A ALMOST IMPOSIBLE PUZZLE

There's something almost poetic about watching the solution! Never even entered my mind the reamaining dark knight is not the one that started in that position.

I think this one was not so difficult because white could only have moved the knights. From that you can tell that the g1 knight had to finish on a dark square.
I think this one is more difficult, reach this position after 4 full moves:

I think you guys need to play something else than chess now, or maybe try to improve your play, cause you seem to have a lack of interest in the real game here lol

why doesn't black take the night in one of his twelve million chances? We don't need to get to that posistion.
#1 Yay, nice! xD
#5 Ehm, I think that one is easy, isn't it? if it's black to move...
...e3 c6, Bc4 e6, Bxe6 dxe6, e4?

If I'm not mistake Herbie, thats the tricky part of the puzzle I think you are to get to that position in 4 full moves, with White left to play.

Oginschile is right, you guys are doing it 3 1/2 moves, not 4. I suppose I could also stipulate the first move to be 1. e4.

I've briefly looked at trying to figure this puzzle out (the second one) but just keep getting distracted by the lack of an n in "A(n) almost impos(s)ible puzzle". Don't mean to be a spelling nazi but its just irking me. I wish there was a way for this to be changed, more than I would like to know the solution to the puzzle. This I believe is a sign for just how tired I am.
Cool puzzles though I would like to see more of these.

I am thinking aloud here re lommis's puzzle: With the stipulation that it began 1. e4 as loomis posted, the White B must have made at least 2 moves [or else a black pawn would be on the f1-a6 diagonal.] That accounts for 3 moves of white, leaving only one to spare. But no other piece of white is misplaced.
So after 1. e4 white's next 3 moves were with the B.
Right so far?

But black's pawns must have made at least 3 moves.
[argument: the white B captured a P on c6 or e6,[*] and was then captured by ..d7xB(c/e)6. So one move with the orig. c-or-e pawn, then after white's pawn-capture one move (PxB) with the orig. d-pawn; the third move being the other pawn moving one step forward - as the problem position shows.]
Again all other black's [non-pawn] pieces are in place - and we have only one black move left to account for.
Hence:
Conclusion:
All 4 black moves were pawn moves. One of them was d7xB; others were non-capture moves.
The four white moves were 1. e4, B moving twice without capturing, and moving once again to capture the [then] c-or-e pawn.
So, unless i have made a mistake in the argument above, or the problem is incorrect; these jigsaw pieces (the statements in the conclusion) can be fitted together.
[*] UPDATE: wrong! the mistake is in the very first clause :) (that's a hint for those who do not yet want to go to page 2 for De Lar's solution).

so the 4th move was, in ambiguous descriptive notation, 4. BxP QPxP.
so, in mixed notation we have:
1.e4 (c/e)6 2. B-_ (e/c)6 3. B-_ 3... ??? 4. BxP d7xB.
now black's 3rd move has no candidate [it has to be a noncapture pawn-move and none is available].
I conclude that loomis's puzzle is impossible.
Where did I go wrong? Or, did I?

normajeanyates, you have a flaw in your logic. :-)
One of your conclusions is that black made 4 pawn moves, but what could they possibly have been? Only 3 of black's pawns have moved and if any of them moved more than once it got to the fifth rank or further and must have been captured. So I don't think it's possible to reach this position in 4 moves with 4 pawn moves by black.
I promise the puzzle has a solution. :-)
This position has to be reached in 4 full moves, its pretty hard.