how can i sort a matrix (all rows and columns are sorted) without using any special function like "sort"?
Afficher commentaires plus anciens
i guess it has to do something with "min" and "max" functions...
for example:
A = [2 3 4 5; 6 9 1 5]; %%%"A" can be of any size %%%
%%%B = sorted A %%%
min(A) = [2 3 1 5]; %%%1st unsorted row of B %%%
max(A) = [6 9 4 5]; %%%2nd unsorted row of B %%%
now i've to sort the rows...
any idea?
2 commentaires
John Petersen
le 5 Déc 2012
Lots of ideas on this at
Babak
le 5 Déc 2012
Matrix A you are providing only has 1 row. How can B have more than 1 row?
Réponses (3)
Jan
le 6 Déc 2012
1 vote
This is a perfect question for an internet research:
or
If you don't want to use sort() then you can write your own sorting algorithm. Like this one (untested):
function B = mysortfunc(A)
if size(A,1)*size(A,2)~=length(A)
errordlg('enter a vector');
return
end
if size(A,2)~=1
A=A';
end
rem_A = A;
for j=1:length(A)
[value,index] = min(rem_A);
B(j) = value;
if index>1
A1 = rem_A(1:index-1);
else
A1=[];
end
if index<length(rem_A)
A2 = rem_A(index+1:end);
else
A2 = [];
end
rem_A = [A1 , A2]
end
3 commentaires
A simplification of your method with pre-allocation:
function B = mysortfunc(A)
A = A(:);
B = zeros(1, numel(A));
for ii = 1:numel(A)
[value, index] = min(A);
B(ii) = value;
A(index) = NaN;
end
However, the scientists have developped much better sorting methods in the last 1000 years. I do not assume that a teacher would be happy to see such a brute-force approach. Therefore I dare to publish this, although it solves a homework question, because submitting it as "solution" is a bad idea.
José-Luis
le 7 Déc 2012
1000 years? Babbage would be impressed...
Jan
le 9 Déc 2012
I'm convinced, that even the scroll of parchment in the Egypt libraries have been sorted with smarter methods 5000 years ago, but I cannot find any resources to prove this.
The optimal sorting machine is still the SFL (spaghetti fork lifter): Cut spaghetti noodles according to the values to be sorted. Lift them up and push them against a wall. The processing time does not depend on the number of elements and even the pre-processing is only O(n).
Pritesh Shah
le 7 Déc 2012
Simple Solution A = [2 3 4 5; 6 9 1 5]
A =
2 3 4 5
6 9 1 5
>> sort(min(A))
ans =
1 2 3 5
1 commentaire
John Petersen
le 7 Déc 2012
You used 'sort' which he specifically requested not to be used, and he also specified that A could be any size.
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