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Sixth row template problem

2009-01-13T09:40:48Z

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As of January 2009 the following problem, initially stated by javerberg and wccanard in the LG forum, is still [[open problems|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 [[Edge templates with one stone|templates]] have been found for sizes up to 5 but none above. Answering with "No" to the former question answers the latter.

== 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 ?

<hex> R7 C11
1:HHHHHVHHHHH
2:_____V_____
</hex>

== 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.)

Here is a start. Just from [[edge template IV1a]] and [[edge template IV1b]], Blue's first move must be one of the following:
<hex>
R7 C19 Q0
1:BBBBBBBBBRBBBBBBBBB
Rj2
Si3 Sj3
Si4
Sg5 Sh5 Si5 Sj5
Sf6 Sg6 Si6 Sj6
Se7 Sf7 Sg7 Sh7 Si7 Sj7
</hex>
Many of these moves will be easy to dismiss. Others will benefit from the [[Parallel ladder]] trick. Of course, symmetry will cut our work in half!

We can dispose of 3 moves on the left (and, using mirror symmetry, the corresponding 3 moves on the right), as follows:

<hex>
R7 C19 Q0
1:BBBBBBBBBRBBBBBBBBB
Rj2
Pg5
Pf6
Pe7
N:on Ri4 Bi5 Rh5 Bg7 Rh6 Bh7
</hex>

At this point, we can use the [[Parallel ladder]] trick as follows:

<hex>
R7 C19 Q0
1:BBBBBBBBBRBBBBBBBBB
Rj2
Pg5
Pf6
Pe7
Ri4 Bi5 Rh5 Bg7 Rh6 Bh7
N:on Rk5 Bj6 Ri6 Bi7 Rl4 Bj5 Rk3
</hex>

Now, let's deal with the remaining intrusions!:

=====One remaining intrusion on the first row=====
<hex>
R7 C19 Q0
1:BBBBBBBBBRBBBBBBBBB
Rj2
Bf7
</hex>

=====The other remaining intrusion on the first row=====
<hex>
R7 C19 Q0
1:BBBBBBBBBRBBBBBBBBB
Rj2
Bg7
</hex>

=====The remaining intrusion on the second row=====
<hex>
R7 C19 Q0
1:BBBBBBBBBRBBBBBBBBB
Rj2
Bg6
</hex>

=====The remaining intrusion on the third row=====
<hex>
R7 C19 Q0
1:BBBBBBBBBRBBBBBBBBB
Rj2
Bh5
</hex>

=====The remaining intrusion on the fourth row=====
<hex>
R7 C19 Q0
1:BBBBBBBBBRBBBBBBBBB
Rj2
Bi4
</hex>

=====The remaining intrusion on the fifth row=====
<hex>
R7 C19 Q0
1:BBBBBBBBBRBBBBBBBBB
Rj2
Bi3
</hex>

====6th row template====
<hex>
R7 C14 Q0
1:BBBBBBBBBRBBBBB
Sa2 Sb2 Sc2 Sd2 Se2 Sf2 Sg2 Sh2 Rj2 Sl2 Sm2 Sn2
Sa3 Sb3 Sc3 Sd3 Se3 Sf3 Sm3 Sn3
Sa4 Sb4 Sc4 Sd4 Sn4
Sa5 Sb5
Sa6
</hex>

====...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 ==
* [[Theory]]
* [[User:Wccanard|Wccanard]]

== External link ==

* The [http://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=339 thread] were the names were associated.

[[category:theory]]
[[category:templates]]

Sixth row template problem

2009-01-13T09:39:15Z

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