It is amazing what complicated variations can be produced in such situation.
Red to play and draw as demonstrated in Main Variation.
Based on the ancient book of collected problems “渊深海阔”, as edited and analysed by leading experts Mr. Liu Guobin and Mr Zhu Baowei.
Main Variation
1. rR=6 C5=4 (variation 1)
2. R6=4 (var2) P6+1
3. R4-2 P5=6
4. K4=5 C4=8 (variation 3)
5. R1=5 B1+3 (variation 4)
6. C1+9 B7+9
7. C1=5 (var5) C8=5
8. C5-2 B3-5
9. R5+2 R4=5
10.R5-6 P6=5
11.K5+1 (draw)
What would happen had Black played 1. ... R4-2 and captured Red Rook? (variation 1)
Variation 1
1. ... R4-2 ?
2. R1=6 R4-2
3. C1+9 B7+9
4. C2+9 checkmate
Again, look at variation 2, had Red played 2. R1=4 instead of the text move.
Variation 2
2. R1=4 ? P5=6
3. K4=5 R4=5 checkmate
Variation within a variation! In the above variation 2, what can happen if Red plays 2. R6+4 instead of 2. R1=4 ?
Variation of a variation
2. R6+4 R4-6
3. R1-4 R4=8
4. C2=3 R8+7
5. R1=3 P6+1
6. R3=4 R8=7 checkmate
Look at variation 3, what can happen if Black played 4. ... R4=5 instead of 4. ... C4=8?
Variation 3
4. ... R4=5
5. K5=6 R5-7
6. R1=4 R5=8
7. C2=5 (note a) P6=5
8. R4+2 P5+1
9. K6+1 R8+7
10.K6+1 R8-1
11.K6-1 C4+5
12.K6=5 K4+1
13.R4+1 K4-1
14.C1+9 R8-7
Black having the advantage.
In the above variation, Red at move 7th (note a) played C2+5 instead of C2=5, now we see what can happen:
Variation 3 (note a)
7. C2+5 P6=5
8. R4=5 R8+2
9. R5-4 R8+1
With two more Bishops, Black is in better situation.
In variation 4, if instead of the correct 5. ... B1+3, Black played 5. ... C8=5?
Variation 4
5. ... C8=5?
6. R5=6 R4-4
7. C1+9 B7+9
8. C2+9 checkmate
(Var 5) Red plays 7. C1=5
7. C1=5
Reinforcing the defence in the centre, threatening to capture the Bishop followed by a backrank checkmate!
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