NEED HELP ON MATRIX MANIPULATIONS

Consider a matrix

A= 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

My objective is to reach from base value 0 to top value 0 as underlined for dynamic time warp algorithm traversal by parsing through the smallest element of each row.

A= 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

The cost assignment can be defined as find the smallest element in each column :

1| 21 14 7 0

0| 18 11 4 3

1| 15 8 1 6

0| 12 5 2 9

0| 9 2 5 12

1| 6 1 8 15

0| 3 4 11 18

1| 0 7 14 21

So cost= 1 0 1 0 0 1 0 1

Cost assignment concept:

assign 1 whenever there's a smallest single unique element in the row .

Assign 0 whenever no smallest element is present or when multiple smallest same elements are present in the same row or column,then first instance should be zero and next to be 1 i.e. for example

1| 20 15 10 5 0

1| 16 11 6 1 4

1| 12 7 2 3 8

0| 8 3 2 7 12

1| 4 1 6 11 16

1 | 0 5 10 15 20

Answer 1

[KSSV]

A = [  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];
                
m = min(A,[],2) ;
 cost = m == 0 | m == 1 ;    

[anshuman mishra]

Thanks, That works fine for matrix A but fails for

1| 20 15 10 5 0

1| 16 11 6 1 4

1| 12 7 2 3 8

0| 8 3 2 7 12

1| 4 1 6 11 16

1 | 0 5 10 15 20

Here the the output from the code is 1 1 0 0 1 1

But i need output 1 1 0 1 1 1 i.e. only the two rows having same smallest element must be assigned value a single value 0 instead of 0 & 0

Im assuming you took only 1 & 0 into consideration instead my question was whenever you detect a single non repeatable smallest element in the rows then assign it as 1 & when same multiple repeatable elements in rows as in the above case then assign 0.