Daily Telegraph (UK)

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Postby ArchieB » Tue Aug 09, 2005 12:15 am

Thanks, tso. I understand and agree with your explanation - my alternative solition must have had a mistake or 2 in it but, unfortunately, I have not kept the original printout to check.

As for the method you use to arrive at the answer, it's basically trial-and-error, i.e. "9 in r7c9 doesn't work so r7c9=5".
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Postby PaulIQ164 » Tue Aug 09, 2005 2:28 pm

Oh God no! Don't start the "are forcing chains trial and error" argument again! I agree with you, but I think we're in the righteous minority round here.
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Postby tso » Fri Aug 12, 2005 6:14 am

ArchieB wrote:As for the method you use to arrive at the answer, it's basically trial-and-error, i.e. "9 in r7c9 doesn't work so r7c9=5".


You're miss-stating the point of the proof which tells us *nothing* about what r7c9 is. You're describing a proof by contradiction, the contrapositive of forcing chains.

The procedure to finding forcing chains varies with the specifics of the puzzle. Here's what I did in this one:

Code: Select all
 .   .   . | .   .   . | .   .   .
 .   .   . | .   .   . |24  49  29
 .   .   . | .   .   . | .   .   .
 ----------+-----------+----------
 .   .   . | .   .   . | .   .   .
 .   .   . | .   .   . |45  14  15
 .   .   . | .   .   . | .   .   .
 ----------+-----------+----------
89   .   . |68   .   . |56   .  59
 .   .   . | .   .   . | .   .   .
89   .   . |68   .   . |26  19  129


I see that r2c789 is a triplet that could ONLY be 249 or 492 -- exactly two possible states.
Similarly, r5c789 is a triplet that could ONLY be 415 or 541.
They each must have a 4 column 7 or 8. This combines all six cells into a single group that can have exactly two states:

249
415

OR

492
541

Since r7c9 must be either 5 or 9, it eliminates the first state which must have BOTH a 5 and a 9 in the 9th column.

From there I just follow the shortest paths to make the forcing chains given.

If you agree that if each of the four cells in r12c12 had two candidates like this:
Code: Select all
[12][13]
[23][34]

... that we could deduce without T&E that r2c2=4 ...

... then you must also agree that given these cells in rows 3, 5 and 7, columns 7, 8 and 9:

Code: Select all
[12][13][35]
[23][--][--]
[34][--][14]


... that we could deduce without T&E that r7c9=1 ...


... and if you do, you must accept this as well:
Code: Select all
[12][13][35]
[24][--][--]
[45][--][15]

... which is essentially the same set of forcing chains that are in the puzzle.

If you're not convinced, please see this post and see if you can draw the line that distinguishes T&E from other logic.
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Postby 2c » Mon Dec 26, 2005 9:36 pm

New here and just a few questions...
First example one I follow:)

But in the 2nd example shouldn't r3c9 be [34] instead of [35]? (just want to make sure I understand your point)

Can the 3rd example even be valid? Regardless of what r3c7 [12] is, a conflict occurs at r7c9.
That is,
if r3c7=1 then finishing out r3 leads to:
.. r3c8=3, r3c9=5, thus making r7c9=1
but finishing out c7 leads to:
.. r5c7=2, r7c7=3 thus making r7c9=5
so r3c7 cannot be 1 since that would create a conflict at r7c9, correct?

or
if r3c7=2 then finishing out r3 leads to:
.. r3c8=1, r3c9=3, thus making r7c9=5
but finishing out c7 leads to:
.. r5c7=4, r7c7=5 thus making r7c9=1
so r3c7 cannot be 2 since that would create a conflict at r7c9 also?

I assume I'm just missing something but would just like to understand.
Thanks for any feedback.


tso wrote:If you agree that if each of the four cells in r12c12 had two candidates like this:
Code: Select all
[12][13]
[23][34]

... that we could deduce without T&E that r2c2=4 ...

... then you must also agree that given these cells in rows 3, 5 and 7, columns 7, 8 and 9:

Code: Select all
[12][13][35]
[23][--][--]
[34][--][14]


... that we could deduce without T&E that r7c9=1 ...

... and if you do, you must accept this as well:
Code: Select all
[12][13][35]
[24][--][--]
[45][--][15]

... which is essentially the same set of forcing chains that are in the puzzle.

If you're not convinced, please see this post and see if you can draw the line that distinguishes T&E from other logic.
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Re: Daily Telegraph (UK)

Postby neil parks » Mon Jul 09, 2007 5:30 pm

Pappocom wrote:1. DT puzzles expect you to guess or use trial-and-error. Indeed, many of them cannot be �solved� unless you do guess.

2. They have been known to have alternative or multiple solutions (12 Feb 2005, four solutions).

While this may have been true in the past, I think the Telegraph puzzles now conform to Pappocom rules. The Telegraph web site claims that there is only one solution and no guessing is required.

Here's where I got stuck playing the online puzzle for 5 July #771. I'm sure the next move is staring me in the face, but I just can't see it. Will someone kindly hit me with a cluestick? Thanks.

Image

(Note: after clicking on the thumbnail to open the full size picture, you may have to click on the full size picture to display it properly.)
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Postby re'born » Mon Jul 09, 2007 9:21 pm

Your puzzle:
Code: Select all
 *-----------------------------------------------------------*
 | 78    5     4     | 1     9     3     | 278   6     27    |
 | 78    1     3     | 27    6     278   | 4     5     9     |
 | 6     2     9     | 47    58    4578  | 3     78    1     |
 |-------------------+-------------------+-------------------|
 | 4     8     2     | 3     7     6     | 1     9     5     |
 | 1     3     5     | 8     24    9     | 6     247*  247-  |
 | 9     7     6     | 5     24    1     | 28-   248*  3     |
 |-------------------+-------------------+-------------------|
 | 2     6     1     | 9     3     47    | 5     47    8     |
 | 3     4     7     | 26    58    58    | 9     1     26    |
 | 5     9     8     | 2467  1     247   | 27    3     2467  |
 *-----------------------------------------------------------*

The only places for a 2 in column 8 are r5c8 and r6c8. Therefore, r6c7<>2 and r5c9<>2. The puzzle solves with singles after this. Alternatively, there are some useful unique rectangles that can solve the puzzle. For instance, the potential deadly pattern in r38c56<58> implies r3c6<>5,8, which solves the puzzle. Or the potential deadly pattern in r56c58<24> creates a pair on <7,8> with r3c8 and thus r7c8<>7, which again solves the puzzle.
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Postby neil parks » Tue Jul 10, 2007 2:16 pm

Thank you very much for the help.

Before reading your response I was not familiar with the term "deadly pattern" so I did a search on these forums and found a thread with detailed explanations of that and other "unique rectangles".

Very interesting stuff. It seems that I have a lot to learn about Sudoku that I didn't know existed. Thanks again for getting me started in the right direction.
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Postby re'born » Tue Jul 10, 2007 7:13 pm

neil parks wrote:Thank you very much for the help.


Your welcome Neil.

neil parks wrote:Very interesting stuff. It seems that I have a lot to learn about Sudoku that I didn't know existed.

A good reference for techniques is Mike Barker's excellent post.

As Claudia says, "Happy solving.":)
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Postby Pat » Wed Jul 11, 2007 11:55 am

neil parks wrote:
I think the Telegraph puzzles now conform to Pappocom rules.



i doubt it very much

the Pappocom software would probably reject many of the DT puzzles, they're Too Tough

~ Pat
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