# Pascal's triangle in Haskell

Posted on**09 July 2017**

(My answer to this Stack Overflow question.)

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
```