# No moves at all, not even to put yourself in check

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]

• If I'm black in the above example, I think I'm happy with the draw. – jafe Oct 19 '18 at 20:24
• 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]. – supercat Oct 19 '18 at 21:18
• @supercat - if you want to post part 2, go for it! – deep thought Oct 19 '18 at 21:23
• @deepthought: I actually did, quite some time ago, but on a different SE board: chess.stackexchange.com/questions/4836/… – supercat Oct 19 '18 at 22:34
• @supercat - ah nice, thanks for the link – deep thought Oct 19 '18 at 22:57

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

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:

• White needs to have promoted 4 pawns, no? There are only 3 white pawns missing. – jafe Oct 19 '18 at 20:05
• Note only one side needs to be stalemated – deep thought Oct 19 '18 at 20:05
• oops, there are too many same coloured bishops. Imagine the G and H files swapped for a more reachable position please :-) – Bass Oct 19 '18 at 20:06
• +1 but still not minimal! – deep thought Oct 19 '18 at 20:11
• I think there's another 6+1 solution, given in my answer. – supercat Oct 20 '18 at 3:22

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

• Before posting the puzzle I proved my solution was unique [OP vanishes in a puff of logic] – deep thought Oct 20 '18 at 15:33
• I'll award a bonus, but it looks like there's a delay enforced. – deep thought Oct 20 '18 at 15:44

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.

• Correct +1, but not minimal! – deep thought Oct 19 '18 at 20:06
• @deepthought got it down to 13 :) – jafe Oct 19 '18 at 20:12
• great! but keep going :-D – deep thought Oct 19 '18 at 20:14

I believe that I have found a better solution based on a (debatable) loophole. Here’s my 4+2 solution. (Or 4+1 if you don’t count the black king, since it is not being used.)