This is a shamless answer,I admit...But the explicit recurrence relation is insanely slow,hope to see you guys solve this problem more efficient.
I prefer g(end+1:end+g(gptr))=gptr; to usage of repmat. My machine to solve 1234567 takes 48msec vs 15.4 sec using repmat. repmat has a performance issue with large column replication. Unfortunately score is code size and not time.
totally agree (not to mention the entire 'growing inside a loop' uglyness), cody style is very far from any reasonable coding standard...
That's very interesting. The time difference on my (presumably much older) version of MATLAB is much less. Your method gives me an average time of about 18.8 sec, while repmat gives me an average time of 19.5 sec.
May you give a short explanation on this solution?
This is the asymptotic expression of nth term based on the golden ratio. See http://en.wikipedia.org/wiki/Golomb_sequence
Find the sum of all the numbers of the input vector
Get the elements of diagonal and antidiagonal for any m-by-n matrix
Maximum value in a matrix
Area of an equilateral triangle
Reindex a vector
Accessing values in a cell
Sums of Multiple Pairs of Triangular Numbers
Return elements unique to either input
Should Mr McNugget even get out of bed today?
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