A Few Math Questions

[Surlethe]

I have two questions.

First: In a game of chess, can you consider the position, assuming no captures and no pawn moves, as an abelian group? If so, can pawn moves and captures be described as transformations from one group to another?

Second: I'm having trouble visualizing functions of complex variables. Is there any good way to do this? I'm afraid that I won't be able to intuitively grasp integration of complex functions if I'm unable to visualize the functions themselves. Also, is it correct to view a function on the complex numbers as a vector field?

Edit for clarification

[Durandal]

If you're talking about the C programming language, no, it's definitely not a vector field. A function in C is basically a procedure which takes 0 or more arguments as input and may or may not return output.

[Surlethe]

Sorry; I should have been clearer. By "C", I mean the complex numbers. I'll clarify the OP.

[Kuroneko]

First: In a game of chess, can you consider the position, assuming no captures and no pawn moves, as an abelian group? If so, can pawn moves and captures be described as transformations from one group to another?

Do you mean the set of positions having group structure? I'm not sure how that would work for chess, but the set of decidable positions, in the sense that it is possible for some player to win no matter what the play, should be representable as a subclass of the Conway games, which do form an Abelian group.

Second: I'm having trouble visualizing functions of complex variables. Is there any good way to do this? I'm afraid that I won't be able to intuitively grasp integration of complex functions if I'm unable to visualize the functions themselves. Also, is it correct to view a function on C as a vector field?

Yes, treating a complex number as a two-dimensional vector is one of the more visual approaches. Functions on C to C would correspond to two-dimensional vector fields over the plane. Another method is more typical of visualizing four-dimensional objects--treat the function topographically, with elevation on the plane representing complex modulus of the value of the function and some other property (color is easiest) for complex argument.

[Surlethe]

Do you mean the set of positions having group structure? I'm not sure how that would work for chess, but the set of decidable positions, in the sense that it is possible for some player to win no matter what the play, should be representable as a subclass of the Conway games, which do form an Abelian group.

Yes, the set of positions having a group structure, with the operation defined as a function from one position to the next in terms of the moves of each individual piece -- that is, in the opening position, could you form a group from the moves of all the pieces which can move? I'm afraid I'm not being clear enough here, since I'm not familiar enough with groups. I'm trying for something more general than forced wins, since those arise relatively rarely in the game.

Yes, treating a complex number as a two-dimensional vector is one of the more visual approaches. Functions on C to C would correspond to two-dimensional vector fields over the plane. Another method is more typical of visualizing four-dimensional objects--treat the function topographically, with elevation on the plane representing complex modulus of the value of the function and some other property (color is easiest) for complex argument.

Then this makes visualizing line integrals easier, I think, since they simply reduce to the work integrals of multivariable calculus. How does one think of function compositions, then, in the context of a two-dimensional vector field? Specifically, I'm trying to work out what sin z does to the z-plane, but I'm having trouble visualizing it in terms of a transformation.

[Durandal]

Sorry; I should have been clearer. By "C", I mean the complex numbers. I'll clarify the OP.

Well in that case my usefulness has been exhausted.

[Ariphaos]

Yes, the set of positions having a group structure, with the operation defined as a function from one position to the next in terms of the moves of each individual piece -- that is, in the opening position, could you form a group from the moves of all the pieces which can move? I'm afraid I'm not being clear enough here, since I'm not familiar enough with groups. I'm trying for something more general than forced wins, since those arise relatively rarely in the game.

Probably should also add the caveat that forced captures should be excluded from membership as well (namely, having a knight put a check on the king, in the initial case).

But you can assign each specific position a symbol in such a case, forming a set, though I have no idea if you could render the operations moving from one symbol to another in traditional linear math a la 2+2=4. I'm not sure if it forms a group since certain moves would be disallowed based on their position.

Brings up an interesting thought (to me anyway) on how to render games like chess or go in such a manner, if it's even possible.

[Kuroneko]

Yes, the set of positions having a group structure, with the operation defined as a function from one position to the next in terms of the moves of each individual piece -- that is, in the opening position, could you form a group from the moves of all the pieces which can move?

