Find the sum of the elements in the diagonal that starts at the top-right corner and ends at the bottom-left corner.
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Problem Comments
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4 Comments
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1 older comment
@bmtran (Bryant Tran)
on 12 Feb 2012
you might want to make a case that's not a magic square, considering the properties of the diagonals of a magic square...
Gaik Tamazian
on 17 Feb 2012
There are only magic squares in test cases, however the problem doesn't specify a type of the matrix.
David Young
on 1 Mar 2012
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.
Kye Taylor
on 7 Apr 2012
The "antidiagonal" is cooler sounding than "second diagonal"
Solution Comments
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2 Comments
Brian Butler
on 14 Nov 2015
I learned about fliplr() in another problem and it is helpful here!
MAHMOUD ABDEL-LATIF
on 12 Jun 2019
Thanks for the hint
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1 Comment
Nicholas Hortance
on 8 Jul 2015
I like you're use of a nested loop here.
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3 Comments
Haresh Varu
on 29 May 2013
The last test suite result should be 14.
Abdullah Caliskan
on 22 Jan 2014
yes ı agree
md.wahid tousif_rahman
on 13 Jun 2017
Trickey ONE indeed !!
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1 Comment
David Young
on 1 Mar 2012
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.
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1 Comment
Stephane Dauvillier
on 7 Mar 2012
it doesn't find the sum of the elements of the second but first diagonal
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