Is this position legal?

Ok h3 is legal to. Black could take a1 rook, promote to knight get to h3. White takes h3, plays rook h1, g1,g6, c6. Black goes dxc6 be6 bc4 ba6.

@Eeyores: Correct! Great job! (The third way that could be considered is black saccing a bishop on c3, the white rook emerges, then black sacs something else on b3, but that's proven impossible as then the c8-bishop couldn't usefully come out.)

@chesssmart82: Getting better; both are legal but less trivially so. In the first white took the f8-bishop (with R, N, Q or even K); in the second the c1- and c8-bishops were captured, and underpromoted pawns made it to g3 and g6. I'll show diagrams as these are a good introduction to basic retro concepts.

You guys still remember my puzzle? Just wanted to say I have reached the absolutely correct amount of moves and proven it correct

40 isn't the lowest. I'm quite terrible with the numbers here (too complicated for my taste) but I can give a guess and a shot.

The guess would be: 1 ply of king move + 9 ply of captures + 3 ply (doublemove) of pawn pushing + 5 ply to promote a pawn and (the guesswork is here) 2 ply x 8 sacrificial lambs + 1 ply pawn sac = 35 ply. Which means the absolute minimum number of moves needed is about 35 moves (since I didn't do it rigorously, I can't say for sure.)

Also. A 36.0 move solve (with inefficiencies; note the extra 2 ply white had to kill.)

@remellion I didn't say that 40 moves would be the lowest possible I just said I found the fastest wayf

I'd like to hear of your fastest way. I tried again and got 32.0 moves, while my retroanalysis suggests the lower bound of 31.0.

Per side, the ending pawns make 9 captures and 3 doublemoves, promoted pawn (a/h) 5 ply to promote and 1 ply to sac itself, the rooks need at least 2 ply apiece, bishops 3 ply total (c1/f8 bishops need 2 each), queen 1 ply, king 1 ply, sacrificial pawn 1 ply and knights at least 3 ply (g1/b8 knights at least 2) for a total of at least 31 ply.

My solution here optimises everything except the knights, which take 4 ply total per side for 32.0 moves total. Which raises the question: Is 31.0 possible? (It is the absolute theoretical minimum.) If not, 31.5 perhaps?

#221 is legal, last move would be bishop moving from being next to the queen.

#222 is obviously illegal, looks cool but way to many pieces on the board!

@remellion I tried many times but couldn't make in 31 moves. 

I counted that it must be 32 moves but when I'm now checking again I got 31. interesting... I'm gonna try to find my notebook, maybe it tells me why

I found my note book. This is what it says:

a-pawn moves 5times+b-pawn once (because black's a-pawn must take it)+ c-pawn 4 times + d-pawn 2 times + e-pawn once + g-pawn twice +h-pawn three times = pawns total 18  ... 18+1 king+ 1 queen + 4 knight moves + 3 original bishop moves + a8 bishop move to c6 + 4 rooks = 32  

Fantastic news. I just got pen and paper and proved 32.0 is the absolute minimum within 2 minutes. I'm an idiot for not trying this visually earlier.

Basically, the calculations from my previous post hold. Except that 4 ply is the minimum to be spent on both knights. Why this is so is obvious once you look at the squares where pieces die, relative to the knights*.

Looking at the squares the white pieces died, they are b4 (the b-pawn), c2, c3, c4, c5, c6, d3, d6 and e4. (It helps to highlight those squares visually.) The white knights started on b1 and g1. Looking at wNg1, it needs to spend 3 ply to get to any of those squares except c3 (2 ply)... which would then force the wNb1 to spend 2 ply itself getting somewhere useful. Thus both knights together need at least 4 ply. The same applies for black (rotate the board.)

Therefore 32.0 is the absolute minimum. Hurrah, we're done! An interesting challenge indeed.

[*Note:] First we actually need to prove that there is an optimal way for the pawns to align themselves to select the squares where pieces die. wPb2 and wPg7 must die on b4 and g5 respectively after a doublemove as that is obviously optimal. Neither should want to promote or go further which would waste time. Therefore wPh4 and bPa5 must be played to capture them. wPa2 and bPh7 must then promote to become food, and after promoting, can be freed to die in 1 ply (along the diagonal) by doublepushing wPg4 and bPb5. This uniquely determines which 6 pawns doublepush and the squares that pieces were captured on.

Yep well said remellion. Istead of listening my chemistry teacher I made a empthy chess board in my chemistry notebook. I used lines to count the pawn moves and then tried to vizualize the fastest way to sacrifice everything to achieve it.  

at last, the great mystery is solved 

The fact that it is possible to solve the minimum number of moves without actually touching any pieces, is interesting indeed.   

can white castle to defend after Qa8?

Nope, since from the diagram white cannot retract anything but a king or rook move.

I have conjured up another pretty little devil. Is the following legal a) as the diagram, b) without the c6-knight?

As usual, here's to hoping it's not cooked.