Assassin 47

Jean-Christophe · Post by **Jean-Christophe** » Tue Apr 24, 2007 3:15 pm

Here is my walkthrough for V 1.5. The difficulty was to spot the starting place, then all the rest flows quite easilly. Very enjoyable

BTW what's the meaning of these "versions" V1.5, V2... ? I'm new on this forum (but not so new to killers

)

Cage 6/2 in N9 = {15|24} = {(2|5)..} = {(1|4)..}
Innies of N9 -> R79C7 = 7 <> {25} = {(1|4)..}
Cage 6/2 & R79C7 froms a complex naked pair on {14} -> no {14} elsewhere in N9
-> Cage 12/2 in N9 <> {48}
8 of N9 locked in R8C789 -> not elsewhere in R8
Cage 20/3 in N9 = {8..} = {389|578}
6 of N9 locked in R79C7 = {16} (NP @ N9, C7)
Cage 6/2 in N9 = {24} (NP @ N9, R7)
-> Cage 12/2 in N7 <> {48}
8 of N7 locked in R79C3
Innies of N7 -> R79C3 = 13 = {58} (NP @ N7, C3)
Cage 12/2 in N7 = {39} (NP @ N7, R7)
Cage 5/2 in N7 = {14} (NP @ N7, R9)
R79C7 = [16]
Cage 15/3 in N7 = {267} (NT @ R8)
Cage 20/3 in N9 = {389} (NT @ N9, R8)
Cage 12/2 in N9 = {57} (NP @ R9)
R79C3 = [58]

3 Cages 22/3 in R1, C4 & C6 = {9..} -> no 9 elsewhere in R1, C4 & C6
BTW Not a necessity, but notice: the two Cages 22/3 in C4 & C6 forms a X-Wing on 9 -> not elsewhere in N25, not in elsewhere R3
R9C5 = 9 (HS @ N8)
R78C4 = 9 = [81]
Cage 12/2 in R78C5 = [75]
R78C6 = [64]

Innies of C1234 -> R29C4 = 6 = [42]
Innies of C4 -> R16C4 = 8 = [53]
R9C46 = [23]
R1C23 = 17 = [89]
Innies of C6789 -> R2C6 = 1
R12C5 = 8 = {26} (NP @ N2, C5)
Innies of C6 -> R16C6 = 9 = [72]
R1C78 = 6 = {24} (NP @ N3, R1)
R12C5 = [62]
R3C5 = 3, R3C46 = [98]

Outies of N3 -> R4C789 = 7 = {124} (NT @ R4, N6)
R4C5 = 8, R56C5 = {14} = 5
Innies of R56 -> R5C46 = 16 = [79]
R4C46 = [65]
R4C123 = {379} (NT @ N4)
Cage 14/3 in R6 -> R6C23 = [56]
Cage 17/3 in R6 -> R6C78= {78}
R6C9 = 9
...
The rest are naked & hidden singles

Glyn · Post by **Glyn** » Tue Apr 24, 2007 5:04 pm

Hi everyone

Para

I think the Pattern matching that we often see come up as a hint in the latest Sumocue is an all-encompassing approach to single digit analysis.

In the Sudocue it would appear to be equivalent to the Nishio option.

I think Ruud implements the Pattern Overlay Method(POM) via Templates(lots of them).

It can be extended to multiple digits as well.

All the best

Glyn

mhparker · Post by **mhparker** » Tue Apr 24, 2007 7:07 pm

Hello Jean-Christophe, and welcome to the forum!

BTW what's the meaning of these "versions" V1.5, V2... ?

Once upon a time, not so long ago, there was just one Assassin every week. As time went on, these puzzles became more difficult. Although this satisfied the hardcore gurus on this forum, obviously some people complained that puzzles had become too tough. Thus, a new trend was started: The Assassins on the main Assassin page on this site (http://www.sudocue.net/weeklykiller.php) were made a notch easier, but - to make up for the loss - Ruud would post a V2 ("Variation 2") around the same time on this forum. The V2's are generally as hard as (or harder than) the originals (now called the V1's) before the split. All versions (V1, V2, ...) use the same cage pattern.

