subsets: how to identify the target and solve?

Post the puzzle or solving technique that's causing you trouble and someone will help

Postby tarek » Thu Sep 18, 2008 12:06 pm

what seems to be crtitical for the 10th placement is this move
Code: Select all
47 is a hidden  double in c9
[r1c9]<>2<>8 [r7c9]<>2<>6


These are the possible intersections prior to that move
Code: Select all
1 forms an intersection in b1/r1
[r1c56]<>1
4 forms an intersection in b8/r9
[r9c89]<>4
6 forms an intersection in b1/r1
[r1c79]<>6
7 forms an intersection in b5/r6
[r6c23]<>7
7 forms an intersection in b9/r7
[r7c23]<>7
5 forms an intersection in b1/r3
[r3c4]<>5


so would
Code: Select all
4 forms an intersection in b8/r9
[r9c89]<>4
47 is a hidden  double in c9
[r1c9]<>2<>8 [r7c9]<>2<>6
be the shortest path to that 10th placement?

tarek
Last edited by tarek on Thu Sep 18, 2008 8:13 am, edited 1 time in total.
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Postby Glyn » Thu Sep 18, 2008 12:12 pm

Thanks Danny. BTW Last time I checked I was a male.:) If you're not familiar with Welsh names it can be confusing.
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Postby daj95376 » Thu Sep 18, 2008 3:42 pm

Glyn wrote:If you're not familiar with Welsh names ...

Not familiar with Welsh names:!::!::!: My last name is Jones.:D:D:D

My mom named me after the song Oh Danny Boy, which was very popular when I was born. If it wasn't a Welsh song, then it was in the neighborhood.
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Postby Riddick51PB » Thu Sep 18, 2008 7:08 pm

A naked triple exists in Box 1. The definition of naked triple is thus:

Link: It is formed by 3 cells that have candidates for only 3 digits (2 or 3 digits) and are collocated in the same house.


from Glyn's post here, it appears a naked triple exists in these 3 cells:
r3,c1=25 (2 digits)
r3,c2=58 (2 digits)
r3,c3=258 (3 digits)



next, define house as Box 1
Code: Select all
.---------------------.---------------------.---------------------.
| 126    3      1268  | 9      1247   128   | 2467   5      24678 |
| 7      9      4     | 158    12     1258  | 236    236    268   |
| 25     58     258   | 4578   3      6     | 2479   249    1     |
:---------------------+---------------------+---------------------:
| 12     17     1279  | 6      8      3     | 249    1249   5     |
| 1256   4      3     | 15     129    125   | 8      7      269   |
| 8      1567   125679| 157    1279   4     | 2369   12369  269   |
:---------------------+---------------------+---------------------:
| 9      1678   1678  | 3      5      18    | 2467   246    2467  |
| 4      57     57    | 2      6      9     | 1      8      3     |
| 3      2      168   | 148    14     7     | 5      469    469   |
'---------------------'---------------------'---------------------'

how then, do i apply? answer: strip 2,5, 8 from all cells in the defined house, which is Box 1, except for the 3 cells in r3,c123.

question: is my move correct here?





which gives the resulting Box 1:
Code: Select all
.---------------------.---------------------.---------------------.
| 16     3      16    | 9      1247   128   | 2467   5      24678 |
| 7      9      4     | 158    12     1258  | 236    236    268   |
| 25     58     258   | 4578   3      6     | 2479   249    1     |
:---------------------+---------------------+---------------------:
| 12     17     1279  | 6      8      3     | 249    1249   5     |
| 1256   4      3     | 15     129    125   | 8      7      269   |
| 8      1567   125679| 157    1279   4     | 2369   12369  269   |
:---------------------+---------------------+---------------------:
| 9      1678   1678  | 3      5      18    | 2467   246    2467  |
| 4      57     57    | 2      6      9     | 1      8      3     |
| 3      2      168   | 148    14     7     | 5      469    469   |
'---------------------'---------------------'---------------------'


the puzzle exhibits a naked triple in c3. defined thusly:
168 in r179,c3.

question: how can i use 168 if i stripped the 8 in the previous naked triple application?
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Postby Pat » Fri Sep 19, 2008 3:49 am

daj95376 wrote:
Code: Select all
57  Naked  Pair   [r8c23]  => [r7c23]<>7
247 Hidden Triple [r7c789] => [r9c89]<>247
269 Naked  Triple [r569c9] => [r127c9]<>269
    Naked  Single          => [r2c9]=8      (10th placement)

Both paths use the same number of Subsets and arrive at alternate assignments


not quite --
    the {2,4,7} "hidden" trio in r7
    does not create exclusions in b9 --

    it becomes a "naked" trio in b9
    and only then can we use it to exclude in that box
      (i.e. altogether this uses 4 subsets, not 3)
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Postby daj95376 » Fri Sep 19, 2008 9:48 am

Pat wrote:not quite --
    the {2,4,7} "hidden" trio in r7
    does not create exclusions in b9 --

    it becomes a "naked" trio in b9
    and only then can we use it to exclude in that box
      (i.e. altogether this uses 4 subsets, not 3)

Hello Pat, long time no split hairs together.

