## Suggest A Move (SAM#1)

Post the puzzle or solving technique that's causing you trouble and someone will help
JUST dont forget the butter whipping cream cinneman
and maple syrup
Some do, some teach, the rest look it up.

StrmCkr

Posts: 1134
Joined: 05 September 2006

Luke451 wrote:
aran wrote:either 245r8c189 holds => <24>r8c46
or UR24 r89c4 must be avoided => <24>r8c4.
Only the overlap of the above is established as true : <24>r8c4.

Thanks for the "why" to my "why not."

Code: Select all
` *--------------------------------------------------------------------*  | 7      128    4      | 18     5      189    | 6      3      129    |  | 58     125    9      | 36     4      36     | 18     7      12     |  | 3      18     6      | 189    2      7      | 5      48     49     |  |----------------------+----------------------+----------------------|  | 59     6      2      | 7      8      49     | 3      145    145    |  | 1      45     7      | 26     3      26     | 49     459    8      |  | 89     48     3      | 49     1      5      | 2      6      7      |  |----------------------+----------------------+----------------------|  | 6      7      8      | 1245   9      124    | 14     1245   3      |  |*24     9      1      |*24 358 6      2348   | 7      2458   45     |  |*24     3      5      |*24 18  7      1248   | 1489   12489  6      |  *--------------------------------------------------------------------* 1. 245r8c189=8r8c8-(8=4)r3c8-(4=9)r3c9-(9=18)r13c4-(18=24)r9c4 : =><24>r8c4 `

If (24) ends the chain we're staring at a Type 1 unique rectangle. That's thinking outside the bun...

I think I've got the handle on chain 2, in a way:
aran wrote:2. 24r9c14=18r19c4-(18=9)r3c4-(9=4)r3c9-(4=8)r3c8-(8=1)r2c7-(1=4)r7c7-(4=5)r8c9 :=><24>r9c78 (memory for the <2> highlighted)

If the chain had ended on the bold (4), that would have been enough for r9c78<>4. Taking the extra step completes the establishment of chain memory (or one's favorite alternate terminology) for 3 of the 4 candidates in r8c8. Only (2) is left, hence r9c8<>2. Rather clever, even if this description isn't.

And AFAIC, waffle all you want. I'll take mine with blueberries.

Luke, you're exactly right about chain 2 : there's no "in a way".
And here's some waffle...
To begin with, I did stop that chain on the "4" with its conclusion r9c78<>4, as you mention. Initially thinking that the <2> elimination resulted from branching of the original chain (so wasn't going to write that up : aside from anything else, branching is "ugly" to present) but then realised that it arose from memory within the same chain. BTW I should have ended that chain one link further with -(458=2)r8c8, but got side-tracked by highlighting the memory.
aran

Posts: 334
Joined: 02 March 2007

storm_norm wrote:
Code: Select all
`.---------------------.---------------------.---------------------.| 7     U128    4     | 18     5      189   | 6      3     U129   || 58    U125    9     | 36     4      36    | 18     7     U12    || 3     *18     6     |*-189   2      7     | 5      48    *49    |:---------------------+---------------------+---------------------:| 59     6      2     | 7      8      49    | 3      145    145   || 1      45     7     | 26     3      26    | 49     459    8     || 89     48     3     | 49     1      5     | 2      6      7     |:---------------------+---------------------+---------------------:| 6      7      8     | 1245   9      124   | 14     125    3     || 24     9      1     | 23458  6      2348  | 7      258    45    || 24     3      5     | 1248   7      1248  | 1489   1289   6     |'---------------------'---------------------'---------------------'`

notice the marked cells contain the deadly pattern on {1,2}.
if the 9 is removed from r9c1 then we know that the resulting UR can be avoided by removing the 1's in r12c2, this would force 1 into r3c2.
in other words, neither the 9 in r1c9 nor the 1 in r3c2 can both be false or the deadly pattern is forced to exist.
creates this inference...

