Sudoku Players' Forum (clone)

by **daj95376** » Fri Aug 05, 2011 9:17 pm

Thanks Ron for the evaluation. I agree that the line must be drawn somewhere on creating UR patterns.

As it stands now, my solver resolves strong links associated with <ab> and the UR cells -- but no other values. Then it makes one more pass using Naked Singles for all values to resolve patterns like UR+4x/2SL. However, if more than one Naked Single exists and can contribute to creating the DP, then my solver uses it. For now, I'm not going to sweat a scenario resulting from multiple Naked Singles.

Thanks again & Regards, Danny

by **David P Bird** » Sat Aug 06, 2011 9:24 pm

Ronk, Some time ago (2008) I ran through MB's list of UR types and found every one I tested them could be expressed as a linear AIC. It was a tedious task and I never completed it. It took some time to find my rough notes but they contain a copy of Mike's original written description of this pattern:

UR+4/2SL: three or four UR cells with extra candidates (Z is optional), plus two strong link and at least one extra cell
--- UR+4x/2SL: "Y" is a single candidate "y", the extra cell "(ab)y" can include "a", and/or can include "b" if it shares a house with "abX" => "b" can be removed from "abX".
Code: Select all
abW-----abX a | |a | aby ab(Z) (ab)y

So it requires that if the cell (ab)y contains "b" it must be in sight of abX. This isn't represented in the diagram but is "implicit" in this representation, which what I meant when I said:

Danny in Mikes figure I believe it's implied that there is a compound strong link between the b's involved ...

(a)r1c4 = (a#2)r1c1,r4c2 â€“[UR]- (b#2)r1c4,r4c1 = (b)r4c46 => r1c4 <> b

I therefore stand by what I said. I accept however that you may have found another logical route where the (ab)y cell is in box 3, but I think you then must require certain ALSs to exist and so require other restrictions.

Please also consider the counter example I gave with a possible solution with b in r1c4 if (ab)Y is moved to r3c9.

It would help if you would notate or illustrate the logical route you had in mind.

by **daj95376** » Sun Aug 07, 2011 4:46 am

David P Bird wrote:Ronk, Some time ago (2008) I ran through MB's list of UR types and found every one I tested them could be expressed as a linear AIC. It was a tedious task and I never completed it. It took some time to find my rough notes but they contain a copy of Mike's original written description of this pattern:
UR+4/2SL: three or four UR cells with extra candidates (Z is optional), plus two strong link and at least one extra cell
--- UR+4x/2SL: "Y" is a single candidate "y", the extra cell "(ab)y" can include "a", and/or can include "b" if it shares a house with "abX" => "b" can be removed from "abX".
Code: Select all
abW-----abX a | |a | aby ab(Z) (ab)y

So it requires that if the cell (ab)y contains "b" it must be in sight of abX. This isn't represented in the diagram but is "implicit" in this representation, which what I meant when I said:
Danny in Mikes figure I believe it's implied that there is a compound strong link between the b's involved ...

(a)r1c4 = (a#2)r1c1,r4c2 â€“[UR]- (b#2)r1c4,r4c1 = (b)r4c46 => r1c4 <> b

I therefore stand by what I said. I accept however that you may have found another logical route where the (ab)y cell is in box 3, but I think you then must require certain ALSs to exist and so require other restrictions.

Please also consider the counter example I gave with a possible solution with b in r1c4 if (ab)Y is moved to r3c9.

It would help if you would notate or illustrate the logical route you had in mind.

David,

I ignored your original reply because I wasn't sure of what you were describing. Later, I decided that you were describing MB's UR as if it were in specific cells of a grid. That was masked by a typo where r4c3 should have been r3c4.

This re-write of your logic appears to be screwed up because cells r1c1, r4c2, r1c4, and r4c1 are in three boxes and don't come close to describing a UR pattern. Finally, the fact that "b" can't be present in both diagonal cells doesn't force "b" true in any other cell in the diagram.

Here is MB's UR in a grid ... and what sequentially happens in my solver if r1c4=b.

