--- title: "Project Euler #1: Multiples of 3 and 5" author: nbloomf date: 2017-03-05 tags: project-euler, literate-haskell --- First some boilerplate. > module ProjectEuler001 where > > import System.Exit [Problem 1](https://projecteuler.net/problem=1) from Project Euler:
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000.
First, let's do the most obvious thing: list the integers below 1000, filter out the ones divisible by either 3 or 5, and sum. For reasons I'll get to later I will parameterize this function on $n$, the upper bound for our numbers. > pe1' :: Integer -> Integer > pe1' n = sum $filter div3or5 [1..(n-1)] > where > div3or5 k = (kmod3 == 0) || (kmod5 == 0) We can verify that pe1' 10 == 23 as claimed. Now we can compute the answer to the given problem like so: haskell$> pe1' 1000 233168  And done. But wait! Anything worth doing is worth overdoing, so let's try something less obvious. Suppose we wanted the sum of the multiples of 3 or 5 less than a bigger bound -- say $n = 10^{10}$. My laptop computes pe1' 1000 in a fraction of a second, but hangs on $10^{10}$. And the reason why is clear: pe1' n constructs a list $n$ items long and deconstructs it. So the time complexity is roughly $n$. We can verify this with a little experiment. In GHCi, say :set +s and the interpreter will report time and memory usage for each computation. The following table summarizes the execution time of pe1' for several inputs on my machine. ----------------------------------- n pe1' n Time (s) ------ ----------- ----- $10^2$ 2318 0.01 $10^3$ 233168 0.02 $10^4$ 23331668 0.06 $10^5$ 2333316668 0.46 $10^6$ 233333166668 4.55 $10^7$ 23333331666668 45.38 $10^8$ my laptop choked! ----------------------------------- Two things jump out at me: first, there's a pattern in the values, which I was not expecting. Second, and more germane, the time seems to increase by a factor of 10 from row to row after $n = 10^4$. (Below this $n$ I suspect the time is dominated by the time required to print the output.) So the complexity is $O(n)$. Can we do better? ----------------- Let's break the problem down a little. We want the sum of all numbers less than $n$ which are divisible by either 3 or 5; let's start with something simpler: the sum of all the numbers less than $n$ which are divisible by 3. The sum of the first $k$ multiples of 3 is $$3 + 6 + 9 + \cdots + 3k = 3(1 + 2 + 3 + \cdots + k) = 3 \frac{k(k+1)}{2},$$ using the formula for the $k$th [triangular number](https://en.wikipedia.org/wiki/Triangular_number). Now how many multiples of 3 are there below $n$? This is exactly what is counted by the quotient in integer division. That is, given positive integers $a$ and $b$, when we decompose $a$ as $a = qb + r$ using the division algorithm with $r >= 0$, that $q$ is precisely the number of numbers between 1 and $a$ (inclusive) which are divisible by $b$. The Haskell function to find this quotient is (shockingly) quot. But there's nothing magic about 3 in that analysis; we might as well replace 3 with any other positive integer, say $t$. Then we can sum the multiples of $t$ below $n$ like so. > sumMultsOfBelow :: Integer -> Integer -> Integer > sumMultsOfBelow t n = t * q * (q+1) quot 2 > where q = (n-1) quot t What is the complexity of sumMultsOfBelow? It's harder to say exactly, because I don't know how quot is implemented. But the schoolbook algorithm for integer division is bounded above by $O(\log(n)^2)$; this is much cheaper than $O(n)$. So we can easily find the sum of the multiples of 3 below 1000 and the multiples of 5 below 1000: haskell $> (sumMultsOfBelow 3 1000) + (sumMultsOfBelow 5 1000) 266333  But this doesn't agree with the more naive (but clearly correct) pe1'. And it's not too hard to see why -- the sets of multiples of 3 and multiples of 5 are not disjoint, so some numbers are being included twice in the sum. Which ones? Well, the numbers that are divisible by both 3 and 5, which are precisely the multiples of their least common multiple, 15. And sure enough, if we account for those... haskell$> (sumMultsOfBelow 3 1000) + (sumMultsOfBelow 5 1000) - (sumMultsOfBelow 15 1000) 233168  Let's wrap this in a definition. > pe1'' :: Integer -> Integer > pe1'' n = (sumMultsOfBelow 3 n) > + (sumMultsOfBelow 5 n) > - (sumMultsOfBelow 15 n) At this point, I was planning to include another table of timings to show how much faster pe1'' is, but I had to go to $n = 10^{85}$ before it took longer than 0.01 seconds (the smallest unit of time reported by GHCi). For reference, pe1'' (10^100000) took 2.03 seconds on my machine. An $O(n)$ algorithm has no hope on inputs that big. And by the way that digit pattern, a 2, followed by 3s, then 1, then 6s, then 8, holds. Before we get too excited, let's verify that pe1' and pe1'' agree, at least on small inputs. > test_pe1 :: Integer -> Bool > test_pe1 n = pe1' n == pe1'' n And then: haskell $> and$ map test_pe1 [1..1500] True  So it is with confidence we say that pe1' and pe1'' are the same function. Then the answer to problem 1 is: > pe1 :: Integer > pe1 = pe1'' 1000 > > main :: IO () > main = do > let success = all test_pe1 [1..1500] > if success > then putStrLn \$ show pe1 > else exitFailure