# euler.lisp @ 7c2948c7ec11

Problem 9

author | Steve Losh <steve@stevelosh.com> |
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date | Sun, 10 Apr 2016 19:31:21 +0000 |

parents | 20c03264a55e |

children | 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)