a Solution to m_b_metcalf's 11.4 puzzle (my variation)

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a Solution to m_b_metcalf's 11.4 puzzle (my variation)

Postby StrmCkr » Mon Feb 19, 2007 9:17 pm

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Re: a Solution to m_b_metcalf's 11.4 puzzle (my variation)

Postby re'born » Mon Feb 19, 2007 9:39 pm

StrmCkr wrote:
r2c3 = 5
R7c7 =6

and i
ran out of time.... (finsihing this when i get home from work.)


Perhaps when you get home you will fix this. I don't think your technique (which I don't yet understand) is giving you correct deductions. Unless Simple Sudoku and Sudo Cue are mistaken, r7c7 <> 6 in the solution to your puzzle.
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Postby RW » Mon Feb 19, 2007 10:12 pm

StrmCkr wrote:when the 5 or the 6 falls inside there respective rectagle a deadly pattern is formed,

This is not true. There's plenty of possibilities to avoid the deadly pattern even if all 5s and 6s fell inside their respective rectangles. This is the first point where your logic fails.

StrmCkr wrote:therefor the 5 and the 6 must be located at the % squares.

Even if what you mentioned at first was true, this statement would be false. There is other possibilities here too.

I'm sorry, but you'll have to start over on this one, there's no deductions to be found from those two rectangles.

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Postby ronk » Mon Feb 19, 2007 10:26 pm

RW wrote:I'm sorry, but you'll have to start over on this one, there's no deductions to be found from those two rectangles.

If StrmCkr had finished his novel before publishing it, he might have found the error himself.:)
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Postby StrmCkr » Tue Feb 20, 2007 6:46 am

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Postby RW » Tue Feb 20, 2007 7:18 am

Sorry, but the typo didn't change a thing. Your technique is still not valid.

StrmCkr wrote:when the 5 or the 6 falls inside there respective rectagle a deadly pattern is formed

Here I added the 5 and 6 to their respective rectangles, perhaps you could explain, step by step, how you arrived at the conclusion above.
Code: Select all
 *-----------------------------------------------------------------------------*
 | 9       48      12      | 378     234678  23468   | 1467    137     5       |
 | 48      7      +6       | 3589    3489    1       | 49      2       34      |
 | 12      45      3       | 579     24679   24569   | 8       179     1467    |
 |-------------------------+-------------------------+-------------------------|
 | 1467    469     8       | 1579    12679   2569    | 3       157     24      |
 | 137     2       179     | 4       13789   3589    | 157     6       178     |
 | 5       346     147     | 1378    123678  2368    | 24      178     9       |
 |-------------------------+-------------------------+-------------------------|
 | 3678    3689   +5       | 2       1389    389     | 1679    4       13678   |
 | 234678  34689   2479    | 1389    5       3489    | 12679   13789   123678  |
 | 2348    1       249     | 6       3489    7       | 259     3589    238     |
 *-----------------------------------------------------------------------------*

Why is the deadly pattern formed here?

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Postby StrmCkr » Tue Feb 20, 2007 7:59 am

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Postby RW » Tue Feb 20, 2007 9:42 am

StrmCkr wrote:then i wondered if perhaps the rectangles generated errors are prevented by placement of the matching numbers of each rectangle into the opposit one.

Yes, of course the possible deadly patterns are prevented by doing that. However, this gives you a valid deduction only when there is no other possible ways to prevent the deadly patterns. In this puzzle there is plenty of them (r1c2/r4c1/r6c3/r8c2/r9c1=4, r6c3=6, r7c256/r8c2378/r9c58=9...). You cannot ignore these options.

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Postby StrmCkr » Tue Feb 20, 2007 9:54 am

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Postby StrmCkr » Tue Feb 20, 2007 10:35 am

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Postby RW » Tue Feb 20, 2007 11:39 am

StrmCkr wrote:so y woudnt it be possible to identify a rectagle buried in an intial set of numbers and determine a way of deducing which of them is the N corrorner to remove. instead of having to slowly reduce numbers to express the rectangle in entirety as singlualry expressed then solve it.

Sure, it could possible to do that. But then you have to find a binding proof for your deduction, which you haven't showed here yet. You just mention two rectangles (that are very far from becoming deadly patterns) and based on that throw out some solved cells. You have to show that if your solved cells are not true then a deadly pattern (or other contradiction) occurs. Until then your deductions are invalid. And remember that it's not a question of "which corner", it's "which corners" will not have the deadly candidates. For both of your rectangles it's very possible that none of the corners will have any of the candidates you mentioned.

Also, in the other example (which I assume should say r8c3=4 and r8c4=1), even if you did determine that those cells cannot hold the candidates of your two rectangles, what happened to the other possible candidates in those cells?

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