'Variant Chess', issue 41, January 2003, page 4

25 years ago, Václav Kotesovec wrote an article in feenschach in which he examined some endings involving grasshoppers. He recently returned to the attack with a modern computer, and he reported his results in the September issue of the Bratislava chess composition journal Pat a mat (issue 38, pages 206-8) and on his web site (http://web.telecom.cz/vaclav.kotesovec). He has kindly given us permission to report some of them in VC.

The grasshopper moves on queen lines, but only when the line is obstructed; it travels along the line until it meets an obstruction, hops over it, and lands on the square beyond. Suppose grasshoppers for queens in the normal game array; then 1.Gd1-b3/d3/f3 are all possible, and Gd1-f3 threatens Gxa8 and Gxf8. Better for actual play is Boyer’s layout (Encyclopedia of Chess Variants, page 130), where the pawns are placed on row 3 (no double-step initial move) and there is a line of grasshoppers across row 2. The grasshopper is a weak piece which gets weaker as the game goes on, and no harm is done by having several of them.

As regards the ending, Václav’s results are as follows.

• Against a bare king, the minimum numbers of pieces needed to force mate are K+4G, K+N+2G, K+B+2G, K+2N+G, and K+2B+G (same-square bishops). Weaker forces may be able to give mate (mating positions can be constructed with K+3G, K+N+G, and K+B+G), but they cannot force mate against an unwilling opponent.
• Against king and grasshopper, K+2N can force mate, and so can the conventional winning forces against a bare king (K+Q, K+R, K+2B with unlike bishops, K+B+N).
• Against king and two grasshoppers, K+Q can force mate, but K+R cannot.
• K+Q can also force mate against king, knight, and grasshopper and against king, bishop, and grasshopper, but the ending of K+Q against king, rook, and grasshopper is not easily characterized.

All these results are for the 8x8 board.

Václav showed K+4G v K to be won in 1977 (the key is that a tight square of grasshoppers is self-defending, because the enemy king cannot approach), and the hardest case against a bare king is K+N+2G. He gives an example.

Ka1 Nb1 Ga2 Gb2 - Kd4: 1.Gc2 Kd3 2.Na3 Kd4 3.Ka2 Kc3 4.Gc4 Kb4 5.Gb5 Kc3 6.Gc2 Kb4 7.Kb2 Ka4 8.Gb1 Kb4 9.Gd3 Kc5 10.Kc3 Kd5 11.Nb5 Kc5 12.Nd4 Kd6 13.Kc4 Ke5 14.Gc5 Kd6 15.Gc3 Ke7 16.Kd5 Kf6 17.Ge3 Kg5 18.Ke5 Kg6 19.Gc3 Kf7 20.Nf5 Ke8 21.Ke6 Kd8 22.Nd4 Kc7 23.Ge5 Kd8 24.Nb5 Kc8 25.Ke7 Kb7 26.Kd6 Kb8 27.Gd7 Kc8 28.Gc7 Kd8 29.Nd4 Ke8 30.Ge7 Kf8 31.Nf5 Kg8 32.Ke5 Kf8 33.Ke6 Ke8 34.Ng7+ Kd8 35.Kd6 Kc8 36.Kc6 Kb8 37.Ne6 Ka7 38.Nc5 Kb8 39.Nb7 Ka8 40.Gc7 Kb8 41.Nd6 Ka7 42.Kb5 Kb8 43.Ge5 Ka7 44.Ga4 Kb8 45.Gc6 Ka7 46.Ka5 Ka8 47.Ka6 Kb8 48.Kb6 Ka8 49.Ne8 Kb8 50.Nc7 mate.

The lion is like the grasshopper except that it can carry on beyond the hurdle; it is not forced to stop on the square immediately beyond.

• Against a bare king, the minimal combination that can force mate is K+3L.
• If lions and grasshoppers are mixed, K+L+2G suffice.

Václav gives examples of both, though the win with K+L+2G is clearly harder than that with K+3L.

