The New Sudoku Players' Forum

[ronk]

The XY-Chain in my first reply accounts for r8c3<>7.

Yes, but this discontinuous nice loop is more in keeping with your 2-digit POM analysis.

r8c3 -7- r8c4 =7= r7c4 =5= r7c1 -5- r1c1 =5= r1c2 =7= r1c3 -7- r8c3 ==> r8c3<>7

In case you remember little of NL notation , that's ... (7)r8c4 = (7-5)r7c4 = (5)r7c3 - (5)r1c1 = (5-7)r1c2 = (7)r1c3 ==> r8c3<>7

That leaves this Kraken Row [r8] on <4> -- which is sufficient for r7c7,r9c6<>4.

Code: Select all

(4-7)r8c2 = r8c4 - r7c4 = (7-4)r7c7
 ||
(4  )r8c3 - r5c3 = r5c7 - (  4)r7c7
 ||
(4  )r8c9               - (  4)r7c7


That's as good as any ... and it only uses two digits. (I added the strong inference symbols for the kraken row.)

My solver think that the fact that r7c6<>1 is a one-step solution, but I don't know if it is?

r7c6=4 is a "singles backdoor", so it's obviously a one-step solution. However, in keeping with the title of this thread, by what deductive technique can the exclusion r7c6<>1 be shown.

[daj95376]

My solver think that the fact that r7c6<>1 is a one-step solution, but I don't know if it is?

If you can eliminate the companion candidate in any of these bivalue cells, then you have a single-stepper solution.
(I did not include all of the bilocation eliminations that also produce a single stepper solution.)

Code: Select all

r1c1    =  5     simple SSTS backdoor
r1c2    =  7     simple SSTS backdoor
r1c3    =  1     simple SSTS backdoor
r2c2    =  4     simple SSTS backdoor
r2c4    =  6     simple SSTS backdoor
r3c1    =  6     simple SSTS backdoor
r3c5    =  4     simple SSTS backdoor
r4c4    =  4     simple SSTS backdoor
r4c9    =  3     simple SSTS backdoor
r5c3    =  6     simple SSTS backdoor
r5c7    =  4     simple SSTS backdoor
r6c1    =  4     simple SSTS backdoor
r6c5    =  6     simple SSTS backdoor
r7c1    =  1     simple SSTS backdoor
r7c2    =  6     simple SSTS backdoor
r7c4    =  5     simple SSTS backdoor
r7c6    =  4     simple SSTS backdoor
r7c7    =  7     simple SSTS backdoor
r7c8    =  2     simple SSTS backdoor
r8c4    =  7     simple SSTS backdoor
r9c6    =  1     simple SSTS backdoor


Your cell, r7c6, is in the list. Now, all you need to know is how your solver found the elimination. My solver used a SIN and lots of memory.

Code: Select all

1r7c6  4r7c7  6r5c7  6r9c3  1r8c3  4r8c2  7r8c4  =>  contradiction [r7]-7


RonK: In a pinch, I can still read NL notation. However, I do appreciate that you saved me the trouble by converting it to Eureka notation. My solver looks for the simplest (to me) solution at every step. An XY-Chain takes preference over a generic chain. However, in analysis mode, it did list this generic chain that uses two digits. Look familiar?

Code: Select all

(X)s    = (X-Y)a    = (Y)b - (Y)c = (Y-X)d    = (X)t     Inverted W-Wing (courtesy of Norm)

(7)r1c3 = (7-5)r1c2 = r1c1 - r7c1 = (5-7)r7c4 = (7)r8c4  =>  r8c3<>7


[SudoQ]

Your cell, r7c6, is in the list. Now, all you need to know is how your solver found the elimination.

There are probably multiple paths. My looks like this:

Code: Select all

1r7c6  4r7c7  5r7c1  4r9c6  7r9c7  4r5c3  6r9c2 => [c3]-6

[pjb]

When I put this puzzle into my solver, it solves the puzzle using the multi-digit method in two steps. It shows how pattern analysis is a useful and overlooked technique.