UR12[(9)r9c1 = (1)r3c2]...

can be used in this chain

(9)r3c4 = (9)r3c9 - UR12[(9)r9c1 = (1)r3c2]; r3c4 <> 1

I have been going back and culling out some of the pearls from this thread. This elimination is one them. However, I remember that at first I found the explanation a little confusing so I thought I would revisit it. The wording: if the 9 is removed from r9c1 then we know that the resulting UR can be avoided by removing the 1's in r12c2, this would force 1 into r3c2. In other words, neither the 9 in r1c9 nor the 1 in r3c2 can both be false or the deadly pattern is forced to exist., especially the 'in other words' part infers that since r3c2=1 would remove the deadly pattern, then r3c2<>1 (if r9c1<>9) would allow the deadly pattern.

At first glance, that didn't seem to gel with me because even if r9c1<> 9 and r3c2<>1 then the 8 in r1c2 or the 5 in r2c2 could still potentially prevent the deadly pattern. IMO, unless I'm missing something, the more important point is that, in this particular grid, r3c2<>1 actually would remove the 8 and the 5 ie. would make r1c2<>8 -> r6c2=4 -> r5c2=5 -> r2c2<>5 and thus, the deadly pattern would exist.

Now I know that storm_norm understands this (in fact, it makes the elimination all that much more clever) and some people probably picked it up right away, but some may not have so...there it is!

This reminded me once again that, particularly with uniqueness techniques, while there are the established patterns such as types 1 to 6, there are still eliminations possible outside the UR by looking a little further.
DonM
2013 Supporter

Posts: 475
Joined: 13 January 2008

DonM wrote:IMO, unless I'm missing something, the more important point is that, in this particular grid, r3c2<>1 actually would remove the 8 and the 5 ie. would make r1c2<>8 -> r6c2=4 -> r5c2=5 -> r2c2<>5 and thus, the deadly pattern would exist..

If UR digits can't be entirely within a UR pattern, at least one UR digit must be outside the pattern. With an x-wing overlay for one of the UR digits, looking outside the pattern usually provides the simplest strong-inference-set. Indeed, for this elimination ...
Code: Select all
`aur(12)r12c29:[r3c2 =1= r1c46] ==> r1c2<>1, r3c4<>1`
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

DonM wrote:
storm_norm wrote:
Code: Select all
`.---------------------.---------------------.---------------------.| 7     U128    4     | 18     5      189   | 6      3     U129   || 58    U125    9     | 36     4      36    | 18     7     U12    || 3     *18     6     |*-189   2      7     | 5      48    *49    |:---------------------+---------------------+---------------------:| 59     6      2     | 7      8      49    | 3      145    145   || 1      45     7     | 26     3      26    | 49     459    8     || 89     48     3     | 49     1      5     | 2      6      7     |:---------------------+---------------------+---------------------:| 6      7      8     | 1245   9      124   | 14     125    3     || 24     9      1     | 23458  6      2348  | 7      258    45    || 24     3      5     | 1248   7      1248  | 1489   1289   6     |'---------------------'---------------------'---------------------'`

notice the marked cells contain the deadly pattern on {1,2}.
if the 9 is removed from r9c1 then we know that the resulting UR can be avoided by removing the 1's in r12c2, this would force 1 into r3c2.
in other words, neither the 9 in r1c9 nor the 1 in r3c2 can both be false or the deadly pattern is forced to exist.
creates this inference...

UR12[(9)r9c1 = (1)r3c2]...

can be used in this chain

(9)r3c4 = (9)r3c9 - UR12[(9)r9c1 = (1)r3c2]; r3c4 <> 1

I have been going back and culling out some of the pearls from this thread. This elimination is one them. However, I remember that at first I found the explanation a little confusing so I thought I would revisit it. The wording: if the 9 is removed from r9c1 then we know that the resulting UR can be avoided by removing the 1's in r12c2, this would force 1 into r3c2. In other words, neither the 9 in r1c9 nor the 1 in r3c2 can both be false or the deadly pattern is forced to exist., especially the 'in other words' part infers that since r3c2=1 would remove the deadly pattern, then r3c2<>1 (if r9c1<>9) would allow the deadly pattern.

