Gérard TAILLE wrote:
Bill Spight wrote:
Third, when the theory can tell you best play with less calculation, why not use it?
That is the point, the theory give us heuristically the best play but when a choice is difficult and the game very close you cannot avoid reading.
Gérard TAILLE wrote:
Bill Spight wrote:
Actually, the theory gives exact results that enable us to restrict reading. Not in all cases, OC, but very often. The problem you just proposed is a case in point. With one exception, you do not have to read the whole branch of the game tree to find the best play. But with your proposed solution, you do have to read the whole branch to find out if one option wins the game. OC, if it does, so will best play. And best play in your problem will take less reading to find, as a rule.
If the estimations of the two best areas are not close OC theory will give the best move, otherwise the probability to find the best move is near from 50%.
Take again the two areas G1 = {8|5||0} and G2 = {6|0}. The miai values are 6¼ and 6. The problem is that ¼ is already too small.
Now add the quasi ideal environment {5½, 5, 4½, 4, 3½, 3, 2½, 2, 1½, 1, ½}
By choosing G1 your score will be 8, and by choosing G2 your score will be 9.
It is quite logical : the result of the game is a naturel number (no decimals). That means that a very small change in the environment will easily add or substract a full 1 point. As a consequence when the miai values of G1 and G2 are close you hardly can expect to make the best choice with more than 50%.
In that case reading is the only possiblity and my proposal is just to choose the easiest reading!
What do you mean by the quasi ideal environment? Does 5½ = {5½|-5½} or {2¾|-2¾}? My guess is the latter, since otherwise the choice will be in {5½|-5½}, hands down. Also, who plays first? OK, I went back and found that you had originally stipulated
Black to play. Fine.
If 5½ = {5½|-5½} then the result of playing in the environment alone is 5½ - 5 + 4½ - . . . + ½ = 3, and that fits with your claim that starting with G2 gives a result of 9.
But in that case your play is in the environment, not G2. If 5½ = {2¾|-2¾} then the result of playing in the environment is 1½ and the result of starting with G2 gives a result of only 7½. For ease of calculation we can write that environment as {5½|0} + {0|-5} + . . . + {½|0}. Doing so adds 1½ to the mean value of the environment and makes the result of starting with G2 9 points for Black. We find that result by calculating
6 - 0 + 5½ - 5 + 4½ - 4 + 3½ - 3 + 2½ - 2 + 1½ - 1 + ½ = 9
If Black plays in G1 we get {8|5} - 0 + 5½ - .... Note that {8|5} + {0|-3} = 5, so we can write the result as
5 - 0 + 5½ - 5 + 4½ - 4 + 3½ - 0 + 0 - 0 + 0 - 0 = 9½
That does not fit your calculations, so I do not know what you meant.
Edited for correctness.
Edit2: How about the environment {2½|-2½} + {2|-2} + (1½|-1½} + {1|-1} + {½|-½}? I don't know the answer, but let's give it a try.
If Black plays is G2 she gets this result:
6 - 0 + 2½ - 2 + 1½ - 1 + ½ = 7½
{8|5} + (1½|-1½} = 6½ so if Black plays in G1 she gets this result:
6½ - 0 + 2½ - 2 + 1 - ½ = 7½
It does not matter, does it?
_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?— Winona Adkins
My two main guides in life:
My mother and my wife.
Everything with love. Stay safe.