The first thing to see is what happens if you barge straight ahead: 1.Kg3 Ke4 2.Kf2 Kf4 3.e3+ Kg4 and Black draws easily. But suppose it were Black's move if the position after 2...Kf4. Then he would be in trouble since 3...Kg4 4.Ke3! Kxh4 (4...Kf5 5.Kf3 Ke5 6.e4 is an easy win for White who simply has to choose his moment to dart in to g5 with the king) 5.Kf4! Kh3 leads to a position we shall reach later.
The position after 2...Kf4 is one of mutual zugzwang: with White to play it's a draw, but with Black to play White wins.
White must play with great subtlety: 1.Kg3! (remarkably, 1.Kg2 Ke4 or 1.Kh2 Kd4! lead only to a draw) 1...Ke4 2.Kg2!! Ke3 (White gets what he wants after 2...Kf4 3.Kf2) 3.Kf1 Ke4 (3...Kd4 4.Kf2 Ke4 5.e3 wins easily for White) 4.Ke1!! Ke3 (or 4...Kf5 5.Kd2! Kg4 6.Ke3) 5.Kd1! Kf4 6.Kd2Ke4 7.e3 Kf3 (or 7...Kf5 8.Kd3 Ke5 9.e4 Kf4 10.Kd4) 8.Kd3 Kg3 9.Ke4 Kg4 10.Ke5 Kxh4 (finally there is no choice) 11.Kf4! Kh3 12.e4 Kg2 13.e5! (13.Kg5? is met by Kg3!) 13...h4 14.e6 h3 15.e7 h2 16.e8(Q) h1(Q) 17.Qe2+ Kg1 (or 17...Kh3 18.Qg4+ Kh2 19.Qg3 mate) 18.Kg3 and wins.
The depressing thing is that if a position with only two kings and three pawns can be so complex, real chess must be very difficult indeed.
Two weeks ago, we posed a little sum: BANANA = ORANGE + APPLES and asked, if each distinct letterrepresents a different digit from 0 to 9, what was the value of GRAPES? Two possible answers are:
632045+288157=920202 so GRAPES=306579 or 4328857.
Several readers, however, have pointed out that apples plus oranges are very unlikely to make bananas. This week, therefore, we pose a different question entirely:
If ORANGE + GRAPES = BANANA, and each letter is a distinct digit from 0 to 9, what is the value of APPLES?
A Chambers Encyclopaedic Dictionary awaits the first correct answer opened on 14 September. Entries to: Saturday Pastimes, the Independent, 1 Canada Square, Canary Wharf, London E14 5DL.