Topological go

Defn 1: a go-space (X,T,S) is a topological space (X,T) together with a subset S of T which does not contain the empty set.

Rule 1: Play is on a go-space (X,T,S) by two players called Black and White.

Rule 2: Each element of X may be coloured black, white or empty.

Defn 2: If an element of S, (s say) is such that all elements of s are coloured a single colour then s is said to be coloured that colour (So an element of S is either black, white, empty, or non-coloured - i.e. mixed).

Defn 3: A stone is an element of S which is coloured black or white.

Defn 4: A legal placement is an element of S which is coloured empty.

Defn 5: A set of stones all of which have the same colour is go-connected if the closure of the union of the stones is T-connected.

Defn 6: A go-connected set of stones G has liberty if there exists a legal placement e, such that if e were coloured with the same colour as the stones of G then {e} union G would be go-connected.

Defn 7: the string of a stone s, or string(s), of colour C is the union over all subsets A of {t in S : t is of colour C} such that A contains s, and A is go-connected.

Defn 8: Clearing a colour C is the process of colouring empty all elements in:

{x in X such that x is of colour C, and for all stones s of colour C in S : x is in s, string(s) does not have liberty}

Defn 9: A board play consists of choosing a legal placement e, colouring all elements of e one's own colour; then clearing the opponent colour, and then clearing one's own colour.

Rule 3: The game starts with all elements of X coloured empty, the players alternate turns (a,b), starting with Black.

Primitive Rules

Defn 10a: A turn is a board play that doesn't repeat an earlier X colouring.

Rule 4a: The first player unable to take their turn is the loser.

Cardinality 'Stone' Scoring Rules

Defn 10b: A turn is either a pass or a board play that doesn't repeat an earlier X colouring.

Defn 11b: A player's score is: |{x in X : x is of that player's colour}|

Rule 4b: The game ends after 2 consecutive passes.

Rule 5b: The player with the higher score (in terms of cardinality) at the end of the game is the winner. Equal scores result in a tie.

Acknowledgements

This is derived from the Tromp-Taylor rules, scoring from primitive rules.

Comments

D1 - S is the set of possible stone placements. S is totally general as long as it doesn't contain the empty set, though I'm trying to work out general principles of choosing (X,T,S) to get a reasonable game (e.g. one in which it's actually possible to put a stone down...). The empty set not being in S is essential unless we are going to put 'non-empty' conditions all over the place. For example, if the empty set is in S with the other definitions as they are now, its impossible to take anything since we can take e the empty set in Defn 6.

R2 - We could define colouring more rigorously as a function from X to {b, w, e}, then D2 as another function defined from this one. However, there's still an element of concepts 'outside of the maths' with players then altering this function as play goes on. Perhaps I'd need to take a look at some game theory in order to work out how to formalise taking turns (and altering the colouring function as 'time' goes on) in a convenient way.

D2 - Note that I'm using the word 'colour' in two different ways, one for colour of elements of X, and one as defined here for colour of sets in S. It should be clear which one I mean, and in the definitions the links take you to the definition of the relevant use of the word 'colour' (R2 or D2).

D3 - If I defined the stones to be the elements of X then I think it would reduce to general graph go however you define connection.

D4 - Note with this and D3 I'm just giving names to coloured elements of S referred to in D2 for convenience later. Interestingly, the definitions for the concepts of stone or legal placement are nearly identical, though the uses are very different.

D5 - This is nice - you don't need a separate idea of connecting by moving horizontally and vertically along grid lines - the topological connectedness property does it for you. I've also considered saying 2 stones are go-connected if the intersection of their closures is non-empty. This doesn't have that nice property, but might be simpler to work with in some cases.

One of the topological definitions of a set U being T-connected (the one I'm using) is if within the induced topology (U,TU) there are no clopen sets other than U and the empty set. TU is the set {U intersect t : t is in T}, a clopen set is a set which is open and closed.

Previously I took the union of the closures of the stones, but this isn't the same as the closure of the union in general infinite cases. The closure of the union is slightly 'bigger'.

D6 - We don't define liberties (plural), since its not...well defined here. You can have situations where you can fit say, 2 legal placements next to a stone, or if you do it differently then 3 legal placements.

D7 - This definition can deal with any amount of infinite strings, boards etc. with no problems - it only uses set theoretic concepts. Previously I defined this with a 'maximal' go-connected subset, which is somewhat unrigorous. There is a proof below that string(s) is go-connected.