At first glance, that didn't seem to gel with me because even if r9c1<> 9 and r3c2<>1 then the 8 in r1c2 or the 5 in r2c2 could still potentially prevent the deadly pattern. IMO, unless I'm missing something, the more important point is that, in this particular grid, r3c2<>1 actually would remove the 8 and the 5 ie. would make r1c2<>8 -> r6c2=4 -> r5c2=5 -> r2c2<>5 and thus, the deadly pattern would exist.

Now I know that storm_norm understands this (in fact, it makes the elimination all that much more clever) and some people probably picked it up right away, but some may not have so...there it is!

This reminded me once again that, particularly with uniqueness techniques, while there are the established patterns such as types 1 to 6, there are still eliminations possible outside the UR by looking a little further.

Don
Storm Norm's version is fine : it relies on the fact that 2 is restricted to r12c2.
Therefore if <9>r1c9, then we have : 12r12c9 + for sure 2 over r12c2; so the only possible avoidance of the deadly pattern 12r12c29 is for 1 to lie outside r12c2 ie at r3c2. Hence 9r1c9=1r3c2.
The use of that in his deduction <1>r3c4 is, I agree with you, a fine move.
aran

Posts: 334
Joined: 02 March 2007

aran wrote:Storm Norm's version is fine : it relies on the fact that 2 is restricted to r12c2.
Therefore if <9>r1c9, then we have : 12r12c9 + for sure 2 over r12c2; so the only possible avoidance of the deadly pattern 12r12c29 is for 1 to lie outside r12c2 ie at r3c2. Hence 9r1c9=1r3c2.
The use of that in his deduction <1>r3c4 is, I agree with you, a fine move.

Between you & ronk, it's now a little clearer. Thanks to both.

IMO, uniqueness-based techniques, other than the basic types, are some of the least understood & underused. One thing I'm glad to see though is that they are finally being used more routinely in solutions without any qualifications about whether uniqueness can be assumed in any given puzzle, something that really BUGged me.

Totally digressing for a moment. Has anyone noticed how bizarre the situation is at the moment when it comes to a person trying to learn how to solve Sudoku using basic & advanced methods? I went into Borders (a book store chain for those not in the U.S.) the other day and just for the heckuvit checked out the Sudoku section. There were easily over 100 Sudoku books with all sorts of names. While right next to them were books on Chess, Bridge with indepth descriptions on how to play them, not one of the Sudoku books had anything more than 1-3 pages on very rudimentary instructions on how to solve a sudoku puzzle (eg. pretty much limited to cross-hatching, perhaps basic naked pairs and the like.) and most of them had nothing. If you were lucky you might come across Paul Stephen's Mastering Sudoku which is probably the best book available for basic methods now that Andrew Stuart's 'Logic of Sudoku' is sadly unavailable now. Otherwise there's virtually nothing that would teach someone how to logically solve a puzzle. Which means that any person trying to get to the level of those on this forum have got a lot of hard-core sleuthing to do.

All the books were just puzzles and virtually all of them in the newspaper-level difficulty and below. Plus, if you were to go by the 'authors' of the books, you would think that Will Shortz was a household sudoku name. Anybody ever seen ole Will ever post anything about Sudoku anywhere? Very wierd!
Last edited by DonM on Wed Feb 25, 2009 10:22 pm, edited 2 times in total.
DonM
2013 Supporter

Posts: 475
Joined: 13 January 2008

Code: Select all
`.------------------.------------------.------------------. | 5     17    38   | 9     18    37   | 2     6     4    | |U2467  9     26   | 18   U24    67   | 5     3     18   | |U246   124   38   | 5    U24    36   | 9     7     18   | :------------------+------------------+------------------: | 8     3     9    | 4     7     5    | 1     2     6    | |24-6   24    126  | 3     16    9    | 8     5     7    | | 67    57    156  | 18    168   2    | 3     4     9    | :------------------+------------------+------------------: | 3     8     7    | 6     5     1    | 4     9     2    | | 9     6     4    | 2     3     8    | 7     1     5    | | 1     25    25   | 7     9     4    | 6     8     3    | '------------------'------------------'------------------'`

and to beat around the dead bush... again...