I presume that you are referring to: ...

Why do you think the POM should be limited to two (digit) layers in this case? Might not the use of additional layers result in shorter chains or simpler patterns?

BTW did your look for the chains or networks that produce your eliminations?

...8239462.3.59178.891.75..792.15.8.83.972.51.153.8297..8.3...93...968..9..28....

1. The POM method first found that all patterns for 5 passed through r1c2 or r7c4. Then patterns of 7 that include these can be eliminated, and the remaining patterns of 7 do not include r8c3, so r8c3 <> 7.

2. Then it found that all patterns for 7 passed through r7c7 or r8c2. Then patterns of 4 that include these can be eliminated, and the remaining patterns of 4 all include r7c6, so r7c6 = 4. It then solved with singles.

3. If I used chains instead, it didn't need networks, just simple chains:
a. r1c2 -5- r1c1 -1- r7c1 -5- r7c4 -7- r7c7 =7= r9c7 => r9c2<> 7
b. r2c2 -6- r2c4 -4- r3c5 -6- r3c1 -4- r6c1 -6- r5c3 =6= r9c3 =7= r9c7 -7- r7c7 -4- r7c6 -1- r9c6 => r9c2 <> 4
c. r1c3 -7- r1c2 -5- r1c1 -1- r7c1 -5- r9c2 -6- r2c2 -4- r2c4 -6- r4c4 =6= r4c7 -6- r5c7 -4- r7c7 =4= r7c6 -4- r9c6 => r9c3 <> 1
d. r1c3 -7- r1c2 -5- r9c2 -6- r2c2 -4- r2c4 -6- r3c5 -4- r3c1 -6- r6c1 -4- r5c3 -6- r5c7 -4- r7c7 -7- r7c4 -5- r7c1 => r1c1, r8c3 <> 1 It then solved with singles.
Currently, I don't look for the shortest chain.

With regard to POM analysis, I have played around with patterns passing through 3 cells, but tooo slow in javascript

pjb

[ronk]

3. If I used chains instead, it didn't need networks, just simple chains:
a. r1c2 -5- r1c1 -1- r7c1 -5- r7c4 -7- r7c7 =7= r9c7 => r9c2<> 7
b. r2c2 -6- r2c4 -4- r3c5 -6- r3c1 -4- r6c1 -6- r5c3 =6= r9c3 =7= r9c7 -7- r7c7 -4- r7c6 -1- r9c6 => r9c2 <> 4
c. r1c3 -7- r1c2 -5- r1c1 -1- r7c1 -5- r9c2 -6- r2c2 -4- r2c4 -6- r4c4 =6= r4c7 -6- r5c7 -4- r7c7 =4= r7c6 -4- r9c6 => r9c3 <> 1
d. r1c3 -7- r1c2 -5- r9c2 -6- r2c2 -4- r2c4 -6- r3c5 -4- r3c1 -6- r6c1 -4- r5c3 -6- r5c7 -4- r7c7 -7- r7c4 -5- r7c1 => r1c1, r8c3 <> 1 It then solved with singles.
Currently, I don't look for the shortest chain.

I can understand not looking for "the" shortest chain in any given pencilmarks, but I don't understand not looking for the shortest chain for the exclusion found. For example, for the above ...

a. -7- r1c2 -5- r1c1 -1- r7c1 -5- r7c4 -7- r7c7 =7= r9c7 => r9c2<> 7 (same chain, 5 strong links)
b. -4- r2c2 -6- r3c1 =6= r6c1 -6- r5c3 =6= r9c3 =7= r9c7 -7- r7c7 -4- r7c6 =4= r9c6 => r9c2 <> 4 (6 SLs instead of 10)
c. r7c1 =1= r7c6 =4= r7c7 -4- r5c7 =4= r5c3 =6= r9c3 => r9c3 <> 1 (4 SLs instead of 11)
d. -1- r1c3 -7- r9c3 =7= r9c7 =3= r4c7 =6= r4c4 -6- r2c4 =6= r2c2 -6- r9c2 -5- r7c1 -1- => r1c1, r8c3 <> 1 (7 SLs instead of 13)

... and even these may not be the shortest.

