Since Cody, I dislike for-loops, but your solution is way simpler than my own algorithm. Mission accomplished...
Nevertheless, I would hope for a non-looped piece of code. Let's see in phase 3.
Make the vector [1 2 3 4 5 6 7 8 9 10]
Determine the square root
Divisible by 5
Find my daddy long leg (No 's')
Write c^3 as sum of two squares a^2+b^2
Wayfinding 1 - crossing
Create a random vector of integers with given sum
Three grind is shipsstraigt
Wayfinding 2 - traversing
A SUBSREF variant that accepts the 'end'-operator.
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