The flow of Komodo eval

[whereagles]

lol... I feel like sourcing-out that vector field and get a quantum field theory description of it. Probably akin to some non-renormalizable ghost-matter interacting theory.

Too bad I forgot how to do it.. I quit physics for optimization

[Laskos]

lol... I feel like sourcing-out that vector field and get a quantum field theory description of it. Probably akin to some non-renormalizable ghost-matter interacting theory.

Too bad I forgot how to do it.. I quit physics for optimization

Come on, I tried to present as intuitively and graphically as it gets. The graphical reading is just "if one starts with Komodo eval of +1.2 pawn advantage at move 30, on average, the game will go to +1.6 pawn advantage to move 40 against an equal opponent to Komodo". Vector field is just a fancy name, follow the arrows for the same purpose, to track the progress of Komodo move by move for different starting evals.

[Laskos]

If the score is above 70 cp on the 10th move, you are trapped in the vector field of doom.

Well, it's more likely that the game will end in Win than in Draw, but just probabilistically.

[Laskos]

I've never been interested in maths, I'm weak at it.

I suspect, from your previous work, that you do possess most high-school math, and it's what I most often use myself.

So first you have a high-school function for the boundary Win-Draw:

eval[x] = 5*0.75/(5 + 0.75*sqrt(30) - 0.75*sqrt(x))

x is the variable move number. This formula gives the eval at move x for the boundary Win-Draw.

Then you asked for that f(x):
f[x] = 0.5 + 0.5 * theta(y[x] - eval[x])

Theta is the step function:
theta[x] is 1 for x>0, 0 otherwise, for our purposes.
You don't need to know anything on fancy generalized functions. If you don't want to use "if" and such, then if your software supports abs or sign operations, theta is very easy to write as:

theta[x] = (1 + abs[x]/x)/2

If your software doesn't know even abs operation, I can concoct an ad hoc "engineering" representation of theta for our purposes, in elementary functions.

Say theta[x] = 1 / (1+exp[-100*x])
It has the accuracy of an eval difference of 0.01, sufficient for us. An overflow might happen, but you can easily manage.

[whereagles]

(...) I tried to present as intuitively and graphically as it gets. (...)

I know.. There is nothing too deep to it. It just reminded me of the good old days

[nimh]

I've never been interested in maths, I'm weak at it.

I suspect, from your previous work, that you do possess most high-school math, and it's what I most often use myself.

So first you have a high-school function for the boundary Win-Draw:

eval[x] = 5*0.75/(5 + 0.75*sqrt(30) - 0.75*sqrt(x))

x is the variable move number. This formula gives the eval at move x for the boundary Win-Draw.

Then you asked for that f(x):
f[x] = 0.5 + 0.5 * theta(y[x] - eval[x])

Theta is the step function:
theta[x] is 1 for x>0, 0 otherwise, for our purposes.
You don't need to know anything on fancy generalized functions. If you don't want to use "if" and such, then if your software supports abs or sign operations, theta is very easy to write as:

theta[x] = (1 + abs[x]/x)/2

If your software doesn't know even abs operation, I can concoct an ad hoc "engineering" representation of theta for our purposes, in elementary functions.

Say theta[x] = 1 / (1+exp[-100*x])
It has the accuracy of an eval difference of 0.01, sufficient for us. An overflow might happen, but you can easily manage.

I think it's the best and easiest if I use the original logistic function you gave me a while ago.

Code: Select all

((1+(EXP(G26/1,2)-EXP(-G26/1,2))/(EXP(G26/1,2)+EXP(-G26/1,2)))/2)

And modify it so that the value changes according to the varying move number. Let's assume the boundary of the vector field of doom is equal to 75.00% - 0.66 cps at move 27 (the mean of the average length of the games in my dataset) and 1.70 cps at move 80, plus there is a smooth exponential or a power function. What would the new formula be?

I'd like to try it out to see if it yields a more reliable strength-accuracy relationship than before.

[mjlef]

Kai,

You do some beautiful work.

Mark

[gordonr]

Kai,

You do some beautiful work.

Mark

+1

[Mathboy]

Thanks, this is wonderful work.

Do you think the vector field looks the same at very different time controls?

John

[fern]

And now, Kai, in plain English, what are the conclusion of that statistical exercize?

Fern