You are correct that 247 doesn't form a Hidden Triple in [b9]. If I had called it a Locked Hidden Triple, which Ruud liked to use, then the eliminations in [b9] would have (probably) qualified.

Bottom Line: a Naked/Hidden Pair/Triple should be allowed, in one operation, to perform eliminations in all units containing it. After all, that's true for Naked/Hidden Singles!!!
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Postby tarek » Fri Sep 19, 2008 10:05 am

This reminds me of the naked subset posts in the superior variants ....

The subset should eliminate in targets that could SEE all of its cells.

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Postby ronk » Fri Sep 19, 2008 10:30 am

daj95376, is there a reason for not using locked candidates ... (4)r7\b9:?:
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Postby daj95376 » Fri Sep 19, 2008 10:47 am

ronk wrote:daj95376, is there a reason for not using locked candidates ... (4)r7\b9:?:

There is a (small) group of us who read the head post to indicate that Subsets were preferred. Probably because that's the focus of the discussion where the puzzle was presented in the Mensa book.

Failing that, there is the or part of the specifications.

Personally, I think using Locked Candidates (as Glyn did) makes more sense when only one candidate value is actually being eliminated. I have a feeling you agree.
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Postby ronk » Fri Sep 19, 2008 3:31 pm

daj95376 wrote:
ronk wrote:daj95376, is there a reason for not using locked candidates ... (4)r7\b9:?:

There is a (small) group of us who read the head post to indicate that Subsets were preferred. Probably because that's the focus of the discussion where the puzzle was presented in the Mensa book.

Ah so, I should have read a few more of the posts, particularly the opener.

Personally, I think using Locked Candidates (as Glyn did) makes more sense when only one candidate value is actually being eliminated. I have a feeling you agree.

Given that the game is to use only subsets and singles, and if I put my amateur mathematician's hat on, I have to agree with Pat.

For a hidden N-tuple: If N candidates of a unit are entirely contained within N cells, all other candidates of those N cells may be eliminated.

In fish-like baseSet\coverSet terms for this instance ... (247)r7\r7c789 ==> r7c789<>135689
This expression has no direct effect on other cells of box 9. A similar argument exists for naked N-tuples.

Failing that, there is the or part of the specifications.

I have no idea what you mean by that.
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Postby daj95376 » Fri Sep 19, 2008 5:02 pm

ronk wrote:
Failing that, there is the or part of the specifications.

I have no idea what you mean by that.

head post wrote:or, if you prefer, just solve the darn thing using your preferred course of action.
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Postby daj95376 » Fri Sep 19, 2008 5:31 pm

ronk wrote:Given that the game is to use only subsets and singles, and if I put my amateur mathematician's hat on, I have to agree with Pat.

For a hidden N-tuple: If N candidates of a unit are entirely contained within N cells, all other candidates of those N cells may be eliminated.

In fish-like baseSet\coverSet terms for this instance ... (247)r7\r7c789 ==> r7c789<>135689
This expression has no direct effect on other cells of box 9. A similar argument exists for naked N-tuples.

First off, I'm sure you and Pat are correct in how Naked/Hidden N-tuples are handled by the Sudoku community. I don't even have a problem accepting it ... up to the point where I can get an ornery discussion going.

Code: Select all
 Random Easy Puzzle Created by Simple Sudoku
 *-----------------------------------------------------------------------------*
 | 359     3578    3579    | 1349    6       349     | 134578  4578    2       |
 | 2369    4       23679   | 139     8       5       | 137     7       137     |
 | 35      1       35      | 2       7       34      | 3458    6       9       |
 |-------------------------+-------------------------+-------------------------|
 | 7       35      1345    | 13458   134     3468    | 2       9       3568    |
 | 3459    6       3459    | 345789  34      234789  | 3578    1       3578    |
 | 359     2       8       | 13579   13      3679    | 3567    57      4       |
 |-------------------------+-------------------------+-------------------------|
 | 8       9       3467    | 347     5       1       | 467     2       67      |
 | 245     57      2457    | 6       9       478     | 14578   3       1578    |
 | 1       357     34567   | 3478    2       3478    | 456789  4578    5678    |
 *-----------------------------------------------------------------------------*

There is a Naked Single in [r2c8]=7. How do you think Simple Sudoku handles it?

a) Performs 3x Naked Single operations; one each for [b3], [r2], and [c8]. -or-
b) Performs 1x Naked Single operation and eliminate 7 in all three units concurrently.

