# randomly select 3 numbers whose sum is 356 and each of these 3 is more than 30

please how can I randomly select 3 numbers whose sum is 356 and each of these 3 is more than 30?

So output should be for example [100, 34, 222] (but not [1,5,350])

I would like to use random module to do this. thank you!

## Answers

import random def rnd_to_sum_with_min(desired_sum=356, numbers=3, minimum=31): selection = [] for i in range(numbers-1): upper_limit = desired_sum - sum(selection) - (numbers - i) * minimum choice = random.randint(minimum, upper_limit) selection.append(choice) selection.append(desired_sum - sum(selection)) return selection

This solution uses a function. This is so that parameters can be modified on-the-fly. This makes for a robust solution.

The method is to select the first N-1 numbers randomly, and determine the last number according to the difference to the desired sum. Each number is determined by first knowing the minimum value desired, and then calculating the upper limit of that number based on the desired sum, and what numbers have already been chosen, making sure to leave enough behind for the next n numbers.

I also wrote a recursive function just for fun (Which is faster):

def recursive_rnd_sum_by_n_with_min(_sum=356, _num=3, _min=31): def f(_sel=[]): left = _num - len(_sel) so_far = sum(_sel) if left == 1: return _sel + [_sum - so_far] return f(_sel + [random.randint(_min, _sum - so_far - left * _min)]) return f()

This question is rather subjective to the definition of random, and the distribution you wish to replicate.

The simplest solution:

- Choose a one random number, rand1 : [30,296]
- Choose a second random number, rand2 : [30, (326-Rand1)]
- Then the third cannot be random due to the constraint so calc via 356-(rand1+rand2)

When choosing the first number, it needs to be 31 or more, and it must leave at least 62 on the table (assuming numbers don't have to be different) since the *other* two numbers will need to 31 or higher as well.

So it needs to come from the inclusive range 31..(356-31-31).

For the *second* number, it has to be 31 or more and it has to leave at least 31 on the table for the last number.

So its inclusive range will be 31..(356-31-first).

Then, the final number is simply 356-first-second and there you have it.

There's already plenty of answers on SO about generating random numbers in a range in Python so I won't rehash that information, but the following code shows how to select the ranges for the two numbers and calculate the third:

import random first = random.randint(31,356-31-31) second = random.randint(31,356-31-first) third = 356 - first - second print first, second, third, (first + second + third)

It *may* be different depending on whether you allow duplicates but this answer should give you the basic approach.