# Project Euler Answers

## Table of Contents

## Summary

My answers to various project Euler problems. I occasionally will be doing these because they are fun and pretty entertaining.

## 1: Multiples of 3 or 5

### Brute Force Answer

Here I use the simple brute force answer, which is to generate a list to 1000 and then get the multiples of three and five in that list using the modulus operator. It's pretty short but I probably can code golf-it and optimize it quite a bit.

(letfn [(divisible? [n div] (zero? (mod n div)))] (->> (range 1000) (filter #(or (divisible? % 3) (divisible? % 5))) (apply +)))

### Optimized Answer

So instead of that approach there's a much faster one using recursion. 1 the n-ary function syntax for Clojure is really nice here, allowing me to use default arguments.

(letfn [(multiples-to ([n mul] (multiples-to n mul [] 1)) ([n mul l i] (let [multiple (* i mul)] (if (> multiple n) l (recur n mul (conj l multiple) (inc i))))))] (->> (concat (multiples-to 1000 3) (multiples-to 1000 5)) distinct (apply +)))

This answer unfortunately only really shows improvements in efficiency when the multiples are around 100 or 300, rather than 3 and 5. In fact, it is actually considerably less efficient than the first "brute-force" answer in the 3-5 case.

## 2: Even Fibonacci Numbers

This just gets even fib numbers. There's not much to optimize here because the process is both fairly simple and (as far as I can tell) cannot be changed.

(letfn [(fib ([max-num] (fib max-num [1 2])) ([max-num lst] (let [next (apply + (take 2 lst))] (if (> next max-num) lst (fib max-num (cons next lst))))))] (->> (fib 4000000) (filter even?) (apply +)))

## 3: Largest Prime Factor

The goal is to find the largest prime factor of a number. To start we will first begin by creating a function that finds a number of primes up to the target value divided by 2. This is because the target value will, ultimately, be at most \(n \times 2\).

### Finding Prime Numbers

To start out we find a list of prime numbers. A neat feature of prime numbers is that to test a number for primality you both only need to test for the division of a number by other prime numbers up to the square root of the value, for the largest number will be at most the square root.

#### Checking for Divisibility

To start off, and to make our code more clear, we define a small predicate called `divisible?`

which checks if a number is divisible by some other number. While it doesn't really shorten the code all that much, it adds clarity to the code

(defn divisible? [num div] (zero? (mod num div)))

#### Sieve of Eratosthenes

The most elegant solution would be the sieve of Eratosthenes. The logic behind it is fairly simple:

- all natural numbers are either primes or a composite of a number of primes
- a number is prime if no prime divisor less than itself exists
- therefore, if one creates a list from \(2...n\) and takes the head of the list and filter out all divisors, add the head to a second list, and continue until the list is emptied the second list should contain the primes from \(2...n\)

As you can see, it follows that the list generated will be entirely comprised of prime numbers.

To solve this we can implement a lazy sieve of Eratosthenes. Lazy in that the resulting sequence will be infinite.

(defn seive ([] (seive (drop 2 (range)))) ([l] (lazy-seq (let [[x & xs] l] (if (empty? xs) [x] (cons x (seive (filter #(not (divisible? % x)) xs)))))))) (def primes (seive))

Unfortunately this does not work for what we want. It is very beautiful, and I wish it did work, but because of the specific features of TCO in Clojure 2 Even with the `recur`

special form. it doesn't actually produce TCO code, and, because it is recursive, will end up causing a stack overflow.

However, because of the lazy list semantics you can take a nearly infinite series of primes from it, assuming you only increase your requested list size by one each time you take from it. This is interesting, though unfortunately not something I can really use.

#### A More Ugly Approach

So in order to solve this we can rewrite the is-prime function to, rather than generating a list from 0 to \(n\) and then filtering it, simply loop until a value at the range of the prime number is found. This ensures that the space complexity of it is \(O(\log(n))\) rather than \(O(n)\), though does not change the time complexity.

First we define an atom to memoize our primes called `found-primes`

, and initialize it with the first prime number, 2.

(def ^:private found-primes "A map of all primes that have been found thus far by the system" (atom #{2})) (def ^:private max-prime (atom 2))

Next we define an `is-prime?`

predicate which will test if a value is prime by iterating over the list of primes that are less than the square root of the new value. This is basically the same as the sieve of Eratosthenes applied to a single element. Below is the function itself, don't worry, we will break it down.

(defn is-prime? ([n] (is-prime? n false)) ([n contiguous?] (let [max-div (int (inc (Math/sqrt n)))] (letfn [(div? [n] (->> @found-primes (take-while #(> max-div %)) (some #(divisible? n %)) not))] (cond (contains? @found-primes n) true (> @max-prime n) false (< @max-prime max-div) (do (doall (map #(is-prime? % true) (range @max-prime max-div))) (if (div? n) (do (swap! found-primes conj n) true) false)) (div? n) (do (if contiguous? (do (swap! found-primes conj n) (reset! max-prime n)) (swap! found-primes conj n)) true) :else false)))))

There are, however, a number of fun tricks being used to speed up the computation. First we generate the maximum divisor using this little bit of code. This isn't really the maximum size of the prime divisor, but rather the maximum size of the minimum element of the pair of divisors produced.

(int (inc (Math/sqrt n)))

Next we have the conditional. First it checks if the number is in the list of found primes, and if it is then it simply returns `true`

. This means that it has \(O(1)\) runtime should the prime be found already. 3 Though in Clojure, since sets are trees, it has more \(O(log(n))\) runtime.

(contains? @found-primes n) true

But, if it is both not in the list of found primes *and* is less than the largest prime, it must be the case that it is not a prime, and therefore it returns false.

(> @max-prime n) false

Following this we check the case that the `max-prime`

is smaller than `max-div`

. In this case we check the primality of every prime from the last largest prime to the `max-div`

. This places each prime number in the set of found primes and ensures that it can run and check if the number is prime. Following that we check that `n`

is prime.

We do add it to the list of primes, but we don't set the `max-prime`

, even though it far exceeds it. This is because the value `max-prime`

represents the maximum prime for which all primes lower than it are known. In this case not all primes lower than it are known.

Unfortunately it is not possible to, for any \(n\) know that it is the next prime without testing the range between n and the previous prime. Therefore, for now, this part of it is the only way to check if previous elements are prime.

(< @max-prime max-div) (do (doall (map #(is-prime? % true) (range @max-prime max-div))) (if (div? n) (do (swap! found-primes conj n) true) false))

Now we check if a number is or is not in fact prime, and in this case we do set the prime in the event it is contiguous because the number is the next number in the set of found primes.

(div? n) (do (if contiguous? (do (swap! found-primes conj n) (reset! max-prime n)) (swap! found-primes conj n)) true)

Once we have this function, we can then begin to look into generating the primes to some value. Like before we want to ensure that we can use as many numbers form our prime list as possible, therefore we implement a simple function that will first attempt to return all the values from our `found-primes`

list. should they not exist, it will then use the side effects of the `is-prime?`

function to generate it.

<<divisible?>> (defn prime-range ([end] (prime-range 2 end)) ([start end] (if (@max-prime) (drop-while)) (->> (range (max start 2) end) (filter is-prime?)))) (defn prime-factors [x] (->> (prime-range (inc (Math/sqrt x))) (filter #(divisible? x %)))) (last (prime-factors 600851475143))