# calculate the minimum probability

9 views (last 30 days)
ELISABETTA BILLOTTA on 19 Jan 2022
Answered: Steven Lord on 19 Jan 2022
I have in one case two matrices and in the other case three matrices of size 2922x1. the values within the matrices are probability values. how do i calculate the minimum probability between these 2 and between these 3 matrices?
##### 2 CommentsShowHide 1 older comment
ELISABETTA BILLOTTA on 19 Jan 2022 then let's take the case of the two matrices: these two matrices are constructed in the same way (200x200 double), that is, made up of 200 rows and 200 columns. I have to create a third matrix characterized by the minimum value between the two on all rows and all columns.
case of the three matrices: there are three matrices with 200 rows and 200 columns, I have to create a fourth matrix that fourth matrix characterized by the minimum value among the three on all rows and all columns.
considering the image:
a3 is the minimum (lowest) value between a1 and a2,
b3 is the minimum value between b1 and b2
and so on
hope to be better explained now .. let me know thanks

Jon on 19 Jan 2022
Say you have three matrices of the same size A,B,C then you can find the minimum as you describe using
D = min(min(A,B),C)

### More Answers (1)

Steven Lord on 19 Jan 2022
Assuming they are the same class and size, you can concatenate them and call min with a dimension input.
A = randi([-5, 5], 4)
A = 4×4
-2 0 3 -1 -1 0 4 5 3 -1 5 -2 0 -3 4 1
B = randi([-5, 5], 4)
B = 4×4
-3 2 -3 1 -4 3 3 -1 -1 5 5 -5 -2 1 2 -1
C = cat(3, A, B)
C =
C(:,:,1) = -2 0 3 -1 -1 0 4 5 3 -1 5 -2 0 -3 4 1 C(:,:,2) = -3 2 -3 1 -4 3 3 -1 -1 5 5 -5 -2 1 2 -1
D1 = min(C, [], 3) % Min along the third dimension
D1 = 4×4
-3 0 -3 -1 -4 0 3 -1 -1 -1 5 -5 -2 -3 2 -1
The concatenation strategy can work with more than two arrays, but for this simple example of two arrays you can just call the binary form of min.
D2 = min(A, B)
D2 = 4×4
-3 0 -3 -1 -4 0 3 -1 -1 -1 5 -5 -2 -3 2 -1