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Albert Yam submitted Solution 33034 to Problem 10. Determine whether a vector is monotonically increasing

on 14 Jan 2013

Albert Yam submitted Solution 133355 to Problem 792. Set some matrix elements to zero

on 31 Aug 2012

Albert Yam submitted Solution 133354 to Problem 796. Removing rows from a matrix is easy - but what about inserting rows?

on 31 Aug 2012

Albert Yam submitted Solution 133353 to Problem 362. Find matching string from a list of strings

on 31 Aug 2012

Albert Yam submitted Solution 105928 to Problem 722. Make a run-length companion vector

on 3 Jul 2012

Albert Yam submitted Solution 105916 to Problem 731. Given a window, how many subsets of a vector sum positive

on 3 Jul 2012

Albert Yam submitted Solution 105849 to Problem 729. Flag largest magnitude swings as they occur

on 3 Jul 2012

Albert Yam submitted Solution 96716 to Problem 639. String substitution, sub problem to cryptoMath

on 7 Jun 2012

Albert Yam submitted Solution 96692 to Problem 362. Find matching string from a list of strings

on 7 Jun 2012

Albert Yam submitted Solution 74343 to Problem 105. How to find the position of an element in a vector without using the find function

on 11 Apr 2012

Albert Yam submitted Solution 74341 to Problem 105. How to find the position of an element in a vector without using the find function

on 11 Apr 2012

Albert Yam submitted Solution 74339 to Problem 105. How to find the position of an element in a vector without using the find function

on 11 Apr 2012

Albert Yam submitted a Comment to Solution 73793

%the part of the code i cut out, should work in theory ..
fliplist = fliplr(list);
idx = triu(ones(length(list)),0);
for j = 1:length(idx)
idx2 = unique(perms(idx(j,:)),'rows');
[a b] = size(idx2);
for i = 1 : a
idxi = boolean(idx2(i,:)');
list2 = list;
list2(idxi,:) = fliplist(idxi,:);
list2 = list2';
list2 = list2(:);
dlist = abs(diff(list2));
val2 = sum(dlist(2:2:end));
if val2 < val
val = val2;
orientation = idxi';
end
if val == 0
return
end
end
end

on 10 Apr 2012

Albert Yam submitted Solution 73753 to Problem 492. Find best placement for ordered dominoes (harder)

on 10 Apr 2012

Albert Yam submitted Solution 73749 to Problem 487. Find perfect placement of non-rotating dominoes (easier)

on 10 Apr 2012

Albert Yam submitted a Comment to Problem 391. Poker Series 11: selectBestHand

On a second thought.. shouldn't there be only one deck?

on 29 Feb 2012

Albert Yam submitted a Comment to Solution 54153

Should work for :
hm1=[1 0 0 0 0 0 0 0 0 0 0 0 1;1 0 0 0 0 0 0 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0 0 0 0];
hm2=[1 0 0 0 0 0 0 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0 0 0 1];

on 29 Feb 2012

Albert Yam submitted a Comment to Problem 391. Poker Series 11: selectBestHand

hm1=[1 0 0 0 0 0 0 0 0 0 0 0 1;1 0 0 0 0 0 0 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0 0 0 0];
hm2=[1 0 0 0 0 0 0 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0 0 0 1];
Is player 1 supposed to win in this case?

on 29 Feb 2012

Albert Yam submitted a Comment to Problem 386. Poker Series 10: bestHand

Test 6 and 7 are the same (no actual problems, just saying).

on 28 Feb 2012