continued from part I, part II and part III. . .

Beancounting revisited.

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.

Move 1.

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.

Move 2.

2.Nxc6

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

Move 3.

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,

five pounds.

Coming up with a "formula" for situations like this seems pretty challenging, and would probably be pretty hard to remember even if you could. There are couple of guidelines that may help, though: I recall hearing (from Heisman I think) that if there are a series of forcing moves (checks, captures) available on the board, it is generally better to be the one who makes such a move first, because the mover then has the initiative and the opponent is forced to respond. In this case, 2.Nxc6 gives the initiative firmly to black, who, with 2...NxQ, forces white to play 3.NxQ, then black is lost, for after 3...NxB, either of white's temporary "equalizing" moves (4.PxN or 4.NxB) loses since black gets to capture *last* as well (after an in-between check in one case to get the knight to safety).

ReplyDeleteSo in summary, if you can capture first and see a way to capture last as well, that's usually a very good sign to start capturing. Obviously the material in question does matter, and there the "count just one side at a time" technique can help.

"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."

ReplyDeleteI really like this- the emphasis on the need to see

whetherthere is any reason to continue the sequence. For many people think there would be a sort of natural law to continue chopping wood until the entire forest is cut to the ground.Concernng your example: Such "parallel chains" are problematic for me as well. What I am trying to do: The chains are parallel as long as the one who initiates it doesn't decide to cut them.

So e.g. we have:

1...Nxe3

2.Nxc6 bxc6 (cut!)

3.fxe3

or 1...Nxe3

2.Nxc6 Nxd1

3.Nxd8 Raxd8 (cut!)

4.Raxd1

and so on. This doesn't help by itself but we know that

whenever the chain is cut off this way the end-result has to be equal.(material in balance)So what we have to look for is the end of the symmetry of the chains, which will inevitably come (be it by giving check as here or by existence of a piece of a different value in the two chains). This means, up to then its only a visualisation problem (+checking out each time if symmetry is upheld) and not at all a counting problem (other than comparing values of each chain-member "locally").

Using this method we would proceed in this example in the following way: Consider the chain initiated by Nxe3: Bishop vs Knight, symmetry is ok, now queen vs queen, symmetry broken only with check on d7, which is nonsense, because queen is gone, so symmetry ok again, now knight versus bishop, symmetry ok, now nothing more to take (except pawn) so look closer for end of symmetry: Yes, end of symmetry. By our law (put in italics in a preceding paragraph) we know that whatever the material balance might look like it has to be equal before our next move (so no need to add up). Now we visualize the situation and realize that both knights are threatened but black can say check and then take white's knight.

kind regards,

svensp

Waaek,

ReplyDeleteAlthough it not may look so, I'm not really interested in the formula's itself. The challenge is to find methods to prevent you from the necessity to see everything before the minds eye. Especially alternating sequences causes an immediate short term memory overload by me. That I seek to prevent.

If I may, I'd like to try to add a constructive comment for a change. :)

ReplyDeleteFor the chains stuff, I too have a problem going back and forth between Whites 1st capture, then Blacks 1st, then Whites 2nd etc.

So, at some point (once I have figured out I am looking at such a position) I look at what the critical sequence is for Black and the critical sequence for White independently (but yes, I keep my eyes open for interactions between them). I count to myself One, Two, Three, Four as I visualize Blacks four moves in your example. And I count One, Two, Three as I visualize Whites three moves in your example. I sum the value of wood captured in each sequence and visualize the final composite image. I stop at Blacks 4th, in this example, as the check stops the chain. I'll also independently verify that the Ne2 is not trapped.

I do that in addition to evaluating the sequence on an alternating black/white moving basis. But looking at and verifying the sequences independently helps me to compose them together.

Using both approaches helps my validation and is useful as a blundercheck.

Svensp,

ReplyDeletegood analysis. ONLY by a duplo-move you can gain wood. In this case Ne2+ with the double fuction escape and check.

Glenn,

ReplyDeleteinteresting idea's, I will have a look at that.

Thanks for the lesson, Mr. Bean.

ReplyDeleteI do exactly what Glenn does. Otherwise I get very confused.

ReplyDelete