Create n x 2n "mirror" matrix of this type:
For n = 2
m = [ 1 2 2 1
1 2 2 1 ]
For n = 3
m = [ 1 2 3 3 2 1
1 2 3 3 2 1
1 2 3 3 2 1 ]
This problem was fun. I was able to find a better way to solve beat my own solution. Not the best way though. But lot better.
What is the next step in Conway's Life?
Implement a ROT13 cipher
Project Euler: Problem 10, Sum of Primes
Determine the number of odd integers in a vector
Sum of series I
Sum of series V
Sum of series III
Sum of series VI
Sum of series VII
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