Code: Select all: +-----------------------------------------------+ | abW / / | abX / / | / / / | <- SL on "a" | . . . | / . . | . . . | | aby . . | ab(Z). (ab)y | . . . | |---------------+---------------+---------------| | . . . | / . . | . . . | | . . . | / . . | . . . | | . . . | / . . | . . . | |---------------+---------------+---------------| | . . . | / . . | . . . | | . . . | / . . | . . . | | . . . | / . . | . . . | +-----------------------------------------------+ ^ |------------------------------- SL on "a"

Code: Select all: r1c4 =b r1c1,r3c4=a (2SL) r3c6 =y (exposed Naked Single) r3c1 =b follows and completes the DP +-----------------------------------------------+ | a / / | b / / | / / / | <- SL on "a" | . . . | / . . | . . . | | b . . | a . y | . . . | |---------------+---------------+---------------| | . . . | / . . | . . . | | . . . | / . . | . . . | | . . . | / . . | . . . | |---------------+---------------+---------------| | . . . | / . . | . . . | | . . . | / . . | . . . | | . . . | / . . | . . . | +-----------------------------------------------+ ^ |------------------------------- SL on "a"

Regards, Danny

by **David P Bird** » Mon Aug 08, 2011 8:30 am

Danny, thanks for your response and sorry for my erroneous notation. I've delayed replying hoping for a response from Ronk.

From my analysis of Mike's pattern, the deduction can only hold if there are no b's in r3c2789 or in r1c56,r2c456 regardless of whether we navigate through a compound link to a split b node or to two ALS nodes in row 3 and box 2. If that's right, then we don't need the ALSs at all, and extra candidates are possible.

However, from your diagrams it looks as if your solver employs single-sided assumptions. This makes me wonder if there's any point in me trying to contribute.

by **ronk** » Mon Aug 08, 2011 12:13 pm

David P Bird wrote:From my analysis of Mike's pattern, the deduction can only hold if there are no b's in r3c2789 or in r1c56,r2c456 regardless of whether we navigate through a compound link to a split b node or to two ALS nodes in row 3 and box 2. If that's right, then we don't need the ALSs at all, and extra candidates are possible.

Hopefully a picture is still worth 1000 words. For the Xsudo image below of a Mike BarkerUR+4/2SL, a=1, b=2 and y=3.

David P Bird, I think you are somehow mixing native and derived links ("inferences"

. Note candidate <2> is in all the locations you mention, but none is a member of either a strong or weak inference. Xsudo necessarily considers only native strong inferences.

____

P.S. Although Mike did so, IMO candidate <1> in r4c6 shouldn't really be part of the pattern ... because locked candidates (1)c4\b2 yields r4c6<>1.

by **daj95376** » Mon Aug 08, 2011 4:26 pm

There is a scenario for UR+4x/2SL that I forgot to include.

Code: Select all: +-----------------------------------------------+ | abW / / | abX / / | / / / | <- SL on "a" | . . . | / . . | . . . | | aby . . | ab(Z). . | . (a)y . | |---------------+---------------+---------------| | . . . | / . . | . . . | | . . . | / . . | . . . | | . . . | / . . | . . . | |---------------+---------------+---------------| | . . . | / . . | . . . | | . . . | / . . | . . . | | . . . | / . . | . . . | +-----------------------------------------------+ ^ |------------------------------- SL on "a"

Mike Barker makes use of it in his "zoo" example.

Code: Select all: ..2.76...7.......8...9....6..5...7.2..4.6.....3....9..8...4...5.....1.9.5.98...1. r9c7 =2 r7c7,r9c2=6 (2SL) r4c2 =1 (exposed Naked Single) r7c2 =2 (follows and completes DP) +-----------------------------------------------------------------------+ | 134 8 2 | 1345 7 6 | 1345 345 9 | | 7 9 6 | 12345 1235 2345 | 12345 2345 8 | | 134 5 13 | 9 8 234 | 1234 7 6 | |-----------------------+-----------------------+-----------------------| | 16 @16 5 | 34 9 348 | 7 348 2 | | 9 7 4 | 123 6 238 | 35 358 13 | | 2 3 8 | 1457 15 457 | 9 6 14 | |-----------------------+-----------------------+-----------------------| | 8 *26+1 137 | 2367 4 9 | 26+3 23 5 | | 36 246 37 | 23567 235 1 | 8 9 347 | | 5 *26+4 9 | 8 23 237 | 26+34 1 347 | +-----------------------------------------------------------------------+ # 89 eliminations remain r79c27 <26> UR via s-link + N_Singles <> 2 r9c7

by **David P Bird** » Mon Aug 08, 2011 5:29 pm

Ronk Firstly thanks for providing the link to Mikes original post and secondly you're right that the strong links for 'a' eliminate all other a's in row 1 and box 2 and (a)r3c6 should have already been eliminated.

Now to the nitty gritty: Your diagram was what I needed and I nearly understand it! - I don't know how to translate the strong inference for cell r1c4 or the inferences between (1)r1c4 and (1)r3c4 link (which are both weak and strong) into a chain or even a truth matrix.

Abandoning working in linear chains removed the scale from my eyes though, and I now believe we can modify the requirements for this pattern.

