30 lines
1.3 KiB
Plaintext
30 lines
1.3 KiB
Plaintext
Provide code that produces a list of numbers which is the n-th order forward difference, given a non-negative integer (specifying the order) and a list of numbers.
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The first-order forward difference of a list of numbers (A) is a new list (B) where B<sub>n</sub> = A<sub>n+1</sub> - A<sub>n</sub>. List B should have one fewer element as a result.
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The second-order forward difference of A will be tdefmodule Diff do
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def forward(arr,i\\1) do
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forward(arr,[],i)
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end
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def forward([_|[]],diffs,i) do
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if i == 1 do
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IO.inspect diffs
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else
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forward(diffs,[],i-1)
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end
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end
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def forward([val1|[val2|vals]],diffs,i) do
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forward([val2|vals],diffs++[val2-val1],i)
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end
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endhe same as the first-order forward difference of B.
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That new list will have two fewer elements than A and one less than B.
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The goal of this task is to repeat this process up to the desired order.
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For a more formal description, see the related [http://mathworld.wolfram.com/ForwardDifference.html Mathworld article].
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Algorithmic options:
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*Iterate through all previous forward differences and re-calculate a new array each time.
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*Use this formula (from [[wp:Forward difference|Wikipedia]]):
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:<math>\Delta^n [f](x)= \sum_{k=0}^n {n \choose k} (-1)^{n-k} f(x+k)</math>
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:([[Pascal's Triangle]] may be useful for this option)
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