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(unofficial) Assassin 98
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sudokuEd
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Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

PostPosted: Sat Apr 19, 2008 7:39 am    Post subject: Reply with quote

A very busy week with these two puzzles! Desperately trying to clear up the loose-ends before getting any further behind!

About uA98 V1 sudokuEd wrote:
Solving is like a lazy tropical river with 16 bridges & many back-waters to explore.
Andrew wrote:
Ed, was this meant to be a hint?
Not a hint - just describing how the solution process seemed with those cages that "bridge" across nonet boundaries, with the 12(3)r6c6 especially helpful to my solution. I used outies n69 rather than Afmob's more productive i/o n69 (his "easier uA98 WT"). I also liked how there were so many i/o possibilities all over the place.

Andrew wrote:
I didn't use any very difficult steps but since it took me a long time to spot the breakthrough move I'll rate uA98 as an Extremely Hard 1.25
I thought there were a good number of clever & difficult steps (complex cage-blocks etc) so felt more like a 1.50 WT. After working through Andrew & Afmob's walk-throughs, I think my rating for uA98 v2 as (easy) 1.50 is high.

About V2 Andrew wrote:
I wonder which, if any, of my moves SS wouldn't be able to find. ....There must be some reason why it gave this puzzle such a high rating. Maybe it's because step 23 analyses a hidden cage?
This is the one. Your step 23 combines the innies r1 = h12(4) with the combinations in 8(3)r1c8 to be powerful. SS can only do this hidden cage-placement with the "45 Extended" routine which is way down the list below "Multiple House". SS does ridiculous contortions with obscure (& unproductive) "Multiple House" elims before it can finally get to "Extended". By moving "Extended" one above "Multiple House" on the routine order, the SS(v3)score is.....you guessed it...1.50!

I've put onto Richard's list to get a "4-cell 45 Extended" routine to sit above "Multiple House". This could make a big difference to the scoring of the 1.45+ puzzles. So, thanks for the question Andrew!

