Quite simply, may White castle in the below position? There is much more than what meets the eye! For the record, White pawns move up and Black pawns move down.

Nikita M. Plaksin & Michel Caillaud, Feenschach 05/1982, 3. Prize

enter image description here

  • $\begingroup$ Harold vs. Kumar, 2004. $\endgroup$
    – Jafe
    Feb 11, 2021 at 7:57
  • 2
    $\begingroup$ @Jafe I don't get the joke... $\endgroup$ Feb 11, 2021 at 16:38
  • $\begingroup$ @RewanDemontay, I think it's a reference to the movie "Harold & Kumar Go to White Castle" $\endgroup$
    – justhalf
    Feb 18, 2021 at 7:55

2 Answers 2


There are 6 missing black pieces, and since the black pawns haven't moved, the bishop at c8 must have been captured by white knight. That gives 5 pawn captures, and so by counting, the third white knight must have promoted at g7.

That means the white knight at h8 must have been there before black pawn got into g6.

Now, consider white knight at h1. It must have been there before the black pawn at g3 takes its position. And the bishop, too, must have been at g1 before black pawn at g3 takes its position. And so does the white rook at f1. And since black knight at h2 couldn't have come from f3 (since it will be a check and the white king would have moved), it must have come from g4 (f1 is impossible since the rook needs to be there first in order for the bishop from h2 to move into g2). So the bottom right position needs to happen before white pawn at g4 takes its position. Also, for all these to happen, white pawn at d2 should have moved once, and at least one of e2 or g2 has moved at least once to free the white bishop at f1.

Now, consider the pawn at h7. It must go to at least h6 before white bishop takes its position at h5. So it must have passed the original h-column black pawn. So the capture h4xQg3 must have occurred. Which means white pawn at g4 has already been there before the white pawn at h7 takes its position. Which means the white bishop at h5 must have been there before black pawn moved into g6.

Now, it means all the white pieces must have been in a locked (and final) position at the moment black moves g7-g6 to unlock the black bishop, king, queen, and rook.

Note that white has at most 10 pawn moves left before we can get a white pawn to f6 to capture a black piece at g7 (2 for a2 to b4, 1 for e3, 2 for d3 to d5, 4 for c2 to f6, and 1 for h6 to h7) and no more extra moves since all white pieces are locked after g6. So we need to be able to get the black queen and rook out of the fortress in 10 moves or under, with white moves first (since black last move would be g6).

Is it possible?

Yes. The black bishop would need to go f8-g7-e5, then black king e8-f8-g7-h6, then , then black queen d8-g8-g7-f6, then black rook a8-g8-g7, which is 10 moves.
The key position before locking with g6: enter image description here

And note that:

By the time we reach the final position, it would be black's turn to move. So in order to let white castle, we need to buy a tempo. But how can the black king buy tempo where there is no place to do triangulation? See picture below to see the problem.
enter image description here
The red squares are squares where the black king cannot move into (excluding the newly promoted knight, since it can be moved as fit), while the non-red squares are okay. We see that every cycle of non-red squares is an even-length cycle, therefore it seems impossible for black king to buy tempo (if we restrict black king buying tempo only after white pawn promotion into knight. So far it seems that black king cannot buy tempo before, since there are the same number of white moves available as the required number of moves by black to get into position.)

So, can the black king buy tempo?

Yes! Note that in a-file, the squares are threatened by white rook, and we can use the white knight to block the white rook (big credits to Albert.Lang for this hint!). Giving space for black king to do triangulation at a6 (which is incidentally, the only place to do triangulation).

So, the answer is:

Yes! White can castle, as shown in the proof game (the desired position is at move 70 after black's turn, and the white castling at move 71)

  • $\begingroup$ Ok, fixed my answer. $\endgroup$
    – justhalf
    Feb 8, 2021 at 4:20
  • $\begingroup$ Ah, you mean my comment? Sorry for that, couldn't edit it now. I have deleted it. Ouch, which part is incorrect? $\endgroup$
    – justhalf
    Feb 8, 2021 at 4:29
  • $\begingroup$ Hmm, I couldn't think of a way to have a white knight free before g6. Maybe I'll let someone else answer it. As for the spoilers in answer, which parts should I hide? Just the final two paragraphs will do? $\endgroup$
    – justhalf
    Feb 8, 2021 at 4:32
  • $\begingroup$ Ok even with your hint I can't figure it out, haha. $\endgroup$
    – justhalf
    Feb 8, 2021 at 4:40
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    $\begingroup$ @justhalf Done. Pleasure to do business with you ;-) $\endgroup$ Feb 18, 2021 at 7:56

This is just a very minor detail on top of @justhalf's work:

How can we trade one tempo to ensure we reach the final position with white to move? White can only move the Nb8 with its in-built parity preservation. The black K doesn't seem to have any triangle it could use to switch parity, so where to get that tempo? Solution: Use the N to block the a file, this makes a6 available to the black K which can now use b5-a6-b6-b5 or similar.

  • 1
    $\begingroup$ Enjoy the bounty! (for historical reference, my answer was missing this final piece, which led me to start a bounty for this question, and Albert.Lang provided the final missing piece, earning the bounty) $\endgroup$
    – justhalf
    Feb 18, 2021 at 7:58

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