Difference between revisions of "Sixth row template problem"
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This would involve placing a stone on the 6th row of a sufficiently wide board, and showing how to always connect to the bottom. (Note this does not necessarily identify the minimal template needed.) | This would involve placing a stone on the 6th row of a sufficiently wide board, and showing how to always connect to the bottom. (Note this does not necessarily identify the minimal template needed.) | ||
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====6th row template==== | ====6th row template==== | ||
<hex> | <hex> | ||
Revision as of 10:24, 13 January 2009
As of January 2009 the following problem, initially stated by javerberg and wccanard in the LG forum, is still open:
Is there a one stone sixth row template that uses no stones higher than the sixth row?
More generally, it is still unknown whether one stone edge templates that use no cell higher than the initial stone) can be found for all heights. Such templates have been found for sizes up to 5 but none above. Answering with "No" to the former question answers the latter.
Contents
Description
Is there a number m such that the game on the board of width m designed as follows, with Blue's turn to play, is won by Red ?
Generalisation
The general problem of knowing if there is n such that there is no one stone edge template on the n^th row<math>n^th</math> is also referred to as the n-th row template problem.
Possible paths to answer
By "hand"...
...answering "Yes"
This would involve placing a stone on the 6th row of a sufficiently wide board, and showing how to always connect to the bottom. (Note this does not necessarily identify the minimal template needed.)
See defending against intrusions in template VI1 for complete proof.
6th row template
...answering "No"
This would involve showing how to connect (in the diagram above) the Blue stones to the right (plus Blue stones on the far right edge) to Blue stones on the left (plus Blue stones on the far left edge), no matter how wide the board is.
Computer Aided demonstration ...
... answering "Yes"
Such a proof would use the computer to find the template and it's carrier. Afterwards it should be easy to manually check that every Blue intrusion does not prevent Red from connecting to bottom.
... answering "No"
TODO
See Also
External link
- The thread were the names were associated.
