Difference between revisions of "99 questions/31 to 41"
RossPaterson (talk  contribs) (tidy up P35 a little) 

Line 137:  Line 137:  
Kind of ugly, but it works, though it may have bugs in corner cases. This uses the factor tree method of finding prime factors of a number. factorPairOf picks a factor and takes it and the factor you multiply it by and gives them to primeFactors. primeFactors checks to make sure the factors are prime. If not it prime factorizes them. In the end a list of prime factors is returned. 
Kind of ugly, but it works, though it may have bugs in corner cases. This uses the factor tree method of finding prime factors of a number. factorPairOf picks a factor and takes it and the factor you multiply it by and gives them to primeFactors. primeFactors checks to make sure the factors are prime. If not it prime factorizes them. In the end a list of prime factors is returned. 

−  Another possibility is to observe that you need not 
+  Another possibility is to observe that you need not ensure that potential divisors are primes, as long as you consider them in ascending order: 
<haskell> 
<haskell> 

primefactors n = primefactors' n 2 where 
primefactors n = primefactors' n 2 where 

primefactors' 1 _ = [] 
primefactors' 1 _ = [] 

−  primefactors' n 
+  primefactors' n factor 
−  +   n `mod` factor == 0 = factor : primefactors' (n `div` factor) factor 

−  +   otherwise = primefactors' n (factor + 1) 

</haskell> 
</haskell> 

Thus, we just loop through all possible factors and add them to the list if they divide the original number. As the primes get farther apart, though, this will do a lot of needless checks to see if composite numbers are prime factors. 
Thus, we just loop through all possible factors and add them to the list if they divide the original number. As the primes get farther apart, though, this will do a lot of needless checks to see if composite numbers are prime factors. 

+  However we can stop as soon as the candidate factor exceeds the square root of n: 

+  <haskell> 

+  primefactors n = primefactors' n 2 where 

+  primefactors' n factor 

+   factor*factor > n = [n] 

+   n `mod` factor == 0 = factor : primefactors' (n `div` factor) factor 

+   otherwise = primefactors' n (factor + 1) 