Code: Select all: aW ------ a(bc)X a | |a | abc aY bc

Candidate 'a' is strongly linked on two adjacent sides. In the unlinked UR cell the candidates 'bc' form an almost naked pair with a cell in sight of the doubly linked UR cell. W X and Y represent any other candidates. Candidates 'b' & 'c' can be eliminated from the doubly linked UR cell. This can also be extended to employ an almost naked triple.

Walkthrough:
If r1c4 doesn't contain 'a' it becomes true at r1c1 and r3c4 and so false in r3c1. To avoid the UR whichever of 'bc' is true in r3c1 must be false in r1c4 while the other one must be true in r3c6 and so also false in r1c4. So either r1c4 holds 'a' or a member of X.

I can't represent that as a linear AIC though. Consequently I'll have to add it to my list of recognisable inferences. I see it as a combination of a UR and an almost naked tuple, but it derserves a name rather than just a type number.

by **ronk** » Mon Aug 08, 2011 9:53 pm

David P Bird wrote:I now believe we can modify the requirements for this pattern.

Code: Select all
aW ------ a(bc)X a | |a | abc aY bc

Candidate 'a' is strongly linked on two adjacent sides. In the unlinked UR cell the candidates 'bc' form an almost naked pair with a cell in sight of the doubly linked UR cell. W X and Y represent any other candidates. Candidates 'b' & 'c' can be eliminated from the doubly linked UR cell.

Eliminating both candidates 'b' and 'c' from the strong corner** on 'a' is OK only if a UR on tokens 'a' and 'c' is possible too. A proper diagram for this should explicitly show all three tokens in all four corners of the UR, with several of them parenthesized to indicate they don't need to be there for the single UR case. Such a diagram would be too complicated IMO.

Besides, it should be understood that Mike's UR+4/2SL can be applied twice, first with UR tokens 'a' and 'b', and then with tokens 'a' and 'c'.

** - IIRC strong corner is a term coined by Myth Jellies.

by **David P Bird** » Tue Aug 09, 2011 9:46 am

Ronk wrote:Eliminating both candidates 'b' and 'c' from the strong corner** on 'a' is OK only if a UR on tokens 'a' and 'c' is possible too. A proper diagram for this should explicitly show all three tokens in all four corners of the UR, with several of them parenthesized to indicate they don't need to be there for the single UR case. Such a diagram would be too complicated IMO.

Besides, it should be understood that Mike's UR+4/2SL can be applied twice, first with UR tokens 'a' and 'b', and then with tokens 'a' and 'c'.

I hear what you say and as you're the patterns man I won't step on your toes.

That said, we have two linked theorems for URs and ARs which dictate that, lacking a given, a rectangle of four cells in two boxes can't resolve to a two digit solution. Those opposed to uniqueness based methods won't be interested in the UR's but the same patterns also provide AR deductions which should be of interest to them.

For those unfamiliar with this distinction between Unique and Avoidable Rectangles the logic goes like this:
If a UR exists in a puzzle it will be an isolated sub-puzzle and it will be impossible to eliminate either of the two digits in the 4 cells. Assuming a unique solution, one of the cells must contain a third digit.
In an AR one of the digits has already been eliminated somewhere in the pattern so we know, rather than having to assume, that the 4 cells aren't isolated and so have a unique solution. Consequently they must contain at least 3 different digits.

Incidentally I'm coming to believe that it's impossible to convert a UR into an AR using only "primitive" exclusions (which just use strong inferences from bilocals and bivalues). If this is so, it means that there's no point looking for them until some more adavanced exclusion has been made.

by **ronk** » Thu Aug 11, 2011 12:11 pm

daj95376 wrote:There is a scenario for UR+4x/2SL that I forgot to include.

Code: Select all
+-----------------------------------------------+ | abW / / | abX / / | / / / | <- SL on "a" | . . . | / . . | . . . | | aby . . | ab(Z). . | . (a)y . | |---------------+---------------+---------------| | . . . | / . . | . . . | | . . . | / . . | . . . | | . . . | / . . | . . . | |---------------+---------------+---------------| | . . . | / . . | . . . | | . . . | / . . | . . . | | . . . | / . . | . . . | +-----------------------------------------------+ ^ |------------------------------- SL on "a" [abX] <> b

Thanks, I was unaware of that variation. Two illustrations might be better than one for this UR+4/2SL, but I hesitate to change Mike's post.

by **surbier** » Wed Aug 31, 2011 7:04 am

David P Bird wrote:Ronk, Some time ago (2008) I ran through MB's list of UR types and found every one I tested them could be expressed as a linear AIC. It was a tedious task and I never completed it. It took some time to find my rough notes but they contain a copy of Mike's original written description of this pattern:
UR+4/2SL: three or four UR cells with extra candidates (Z is optional), plus two strong link and at least one extra cell
--- UR+4x/2SL: "Y" is a single candidate "y", the extra cell "(ab)y" can include "a", and/or can include "b" if it shares a house with "abX" => "b" can be removed from "abX".
Code: Select all
abW-----abX a | |a | aby ab(Z) (ab)y

So it requires that if the cell (ab)y contains "b" it must be in sight of abX. This isn't represented in the diagram but is "implicit" in this representation, which what I meant when I said:
...

