I've been back to Edinburgh for a family reunion and got to spend a

morning with my dad who's a maths teacher by trade â€” I always thought

he might be interested in haskell but never got the time to really look

at some stuff in a bit of detail. I'd been pointed at Project Euler and thought it would be an interesting thing to look at together.

We worked through 5. What is the smallest number divisible by each of the numbers 1 to 20?

We started off with factorial of 20, and then the product of the

primes in the range. Which meant we needed a sieve. I'd written a sieve

in Perl (using `HOP::Stream`

sub pf { my $s = shift; my $head = drop $s; my $s2 = filter { $_[0] % $head } $s; return node($head, promise { pf($s2) } ); }

I still don't find this technique (wrapping filtering coderefs

around each other in a lazy stream) natural, and it took me a while to

translate this haskellish technique back into haskell…

> makeSieve n=filter (\i->i `mod` n /= ) > primes[]=[]> primes (x:xs)=x : (primes $ (makeSieve x) xs)

Which is all very nice, but of course this is not the right number:

> *Main> primes[2..20]>[2,3,5,7,11,13,17,19]> *Main> product $ primes[2..20]> 9699690

The right answer is the product of all the numbers 1..20 except for

the ones that are already included in another number. So for example

20*19*18*… but not including 10,5,4, as they're already factors of 20.

We had the idea that we needed to get the factors of each number, so

that we could then throw out those whose factors have already been used.

So here's a nice prime factorization technique:

> factorize n=factorize' n (primes[2..])[]> > factorize' n (p:px) results >|n == 1 >=results >|n `mod` p == >=factorize' (n `quot` p) (p:px) (p:results) >|otherwise >=factorize' n px results

Used like this

> *Main> factorize 18 >[3,3,2]

But now we want to know the powers of each factor: e.g. in this case, 3^2 and 2^1.

> powf n=map (\l->(head l, length l) ) $ groupBy (==) n

It made sense for the result to come back as tuples of `(value, power)`:

> *Main> powf $ factorize 18 >[(3,2),(2,1)]

And we can get the whole lot like so:

> *Main> map (powf.factorize)[1..20]>[[],[(2,1)],[(3,1)],[(2,2)],[(5,1)], ....

But now we just want the maximum power for each one. I tried to get

my brain in gear to work out how to do this recursively, then gave up

and decided to just sort them…

We sort by value and the descending power. It made sense to use the library function sortBy, which takes a sorting function:

> cmp (a,a') (b,b')|a == b=compare (b') (a') >|a < b=LT >|otherwise=GT

(compare is the inbuilt function that compares two Ord values)

> *Main> sortBy cmp $ concatMap (powf.factorize)[1..20]

Now we want the top power for each value. The way that came to mind was to group by the value:

> groupBy (\(x,_) (y,_)->x == y) $ > sortBy cmp $ concatMap (powf.factorize)[1..20]

Now that we've grouped we just want the maximum ord for each:

> map (\l->head l) $ > groupBy (\(x,_) (y,_)->x == y) $ > sortBy cmp $ concatMap (powf.factorize)[1..20]

Giving us (value,power) pairs like so:

>[(2,4),(3,2),(5,1),(7,1),(11,1),(13,1),(17,1),(19,1)]

And then of course we want to exponentiate these value:

> let factors=map (\(n,p)->n ^ p ) $ > map (\l->head l) $ > groupBy (\(x,_) (y,_)->x == y) $ > sortBy cmp $ concatMap (powf.factorize)[1..20]

Now it's just a matter of

> *Main> product factors

And Bob's your uncle. What could be simpler? ;-)

I clicked through to the Project Euler discussion forums to see a nice Ruby version:

num = (1..20).inject(1) { |result, n| result.lcm n }

OK, Ruby still seems weird to me, but that's because I haven't

learnt it. The point is that this seems rather more compact. As soon as

I pointed this to my Dad, he said, â€œOf course it's a least common

multiple problemâ€. Which of course is more or less the approach that

we'd taken, but rather more verbosely than the Ruby version.

So I googled â€œhaskell lcmâ€ to check if there was a function with a similar name somewhere. There was indeed a function `lcm`… in the Prelude itself.

â€œOh,â€ I said, â€œmaybe this will work…â€ and typed the following into ghci:

> foldl1 lcm[1..20]

This of course is equivalent to the above… but 20 times shorter.

This is both humbling and rather interesting. I think it was worth

going through the entire process to get to this point, as I learnt a

lot. Also, now having an (inefficient) prime sieve and a prime

factoriser, I can easily solve a number of other questions:

3: Find the largest prime factor of 317584931803.

> *Main> factorize 317584931803

7: Find the 10001st prime number

> *Main> show $ last (take 10001 $ primes[2..])

(This one takes quite a long time – about 12 seconds – to complete.

Apparently there are much better sieves, but we're still comfortably

within the project's â€œminute ruleâ€ so I'm not going to prematurely

optimize yet!

**Update: **bluestorm points out that what I believed to be a Sieve of Eratosthenes actually isn't, citing The Genuine Sieve of Eratosthenes, a very readable paper on the topic.