Chess Puzzle
John Woolley
may have thought of the same thing.
Can a situation exist in which the only correct move is to
move a Pawn to the eighth rank and promote it to a Rook? To a
Bishop? If so, construct such a situation. If not, prove it.
I'll post my solution(s) in a few days.
jrw
>From: j...@evolving.com (John Woolley)
>Subject: Chess Puzzle
>Date: Mon, 29 Aug 1994 22:59:38 GMT
SPOILER BELOW
A quick guess -- when promoting the pawn to a queen would result in a
stalemate, same for a rook. I can't remember if you are required to promote a
pawn on the eighth rank.
STRIJBOS M.M.
> I'll post my solution(s) in a few days.
>
By the way , CAN you play chess at all??
Michael Wolfe
>In article <1994Aug29.2...@evolving.com> j...@evolving.com (John Woolley) writes:
>>From: j...@evolving.com (John Woolley)
>>Subject: Chess Puzzle
>>Date: Mon, 29 Aug 1994 22:59:38 GMT
>
>>This is an original puzzle (i.e. I made it up). Someone else
>>may have thought of the same thing.
>
>>Can a situation exist in which the only correct move is to
>>move a Pawn to the eighth rank and promote it to a Rook? To a
>>Bishop? If so, construct such a situation. If not, prove it.
>
>
The proposed answer, along with my possible spoiler, appear below.
>A quick guess -- when promoting the pawn to a queen would result in a
>stalemate, same for a rook. I can't remember if you are required to promote a
>pawn on the eighth rank.
As usual, there is always some ambiguity in the problem statement. What
does "the only correct move" mean? If you take it to mean "the
BEST move," then there is at least one scenario in which promotion to a
rook would be better than promotion to a queen:
Black King on a7 (sorry if you don't like the notation).
White King on c6, White Pawn on c7.
This is a situation in which promotion to a queen would produce a
stalemate. By promoting to a rook (c7-c8R), black's only response is
a7-a6, and white mates with c8-a8. Promoting to a rook, as it turns out,
is white's fastest route to mate in this situation, but it is NOT the
only route. For example, white could promote the pawn to a queen after
moving the king c6-d7. Again, it depends on how you interpret "correct."
I have been trying to concoct a similar situation for promotion to a
bishop but have been unsuccessful so far. I am beginning to believe that,
because there has to be another piece on the board (otherwise, promoting
to a bishop would be useless), it might always be better to avert a
stalemate first by moving the other piece and then to queen the pawn. Of
course, I haven't been able to prove this either, so I am eagerly awaiting
more solution ideas.
--Michael Wolfe
--The Institute for the Learning Sciences
--Northwestern University
--wo...@ils.nwu.edu
Dick Adams
> Can a situation exist in which the only correct move is to
> move a Pawn to the eighth rank and promote it to a Rook? To a
> Bishop? If so, construct such a situation. If not, prove it.
Analysis follows:
ANALYSIS:
First, the legal moves of Rooks and Bishops are subsets of the
set of legal moves of the Queen. Therefore, one intuitively
obvious reason for promoting to a piece with a lesser set of
moves would be to avoid stalemate.
Second, the legal moves a Knight are not included in the set
of legal moves of the Queen. Therefore, it should be possible
to construct a situation where promotion to a Knight could
cause or directly contribute to a Checkmate.
Dick
Bram Cohen
>
> Can a situation exist in which the only correct move is to
> move a Pawn to the eighth rank and promote it to a Rook? To a
> Bishop? If so, construct such a situation. If not, prove it.
Lightning just struck, so here's my solution (by *far* the best chess puzzle
I've ever come up with.)
White pawns on a7 and f5. White bishop on h1 and white king on h8.
Black king on f8 and black queen on d8. White to play and win.
This is a really interesting puzzle, so you might want to analyze it before
reading the solution below.
If white now does the obvious 1. a8=Q?? then black plays 1 ... Kf7+!! which
is followed by either 2. Qd8 stalemate or 2. Kh7?? Qh4 mate
The only other ways of avoiding 1 ... Qh4 mate are f6, Kh7, Bf3, and a8=R.
f6 and Bf3 allow black to mate easily. If white plays 1. Kh7 then
1 ... Qh4+ 2. Kg6 Qh1 prevents the a-pawn from promoting.
a8=R is the best move. White can then easily trade the rook and queen, which
yields a pawn and bishop versus king, which is an easy win.
