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February 5, 2016 08:29
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Algorithm to get the maximum size of n squares that fit into a rectangle with a given width and height (http://math.stackexchange.com/questions/466198/algorithm-to-get-the-maximum-size-of-n-squares-that-fit-into-a-rectangle-with-a) #WIP
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int main(){ | |
double x=5, y=6, n=7;//values here | |
double px=ceil(sqrt(n*x/y)); | |
double sx,sy; | |
if(floor(px*y/x)*px<n) //does not fit, y/(x/px)=px*y/x | |
sx=y/ceil(px*y/x); | |
else | |
sx= x/px; | |
double py=ceil(sqrt(n*y/x)); | |
if(floor(py*x/y)*py<n) //does not fit | |
sy=x/ceil(x*py/y); | |
else | |
sy=y/py; | |
printf("%f",MAX(sx,sy)); | |
return 0; | |
} |
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def values(x = 5.0, y = 6.0, a = 7.0) | |
n = (a / 4.0).ceil | |
px = ( Math.sqrt( n * x / y) ).to_f.ceil | |
sx = 0.0 | |
sy = 0.0 | |
if (px * y / x).floor * px < n # does not fit, y/(x/px)=px*y/x | |
sx = y / (px * y / x).to_f.ceil | |
else | |
sx = x / px | |
end | |
py = ( Math.sqrt(n * y / x) ).to_f.ceil | |
if (py * x / y).floor * py < n # does not fit | |
sy = x / (x * py / y).to_f.ceil | |
else | |
sy = y / py | |
end | |
# printf "%f", [sx, sy].max | |
v = [sx.round(2), sy.round(2)].max | |
printf "%f", v | |
return v | |
end |
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