memory overflow with double factorial function
14 次查看(过去 30 天)
显示 更早的评论
I am trying to create a recursive formula to calculate the double factorial of a number, n. The code I have come up with currently is as follows:
function [DFact] = DFactorialRec(n)
DFact = n*DFactorialRec(n-2)
end
This clearly does not work as it will recursively continue for negative values of n. Please could someone suggest an alteration to stop the memory overflow error?
================================================================
Out of memory. The likely cause is an infinite recursion within the program.
Error in DFactorialRec (line 2)
DFact = n*DFactorialRec(n-2)
================================================================
Thanks!
5 个评论
Walter Roberson
2020-10-9
The code I posted worked for me provided that n is a scalar. If it is a non-scalar then you have the problem that your if is comparing a non-scalar to a scalar, and that comparison might be true for some elements but false for others...
采纳的回答
John D'Errico
2020-10-9
编辑:John D'Errico
2020-10-9
Recursive formulas like this always seem useful, but they are terrible in terms of coding, in terms of efficiency. You don't want to write it that way! Yes, the code you wrote looks pretty. That it will not work as you wrote it is a problem of course.
Think about it. For every recursive call, you need to allocate a new workspace, you incur function call overhead. This is a bad thing, and done for absolutely no good reason. Instead, a simple loop is all you need. Start at the bottom.
But better even yet is to not bother with an explicit loop. You can write this function in not much more than one line of code. Just use prod. The rest of what I did was mostly to make it friendly.
function DFact = DFactorial(n)
% Double factorial function, n!!
% https://en.wikipedia.org/wiki/Double_factorial
%
% n!! = 1 for both n == 1 and n == 0.
if isempty(n) || (numel(n) > 1) || (n < 0) || (mod(n,1) ~= 0)
error('The sky is falling. n must be scalar, non-negative, integer.')
end
DFact = 1;
if n > 1
start = 1 + mod(n + 1,2); % caters for either parity of n
DFact = prod(start:2:n);
end
end
Will this code fail for large values of n? Of course. It is easy to overwhelm the dynamic range of a double. But that will happen for any code that uses double precision. An exact integer will not be produced for n > 29 when done in double precision, and inf will result above 300. But again, that will happen in any case.
It is also easy to write a version that handles vector or array arguments for n.
2 个评论
Nate Crummett
2022-8-7
Vectorized code from above:
function DFact = DFactorial(n)
% Double factorial function, n!!
% https://en.wikipedia.org/wiki/Double_factorial
%
% n!! = 1 for both n == 1 and n == 0
% Inputs must be vectors of positive, real integers
%
% Designed by John D'Errico on 9 Oct 2020, taken from the MATLAB forum
% Vectorized by Robert Nate Crummett 30 July 2022
DFact = zeros(size(n));
idx_zeros = find(n == 0);
idx_ones = find(n == 1);
DFact([idx_ones, idx_zeros]) = 1;
n_end = n(~DFact);
start = 1 + mod(n_end + 1,2); % caters for either parity of n
prod_results = zeros(size(n_end));
for i = 1:length(n_end)
prod_results(i) = prod(start(i):2:n_end(i));
end
DFact(~DFact) = prod_results;
end
更多回答(1 个)
SteveC
2024-7-22
编辑:Walter Roberson
2024-7-22
The double factorial of an integer -1 and larger is defined as in
and generalized in the Pochhammer symbol
It appears in electromagnetic expansions and in special functions.
if the argument is an integer -1 and larger, it is written in Matlab as
fDF = @(bb) prod(bb:(-1):1)
For a vector example
>> arrayfun(fDF,-1:5)
1 1 1 2 6 24 120
If the application manages the validity of the argument, a single line without parity check is OK to define the function.
For big args, a Stirling form...
1 个评论
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Argument Definitions 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!