Techy odds and ends

The "N Queens" problem is a puzzle where you take an NxN chequers or chess board (it's usually 8x8 but this works for the general case) on which you have to place N queens in such a way that none of them threaten or are threatened by any of the others. Reminder: in the game of chess, a queen can move as many squares as she likes horizontally, vertically or diagonally but cannot jump over any of the other pieces on the board like a knight. This means that any piece in her direct line of sight on the same line, in the same column or on a diagonal to her is fair game. With this in mind, it is clear that there is only one solution to the problem on a 1x1 board, i.e. a single square (you just plonk a queen on it and you're done), and that there is no way to organise two queens on a 2x2 board or three queens on a 3x3 board without them threatening each other. For a 4x4 or greater board, there is more than one solution. I thought I'd have a go at solving th...

The program downloadable from this page is a numerical solver. It attempts to find the zeroes of a function $f(x)$, i.e. it attempts to find $x$ such that $f(x)=0$ There are several methods for this. One that requires little computing power is Newton's Method, or the Newton-Raphson method, a description of which can be found on Wikipedia . We have no way of knowing the exact value of $f'(x)$ on the HP-41CX used here because that machine has no symbolic math capabilities, so we calculate an approximation of $f'(x)$ by taking a small value $\epsilon$ and calculating: \[ f'(x) \approx \frac {f(x+\epsilon)-f(x)} \epsilon \] This then allows us to calculate the next value of $x$ to use and display so that we see the calculator homing in on the zero that we're looking for: \[ x_{n+1} = x_n - \frac {f(x_n)} {f'(x_n)} \] Once we find a value of $x$ such that $|f(x)| \lt \epsilon$, we stop and that value of $x$ is deemed to be a zero of $f(x)$. The s...