asura wrote:

Bill Spight wrote:

By *reverse values* I take it you mean the negatives of the values.

Yes, that's what I meant. Using the word "reverse" instead of "negativ" was a bad choice, because "reverse" already has another meaning in this domain.

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On Numbers and Games,

Winning Ways, and

Mathematical Go all address how to simplify games.

You delete dominated options and reverse through reversible options. Sensei's Library explains that, too, but maybe not as well as the textbooks, as Sensei's Library is not aimed at mathematicians.

I've read either "On numbers and Games" or "Winning Ways" some years ago, but I can't remember wich one it was. Probably this is a sign to (re)read them both one more time.

Actually I believe I was only interested in the stuff about the (surreal) numbers at that time and I know about the Up, Down, Tiny, Miny stuff only from senseis library.

Because I don't have access to these books at the moment, I want to ask one more question.

For removing dominated options I only know that one can (and should) remove all options that are

*clearly* worse than another one, like in {1,2|3} Black would never move to the 1, because 2 will be always better (or at lest equal) than 1 for black.

And under removing reversals I understand something like forcing black to continue, if after a black move followed by a white move, white can move from here to a better position for her than she could get from moving in the initial position. (probably an equal position could/should be used, too, to qualify a play as an reversal)

Not quite right. We start with game G. Black has a move to GL. White has a move from GL to GLR. We compare GLR with G. If GLR ≤ G, then we can replace GL in G with all of the left options of GLR.

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But I cannot see, how this helps in this case.

In Y={0,*|*} it's not clear wether 0 or STAR is better for black.

Right. They are

*confused*.

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The best black play depends on if there is an additional STAR or not.

Assumes that in Z = Y + * Black should move in Y. OC, Black may move in Y, but simplification removes options that might be playable in certain circumstances. It is just that they are not necessary. You may verify that Black wins in Z if Black plays first in *.

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When I try to solve it with my 'own' knowlege, the only way I see is to consider both szenarios:

1) Y = {0,*|*} (black wins)

Right. And Black to play wins by playing to 0, but loses by playing to *. That raises the question of whether the Black option to move to * can be removed. So we can check whether Y = Up. (If we do that, we don't have to check for reversal.

)

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2) Y+* = {0,*|*} + * = {*,*+*|*+*} = {*,0|0} (=X) (first player wins)

Not a correct derivation. Y + * = {*,0,Y|0,Y}

Removing dominated options we get

Y + * = {*,Y|0}

Black's Y option is reversible. The White option of Y is *, which is less than Y + *.

* - (Y + *) = −Y , and

−Y < 0.

Replacing the Y options with the Black option of * we get

Y + * = {*,0|0}

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Probably you already knew that {0,*|*} = {0|*}, but for me it looks a bit like

{0,*|*} = ??? = {0|*}.

So I'd like to ask you if you could write down just one more step inbetween?

As I indicated above, since {0,*|*} is a Black win, we may be suspicious of the * option for Black, and then compare {0,*|*} with {0|*} directly. But we can also show that the * option for Black is reversible.

The White option of * is 0. 0 < {0,*|*}, as we already know. So the * option is reversible, and we may replace it with all of the Black options of 0. There are none, of course. So:

{0,*|*} = {0|*}