Life In 19x19

Click Here To Show Diagram Code

[go]$$c Go problem.
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$$ | . . . . . . j i h g . . . . . . . . . |
$$ | . . . . . . k X . f . . . . . . . . . |
$$ | . . . , . . l . . e . . . . . , . . . |
$$ | . . . . . . m . . d . . . . . . . . . |
$$ | . . . . . . n a b c . . . . . . . . . |
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$$ ---------------------------------------[/go]

I don't stand a chance of working this out but an interesting question from the point of view of mathematical simplicity is the following:

In the above go board a white piece is placed at "a" and its "value" determined. It is then placed at "b" and a further value determined. The test white stone follows a closed path ending at "n". All the values for each position are then added up.

The question is. Do all the values for white add up to zero??????

This isn't trivial.. those with a mathematical background will see this as a test for a conservative field. If the sum is zero then the maths becomes "easier"... actually still stink hard but all the same... Ant takers???

What values?

see:

viewtopic.php?f=15&t=3768

BobC wrote:The question is. Do all the values for white add up to zero??????

Extremely unlikely. Your rectangle is a very arbitrary set of points. And there are other larger rectangles which have obviously more points.

those with a mathematical background will see this as a test for a conservative field.

My mathematical background tells me that this is completely unrelated to conservative fields. The test for conservative fields is a line integral in a vectorfield. This is a scalar field. And your definition of a boundary volume is completely arbitrary.

BobC wrote:The question is. Do all the values for white add up to zero??????

Every point there has positive value for white (having a stone is better than not), so I don't see how the values could possibly sum to zero.

There are positions where white would like to take a white stone off the board, but they occur in shortage of liberty situations, not on boards this empty.

There are many more positions where white would like to remove a white-black pair of stones (i.e., undo a bad forcing move).

(Apologies if the above is obvious to you (you don't have a rank listed) and I completely miss your point. Maybe you mean something I don't expect by "value"?)

BobC wrote:

$$c Go problem.
$$ ---------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . j i h g . . . . . . . . . |
$$ | . . . . . . k X . f . . . . . . . . . |
$$ | . . . , . . l . . e . . . . . , . . . |
$$ | . . . . . . m . . d . . . . . . . . . |
$$ | . . . . . . n a b c . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
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$$ ---------------------------------------

Click Here To Show Diagram Code

[go]$$c Go problem.
$$ ---------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . j i h g . . . . . . . . . |
$$ | . . . . . . k X . f . . . . . . . . . |
$$ | . . . , . . l . . e . . . . . , . . . |
$$ | . . . . . . m . . d . . . . . . . . . |
$$ | . . . . . . n a b c . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
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$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ ---------------------------------------[/go]

I don't stand a chance of working this out but an interesting question from the point of view of mathematical simplicity is the following:

In the above go board a white piece is placed at "a" and its "value" determined. It is then placed at "b" and a further value determined. The test white stone follows a closed path ending at "n". All the values for each position are then added up.

The question is. Do all the values for white add up to zero??????

This isn't trivial.. those with a mathematical background will see this as a test for a conservative field. If the sum is zero then the maths becomes "easier"... actually still stink hard but all the same... Ant takers???

No, the values do not add to zero. Not by a long shot.

Edit: Your post has some ambiguity. Are we to evaluate the position, or the play of the White stone?

My answer was based on thinking that you meant to evaluate the play of the stone, but perhaps you meant to evaluate the position with one Black stone and one White stone. In that case the sum of the values is still not zero, but it is closer to zero. If they were zero, then either in all cases the values of the White play and Black play are equal, or some of the White values are better than the value of the Black stone. In that case, Black had a better play. But, in fact, the Black play is considered better than most of the White plays, and roughly equal to some of them. I. e., the sum of the positional values should be positive, from the standpoint of Black.

thanks for that Bill... I was just catching up on your work on thermography with Elwyn Berlekamp.. interesting stuff... Also tripped over Fearnly doing this sort of again down at the dept Engineering Oxford... very small world..

Interesting at Berkeley theres a plasma group led by Birdsall who does/did particle in cell codes.. a bit like I was alluding to above... There are a lot of similarities/ideas common between go and plasma modelling.. Action at a distance ... etc..

Having said all that... I'm not sure pic codes actually gave us any answers in plasma physics.. so don't expect a "solution" for go

Li Kao wrote:

BobC wrote:The question is. Do all the values for white add up to zero??????

Extremely unlikely. Your rectangle is a very arbitrary set of points. And there are other larger rectangles which have obviously more points.

those with a mathematical background will see this as a test for a conservative field.

My mathematical background tells me that this is completely unrelated to conservative fields. The test for conservative fields is a line integral in a vectorfield. This is a scalar field. And your definition of a boundary volume is completely arbitrary.

here goes..
its a closed loop - it can be arbitrary..... thats the whole point of a conservative field.

Not wise to swap mathematical backgrounds on this

daniel_the_smith wrote:

BobC wrote:The question is. Do all the values for white add up to zero??????

Every point there has positive value for white (having a stone is better than not), so I don't see how the values could possibly sum to zero.

There are positions where white would like to take a white stone off the board, but they occur in shortage of liberty situations, not on boards this empty.

There are many more positions where white would like to remove a white-black pair of stones (i.e., undo a bad forcing move).

(Apologies if the above is obvious to you (you don't have a rank listed) and I completely miss your point. Maybe you mean something I don't expect by "value"?)

Good points. This MIGHT be akin to measuring an electric field using a test "charge". You assume that the test charge has no effect on the field. Likewise one might assume that the white stone has no intrinsic value....

Ah, Bill's edit helped me understand what you are probably asking. I thought you were asking about the value of a white move. If you are instead asking about the value of the position as a whole, it makes more sense.

I think it's safe to say that for any given closed loop like the one you mark, it's almost certain that the values of the resulting positions will not sum to zero. An equivalent question is, does every point on the loop have a corresponding point with an exactly opposite value? For non symmetrical positions I'm pretty confident the answer is nearly always "no".

EDIT: if this is an attempt to figure out how much "influence" is worth or something like that, it's not likely to succeed. The value of influence is almost entirely determined by the specific flaws and distances present in a position. It might superficially look like a field effect, but I don't think it is very similar.

daniel_the_smith wrote:if this is an attempt to figure out how much "influence" is worth or something like that, it's not likely to succeed.

For early stones in the corner, it is easier. Early stones on an empty side or in the center have so much empty space around them that evaluating influence in terms of territory is still too hard even for me.

no ..it wasn't an attempt to calculate the worth on influence... although that might be a by product.

The point was to cast a go board grid as a type of particle in cell code problem. as per:

http://en.wikipedia.org/wiki/Particle-in-cell

which is an extension of the lattice Boltzman approach

http://en.wikipedia.org/wiki/Lattice_Boltzmann_methods

Before you do this you need help. If "stones" were viewed as a source of "potential" (from which we calculate a field (hopefully)). Then your life is easier if the field generated is conservative.

Although I think it is "doable" and from a maths/physics point of view, instinctively it seems more robust as an idea than "temperature... I really have no idea what the value would be. You "might" be able to identify urgent moves in terms of local potential.... the lower the potential the more urgent the move.... on the other hand you might end up with a mathematical mess.....

There are some interesting observations... Surrounded territory.. where no life is possible... would imply no potential for an opposing colour of stone. This has a parallel with there being no field inside a hollow metal sphere....

Just need funding for a PhD student... and three years to think about it... Any takers?