D8 - I think its better to define the colouring on X rather than on S, in order to allow a little more generality. If S contains non-connected sets then they generally die as soon as they are placed, unless there happens to be a 'liberty placement' which touches all parts of the stone. With this definition you can place the 'stone' and each part of it can act as a separate entity if there happens to be an element of S which works like one of the parts of the played stone. Basically it allows topological go to encompass variants in which you play more than one stone at a time. We don't get any weird effects such as bits of stone being left on the board whilst not being capturable or otherwise acting like normal stones, because of this result.

D10a,b - In effect, D10a is primitive ko, and D10b is positional super ko

R4a,b - The game might never end. Even if we make the players take each move in half the time of the previous move, if there are an uncountable number of elements in S then it's possible for the game to still have an uncountable number of moves left. The problem with uncountable numbers of moves is that black and white are supposed to alternate and effectively list their moves, but then the uncountable would be countable. Maybe there's a way round somehow...axiom of choice?

Scoring - Cardinality 'stone' scoring will give a draw in many games with X infinite. Primitive rules do not, but the game is somewhat different from most regular go rule sets when we specialise to normal 2d go.

Examples

1) Regular go with primitive rules

If we take X to be [0.5,19.5]x[0.5,19.5], T to be the usual Euclidean topology (well, the induced topology on X), S to be sets of the form:

{(x,y) : (x-p)^2+(y-q)^2<1/4, for p, q, integers in [0,19]}

...then we get the usual 2 dimensional game of go with primitive rules (no passes). Cardinality 'stone' scoring doesn't work (it's well defined but will not distinguish a winner) since if a player has a single live stone at the end of the game they have score uncountably infinite.

2) General graph go

...is a specialisation of Topological go in the following way (though there are other ways):

If the graph is a set of vertices V, with a set of edges E, then define:

X = {xv : v is in V} union {yuv : an edge joining u and v is in E}

...where the xv and yuv are all distinct. (note the terminology for yuv gives yuv=yvu)

T has a base:

{{xv} : v is in V} union {{xv, yuv, xu} : u,v are in V and joined by an edge}

S is the sets:

{xv} where v is in V

With these definitions we should have go on a general graph. I may get around to proving this statement fully, using this definition of general graph go - see below.

With a general graph we can of course get another go space which acts like regular go, though we can easily specify to diamond structure instead.

Some Results

1) The specialisation to General Graph go

For now just a check to see if adjacient stones are go-connected, whereas non-adjacient stones are not:

Suppose xu, xv are in X as above, and u is adjacient to v (i.e. yvu is in X). First we need to work out the closure of {xu}. This is defined to be the intersection of the closed sets containing {xv}. Equivalently p is in the closure of xv if p is not in any open set not containing xv.

The elements in X satifying this are xv itself, and yvw, for w some vertex. Similarly for xu. So the union of the closures is U := {xv, xu, yuv} union {yvw, yuw : w is in V, if such y's exist in X}.

Claim: This set is T-connected.

Proof: We show that there are no clopen sets in U other than the empty set and U itself. Consider the element yuv. We wish this element to be in an open set - but any open set which contains yuv must also contain xu and xv. If there are no other elements in U then we are done. Suppose (without loss of generality) there exists a yuw in U, where w =/= v. Suppose for contradiction we can find an open set containing yuw other than the set containing yuv. In the induced topology on U, if yuw is in an open set, then xu is also in that set. So yuw must be in the set also containing yuv. Contradiction. So there are no clopen sets in U other than U and the empty set, so U is connected. QED.

In the case where u and v are not adjacient in the graph, the union of the closures is: U := {xv, xu} union {yvw, yuw : w is in V\{u, v}, if such y's exist in X}. Then a clopen set in this is: {xv} union {{yvw : w is in V\{u, v}, if such y's exist in X}, since {xu} union {{yuw : w is in V\{u, v}, if such y's exist in X} is also open. So U is not connected.

2) Proposition: after a turn in a finite game, any coloured element of X is in a stone.

Proof: by induction on the number of turns played.

Base: at the start of the game, by Rule 3, there are no coloured elements of X, so they are all inside stones.

Inductive step: Defn 9 tells us how the board is changed in a turn (either definition of turn). When we have just coloured all elements of e one's own colour, all coloured elements outside of e are still in stones by induction. All elements inside e are inside e, which is a stone.

When we have just cleared a colour C, we have coloured empty all elements in {x in X such that x is of colour C, and for all stones s of colour C in s : x is in s, string(s) does not have liberty}. If x in X is still coloured after this, then x was not in that set. I.e. x is not of colour C so it's in a stone by induction, or x is of colour C and there exists a stone s : x is in s, and string(s) has a liberty. I.e. x is in a stone. []

I'm not sure this result extends to games with countably infinite numbers of moves. I'll try to think up a counterexample.