The puzzle you've just done was designated as a V1.5 (the first time I've seen this) - in other words, harder than the V1, but not hard enough to warrant a "V2" tag.

Once in a while, you might see a V3 as well. These are not necessarily any harder than the V2's.

Last week, another trend was started. The V1 was obviously a bit too difficult for Ruud's liking, so (instead of publishing a V2), he published a "Lite" version, this being somewhat easier than the original.

Hope that helps.

Para · Post by **Para** » Tue Apr 24, 2007 9:02 pm

mhparker wrote: Thus, a new trend was started: The Assassins on the main Assassin page on this site (http://www.sudocue.net/weeklykiller.php) were made a notch easier, but - to make up for the loss - Ruud would post a V2 ("Variation 2") around the same time on this forum.

Well not just Ruud. Ed has few of them on his name, but generally i think anyone who feels like it can create a V2. This V2 either has a different solution or we keep the original puzzle and play a little with the cages. Combine 2 cages or steal one cell from another cage. Usually it is the idea to block/change the opening of the puzzles. They tend to be solvable, but sometimes accidently one sneaks in that is unbelievably hard(like Ed's Ass 42V2 or Ed's Blackhole-3 killer).

greetings

Para

Andrew · Post by **Andrew** » Tue Apr 24, 2007 9:34 pm

Welcome to the forum Jean-Christophe! It's good to have a "new face".
[Edited, now that I know that you are the developer of JSudoku].

Para has got his reply to Mike's message in before me but I think I can still add to it.

We should give Ed credit for introducing V2s. I think the first one was for Assassin 17 and, if I remember correctly, it was because he'd missed the easy way to get started so he created a V2 which could only be started the way that he had started the original Assassin.

Para gave an excellent explanation of how variants differ from the original puzzle. Also, as he pointed out, where there are V2 and V3 one cannot tell which one will be harder. They can be either way round. They are both harder versions of the original puzzle, usually a lot harder.

Mike wrote:As time went on, these puzzles became more difficult. Although this satisfied the hardcore gurus on this forum, obviously some people complained that puzzles had become too tough. Thus, a new trend was started: The Assassins on the main Assassin page on this site were made a notch easier ...

I'll agree with most of that except for the final part. Yes some recent Assassins have been slightly easier but averaged over several puzzles the standard hasn't become easier.

I first started Assassins in August last year and did the earlier ones before doing them as they came out. The early ones were much easier than any we get now. The first difficult ones were, if I remember correctly, Assassins 18 and 24. Some of the more recent ones have been much harder and I think there have been one or two that were only solved jointly on the forum (tag solutions).

Para mentioned a couple of hard ones from Ed. Others that I remember were Ed's first two tag killer-Xs ULURU and UTA. For the latter he introduced a slightly easier UTA2 part way through after we had all got stuck. I believe that Richard's solver has since managed to do UTA although I haven't yet looked at the output for this one. Probably solved by multiple innies or outies and some clever combination work.

CathyW · Post by **CathyW** » Tue Apr 24, 2007 9:43 pm

Some of us know JC from another Killer Sudoku forum! I can't solve most of his puzzles without help from his software or awaiting his walkthroughs. It will be good to see your take on Ruud's Assassins JC.

Andrew · Post by **Andrew** » Wed Apr 25, 2007 3:40 am

I like the quotation at the end of JC's message. "When you have eliminated the impossible, whatever remains, however improbable, must be the truth." Sherlock Holmes. Very appropriate for sudokus!

I found that V1.5 went very smoothly right from the beginning. If anyone used a solver to do the preliminary eliminations, they probably found it harder to get started. Unusually the methodical start, that I've been using for the past two to three months, proved to be just right for this puzzle.

It was no harder than Assassin 47, in fact probably slightly easier because there wasn't a difficult key move.

As soon as I saw the diagram I spotted from the two vertical 22(3) cages that there must be 9 in R3C46. In the walkthrough that comes out in steps 8 to 11 without needing to see that X-wing.

