Find the sum of the elements in the diagonal that starts at the top-right corner and ends at the bottom-left corner.
you might want to make a case that's not a magic square, considering the properties of the diagonals of a magic square...
There are only magic squares in test cases, however the problem doesn't specify a type of the matrix.
You should definitely strengthen the test suite. Lots of solutions rely on both diagonals being the same, even though this is not a property of the stated problem.
The "antidiagonal" is cooler sounding than "second diagonal"
I learned about fliplr() in another problem and it is helpful here!
Thanks for the hint
I like you're use of a nested loop here.
The last test suite result should be 14.
yes ı agree
Trickey ONE indeed !!
But you can't rely on x always being a magic square - that's just an accident of the inadequate test suite, isn't it? So although this *looks* like a good solution, it isn't really. I think it's bad practice to take advantage of an inadequate test suite.
it doesn't find the sum of the elements of the second but first diagonal
Change the sign of even index entries of the reversed vector
07 - Common functions and indexing 4
Omit columns averages from a matrix
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