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generate a sequence from matrix

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anshuman mishra
anshuman mishra on 29 Jul 2019
Commented: dpb on 30 Jul 2019
given matrix:
new =[
35 28 21 14 7 0
30 23 16 9 2 5
25 18 11 4 3 10
20 13 6 1 8 15
15 8 1 6 13 20
10 3 4 11 18 25
5 2 9 16 23 30
0 7 14 21 28 35]
i need to evaluate from 8th row,1st column.
i need to traverse from 0,2,3,1,1,3,2,0 (least element in each row)
and assign values 1 0 1 1 1 0 1 ( 1 when traversing diagonally and 0 when traversing sideways or upward but not backward or downward)

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Accepted Answer

dpb
dpb on 29 Jul 2019
Interesting...kinda' cute solution...altho note that the suggested answer is wrong; the minimum for row 7 is the 3 in column 2, not 4 in colum 3
[~,j]=min(flipud(new),[],2);
ix=diff(j)~=0;
Above returns
ix =
7×1 logical array
1
0
1
1
1
0
1
>>

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dpb
dpb on 29 Jul 2019
Just pass the desired subset of the array...
[~,j]=min(flipud(new(:,2:end)),[],2);
ix=diff(j)~=0;
In general, the column expression can be any general subscripting vector desired.
anshuman mishra
anshuman mishra on 30 Jul 2019
okay,thanks. i have a new issue:
given matrix
a=
16 8 0
13 5 3
10 2 6
7 1 9
4 4 12
1 7 15
2 10 18
5 13 21
when im traversing from 1 to 4 ( as underlined) it should go from 1 to 4 diagonally not vertically(The row above 1 has two 4's, so to avoid conflict i want to move diagonally instead of vertically).The previous code like it assigns 1 for diagonal movement should also work here,because since it goes vertically it gives me 0, whereas i want it to go diagonally,and get 1.
dpb
dpb on 30 Jul 2019
Well, you've just broken any chance for a one-liner I can see... :)
You'll have to do some preprocessing to eliminate the duplicated minima where they exist before calling the above algorithm or have to go back afterwards and check for duplicates and clean up the result if founc.
Or, just loop from the git-go and don't try to be fancy and find each next location one-at-a-time accounting for the duplicates as you go.

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