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Stacks of coins
Difficulty: Advanced
Keywords: EndGame
A very simple endgame model for Go. Stacks of coins stand on a table. The two players alternately take one stack, until the table is empty. This is a useful starter model. One stack on its own is entirely trivial, so anything you can get out of this shows the use of disjunctive sums of games, a piece of general theory. It illustrates the idea that you look to play big points when everything is gote. The best play is to take the largest stack, and if you know enough to ask 'how do you prove that?', you know enough to prove it. It also illustrates miai, namely two stacks of the same value. They might as well not be there, since they will be divided one to Black and one to White if play is correct. If the sizes of the stacks are s, t, u, v ... in decreasing order, the value of starting is s - t + u - v + ... which is at most s. If we aren't given full information about the stacks, we can guess this value as s/2. That's a more meaty use of the model, and requires some discussion. The value depends upon how many stacks of coins there are. If the number is even then s/2 is the best guess. If the number is odd it will vary between s for small numbers and s/2 at the limit for large numbers. -- Saesneg Charles Matthews: The discussion depends on what one accepts as a decent probability model. Look at the hypothesis that the number of stacks of a given value v is equally likely to be even or odd. If this is OK up to v = s then by removing miai (pairs of stacks of equal size) one can look at cases with at most one stack of each value, and do a calculation. Which gives the answer s/2 as expected value. I'm reasonably happy with that hypothesis as a model of small endgame plays. I don't see that it is so good for the larger endgame plays - given there is one of (deiri) value 15, there is quite a good chance there are no others (tedomari). So I'd have thought this calculation has a certain, limited application, because the 19x19 board has only restricted scope. This argument then affects what one thinks about reverse sente. victim: I disagree with Saesneg. The value will be somewhere between 0 and s, but it won't be 0 if it's odd and won't be s if it's even. s/2 is the expected value, and you'll get extreme values if the stacks are miai-heavy. E.g. if there are five stacks, s=t and u=v and w=1, the value is 1. For five stacks where t=u and v=w, the value is s. For four stacks, s=t and u=v, the value is 0. For four stacks where t=u and v=1, the value is s-1. Robert Pauli: Even (seemingly) simple games deserve rules without ambiguties, Charles. So, before someone starts a rule debate by taking more than one coin in his turn (takes a pile of coins), how about:
Charles Don't understand - you do take one whole pile. BobMcGuigan: The analogy with all-gote go endgames is that the number of coins in a pile equals the point value of the endgame move. Right, Charles? Charles Sure. You can make the coins be of different values, instead of having stacks. But the latter would seem to be more graphic. And you can say, lining up the stacks by size to get a histogram, that the steps down in size are the key quantities in the theory. As part of (the much deeper) atomic weight theory, one can worry about the type of stack in which one player is able to take the top coin only, while the other can only remove the whole stack. But that game should be played under the Nim?-style ending condition typical of CGT: last to have a play wins. You aren't looking at the accumulated coins for each side.
Robert Pauli:
(a rather trivial game) However, I came here from EquivalentGameToGo, quote (my format): The first stack is (top to bottom): 25 cents, 10 cents, 5 cents, 1 cent, 1 cent. The Second stack is: 10 cents, 5 cents, 5 cents, 1 cent. The third stack is 25 cents 10 cents, 1 cent. 1 cent. Looks pretty much in my direction - covered coins being follow-up's (and coin rims should identify values for total information). I certainly would prefer this to be the Stacks Of Coins Game. Charles It's a more realistic endgame model, in that a play leaves a follow-up play. But sometimes you need to give people the absolutely basic model of gote plays.
Robert Pauli: "Stacks of coins" - anybody read this slowly and tell me what he imagines. Does he assume each stack to be made of one coin? Of course not. Does he assume all coins to be of the same value? Of course not.
"Stacks on a table" - please "take a pile". Really wonder what will happen. Maybe he'll take all stacks at once for a very nice pile, maybe she'll be a little shy and just take some upper part of a single stack. "One stack on its own is entirely trivial." Come on, several stacks as well - if whole stacks are at our disposal. "It also illustrates miai." Come on, you need to know absolutely nothing about symmetry to play coins on a table - to give it a name that fits - just grab one of the highest valued coins. Difficulty = Advanced? Well, maybe :-) Charles Robert, I find you a bit of a pedant, really. You are also missing the point. Simple stacks is enough to show that you don't always double deiri for reverse sente. Which has puzzled many people.
Robert Pauli: It most likely shows in two cases:
You managed to trigger both. ;-) For the benefit of others let me edit the top to make things clear. Hope you can bear it. My part of the compromise is seeing a broad term being wasted for a special (trivial) case of its meaning. OK? Charles Robert, please edit as you see fit. This is a copy of the living page "Stacks of coins" at Sensei's Library. ![]() |