Here is my walkthrough.

1. R7C12 = {39/48/57}, no 1,2,6

2. R78C5 = {39/48/57}, no 1,2,6

3. R7C89 = {15/24}

4. R9C12 = {14/23}

5. R9C89 = {39/48/57}, no 1,2,6

6. R1C234 = 9{58/67}, 9 locked for R1

7. R234C3 = {128/137/146/236/245}, no 9

8. R345C4 = 9{58/67}, 9 locked for C4
8a. 9 in R1 locked in R1C23, locked for N1

9. R345C6 = 9{58/67}, 9 locked for C6

10. 9 in N8 locked in R789C5, locked for C5

11. 9 in N2 locked in R3C46, locked for R3

12. R8C789 = {389/479/569/578}, no 1,2

13. 13(4) cage in N2 = {1237/1246/1345} = 1{237/246/345}, no 8, 1 locked for N2

14. 45 rule on N7 2 innies R79C3 = 13 = {49/58/67}, no 1,2,3

15. 45 rule on N9 2 innies R79C7 = 7 = {16/34} (cannot be {25} which clashes with R7C89), no 2,5,7,8,9
[I didn’t spot that this makes a killer pair {14} in R79C7 and R7C89, as in Jean-Christophe’s walkthrough. This didn’t matter because of my next move. Maybe I was already looking ahead to it.]

16. 2 in N9 locked in R7C89 = {24}, locked for R7 and N9, clean-up: no 8 in R7C12, no 8 in R8C5, no 9 in R9C3 (step 14), no 3 in R79C7 (step 15), no 8 in R9C89
16a. Naked pair {16} in R79C7, locked for C7 and N9

17. 8 in N9 locked in R8C789, locked for R8, R8C789 = 8{39/57}

18. 45 rule on R1 3 innies R1C159 = 10 = {127/136/145/235}, no 8

19. 45 rule on R8 3 innies R8C456 = 10 = {127/136/145/235}, no 9, clean-up: no 3 in R7C5
19a. 1,2 only in R8C46 -> no 7 in R8C46

20. 8 in N7 locked in R79C3 = {58} (step 14), locked for C3 and N7, clean-up: no 7 in R7C12
20a. R7C12 = {39}, locked for R7 and N7, clean-up: no 2 in R9C12
20b. R9C12 = {14}, locked for R9 and N7 -> R9C7 = 6, R7C7 = 1

21. R8C4 = 1 (hidden single in R8)
21a. R7C34 = 13 = {58}, locked for R7 -> R7C56 = [76] -> R8C5 = 5, R8C6 = 4 (hidden single in N8), R7C34 = [58], R9C3 = 8, R9C5 = 9 (hidden single in R9), clean-up: no 3 in R9C89
21b. R9C89 = {57}, locked for N9

22. R345C6 = {589}, no 7, locked for C6

23. R345C4 = {679}, no 5, locked for C4 -> R1C4 = 5 -> R1C23 = 17 = [89]

24. 45 rule on C5 2 remaining innies R12C5 = 8 = {26}, locked for C5 and N2
24a. R2C46 = 5 = [41]
24b. 8 in N2 locked in R3C56, locked for R3

25. 2 in N5 locked in R6C46, locked for R6

26. R1C159 (step 18) = {127/136} = 1{27/36}, no 4, R1C5 = {26} -> no 2,6 in R1C19, 1 locked in R1C19 for R1

27. R1C678 = 4{27/36}
27a. R1C6 = {37} -> no 3,7 in R1C78
27b. 4 in N3 locked in R1C78, locked for N3

28. 18(3) cage in N14, no 8,9 in R3C1 -> no 1 in R4C12

29. R6C234 = {239/257/347/356}, no 1, no 3,6 in R6C2

30. R6C678 = {179/269/278/359/368/467} (cannot be {458} because no 4,5,6 in R6C6), no 3 in R6C7, no 3,4 in R6C8

31. 45 rule on N1 3 remaining outies R4C123 = 19, no 1
31a. Max R4C3 = 7 -> min R4C12 = 12, no 2 in R4C12

