# generating stripboard layouts

Posted on Feb 7, 2014

All you have to do is insert the pieces so that things that should connect are on the same lane. To make things even easier, you can break lanes, or join lanes with a wire link. Right, this brings me to the meat of this post. How can we automate this, to get an optimal and flawless translation from what is basically a graph, to a set of lanes In

he N-Queens problem, N queens must be arranged on an NxN board, so that no queen attacks another queen. This is implemented using all-difffd 2, which forces the queens to all be on different rows, columns and diagonals. For starters, lets define lanes as a finite domain of numbers. This allows us to use all-difffd, but gets us in trouble when we need to express broken lines. In this simplified version, I also just defined components as a list of points they are connected to. Components that are connected, use the same fresh variable. This also gets us into trouble when we insert wire links. > (run 1 (q) (fresh (l1 l2 l3 l4) (infd l1 l2 l3 l4 (range 0 10) (= q (list (list `r1 l1 l2) (list `r2 l3 l4) (list `fet l1 l3 l2) (distinctfd (list l1 l2 l3 l4) (r1 0 1) (r2 2 3) (fet 0 2 1) (define lanes (range 0 50) (define lane (lambda (l) (infd l lanes) (define resistor (lambda (l1 l2 r) (fresh (l2+) (lane l1) (lane l2) (lane l2+) (plusfd l2 2 l2+) ( (run 5 (q) (fresh (l1 l2) (resistor l1 l2 q) (r 3 0) (r 4 0) (r 4 1) (r 5 0) (r 5 1) > (run 1 (q) (fresh (l1 l2 l3 l4 r1 r2) (infd l1 l2 l3 l4 lanes) (resistor l1 l2 r1) (resistor l3 l4 r2) (= q (list r1 r2 (list `fet l1 l3 l2) (distinctfd (list l1 l2 l3 l4) (r 3 0) (r 4 1) (fet 3 4 0)

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