+  </haskell> 

== Problem 36 == 
== Problem 36 == 
Revision as of 12:59, 13 December 2006
These are Haskell translations of Ninety Nine Lisp Problems.
If you want to work on one of these, put your name in the block so we know someone's working on it. Then, change n in your block to the appropriate problem number, and fill in the <Problem description>,<example in lisp>,<example in Haskell>,<solution in haskell> and <description of implementation> fields.
Arithmetic
Problem 31
Determine whether a given integer number is prime.
Example: * (isprime 7) T Example in Haskell: P31> isPrime 7 True
Solution:
isPrime :: Integral a => a > Bool
isPrime p = all (\n > p `mod` n /= 0 ) $ takeWhile (\n > n*n <= x) [2..]
Well, a natural number p is a prime number iff no natural number n with n >= 2 and n^2 <= p is a divisor of p. That's exactly what is implemented: we take the list of all integral numbers starting with 2 as long as their square is at most p and check that for all these n there is a remainder concerning the division of p by n.
Problem 32
(**) Determine the greatest common divisor of two positive integer numbers.
Use Euclid's algorithm.
Example: * (gcd 36 63) 9 Example in Haskell: [myGCD 36 63, myGCD (3) (6), myGCD (3) 6] [9,3,3]
Solution:
gcd' 0 y = y
gcd' x y = gcd' (y `mod` x) x
myGCD x y  x < 0 = myGCD (x) y
 y < 0 = myGCD x (y)
 y < x = gcd' y x
 otherwise = gcd' x y
The Prelude includes a gcd function, so we have to choose another name for ours. The function gcd' is a straightforward implementation of Euler's algorithm, and myGCD is just a wrapper that makes sure the arguments are positive and in increasing order.
Problem 33
(*) Determine whether two positive integer numbers are coprime. Two numbers are coprime if their greatest common divisor equals 1.
Example:
* (coprime 35 64) T
Example in Haskell:
* coprime 35 64 True
Solution:
coprime a b = gcd a b == 1
Here we use the prelude function for computing gcd's along with a test of the result's equality to one.
Problem 34
(**) Calculate Euler's totient function phi(m). Euler's socalled totient function phi(m) is defined as the number of positive integers r (1 <= r < m) that are coprime to m. Example: m = 10: r = 1,3,7,9; thus phi(m) = 4. Note the special case: phi(1) = 1.
Example: * (totientphi 10) 4 Example in Haskell: * totient 10 4
Solution:
totient 1 = 1
totient a = length $ filter (coprime a) [1..a1]
where coprime a b = gcd a b == 1
We take coprime from the previous exercise and give it to filter, which applies it to each element of a list from 1 to one less than the number, returning only those that are true. lenght tells us how many elements are in the resulting list, and thus how many elements are coprime to n
Problem 35
(**) Determine the prime factors of a given positive integer. Construct a flat list containing the prime factors in ascending order.
Example: * (primefactors 315) (3 3 5 7) Example in Haskell: * primeFactors 315 [3, 3, 5, 7]
Solution:
primeFactors :: Integer > [Integer]
primeFactors a = let (f, f1) = factorPairOf a
f' = if prime f then [f] else primeFactors f
f1' = if prime f1 then [f1] else primeFactors f1
in f' ++ f1'
where
factorPairOf a = let f = head $ factors a
in (f, div a f)
factors a = filter (isFactor a) [2..a1]
isFactor a b = rem a b == 0
prime a = (length $ factors a) == 0
Kind of ugly, but it works, though it may have bugs in corner cases. This uses the factor tree method of finding prime factors of a number. factorPairOf picks a factor and takes it and the factor you multiply it by and gives them to primeFactors. primeFactors checks to make sure the factors are prime. If not it prime factorizes them. In the end a list of prime factors is returned.
Another possibility is to observe that you need not ensure that potential divisors are primes, as long as you consider them in ascending order:
primefactors n = primefactors' n 2 where
primefactors' 1 _ = []
primefactors' n factor
 n `mod` factor == 0 = factor : primefactors' (n `div` factor) factor
 otherwise = primefactors' n (factor + 1)
Thus, we just loop through all possible factors and add them to the list if they divide the original number. As the primes get farther apart, though, this will do a lot of needless checks to see if composite numbers are prime factors. However we can stop as soon as the candidate factor exceeds the square root of n:
primefactors n = primefactors' n 2 where
primefactors' n factor
 factor*factor > n = [n]
 n `mod` factor == 0 = factor : primefactors' (n `div` factor) factor
 otherwise = primefactors' n (factor + 1)
Problem 36
(**) Determine the prime factors of a given positive integer.
Construct a list containing the prime factors and their multiplicity.
Example: * (primefactorsmult 315) ((3 2) (5 1) (7 1)) Example in Haskell: *Main> prime_factors_mult 315 [(2 3), (1 5), (1 7)]
