In both parts we know one to be true, which means it suffices to disprove one possibility for each and that settles the proof.
Part one:
Let's start by assuming the pawn on f2
is white and work our way up to see if we hit any inconsistencies. So here's our diagram:
[title "Assuming f2 pawn is white"]
[fen "r2qk2r/1ppbp1p1/1pn2n2/b2pP3/3P4/2N3P1/PPP2P2/R2QK2R w - - 0 1 "]
First trivial observation is the fact that the black bishop on a5 is a promoted piece as the original dark square bishop could never have escaped from f8
. This means either the h
or f
pawn has promoted on g1
into a bishop that is now on a5
. Having a white pawn on f2
also implies that the g3
pawn isn't the h
-pawn having made a capture on g3
or else the promoted bishop could never have escaped white's kingside and reach a5
, thus, g2-g3
is played at some point, and it was on g2
when black made the promotion (else no escape route). This in turn means the promoted black pawn must have been the h
-pawn. But how did it get passed the h
pawn? We know that white is down three pieces, the two bishops and the h
-pawn. Additionally, we know that one piece was captured on b6
and one on g1
(for black's h-pawn to promote). Now both b6
and g1
are dark squares, but 'h2' pawn couldn't have been captured on either, nor could have the f1
bishop as it is a light square piece. Therefore, white must also have a promoted piece, as nothing else accounts for the missing captured piece, for instance a knight (or any other non-pawn) being captured on g1
by the h7
pawn and the h2
having promoted into a knight. So clearly the h
pawn hasn't been captured but has instead been promoted. But therein lies the conflict: one of the h
-pawns must have made at least two captures to get passed the other h
-pawn. It couldn't have been the white pawn as black has only lost one piece so far, but it cannot be the black h
-pawn either, because it must have then made 3 captures, 2 to get passed the white h
-pawn and one last capture on g1
to promote into the a5
bishop. Knowing a capture on b6
has also taken place, this accounts for a total of 4 captures, but white has only lost 3 pieces at most. Thus, neither h
-pawns could have legally promoted given that the f2
pawn is white, so our starting assumption must be wrong. We conclude the f2
pawn must be a black pawn.
Notice that throughout the analysis, the position of the knight either f3/f4,
would have made no discernible difference on the resolution of the f
-pawn's conflict, so we were safe to do the analysis independently. The reverse is not true, namely, the knight's position is very much dependent on the fact that the pawn on f2
is black.
Part two: Knowing now that the f2
pawn is a black pawn, let's take the most restrictive case and assume the knight is on f3
and perform a similar analysis to see if we hit any walls. So here's the corresponding diagram:
[title "Assuming the knight is on f3"]
[fen "r2qk2r/1ppbp1p1/1pn2n2/b2pP3/3P4/2N2NP1/PPP2p2/R2QK2R w - - 0 1 "]
It is the most restrictive consideration as it forces the f7
pawn to have made at least 2 captures to reach f2,
because the white knight is assumed to be on f3
, which means that f7
pawn must have just made a capture on f2
from e3
(that it reached from yet another capture). So far we have accounted for 4 white pieces having been captured and the h7
and h2
having promoted for black and white respectively. The only1 way the latter could have happened is if the h7
pawn (i) made two captures to go around the white h
-pawn, or (ii) made only one capture from h3
(so taking something on g2
) while white's h2
pawn still untouched. (i) is impossible by just considering the piece count, because so far there has been one capture on b6
, two captures by f7
pawn and two by the h7
pawn, and that is already more captures than the number pieces white has lost. So only option is (ii): we know that the h7
pawn must have captured a non-pawn piece on g2
(as the g3
white pawn couldn't have been from either the h
pawn which promoted, or the f2
pawn as black only lost one piece and that was the original f8
bishop on its original square). This in turn implies that g2-g3
must have been played by white before h7
captured on g2
. But that means the promoted black bishop could not have escaped white's kingside via the g3-e5
diagonal anymore, therefore, it must have escaped from the f2-d4
diagonal instead, which settles the fact that the f7
pawn couldn't have captured a pawn on f2
(from e3
), or else the bishop could never have reached a5
, had there still been a pawn on f2
just presumably captured by black's f7
pawn.
Now here's the dilemma: what on earth happened to white's f2
pawn in all this? Can we logically account for it knowing black just took a non-pawn piece on f2
and there's a white knight lying on f3
(starting assumption)? It definitely couldn't have been the piece that the f7
pawn captured to reach the e-file
either as white only made one capture and that was on f8
, so f2
-white-pawn couldn't have magically been on the e
-file. Trivially, it couldn't have been captured on b6
or g2
either. It couldn't have been promoted either as it would have forced black's king to move (disallowed in the starting description). So f2
pawn couldn't have been promoted, or been accounted for by any of the four captures performed by black, but somehow it isn't on the board, thus, we can conclude that we've reached an inconsistency again, meaning there's no logical set of events that would legally reach the position given above with the knight standing on f3
which was our original assumption, so the knight must be on f4.
Come to think of it, there's a much shorter way of realising that the knight cannot be on f3
without focusing on the fate of the white f2
-pawn: quick recap of captures: we know a dark square piece has been captured on b6, two non-pawn pieces captured one on e-file and one on f2,
one non-pawn piece also captured on g2
. But how many non-pawn pieces has white lost? The two bishops and another non-pawn (we cannot know which because we don't know what piece the h2
pawn promoted into) on the e
-file. That's 3, so there's no other non-pawn left for black to have just captured on f2
(we already know it couldn't have been a pawn on f2
captured). And that's another formulation of the same inconsistency. Notice that if we are to assume that the non-pawn captured on f2
was a piece that resulted from the promotion of the f2
-pawn, then it implies that the f2
pawn reached f7
with a check and was not captured by anything thus allowed to promote, but that's impossible as we know neither kings have moved yet, in other words, if the f2
pawn were to ever promote it must have gone passed the f7
square without being captured, which is impossible. Anyhow, let's end with the final correct diagram:
[title "Solution"]
[fen "r2qk2r/1ppbp1p1/1pn2n2/b2pP3/3P1N2/2N3P1/PPP2p2/R2QK2R w - - 0 1 "]
If you're interested in these kinds of discussions, have a look into the books of Raymond Smullyan, I even recall similar puzzles in one of his books but I have no copies at my disposal at the moment to check for you, maybe someone else can verify.
1: as the white pawn couldn't have made two captures to go around the black h
pawn, because white has only made one capture in this game and that has been the f8
bishop on its original square.