the marked cells form a UR on {2,4}...
normally you would look for the common candidates and form a naked pair with the {6,7} in r6c1 to eliminate the 6 in r5c1...

but...

you can also avoid the deadly pattern by simply seeing that both the 2 and the 4 must exist in r5c1 to avoid the deadly pattern
in other words if 6 were true for r5c1 the deadly pattern is forced to exist and thus can be victimized.
storm_norm

Posts: 85
Joined: 27 February 2008

one more on the same line of thinking...

Code: Select all
`.---------------.---------------.---------------.| 6    34   28  | 5   U78   1   |U278  9    34  ||*38   7    9   | 2    6    4   | 3*8  1    5   || 25   45   1   | 3   U78   9   |U278  47   6   |:---------------+---------------+---------------:| 9    8    7   | 4    3    2   | 6    5    1   || 34   2    34  | 1    5    6   | 9    8    7   || 1    6    5   | 7    9    8   | 4    3    2   |:---------------+---------------+---------------:| 25   359  23  | 8    4    7   | 1    6    39  ||*4-78 39   48  | 6    1    5   |3*7   2    349 || 47   1    6   | 9    2    3   | 5    47   8   |'---------------'---------------'---------------'`

the UR {7,8} tells us that either the 7 in r2c7 or the 8 in r8c7 must be true. if both were false then the deadly pattern must exist. right?
so the strong inference can be made between them
and there just so happens to be a strong inference on 8's in clolumn 1 that sees it too.

UR78[(7)r8c7 = (8)r2c7] - (8)r2c1 = (8)r8c1; r8c1 <> 7

and cracks the puzzle.

sorry, done now
storm_norm

Posts: 85
Joined: 27 February 2008

storm_norm wrote:one more on the same line of thinking...

Code: Select all
`.---------------.---------------.---------------.| 6    34   28  | 5   U78   1   |U278  9    34  ||*38   7    9   | 2    6    4   | 3*8  1    5   || 25   45   1   | 3   U78   9   |U278  47   6   |:---------------+---------------+---------------:| 9    8    7   | 4    3    2   | 6    5    1   || 34   2    34  | 1    5    6   | 9    8    7   || 1    6    5   | 7    9    8   | 4    3    2   |:---------------+---------------+---------------:| 25   359  23  | 8    4    7   | 1    6    39  ||*4-78 39   48  | 6    1    5   |3*7   2    349 || 47   1    6   | 9    2    3   | 5    47   8   |'---------------'---------------'---------------'`

the UR {7,8} tells us that either the 7 in r2c7 or the 8 in r8c7 must be true. if both were false then the deadly pattern must exist. right?
so the strong inference can be made between them
and there just so happens to be a strong inference on 8's in clolumn 1 that sees it too.

UR78[(7)r8c7 = (8)r2c7] - (8)r2c1 = (8)r8c1; r8c1 <> 7

and cracks the puzzle.

sorry, done now

Storm Norm
Technically UR78 doesn't really exist (2r13c7 being forced).
7r8c7=8r2c7 holds independently in any case.
aran

Posts: 334
Joined: 02 March 2007

Technically UR78 doesn't really exist (2r13c7 being forced).
7r8c7=8r2c7 holds independently in any case.

I think its valid...
but this forces me to find a better example.
unless you find one before me, aran.
storm_norm

Posts: 85
Joined: 27 February 2008

storm_norm wrote:
Technically UR78 doesn't really exist (2r13c7 being forced).
7r8c7=8r2c7 holds independently in any case.

I think its valid...
but this forces me to find a better example.
unless you find one before me, aran.

Here's one contender Storm Norm :
UR39 avoidance r78c29 =>5r7c2=4r8c9
then : 5r7c2=4r8c9-(4=7)r9c8-(7=3)r8c7-(3=9)r8c2 : =><9>r7c2
aran

Posts: 334
Joined: 02 March 2007

aran wrote:
storm_norm wrote:
Technically UR78 doesn't really exist (2r13c7 being forced).
7r8c7=8r2c7 holds independently in any case.

