# Euler problem 5

I am starting to learn Rust and it seems pretty awesome but it's way different than any C based language. I want to know how can I make this code more idiomatic and also how can I improve my solution itself.

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

fn near_pow(number:f64, exponent:f64)-> f64
{
exponent.powf(number.log(exponent).floor())
}

fn is_prime(number:i32) -> bool
{
use std::ops::Rem;

if number == 2 { return true; }
if number.rem(2)== 0 { return false; }

let mut i = 3;
while (i*i) <= number
{
if number.rem(i) == 0 {return false;}
i+= 2;
}
true
}

fn euler_problem5(to:i32) -> f64
{
(1..to+1).filter(|&x| is_prime(x))
.fold(1f64, |p , x|p * near_pow(to as f64, x as f64))
}
fn main() {

let x = euler_problem5(20);
println!("{}",x);

}


• You changed the code after you posted it. That makes it kind of difficult to give feedback... – Shepmaster Sep 28 '15 at 15:29
• yes i wasn't sure if i should answer myself or edit. Although i didn't change anything that significant. I am sorry about that. – MAG Sep 28 '15 at 15:32
• anything that significant — except you invalidated one of my points. There's no rush to post your question to Code Review; take your time and make sure that the code you submit the first time is the code you want reviewed. – Shepmaster Sep 28 '15 at 15:36
• FWIW, the filter is better though! – Shepmaster Sep 28 '15 at 15:37
• yes sir you are absolutely right, I will keep that in mind next time :). – MAG Sep 28 '15 at 15:42

1. Rust uses "Egyptian braces" in the majority of cases.

if condition {
// block
} else {
// another block
}

2. Use a space after a : in a type.

fn foo(value: type)

3. Use spaces around operators like ==, +, and * and around symbols like ->.

i += 2;

4. Use a space after a ,.

println!("{}", x);

5. There's no need to use a ; in one-line early return statements (guard clauses).

if number == 2 { return true }

6. Use the % operator instead of calling the rem method:

if number % 2 == 0 { return false }

7. Write the closure to fold inline, especially since you aren't able to give it a better name than fold_op (OP changed the code I am referring to; look in the revision history to see what I'm talking about).

(1..to + 1).fold(1f64, |p, x| {
// stuff
})

8. Whenever possible, try to drive out mutability. I also dislike return statements inside of loop bodies (or really anywhere that aren't guard clauses). Try to use iterators instead.

All together:

fn near_pow(number: f64, exponent: f64) -> f64 {
exponent.powf(number.log(exponent).floor())
}

fn is_prime(number: i32) -> bool {
if number == 2 { return true }
if number % 2 == 0 { return false }

(0..)
.map(|v| 3 + 2 * v) // Can use Range::step_by when stable
.take_while(|i| i * i <= number)
.all(|i| number % i != 0)
}

fn euler_problem5(to: i32) -> f64 {
(1..to + 1).fold(1f64, |p, x| {
if is_prime(x) {
p * near_pow(to as f64, x as f64)
} else {
p
}
})
}

fn main() {
println!("{}", euler_problem5(20));
}


I'm not well-versed on how to do the Euler problems in an efficient way, so hopefully someone else will chime in there.

Update after the original code changed

fn euler_problem5(to: i32) -> f64 {
(1..to + 1)
.filter(|&v| is_prime(v))
.fold(1f64, |p, x| p * near_pow(to as f64, x as f64))
}


user5402 argues to replace floating point with integer operations, but either way you should probably use unsigned integers as you don't need to support negative numbers. This allows you a bit more upper bound to your values as well.

# f64

My biggest critique is that you are using floating point arithmetic to solve an integer problem.

Given the possibility for round off error and inexact answers, I would opt for a simple while-loop to compute nearest_pow:

(Sorry - this is Python, I'm not a Rust programmer yet)

def nearest_pow(p,n):
a = 1
while a*p <= n:
a = a * p
return a


Rust has fold and take_while, so I'm sure you can implement this in a functional manner.

This is efficient - it only performs log n iterations - and won't suffer from any round-off errors.

Moreover, it also works for big integers which might come in handy for other number theoretic problems.

Also, I would return the answer as an integer value - perhaps a i64.

• A reasonable suggestion, integer operations will almost always be faster than floating point. However, a literal translation of your sample code seems to differ in results...? Maybe the argument order changed? – Shepmaster Sep 28 '15 at 17:16
• You also need to handle the edge cases of 0 and 1. – Shepmaster Sep 28 '15 at 17:37
• The OP's is_prime function is returning true for p = 1 which could be considered a problem. Alternatively the main loop could just start at 2. – ErikR Sep 28 '15 at 18:53
• Yea you are right forgot about that edge case and I should have started at 2. I knew there was gonna be a lost in precision for big numbers possibly giving a power nearest to x so that x <= b^e, but would this happen in the range of 32 integers? Also could you say my near_pow function is O(1) ? (I am guessing its probably not). – MAG Sep 28 '15 at 23:55