Chess Strategy. Edward Lasker
consideration for the present and to devote my attention first to such combinations as involve questions of material. Let us master a simple device that makes most combinations easy both for attack and defence. It amounts merely to a matter of elementary arithmetic, and if the beginner neglects it, he will soon be at a material disadvantage.
Diagram 4 may serve as an example :
Diag. 4.
It is Black’s move, and we will suppose he wishes to play P—K 4. A beginner will probably calculate thus : I push on my pawn, he takes with his pawn, my Knight takes, so does his, then my Bishop takes, and so on. This is quite wrong, and means waste of time and energy.
When the beginner considers a third or fourth move in such a combination, he will already have forgotten which pieces he intended to play in the first moves. The calculation is perfectly simple upon the following lines : I play P—K 4, then my pawn is attacked by a pawn and two Knights, a Bishop and two Rooks, six times in all. It is supported by a Bishop, two Knights, two Rooks and a Queen, six times in all. Therefore I can play P—K 4, provided the six units captured at K 4 are not of greater value than the six white units which are recaptured. In the present instance both sides lose a pawn, two Knights, two Rooks, and a Bishop, and there is no material loss. This established, he can embark on the advance of the K P without any fear.
Therefore: in any combination which includes a number of exchanges on one square, all you have to do is to count the number of attacking and defending units, and to compare their relative values ; the latter must never be forgotten. If Black were to play Kt × P in the following position, because the pawn at K 5 is attacked three times, and only supported twice, it would be an obvious miscalculation, for the value of the defending pieces is smaller.1
Diag.5.
Chess would be an easy game if all combinations could be tested and probed exhaustively by the mathematical process just shown. But we shall find that the complications met with are extremely varied. To give the beginner an idea of this, I will mention a few of the more frequent examples. It will be seen that the calculation may be, and very frequently is, upset by one of the pieces involved being exchanged or sacrificed. An example of this is found in Diagram 6 ; Kt × P fails on account of R × B ; this leaves the Knight unprotected, and White wins two pieces for his Rook. Neither can the Bishop capture on K 5 because of R × Kt, leaving the Bishop unprotected, after which B × Kt does not retrieve the situation because the Rook recaptures from B 6.
Diag. 6.
Diag.7.
A second important case, in which our simple calculation is of no avail, occurs in a position where one of the defending pieces is forced away by a threat,the evasion of which is more important than the capture of the unit it defends. In Diagram 7, for instance, Black may not play Kt × P, because White, by playing P—Q 6, would force the Bishop to Kt 4 or B 1, to prevent the pawn from Queening and the Knight would be lost. A further example of the same type is given in Diagram 8. Here a peculiar mating threat, which occurs not infrequently in practical play, keeps the Black Queen tied to her K P 2 and unavailable for the protection of the B at B 1.
Diag. 8.
White wins as follows :
1 Kt × B, Kt × Kt; 2 R × Kt, Q × R; 3 Kt—B 7, ch K—Kt 1; 4 Kt—R 6 double ch, K—R 1; 5 Q—Kt 8 ch, R × Q; 6 Kt—B 7 mate.
We will now go a step further and turn from “acute” combinations to such combinations as are, as it were, impending. Here, too, I urgently recommend beginners (advanced players do it as a matter of course) to proceed by way of simple arithmetical calculations, but, instead of enumerating the attacking and defending pieces, to count the number of possibilities of attack and defence.
Let us consider a few typical examples. In Diagram 9, if Black plays P—Q 5, he must first have probed the position in the following way. The pawn at Q 5 is attacked once and supported once to start with, and can be attacked by three more White units in three more moves (1 R—Q 1, 2 R (B 2)—Q 2, 3 B—B 2) Black can also mobilise three more units for the defence in the same number of moves (1 Kt—B 4 or K 3, 2 B—Kt 2, 3 R—Q 1). There is, consequently, no immediate danger, nor is there anything to fear for some time to come, as White has no other piece which could attack the pawn for the fifth time.
Diag.9.
It would be obviously wrong to move the pawn to Q? after White’s R—Q 1, because White could bring another two pieces to bear on the P, the other Rook and the Knight, whilst Black has only one more piece available for the defence, namely, his Rook.
The following examples show typical positions, in which simple calculation is complicated by side issues.
In Diagram 10, the point of attack, namely, the Black Knight at K B 3, can be supported by as many Black units as White can bring up for the attack, but the defensive efficiency of one of Black’s pieces is illusory, because it can be taken by a White piece. The plan would be as follows : White threatens Black’s Knight for the third time with Kt—K 4, and Black must reply Q Kt—Q 2, because covering with R—K 3 would cost the “exchange,” as will appear from a comparison of the value of the pieces concerned. The “exchange” is, however, lost for Black on the next move, because White’s further attack on the Knight by Q—B 3 forces the Rook to defend on K 3, where it gets into the diagonal of the Bishop, which at present is masked by White’s Knight. The sequel would be 3 Q Kt × Kt ch, R × Kt (not B × Kt on account of B × R winning a whole Rook), 4 B × R, and so on.
Diag. 10.
A similar case is shown in Diagram 11.
Diag. 11.
Here, too, there is a flaw in the simple calculation, because the defending units are not secure. Beginners should devote special attention to this position, which is in practice of frequent occurrence.
It can be easily perceived that the Bishop cannot capture the pawn at B 7 on account of P—Q R 3. But to take with the Knight would also be an error, because Black would then keep chasing away the covering Bishop.
1 P—Kt 4 ; 2 B—Q 6, K—B 3 ; 3 Kt—K 8, B—B 2 ; and wins one of the pieces.
Finally, one more example, in which one of the defending pieces being pinned makes simple calculation impracticable.
In Diagram 12 it seems at first sight as if Black could play Kt × P : although White can pin the Knight with R—K 1 and then attack it once more with his Knight, Black would appear to have sufficient protection available, with his Kt and B. White has no time to double Rooks, because if he does so, after his R—K 2 Black would play the King away from his file and allow the Knight to escape.
Diag. 12.
But White can, by a simple sacrifice, bring the slumbering R at R 1 into sudden action :
1...Kt × P; 2 R—K 1, B—B 4; 3 Kt—B 3, Kt—Q 3; 4 R × Kt, Kt × R ; 5 R—K 1, and White wins two pieces for his Rook.
These illustrations will be sufficient