Solution 4286

Submitted on 26 Jan 2012 by David Hruska

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Brett Shoelson on 11 May 2012

Nice, David!

Jon Agustsson on 6 Jul 2012

Nice indeed.
I am surprised that it is more efficient to solve this with a recursive function than a while loop.
This solution has size 43:

while n(end) > 1
n = [n mod(n(end),2)*(n(end)*2.5+1) + n(end)/2];
end

Is the loop overhead greater than the recursion overhead? Or is the weak point having to go to n(end) at each iteration?

Jon Agustsson on 6 Jul 2012

Strange I can not delete my first comment. I wanted to format it a bit nicer :)

Chien on 14 Aug 2013

Good solution, David!:)

Test	Status	Code Input and Output
1	Pass	%% n = 1; c_correct = 1; assert(isequal(collatz(n),c_correct))
2	Pass	%% n = 2; c_correct = [2 1]; assert(isequal(collatz(n),c_correct))
3	Pass	%% n = 5; c_correct = [5 16 8 4 2 1]; assert(isequal(collatz(n),c_correct))
4	Pass	%% n = 22; c_correct = [22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1]; assert(isequal(collatz(n),c_correct))

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Problem 21. Return the 3n+1 sequence for n