32. 45 rule on N14 2 remaining innies R6C23 = 11 -> R6C4 = 3, R9C46 = [23], R1C6 = 7, R6C6 = 2, R3C456 = [938]
32a. R6C23 = {47}/[56], no 9

33. R1C6 = 7 -> R1C78 = 6 = {24}, locked for R1 and N3 -> R12C5 = [62]

34. R6C6 = 2 -> R6C78 = 15 = {78}/[96], no 1,4,5, no 9 in R6C8

35. R234C3 = {137/146/236}, 1 only in R3C3 -> no 4,7 in R3C3

36. 4 in N1 locked in R3C12 -> no 4 in R4C2 which can “see” both of R3C12

37. 4 in C3 locked in R456C3, locked for N4
[This would have eliminated 4 from R4C2 but I spotted step 36 first.]
37a. R6C23 = [56/74], no 7 in R6C3

38. 45 rule on N1 3 remaining innies R23C3 + R3C1 = 10 = {127/136/235} (cannot be {145} because no 1,4,5 in R2C3), no 6 in R2C3, no 4,7 in R3C1

39. R3C2 = 4 (hidden single in N1) -> R9C12 = [41]

40. R234C3 (step 35) = {137/236} (cannot be {146} because no 1,4,6 in R2C3) = 3{17/26}, no 4, 3 locked for C3

41. 45 rule on N3 3 remaining innies R23C7 + R3C9 = 21 = {579/678} = 7{59/68}, no 1,3, 7 locked for N3
41a. 9 only in R2C7 -> no 5 in R2C7
41b. 6 only in R3C9 and 8 only in R2C7 -> no 7 in R2C7
41c. R234C7 = [952] ([853] isn’t consistent with step 41) -> R3C9 = 7 (step 41)

42. 12(3) cage in N36 R3C9 = 7 -> R4C89 = 5 = {14}, locked for R4 and N6

43. 18(4) cage in N1 cannot have three odd numbers in R1C1 + R2C12 -> 6 locked in R2C12, locked for R2 and N1
[This also comes from the remaining combinations {1467/3456} = 46{17/35}]

44. Naked pair {12} in R3C13, locked for R3 and N1 -> R1C1 = 3, R2C3 = 7, R3C8 = 6

45. Naked pair {38} in R2C89, locked for N3 -> R1C9 = 1

46. A few naked singles to clear up R1C78 = [42], R4C89 = [14], R7C12 = [93], R7C89 = [42], R9C89 = [75], R6C78 = [78], R5C7 = 3, R6C2 = 5, R6C3 = 6 (cage sum)

and the rest is naked singles

mhparker · Post by **mhparker** » Wed Apr 25, 2007 9:04 am

Andrew wrote:We should give Ed credit for introducing V2s.

Yes, indeed, sorry about that, Ed!

Andrew wrote:Unusually the methodical start, that I've been using for the past two to three months, proved to be just right for this puzzle.

Absolutely. This shows that the methodical approach is sometimes the best one. Thanks for the walkthrough, Andrew.

Andrew wrote:If anyone used a solver to do the preliminary eliminations, they probably found it harder to get started.

On the contrary, Andrew, automated solvers also take the methodical approach. I believe the V1.5 was rated as harder than the V1 simply because it required conflicting combinations. However, in this case, the only place this was necessary was the one you mentioned in step 15 of your walkthrough. You are probably right, in that many people will see this much more quickly than finding a move to break through the gridlock in the V1. (Incidentally, there is an easy and effective way of handling the gridlock in the V1 that nobody has seen yet...

)

CathyW · Post by **CathyW** » Wed Apr 25, 2007 3:08 pm

mhparker wrote:(Incidentally, there is an easy and effective way of handling the gridlock in the V1 that nobody has seen yet... )

You will share once no. 48 is issued won't you?

Andrew · Post by **Andrew** » Wed Apr 25, 2007 7:52 pm

mhparker wrote:(Incidentally, there is an easy and effective way of handling the gridlock in the V1 that nobody has seen yet... )

CathyW wrote:You will share once no. 48 is issued won't you?

