| |
Main
Date: 06 Aug 2008 23:01:10
From: John Salerno
Subject: Help me understand this scenario re: undermining
|
Here's the situation from the book "Learn Chess" (p. 36):
White: R on e1, B on b5, pawn on e5
Black: B on d7, B on g7, N on c6
It is assumed that White has just played 1 dxe5, which is where the pawn
stands now. The example is meant to show two different methods of
recapturing the pawn on e5 in order to illustrate the idea of undermining.
The first method is:
1 ... Bxe5
2 Bxc6 Bxc6
3 Rxe5
(White "wins a piece")
The second method is:
1 ... Nxe5
2 Rxe5 Bxe5
3 Bxd7
(White "wins two pieces for a rook")
The book goes on to say "Notice that it pays to sacrifice the exchange
temporarily in this second variation ... and that this sacrifice only
gains material since the capture by the rook protects the white KB."
But I'm confused. The above quote seems to suggest that the second
method is better for White. But doesn't White actually end up losing
material this way? Yes, he is still ahead in the end, but by a lesser
amount.
The way I see it is this: in the first method there's an exchange of
bishops and White wins a knight, an advantage of about 3.5 units. In the
second method, White captures a bishop and knight (3.5 + 3.5 = 7) but
loses a rook (-5) for a net gain of 2 units.
Am I missing something here? Am I wrong in thinking that the authors are
promoting the second method, or is there more to the second method that
I'm missing? Am I missing something in my analysis or calculations?
It seems that even though with the second method you gain two pieces for
the price of one, the *value* of those pieces makes the first method
more desirable for White. Because in fact, aren't you also winning two
for one in the first method? I don't understand why the authors
distinguish the second method by saying White has won two pieces for a
rook (as I quoted beneath the moves). Couldn't you just as easily say
about the first method that White has won two pieces for a bishop?
Thanks!
|
|
| |
Date: 06 Aug 2008 22:06:05
From: help bot
Subject: Re: Help me understand this scenario re: undermining
|
On Aug 6, 11:01=A0pm, John Salerno <[email protected] > wrote:
> Here's the situation from the book "Learn Chess" (p. 36):
>
> White: R on e1, B on b5, pawn on e5
> Black: B on d7, B on g7, N on c6
>
> It is assumed that White has just played 1 dxe5, which is where the pawn
> stands now. The example is meant to show two different methods of
> recapturing the pawn on e5 in order to illustrate the idea of undermining=
.
>
> The first method is:
>
> 1 ... =A0Bxe5
> 2 Bxc6 Bxc6
> 3 Rxe5
> (White "wins a piece")
>
> The second method is:
> 1 ... =A0Nxe5
> 2 Rxe5 Bxe5
> 3 Bxd7
> (White "wins two pieces for a rook")
>
> The book goes on to say "Notice that it pays to sacrifice the exchange
> temporarily in this second variation ... and that this sacrifice only
> gains material since the capture by the rook protects the white KB."
>
> But I'm confused. The above quote seems to suggest that the second
> method is better for White. But doesn't White actually end up losing
> material this way? Yes, he is still ahead in the end, but by a lesser
> amount.
I don't think that is what the writer meant.
*Black* made the choice to capture such
that White had to sacrifice the Rook in
order to get his two Bishops. It "pays" to
give up the Rook, because in return you
gets both Bishops.
> The way I see it is this: in the first method there's an exchange of
> bishops and White wins a knight, an advantage of about 3.5 units. In the
> second method, White captures a bishop and knight (3.5 + 3.5 =3D 7) but
> loses a rook (-5) for a net gain of 2 units.
Math trouble: if you use a scale in which
a Knight is worth a whopping 3.5 points,
then you need to stretch the Rook's value
to that same scale. In fact, you also need
to stretch the scale to match the Bishop's
value relative to the Knight.
Moreover, you used a very low value of
5.0 for the Rook, but the exceedingly high
value of 3.5 for the Knight, and these just
don't work /together/. The R=3D5 fits the
scale where a Q=3D9 and B=3D3, N=3D3. On
any scale which puts a Knight at 3.5, all
the other /pieces/ must be re-calibrated
higher to match.
> Am I missing something here? Am I wrong in thinking that the authors are
> promoting the second method, or is there more to the second method that
> I'm missing? Am I missing something in my analysis or calculations?
I would say that the authors crafted a rather
poor example. First of all, they need to place
two Kings somewhere on the board. Next,
the loss of a pawn is *not* trivial here--
especially if it is the last pawn!
