Matrix value comparision and manipulation

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

Answer 1

[KALYAN ACHARJYA]

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]

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]

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]

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