It could be posted now in tiny text. You've got me intrigued Mike.

I must admit that, once I've managed to solve an Assassin or a forum puzzle, I don't usually go looking for alternative better steps although I do sometimes find a few while checking my walkthroughs even when I don't post them. I usually work on the basis that if there are interesting and better steps, they will be posted in a walkthrough.

Andrew wrote:If anyone used a solver to do the preliminary eliminations, they probably found it harder to get started.

mhparker wrote:On the contrary, Andrew, automated solvers also take the methodical approach.

Maybe I shouldn't comment on automated solvers, since I haven't yet downloaded any, although I have run weekly extreme jigsaws on www.sudoku.org.uk through Andrew Stuart's jigsaw solver which is linked from the extreme jigsaw solutions. I agree that they take a methodical approach. However I was under the impression that one can click a button at the start, or at any intermediate position, to tell the solver to eliminate candidates that no longer go in a particular cage; first time round that would do the preliminary steps but not any interactions.

One reason why I haven't yet downloaded SumoCue is that I understand it doesn't yet have a Save facility. That's why I'm happy to continue solving puzzles on an Excel spreadsheet.

mhparker · Post by **mhparker** » Wed Apr 25, 2007 10:57 pm

Andrew wrote:It could be posted now in tiny text. You've got me intrigued Mike.

OK, here's my walkthrough. I've included a candidate grid at the point of interest for convenience.

Hope I haven't set people's hopes too high!

Walkthrough - Assassin 47 (http://www.sudocue.net/weeklykiller.php?id=47)

1. 9/3 at R1C2: no 7,8,9

2. 11/3 at R3C1: no 9

3. 19/3 at R3C4: no 1

4. 8/3 at R3C6 = {1(25|34)} -> no 1 elsewhere in C6

5. 21/3 at R3C9: no 1,2,3

6. 27/4 at R5C1 = {(378|468|567)9} -> no 9 elsewhere in N4

7. 12/4 a R5C7 = {12(36|45)} -> no 1,2 elsewhere in N6

8. 9/3 at R6C2: no 7,8,9

9a. 23/3 at R6C6 = {689} -> no 6,8,9 elsewhere in R6
9b. 9 in N4 locked in R5 -> no 9 elsewhere in R5
9c. 9/3 at R6C2 = {3(15|24)} -> no 3 elsewhere in R6

10. 7 in N6 locked in R4 -> no 7 elsewhere in R4

11. 7 in N4 locked in 27/4 at R5C1 = {(38|56)79} (no 4)

12. 15/2 at R7C1 = {69|78}

13. 19/3 at R7C3: no 1

14. 4/2 at R7C5 = {13} -> no 1,3 elsewhere in C5 or N8

15. 11/3 at R7C6: no 7,9 ({137} no longer available due to step 14)

16. 8/2 at R7C8: no 4,8,9

17. 21/3 at R8C7: no 1,2,3

18. 33/5 at R9C3 = {(36|45)789} -> no 7,8,9 elsewhere in R9

19. 3/2 at R9C8 = {12} -> no 1,2 elsewhere in R9 or N9

20. Hidden Single (HS) in R7 at R7C5 = 1 -> R8C5 = 3

21. 8/2 at R7C8 = {35} (only remaining combination) -> no 3,5 elsewhere in R7 or N9

22. 11/3 at R7C6 = {245} (only remaining combination) -> R8C6 = 5, R7C6 = 2, R7C7 = 4

23. 873 at R3C6 = {134} -> no 3,4 elsewhere in C6

24. (Sterile) Naked Quad on {6789} in R7 at R7C1234
-> sum of R7C1234 = 30 -> sum of R7C34 = 15 -> R8C4 = 4

