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The Independent Online
Not many players realise - though they all would if they stopped to think about it - that king and one black-squared bishop win against king and 32 black-squared bishops (and you can even add 28 rooks in the latter case, as long as you leave the empty square next to the overburdened king).

The total uselessness of 32 black-squared bishops raises the question of what precisely is the optimum number of same-coloured bishops. The above study by Alexei Troitsky (first published in 1905) gives a surprising example in which five of them come in handy. It is White to play and win.

At first glance, it looks impossible, because White has only his king that can control a white square.

There is only one hope: to persuade Black to advance his pawn to a2, then force him to play Ka1 when White has a bishop on the long diagonal and his king on c3. White will then be able to play Kc2 mate. For this to work, however, he will need a bishop on a1 to prevent the black pawn from advancing there itself.

OK, here goes: 1.B(c7)e5 a5 (or 1...Ka2 2.Kc2 a5 3.Ba1 a4 4.B(b8)e5 a3 5.Kc3 leading back to the main line) 2.Ba1 a4 3.B(b8)e5 Ka2 (best) 4.Kc2 a3 5.Kc3! (necessary to prevent stalemate) 5...Kxa1 6.Kb3+ Kb1 7.Ba1 a2 8.Kc3 Kxa1 9.Kc2 mate!

The remarkable thing is that White needs all five of his bishops: two have to be sacrificed on a1, one is needed to cover the c1 square, a fourth has to lurk on the a3-f8 diagonal to prevent Black's king from moving to a3 at a crucial moment, and the last one has to be in place on the long diagonal to deliver the final discovered mate.

We all know about the power of major pieces tripled on an open file. This position is the only known example of bishops tripled on an open diagonal.