So where do we go from here?
I'm assuming a rather different idea about the role of ladder breakers.
Edit: OK, I think I see your point, mine isn't a ladder. So we can't really get away from the first line?
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| Author: | SmoothOper [ Wed Dec 11, 2013 3:04 pm ] |
| Post subject: | Fewest stones needed to break all ladders |
I wonder if any one has solved the fewest stones to break all ladders problem. This is probably pretty vague, maybe just any ladder from the third or forth line going inwards. I am thinking sort of analogously to the queens problem, placing 8 queens such that they weren't mutually attacking. |
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| Author: | mitsun [ Wed Dec 11, 2013 3:06 pm ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
Useful factoid: san-ren-sei (three star points) breaks all ladders. |
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| Author: | HermanHiddema [ Wed Dec 11, 2013 3:10 pm ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
mitsun wrote: Useful factoid: san-ren-sei (three star points) breaks all ladders. Yes. The handicap points are exactly at the maximum distance from each other so that they break all ladders. If the space between them were one wider, a ladder could pass though the gap. Therefore, 8 stones on the corner and side star points break all ladders going up from anywhere. Note that the stones need to be on the fourth line for this to work. On the third line: |
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| Author: | SmoothOper [ Wed Dec 11, 2013 3:17 pm ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
mitsun wrote: Useful factoid: san-ren-sei (three star points) breaks all ladders. breaks all ladders going one direction. |
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| Author: | DrStraw [ Wed Dec 11, 2013 4:52 pm ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
So an 8 stone handicap breaks all ladders? If W plays tengen first does that enable them all? |
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| Author: | Solomon [ Wed Dec 11, 2013 9:24 pm ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
| Author: | Bill Spight [ Wed Dec 11, 2013 10:14 pm ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
| Author: | ez4u [ Fri Dec 13, 2013 3:07 am ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
Probably we can all agree the first diagram is a ladder. If so, we can't use stones on the third line.... If the next diagram is also a 'ladder' we can't use the 3-2 point. However, it seems to my eye that filling the second row with alternate stones as below will break all ladders. This requires four rows of eight stones or 32 stones in total. |
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| Author: | Bill Spight [ Fri Dec 13, 2013 6:48 am ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
| Author: | ez4u [ Fri Dec 13, 2013 7:10 am ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
So where do we go from here? I'm assuming a rather different idea about the role of ladder breakers. Edit: OK, I think I see your point, mine isn't a ladder. So we can't really get away from the first line? |
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| Author: | SmoothOper [ Fri Dec 13, 2013 7:25 am ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
I was thinking about posing a subtly different question. What placement of stones gives maximal ladder breakage for X stones. The easy case I think would be for one stone, at tengen. Two stones? Three stones etc. |
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| Author: | hyperpape [ Fri Dec 13, 2013 7:37 am ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
We could consider some kind of minimal ladders, for instance a bend, block, bend pattern. I'm not saying that's the "true" meaning of ladder or anything like that. But it might be an interesting question, and perhaps we can avoid having to put as many stones on the board as Bill had to 
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| Author: | SmoothOper [ Fri Dec 13, 2013 8:36 am ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
For more stones I think directionality and edge cases might come into play, as well as overlap. |
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| Author: | SmoothOper [ Fri Dec 13, 2013 10:36 am ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
Here is a configuration for three with minimum overlap, I would have to count to figure out the optimum centering. |
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| Author: | SmoothOper [ Fri Dec 13, 2013 10:49 am ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
There that is better. Every position is covered for ladders in some direction, a majority of the board is covered in two directions. I suspect tengen would work its way back into the picture at five stones and triple directional coverage. |
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| Author: | ez4u [ Fri Dec 13, 2013 5:33 pm ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
@SmoothOper: Please define 'ladder'. 
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| Author: | SmoothOper [ Fri Dec 13, 2013 7:49 pm ] |
| Post subject: | Re: Fewest stones needed to break all ladders |
I guess 7-4's would do pretty well, but not any more efficiently than star points. |
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