Distance between openings

 Yeah, right, MapQuest on the space of all openings and chess positions.

Okay, I'll continue by calculating distances between main lines in Najdorf, Ruy Lopez and French after 10 moves.

Dividing the height into white and black parts, one gets:

height(Najdorf)=12+9

height(Ruy Lopez)=7+11

height(their common stem)=2+7

So, by the formula in post #28, we obtain after 10 moves

distance(Najdorf,Ruy Lopez)=(12+9) + (7+11) - 2x(2+7) = 15 + 6 = 21

Main line in Najdorf :

1.e4 (h=2+0) c5 (h=2+2) 2.Nf3 d6 (h=2+3) 3.d4 (h=4+3) cxd4 (h=4+5) 4.Nxd4 (h=5+5) Nf6 5.Nc3 a6 (h=5+6) 6.Bg5 e6 (h=5+7) 7.f4 (h=7+7) Be7 8.Qf3 Qe7 9.O-O-O (h=10+7) Nbd7 10.g4 (h=12+7) b5 [21=12+9]

Main line in French (Winawer, Advance) :

1.e4 (h=2+0) e6 (h=2+1) 2.d4 (h=4+1) d5 (h=4+3) 3.Nc3 Bb4 4.e5 (h=5+3) c5 (h=5+5) 5.a3 (h=6+5) Bxc3 (h=6+6) 6.bxc3 (h=8+6) Ne7 7.Qg4 O-O (h=11+6) 8.Bd3 Nbc6 9.Qh5 Ng6 10.Nf3 Qc7 [17= 11+6]

Their common irreversible stem (CIS) has height 8=4+4 :

So, by the formula in post #28, we obtain after 10 moves :

distance(Najdorf,Winawer)=(12+9) + (11+6) - 2x(4+4) = 15 + 7 = 22

CIS(Ruy Lopez,Winawer) is the same as CIS(Najdorf,Ruy Lopez,Winawer) and has height 2+4 :

distance(Ruy Lopez,Winawer)=(7+11) + (11+6) - 2x(2+4) = 14 + 9 = 23

Conclusion: Main lines in Najdorf, Ruy Lopez and Winawer after 10 moves are approximately equidistant from each other with the distance in the range 21-23.

Gambitking: could U please specify this "King's Gambit, Schallopp Defence, rook sac".

First of all, I'll calculate heights of these gambits. The height of taking the white rook is counted as 1+5 (1 cuz white lose the ability of short castling and 5 is the value of rook).

1. e4 (h=2+0) e5 (h=2+2) 2. Nf3 f5 (h=2+4) 3. Bc4 fxe4 (h=2+6) 4. Nxe5 (h=3+6) Qg5 5. d4 (h=5+6) Qxg2 (h=5+7) 6. Qh5+ g6 (h=5+8) 7. Nxg6 (h=6+8) Qxh1+ (h=7+13) 8. Ke2 [22=9+13]

1. e4 (h=2+0) e5 (h=2+2) 2. f4 (h=4+2) exf4 (h=4+4) 3. Nf3 Nf6 4. e5 (h=5+4) Nh5 5. d4 (h=7+4) g5 (h=7+6) 6. h4 (h=9+6) g4 (h=9+7) 7. Ng5 Ng3 8. Bc4 Nxh1 [22=10+12]

Common irreversible stem (CIS) is of height 5+8 :

So, we get

distance(Latvian/poisoned,Schallop/rook sac)=(9+13) + (10+12) - 2x(5+8) = 9 + 9 = 18

LoL My respect to that genial inventor Leo da Vinci!

I'm doing some easy but fancy math here studying the properties of HFZ metric on da space of all chess games.

Yeah, it could be done on a globe of dimension 16 (corresponding to 16 pawns with their irreversible moves). My genius is usually fixed on fancy but useless things LoL.

Geometrical square: take 2 tesseracts and connect all vertices to obtain a hypercube of dimension 16.

Currently I'm playing several (counter)gambit tournaments: Albin, Faulkbeer, double Göring and Smith Morra; Benko and Budapest coming soon. It looks like our common interest.

Distance zero, of course. Well, good point about counting heights in this case. If a pawn is taken en passant, the height of its own move is 1 and not 2.

Linksspringer: First of all, the computation of heights :

1.e4 (h=2+0) e6 (h=2+1) 2.d4 (h=4+1) d5 (h=4+3)3.e5 (h=5+3) f6 (h=5+4) 4.exf6 (h=7+4) Nxf6 (h=7+5) and

1.e4 (h=2+0) Nc6 2.Nf3 f5 (h=2+2) 3.exf5 (h=4+2) d5 (h=4+4) 4.d4 (h=6+4) Bxf5 (h=6+5) 5.Bb5 e6 (h=6+6)

It's an indirect measure of how much idle time you have on your hands.

HUH???

Linksspringer: the distance is zero cuz 2 positions have common irreversible stem (forgeting 2 exchanged pawns) and can be linked by reversible moves. My computations through heights should be modified in this case. I'll think about it later. Thanx for this nice counter-example!

U can say this also about playing chess or other games in general.