# Matrix value comparision and manipulation

2 views (last 30 days)
anshuman mishra on 16 Dec 2019
Edited: anshuman mishra on 16 Dec 2019
i have a matrix C=
21 14 7 0
18 11 4 3
15 8 1 6
12 5 2 9
9 2 5 12
6 1 8 15
3 4 11 18
0 7 14 21
smallest element amongst the rows: D= 0 3 1 2 2 1 4 0
Smallest element amongst columns: E= 0 1 1 0
whenever the 1st element of E is matched with first element of D ,then assign 1, then the 2nd element of E should match 2nd element of D, if not assign 0.then 2nd element of E which is not yet matched is compared with 3rd element of E & so on.
FINAL OUTPUT : 1 0 1 0 0 1 0 1

KALYAN ACHARJYA on 16 Dec 2019
Edited: KALYAN ACHARJYA on 16 Dec 2019
C=[21 14 7 0
18 11 4 3
15 8 1 6
12 5 2 9
9 2 5 12
6 1 8 15
3 4 11 18
0 7 14 21];
D=min(C')
E=min(C);
result=ismember(D,E)
#Final_Output:
result =
1×8 logical array
1 0 1 0 0 1 0 1

anshuman mishra on 16 Dec 2019
sir, Your answer is correct only for this case
if C=
20 15 10 5 0
16 11 6 1 4
12 7 2 3 8
8 3 2 7 12
4 1 6 11 16
0 5 10 15 20
smallest element amongst the rows: D= 0 1 2 2 1 0
Smallest element amongst columns: E= 0 1 2 1 0
So final output should be : 1 1 1 0 1 1 not 1 1 1 1 1 1
Steven Lord on 16 Dec 2019
Why is the fourth element of the output 0 instead of 1? The minimum element in the fourth row of C is the 2 in the third column, which happens to be one of the instances of the minimum element in the third column?
What rule are you following that makes the answer [1 1 1 0 1 1] instead of [1 1 1 1 1 1] or [1 1 0 1 1 1]?
anshuman mishra on 16 Dec 2019
smallest element amongst the rows: D= 0 1 2 2 1 0
Smallest element amongst columns: E= 0 1 2 1 0
So here
0 matches with 0 , so we assign 1
1 matches with 1 , so we assign 1
2 matches with 2 , so we assign 1
1 doesnt match 2, assign =0
so here before going to next element of E after 1 i.e 0, we need to match 1 of E with next element of D i.e 1
1 matches with 1, assign 1
0 matches with 0, assign 1
So final cost = 1 1 1 0 1 1

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