After the next Naked Single [r6c8]=5, Simple Sudoku finds a Hidden Single in [r1c2]=8. How do you think Simple Sudoku handles it?

a) Performs [r1c2]<>357 in [b1] and then perform a separate Naked Single operation for [r1c78]<>8 in [r1]. -or-
b) Performs [r1c2]<>357 and [r1c78]<>8 concurrently.

In both cases, multiple units are affected by one operation.

Why can't the Naked Pair in [r3c13]=35 affect [b1] and [r3] in one operation?

In the 247 Hidden Triple that I listed previously, I first found the Hidden Triple, and then I performed eliminations in all units containing the Hidden Triple. This is the same logic used for Naked/Hidden Singles. (Yes, I forgot to list the eliminations in the cells containing the Hidden Triple, but they were of little consequence.)

It's my belief that the Sudoku community dropped the ball when they stopped using multiple unit eliminations past Naked/Hidden Singles.
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Postby Pat » Sun Sep 21, 2008 4:17 am

daj95376 wrote:[ in a different example than previously discussed ]

a Naked Single r2c8=7
How do you think Simple Sudoku handles it?
    a) Performs 3x Naked Single operations; one each for [b3], [r2], and [c8] -or-
    b) Performs 1x Naked Single operation and eliminate 7 in all three units concurrently
a Hidden Single in r1c2=8
How do you think Simple Sudoku handles it?
    a) Performs [r1c2]<>357 in [b1] and then perform a separate Naked Single operation for [r1c78]<>8 in [r1] -or-
    b) Performs [r1c2]<>357 and [r1c78]<>8 concurrently
In both cases, multiple units are affected by one operation.

Why can't the Naked Pair in r3c13=35 affect [b1] and [r3] in one operation?


[ going back to the original example ]

In the 247 Hidden Triple that I listed previously, I first found the Hidden Triple, and then I performed eliminations in all units containing the Hidden Triple. This is the same logic used for Naked/Hidden Singles. (Yes, I forgot to list the eliminations in the cells containing the Hidden Triple, but they were of little consequence.)


It's my belief that the Sudoku community dropped the ball when they stopped using multiple unit eliminations past Naked/Hidden Singles.



this is what happens when i vanish for a couple of days---

daj95376, it seems you're not satisfied with ronk's explanation; thus you force me to spell out some basic definitions, please tell me if we differ on the definitions.


A. "naked single" r2c8=7
    the term "naked single" means that we've identified a cell in which only one possible digit can be placed.

    the "naked single" operation is to place this digit in this cell;
    and as with any placement, this is immediately followed by the Basic Exclusions --
    excluding this digit in all the peer cells
    (your option b).

      (a "naked single" never involves any unit whatsoever,
      so the question of which unit is affected seems to me inappropriate in this context
      )
B. "hidden single" b1 8
      (here you should notice that i dislike with your phrasing -- "a Hidden Single in [r1c2]=8" --
      i find a "hidden single" in a unit and not in a cell
      )
    the term "hidden single" means that, within a specific unit,
    we've identified a digit for which only one possible cell is available.

    the "hidden single" operation is to place this digit in this cell;
    and as with any placement, this is immediately followed by the Basic Exclusions --
    excluding this digit in all the peer cells
    (your option a, except that what you call "a separate Naked Single operation" is the Basic Exclusions relevant to the placement, no reason to even mention "Naked Single" here).

      (the "hidden single" was found in one unit, b1 in this example;
      therefore the relevant peers would be in the other 2 units -- r1 and c2 in this example
      )
C. "naked" duo r3c13=35
    the term "naked" duo means that, within a specific unit,
    we've identified 2 cells for which only 2 possible digits remain.

    thus r3c13 is a "naked" duo in r3
    and also (at the very same time) a "naked" duo in b1.

    when counting "steps" of solving,
    these 2 duos obviously belong in the same "step";
    but when counting the subsets i've used,
    i'd have to count them as 2 different duos.
D. the {2,4,7} trios [ of the original example ]
    you say "I forgot to list the eliminations in the cells containing the Hidden Triple, but they were of little consequence" -- well i strongly disagree -- these exclusions which you so casually dismiss are in fact the exclusions which create the "naked" trio in b9, without them you'd never have the "naked" trio in b9 (and you'd have no excuse for making the exclusions in b9\r7).
      your earlier suggestion -- that the 2 "naked" duos exist in the same 2 cells and at the same time, and might therefore be considered as 1 -- is almost acceptable;
      but here we have a more serious problem -- the 2 trios exist in the same 3 cells but at different times -- going from the "hidden" trio to the "naked" trio is definitely an extra "step" in solving -- so, definitely 2 different trios and also 2 separate "steps" in solving.

