The New Sudoku Players' Forum

[daj95376]

_

It appears that we have a case of missing the forest for the trees. When solving Dan's puzzles, no one quantifies the basic steps leading up to the first advanced step. Take the following for example:

Code: Select all

 +-----------------------+
 | . . . | 2 . . | . 8 1 |
 | 6 . . | . . . | . . . |
 | . . 4 | 1 . . | 6 . . |
 |-------+-------+-------|
 | . 3 . | . . 7 | . . 9 |
 | 5 . . | . 4 . | . . 8 |
 | 1 . . | 3 . . | . 2 . |
 |-------+-------+-------|
 | . . 6 | . . 8 | 7 . . |
 | . . . | . . . | . . 5 |
 | 9 2 . | . . 5 | . . . |
 +-----------------------+

 start   Givens:
         Naked  Single                   =  1    r1c9,r3c4,r6c1
         Naked  Single                   =  2    r1c4,r6c8,r9c2
         Naked  Single                   =  3    r4c2,r6c4
         Naked  Single                   =  4    r3c3,r5c5
         Naked  Single                   =  5    r5c1,r8c9,r9c6
         Naked  Single                   =  6    r2c1,r3c7,r7c3
         Naked  Single                   =  7    r4c6,r7c7
         Naked  Single                   =  8    r1c8,r5c9,r7c6
         Naked  Single                   =  9    r4c9,r9c1
 end     Givens:

         Hidden Single                   =  5    r7c2

     b5  Naked  Pair                     <> 69   r4c45,r5c6,r6c5
         Hidden Single                   =  6    r4c8
         Hidden Single                   =  6    r9c9

     b5  Naked  Pair                     <> 58   r4c5

   c4    Hidden Pair                     =  58   r24c4
   c6    Hidden Pair                     =  12   r58c6

   c5b8  Locked Candidate 1              <> 3    r123c5
   c7b6  Locked Candidate 1              <> 5    r12c7
   c5b2  Locked Candidate 1              <> 7    r89c5

 +-----------------------------------------------------------------------------------------+
 |  37       79       3579     |  2        5679     3469     |  349      8        1        |
 |  6        1789     1235789  |  58       5789     349      |  2349     34579    2347     |
 |  2378     789      4        |  1        5789     39       |  6        3579     237      |
 |-----------------------------+-----------------------------+-----------------------------|
 |  248      3        28       |  58       12       7        |  145      6        9        |
 |  5        679      279      |  69       4        12       |  13       137      8        |
 |  1        46789    789      |  3        58       69       |  45       2        47       |
 |-----------------------------+-----------------------------+-----------------------------|
 |  34       5        6        |  49       1239     8        |  7        1349     234      |
 |  3478     1478     1378     |  4679     12369    12       |  123489   1349     5        |
 |  9        2        1378     |  47       13       5        |  1348     134      6        |
 +-----------------------------------------------------------------------------------------+
 # 123 eliminations remain


So why are there those who feel that "stte" and "lclste" are necessary?

If you think about it, every "stte" consists of numerous Naked/Hidden Single steps. Are we now to start counting the trailing Naked/Hidden Single steps and bragging about having fewer than those in some other solution's trailing steps? IIRC, solutions on the Eureka! site would count the Singles and list every occurrence of Subsets. Are we ready to do the same here?

If a step cracks the puzzle, then including ";Basics" should be the most additional information that anyone needs to add.

[DonM]

_

It appears that we have a case of missing the forest for the trees. When solving Dan's puzzles, no one quantifies the basic steps leading up to the first advanced step....

So why are there those who feel that "stte" and "lclste" are necessary?

If you think about it, every "stte" consists of numerous Naked/Hidden Single steps. Are we now to start counting the trailing Naked/Hidden Single steps and bragging about having fewer than those in some other solution's trailing steps? IIRC, solutions on the Eureka! site would count the Singles and list every occurrence of Subsets. Are we ready to do the same here?

If a step cracks the puzzle, then including ";Basics" should be the most additional information that anyone needs to add.

A case of missing the forest for the trees? How about creating a misleading smokescreen? As long as everyone starts at the same place, it really doesn't matter where the puzzle solving starts. In these threads, the practice is often to start after basics. On the Eureka forum and sometimes here, it is/was after the Simple Sudoku Technique Set (SSTS). So, you and I know, no one is suggesting listing opening basic moves.

While it may serve the purpose of embellishing the argument by making 'singles' sound like something complicated, most of us find them to be inconsequential. And as for the Eureka site, some solvers listed the number of singles, others simply said 'several singles'. I don't recall anyone finding the practice onerous. IMO, overall, counting singles at the end of a solution is less important than showing basic moves.

In my mind a step that cracks a puzzle with only singles left does so with more elegance and efficiency than one that leaves a number of basic moves. Further and more importantly, I suspect that any manual solver here who happens to come up with a step that still leaves basic methods knows that he/she has been outdone by someone who comes up with a single step that ends in singles only.

Since you mention 'bragging rights', perhaps the competitive aspect of solving is not so present here though I hope it is. However, I find that the drive to come up with the most efficient solution makes one a better solver. I suspect that that perspective is different depending on whether one is solving manually or using a computer solver.

[SteveG48]

Steve G48 wrote : -4r1c7 = (379)r1c127 - (379)r1c356 = (5)r1c3 - (5=6)r1c5 - (6)r1c6 = (6-9)r6c6 = (9)r5c4 -(9=4)r7c4 -(4=3)r7c1 -(3)r1c1|r7c9 = (3)r23c9 - (3)r1c7 => r1c7 = 4.

