index of a sequence

Hello Everyone,

 i have a small problem regarding Johnson's Algorithm, Scheduling (2 machines(a,b), 5 jobs(5 columns for each )

%a=[50 150 80 200 30];
%b=[60 50 150 70 200];

 my idea is to find the min(min(a),min(b)) (application : min(30,50) = 30) in a FOR LOOP (i:5) then construct a sequence of the two machines (a,b) , regarding each side of the machines (a,b) , for example if the first min is in (a) then put it LEFT otherwise put it RIGHT after that SORT each side separately , the final result i get is : 30 50 80 70 50 , what i want to get is the INDEX of this sequence, for example the index of this sequence "30 50 80 70 50" should be "5 1 3 4 2" .

%-------------the code-------------------
clear all;clc
a=[50 150 80 200 30];
b=[60 50 150 70 200];
comp01=[];
comp02=[];
for i = 1:5
  t= min((a(i)),(b(i)));
  if a(i)== t
        comp01 = [t,comp01];
        left = comp01;
        left = sort(left,'ascend');
        %[left,x1] = sort(left);
  else
      comp02 = [t,comp02];
      right = comp02;
      right = sort(right,'descend');
      %[right,x2] = sort(right);
 end
end
Sequence_in_numbers = [left right]
%Sequence_in_index = [x1 x2]
%-------------end of the code-------------------

Answer 1

[Matt Kindig]

You don't really need the loops:

a=[50 150 80 200 30];
b=[60 50 150 70 200];
[d,rows]= min([a;b], [], 1);    %get lowest value in each column
lcols = find(rows==1);          % columns where a is smaller
rcols = find(rows==2);          % columns where b is smaller
[left,lorder] = sort(d(lcols), 'ascend');     %left elements
[right,rorder] = sort(d(rcols),'descend');    %right elements
Sequence_in_numbers = [left right];           %sequence of numbers
Sequence_in_index = [lcols(lorder), rcols(rorder)];  %index of the sequence

[Hajar Sadki]

i have two machines Ai Bi and i=1...n . i'm looking for sequence of numbers and index of the sequence. it is same code ? what can i change ?

Answer 2

[Andrei Bobrov]

[v,ii]= min([a;b]);
t = accumarray(ii',(1:numel(a))',[],@(x){x});
[v2,i2]=cellfun(@(x)sort(v(x)),t,'un',0);
sn = [v2{1},v2{2}(end:-1:1)];
si = [t{1}(i2{1});t{2}(i2{2}(end:-1:1))]';

[sami]

thank you too man

Answer 3

[sami]

thank you man, it was very very helpful