# Difference between revisions of "Euler problems/171 to 180"

m (import) |
DonStewart (talk | contribs) (Unrelated content) |
||

Line 41: | Line 41: | ||

== [http://projecteuler.net/index.php?section=problems&id=174 Problem 174] == |
== [http://projecteuler.net/index.php?section=problems&id=174 Problem 174] == |
||

Counting the number of "hollow" square laminae that can form one, two, three, ... distinct arrangements. |
Counting the number of "hollow" square laminae that can form one, two, three, ... distinct arrangements. |
||

− | |||

− | Solution: This was my C++ code, published here without my permission nor any attribution, shame on whoever put it here. [[user:henk263|henk263]] |
||

== [http://projecteuler.net/index.php?section=problems&id=175 Problem 175] == |
== [http://projecteuler.net/index.php?section=problems&id=175 Problem 175] == |

## Revision as of 08:42, 26 February 2008

## Contents

## Problem 171

Finding numbers for which the sum of the squares of the digits is a square.

Solution: this was Antoine Celeriers C code pasted from the Project Euler Forum.

## Problem 172

Investigating numbers with few repeated digits.

Solution:

```
factorial n = product [1..toInteger n]
fallingFactorial x n = product [x - fromInteger i | i <- [0..toInteger n - 1] ]
choose n k = fallingFactorial n k `div` factorial k
-- how many numbers can we get having d digits and p positions
p172 0 _ = 0
p172 d p
| p < 4 = d^p
| otherwise =
(p172' p) + p*(p172' (p-1)) + (choose p 2)*(p172' (p-2)) + (choose p 3)*(p172' (p-3))
where
p172' = p172 (d-1)
problem_172= (p172 10 18) * 9 `div` 10
```

## Problem 173

Using up to one million tiles how many different "hollow" square laminae can be formed? Solution:

```
problem_173=
let c=div (10^6) 4
xm=floor$sqrt $fromIntegral c
k=[div c x|x<-[1..xm]]
in sum k-(div (xm*(xm+1)) 2)
```

## Problem 174

Counting the number of "hollow" square laminae that can form one, two, three, ... distinct arrangements.

## Problem 175

Fractions involving the number of different ways a number can be expressed as a sum of powers of 2. Solution:

```
sternTree x 0=[]
sternTree x y=
m:sternTree y n
where
(m,n)=divMod x y
findRat x y
|odd l=take (l-1) k++[last k-1,1]
|otherwise=k
where
k=sternTree x y
l=length k
p175 x y=
init$foldl (++) "" [a++","|
a<-map show $reverse $filter (/=0)$findRat x y]
problems_175=p175 123456789 987654321
test=p175 13 17
```

## Problem 176

Rectangular triangles that share a cathetus. Solution:

```
--k=47547
--2*k+1=95095 = 5*7*11*13*19
lst=[5,7,11,13,19]
primes=[2,3,5,7,11]
problem_176 =
product[a^b|(a,b)<-zip primes (reverse n)]
where
la=div (last lst+1) 2
m=map (\x->div x 2)$init lst
n=m++[la]
```

## Problem 177

Integer angled Quadrilaterals.

Solution: This C++ solution is stolen from balakrishnan. Check out the forum if you want to see his solution to the problem

## Problem 178

Step Numbers

Count pandigital step numbers.

```
import qualified Data.Map as M
data StepState a = StepState { minDigit :: a
, maxDigit :: a
, lastDigit :: a
} deriving (Show, Eq, Ord)
isSolution (StepState i a _) = i == 0 && a == 9
neighborStates m s@(StepState i a n) = map (\x -> (x, M.findWithDefault 0 s m)) $
[StepState (min i (n - 1)) a (n - 1), StepState i (max a (n + 1)) (n + 1)]
allStates = [StepState i a n | (i, a) <- range ((0, 0), (9, 9)), n <- range (i, a)]
initialState = M.fromDistinctAscList [(StepState i i i, 1) | i <- [1..9]]
stepState m = M.fromListWith (+) $ allStates >>= neighborStates m
numSolutionsInMap = sum . map snd . filter (isSolution . fst) . M.toList
numSolutionsOfSize n = sum . map numSolutionsInMap . take n $ iterate stepState initialState
problem_178 = numSolutionsOfSize 40
```

## Problem 179

Consecutive positive divisors.

## Problem 180

Rational zeros of a function of three variables. Solution:

```
import Data.Ratio
{-
After some algebra, we find:
f1 n x y z = x^(n+1) + y^(n+1) - z^(n+1)
f2 n x y z = (x*y + y*z + z*x) * ( x^(n-1) + y^(n-1) - z^(n-1) )
f3 n x y z = x*y*z*( x^(n-2) + y^(n-2) - z^(n-2) )
f n x y z = f1 n x y z + f2 n x y z - f3 n x y z
f n x y z = (x+y+z) * (x^n+y^n-z^n)
Now the hard part comes in realizing that n can be negative.
Thanks to Fermat, we only need examine the cases n = [-2, -1, 1, 2]
Which leads to:
f(-2) z = xy/sqrt(x^2 + y^2)
f(-1) z = xy/(x+y)
f(1) z = x+y
f(2) z = sqrt(x^2 + y^2)
-}
unique :: Eq(a) => [a] -> [a]
unique [] = []
unique (x:xs) | elem x xs = unique xs
| otherwise = x : unique xs
-- Not quite correct, but I don't care about the zeros
ratSqrt :: Rational -> Rational
ratSqrt x =
let a = floor $ sqrt $ fromIntegral $ numerator x
b = floor $ sqrt $ fromIntegral $ denominator x
c = (a%b) * (a%b)
in if x == c then (a%b) else 0
-- Not quite correct, but I don't care about the zeros
reciprocal :: Rational -> Rational
reciprocal x
| x == 0 = 0
| otherwise = denominator x % numerator x
problem_180 =
let order = 35
range :: [Rational]
range = unique [ (a%b) | b <- [1..order], a <- [1..(b-1)] ]
fm2,fm1,f1,f2 :: [[Rational]]
fm2 = [[x,y,z] | x<-range, y<-range,
let z = x*y * reciprocal (ratSqrt(x*x+y*y)), elem z range]
fm1 = [[x,y,z] | x<-range, y<-range,
let z = x*y * reciprocal (x+y), elem z range]
f1 = [[x,y,z] | x<-range, y<-range,
let z = (x+y), elem z range]
f2 = [[x,y,z] | x<-range, y<-range,
let z = ratSqrt(x*x+y*y), elem z range]
result = sum $ unique $ map (\x -> sum x) (fm2++fm1++f1++f2)
in (numerator result + denominator result)
```