continued from part I, part II and part III. . .
Let me try to reformulate my method of beancounting in more understandable english.
Goal of the beancounting formula: To evaluate if you should start a sequence of capture and recapture.
- If you start to capture or in this case you evaluate whether to capture or not, start with your piece with the lowest value and work upwards. Your opponent will do the same.
- Situation: You have attackers and start to capture. The opponent has a victim and defenders of that victim.
- Take the sum of the value of your attackers. If there are more attackers than there are defenders, then there is no need to take the value of all your attackers. Since not every attacker will be involved in the trade. Add up the value from as much attackers as there are defenders. Example: 5 attackers, 3 defenders = take the sum of your 3 lowest valued attackers.
- Take the sum value of all the pieces who take part in the defense.If there are more defenders than there are attackers, there is no need to take the value of all your defenders. Since not every defender will be involved in the trade. Add up the value from as much defenders as there are attackers. Example: 5 defenders, 3 attackers = take the sum of the 3 lowest valued defenders (start with the victim!).
- If the value of your opponents pieces exceeds the value of your attackers you can safely start the capture sequence.
- If your first attacker has a lower value than the victim you will gain wood with the very first move. After that you can stop the sequence and re-evaluate if there is any reason to continue with the capture sequence.
- Your opponent will stop to recapture when he sees that he cannot hold his victim. So you will never gain more than the value of the victim.
- When a queen is in front of her rooks or her bishop, the rear piece has no direct access to the square of the victim. This makes that the formula above cannot be applied. This is common knowledge of which everybody is aware (don't put your queen in front of your rooks). In stead of complicating the simple rules above you have to think for yourself here. Don't worry, it concerns probably less than 1% of the cases.
- When the victim is a pawn defended by a pawn. This is the only case where your opponent can stop the sequence short and it costs you wood. The reason for this is that if your attacker is at least a light piece (value=3), the sum value of the victim plus its first defender is only 2. Again not something to worry about. Since you will not have the inclination to take a pawn that is protected by a pawn with a light or heavy piece without good reason anyway.
Chains of capture.
We have found how the beancounting formula can be applied in situations where there is more than one theatre of action. Now I want to investigate if the application of the formula can be
extended to positions where there are chains of captures on both sides. I don't know beforehand if this is going to lead somewhere so please bear with me. This is the startposition which I composed:
Black to move.
Black plays the best move according to Rybka 1. ... Nxe3
How should white proceed?
Of course we see in a flash according to the beancounting method that the white bishop on e3 was well protected. A grandmaster wouldn't have counted his beans faster:)
There are 2 moves to consider for white:
- Immediate takeback with 2.fxe3
- Counterattack with 2.Nxc6
Nxc6-Nxd8-Nxe6-Nxf8 collected 20 points of wood, investment 1 knight.
For black: Nxe3-Nxd1-Nxc3 collected 15 points, investment 1 knight.
But this kind of beancounting doesn't work this way. Let's see what happens.
Black has 3 choices:
- Retract the knight, e.g. with 2. ... Nd5
- Takeback the white knight with 2. ... bxc6
- Continue the attack with 2. ... Nxd1
Theoretically white has 3 choices:
- Retract the knight, e.g. with 3.Ne5
- Takeback the black knight, e.g. with 3.Rxd1
- Continue the attack with 3.Nxd8
Now black has 3 options:
- Retract the knight with 3. ... Nxb2
- Takeback the white knight with 3. ... Rxd8
- Continue the attack with 3. ... Nxc3
White has 3 options:
- Retract the knight with 4.Nxb7
- Takeback the black knight with 4.bxc3
- Continue the attack with 4.Nxe6
Now black all of a sudden wins a piece with 4. ... Ne2+ 5.Kh1 fxe6
Wow, where does that came from?
This kind of positions I find difficult to handle in practice. There are two things killing me when I try to find the solution by projecting both chains before the minds eye:
- Every ply adds 3 possibilities, so the total amount of possibilities adds up to 15 moves between which you have to decide.
- I have to alternate my attention between two different sides of the board, and doing two book keepings.
If I look at the moves, which time and again have 3 options (retract, takeback and continue), there seems to be an indication that there is some kind of system hidden in it. Let's see what we can find.
Black takes first. White counterattacks and and restores the material balance. Although the chain of black is shorter, he can end his chain with a duplo move: 4. ... Ne2+
This move accoplishes two things:
- It saves the black knight.
- It checks the king.
So the parameters of generalisation are:
- The lenght of both chains.
- The values of the pieces in the chain
- Are the pieces in the chain protected (beancounting!)
- What influences the decision to retract, to takeback or to continue the chain
- The availability of a duplomove along the way. This decides if wood is going to be won.
to be continued. . .
To stop the train,
in cases of emergency,
pull on the chain,
pull on the chain,
penalty for improper use,