25. 1,2 in R8 locked in 12/3 at R8C1 = {129} -> no 9 elsewhere in R8 or N7

26. 15/2 at R7C1 = {78} -> no 7,8 elsewhere in R7 or N7

27. Naked Single (NS) at R7C3 = 6 -> R7C4 = 9

28. 21/3 at R8C7 = {678} -> no 6,7,8 elsewhere in N9 -> R9C7 = 9

29. Innie N7: R9C3 = 3

30. 18/3 at R2C3 = {(18|27|45)9} (3,6 unavailable) -> 9 locked in R23C3
-> no 9 elsewhere in C3 or N1

31. 19/3 at R3C4 = {568} (only remaining combination - 4,9 unavailable)
-> no 5,6,8 elsewhere in C4 -> R9C4 = 7

32a. Innie C1234: R2C4 = 1
32b. Split cage 20/3 at R12C5: no 2

33a. 7 in N5 locked in R56 innies (R5C456+R6C5) = 19/4 -> no 7 elsewhere in C5

At this point, the grid is as follows:

Code: Select all

.-----------.-----------------------------------.-----------.-----------------------------------.-----------.
| 12345678  | 123456      1245        23        | 45689     | 6789        1235678     123456789 | 123456789 |
|           '-----------.-----------.-----------'           '-----------.-----------.-----------'           |
| 2345678     2345678   | 245789    | 1           45689       6789      | 235678    | 23456789    23456789  |
&#58;-----------.           |           &#58;-----------.-----------.-----------&#58;           |           .-----------&#58;
| 12345678  | 12345678  | 1245789   | 568       | 245689    | 34        | 1235678   | 123456789 | 456789    |
|           '-----------&#58;           |           |           |           |           &#58;-----------'           |
| 1234568     1234568   | 12458     | 568       | 245689    | 134       | 35678     | 456789      456789    |
&#58;-----------------------'-----------&#58;           |           |           &#58;-----------'-----------------------&#58;
| 356789      356789      578       | 568       | 245678    | 134       | 12356       123456      123456    |
|           .-----------------------'-----------&#58;           &#58;-----------'-----------------------.           |
| 57        | 12345       1245        23        | 2457      | 689         68          689       | 1245      |
&#58;-----------'-----------.-----------------------+-----------+-----------------------.-----------'-----------&#58;
| 78          78        | 6           9         | 1         | 2           4         | 35          35        |
&#58;-----------------------'-----------.           |           |           .-----------'-----------------------&#58;
| 129         129         12        | 4         | 3         | 5         | 678         678         678       |
&#58;-----------------------.-----------'-----------'-----------'-----------'-----------.-----------------------&#58;
| 45          45        | 3           7           68          68          9         | 12          12        |
'-----------------------'-----------------------------------------------------------'-----------------------'

--- the next move is the key one for this puzzle, as it both ---
--- breaks the gridlock and provides for some rapid placements, ---
--- without requiring any heavy computational work ---

33b. This hidden 19/4 R56 innie cage cannot also contain an 8 (would imply {1378},
impossible because only 1 innie cell (R5C6) has either of {13}) -> no 8 in R5C45
33c. 8 in R5 now locked in 27/4 in N4 = {3789} -> no 8 elsewhere in N4
33d. 3 in 27/4 at R5C1 locked in R5C12 -> no 3 elsewhere in R5 or N4

34. NS at R6C1 = 7

35. NS at R5C3 = 8

36. 15/2 at R7C12 = [87]

37. HS in R5 at R5C5 = 7

38. HS in R6 at R6C4 = 3

39. NS at R1C4 = 2

40a. HS in C6/N2 at R3C6 = 3
40b. 4 in 8/3 at R3C6 locked in N5 at R45C6 -> no 4 elsewhere in N5

41a. HS in R4 at R4C7 = 3
41b. 7 in N6 now locked in R4C89 -> 21/3 at R3C9 = {7(59|68)} (no 4), no 7 in R3C9
41c. Split 12/2 cage at R23C7 = {57} (4,9 unavailable) -> no 5,7 elsewhere in C7 or N3

42. HS in C7 at R5C7 = 2

43. HS in C7 at R1C7 = 1

44. HS in R1 at R1C6 = 7 -> R1C8 = 8 (last digit in cage)

45. 6 in N4 locked in R4C12 -> no 6 elsewhere in R4, 11/3 cage at R3C1 = {146} (3 unavailable), no 6 in R3C1

46. 7 in C3/N1 locked in 18/3 at R2C3 = {79}[2]

47. NS at R8C3 = 1

48. HS in C5 at R6C5 = 2

49. Innie R6: R6C9 = 4 -> Split 6/2 cage at R5C89 = {15} (no 6) -> no 1,5 elsewhere in R5 or N6

From now on, the puzzle can be completed via Singles only.