Groups need a binary operation, so this would mean that two different games (positions) are 'combined' in some manner. This has a straightforward meaning in something like Go, in which endgames can be seen as collections of isolated subgames, but what this would mean in chess is less clear. This is definitely possible formally in terms of Conway games (except that possibly some elements that have no corresponding chess positions might be introduced), but doing so 'naturally' is a more interesting problem. A group is easy to make, even an Abelian one.

I see your concern about the pawns--if there are no pawns and both castling and capture are disallowed, then every move is immediately reversible, although castling is not so much of a problem if sequences of moves are allowed to reverse positions. However, if we identify the opposite edges of the chessboard with likewise direction, so that the board now has the topology of a torus (equivalently, take an infinite chessboard and identify coordinates mod 8), the pawns are no longer a problem. To make this work, also identify a given position with the equivalence class of all relative sequences of moves from the initial position that result in that given position. By 'relative', I mean, "rook 1 up two units" rather than "Ra1a3". Obviously, not all positions are legal in this way (in particular, each column must have exactly one pawn, if pawns are in the game), but this gives a straightforward way to do a sum of positions. Given positions X and Y, with representative sequences x and y, simple append the sequence of moves to one another. If conflicts occur (such as one piece moving on top of another), then there must be alternative sequences x', y' which make the appropriate detours and still result in the given positions X and Y, and it is not possible to leave the board because the chessboard is now a torus. Thus, X+Y is well-defined in all instances. For the same reason, X+Y = Y+X, and thus the group is Abelian.

You can remove the pawns if you wish. The fact that the actual chessboard is not a torus is irrelevant--this is a group on board positions; the use of illegal moves to generate them (illegal on non-torii boards) is of no consequence, as once they are generated on the torus, they can be mapped directly back to the real chessboard. Again, the only real constraint is that there are no captures.

I'm afraid I'm not being clear enough here, since I'm not familiar enough with groups. I'm trying for something more general than forced wins, since those arise relatively rarely in the game.

It's not at all clear that the initial position itself isn't a forced win situation--for good enough play. It may just be that 'good enough' is simply too far above human ability, not that it is impossible in principle.

Then this makes visualizing line integrals easier, I think, since they simply reduce to the work integrals of multivariable calculus.

Not quite. Given a vector field F:R²→R² and a path σ:[a,b]→R², the line integral is Int_σ[ F⋅dσ ] = Int_a^b[ F(σ(t))⋅σ'(t) dt ]. If F and σ are interpreted as complex-valued, this is the real part of Int_a^b[ conj[F(σ(t))] σ'(t) dt ], where conj() is complex conjugate. In other words, the work integral is the real part of the corresponding line integral of the conjugate 'field'. There is indeed a lot of similarity with the work integral, but it would be a mistake to interchange the two simply because they have similar form.

How does one think of function compositions, then, in the context of a two-dimensional vector field?

I don't think this has a good visualization, but a multi-step one is easy enough (but it's not useful).

Specifically, I'm trying to work out what sin z does to the z-plane, but I'm having trouble visualizing it in terms of a transformation.

Well, by basic trigonometry, we have sin(u+iv) = sin u cosh v + i cos u sinh v. Thus, if z = x+iy and w = u+iv, the mapping z = sin w is a maps between a vertical strip and the entire plane. Lines of constant v in the w-plane are mapped to ellipses with foci (±1,0), whereas lines of constant u are mapped to hyperbolas with the same foci. This mapping is actually very useful since sine is meromorphic; for example, this an ideal coordinate transformation to represent a cross-section of the gravitational or electric field of an elliptically-symmetric mass or charge, since it maps the field of a sphere conformally into ellipsoids (interpret (u,v) as (θ,r), so that a field of a circle is mapped conformally onto an ellipse).

[drachefly]

Kuroneko alluded to certian properties of groups without actually defining them, so I'll fill in.

