NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/rec/games/chess/12000

Internet Message Format | 1992-12-26 | 3.7 KB

Path: sparky!uunet!usc!zaphod.mps.ohio-state.edu!malgudi.oar.net!uoft02.utoledo.edu!dcrosgr
From: [e-mail]
Newsgroups: rec.games.chess
Subject: Re: It is legal to play chess in Yugoslavia
Message-ID: <[e-mail]>
Date: 25 Dec 92 23:24:03 EST
References: <[e-mail]> <[e-mail]> <[e-mail]> <[e-mail]>
Organization: University of Toledo, Computer Services
Lines: 59

In article <[e-mail]>, [e-mail] (John Black) writes:
> In article <[e-mail]> [e-mail] writes:
>>In article <[e-mail]>, [e-mail] (John Black) writes:
>>> In article <[e-mail]> [e-mail] (Paul Rubin) writes:
>>>>    > Obligatory chess question:  Does there exist a chess position in
>>>>    > which Black has enough material to mate White in theory, but Black
>>>>    > cannot mate White even with White's cooperation, while on the other
>>>>    > hand White can mate Black without Black's cooperation?
>>>>
>>However, it depends on "has enough pieces". Obviously, neither side has enough
>>pieces to mate the other, and yet is they were in a different position, they
>>could mate. (So to speak.) Interesting problem. Any other solutions?
>>
>
> Well, mine wasn't truly a solution since White could not force a win.  But
> maybe the idea (Paul Rubin's, I think?) of giving Black two same-colored bishops
> will work.  Try this: Black has two black-squared bishops and White has two
> queens.  Now, Black has enough *material* to mate White in theory (i.e. if the
> bishops were on different colored squares and White were kind enough to keep
> his queens out of the way, then a mate could be arrived at).  But with the
> bishops both on the black squares, no mate can be constructed even with White's
> help.  And it's pretty obvious that even without Black's cooperation in this
> position, White will force a win (assuming a starting position where White does
> not immediately lose material).
>
> john//
> [e-mail]

I dunno. I think people are relying too much on math (points required for a
mate to be possible) with pieces required for a mate.

Look at it another way. 32 degrees F. is NOT the freezing point of water, it is
the melting point of ice. Important distinction, as you can have pure water
supercooled to -20F. (Provided nothing disturbs it. If you tap the beaker
however, ice crystals start to form along the shock wave and, from there, the
whole thing starts to solidify. Therefore, you have a range of about 52F where
you might have water, and you might have ice. Without looking at the beaker,
you don't know which you have.

These chess hypos are similar. For example, while both sides have enough
'points' to possibly force a mate, or even be subject to one, the board does
not permit it. In short, the reality of the board over-rides the theoretical
possibility of there being a mate. Therefore, there is a range of points a
player may have, where, depending upon the exact type of pieces, and their
poistions either a mate (pursuant to your rules of this problem) is possible,
or it is not possible. The system of assigning points to the pieces and stating
that you need this many to mate the other side is like the ice range above.
Above X points (I'll let another figure it out...) and provided your opponent
'helps' there is no possibility of not being able to mate.

Of course, if you have less than five points, not even God can help you win.
(Barring the damned pawn promotion.)

Your situation with all the pawns offset is, I think, the upper limit. (of
course, your opponent must also have the same number of pawns.)