Re: [Bug-gnubg] Even, odd and half plies (WAS: New Contact Net Error Rates)

[Nis]

--On Friday, February 28, 2003 08:15 -0800 David Montgomery 
<address@hidden> wrote:

> From: "Nis" <address@hidden>
> > 1) Average between 12 rolls at 1-ply and 24 times the static 0-ply eval.
> > 2) Average between 12 rolls at 2-ply and 24 at 1-ply.
> >
> > Perhaps we can implement both - using 1) for reduced 1-ply, and 2) for
> > higher plies.
>
> Yes.  But to me, these are the same thing.

The rest of your message clearly disagrees with this.

> Well, I guess I'm still not sure I understand. Let me try again.
> For 2-ply, of course, there are 1296 terminal nodes (looking at just
> the selected lines of play). So my guess for the best way to
> do 2) is:
>
> 2A) Do a weighted average of 432 2-ply nodes selected by striated
>     sampling with the result of a full 1-ply evaluation, using
>     empirically derived weights.
>
> It appears to me you are suggesting
>
> 2B) Evaluate 12 of the 1-ply continuations at 2-ply, and 24 at 1-ply.
>     Add these up and dived by 36.

Or to harmonize wordings:

2B) Do a weighted average of the 432 2-ply nodes resulting from 12 of
   the initial rolls and the 1-ply evaluations of the 24 other rolls (each
   with weight 36)

The idea is that each 2-ply leaf node will have the same weight in the 
average - either as part of a 1-ply or 2-ply evaluation. In your scenario, 
the 432 selected leaf nodes are treated as special - possibly introducing 
noise.

As an example, take the situation where I will win, unless I roll 2-1 and 
opponent rolls 6-6:

In this case, if all our evaluations are exact, we will have

1-ply = 2-ply = 1294/1296

while 2A) gives either 1292/1296 or 1295/1296 depending on whether the 
winning sequence is included in the 432 evaluated on 2-ply.

(Here I am using a weight of 1/3 for simplicity)

2B) doesn't have this problem.

> 2A) is taking the current implementation (as I understand it)
> and creating a weighted average with the 1-ply evaluation.
>
> 2B) requires a different implementation, because you select the
> nodes for deeper evaluation at the first ply, rather than spreading
> the sampling evenly over the leafs.
>
> Both 2A) and 2B) evaluate 432 nodes at the deepest level.  2A)
> evaluates 12 nodes for each of the 36 1-ply rolls, so you get
> 2-ply information for every 1-ply continuation.  2B) evaluates
> 36 nodes for each of 12 1-ply rolls, and 0 nodes for the other
> 24.
>
> In fact, 2A) is exactly the 1-ply lookahead version of
>
> > 1) Average between 12 rolls at 1-ply and 24 times the static 0-ply eval.
>
> averaging 36 of these evaluations.
>
> 2B) is not.  It's a mixture of
>
> Average between 36 rolls at 1-ply and  0 times the static 0-ply eval.
> Average between  0 rolls at 1-ply and 36 times the static 0-ply eval.

All correct - although I prefer "average" rather than "mixture" to describe 
the last calculation :-)

I agree with you that 2A is a lot easier to implement, while keeping a lot 
of the benefits from the approach. But 2B seems to me to be "the right way 
(TM)" - and thus I will keep working for it's implementation - even if that 
means I have to learn to write C :-)

Out of interest: Do we have an agreed way to compare different evaluation 
settings? Something like "ppg in a money session of N games"

--
Nis Jorgensen
Greenpeace
Amsterdam