Hi ronk,

I agree. When I once read the original post, I interpreted the not quite correct english term "and/or"
as two distinct cases:

Code: Select all: abW-----abX a | |a | aby ab(Z) (ab)y | case 1: (ab)y cell sees abX aby ab(Z) | (a)y case 2: (a)y cell does not see abX

IMO MBÂ´s definition of the UR+4x/2SL covers both variants. There is no need to modify
MBÂ´s post in this concern.

by **surbier** » Mon Jan 16, 2012 1:19 pm

Hi, a comment on the UR+4X/3SL

I was searching for examples for the UR+4X/3SL. I applied my code over
the last weekend to a huge number of grids without any success.
The priority of applied methods I used for this test is:

- hidden/naked singles
- line-block interaction.
- hidden/naked subsets (pairs/triples/quads)
- x-wing
- UR+4X/3SL
- other methods

Today I decided to switch on my brain instead my computer

and was reading the description more carefully:

UR+4X/3SL: the links with equal labels are disjoint which forms a
continuous nice loop => X-wing eliminations if not previously performed
("a" can be removed from cells common to "abX" and "abY" and cells common
to "abZ" and "abW", "Z" can be removed from "abZ", "W" can be removed
from "abW" which reduces the problem to UR+2B/1SL so "b" can be removed
from "abX" and "abY"

Code: Select all: a-bX-----ab-Z a | b| a | a-bY-----ab-W

It turns out easily that after all "a"s cells common to "abZ" and
"abW" have been removed via x-wing, there appears a hidden pair via
the required strong link in "b" to remove Z and W. So the UR+4X/3SL
can be decomposed to a x-wing, a hidden pair and a UR+2B/1SL.

I will discard this method from my solver.

by **surbier** » Mon Jan 16, 2012 3:18 pm

Here an example, where the UR+4X/3SL is applied before hidden pairs are applied:

Code: Select all: 050803040006702100301000709500020004000000000900040006403000907005301400070904030 +-----------------------+-----------------------+-----------------------+ | 7 5 9 | 8 1 3 | 6 4 2 | | 8 4 6 | 7 9 2 | 1 5 3 | | 3 2 1 | 4 56 56 | 7 8 9 | +-----------------------+-----------------------+-----------------------+ | 5 1368 78 | 16 2 *(79)(Y=8) | 38 *(79)(X=1) 4 | | 126 68 4 | 56 3 *(79)(W=8) | 258 *(79)(Z=2) 15 | =(b=7)= | 9 138 278 | 15 4 78 | 2358 127 6 | +-----------------------+-----------------------+-----------------------+ | 4 168 3 | 2 568 56 | 9 16 7 | | 26 9 5 | 3 7 1 | 4 26 8 | | 126 7 28 | 9 68 4 | 25 3 15 | +-----------------------+-----------------------+-----------------------+

UR+4X/3SL (97)[r4c6|r5c8]

There is a solved x-wing pattern in 9 and a strong link b=7 in row 5.

first part of the UR+4X/3SL :
(W=8)r5c6 (Z=2)r5c8 (or vice versa due to symmetry), both W,Z can be eliminated

or alternatively (much easier) a hidden pair: (79)r5c6 and (79)r5c8.

second part of the UR+4X3SL eliminations that are equivalent to a UR+2B/1SL:
(X=8)r4c6 (Y=1)r4c6 (or vice versa due to symmetry), b=7 can be removed in r4c6 and in r4c8

by **daj95376** » Mon Jan 16, 2012 7:26 pm

Code: Select all: UR+4X/3SL: abX-----abZ a | b| a | abY-----abW After X-Wing: abX-----abZ | a | |a b|a | a | abY-----abW After Hidden Pair: abX-----ab | a | |a b|a | a | abY-----ab UR Type 4: (UR+2X/1SL) abX ab | |a | abY ab

by **surbier** » Mon Jan 16, 2012 8:34 pm

I agree,

The UR+2X/1SL matches the constellation after the HP even better than the UR+2B/1SL.

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