> It seems that even though with the second method you gain two pieces for
> the price of one, the *value* of those pieces makes the first method
> more desirable for White. Because in fact, aren't you also winning two
> for one in the first method? I don't understand why the authors
> distinguish the second method by saying White has won two pieces for a
> rook (as I quoted beneath the moves). Couldn't you just as easily say
> about the first method that White has won two pieces for a bishop?
They seem to be quite clumsy in that
regard. There is a significant difference
between trading, say, two Knights for
a Rook and pawn, and trading some
other combination of "pieces". The
positions of the Kings is important in
such endings, and it is crucial to know
exactly what is left on the board, and
what is the theoretical outcome of
such a combination.
If this is not an ending, but rather a
tactical exchange in a completely
unknown situation, then they still erred
in not counting the pawn in their
summations (i.e. White "wins two
pieces for a rook"). If the two pieces
are Knights, you may well have "won"
nothing at all in trading for a Rook and
pawn-- depending on the position. And
in some cases, this could also apply
to Bishops.
-- help bot
|
| | |
Date: 08 Aug 2008 18:49:14
From: help bot
Subject: Re: Help me understand this scenario re: undermining
|
On Aug 8, 6:25=A0pm, John Salerno <[email protected] > wrote:
> >> I was using the books numbers. They assign 5 to rooks and 3.5 to bisho=
ps
> >> and knights.
>
> > =A0 Then they are as loony as a three dollar bill.
> > Here is the simplest scale:
>
> > pawn =3D 1
>
> > Knight =3D 3
>
> > Bishop =3D 3
>
> > Rook =3D5
>
> > Queen =3D 9
>
> Well thanks for that. It's easier to add up the whole numbers anyway! :)
Note that "beginners" tend to have trouble
in stopping passed pawns from promoting
so the simple scale above, which tends to
over-value pawns a bit, works out nicely.
Beginners also tend to overlook Knight-
forks, so it's okay to treat Bishops and
Knights as equal in value... even though
the Bishop is really superior.
Anyhow, the real key to improvement is
in "seeing" the tactics, not in perfect math
or in perfect valuations. I am thinking of a
famous world championship contest in
which Tigran Petrosian was playing Boris
Spassky, and one of them won by offering
the exchange, which the other grabbed; in
that game, had the eventual loser merely
captured in reverse order -- with his more
valuable man first -- he would have won a
pawn for nothing! Instead, he captured in
what must have seemed the logical way,
less-valuable man first, gained material
but lost the game because of his
resulting bad/difficult position. This was
all of two moves deep, yet the two
strongest players in the world *both*
mucked it up! Thus, there is hope for us
mere mortals, after all.
-- help bot
|
| | |
Date: 08 Aug 2008 17:04:20
From: SBD
Subject: Re: Help me understand this scenario re: undermining
|
Yet, B+P very often = R in the ending, whereas more often N+2P is
needed to =R.
Maybe Fischer's 3.25 for B makes sense then, since anything less than
+1.0 is difficult to convert, particularly late in the game.
|
| | |
Date: 07 Aug 2008 20:45:14
From: help bot
Subject: Re: Help me understand this scenario re: undermining
|
On Aug 7, 6:38=A0pm, John Salerno <[email protected] > wrote:
> I was using the books numbers. They assign 5 to rooks and 3.5 to bishops
> and knights.
Then they are as loony as a three dollar bill.
Here is the simplest scale:
pawn =3D 1
Knight =3D 3
Bishop =3D 3
Rook =3D5
Queen =3D 9
----------------------------------------------------------
Any scale which values the Knight at 3.5
is s-t-r-e-t-c-h-i-n-g things just a bit, and in
the process giving pawns short shrift. On
such a scale the Rook could not possibly
be worth only 5.0 points. Note that three
minor pieces need to add up to the value of
the Queen, thus: 3.5 + 3.5 + 3.5 =3D 10.5.
That would mean the Queen is worth more
than two Rooks, which is nonsense unless
you have a certain type of position.
A long time ago, I discovered why it was
that world champion Tigran Petrosian so
often sacrificed the exchange. It turns out
that he used a scale from some ancient
manuscript that put the Rook's value at
only 4.5 pawns! So you see, when he did
his "sacrificing", he wasn't giving up any
material on /his/ scale of piece values. It
is so much easier to sacrifice a small
(4.5) Rook, than a big (5.5) one, you see.
