Random number that is divisible by N

Question

I want to generate a random number from range [a,b] that is dividsible by N (4 in my case). I have the solution, but is there a better (more elegant) way to do it?

result = random.randint(a, b)
result = math.ceil(result / 4) * 4

Solutions from here: Python: Generate random number between x and y which is a multiple of 5 doesn't answer my question since I'll have to implement something like:

random.randint(a, b) * 4;

I'll have to divide original range by 4 and it's less readable then my original solution

Your solution is wrong. For e.g. if a=1, b=15, your solution may give 16 which is wrong — Sabareesh, May 14 '19 at 05:53
"a number that divides by N" isn't legal English, and "a number that divides N" means the other thing. I think you mean **"a number that is divisible by N"** — smci, May 14 '19 at 05:55
@DYZ please see my comment regarding duplicate solution you've mentioned — KocT9H, May 14 '19 at 07:08

balderman · Accepted Answer · 2019-05-14T07:23:59.990

2

A generic solution and an example

import random

def divisible_random(a,b,n):
    if b-a < n:
      raise Exception('{} is too big'.format(n))
    result = random.randint(a, b)
    while result % n != 0:
      result = random.randint(a, b)
    return result
# get a random int in the range 2 - 12, the number is divisible by 3
print(divisible_random(2,12,3))

edited May 14 '19 at 07:23

answered May 14 '19 at 06:52

balderman

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Simple and generic, thanks! – KocT9H May 14 '19 at 07:11
It's probably worth mentioning that this *can* be really slow, even causing an infinite loop as an extreme edge case. – iz_ May 14 '19 at 07:12
1

This can easily get stuck in an infinite loop, `print(divisible_random(2,5,3))` @KocT9H – Devesh Kumar Singh May 14 '19 at 07:18
I agree about the option to get stuck in infinite loop. Extra protection should be added to the function. – balderman May 14 '19 at 07:20
Thanks for pointing it out. I've added a counter to stop the loop and return (result // 4 * 4) in case counter is 0. – KocT9H May 14 '19 at 08:04
IMO, when you can extend already provided functions by `random` module in a single line, not worth the effort to keep choosing numbers in a loop and wasting cycle time! – Devesh Kumar Singh May 14 '19 at 08:14

Divyanshu Srivastava · Answer 2 · 2019-05-14T06:11:41.527

1

The first thing coming to my mind is creating a list of all the possible choices using range in the given interval, followed by randomly choosing one value using choice.

So, in this case, for a given a and b,

random.choice(range(a + 4 - (a%4), b, 4))

If a is a perfect multiple of 4, then

random.choice(range(a, b, 4))

Would give you the required random number.

So, in a single generic function, (as suggested in comments)

def get_num(a, b, x):
    if not a % x:
        return random.choice(range(a, b, x))
    else:
        return random.choice(range(a + x - (a%x), b, x))

where x is the number whose multiples are required.

edited May 14 '19 at 06:11

answered May 14 '19 at 05:59

Divyanshu Srivastava

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I was about to post this. +1 – gmds May 14 '19 at 06:00
3

The solution is wrong: `random.choice(range(5 + (5%4), 7, 4))` yields 6. It is also very inefficient. – DYZ May 14 '19 at 06:00
@DYZ missed that, corrected now – Divyanshu Srivastava May 14 '19 at 06:05
1

You probably want to wrap this logic in a single function then... – Julien May 14 '19 at 06:06
I see that you took the approach `a+4-(a%4)` from my answer, please consider upvoting my answer @DivyanshuSrivastava :) – Devesh Kumar Singh May 14 '19 at 06:17
@DeveshKumarSingh Nope, I didn't. – Divyanshu Srivastava May 14 '19 at 06:21
1

@DYZ: Since Divyanshu is wrong about `range` creating a list, the efficiency is actually fine. – user2357112 May 14 '19 at 06:53

Devesh Kumar Singh · Answer 3 · 2019-05-14T08:13:09.410

1

Use random.randrange with a step size of n, using a+n-(a%n) as start if a is non-divisible by n, else use a as start

import random

def rand_n(a, b,n):

    #If n is bigger than range, return -1
    if n > b-a:
        return -1

    #If a is divisible by n, use a as a start, using n as step size
    if a%n == 0:
        return random.randrange(a,b,n)

    # If a is not divisible by n, use a+n-(a%n) as a start, using n as step size
    else:
        return random.randrange(a+n-(a%n),b, n)

edited May 14 '19 at 08:13

answered May 14 '19 at 06:02

Devesh Kumar Singh

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Now what's wrong with this ? – Devesh Kumar Singh May 14 '19 at 06:04
wrong as below, e.g. if a=0, 0 is never chosen although acceptable. – Julien May 14 '19 at 06:04
Got an error def rand_n(a, b, 4): ^ SyntaxError: invalid syntax – Divyanshu Srivastava May 14 '19 at 06:23
You are interchangeably using `n` and 4 in your function. – Divyanshu Srivastava May 14 '19 at 06:24
Still error. Let me fix it for you. – Divyanshu Srivastava May 14 '19 at 06:25
No issues, I fixed it, thanks for letting me know @DivyanshuSrivastava Please upvote if it works now for you :) – – Devesh Kumar Singh May 14 '19 at 06:26

szmate1618 · Answer 4 · 2019-05-14T07:26:27.010

As the others have pointed out, your solution might produce out of range results, e.g. math.ceil(15 / 4) * 4 == 16. Also, be aware that the produced distribution might be very far from uniform. For example, if a == 0 and b == 4, the generated number will be 4 in 80% of the cases. Aside from that, it seems good to me, but in Python, you can also just use the integer division operator (actually floor division, so it's not equivalent to your examlpe):

result = random.randint(a, b)
result = result // 4 * 4

But a more general albeit less efficient method of generating uniform random numbers with specific constraints (while also keeping the uniform distribution) is generating them in a loop until you find a good one:

result = 1
while result % 4 != 0:
    result = random.randint(a, b)

Random number that is divisible by N

4 Answers4