[daj95376]

I can understand not looking for "the" shortest chain in any given pencilmarks, but I don't understand not looking for the shortest chain for the exclusion found. For example, for the above ...

a. -7- r1c2 -5- r1c1 -1- r7c1 -5- r7c4 -7- r7c7 =7= r9c7 => r9c2<> 7 (same chain, 5 strong links)
b. -4- r2c2 -6- r3c1 =6= r6c1 -6- r5c3 =6= r9c3 =7= r9c7 -7- r7c7 -4- r7c6 =4= r9c6 => r9c2 <> 4 (6 SLs instead of 10)
c. r7c1 =1= r7c6 =4= r7c7 -4- r5c7 =4= r5c3 =6= r9c3 => r9c3 <> 1 (4 SLs instead of 11)
d. -1- r1c3 -7- r9c3 =7= r9c7 =3= r4c7 =6= r4c4 -6- r2c4 =6= r2c2 -6- r9c2 -5- r7c1 -1- => r1c1, r8c3 <> 1 (7 SLs instead of 13)

... and even these may not be the shortest.

Possibly shortest (and most efficient) for eliminations in list above. Note: at each step, I only performed the elimination(s) in the list. That's why r9c3<>4 appears twice.

Code: Select all

4-SIS: (7=5)r1c2 - r1c1 = r7c1 - (5=7)r7c4 - r7c7 = (7)r9c7  =>  r9c2<>7
4-SIS: (4)r8c23 = r8c9 - (4=7)r7c7 - r9c7 = r9c3 - (17=4)r18c3  =>  r8c9,r9c23<>4
4-SIS: (3)r4c7 = (3-7)r9c7 = (7-6)r9c3 = r5c3 - r5c7 = (6)r4c7 - loop  =>  r9c3<>1; r4c7,r9c37<>4
6-SIS: (1)r7c1 = (1-4)r7c6 = r7c7 - (4=6)r5c7 - r5c3 = r9c3 - (6=5)r9c2 - (5=1)r7c1  =>  r1c1,r7c6,r8c3<>1


[ronk]

Possibly shortest (and most efficient) for eliminations in list above. Note: at each step, I only performed the elimination(s) in the list. That's why r9c3<>4 appears twice.

Code: Select all

4-SIS: (7=5)r1c2 - r1c1 = r7c1 - (5=7)r7c4 - r7c7 = (7)r9c7  =>  r9c2<>7
4-SIS: (4)r8c23 = r8c9 - (4=7)r7c7 - r9c7 = r9c3 - (17=4)r18c3  =>  r8c9,r9c23<>4
4-SIS: (3)r4c7 = (3-7)r9c7 = (7-6)r9c3 = r5c3 - r5c7 = (6)r4c7 - loop  =>  r9c3<>1; r4c7,r9c37<>4
6-SIS: (1)r7c1 = (1-4)r7c6 = r7c7 - (4=6)r5c7 - r5c3 = r9c3 - (6=5)r9c2 - (5=1)r7c1  =>  r1c1,r7c6,r8c3<>1


Good show, especially the continuous loop!

On the second one, however, I disagree with your SIS count. An ALS consisting of two cells should be counted as two strong inferences. It's only notational convention that makes it appear to be one.

On the fourth one, I disagree with your conclusion. The derived strong inference is (1)r7c1 = (1)r7c1, directly implying the inclusion r7c1=1. Your listed exclusions, are directly due to this inclusion, but indirectly due to the chain itself.

