TITLE: Fold, Gather, Cut
QUESTION [9 upvotes]: Here's a mathematical puzzle I've been thinking about.  Let's say you have a strip of fabric, of length $N$ units ($N$ being an integer), which has regular markings on it every 1 unit along its length.  Your task is to cut the fabric into $N$ lengths of 1 unit each, but to do so using the fewest operations possible.  The operations available to you are:

Cut - you may cut through any number of overlapping layers in a single operation.  All cuts must be through the 1-unit markings on the fabric, and the cut must be in a straight line.  (No using a wavy cut to do the whole thing in 1 operation.)
Gather - you may gather any number of already-cut lengths together in a bundle.  They must all line up on one end.  (If they are the same lengths, they will of course line up on both ends.)
Fold - You may fold any number of layers of fabric in either direction.  Multiple folds can only be counted as 1 operation if the places to be folded are already lined up, either via a previous Gather or another Fold step.  Unlike Cuts, a Fold may occur between the 1-unit markings.  Unfolding does not require an additional operation.

Now it can be trivially shown that for any $N$ which is $2^k$, an ideal solution would be to have $k$ Cuts alternating with $k-1$ Gathers, for a total of $2k - 1$ operations.  This is by no means the only ideal solution.  For example, consider $N=8$.  The steps could be:

Cut into 2 lengths of 4.
Gather lengths.
Cut into 4 lengths of 2.
Gather lengths.
Cut into 8 lengths of 1.

Or you could do the following:

Fold at the 3rd marking.
Fold at the 5th marking the other direction.
Cut down the middle to get 4 lengths of 2 (2 of which are folded).
Gather lengths.
Cut into 8 lengths of 1.

Still 5 steps either way.  It gets trickier when you consider other numbers however.  Say, for $N=9$, you could take any $N=8$ solution, and then add 1 more cut for that last piece that will be 1 unit too long.  But you can do 9 in 5 steps as well:

Fold in half.
Cut at the 3rd and 6th markings (which should be lined up) to get 3 lengths of 3.
Gather lengths.
Fold all 3 in half again.
Cut into 9 lengths of 1.

So my question is, with the given operations, can you compute the minimum number of operations for any given $N$?

REPLY [2 votes]: Here is an algorithm that does it in $2+\lceil\log_2 n\rceil$ steps.  This doesn't match the OP's algorithm for $n=9$, unfortunately, but it's "k+2" rather than "2k-1" when $n=2^k$.

Fold the fabric into a pile that is exactly 2 inches wide.  This takes $\lceil\log_2 n\rceil-1$ steps.
Cut down the middle.  This will leave many pieces of fabric, almost all 2 inches wide (perhaps one piece 1 inch wide).  
Gather it all into a single pile, 2 inches wide.
Cut.

For example, if $n=16$.  Fold (8"), fold (4"), fold (2").  Cut, gather, cut.  6 steps.
Second example, if $17\le n\le 32$.  Pretend it's $32$, the next higher power of 2.  Fold (16"), fold (8"), fold (4"), fold (2").  Cut, gather, cut.  7 steps.