Of course, everything depends on the
position-- just as with Marvin's gamma
ray gun on Mars.
-- help bot
|
| | | |
Date: 08 Aug 2008 18:25:49
From: John Salerno
Subject: Re: Help me understand this scenario re: undermining
|
help bot wrote:
> On Aug 7, 6:38 pm, John Salerno <[email protected]> wrote:
>
>> I was using the books numbers. They assign 5 to rooks and 3.5 to bishops
>> and knights.
>
> Then they are as loony as a three dollar bill.
> Here is the simplest scale:
>
> pawn = 1
>
> Knight = 3
>
> Bishop = 3
>
> Rook =5
>
> Queen = 9
Well thanks for that. It's easier to add up the whole numbers anyway! :)
|
| | |
Date: 07 Aug 2008 18:38:37
From: John Salerno
Subject: Re: Help me understand this scenario re: undermining
|
help bot wrote:
> Math trouble: if you use a scale in which
> a Knight is worth a whopping 3.5 points,
> then you need to stretch the Rook's value
> to that same scale. In fact, you also need
> to stretch the scale to match the Bishop's
> value relative to the Knight.
>
> Moreover, you used a very low value of
> 5.0 for the Rook, but the exceedingly high
> value of 3.5 for the Knight, and these just
> don't work /together/. The R=5 fits the
> scale where a Q=9 and B=3, N=3. On
> any scale which puts a Knight at 3.5, all
> the other /pieces/ must be re-calibrated
> higher to match.
I was using the books numbers. They assign 5 to rooks and 3.5 to bishops
and knights.
> I would say that the authors crafted a rather
> poor example. First of all, they need to place
> two Kings somewhere on the board. Next,
> the loss of a pawn is *not* trivial here--
> especially if it is the last pawn!
That's my fault! I left out all the other pieces on the board because it
didn't seem relevant. They don't come into play in the details of the
example, but perhaps it's wrong not to list them as well.
|
| | | |
Date: 08 Aug 2008 08:17:56
From: Chess One
Subject: Re: Help me understand this scenario re: undermining
|
"John Salerno" <[email protected] > wrote in message
news:[email protected]...
> help bot wrote:
>
>> Math trouble: if you use a scale in which
>> a Knight is worth a whopping 3.5 points,
>> then you need to stretch the Rook's value
>> to that same scale. In fact, you also need
>> to stretch the scale to match the Bishop's
>> value relative to the Knight.
>>
>> Moreover, you used a very low value of
>> 5.0 for the Rook, but the exceedingly high
>> value of 3.5 for the Knight, and these just
>> don't work /together/. The R=5 fits the
>> scale where a Q=9 and B=3, N=3. On
>> any scale which puts a Knight at 3.5, all
>> the other /pieces/ must be re-calibrated
>> higher to match.
>
> I was using the books numbers. They assign 5 to rooks and 3.5 to bishops
> and knights.
Its said that Fischer used Greg's system above of R=5, N=3.0 but with this
adjustment; B=3.25
Of course, this is just rough calculation, ie what is the value of having 2
bishops [is the worth of the bishops greater then?], or 2 bishops against 2
Knights? What about having bishops against knights when the position is
closed?
Sometimes bishops are 'bad', ie; its hard for them to get into the game, and
effectively they serve no more use than a pawn=1.
But Greg is correct, if the R=5, then 3.5 for the bishop [where the pawn =
1] is considered too much.
Other rules-of-thumb which have numeric values on Greg's system are for
example, 3 moves [or tempi] = 1 pawn. Some people would evaluate having the
initiative as worth 0.5 to 1.0 pawn.
But all other things being equal, then in the late middle game, what is the
value of passed center pawns compared with passed rook pawns? Formations
like a supported passed center pawn on the 6th or 7th rank also adjust the
worth of the pawn.
As the game progresses the nature of the position tends to be more important
than initial values of the pieces, and adjusts their values, sometimes
radically. Therefore Greg's suggested values are usually good up to the
middle of the middle game, and therefore, swapping 2 bishops for a rook and
two pawns is usually considered to be losing a pawn or a point.
These evaluations with some adjustments are probably used by all
chess-players - and give chess programmers big headaches, since the scale is
/only/ rule of thumb, and as the game progresses and the overall position
changes the values, evaluating for position /too/ is far more complicated.
Back to the top: 2 popular tactical motifs for 'undermining' go by the names
(a) Deflection/Removing the Guard, and also (b) Decoys.
Phil Innes
|
|