daj95376 wrote:I think using Locked Candidates (as Glyn did) makes more sense when only one candidate value is actually being eliminated.

yes, the natural solution-path would be shorter, using box-line interactions.
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Postby daj95376 » Sun Sep 21, 2008 9:40 am

Pat wrote:A. "naked single" r2c8=7
    the term "naked single" means that we've identified a cell in which only one possible digit can be placed.

    the "naked single" operation is to place this digit in this cell;
    and as with any placement, this is immediately followed by the Basic Exclusions --
    excluding this digit in all the peer cells
    (your option b).

      (a "naked single" never involves any unit whatsoever,
      so the question of which unit is affected seems to me inappropriate in this context
      )
B. "hidden single" b1 8
      (here you should notice that i dislike with your phrasing -- "a Hidden Single in [r1c2]=8" --
      i find a "hidden single" in a unit and not in a cell
      )
    the term "hidden single" means that, within a specific unit,
    we've identified a digit for which only one possible cell is available.

    the "hidden single" operation is to place this digit in this cell;
    and as with any placement, this is immediately followed by the Basic Exclusions --
    excluding this digit in all the peer cells
    (your option a, except that what you call "a separate Naked Single operation" is the Basic Exclusions relevant to the placement, no reason to even mention "Naked Single" here).

      (the "hidden single" was found in one unit, b1 in this example;
      therefore the relevant peers would be in the other 2 units -- r1 and c2 in this example
      )

Pat wrote:C. "naked" duo r3c13=35
    the term "naked" duo means that, within a specific unit,
    we've identified 2 cells for which only 2 possible digits remain.

    thus r3c13 is a "naked" duo in r3
    and also (at the very same time) a "naked" duo in b1.

    when counting "steps" of solving,
    these 2 duos obviously belong in the same "step";
    but when counting the subsets i've used,
    i'd have to count them as 2 different duos.
D. the {2,4,7} trios [ of the original example ]
    you say "I forgot to list the eliminations in the cells containing the Hidden Triple, but they were of little consequence" -- well i strongly disagree -- these exclusions which you so casually dismiss are in fact the exclusions which create the "naked" trio in b9, without them you'd never have the "naked" trio in b9 (and you'd have no excuse for making the exclusions in b9\r7).
      your earlier suggestion -- that the 2 "naked" duos exist in the same 2 cells and at the same time, and might therefore be considered as 1 -- is almost acceptable;
      but here we have a more serious problem -- the 2 trios exist in the same 3 cells but at different times -- going from the "hidden" trio to the "naked" trio is definitely an extra "step" in solving -- so, definitely 2 different trios and also 2 separate "steps" in solving.

I find it interesting that you changed the wording when you went from Naked/Hidden Single to the Naked/Hidden Pair/Triple. Lets see how the Naked Pair would read if we used your wording from the Naked Single.

Pat wrote:the term "naked single" means that we've identified a cell in which only one possible digit can be placed.

I claim that I've identified the Naked Pair [r3c13]=35. Since you didn't identify a specific unit containing the Naked Single, I don't specify a unit for the Naked Pair, either.

Pat wrote:the "naked single" operation is to place this digit in this cell;
and as with any placement, this is immediately followed by the Basic Exclusions --
excluding this digit in all the peer cells
(your option b).

Interesting that you didn't define a Naked/Hidden Pair/Triple operation that also ignores the units where the peer eliminations occur.

To me, a Naked Pair is best exemplified by a forcing chain on one of the bivalue cells. When you do this, the Naked Pair interpretation extends past an individual unit -- into the peer cells. There are 13 peer cells that see the Naked Pair in [r3c13]=35. Extending your use of peer eliminations as part of the operation, there are 7 peer cells where eliminations occurred for my Naked Pair in [r3c13]=35.

Qualifier: The forcing chain analogy is limited to the candidates in the Naked Pair, and the peer cells common to the cells containing the Nakd Pair.
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Postby tarek » Sun Sep 21, 2008 3:45 pm

In the superior variants thread, I wrote:Naked subsets are actually not bound by unit constraint.
if 3 cells that see each other have only 3 candidates, then these 3 candidates must be in these 3 cells, therfore the cells that can see all of these 3 cells cannot house these candidates.


then Pat wrote:
the term "naked" subset
refers to cells which are a subset of some set (unit, sector);
your new "naked" beast
it is no longer a subset,
and we should find a new term to reflect this;
also, it would be good to have a few words by way of proof.
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