November 5th, 2007
As usual, solution related comments go on this page.
Pages: 1 2
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November 8th, 2007 at 12:03 am
If n is the “height” of a single peon above the line (n=1 being the first square above the red line), then the first few terms are
n=0 –> 0 (no peons are needed)
n=1 –> 2 (the first example)
n=2 –> 4 (the second example)
n=3 –> 8 (since you first need a n=2 peon and need to reconstruct the “cannon” below the line)
It would of course be esthetical to have n=4 –> 16 but this is clearly not the case. I have found a n=4 –> 21 solution, so it is clear that the solution for n=5 is at least higher than 29.
November 22nd, 2007 at 3:18 am
Well, I think its time I posted a hint.
WARNING – SEMI-SPOILER!
It is actually impossible to achieve the required task!
You cannot move a peon to 5 squares above the line.
Now try to prove that…
PS – Seb, you’re numbers are pretty close. I do not really remember, but I think that the required number for getting to 4 squares above the line is 20. Anyways, proving 5 is impossible is much nicer than calculating the required number for 4.
May 16th, 2008 at 2:27 pm
The core of the solution is this: Assume your target is (c,5) for some c. To each square assign “potential” function p(x,y). The potential of the board is the sum of p(x,y) for (x,y) which have a pawn on them. The potential function must be carefully-chosen, you need to show that the potential of the board stays constant (or decreases) for each move. Now, you’re left with showing that p(c,5) is greater than the original potential (or exactly equal, since the number of steps must be finite).
The initial thought is to set p(x,y) with something that resembles Fibonacci series, since p(x,y)>=p(x,y+1)+p(x,y+2). But this results with an alternating and unbounded potential
. That thought must be a tad-bit refined, and you’ll have the solution.
Great riddle, I’m putting it on my page