Here is a simple puzzle. There is a number of pancakes all of different sizes.
Pancakes are stacked on top of each other. You are allowed to slip a flipper under one
of the pancakes and flip over the whole stack above the flipper. The purpose is to arrange
pancakes according to their size with the biggest at the bottom. In the simulation below you should just click on the pancake under which, in the real world,
you would slip a flipper.

Actually, a physical, mundane activity like flipping pancakes may at best serve only as a starting point for a puzzle. Other moves could be envisaged that
would require very special training to carry out. For example, we may allow to only flip "end points" of the stack. This is controlled
with the "Scarce" checkbox. In a second modification, we fix the interval of the stack to which
a move applies. It may apply to intervals of length 2,3,4, etc. (with adjustments towards the top of the stack.)

We may draw several conclusions. But, instead of dealing with simulated pancakes, let's agree
to permute (flip) elements of the set {1,2,.,n}. The allowed moves are represented by permutations

M_{i} = {1 . (i-1) n (n-1) (n-2) . (i+1) i},

where M_{i} amounts to the flipping of the tail of the set {1,2,.,n} starting with the index i. The fact that the puzzle
is solvable for any permutation from S_{n} implies that the set {M_{1},M_{2},.,M_{n-1}}
generates S_{n}. (Why has M_{n} been omitted?). In particular, any transposition can be
expressed as a product of M_{i}'s. For example, the transposition (i i+1) that exchanges elements i and (i+1) can be written as

(i i+1) = M_{i+1}M_{i}M_{i+1}M_{i+2},

where, by convention, we first flip M_{i+1}, then M_{i}, and so on, left to right. (When indices get out
of scope simpler formulas apply. For example, (n-1 n) = M_{n-1}.) It's an interesting exercise to find
such representation for a general transposition (i j).

There is another interesting question. What if, on any move, we were only allowed to flip a fixed number k
of elements? For k = 2, the only allowed moves would be (1 2), (2 3), ., (n-1 n).
For k = 3, the set of moves would consist of (1 3), (2 4), ., (n-2 n).
For k = 4, we would have only {4 3 2 1}, {5 4 3 2}, ., {n n-1 n-2 n-3} (we might also add {n n-1 n-2) and (n n-1),
should this be necessary.)

The assertion for k = 3 is obviously wrong. Right? What about other k's?

A few words of the "scarce" variant of the puzzle. Surprisingly, the constructive strategy
that worked before works now as well. It then follows, that any permutation can be represented
as a product of transpositions all of which transpose a fixed index ("top" index in this case.)
This result has also been established quite directly as Theorem 1 in a discussion on transpositions.