For any population p of positive integers between pmin and pmax, we can calculate histogram count n for a list of evenly dividing divisors d. Here's how it works.
Consider the population p = [10, 12, 15, 11, 20, 11]. For the list of divisors d = [2, 3, 5] we get a histogram count n of [3, 2, 3]. That is, three of the numbers in the population are evenly divisible by 2, two are evenly divisible by 3, and three are evenly divisible by 5.
You will face the inverse problem. Given a histogram count corresponding to a list of divisors, produce a population of positive integers that fits the distribution. Your answer will not be unique and can be provided in any order. It just has to match the distribution. The elements of p can be no less than pmin and no greater than pmax.
Example:
pmin = 5
pmax = 10
d = [1 2 3 4 5]
n = [8 3 4 2 2]
One answer (of many possible): p = [5 5 6 8 8 9 9 9]
Solution Stats
Problem Comments
3 Comments
Solution Comments
Show comments
Loading...
Problem Recent Solvers52
Suggested Problems
-
3843 Solvers
-
Rosenbrock's Banana Function and its derivatives
165 Solvers
-
Remove the two elements next to NaN value
703 Solvers
-
Lychrel Number Test (Inspired by Project Euler Problem 55)
111 Solvers
-
Digit concentration in Champernowne's constant
125 Solvers
More from this Author54
Problem Tags
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
It looks like there's an error in the test suite; test cases 3 and 4 have different length d and n vectors.
Oof! I hate it when that happens. Fixed it. Thanks for the note.
Great head-scratcher!