Pascal Triangle in Haskell

Question

I'm new to Haskell and I really need some help!

I have to write a program that includes a recursive function to produce a list of binomial coefficients for the power n=12 using the Pascal's triangle technique.

I have some ideas in my head but because I'm just getting started I have no idea how to implement this to haskell?!

Could someone please help me out??

first row: (a+b)^0 = 1
second row: (a+b)^1 = 1a+1b
third row: (a+b)^2 = 1a^2+2ab+1b^2

and so on...this is my main idea. But I cant even try this out because I have no idea how I put this in Haskell..getting errors all the time

Answer 1

Start with the base case.

pascal 0 0 = 1

Then handle the edge cases

pascal n 0 = 1
pascal n r | n == r = 1

Now expand with the recursive step

pascal n r = pascal (n - 1) (r - 1) + pascal (n - 1) r

If you want the list for a specific row, write a wrapper

binom n = map (pascal n) [0..n]

Figuring out the types shouldn't be hard

pascal :: Integral a => a -> a -> a
binom :: Integral a => a -> [a]

Answer 2

Start by assigning an index to each element in the triangle:

  | 0   1   2   3   4   5   6
--+--------------------------
0 | 1
1 | 1   1
2 | 1   2   1
3 | 1   3   3   1
4 | 1   4   6   4   1
5 | 1   5  10  10   5   1
6 | 1   6  15  20  15   6   1

Here I've simply put the triangle on its side so that we can number them. So here I'd say that the element at (6, 4) is 15, whereas (4, 6) doesn't exist. Now focus on writing a function

pascal :: Integer -> Integer -> Integer
pascal x y = ???

Such that you can generate this version of the triangle. You can start by writing

pascal x y
    | x == 0 = 1
    | x == y = 1
    | x <  y = error "Not a valid coordinate for Pascal's triangle."
    | otherwise = pascal ? ? + pascal ? ?

Note that here, instead of figuring out which elements should be added together by diagonals, you can do it via rectangular coordinates. Here, you'll note that y is which row in the triangle you're on and x is the position of the element in that row. All you need to do is figure out what goes in place of the ?s.

Once you get that working, I've got a one-liner for this triangle that is more efficient and can generate the entire triangle all at once while still using recursion:

import Data.List (scanl1)

pascals :: [[Integer]]
pascals = repeat 1 : map (scanl1 (+)) pascals

Don't try turning this solution in to your professor, it's not what they're looking for and it would make it pretty obvious someone gave you this solution if you've only been doing Haskell for a week. However, it really shows how powerful Haskell can be for this sort of problem. I would show how to index pascals to get a given (n, k) value, but doing so would also give you too many hints for solving the naive recursion.

Since there's been some confusion, the reason why I gave this solution is to draw a parallel between it and the often shown lazy implementation for the Fibonacci sequence:

fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

Compared to

fib 0 = 1
fib 1 = 1
fib n = fib (n - 1) + fib (n - 2)

This definition generates an infinite list of all the Fibonacci numbers, and does so quite efficiently (from the point of view of the CPU, RAM is a different story). It encodes in its first 2 elements the base case, then a recursive expression that can calculate the rest. For the Fibonaccis, you need 2 values to start you off, but for Pascal's triangle, you only need one value, that value just happens to be an infinite list. There is an easy to see pattern going across the columns in the grid I posted above, the scanl1 (+) function just takes advantage of this pattern and allows us to generate it very easily, but this is generating the diagonals of the triangle rather than the rows. To get the rows, you can index this list, or you can do some fancy tricks with take, drop, and other such functions, but that's an exercise for another day.

Answer 3

Start out with the triangle itself:

     1
    1 1
   1 2 1
  1 3 3 1
 1 4 6 4 1
    ...

You should notice that to write down the next row, you must apply this rule: sum the previous rows' adjacent elements, using a 0 for the lonely edge elements. Visually:

    0   1   0
     \+/ \+/
  0   1   1   0
   \+/ \+/ \+/
0   1   2   1   0
 \+/ \+/ \+/ \+/
  1   3   3   1
       ...

Operationally, that looks like this:

For row 0:
[1]  (it's a given; i.e. base case)

For row 1:
[0, 1]   <- row 0 with a zero prepended ([0] ++ row 0)
 +  +
[1, 0]   <- row 0 with a zero appended  (row 0 ++ [0])
 =  =
[1, 1]   <- element-wise addition

For row 2:
[0, 1, 1]
 +  +  +
[1, 1, 0]
 =  =  =
[1, 2, 1]

Generally, for row N:

element-wise addition of:
  [0] ++ row(N-1)
  row(N-1) ++ [0]

Remember that element-wise addition of lists in Haskell is zipWith (+).

Thus we arrive at the following Haskell definition:

pascal 0 = [1]
pascal n = zipWith (+) ([0] ++ pascal (n-1)) (pascal (n-1) ++ [0])

Or in a fashion similar to the famous "lazy fibs":

pascals = [1] : map (\xs -> zipWith (+) ([0] ++ xs) (xs ++ [0])) pascals

Answer 4

Another possible solution (more suitable for beginners in my opinion):

pascal :: Integer -> [Integer]
pascal 0 = [1]
pascal 1 = [1, 1]
pascal n = let p = pascal (n - 1)
    in [1] ++ pascalStep p ++ [1]

pascalStep :: [Integer] -> [Integer]
pascalStep [] = []
pascalStep [_] = []
pascalStep (x:y:xs) = x + y : pascalStep (y : xs)

Using let to avoid more space usage. pascal is calling recursively to find all previous rows, using them to get the next row, until getting to the desired row.

Output:

*Main> pascal 3
[1,3,3,1]
*Main> pascal 4
[1,4,6,4,1]
*Main> pascal 5
[1,5,10,10,5,1]

Answer 5

I'm on my phone so please excuse the mistakes, but you can use Haskell's lazy evaluation in a really cool way here.

pascals :: [[Int]]
pascals = [1]:map (\r -> zipWith (+) (0:r) (r++[0])) pascals

Which you could make point free with a fork but it's rather esoteric.

pascals :: [[Int]]
pascals = [1]:map ((zipWith (+) -<) (0:) (++[0])) pascals

But I personally really like this code, and thinks it's worth being readable-

pascals :: [[Int]]
pascals = [1]:map next pascals
    where next = (zipWith (+) -<) (0:) (++[0])

But combinators like that can get a bit confusing, no matter how much I like point free programming.