Jean-Christophe · Post by **Jean-Christophe** » Thu Apr 26, 2007 5:21 am

I can't agree with you, Mike:

At step 33b you didn't consider hidden 19/4 R56 innie = {1468|2368|2458}. Therefore you can't eliminate 8 from R5C45
Edit : I was wrong

Regards,

mhparker · Post by **mhparker** » Thu Apr 26, 2007 6:22 am

Jean-Christophe wrote:I can't agree with you, Mike

Au contraire, JC, you seem to have overlooked something here:

We know that the hidden R56 19/4 innie already contains a 7, which is locked for N5 in R56C5 (see step 33a), therefore the combinations you mentioned ({1468|2368|2458}) do not apply.

Jean-Christophe · Post by **Jean-Christophe** » Thu Apr 26, 2007 6:32 am

mhparker wrote:Au contraire, JC, you seem to have overlooked something here:

You'r right, I was reading too fast.

mhparker · Post by **mhparker** » Thu Apr 26, 2007 2:10 pm

Hi folks,

This has certainly been an eventful and enjoyable Assassin week, with many highlights. To round it off, I decided to dust off some old software I wrote some time back, and run both of the two Assassin 47's (V1 and V1.5) through it. The engine is a bit limited, to say the least (more like a japanese rickshaw than a Rolls Royce!), but it "masters" the basic techniques, which are needed for just about any Killer:

* Naked singles
* Hidden singles
* Locked candidates
* Simple combinations (where no attempt is made to see if the combination can actually be placed)
* Simple innies and outies (where all cells are peers of each other)

Guess what?: This simple program almost solved both puzzles! In other words, using no naked/hidden subsets, no killer pairs, no complex (non-peer) innies and outies, no innie/outie difference, no permutations (not even in the trivial case for two-cell cages), etc. It's just as well, since it can't handle most of these anyway...

In both cases (V1 and V1.5), the program only needed a single manual "prod" from me before continuing on its merry way and completing the puzzle using only the above-listed elementary techniques.

In the V1 case, the program (as expected) got stuck at the position roughly corresponding to the candidate grid I posted above. As explained in my walkthrough, the only thing I had to do was to manually eliminate the two occurrences candidate 8 in the 19/4 R56 innie cage. Thereafter, it could complete the puzzle on its own. So this move can rightly be regarded as the key move in my solving path.

In the V1.5 case, the program got stuck without being able to place a single digit. However, the only thing it was missing was not being able to perform the move described in step 15 of Andrew's walkthrough. Once I had manually eliminated {25} from the 2 N9 innie cells (R79C7), it could then complete the puzzle with no further problem. Therefore, this fairly straightforward conflicting combination was already the key move! No wonder Andrew marched straight through wondering when the difficult hurdle (that never came) would come!

What does this tell us? Firstly, it shows that the observation that Andrew made about the V1.5 being no harder than the V1 (and possibly even easier) is correct. It also tells us - as we suspected - that both of these Assassins were One-Trick Ponies, where only a single (slightly) difficult move was required. Lastly, it tells us that both of these Assassins were relatively straightforward, provided one proceeded very meticulously.

Having said that, we managed to use our more advanced techniques regardless, and even some new ones like coloring and multi-coloring (thanks, Cathy!), despite this being a bit like using an atom bomb to kill a canary! At least we had fun along the way!

Nevertheless, it will be interesting to see what difficulty level awaits us with Assassin 48, especially in view of the fact that we now have Jean-Christophe on the team.

T minus 8 hours and counting...