# Euler problems/151 to 160

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< Euler problems(Difference between revisions)

(Added problem_152) |
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== [http://projecteuler.net/index.php?section=view&id=152 Problem 152] == |
== [http://projecteuler.net/index.php?section=view&id=152 Problem 152] == |
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Writing 1/2 as a sum of inverse squares |
Writing 1/2 as a sum of inverse squares |
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+ | |||

+ | Note that if p is an odd prime, the sum of inverse squares of |
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+ | all terms divisible by p must have reduced denominator not divisible |
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+ | by p. |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | problem_152 = undefined |
+ | import Data.Ratio |

+ | import Data.List |
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+ | |||

+ | invSq n = 1 % (n * n) |
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+ | sumInvSq = sum . map invSq |
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+ | |||

+ | subsets (x:xs) = let s = subsets xs in s ++ map (x :) s |
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+ | subsets _ = [[]] |
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+ | |||

+ | primes = 2 : 3 : 7 : [p | p <- [11, 13..83], |
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+ | all (\q -> p `mod` q /= 0) [3, 5, 7]] |
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+ | |||

+ | -- All subsets whose sum of inverse squares, |
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+ | -- when added to x, does not contain a factor of p |
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+ | pfree s x p = [(y, t) | t <- subsets s, let y = x + sumInvSq t, |
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+ | denominator y `mod` p /= 0] |
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+ | |||

+ | -- Verify that we need not consider terms divisible by 11, or by any |
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+ | -- prime greater than 13. Nor need we consider any term divisible |
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+ | -- by 25, 27, 32, or 49. |
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+ | verify = all (\p -> null $ tail $ pfree [p, 2*p..85] 0 p) $ |
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+ | 11 : dropWhile (< 17) primes ++ [25, 27, 32, 49] |
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+ | |||

+ | -- All pairs (x, n) where x is a rational number whose reduced |
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+ | -- denominator is not divisible by any prime greater than 3; |
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+ | -- and n>0 is the number of sets of numbers up to 85 divisible |
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+ | -- by a prime greater than 3, whose sum of inverse squares is x. |
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+ | only23 = foldl f [(0, 1)] [13, 7, 5] |
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+ | where |
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+ | f x p = collect $ concatMap (g p) x |
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+ | g p (x, n) = map (\(a, b) -> (a, n * length b)) $ pfree (terms p) x p |
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+ | terms p = [n * p | n <- [1..85`div`p], |
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+ | all (\q -> n `mod` q /= 0) [5, 7, 11, 13, 17]] |
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+ | collect = map (\z -> (fst $ head z, sum $ map snd z)) |
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+ | . groupBy cmpFst . sort |
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+ | cmpFst x y = fst x == fst y |
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+ | |||

+ | -- All subsets (of an ordered set) whose sum of inverse squares is x |
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+ | findInvSq x y = f x $ zip3 y (map invSq y) (map sumInvSq $ init $ tails y) |
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+ | where |
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+ | f 0 _ = [[]] |
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+ | f x ((n, r, s):ns) |
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+ | | r > x = f x ns |
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+ | | s < x = [] |
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+ | | otherwise = map (n :) (f (x - r) ns) ++ f x ns |
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+ | f _ _ = [] |
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+ | |||

+ | -- All numbers up to 85 that are divisible only by the primes |
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+ | -- 2 and 3 and are not divisible by 32 or 27. |
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+ | all23 = [n | a <- [0..4], b <- [0..2], let n = 2^a * 3^b, n <= 85] |
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+ | |||

+ | problem_152 = if verify |
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+ | then sum [n * length (findInvSq (1%2 - x) all23) | |
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+ | (x, n) <- only23] |
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+ | else undefined |
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</haskell> |
</haskell> |
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## Revision as of 13:52, 20 September 2007

## Contents |

## 1 Problem 151

Paper sheets of standard sizes: an expected-value problem.

Solution:

problem_151 = undefined

## 2 Problem 152

Writing 1/2 as a sum of inverse squares

Note that if p is an odd prime, the sum of inverse squares of all terms divisible by p must have reduced denominator not divisible by p.

Solution:

import Data.Ratio import Data.List invSq n = 1 % (n * n) sumInvSq = sum . map invSq subsets (x:xs) = let s = subsets xs in s ++ map (x :) s subsets _ = [[]] primes = 2 : 3 : 7 : [p | p <- [11, 13..83], all (\q -> p `mod` q /= 0) [3, 5, 7]] -- All subsets whose sum of inverse squares, -- when added to x, does not contain a factor of p pfree s x p = [(y, t) | t <- subsets s, let y = x + sumInvSq t, denominator y `mod` p /= 0] -- Verify that we need not consider terms divisible by 11, or by any -- prime greater than 13. Nor need we consider any term divisible -- by 25, 27, 32, or 49. verify = all (\p -> null $ tail $ pfree [p, 2*p..85] 0 p) $ 11 : dropWhile (< 17) primes ++ [25, 27, 32, 49] -- All pairs (x, n) where x is a rational number whose reduced -- denominator is not divisible by any prime greater than 3; -- and n>0 is the number of sets of numbers up to 85 divisible -- by a prime greater than 3, whose sum of inverse squares is x. only23 = foldl f [(0, 1)] [13, 7, 5] where f x p = collect $ concatMap (g p) x g p (x, n) = map (\(a, b) -> (a, n * length b)) $ pfree (terms p) x p terms p = [n * p | n <- [1..85`div`p], all (\q -> n `mod` q /= 0) [5, 7, 11, 13, 17]] collect = map (\z -> (fst $ head z, sum $ map snd z)) . groupBy cmpFst . sort cmpFst x y = fst x == fst y -- All subsets (of an ordered set) whose sum of inverse squares is x findInvSq x y = f x $ zip3 y (map invSq y) (map sumInvSq $ init $ tails y) where f 0 _ = [[]] f x ((n, r, s):ns) | r > x = f x ns | s < x = [] | otherwise = map (n :) (f (x - r) ns) ++ f x ns f _ _ = [] -- All numbers up to 85 that are divisible only by the primes -- 2 and 3 and are not divisible by 32 or 27. all23 = [n | a <- [0..4], b <- [0..2], let n = 2^a * 3^b, n <= 85] problem_152 = if verify then sum [n * length (findInvSq (1%2 - x) all23) | (x, n) <- only23] else undefined

## 3 Problem 153

Investigating Gaussian Integers

Solution:

problem_153 = undefined

## 4 Problem 154

Exploring Pascal's pyramid.

Solution:

problem_154 = undefined

## 5 Problem 155

Counting Capacitor Circuits.

Solution:

problem_155 = undefined

## 6 Problem 156

Counting Digits

Solution:

problem_156 = undefined

## 7 Problem 157

Solving the diophantine equation 1/a+1/b= p/10n

Solution:

problem_157 = undefined

## 8 Problem 158

Exploring strings for which only one character comes lexicographically after its neighbour to the left.

Solution:

problem_158 = undefined

## 9 Problem 159

Digital root sums of factorisations.

Solution:

problem_159 = undefined

## 10 Problem 160

Factorial trailing digits

Solution:

problem_160 = undefined