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stalemate max

Here, white is stalemated. The king is not in check, but white has no legal moves. Not because any move white makes would put the king in check -- which is how stalemates normally happen -- but because no white piece can move at all, even if we temporarily ignore the no-self-check rule.

(Black is not stalemated: king takes bishop.)

Now, this example is a bit wasteful, 62 white pieces plus the two kings. Also, it can't be reached from the starting position.

Come up with an example, using the fewest pieces (total both sides). Can it actually be reached from the starting position?

Small hint:

there is essentially only one answer. [EDIT: OP screwed up]

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    $\begingroup$ If I'm black in the above example, I think I'm happy with the draw. $\endgroup$
    – Jafe
    Commented Oct 19, 2018 at 20:24
  • $\begingroup$ Now for another challenge: produce a legal position where both sides are likewise stalemated [so that even being forced to move into check would be a loss, and a player could pass if no pieces could move at all, the game would be drawn]. $\endgroup$
    – supercat
    Commented Oct 19, 2018 at 21:18
  • $\begingroup$ @supercat - if you want to post part 2, go for it! $\endgroup$
    – deep thought
    Commented Oct 19, 2018 at 21:23
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    $\begingroup$ @deepthought: I actually did, quite some time ago, but on a different SE board: chess.stackexchange.com/questions/4836/… $\endgroup$
    – supercat
    Commented Oct 19, 2018 at 22:34
  • $\begingroup$ @supercat - ah nice, thanks for the link $\endgroup$
    – deep thought
    Commented Oct 19, 2018 at 22:57

3 Answers 3

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Here are my first idea (both sides are essentially the same answer, so the hint fits too):

enter image description here

Both positions seem to be independently reachable by a legal game. It might be possible to find a legal game leading to the whole position too, but that would take a bit of time.

Before that, I'm going to double check for any simpler solutions. :-)

Since OP commented that only one side needs to be stalemated in this rigorous fashion, this should do the trick:

enter image description here

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    $\begingroup$ White needs to have promoted 4 pawns, no? There are only 3 white pawns missing. $\endgroup$
    – Jafe
    Commented Oct 19, 2018 at 20:05
  • $\begingroup$ Note only one side needs to be stalemated $\endgroup$
    – deep thought
    Commented Oct 19, 2018 at 20:05
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    $\begingroup$ oops, there are too many same coloured bishops. Imagine the G and H files swapped for a more reachable position please :-) $\endgroup$
    – Bass
    Commented Oct 19, 2018 at 20:06
  • $\begingroup$ +1 but still not minimal! $\endgroup$
    – deep thought
    Commented Oct 19, 2018 at 20:11
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    $\begingroup$ I think there's another 6+1 solution, given in my answer. $\endgroup$
    – supercat
    Commented Oct 20, 2018 at 3:22
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How about

enter image description here

I think that's an alternative 6+1-piece solution.

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  • $\begingroup$ Before posting the puzzle I proved my solution was unique [OP vanishes in a puff of logic] $\endgroup$
    – deep thought
    Commented Oct 20, 2018 at 15:33
  • $\begingroup$ I'll award a bonus, but it looks like there's a delay enforced. $\endgroup$
    – deep thought
    Commented Oct 20, 2018 at 15:44
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The general idea is

To fill the 8th rank with major pieces which block pawns from advancing, and fill the 7th rank with pawns which block the pieces from moving. We can't have knights on the 8th, though, because they would still have 6th rank squares available.

Here's a solution with 14 pieces total.

Promote three pawns to two rooks and a bishop.

enter image description here

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  • $\begingroup$ Correct +1, but not minimal! $\endgroup$
    – deep thought
    Commented Oct 19, 2018 at 20:06
  • $\begingroup$ @deepthought got it down to 13 :) $\endgroup$
    – Jafe
    Commented Oct 19, 2018 at 20:12
  • $\begingroup$ great! but keep going :-D $\endgroup$
    – deep thought
    Commented Oct 19, 2018 at 20:14

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