Difference between revisions of "Sixth row template problem"

From HexWiki
Jump to: navigation, search
(changed wording in introduction to match new title)
m (... answering "Yes": fixed typo)
(15 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
As of January 2009 the following problem, initially stated by javerberg and wccanard in the LG forum, is still [[open problems|open]]:
 
As of January 2009 the following problem, initially stated by javerberg and wccanard in the LG forum, is still [[open problems|open]]:
  
Is there any one stone sixth row [[template]] ?
+
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 can be found for every 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.
+
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 ==
 
== Description ==
Line 17: Line 17:
  
 
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.
 +
 +
 +
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.
 +
 +
Here is an example for <i>n</i> = 7
 +
<hex> R8 C11
 +
1:HHHHHVHHHHH
 +
2:+++++V+++++
 +
3:____HV_____
 +
</hex>
 +
 +
 +
For now it seems like there is no solution for above example.
  
 
== Possible paths to answer ==
 
== Possible paths to answer ==
Line 23: Line 36:
 
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.)   
  
Here is a start.  Just from [[edge template IV1a]] and [[edge template IV1b]], Blue's first move must be one of the following:
+
See [[defending against intrusions in template VI1]] for complete proof.
 +
====6th row template====
 
<hex>
 
<hex>
R7 C19 Q0
+
R7 C14 Q0
1:BBBBBBBBBRBBBBBBBBB
+
1:BBBBBBBBBRBBBBB
Rj2
+
Sa2 Sb2 Sc2 Sd2 Se2 Sf2 Sg2 Sh2 Rj2 Sl2 Sm2 Sn2
Si3 Sj3
+
Sa3 Sb3 Sc3 Sd3 Se3 Sf3 Sm3 Sn3
Si4
+
Sa4 Sb4 Sc4 Sd4 Sn4
Sg5 Sh5 Si5 Sj5
+
Sa5 Sb5
Sf6 Sg6 Si6 Sj6
+
Sa6
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!
 
  
 
====...answering "No" ====
 
====...answering "No" ====
Line 41: Line 53:
 
=== 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 56: 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.