I think its valid...
but this forces me to find a better example.
unless you find one before me, aran.

Here's one contender Storm Norm :
UR39 avoidance r78c29 =>5r7c2=4r8c9
then : 5r7c2=4r8c9-(4=7)r9c8-(7=3)r8c7-(3=9)r8c2 : =><9>r7c2

aran, that's another Type 6 UR as mentioned on page 1, so you can add <9>r8c9.

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

DonM wrote:Totally digressing for a moment. Has anyone noticed how bizarre the situation is at the moment when it comes to a person trying to learn how to solve Sudoku using basic & advanced methods? I went into Borders (a book store chain for those not in the U.S.) the other day and just for the heckuvit checked out the Sudoku section. There were easily over 100 Sudoku books with all sorts of names. While right next to them were books on Chess, Bridge with indepth descriptions on how to play them, not one of the Sudoku books had anything more than 1-3 pages on very rudimentary instructions on how to solve a sudoku puzzle (eg. pretty much limited to cross-hatching, perhaps basic naked pairs and the like.) and most of them had nothing. If you were lucky you might come across Paul Stephen's Mastering Sudoku which is probably the best book available for basic methods now that Andrew Stuart's 'Logic of Sudoku' is sadly unavailable now. Otherwise there's virtually nothing that would teach someone how to logically solve a puzzle. Which means that any person trying to get to the level of those on this forum have got a lot of hard-core sleuthing to do.

All the books were just puzzles and virtually all of them in the newspaper-level difficulty and below. Plus, if you were to go by the 'authors' of the books, you would think that Will Shortz was a household sudoku name. Anybody ever seen ole Will ever post anything about Sudoku anywhere? Very wierd!

So true. No one has published a truly advanced book on the subject. Wizards, there's a void out there waiting for one of you to fill!

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

me me me... i'll get on it...

hahaha

and no one would understand what I'm saying or trying to explain

id be just like every other day.
Some do, some teach, the rest look it up.

StrmCkr

Posts: 1134
Joined: 05 September 2006

Luke451 wrote:
DonM wrote:Totally digressing for a moment. Has anyone noticed how bizarre the situation is at the moment when it comes to a person trying to learn how to solve Sudoku using basic & advanced methods? I went into Borders (a book store chain for those not in the U.S.) the other day and just for the heckuvit checked out the Sudoku section. There were easily over 100 Sudoku books with all sorts of names. While right next to them were books on Chess, Bridge with indepth descriptions on how to play them, not one of the Sudoku books had anything more than 1-3 pages on very rudimentary instructions on how to solve a sudoku puzzle (eg. pretty much limited to cross-hatching, perhaps basic naked pairs and the like.) and most of them had nothing. If you were lucky you might come across Paul Stephen's Mastering Sudoku which is probably the best book available for basic methods now that Andrew Stuart's 'Logic of Sudoku' is sadly unavailable now. Otherwise there's virtually nothing that would teach someone how to logically solve a puzzle. Which means that any person trying to get to the level of those on this forum have got a lot of hard-core sleuthing to do.

All the books were just puzzles and virtually all of them in the newspaper-level difficulty and below. Plus, if you were to go by the 'authors' of the books, you would think that Will Shortz was a household sudoku name. Anybody ever seen ole Will ever post anything about Sudoku anywhere? Very wierd!

So true. No one has published a truly advanced book on the subject. Wizards, there's a void out there waiting for one of you to fill!

this is human nature tho!

think about it. nothing gets the juices flowing more than a competitive game against your arch rival.

the demand for books on chess, bridge, poker, whatever the game is is going to be high because the incentive is there to read them.
gain knowledge, etc.

last time I checked, you can't lose or win at sudoku.

sorry to say this, but a sudoku book with a plethora of sudoku techniques would still ONLY be a "self help" book, with a spot reserved next to the cooking and home improvement books....
storm_norm

Posts: 85
Joined: 27 February 2008

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