# euler.lisp @ 7c2948c7ec11

`Problem 9`
author Steve Losh Sun, 10 Apr 2016 19:31:21 +0000 20c03264a55e ef04c7b3d0b8
```(in-package #:euler)

;;;;
(defun digits (n)
"Return how many digits `n` has in base 10."
(values (truncate (1+ (log n 10)))))

(defun definitely-palindrome-p (n)
"Return whether `n` is a palindrome (in base 10), the slow-but-sure way."
(let ((s (format nil "~D" n)))
(string= s (reverse s))))

(defun palindrome-p (n)
"Return whether `n` is a palindrome (in base 10)."
(assert (>= n 0) (n) "~A must be a non-negative integer" n)
;; All even-length base-10 palindromes are divisible by 11, so we can shortcut
;; the awful string comparison. E.g.:
;;
;;   abccba =
;;   100001 * a +
;;   010010 * b +
;;   001100 * c
(cond
((zerop n) t)
((and (evenp (digits n))
(not (dividesp n 11))) nil)
(t (definitely-palindrome-p n))))

(defun range (from below)
(loop :for i :from from :below below
:collect i))

(defun square (n)
(* n n))

;;;; Problems
(defun problem-1 ()
;; 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.
(loop :for i :from 1 :below 1000
:when (or (dividesp i 3)
(dividesp i 5))
:sum i))

(defun problem-2 ()
;; Each new term in the Fibonacci sequence is generated by adding the previous
;; two terms. By starting with 1 and 2, the first 10 terms will be:
;;
;;     1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
;;
;; By considering the terms in the Fibonacci sequence whose values do not
;; exceed four million, find the sum of the even-valued terms.
(loop :with p = 0
:with n = 1
:while (<= n 4000000)
:when (evenp n) :sum n
:do (psetf p n
n (+ p n))))

(defun problem-3 ()
;; The prime factors of 13195 are 5, 7, 13 and 29.
;;
;; What is the largest prime factor of the number 600851475143 ?
(apply #'max (prime-factorization 600851475143)))

(defun problem-4 ()
;; A palindromic number reads the same both ways. The largest palindrome made
;; from the product of two 2-digit numbers is 9009 = 91 × 99.
;;
;; Find the largest palindrome made from the product of two 3-digit numbers.
(let ((result (list)))
(loop :for i :from 0 :to 999
:do (loop :for j :from 0 :to 999
:for product = (* i j)
:when (palindrome-p product)
:do (push product result)))
(apply #'max result)))

(defun problem-5 ()
;; 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?
(let ((divisors (range 11 21)))
;; all numbers are divisible by 1 and we can skip checking everything <= 10
;; because:
;;
;; anything divisible by 12 is automatically divisible by 2
;; anything divisible by 12 is automatically divisible by 3
;; anything divisible by 12 is automatically divisible by 4
;; anything divisible by 15 is automatically divisible by 5
;; anything divisible by 12 is automatically divisible by 6
;; anything divisible by 14 is automatically divisible by 7
;; anything divisible by 16 is automatically divisible by 8
;; anything divisible by 18 is automatically divisible by 9
;; anything divisible by 20 is automatically divisible by 10
(loop :for i
:from 20 :by 20 ; it must be divisible by 20
:when (every (lambda (n) (dividesp i n))
divisors)
:return i)))

(defun problem-6 ()
;; The sum of the squares of the first ten natural numbers is,
;;   1^2 + 2^2 + ... + 10^2 = 385
;;
;; The square of the sum of the first ten natural numbers is,
;;   (1 + 2 + ... + 10)^2 = 55^2 = 3025
;;
;; Hence the difference between the sum of the squares of the first ten
;; natural numbers and the square of the sum is 3025 − 385 = 2640.
;;
;; Find the difference between the sum of the squares of the first one hundred
;; natural numbers and the square of the sum.
(flet ((sum-of-squares (to)
(loop :for i :from 1 :to to
:sum (square i)))
(square-of-sum (to)
(square (loop :for i :from 1 :to to
:sum i))))
(abs (- (sum-of-squares 100) ; apparently it wants the absolute value
(square-of-sum 100)))))

(defun problem-7 ()
(nth-prime 10001))

(defun problem-8 ()
(let ((digits (map 'list #'digit-char-p
"7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450")))
(loop :for window :in (n-grams 13 digits)
:maximize (apply #'* window))))

(defun problem-9 ()
(flet ((pythagorean-triplet-p (a b c)
(= (+ (square a) (square b))
(square c))))
(block search
(loop :for c :from 998 :downto 1 ; they must add up to 1000, so C can be at most 998
:do (loop :for a :from (- 999 c) :downto 1 ; A can be at most 999 - C (to leave 1 for B)
:for b = (- 1000 c a)
:when (pythagorean-triplet-p a b c)
:do (return-from search (* a b c)))))))

;;;; Tests
(def-suite :euler)
(in-suite :euler)

(test p1 (is (= 233168 (problem-1))))
(test p2 (is (= 4613732 (problem-2))))
(test p3 (is (= 6857 (problem-3))))
(test p4 (is (= 906609 (problem-4))))
(test p5 (is (= 232792560 (problem-5))))
(test p6 (is (= 25164150 (problem-6))))
(test p7 (is (= 104743 (problem-7))))
(test p8 (is (= 23514624000 (problem-8))))
(test p9 (is (= 31875000 (problem-9))))

; (run! :euler)```