Asked by Matt Fig
on 30 Mar 2011

This challenger is very easy to state:

Given an array of random size and dimensions whose only nonzero element is the first element, find the locations of the nonzero elements of the array returned by a random call to the REPMAT function.

Now for the details. Provided below is a test code for use in testing your solution to the challenger. In the code an array A is created, with random size and number of dimensions held in the vector SA. Another vector RA is created to be used as the second argument to REPMAT. Then a call is made to find the locations of the non-zero elements of the repmatted array using:

I = find(repmat(A,RA));

The challenge is to obtain the same vector as I, following two rules:

- Your function must not reproduce the repmatted array, or any array of equal or larger size, either implicitly or explicitly, in any form. What this means practically is that your function will succeed when REPMAT fails due to either an out of memory problem or a maximum variable size exceeded problem.
- Your function should be faster than the call to FIND and REPMAT shown above.

For those who are indexing experts this may not be much of a challenge, but for those who are still learning this could be very challenging indeed. I think one could learn a lot by completing the challenge, even if it is difficult, so keep trying! Study the test function below to understand the problem better and see how your code will be used (input args, etc). Remember, when testing your solution run the test function more than once in a row to get good timing results. Good luck. And don't forget to vote up good answers! .

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function [] = test_solutions() % Use to test an Answer for the Hump-Day Challenger.

N = 300; % The number of loop iterations. Tm = [0 0]; % To hold the timings. bf = 0; % Flag to check if user code passed. Z = 6; % If you have consistent memory problems, lower this...

for ii = 1:N % First make our array, and decide how to repmat it. SA = ceil(rand(1,ceil(rand*Z))*(Z+1)); % The size of A. A = false(SA); A(1) = true; RA = ceil(rand(1,ceil(rand*Z))*(Z+1)); % The vector for REPMAT.

% Now compare speeds. tic I = find(repmat(A,RA)); Tm(1) = Tm(1) + toc;

tic Imf = hdchallenger(SA,RA); % Your func should take only SA and RA. Tm(2) = Tm(2) + toc;

if ~isequal(I(:),Imf(:)) disp(' Unequal results, please try again.') bf = 1; break end

end

RT = Tm/min(Tm); % The relative timing.

if RT(2)==1 && ~bf fprintf('You passed! You beat FIND by a factor of: %.1f!\n',RT(1)) elseif ~bf fprintf('Your code returns good results, but is slow. Keep trying!\n') end

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Answer by Sean de Wolski
on 30 Mar 2011

Accepted answer

function idx = hdchallenger(SA,RA) %% SCd 03/30/2011 %Updated %Adjust for scalars, zeros, and different lengths (pad with ones) if isscalar(RA) RA = [RA RA]; end if isscalar(SA) SA = [SA SA]; end lenRA = length(RA); lenSA = length(SA); if lenRA<lenSA RA(lenSA) = 1; RA(RA==0) = 1; end

%Engine n = prod(RA); didx = SA(1)*ones(n-1,1); %preallocate for ii = 2:lenRA; skip = n/prod(RA(ii:end)); if ii > lenSA; %Don't exceed SA! break end mpart = zeros(1,ii); mpart(ii) = -1; %remove present 1 dim from count didx(skip:skip:end) = prod([RA(1:ii-1),(SA(1:ii)+mpart)])+didx(skip:skip:end); end idx = cumsum([1; didx]); %integrate!

Andrew Newell
on 30 Mar 2011

It's faster on my machine too. I get killed by the repmat's in my code.

Answer by Andrew Newell
on 31 Mar 2011

I did a lot of tweaking to get rid of `repmat`, preallocate `II`, and replace `sub2ind` by more efficient code. Alas, it is still slower than @Sean de's offering.

function I = hdchallenger1(SA,RA)

% repmat(A,m) = repmat(A,m,m) if length(RA)==1 RA = [RA RA]; end

% false(m) = false(m,m) if length(SA)==1 SA = [SA SA]; end

if length(RA)>length(SA) SA = [SA ones(1,length(RA)-length(SA))]; else RA = [RA ones(1,length(SA)-length(RA))]; end

% Calculate subscripts for ones II = zeros(prod(RA),length(RA)); m1 = [1 cumprod(RA)]; m2 = m1(end)./m1; for j=1:length(RA) JJ = 1 + cumsum([zeros(m1(j),1) SA(j)*ones(m1(j),RA(j)-1)],2); JJ = JJ(:); JJ = JJ(:,ones(1,m2(j+1))); II(:,j) = JJ(:); end

% Convert to linear index (cf. sub2ind) siz = RA.*SA; k = [1 cumprod(siz(1:end-1))]; I = 1+ (II-1)*k';

Sean de Wolski
on 31 Mar 2011

Andrew Newell
on 31 Mar 2011

Answer by Andrew Newell
on 30 Mar 2011

function I = hdchallenger(SA,RA)

% repmat(A,m) = repmat(A,m,m) if length(RA)==1 RA = [RA RA]; end

% false(m) = false(m,m) if length(SA)==1 SA = [SA SA]; end

if length(RA)>length(SA) SA = [SA ones(1,length(RA)-length(SA))]; else RA = [RA ones(1,length(SA)-length(RA))]; end

II = 1; for j=1:length(RA) JJ = (1:SA(j):(SA(j)*(RA(j)-1)+1)); JJ = repmat(JJ,size(II,1),1); if j==1 II = JJ(:); else II = [repmat(II,RA(j),1) JJ(:)]; end end

Ic = num2cell(II,1); I = sub2ind(RA.*SA,Ic{:});

if SA(1)*RA(1)==1 I = I(:); end

Show 2 older comments

Sean de Wolski
on 30 Mar 2011

What causes find to return a row vector? I can't seem to 'find' any pattern!

Answer by Andrew Newell
on 31 Mar 2011

I tried to finesse this problem using the function `sptensor` from the Tensor Toolbox, but alas - `repmat` doesn't work properly because `sptensor` doesn't use matrix indices the way that ordinary multidimensional arrays do (see my question on Indexing arrays with matrices).

Opportunities for recent engineering grads.

## 1 Comment

## Matt Fig (view profile)

Direct link to this comment:http://nl.mathworks.com/matlabcentral/answers/4359#comment_17526

Thanks to Sean de and Andrew Newell both!