Note that underpromotion is the *only* winning move.
The position also looks fairly natural, so a puzzle could probably be
constructed where this position arises after a prior combination, making
the underpromotion much less obvious.
I couldn't find a solution for the bishop version.
--
___ ___ / \____ ############################
/ \_/ \ \_ __ \ # Bram Cohen #
/ \_/ / \/ # bco...@acsu.buffalo.edu #
/\ ___/ # Just trying to be myself #
Mintardjo W.
John Woolley <j...@evolving.com> wrote:
>This is an original puzzle (i.e. I made it up). Someone else
>may have thought of the same thing.
>
>Can a situation exist in which the only correct move is to
>move a Pawn to the eighth rank and promote it to a Rook? To a
>Bishop? If so, construct such a situation. If not, prove it.
>
I think the question is more like if there is any position in which one
can win by promoting a pawn to either a rook or a bishop, then one
inherently can also win by promoting a pawn to a queen or none at all.
Prove that it is true. If not, construct the position that gives the
counterexample.
Here is what I think a position in which you can only win by promoting
pawn to rook:
+ * + * + * + *
* P * + * + * +
+ * + * + * + *
* + * + * + * +
p k + * + * + *
* + * + * + * +
+ p + K + * + *
* + * + * + * +
If black plays 1. ... b1=Q, then white should play 2. b8=R. If white promotes
pawn to queen, then black can reply with 2. ... Ka3 and white loses the
chance of taking black's queen at b1
Here's a position where promoting pawn to bishop is needed:
+ * + * + * + k
* + * + * + * P
+ * + * + P + K
* + * + p b * +
+ * + * N * + *
* + * + * + * +
+ * + * + * + *
* + * + * + * +
If black doesn't play 1. ... Bxh7, white can force a win by promoting pawn
to queen or by checkmate at f7. Otherwise, white plays 2. f7 and here black
forces white to promote pawn to bishop with 2. ... Bg8. The next three
white moves are: 3. f8=B, 4. Nf6 and 5. Bg7++ (ignoring black's moves)
In both cases, it still seems to me that one could have played with something
much more accurate and thus avoid these setup. I would like to see a position
that comes from any tournament games. Does it ever happen?
---
Mintardjo W.
Noam Elkies
In article <33u3ea$2...@news.xs4all.nl> abi...@xs4all.nl (Martian) writes:
>j...@evolving.com (John Woolley) writes:
>
>==Can a situation exist in which the only correct move is to
>==move a Pawn to the eighth rank and promote it to a Rook? To a
>==Bishop? If so, construct such a situation. If not, prove it.
>
>If you define "correct move" as a move which doesn't lose, or
>not giving up a winning position, there are many examples.
>A famous problem is a problem where white can promote a pawn on
>b8 (or c8) to queen, rook, bishop (knight too? I don't have the
>problem in front of me) and the only defence for black is promote
>a pawn on b1 to the same rank.
Yes, Knight too, the famous Babson Task. But that's to achieve
mate in 4, not merely win; a full Babson in a win or draw study
has yet to be achieved. However, Rook and Bishop promotions
are known; here's an article I posted several times about this:
jpk2581 [Joe], in article <1991Oct24...@tesla.njit.edu>:
>Everybody has seen positions where the correct
>move is to promote a pawn to a queen, and it's
>also pretty common to need to promote to a knight,
>but can you come up with two positions, in the
>first the ONLY winning move being to promote to
>a rook and the second the ONLY winning move being
>to promote to a bishop. The rook problem is easy
>and the bishop problem is moderately difficult.
The key in both cases is stalemate --- the only circumstance
where one might prefer to have less power (B or R as opposed to Q).
This can be either to get the underpromoter stalemated (by bottling
up or pinning the promoted piece) or, more commonly, to avoid giving
stalemate.
