Difference between revisions of "Sixth row template problem"

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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.
 
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:
+
One of the way to prove if there is such an <i>n</i> is to prove if there is such <i>n-1</i> for which an <i>(n-1)</i>-row-template with one defender stone originaly placed next to attacker stone in the same row. Of course if such template exists <i>n</i>-row-template is still not proven to exist.
<hex>
+
 
R7 C19 Q0
+
Here is an example for <i>n</i> = 7
1:BBBBBBBBBRBBBBBBBBB
+
<hex> R8 C11
Rj2
+
1:HHHHHVHHHHH
Si3 Sj3
+
2:+++++V+++++
Si4
+
3:____HV_____
Sg5 Sh5 Si5 Sj5
+
Sf6 Sg6 Si6 Sj6
+
Se7 Sf7 Sg7 Sh7 Si7 Sj7
+
 
</hex>
 
</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>
+
For now it seems like there is no solution for above example.
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:
+
== Possible paths to answer ==
 
+
===By "hand"...===
<hex>
+
====...answering "Yes" ====
R7 C19 Q0
+
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.) 
1:BBBBBBBBBRBBBBBBBBB
+
Rj2
+
Pg5
+
Pf6
+
Pe7
+
Ri4 Bi5 Rh5 Bg7 Rh6 Bh7
+
N:on Rk5 Bj6 Ri6 Bi7 Rl4 Bj5 Rk3
+
</hex>
+
  
 +
See [[defending against intrusions in template VI1]] for complete proof.
 
====6th row template====
 
====6th row template====
 
<hex>
 
<hex>
 
R7 C14 Q0
 
R7 C14 Q0
1:BBBBBBBBBBBRBBB
+
1:BBBBBBBBBRBBBBB
 
Sa2 Sb2 Sc2 Sd2 Se2 Sf2 Sg2 Sh2 Rj2 Sl2 Sm2 Sn2  
 
Sa2 Sb2 Sc2 Sd2 Se2 Sf2 Sg2 Sh2 Rj2 Sl2 Sm2 Sn2  
 
Sa3 Sb3 Sc3 Sd3 Se3 Sf3 Sm3 Sn3
 
Sa3 Sb3 Sc3 Sd3 Se3 Sf3 Sm3 Sn3
Line 70: Line 46:
 
Sa5 Sb5
 
Sa5 Sb5
 
Sa6
 
Sa6
</hex>
 
 
</hex>
 
</hex>
  
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=== Computer Aided demonstration ... ===
 
=== Computer Aided demonstration ... ===
 
==== ... answering "Yes" ====
 
==== ... 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.
+
Such a proof would use the computer to find the template and its [[carrier]]. Afterwards it should be easy to manually check that every Blue intrusion does not prevent Red from connecting to bottom.
  
 
==== ... answering "No" ====
 
==== ... answering "No" ====
Line 93: Line 68:
 
[[category:theory]]
 
[[category:theory]]
 
[[category:templates]]
 
[[category:templates]]
{{stub}}
 

Revision as of 17:27, 23 February 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.

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.


One of the way to prove if there is such an n is to prove if there is such n-1 for which an (n-1)-row-template with one defender stone originaly placed next to attacker stone in the same row. Of course if such template exists n-row-template is still not proven to exist.

Here is an example for n = 7


For now it seems like there is no solution for above example.

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 its 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.