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All possible boxes inside a cube recursively
Could you come up with recursive function to calculate all the boxes inside an N by N by N cube? By looping I get this: N -> boxes 1 -> 0 2 -> 18 3 -> 180 4 -> 900 5 -> 3150 The recursive function may be somethig like: f(N) => ... + f(N-1) Here's the recursion to calculate rectangles inside N by N square as a 2D analogy: return N<2? 0 : (Math.pow(N,3) - Math.pow(N,2)) + rec(N-1);
2 ответов
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Definitely possible, would take me quite some time though and the recursion would become increasingly ram hungry as the values get bigger
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AnonyMouse
If you manage you are also encouraged to come up with the recursive function to calculate all possible hyper boxes inside 4-dimensional hypercube with side N.
By looping I got:
N -> hyperboxes
1 -> 0
2 -> 64
3 -> 1198
4 -> 9646
5 -> 49646
And see if you can generalize the recursion to calculate all possible rectangles/boxes/hyperboxes inside a square/cube/hypercube with side N.
For consideration, all possible square/cube/hypercube inside another square/cube/hypercube can be recursively calculated as follows:
return x<2 ? 1 : Math.pow(x,y) + sqr(x-1,y);
where x is the side of square/cube/hypercube and y is 2 for square, 3 for cube and 4 for hypercube (the dimensions).
I was wondering if rectangles inside square and its higher dimensional analogs can be recursively calculated by a general function as well