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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.

Bill: How about an example?

The other possibility for gaining tedomari is to play a move that gives you two independent follow-ups at the local temperature.

Bill: I think that he means that they are miai, and so cancel out.

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:
Thanks for your reply, Bill, even if it crushes my beliefs. Somehow my impression was that the binary tree used to compute miai values was made up of locally, say, minimax-best plays, but that doesn't seem to be the case. What you're saying sounds like miai values themself ranking the moves to be used in the tree. Boy, I'm really confused. Not that I'm new to recursion, but where's the ground case here? Temperature zero? At least it can be checked with minimax, and since nothing can be gained the miai value of each move should be zero in that case. Then I can slowly work my way back...let me instanly hurry to my chamber and contemplate... :-)

Bill: This may help. It's from my talk at the Computers and Games 2002 conference. The slides are [ext] on my home page.

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:
Wow. Be sure - I don't get a single word. Dummy me thought that it simply would go by propagating averages and then to look at the distance between root and one of its daughters:

            = -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 W1 on the 2-3 point. That was obvious to me. The line of play to use when play is restricted to the corner is W1 on the 3-4 point, solidifying the wall. (That is also the one to use at temperature zero, OC.)

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

Robert Pauli:

I admit that, even reading a bit thermography, I still can't decipher your tree decorated with t's:

  • why do some nodes get no t?
  • why do some nodes get more t's?
  • why does t's sign differ?
  • -3-t seems to be the average of -2-2t and -4, but why is it joined with -4?
  • where the heck is -5 from??

Bill: Wow, that's a lot, Robert! Our disucssion has prompted me to think that I should create an Endgame Calculations without Tears? page, but I really can't do much on that until December. Meanwhile, let me give some brief replies.

The coefficient of the t's at the terminal nodes is w - b, where w is the number of White plays starting from the root and b is the number of Black plays. Other coefficients are derived from these.

In min(-3-t, -4) you are comparing the gote and sente possibilities.

-5 is the count if the node is gote. It is derived from solving the system,

    c = -7 + t
    c = -3 - t

If this is some alternate method to find miai values, great,

Bill: It does that, and more. :-)

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.

Bill: Japanese yose theory talks about the miai value of a play; CGT talks about the temperature of a game. When the play is orthodox without regard to the ambient temperature, its miai value equals the temperature of the game. Somewhere Charles Matthews and I had a discussion about whether an inferior play had a miai value, and we each seem to have convinced the other to change sides. ;-)

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... :-)

Bill: There are three basic cases:

  • First, gote, where the count is the mean of B and C.
  • Second, sente, where the count is D or E.
  • Third, reversal, where the answer lies further down the tree.

In a way, thermography simplifies your task because you do not have to worry about these distinctions. More on the calculation page when I get around to it.

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.

Bill: Include every move for each player at every node. (That's often impractical. You can usually eliminate bad plays.)

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.

Bill: It's not so bad.

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):

  • look at each position Black/White can reach from it in one move
  • recursively compute its miai value (or average score)
  • choose one with the highest/lowest value

Is this OK, sensei Bill?

Bill: I don't think that takes reversal into account. For an example of reversal, see hanetsugi.

Robert Pauli:
Wherever I stand, the carpet is pulled. Binary trees? No. Average? No. Depth limit? No. :-) Even the method of multiples doesn't seem to be general: kos.

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.
(OC) 2004 the Authors, published under the OpenContent License V1.0.