For Rook promotions the canonical example is the justly famous
Saavedra position, which concludes thus: White King on c2 and pawn
on c7; Black King on a1 and Rook on d4. White to move must avoid
promoting a Queen because then Rc4+! forces 2.Qxc4 stalemate! Further-
more, 1.Kc3 Rd1! 2.Kc2 Rd4 and 1.Kb3 Rd3+ 2.Kc2 Rd4! only repeat the
position. But the miraculous 1.c8=R!! works: Black no longer has
the stalemate resource, and despite the material equality soon
succumbs to White's threats: 1...Ra4 (against 2.Ra8 mate) 2.Kb3!
and if Black saves his Rook White mates with 3.Rc1.
An example of a Bishop promotion is: White King d3, Knight c1, pawn c7;
Black King a1, Rook b8. White to move must capture on b8 and promote,
but either Queen or Rook will produce stalemate, and a second Knight
will not suffice to mate. So 1.cxb8=B!! and the Bishop and Knight
barely prevail against Black's lone King.
And now the punchline: remarkably, both the Rook and the Bishop
promotions above arise in the analysis of the same position!
This is Liburkin's classic 1931 endgame: White Kd1, Na2, pawns b5,c6;
Black Ka1, Rc5. White plays 1.Nc1! and now if Rxb5 2.c7 Black
cannot stop the pawn with Rc5? (3.Nb3+) nor Rb8?! (3.Nb3+! again,
breaking the stalemate --- White does not need to underpromote yet).
So, 2...Rd5+ anticipating 3.Ke1 Re5+ and Re8, or 3.Kc2 Rc5+, but
instead 3.Nd3! Rxd3+ 4.Kc2! and Black must try Rd4! producing
the Saavedra position, which White wins with 5.d8R!! Alternatively
after 1.Nc1 Black could try the immediate Rd5+ 2.Kc2 (Ke1? R:b5
and Re5+) Rc5+ and now 3.Kd2? permits the new stalemate resource
Rxb5 4.c7 Rb2+ 5.Kd1 Rc2!! drawing, so 3.Kd3! follows, and after
Rxb5 4.c7 Rb8! (Rd5+ 5.Ke4 Rc5 6.Nb3+) White no longer has 5.Nb3+?
(Rxb3 is check!) but still wins with 5.cxb8=B!!
Bonus question: find a credible *middlegame* position showing
a beneficial Knight promotion --- *without check*.
--Noam D. Elkies (elk...@zariski.harvard.edu)
Department of Mathematics, Harvard University
Derek Holt
I can easily concoct a situation where white's only legal move is to promote
a pawn (he has to take black's queen, which is checking his own king,
which has noweher to go), and at the same time I can make it so that
promoting to queen puts black in stalemate, and the same for exactly
one of rook or bishop. So his best move is to promote to bishop or rook
(in the two cases). The snag is that in the situations I have concoted,
he could also promote to knight and still have a winning position
(although the win would be more difficult), so I haven't quite solved
the problem.
To solve the problem (for rook say), we need following conditions:
(i) white must promote a pawn (to avoid check himself),
(ii) promoting to queen or bishop puts black in stalemate,
(iii) promoting to knight results in drawn position.
So far, I've managed (i) and (ii), but not (iii), but I expect that is
possible too.
Derek Holt.
Andrey Tsouladze
: This is an original puzzle (i.e. I made it up). Someone else
: may have thought of the same thing.
: Can a situation exist in which the only correct move is to
: move a Pawn to the eighth rank and promote it to a Rook? To a
: Bishop? If so, construct such a situation. If not, prove it.
The answer depends on what you mean by _correct_ move.
1. Correct=good. There are many problems where the only way to win is to
promote a pawn to a rook, or a bishop (promoting to a queen would lead to
the stallmate), or to a knight. I have seen even beautiful
mate-in-three problems where a white pawn had to get promoted in all four
possible pieces in different variants. I do not believe the same can be
done for a black pawn.
2. Correct=legal. Here, the answer is no. Promoting is a free choice. If
the move itself is legal there is nothing to prevent the pawn to be
promoted to any of four pieces. However, there is a class of problems
which employ retrospective analysis. In these problems, one has to find
out how the given position (illegal at the first glance) had appeared. Very
often, such retroanalysis involves "weak" promotions.