[pjb]

3. If I used chains instead, it didn't need networks, just simple chains:
a. r1c2 -5- r1c1 -1- r7c1 -5- r7c4 -7- r7c7 =7= r9c7 => r9c2<> 7
b. r2c2 -6- r2c4 -4- r3c5 -6- r3c1 -4- r6c1 -6- r5c3 =6= r9c3 =7= r9c7 -7- r7c7 -4- r7c6 -1- r9c6 => r9c2 <> 4
c. r1c3 -7- r1c2 -5- r1c1 -1- r7c1 -5- r9c2 -6- r2c2 -4- r2c4 -6- r4c4 =6= r4c7 -6- r5c7 -4- r7c7 =4= r7c6 -4- r9c6 => r9c3 <> 1
d. r1c3 -7- r1c2 -5- r9c2 -6- r2c2 -4- r2c4 -6- r3c5 -4- r3c1 -6- r6c1 -4- r5c3 -6- r5c7 -4- r7c7 -7- r7c4 -5- r7c1 => r1c1, r8c3 <> 1 It then solved with singles.
Currently, I don't look for the shortest chain.

I can understand not looking for "the" shortest chain in any given pencilmarks, but I don't understand not looking for the shortest chain for the exclusion found. For example, for the above ...

a. -7- r1c2 -5- r1c1 -1- r7c1 -5- r7c4 -7- r7c7 =7= r9c7 => r9c2<> 7 (same chain, 5 strong links)
b. -4- r2c2 -6- r3c1 =6= r6c1 -6- r5c3 =6= r9c3 =7= r9c7 -7- r7c7 -4- r7c6 =4= r9c6 => r9c2 <> 4 (6 SLs instead of 10)
c. r7c1 =1= r7c6 =4= r7c7 -4- r5c7 =4= r5c3 =6= r9c3 => r9c3 <> 1 (4 SLs instead of 11)
d. -1- r1c3 -7- r9c3 =7= r9c7 =3= r4c7 =6= r4c4 -6- r2c4 =6= r2c2 -6- r9c2 -5- r7c1 -1- => r1c1, r8c3 <> 1 (7 SLs instead of 13)

... and even these may not be the shortest.

Thanks for all the feedback. I can see I have heaps of 'optimizing' to go. So little time, so much to do!
pjb

[daj95376]

Possibly shortest (and most efficient) for eliminations in list above. Note: at each step, I only performed the elimination(s) in the list. That's why r9c3<>4 appears twice.

Code: Select all

4-SIS: (7=5)r1c2 - r1c1 = r7c1 - (5=7)r7c4 - r7c7 = (7)r9c7  =>  r9c2<>7
4-SIS: (4)r8c23 = r8c9 - (4=7)r7c7 - r9c7 = r9c3 - (17=4)r18c3  =>  r8c9,r9c23<>4
4-SIS: (3)r4c7 = (3-7)r9c7 = (7-6)r9c3 = r5c3 - r5c7 = (6)r4c7 - loop  =>  r9c3<>1; r4c7,r9c37<>4
6-SIS: (1)r7c1 = (1-4)r7c6 = r7c7 - (4=6)r5c7 - r5c3 = r9c3 - (6=5)r9c2 - (5=1)r7c1  =>  r1c1,r7c6,r8c3<>1


Good show, especially the continuous loop!

Thanks!

On the second one, however, I disagree with your SIS count. An ALS consisting of two cells should be counted as two strong inferences. It's only notational convention that makes it appear to be one.

As far as I'm concerned, most ALS terms are shorthand for a network, and assigning a SIS count to the ALS is purely academic and left to personal interpretation. In addition, there are a lot of "loose ends" about ALS notation. Written as (17=4)r18c3, it should imply that eliminations can only occur in the unit/house containing the two cells; i.e., [c3]. This would actually preclude an elimination in r9c2. My solver actually builds the ANS() as (17)r18c3=(4)r8c3. This allows all peers of r8c3 to be considered for <4>. Thus, the elimination in r9c2. I use the latter ANS() format because I've seen many others derive eliminations using the same logic. However, I (manually) altered my solver's output to use the more compact notation to prevent others from telling me that my solver's notation could be shortened.

