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!
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
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Tick. Tock. Tick. Tock. Tick. Tock. Tick. Tock. Tick. Tock.
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