A group is a set, and a binary operation.
There are a few defining properties of groups, which I will show using 'x' as the operator and 'a', 'b', etc. as symbols for arbitrary elements of the set, which I denote as S.

"closure"
a x b exists and is in S

"identity"
there exists an element i such that
i x a = a

"associativity"
(a x b) x c = a x (b x c)

Abelian groups also have the property that
a x b = b x a

So, if your set is the set of chess positions, then it's not clear how you will represent the moves; and if your set is the set of chess moves, then it'll keep changing from turn to turn and anyway no group action makes sense.



As for complex numbers, that's tricky. For C -> C, I generally just look at the poles and jump to the algebra. For R -> C, it's three dimensional, so that's easy -- e^ikx becomes a corkscrew, and so forth.

[Jaepheth]

Since you can't pass in Chess as you can in Go, how do you have an identity?

[Surlethe]

<snip>

So, if your set is the set of chess positions, then it's not clear how you will represent the moves; and if your set is the set of chess moves, then it'll keep changing from turn to turn and anyway no group action makes sense.

Here's how I'm defining the set and operations now: let the number of pieces and the pawn structure remain constant (i.e., no captures and no pawn moves); let Ge consist of the set of all the legal positions on the board. Then the operation is simply going to be a transformation of the piece positions: b*a is the rearrangement of pieces, by legal moves, from the position in a to the position in b (similar to the way operations are defined in the dihedral group). I'm not entirely sure how to show this definition is even well-defined, though; it may be necessary to specify an initial condition.

[Surlethe]

Since you can't pass in Chess as you can in Go, how do you have an identity?

I'm not sure; that struck me as one of the biggest problems after I posted this. The thing is, by limiting it to piece moves, after any even number of legal moves, you can always repeat a position, so even if there's no actual identity in the set, it can appear after a certain set of moves.

Perhaps by defining a move to be both white and black playing, it is possible to have an identity without giving up the order of turns, thus satisfying the rules and satisfying the group axiom.

[drachefly]

b*a is the rearrangement of pieces, by legal moves, from the position in a to the position in b (similar to the way operations are defined in the dihedral group)

This is not similar to the way operations are defined in the dihedral group!

b is the rearrangement of pieces, by whatever means necessary, from the position in 0 (the origin -- the original setup).
a*b could then be doing the moves to get to a, then doing the moves to get to b. Or you could define * as the composition operator going the more usual way (b, then a)

THAT would be similar to the dihedral group.

[drachefly]

Incidentally, my list of properties of groups was incomplete (never claimed it was). There is another important property!

INVERSES!

for all a, there exists an element a' such that
a * a' = i

so,
b*a' would be the set of moves needed to get from a to b:

start at a. Then apply these moves to it...
b*a'*a = b*i = b

[Surlethe]

b*a is the rearrangement of pieces, by legal moves, from the position in a to the position in b (similar to the way operations are defined in the dihedral group)

This is not similar to the way operations are defined in the dihedral group!

b is the rearrangement of pieces, by whatever means necessary, from the position in 0 (the origin -- the original setup).
a*b could then be doing the moves to get to a, then doing the moves to get to b. Or you could define * as the composition operator going the more usual way (b, then a)

THAT would be similar to the dihedral group.

Right. That's what I was getting at. Is the difference -- where I was mistaken -- using legal moves instead of any moves necessary?

[Kuroneko]

I should probably add that if W_σ[F] is the work integral over a vector field F:R²→R² identified with the complex function f:C→C, then given a path σ, Int_σ[ f(z)dz ] = W_σ[conj(F)] + i W_σ[ refl(F) ], where conj is the conjugate ([x;y]↦[x;-y]) and refl is a reflection about y = x ([x;y]↦[y,x]). This makes the connection explicit.

Since you can't pass in Chess as you can in Go, how do you have an identity?

In the group I've defined, the elements are chess positions rather than moves, so that the identity is the initial position. It is identified with the equivalence class of all sequences of moves from the initial position that end with the initial position.

