I found this in my inbox this afternoon:

On Mon, Jan 17, 2011 at 1:37 PM, John Linton wrote:

While in between studying for Speed School today, I was considering a chess puzzle when I suddenly realized how physical distance - as opposed to "chess distance" - applies to the game.

If one considers a 3x3 section of the board, say bounded by (a1,a3,c3,c1), and one considers a rook check given at a one-square distance, meaning say Rb3 Kb1 is the setup, the king can attack the rook within one move by either Ka2 or Kc2.

However, consider the bishop check where Ba3 Kc1 is the setup. In that case, the bishop is physically farther from the king - instead of just 2.0 units as with the rook check, the bishop is 2*sqrt(2) squares away! This farther physical distance - even though the check is still "one square between" the king and bishop - forbids the king from making a sideways motion that threatens the bishop in one move. Hence Kc2 or Kb1 or both equally useless in attacking the bishop.

Further extending this analysis, if Nb3 checks the Kc1, we can see that the knight is just physically close enough at sqrt(5) squares for the king to attack by both avenues: Kc2 or Kb2!

Of course one might note that in the bishop case King cannot go to Kb2 because that square is checked by the bishop to begin with - but it is precisely the physical distance that gives the king only the check square to attack the bishop along - where as with the rook -- the more powerful piece! -- the king can more easily attack due to physical distance!

John L.

Bobby Amback shot back with: "I always thought that given a choice on how to deliver a mate. The mate delivered with minimal space excluding adjacent squares was the best insult. This of course is second to delivering mate with the lowest ranking piece with the pawn being the greatest insult. Is this similar to the slam dunk?"

Then JDD sent this:

In math there is something called a "metric space". The notion of "distance" is allowed to be more general than the usual Euclidean distance, sqrt (x^2 +y^2). In this case the JL's "chess distance" is different than the Euclidean distance. Is it a metric space? We will need to define a metric space.

But first, let us define "chess distance" as the minimum number of king moves to "reach" a destination (in this case for the king to capture (rather than just attack) the attacking piece as in JL's examples). In the case of the bishop attack, the Euclidean distance was 2*sqrt(2) as JL noted. The chess distance was 3, if check is to be avoided along the way. If no check avoidance, the distance is 2 along the same diagonal. It would seem for consistency, the chess distance should either always use the no check rule or not. Since the check is a somewhat artificial consideration geometrically, we will use the non check avoidance distance. So, is this chess distance a metric space ? (we might later ask about the no check distance).

To be a metric space, several axioms apply.

We have points with coordinates and a distance function D, defining the distance between any 2 points, D(x,y), with x and y as points (not the usual x and y coordinates in this case). Each point has a pair of coordinates.

Then for a metric space,

D(x,y) => 0

D(x,y)=0 if and only if x=y.

D is symmetrical with D(x,y)=D(y,x)

The least trivial axiom is the "triangle inequality" where

D(x,z) =< D(x,y) +D(y,z), where D is the distance function (implicit in the above chess examples) and x, y and z are three sets of coordinates.

What is the D function for the chess distance between 2 arbitrary points (chess squares)? From all this, is "chess distance" a metric space ?

JDD

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