's simplified WT:

Code:
.-----------------------.-----------------------.-----------.-----------------------.-----------------------.
| 123479      123479    | 12346       13456     | 1367      | 345789      345789    | 23456789    23456789  |
|           .-----------+-----------------------:           :-----------------------+-----------.           |
| 23479     | 5         | 79          24        | 3479      | 68          68        | 1         | 23479     |
:-----------'           :-----------------------+-----------+-----------------------:           '-----------:
| 2368        2368      | 12346       123456    | 23456     | 79          79        | 23456       23456     |
|           .-----------:                       |           :-----------------------+-----------.           |
| 367       | 12346789  | 234567      23467     | 2345689   | 23456       12345     | 23456789  | 56789     |
:-----------'           :-----------------------:           :-----------------------:           '-----------:
| 123456789   12346789  | 456789      456789    | 234689    | 1234679     1346789   | 23456789    123456789 |
|           .-----------+-----------------------+-----------+-----------------------+-----------.           |
| 123456    | 2789      | 5689        5689      | 1278      | 12346       13456     | 2789      | 123456    |
:-----------'           :-----------------------:           :-----------------------:           '-----------:
| 123456789   12346789  | 23456789    123456789 | 1278      | 1234679     1346789   | 23456789    123456789 |
|           .-----------:                       :-----------+-----------------------+-----------.           |
| 123456789 | 346789    | 123456789   123456789 | 123456    | 2345679     2456789   | 2346789   | 123456789 |
:-----------'           :-----------------------:           :-----------------------:           '-----------:
| 3456789     346789    | 3467        3467      | 345678    | 12          12        | 3456789     346789    |
'-----------------------'-----------------------'-----------'-----------------------'-----------------------'


The moves spotted by JSudoku are as follows:

26. Killer XY-Chain on {12}:
26a. r2c4<>2 -> r2c3<>9 -> r2c3=7 -> r3c3<>1 (due to no combo for h27(5)n1 innie cage with both of {17}) -> r3c4=1
26b. r2c4=2 -> r3c4<>2
26c. Either way, no 2 in r3c4

27. Only 2 places remaining for digit 2 in n2 (r2c4 and r3c5)
27a. 2 in r2c4 -> 9 in r2c3
27b. 2 in r3c5 -> 17(3)n12 = {269} -> 9 within r45c5
27c. whichever alternative is true -> no 9 in r2c5
27d. -> no 1 in r1c5

This opens up a whole new range of moves we never covered. Both moves correspond to newly-implemented techniques that are both interesting in their own right.
_________________
Cheers,
Mike
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