Solution:
prime_factors_mult n = encode $ prime_factors_mult 2 n []
prime_factors i n xs = if i*i > n then n:xs else if i `divides` n then prime_factors i (n `div` i) (i:xs) else prime_factors (i+1) n xs
divides a b = (b `div` a)*a == b
We iterate through all numbers up to the squareroot of n, and add them to our list, if they divide n. The function 'encode' is the solution to problem 10. It takes a list of numbers, and compresses it to a list of numbers paired with their multiplicity.
Problem 37
(**) Calculate Euler's totient function phi(m) (improved).
See problem P34 for the definition of Euler's totient function. If the list of the prime factors of a number m is known in the form of problem P36 then the function phi(m) can be efficiently calculated as follows: Let ((p1 m1) (p2 m2) (p3 m3) ...) be the list of prime factors (and their multiplicities) of a given number m. Then phi(m) can be calculated with the following formula:
phi(m) = (p1  1) * p1 ** (m1  1) + (p2  1) * p2 ** (m2  1) + (p3  1) * p3 ** (m3  1) + ...
Note that a ** b stands for the b'th power of a. Note: Actually, the official problems show this as a sum, but it should be a product.
Solution: Given prime_factors_mult from problem 36, we get
totient m = product [(p  1) * p ^ (c  1)  (c, p) < prime_factors_mult m]
This just uses a list comprehension to build each term of the product in the formula for phi, then multiplies them all.
Problem 38
<Problem description>
Example: <example in lisp> Example in Haskell: <example in Haskell>
Solution:
<solution in haskell>
<description of implementation>
Problem 39
A list of prime numbers.
Given a range of integers by its lower and upper limit, construct a list of all prime numbers in that range.
Example in Haskell: P29> primesR 10 20 [11,13,17,19]
Solution 1:
primesR :: Integral a => a > a > [a]
primesR a b = filter isPrime [a..b]
If we are challenged to give all primes in the range between a and b we simply take all number from a up to b and filter the primes out.
Solution 2:
primes :: Integral a => [a]
primes = let sieve (n:ns) = n:sieve [ m  m < ns, m `mod` n /= 0 ] in sieve [2..]
primesR :: Integral a => a > a > [a]
primesR a b = takeWhile (<= b) $ dropWhile (< a) primes
Another way to compute the claimed list is done by using the Sieve of Eratosthenes. The primes
function generates a list of all (!) prime numbers using this algorithm and primesR
filter the relevant range out. [But this way is very slow and I only presented it because I wanted to show how nice the Sieve of Eratosthenes can be implemented in Haskell :)]
Problem 40
(**) Goldbach's conjecture. Goldbach's conjecture says that every positive even number greater than 2 is the sum of two prime numbers. Example: 28 = 5 + 23. It is one of the most famous facts in number theory that has not been proved to be correct in the general case. It has been numerically confirmed up to very large numbers (much larger than we can go with our Prolog system). Write a predicate to find the two prime numbers that sum up to a given even integer.
Example: * (goldbach 28) (5 23) Example in Haskell: *goldbach 28 (5, 23)
Solution 1:
goldbach a = head $
filter (\e > (isPrime $ fst e) && (isPrime $ snd e)) $
map (\e > (e, a  e)) [1,3..div a 2]
where
factors a = filter (isFactor a) [2..a1]
isFactor a b = rem a b == 0
isPrime a = (length $ factors a) == 0
Woohoo! I've solved the goldbach conjecture! Where do I collect my prize? This does the obvious thing. It makes a list of odd numbers and then uses it to make up pairs of odd numbers that sum to n. Then it looks for a pair of odd numbers where both are prime and returns it as a tuple.
Solution 2:
 using the previous problem...
goldbach n = head [(x,y)  x < pr, y < pr, x+y==n]
where pr = takeWhile (< n) primes
Problem 41
Given a range of integers by its lower and upper limit, print a list of all even numbers and their Goldbach composition.
In most cases, if an even number is written as the sum of two prime numbers, one of them is very small. Very rarely, the primes are both bigger than say 50. Try to find out how many such cases there are in the range 2..3000.
Example: * (goldbachlist 9 20) 10 = 3 + 7 12 = 5 + 7 14 = 3 + 11 16 = 3 + 13 18 = 5 + 13 20 = 3 + 17 * (goldbachlist 1 2000 50) 992 = 73 + 919 1382 = 61 + 1321 1856 = 67 + 1789 1928 = 61 + 1867
Example in Haskell:
*Exercises> goldbachList 9 20 [(3,7),(5,7),(3,11),(3,13),(5,13),(3,17)] *Exercises> goldbachList' 4 2000 50 [(73,919),(61,1321),(67,1789),(61,1867)]
Solution:
goldbach n = head [(x,y)  x < pr, y < pr, x+y==n]
where pr = takeWhile (< n) primes
goldbachList lb ub = map goldbach $ filter even [max(lb,4)..ub]
goldbachList' lb ub mv = filter (\x@(a,b) > a > mv && b > mv) $
goldbachList lb ub