The only elimination I can see here is r1c7 <> 3

Leren

You don't see 3 eliminations in the steps -(379)r1c356 and -(3)r1c7?

[Marty R.]

Repeating myself: In my book if a solver provides a main move that then falls apart with singles, that is simply better than another main move, similar in complexity, that doesn't fall apart until various locked cells/locked sets are solved. And if just one or two of those basic moves are left, then that solution is better than if 3 or 4 basic moves are left and so on.

A question not just for you, Don, but for others as well: after finding your "one-stepper" that leaves an lclste or numerous locked sets, do you then look for another solution that leaves fewer sets or singles?

I suspect that any manual solver here who happens to come up with a step that still leaves basic methods knows that he/she has been outdone by someone who comes up with a single step that ends in singles only.

I am a manual solver who probably is the least advanced player who has been regularly doing Dan's puzzles ever since he started posting them. Maybe it's due to my ignorance, but when I find the one-step chain that solves the puzzle, I don't care at all how much basics are left. I'm just thrilled to post a one-stepper and feel like I'm on a par with the other guys, even though I know that I put in more "blood, sweat and tears" before finding my solution.

[David P Bird]

It's very easy to start chasing one's own tail when it comes to finding 'best' solutions.

Many of the trivial eliminations we make at the start are insignificant and will result automatically when assignments are made. So, if the objective to find the minimum number of steps needed using methods restricted to certain classes, we should focus not only on how a puzzle is to be completed after a deciding step, but also on the preliminary steps needed to pave the way.

In the Eureka days I often pruned out the earlier steps that particular key steps didn't depend on, but those puzzles typically needed 3 or 4 key steps. Often they could be made in various orders, and there was usually one that was better than the others.

In those days (working manually) I optimised all steps higher than singles, which took quite a while. It would be possible (Leren?) to code a program to do this which would be quite challenging.

The approach to adopt can only be reached by consensus, and I applaud the progress you've made over this regarding where the problem starts, although it bewilders me why Xwings and Swordfish aren't allowed.

DPB

[ArkieTech]

A question not just for you, Don, but for others as well: after finding your "one-stepper" that leaves an lclste or numerous locked sets, do you then look for another solution that leaves fewer sets or singles?

I consider a puzzle cracked when there are no further advanced (non lclste) moves required. I don't count the basic moves nor try to reduce the number of them.

[DonM]

A question not just for you, Don, but for others as well: after finding your "one-stepper" that leaves an lclste or numerous locked sets, do you then look for another solution that leaves fewer sets or singles?

I do, but as DPB infers above, one can drive oneself a little crazy carrying it too far. After all, there may not be a better solution. Still, it is a personal decision. I'm just saying that basic moves that follow the last main move should be mentioned. That can be as simple as saying, 'a couple of linebox locked candidates and a naked pair'.

[Leren]

Steve G48 wrote: You don't see 3 eliminations in the steps -(379)r1c356 and -(3)r1c7?

OK, I get it now - congratulations - the only stte move posted.

Leren

[JC Van Hay]

Steve G48 wrote: You don't see 3 eliminations in the steps -(379)r1c356 and -(3)r1c7?

OK, I get it now - congratulations - the only stte move posted.

Leren

Good news

Even though SteveG48's style and way of solving are looking very familiar to me , here is an alternative presentation of his solution :
[(3579=6)r1c1235 OR 4r1c7=4r1c6]-6r1c6=(6-9)r6c6=9r5c4-(9=4)r7c4-(4=3)r7c1-[3r7c9=3r23c9 AND (3=79)r1c12] :=> -39r1c7(=4)

[daj95376]

_

An alternate interpretation to a network solution from my solver. Results the same as SteveG48.

Code: Select all

 +-----------------------------------------------------------------------------------------+
 |  37       79       3579     |  2        5679     3469     |  349      8        1        |
 |  6        1789     1235789  |  58       5789     349      |  2349     34579    2347     |
 |  2378     789      4        |  1        5789     39       |  6        3579     237      |
 |-----------------------------+-----------------------------+-----------------------------|
 |  248      3        28       |  58       12       7        |  145      6        9        |
 |  5        679      279      |  69       4        12       |  13       137      8        |
 |  1        46789    789      |  3        58       69       |  45       2        47       |
 |-----------------------------+-----------------------------+-----------------------------|
 |  34       5        6        |  49       1239     8        |  7        1349     234      |
 |  3478     1478     1378     |  4679     12369    12       |  123489   1349     5        |
 |  9        2        1378     |  47       13       5        |  1348     134      6        |
 +-----------------------------------------------------------------------------------------+
 # 123 eliminations remain

 A discontinuous loop:   (4)r1c7=...=(4)r1c7

 (4)r1c7=(4-6)r1c6=(6-9)r6c6=r5c4-(9=4)r7c4-(4=3)r7c1-(3  )r7c9=r23c9    -(3)r1c7 \
                                                     -(3=7)r1c1-(7=9)r1c2-(9)r1c7  =(4)r1c7


[keith]

Yes,

"Singles" should properly include X-wings, swordfish, higher fish, and coloring. Single-digit elimination techniques.

If you want to get technical, a pair is not a "single".

I have always thought that "STTE" is an inappropriate term for "only basics remain".

All basics are not singles. All singles are not basics.

Keith