direct-link candidates
Finally, a really nice move from Andrew's uA98 V2 walk-through. His step 21 blandly says
Andrew wrote:
21. 14(3) cage in N7 (step 7) = {149/239/347/356} (cannot be {158/167} because then no place for 9 in R9)
.
Code:
.-----------------------------------------------------------------------.
|(17)   :       |(21)   :       :       |(12)   :       |( 8)   :       |
|  123  :  1 3  |       :       :       |    3  :    3  |  123  :  123  |
|  456  :       |  456  :  4 6  :  4 6  |  45   :  45   |  45   :  45   |
|       :       |   8   :  789  :  789  |  789  :  789  |       :       |
|.......--------|-----------------------|---------------|-------:.......|
|       |(21)   |(19)   :       :       |(21)   :       :       |       |
|  123  |   2   |  123  :  123  :  123  |  123  :   23  :       |  123  |
|  456  |  456  |  456  :  456  :  456  |  456  :  456  :   56  |   5   |
|  789  |  789  |   8   :  789  :  789  |  789  :  789  :  789  |       |
|.......|.......|---------------:.......|.......----------------|-------|
|       |       |(17)   :       |       |       |( 9)   |( 7)   |(28)   |
|  123  |   2   |       :       |  123  |  123  |   23  |   23  |       |
|  456  |  456  |   56  :   56  |  456  |  456  |  456  |  456  |  456  |
|  789  |  789  |   8   :  789  |  789  |  789  |  78   |       |  789  |
|-------|.......|-------:.......|-------|-------|.......|.......|.......|
|(18)   |       :       |       |(13)   :       |       |       |       |
|  123  |   2   :  123  |   23  |       :       |  123  |  123  |       |
|  4    |  456  :  4    |  4    |  456  :  456  |  456  |  45   |  456  |
|       |  789  :       |       |  789  :  789  |  7    |       |  789  |
|.......|---------------|-------|---------------|-------|-------|.......|
|       |( 7)   :       |(14)   :       :       |( 9)   :       |       |
|       |       :  123  |  123  :  123  :  123  |   23  :    3  |       |
|    6  |  456  :       |  45   :  45   :  45   |  456  :  456  |    6  |
|  789  |       :       |   89  :   89  :   89  |       :  7    |  789  |
|.......|---------------|-----------------------|---------------|.......|
|       |( 9)   |(16)   |(13)   :       |(16)   |(18)   :       |       |
|  123  |   2   |       |       :       |  123  |  123  :  123  |       |
|  456  |  45   |       |  456  :  456  |       |  456  :  456  |   56  |
|  789  |  7    |  7 9  |  789  :  789  |       |       :  789  |  789  |
|.......|.......|.......|---------------|.......|-------:.......|-------|
|       |       |       |(17)   |(21)   |       :       |       |(18)   |
|  123  |   2   |       |  123  |  123  |       :       |  123  |  123  |
|  456  |  45   |       |  456  |  456  |  456  :  456  |  456  |  456  |
|  789  |  7    |  7 9  |       |  789  |  789  :  789  |  789  |       |
|-------|-------|-------|.......|.......|---------------|.......|.......|
|(14)   |       :       :       |       :       :       |       |       |
|   2   |       :  123  :  123  |  123  :  123  :  123  |  123  |  123  |
|  456  |       :  456  :  456  |  456  :  456  :  456  |  456  |  456  |
|       |   89  :       :       |  789  :  789  :  789  |  789  |       |
|.......|-----------------------|-----------------------|-------|.......|
|       :       |(10)   :       |(12)   :       :       |       :       |
|       :  1 3  |  123  :   2   |  123  :  123  :  123  |       :   23  |
|   56  :       |  4 6  :  4 6  |  456  :  456  :  456  |  456  :  456  |
|  789  :       |   8   :  789  |  7 9  :  7 9  :  7 9  |  78   :  7    |
.-----------------------------------------------------------------------.

There is a direct link between candidates 1 & 9 in r9c3..9 through combinations in the 10(2) & 12(3) cages (={19} or {129} respectively) . In other words, 9 is linked with 1 in these two cages -> if r9c12 has 1 it must also have 9 [edited to get more accurate, see why below] -> {158/167} blocked from 14(3) since 1 can only be in r9c2. Nice find Andrew!

But the same is true for {239} in 14(3)n7. [edit: not true. Tricked. 9 in r9c1 -> 1 in r9 can be in 12(3) = 1{47/56}]. Since 9 in {239} can only be in r9c1 -> r9c2 must have a 1 which {239} doesn't. In summary, since 1 & 9 are directly linked in every other position in r9 -> r9c12 must have (19) both in or both out [Correction: the direct link is only on the 9 in r9..since 9 in 10(2) or 12(3) must have 1 -> 1 in r9c2 must have 9 in r9c1]

Still feeling very chuffed about that 4 cell overlap Cool [edit: pride goeth before.... Embarassed ].
Ed
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mhparker
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Joined: 20 Jan 2007
Posts: 345
Location: Germany

PostPosted: Mon Apr 28, 2008 7:10 pm    Post subject: Reply with quote

sudokuEd wrote:
There is a direct link between candidates 1 & 9 in r9c3..9 through combinations in the 10(2) & 12(3) cages (={19} or {129} respectively) . In other words, 9 is linked with 1 in these two cages -> if r9c12 has 1 it must also have 9 -> {158/167} blocked from 14(3) since 1 can only be in r9c2. Nice find Andrew!

Ed, haven't got time to go into detail right now (if I do, I'd prefer to do that on Richard's site anyway), but isn't this the same technique that Ruud mentioned in the Talking Shop thread, but extended to three cages?

Still, it's nice to have a practical real-life example for the technique books, as it were. Thanks very much, Andrew, for finding it! Very Happy
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Cheers,
Mike
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