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Tedomari / Discussion
Sub-page of Tedomari
Here is my (24.59.96.237) best understanding of how to get tedomari, in practical terms: If you are to play, and the number of plays at the current temperature is odd, you have tedomari. If it is even, you do not have tedomari. In order to get it, you have to change the parity of the number of plays available. This can mean playing a move that denies your opponent two moves. This means you have "taken" three moves with your play, so an odd number of plays becomes even. In practical terms, usually you deny your opponent one move simply by playing where he would like to play (Your Opponent's Good Move Is Your Good Move). So you need to also deny him a potential follow-up move, and you will grab tedomari.
The other possibility for gaining tedomari is to play a move that gives you two independent follow-ups at the local temperature. I'm sure there are other tricks, anybody want to post some? Charles Matthews When it's down to mutual damage, you can call it 'who blinks first'. If after some plays each advancing into the opponents' territory, Black has a threat worth t and White has a threat worth t', Black hopes t is worth noticeably more than t'. Then if it's White's turn, she can choose between (a) stop Black's threat, and so Black's play stopping White's threat gets tedomari, or (b) play the move setting up the threat but allow Black's threat too - White carrying out the threat is tedomari all right but White has already lost too much. (Discussion following Robert Pauli's example on main page.)
Robert Pauli:
Bill: This may help. It's from my talk at the Computers and Games 2002 conference. The slides are Suppose that we have the following game tree, with / indicating a play by Left (Black) and \ indicating a play by Right (White): G / \ / \ H -7 / \ / \ -2 -4 Suppose that Left (Black) plays first. We get the following backed-up minimax values. -4 / \ / \ -4 -7 / \ / \ -2 -4 For calculating the miai value of G, we add the parameter, t. G / \ / \ H -7+t / \ / \ -2-2t -4 Then we get these backed up thermographic values: max(min(-4,-3-t),-5) / \ / \ min(-4,-3-t) -7+t / \ / \ -2-2t -4 The top value represents the left wall of the thermograph of G. Solving for t in min(-4,-3-t) = -7+t we get t = 2 So the miai value of G is 2. That's kind of sketchy, but I hope you get the idea. (BTW, we are going to have to move some of this discussion. :-))
Robert Pauli: = -5 - / \ | = 2 / \ | = -3 -7 - / \ / \ -2 -4 Just an incident that it turns out to be the same ? Do I need to swallow thermography to get miai values ? (Move it whereever you want, Bill, I'm lost anyway. :-) Bill: Sorry, Robert. I was way too sketchy. You do not need to know thermography to figure out miai values. People, including myself, have been doing that for many, many years. :-) However, the thermograph (TG) gives more information than just the count and miai value. One thing it can indicate is which plays (each line in the TG is associated with at least one line of play) to choose under which circumstances. That is one question in the DGZ problem. To answer this kind of question you have to look at the walls of the TG. You brought up the question of minimax. The walls of the TG can be expressed in minimax terms, and that is what I showed.
Going back to the problem. The line of play to use when White plays first to figure out the count and miai value for the bottom right corner is The right wall of the TG for the corner tells us that, as a rule, the time to switch between those plays is at temperature 1. We could work that out without the TG by considering that playing on the 2-3 allows Black to make a 3 point reverse sente, while playing on the 3-4 is only 2 points worse, on average, than playing on the 2-3. We can gain one point by playing on the 3-4, at the cost of one move. Now, since the remaining play on the board has a miai value of one point, thermography does not give us a clue which play is right. However, as it happens, the play that is correct under most circumstances is still correct for the problem. The thing is, when you are looking for the best play in an area of the board, you do not as a rule look for the best play at temperature 0. That is not realistic in a real game. Usually you want the play that defines the miai value. That's why my eye immediately went to the 2-3 point. As for figuring miai values by taking averages, you realize that that does not work in the following example. = -3.5 - / \ | = 3.5 / \ | = 0 -7 - / \ / \ +4 -4 I admit that, even reading a bit thermography, I still can't decipher your tree decorated with t's:
c = -7 + t c = -3 - t If this is some alternate method to find miai values, great,
but my main problem is simply to set up the (binary) tree. Did a bit homework, reading some DGoZ 2001 articles by Bernhard Herwig about miai values (great job, Bernhard). He explains it via perfect play on N copies of the local position, dividing the total score by N and looking what happens for growing N. (Not a word about my beloved binary trees, however.) With that I get why averaging a single tree doesn't always work: the context isn't rich enough. Your counter-example completed would then be = -4 (not my naive -3.5) / \ = 0 -7 / \ +4 -4 because with, say, 3 copies and Black starting it's better for White (going right and for low) always to answer, leading to -4 -4 -4 instead of +4 -7 -7. Naturally one calls this sente, but that explains nothing. (I know I'm boring you - sorry.) See Method of Multiples. Another thing I noticed was that there's a difference between the miai value of positions and of moves. To make things plain, I would call them average score and average gain - and, by the way, I would have leaves increasing to the right.
Back to a single tree. At Miai values list / Discussion you told Karl that a node's value only is the average of its daughters (values) if it stays between the granddaughters D and E: A / \ / \ B C / \ / \ D > E = Otherwise A is either D or E, whatever (B+C)/2 is nearer to. This seems to suggest that one can do without perfect play on N copies and simply propagate up A = T if T <= D and T >= E = D if there's D and T > D = E if there's E and T < E with T = (B+C) / 2 provided each leaf is in the turn-is-worthless state (or is at temperature zero) and is assigned the score reached by perfect play no matter who starts. Hope that now works (else I'm at tears... :-)
Ok, but how do I set up the tree in the first place? I mean, neither side just has one move, not even in a local position.
This is where I originally went wrong, choosing the follow-up position for each side that evaluated best in the empty context. Now I rather guess that positions have to be evaluated in a context crowded with copies of themselves.
So, to find the left/right daughter of a node that is no leaf, that isn't in the turn-is-worthless state (for simplicity I'll assume moves bringing positions nearer to end):
Is this OK, sensei Bill?
Robert Pauli: Nevertheless, and how impractical it might be, it's the best I currently have. You don't depend on dubious arguments about sente and gote - you simply turn off intuition and proceed. That's what I'm after: not a bunch of tricks for the tournament, but an objective method for the aesthetic pleasure. I'll be happy to see any page that addresses this. This is a copy of the living page "Tedomari / Discussion" at Sensei's Library. ![]() |