Euler problems solved using a functional Java library I created called jFunc.
Euler 06
The sum of the squares of the first ten natural numbers is,
12 + 22 + ... + 102 = 385The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)2 = 552 = 3025Hence 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.
F1<FList<Long>, Long> sqrs = F1.compose(Longs::sum, map(i -> i * i));
F1<FList<Long>, Long> sums = F1.compose(i -> i * i, Longs::sum);
F1<FList<Long>, Long> go = l -> sums.call(l) - sqrs.call(l);
return go.call(range(1, 100));
This problem is broken down into three parts: finding the sum of the squares, finding the square of the sums and finding their difference. Since there are three distinct pieces of the problem it is very natural to create three distinct functions(sqrs, sums, go).
sqrs and sums can be created by composing other functions.
F1<A,C> F1.compose(F1<B,C> f2, F1<A,B> f1) composes functions, if we observe the types closely we can see that the functions given to F1.compose are composed from right to left.
Longs::sum is a function of arity two that sums two Long values.
F1<FList<A>, FList<B>> map(F<A,B> f) takes a function from A -> B and returns a function from FList<A> -> FList<B>.
And go simply combines the two functions sqrs and sums into one neat package.
Euler 07
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. What is the 10 001st prime number?
return Numbers.primes.get(10000);
jFunc defines an infinite list of primes, Numbers.primes. Because jFuncs list type is 0-based indexed, the number 10000 is used instead of 10001.
Euler 08
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949495459501737958331952853208805511 12540698747158523863050715693290963295227443043557 66896648950445244523161731856403098711121722383113 62229893423380308135336276614282806444486645238749 30358907296290491560440772390713810515859307960866 70172427121883998797908792274921901699720888093776 65727333001053367881220235421809751254540594752243 52584907711670556013604839586446706324415722155397 53697817977846174064955149290862569321978468622482 83972241375657056057490261407972968652414535100474 82166370484403199890008895243450658541227588666881 16427171479924442928230863465674813919123162824586 17866458359124566529476545682848912883142607690042 24219022671055626321111109370544217506941658960408 07198403850962455444362981230987879927244284909188 84580156166097919133875499200524063689912560717606 05886116467109405077541002256983155200055935729725 71636269561882670428252483600823257530420752963450
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
F1<String,Long> go = showStr() // convert String to FList<Character>
// Converts each Character into an Integer
.then(fc -> fc.map(c -> Character.getNumericValue(c)))
// Converts each Integer to Long
.then(fi -> fi.map(Integer::longValue))
// A sliding window that creates a FList<FList<Long>>
.then(fl -> fl.window(13))
// Converts each FList<Long> to a product.
.then(ffl -> ffl.map(Longs::product))
// Find the maximum product within the FList.
.then(f -> f.maximum(Long::compare));
Try<Long> digits = Try.with(() -> Euler08.class.getResource("euler08.txt").openStream())
.bind(IO.readFileFromStream)
.map(go);
return stream.get();
If we break the problem into a series of steps we end up with:
- Obtain 1000 digit String from file.
- Convert String into a FList of Characters.
- Convert each Character into an Integer
- Convert each Integer into a Long
- Convert FList
to FList<FList > using a sliding window to break the FList of longs into smaller FLists of size 13 - Find the product of each FList.
- Find the largest Long within the FList.
The Function go is a composition of the steps from 2 to 7.
Step 1. Is accomplished with a Try. A Try is used because reading the 1000 digits from a file throws a checked Exception.
Euler 09
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2 For example, 32 + 42 = 9 + 16 = 25 = 52. There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.
FList<Integer> list =
start(1)
.For( a -> range(a + 1, 1000)
// After genertating a and b, c is limited to only one possiblilty.
,(a, b) -> flist(1000 - b - a)
,(a, b, c) -> guard(a*a + b*b == c*c)
.semi(flist(a*b*c)));
return list.head();
The For method sequences monadic operations and is analogous to a do-block in Haskell.
(a, b) -> flist(1000 - b - a) after genertating a and b, c is limited to only one possiblilty.
(a, b, c) -> guard(a*a + b*b == c*c) filters any value from the FList where the statement a*a + b*b == c*c is false.
Because there exits only one Pythagorean triplet where a + b + c = 1000 we only have to return the head of the generated list.
Euler 10
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million.
FList<Long> p = Numbers.primes
.takeWhile(i -> i < 2000000L);
return Longs.sum(p);
Since jFunc comes with an infinite FList of primes we only have to filter the list of all the values above 2000000, then sum the remaining values. The method takeWhile(P1<A> predicate) does the filtering and Longs.sum sums the remaining values.