Quickest Way to Pawn-Filled File

Starting with the initial board, what is the quickest way you can come up with where there are six pawns in the same file of the same color?

I keep trying and failing this...

edit:

14 moves!

15!! I believe. Keep getting 15.

Cool! To be honest, I never even tried doing it for white pawns. I think it's easier for the black pawns though. Try reversing the colors and make it a 13 mover! It might work... But probably not. 

Still worth a try!

Without showing a specific answer, can anybody prove what the theoretical minimum is?

13 I believe. How long it takes all of the black pawns to reach their desired squares if they could go diagonally without capturing. The problem lies in getting the white pieces to let them capture. White needs to get pieces to (if you're using the d file. I think it's best because it has the queen. But it's the same thing but on the other side of the board when you're going for the e file.) d6, d5, c5, d4, e4, d3, b4, c3, and d2.

The key is figuring our how white can get pieces onto those squares fast enough. So far the best I've gotten is one move above "par" (by "par" I mean the theoretical minimum.)

 

 

edit: Apparently I was wrong about the d file thing. Congratulations Harry_Li!!

So the fastest so far is on black's 13th move -- anybody faster? Or anybody prove that this is the fastest?

Careful with such a proof though -- the easiest way to disprove a "proof" is by counter-example! :-)

From just looking at the start, say we choose a middle row, d or e. Now the best way (their may be others equally as good) uses all black's moves as pawn moves (not white because white will need the tempo to get pieces to get taken...? Or maybe its for white and black!). This way is that the outer pawns go the furthest down, and thats the 7th and 6th row, the 2nd outer goes 5th and 4th w/e. Now for the outer pawns of the d row, to use the most efficient way, will be the a and f pawn. The a pawn will take 4 tempo to get to the 7th and the f pawn 3 to the 6th.
Thats 7 tempo there. Sorry gotta go bye.

Edit: Some1 work out the temp for the other "outer pawns" and add them to 7 to get the minimum.

Edit 2: all that was giberish. You can just work it out easily by using the outer pawns just count your way to the rows, don't bother with all that nonsense up there (its basically doing the same thing).

It takes 11 moves to get the black pawns in place on the e file.

I think the theoretical limit is 12, the idea being to maximize the number of two-square pawn advances. As the pawns on the second, third and fourth rank cannot have two-square advances, the maximum number of two-square advances is three...as shown below. I may be missing some other factors though, which may allow something even faster...

Now the next question is whether all the pieces to be captured can be moved into place in fewer moves than that...is the "rate-determining factor" the number of pawn moves, or the number of "pieces to be captured" moves?

On a side note, that question will also decide whether it's faster to set up using black pawns or white pawns, given that white moves first. There should be one half-move's difference between the variations...

tyzebug, good job simplifying the problem. Also, good job in knowing that there are separate issues involved. But it seems that you've shown that it cannot be any quicker than on white's 12th move.

Currently I am having massive difficulties getting the "pieces to be captured" into place...basically there is only an "allowance" of three moves, i.e. all but three moves must get a piece into the correct position to be captured. With thirteen moves there are a few ways (as Harry has shown neatly), but with twelve it becomes extremely restrictive. EDIT: No, wait, it's only two moves if you want it to end exactly on White's 12th move. Gaaaargh. This keeps getting tougher. Although you can get three moves of "allowance" if you set up with black pawns instead...but that would be slightly slower, eh?

 

A note for anyone else who read my previous post though, don't use that setup for the capturable pieces, because the piece on the c file seems impossible to get into place (there are nine captures, meaning that if the rooks aren't used (which seems likely) then the captures are two bishops, two knights, the queen and four pawns. But if that's the case the only piece that can be on the c-file is a pawn, which wastes a lot of moves marching it up the board). Instead, this formation (reflected or translated horizontally...there are six possible ways to place it, I think) seems to work better, by allowing the four pawns to reach "capturable" positions within single moves.

 

I could be totally mistaken though, especially if the solution involves rooks. Hmmm...come to think of it, I should give that some thought.

The most complete proof would be to program a computer to reach that position. It shouldn't take long to calculate fully to 13-ply, especially since so many moves could be immediately eliminated.

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Awww...but that would spoil the fun! XD I mean, computers would solve mate-in-twos in fractions of a second...puzzles are for us to have fun thinking over.

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Sure, but computers are a great way to check the solution once you've found it (it always bothers me when people post puzzles with obvious cooks that could have been found in three seconds of computer analysis).

Do you have an answer ozzie_c_cobblepot or is this something you have pondered over yourself?

Also you say nothing about the rules of chess in this problem therefore I could do it in 5 moves by moving the pawns directly to the required squares - just thinking outside the box.

Yeah, and throwing all the rules out of the window, it could be done in zero moves by setting up the position like that in the first place. -.-

 

Anyway, yes, I'm curious...ozzie, do you have a solution to this question already? You have dropped some tantalizing hints so far, so do you already have an answer? (Don't spoiler it for me yet though. XD)