[Kuroneko]

So, if your set is the set of chess positions, then it's not clear how you will represent the moves; ...

I represent a position as the equivalence class of all sequences of relative moves from the initial position to that given position.

b is the rearrangement of pieces, by whatever means necessary, from the position in 0 (the origin -- the original setup).
a*b could then be doing the moves to get to a, then doing the moves to get to b. Or you could define * as the composition operator going the more usual way (b, then a)

That's exactly what I did, except that I combine the sequences of moves on a toroidal board, so that it is impossible for a move to prescribe leaving the board.

Incidentally, my list of properties of groups was incomplete (never claimed it was). There is another important property! INVERSES!

Yes, and the identity should be two-sided, i×a = a×i = a, since groups in are not necessarily Abelian.

[skotos]

Here's how I'm defining the set and operations now: let the number of pieces and the pawn structure remain constant (i.e., no captures and no pawn moves); let Ge consist of the set of all the legal positions on the board. Then the operation is simply going to be a transformation of the piece positions: b*a is the rearrangement of pieces, by legal moves, from the position in a to the position in b (similar to the way operations are defined in the dihedral group). I'm not entirely sure how to show this definition is even well-defined, though; it may be necessary to specify an initial condition.

One problem I see with this is your definition of Ge. Ge cannot be the set of all legal positions on the board. Specifically, you need to exclude "dead-end" positions, by which I mean: checkmate positions, stalemate positions, and any position from which all of the moves lead to "dead-end" positions. These positions are all irreversible, even though they do not involve captures or pawn pushes.

On a similar note, you should consider castle eligibility.

Once these restrictions are in place, you could define the group by simply enumerating all of the positions in Ge. Once these positions were numbered (so that, for example, I(p) is the number assigned to position p during the enmueration), then the operation a * b could simply be the position corresponding to (I(a) + I(b)) mod N, where N is the total size of Ge. Unfortunately, that operation has no obvious correspondence to the moves needed to reach positions a or b.

[drachefly]

So, if your set is the set of chess positions, then it's not clear how you will represent the moves; ...

I represent a position as the equivalence class of all sequences of relative moves from the initial position to that given position.

Yes, I noticed. But how do you define what a primitive, single, legal move is?

As for the rest, I was basically rehashing what you did but going slower for the rest of the audience.

Incidentally, my list of properties of groups was incomplete (never claimed it was). There is another important property! INVERSES!

Yes, and the identity should be two-sided, i×a = a×i = a, since groups in are not necessarily Abelian.[/quote]

I already covered that by making the inverses work both ways --

a x a' x a = i x a = a x i

[Grog]

Supose that the inverses and identities are onesided:

i x a = a
a' x a = i
c = a x a' = (a x a') x (a x a') = c x c
i = c' x c = (c' x c) x c = c = a x a'
a x i = a x (a' x a) = i x a = a

twosidedness follows or maybe I just need some sleep.

[drachefly]

For that to work, you needed inverses to be two-sided.

[Grog]

For that to work, you needed inverses to be two-sided.

I dont think so. It has to work from the same side on all elements tough (I'm not sure if this requirement can be removed). Then the above proves that it is two-sided.
To me it seams that way and this time I can't blame my lack of sleep so maybe I'm just stupid. Where do I use two-sidedness of the inverse?

[drachefly]

a x i = a x (a' x a) = i x a = a

The second statement here uses i = a' x a, which is the reverse of the way you defined inverses, a x a' = i

[Grog]

a x i = a x (a' x a) = i x a = a

The second statement here uses i = a' x a, which is the reverse of the way you defined inverses, a x a' = i

I thought the line above that proved that a' x a = a x a'

[drachefly]

Hmm... now that I've re-examined it more closely, it looks valid. I wonder, does this work if instead of
i x a = a
a' x a = i

we take

i x a = a
a x a' = i
and do not assume that (a')' = a?
then we can't put together line 3... instead, we get

c = a x a' = a x i x a' = a x (a x a') x a' = a x (a' x (a')') x a' -> not very useful

Interesting...