On the fourth one, I disagree with your conclusion. The derived strong inference is (1)r7c1 = (1)r7c1, directly implying the inclusion r7c1=1. Your listed exclusions, are directly due to this inclusion, but indirectly due to the chain itself.

I was "slapped down" once by aran (I believe) because I asserted that an AIC was just an abbreviated forcing chain with two streams. He asserted that the logic of an AIC is based on the elimination of any candidate that forces both ends of the AIC to be false. Since no one countered his assertion, I updated my interpretation of an AIC. If you review my elimination list, you'll see that evey candidate listed does just that ... forces both ends of the AIC to be false!

[ronk]

My solver actually builds the ANS() as (17)r18c3=(4)r8c3. This allows all peers of r8c3 to be considered for <4>. Thus, the elimination in r9c2. I use the latter ANS() format because I've seen many others derive eliminations using the same logic. However, I (manually) altered my solver's output to use the more compact notation to prevent others from telling me that my solver's notation could be shortened.

What's an "ANS"?

[daj95376]

What's an "ANS"?

Almost Naked Subset: A Naked Subset with one or more additional candidates of the same value.

Code: Select all

 +-----------------------------------------------+
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .  17   .   |   .   .   .   |   .  17   .   |   Naked Subset
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |---------------+---------------+---------------|
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .  17   .   |   .   .   .   |   .  178  .   |   Almost Naked Subset
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |---------------+---------------+---------------|
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .  17   .   |   .   .   .   |   .   78  .   |   ALS but not ANS
 |   .   .   .   |   .   .   .   |   .   .   .   |
 +-----------------------------------------------+


When I use ANS:(8)r5c8=(17)r5c28, I'm saying that the absence of <8> leaves a Naked Pair in those two cells. Precise.

Whereas ALS:(8=17)r8c28 could imply that a Naked Pair is left in those two cells ... or something like r8c8=7 and r8c2=1 could occur. I'm not interested in dealing with all the permutations of interpreting something like ALS:(1=2345)r9c2468.

[ronk]

When I use ANS:(8)r5c8=(17)r5c28, I'm saying that the absence of <8> leaves a Naked Pair in those two cells. Precise.

Whereas ALS:(8=17)r8c28 could imply that a Naked Pair is left in those two cells ... or something like r8c8=7 and r8c2=1 could occur.

For me, that's a distinction without a difference. How is that distinction useful for you?

I'm not interested in dealing with all the permutations of interpreting something like ALS:(1=2345)r9c2468.

Sound like the distinction is not extensible then, at least not from a practical POV.

[RW]

As a manual solver, my approach would be quite different. The great amount of 46 bivalues draws my attention and I color the cells:

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 +--------------------------------------------------------------+
 |  15    57    17    |  8     2     3     |  9     4     6     |
 |  2    a46    3     | b46    5     9     |  1     7     8     |
 | b46    8     9     |  1    a46    7     |  5     23    23    |
 |--------------------+--------------------+--------------------|
 |  7     9     2     | a46    1     5     |  346   8     34    |
 |  8     3    b46    |  9     7     2     | a46    5     1     |
 | a46    1     5     |  3    b46    8     |  2     9     7     |
 |--------------------+--------------------+--------------------|
 |  15    26    8     |  57    3     14    |  47    26    9     |
 |  3     2457  147   |  57    9     6     |  8     12    245   |
 |  9     4567  1467  |  2     8     14    |  347   136   345   |
 +--------------------------------------------------------------+

r7c7=4 => r9c7=7
       => r9c6=4
       => r7c6=1 => r7c1=5 => r9c2=6

In other words, if r7c7=4, then cells colored 'a' can see both 4 and 6 => r7c7<>4

RW

[pjb]

