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Zhang-Suen thinning algorithm: Difference between revisions
Updated D entry
(Go solution) |
(Updated D entry) |
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=={{header|D}}==
This uses the module from the Bitmap Task. And it performs no heap allocations.
<lang d>import std.stdio, std.algorithm, std.string, std.functional,
std.typecons, std.typetuple, bitmap;
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/// Zhang-Suen thinning algorithm.
Img zhangSuen(Img image1, Img image2) pure
in {
assert(image1.image.all!(x => x == Img.black || x == Img.white));
assert(image1.nx == image2.nx && image1.ny == image2.ny);
} out(result) {
assert(result.nx == image1.nx && result.ny == image1.ny);
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} body {
/// True if inf <= x <= sup.
static inInterval(T)(in T x, in T inf, in T sup) pure nothrow @safe @nogc {
return x >= inf && x <= sup;
}
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/// Return 8-neighbours+1 of point (x,y) of given image, in order.
static void neighbours(in Img I, in size_t x, in size_t y,
out Neighbours n) pure nothrow @safe @nogc {
n = [I[x,y-1], I[x+1,y-1], I[x+1,y], I[x+1,y+1], // P2,P3,P4,P5
I[x,y+1], I[x-1,y+1], I[x-1,y], I[x-1,y-1], // P6,P7,P8,P9
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}
if (image1.nx < 3 || image1.ny < 3) {
return image2;
}
immutable static zeroOne = [0, 1]; //**
Neighbours n;
bool hasChanged;
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(!n[ab[0]] || !n[4] || !n[6]) && // Cond. 4
(!n[0] || !n[2] || !n[ab[1]]) && // Cond. 3
//n[].count([0, 1]) == 1 &&
n[].count(zeroOne) == 1 && // Cond. 2
// n[0 .. 8].sum in iota(2, 7)) {
inInterval(n[0 .. 8].sum, 2, 6)) { // Cond. 1
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img.textualShow(/*bl=*/ '.', /*wh=*/ '#');
"\nTo thinned:".writeln;
img.zhangSuen(img.dup).textualShow(/*bl=*/ '.', /*wh=*/ '#');
writeln;
}
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