Kb2 Ga1 Gb1 Lion a2 - Kc4: 1.Gc3 Kd5 2.Kb3+ Kc6 3.Gb4 Kc5 4.Gb2 Kd6 5.Kc4 Ke5 6.Gd4+ Kd6 7.Ld5+ Ke6 8.Gc5 Kd7 9.Lb3 Kc6 10.Gb4 Kd7 11.Kd5 Ke7 12.Le6 Kf6 13.Lc4 Kg5 14.Gd4 Kf6 15.Ge5+ Kf5 16.Kd6 Kg6 17.Ke6 Kg5 18.Gc3 Kg6 19.Gc5 Kg5 20.Ge3 Kg6 21.Lf4 Kh5 22.Kf5 Kh4 23.Gg5 Kg3 24.Lf6 Kh2 25.Kf4 Kg1 26.Gg4 Kg2 27.Gg3+ Kg1 28.Kf3 Kh2 29.Lf2 Kh3 30.Kf4 Kh2 31.Lf5 Kh1 32.Ke3 Kg1 33.Kf3 Kh2 34.Kf2 Kh1 35.Ge6 Kh2 36.Ld7 Kh1 37.Lg4 Kh2 38.Kf3 Kh1 39.Ke2 Kh2 40.Kf2 Kh1 41.Gh3 Kh2 42.Gf3 Kh1 43.Ld1 Kh2 44.Lh5 Kh1 45.Gh3 mate.

These are analogies of the XiangQi cannon. The Pao is exactly the same as the cannon (it moves along rook lines, but captures only if there is one man between it and its target), and the Vao and Leo are the equivalents on bishop and queen lines.

• Against a bare king, K+2Le and K+3Pa can force mate, but K+4Va cannot.
• With mixed forces, K+Le+Pa, K+Le+Va, K+2Pa+Va, and K+Pa+2Va all suffice, the last case even if the Vaos run on squares of the same colour.

The hardest cases appear to be K+Le+Va and K+Pa+2Va (same colour).

Kb1 Leo a1 Vao a2 - Kb3: 1.Lb2 Ka3 2.Vc4 Kb4 3.Vf1 Kc3 4.Kc1 Kb3 5.Le5 Kc4 6.Kd2 Kd5 7.Lh5 Ke4 8.Ld1 Kf4 9.Kd3 Ke5 10.Ke3 Kd5 11.Ve2 Ke6 12.Ke4 Kd6 13.Ld2 Ke7 14.Ke5 Kd7 15.Ld1 Kc6 16.Kd4 Kb5 17.Kc3 Kc6 18.Vd3 Kc7 19.Kc4 Kc6 20.Ld2 Kb7 21.Kb5 Kc7 22.Kc5 Kb7 23.Lc3 Ka6 24.Kb4 Kb7 25.Vc4 Kb6 26.Lc2 Ka7 27.Ka5 Kb7 28.Kb5 Ka7 29.Lb3 Ka8 30.Ka6 Kb8 31.Lc2 Ka8 32.Kb6 Kb8 33.Lc3 Ka8 34.Lh8 Kb8 35.Vg8 mate.

Kb1 Pao a2 Vao a1 Vao b2 - Ka4: 1.Vc3 Kb3 2.Pb2 Kc4 3.Kc2 Kd4 4.Pb3+ Ke4 5.Kd2 Kd5 6.Kd3 Ke6 7.Kc4 Kf5 8.Vd4 Kf4 9.Vac3 Ke4 10.Kc5 Kf5 11.Kd5 Kg4 12.Ke5 Kg5 13.Pb4 Kg6 14.Ke6 Kg5 15.Pc4 Kg6 16.Ve5 Kh5 17.Vcd4 Kg5 18.Pb4 Kh6 19.Kf5 Kh7 20.Kf6 Kg8 21.Ve3 Kh7 22.Kf7 Kh6 23.Pb5 Kh7 24.V5f4 Kh8 25.Vh6 Kh7 26.Pg5 Kh8 27.Ph5 mate.

Václav’s web site gives a great deal more, including numerical tables and further specimen wins.