In order to "promote" the thread, I would like to present a problem which
I made a couple of minutes ago. It is not so difficult ... if you do not
take it _too_ seriously. Here goes:
W: Ka3, Rc7, Bh2, Nc8, pawns b5, b7, c4.
B: Ka5.
Mate in two.
Andrey
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
* Andrey Tsouladze * *
* Department of Biology * You *
* Technion - Israel Institute of Technology * have *
* Haifa 32000 * been *
* Israel * warned... *
* E-mail: ts...@aluf.technion.ac.il * *
* E-mail: ts...@techunix.technion.ac.il * *
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
John Woolley
> Can a situation exist in which the only correct move is to
> move a Pawn to the eighth rank and promote it to a Rook? To a
> Bishop? If so, construct such a situation. If not, prove it.
The answer is yes, such situations can exist.
First, the rook.
........ White (upper-case letters) to move. If 1. K
..P..... moves or 1. P-R4, R-R1 picks off both pawns at
........ leisure and wins easily. If 1. P-B8 (queen),
........ R-B5 check, 2. QxR stalemate. (If 2. K moves,
r....... RxQ wins easily.) If 1. P-B8 (knight or bishop),
.......P R-B5 check wins the piece and the game. But
..K..... after 1. P-B8 (rook) Black has no non-losing move.
k....... If 1. ..., R-B5 check, 2. RxR wins. If 1. ...,
R-R7 check, 2. K-B1 or K-N3, and White can run
the other pawn down at leisure, trading rooks if the occasion
arises. If 1. ..., R-R5 (or anywhere off the QR file), 2.
R-QR8 check, R-QR5, 3. RxR checkmate.
Now the bishop, a similar idea but a little trickier.
r....... White to move. If 1. B-Q4, R-QN1 stops the pawn
.P...... and wins pretty soon. If 1. B-R7, RxB, 2. K-N1,
........ RxP wins. (Or if 2. P-N8 (queen), R-R8 check, 3.
........ Q-N1, RxQ checkmate.) If 1. B-B2 (or B-K3 or
........ B-B5 or B-N6), R-R8 check, 2. B-N1, B-K5
...b...k checkmate. (If 2. B interposes at K1 or B1, RxB
.......P checkmate.) If 1. P-N8 (queen) (or whatever),
......BK RxQ wins soon. If 1. PxR (knight or rook), B-K5
checkmate. If 1. PxR (queen), B-K5 check, 2. QxB
stalemate. BUT if 1. PxR (bishop), White has a pair of
unmatched Bishops and a Pawn, against Black's Bishop. He can
trade either Bishop or the Pawn for Black's Bishop, and still
win (if he knows how to checkmate with two Bishops, that is).
-- Fr. John Woolley
Timothy Y. Chow
me of a puzzle I once thought of: create a (legal) position with the maximum
number of pieces such that *both* sides are stalemated. I tried for a while
but wasn't able to use all the pieces. Can you do better?
--
Tim Chow tyc...@math.mit.edu
Where a calculator on the ENIAC is equipped with 18,000 vacuum tubes and weighs
30 tons, computers in the future may have only 1,000 vacuum tubes and weigh
only 1 1/2 tons. ---Popular Mechanics, March 1949
Andrey Tsouladze
No it is not _always_ true. Consider the following position which I made
up as an example.
W: Kc6, Bf1, pawn h7
B: Ka7, Qg8.
The only winning move is to take the black queen and promote the pawn to
a knight. Promoting to a queen or a rook would lead to a stalemate, and
promoting to a bishop would produce two bishops which move along
diagonals of the same color. And certainly white cannot wait.
If you remove Bf1 from the board, and add Nb4 instead then only promotion
to a bishop wins the game.
I remember an extremely beautiful problem made by a famous chess problem
composer. Unfortunately his name slipped my mind. Here it is:
W: Kb5, pawn c6
B: Ka1, Rd6.
White to start and win.
If nobody finds the solution I'll post it in a week or more. Those who
want it before can email me.
Good luck,