3. If I used chains instead, it didn't need networks, just simple chains:
a. r1c2 -5- r1c1 -1- r7c1 -5- r7c4 -7- r7c7 =7= r9c7 => r9c2<> 7
b. r2c2 -6- r2c4 -4- r3c5 -6- r3c1 -4- r6c1 -6- r5c3 =6= r9c3 =7= r9c7 -7- r7c7 -4- r7c6 -1- r9c6 => r9c2 <> 4
c. r1c3 -7- r1c2 -5- r1c1 -1- r7c1 -5- r9c2 -6- r2c2 -4- r2c4 -6- r4c4 =6= r4c7 -6- r5c7 -4- r7c7 =4= r7c6 -4- r9c6 => r9c3 <> 1
d. r1c3 -7- r1c2 -5- r9c2 -6- r2c2 -4- r2c4 -6- r3c5 -4- r3c1 -6- r6c1 -4- r5c3 -6- r5c7 -4- r7c7 -7- r7c4 -5- r7c1 => r1c1, r8c3 <> 1 It then solved with singles.
Currently, I don't look for the shortest chain.

I can understand not looking for "the" shortest chain in any given pencilmarks, but I don't understand not looking for the shortest chain for the exclusion found. For example, for the above ...

a. -7- r1c2 -5- r1c1 -1- r7c1 -5- r7c4 -7- r7c7 =7= r9c7 => r9c2<> 7 (same chain, 5 strong links)
b. -4- r2c2 -6- r3c1 =6= r6c1 -6- r5c3 =6= r9c3 =7= r9c7 -7- r7c7 -4- r7c6 =4= r9c6 => r9c2 <> 4 (6 SLs instead of 10)
c. r7c1 =1= r7c6 =4= r7c7 -4- r5c7 =4= r5c3 =6= r9c3 => r9c3 <> 1 (4 SLs instead of 11)
d. -1- r1c3 -7- r9c3 =7= r9c7 =3= r4c7 =6= r4c4 -6- r2c4 =6= r2c2 -6- r9c2 -5- r7c1 -1- => r1c1, r8c3 <> 1 (7 SLs instead of 13)

... and even these may not be the shortest.

Being inspired and challenged by your responses, I modified my solver to look for progressively longer chains starting with length 4, and bingo got a completely better result:
a. r1c3 -1- r1c1 -5- r7c1 =5=r 7c4 -5- r8c4 => r8c3 <> 7
b. r7c7 -7- r7c4 -5- r7c1 -1- r8c3 => r8c9 <> 4 then line-box r9c23 <> 4
c. r7c1 -5- r7c4 -7- r8c4 -5- r8c9 -2- r8c8 => r8c3 <> 1 then solves with singles

pjb

[daj95376]

When I use ANS:(8)r5c8=(17)r5c28, I'm saying that the absence of <8> leaves a Naked Pair in those two cells. Precise.

Whereas ALS:(8=17)r8c28 could imply that a Naked Pair is left in those two cells ... or something like r8c8=7 and r8c2=1 could occur.

For me, that's a distinction without a difference. How is that distinction useful for you?

I build an SI table and then search for AICs that can be constructed from its entries. ANS:(8)r5c8=(17)r5c28 constitutes a single entry in the SI table. ALS:(8=17)r8c28, on the other hand, would constitute two entries: (8)r8c8=(7)r8c8 and (8)r8c8=(1)r8c2. In the case of example ALS:(1=2345)r9c2468, there would be an entry for (1=2), (1=3), (1=4), and (1=5).

In my tables, I don't have a way to maintain the original cells of the ALS. I only maintain the cells for each side of the SI. Thus, I would only maintain the cells containing <1> and <5> in the SI entry for (1=5) in ALS:(1=2345)r9c2468.

Since I solve for all chains present, within a maximal chain length, the search time grows exponentially as the size of the SI table grows. As it is, the execution time is questionably acceptable.