# Pascal's Triangle in Python

### Περιεχόμενα

You need to generate Pascal's Triangle in Python, and you're lazy (an admirable trait). Alternatively, you're looking for a Pascal's Triangle generator that can show really high-ranking rows, ones with multi-hundred-digit (or multi-million-digit) coefficients.

## Solution (pre Python 3)

```def pascal(n):
"""
Yield up to row ``n`` of Pascal's triangle, one row at a time.

The first row is row 0.
"""
def newrow(row):
"Calculate a row of Pascal's triangle given the previous one."
prev = 0
for x in row:
yield prev + x
prev = x
yield 1

prevrow = [1]
yield prevrow
for x in xrange(n):
prevrow = list(newrow(prevrow))
yield prevrow
```

## Discussion

It all revolves around the auxilliary function `newrow()`, which generates a row of Pascal's triangle given the previous one. We bootstrap it with the 0-th row, `[1]`, and then we iterate, cascading out new rows.

The return value of `pascal()` is a generator of lists, one list to a row, starting with row 0. If you'd rather have a list of lists, you simply call `list(pascal(x))`.

Python generators are speedy beasts. On my desktop computer, `pascal(1000)` runs in 0.3 seconds, yielding the first 1,001 rows of the triangle — and remember, you need Arbitrary-precision arithmetic (aka bignums) to calculate the coefficients of most of these rows.

In fact, `pascal(35)` is the biggest triangle a C implementation using 32-bit `unsigned int` numbers can generate. Even with 64-bit integers, C can handle up to `pascal(68)`. The beauty of Python for numerical recipes like this is that it switches seamlessly to bignums whenever it's necessary. This is a serious boon when it comes to factorially-exploding results like this one. For your information, the highest coefficient in `pascal(1000)` is a 300-digit integer.

## Extensions

This version includes an example (and doctest), and checks the value of `n` for correctness:

```def pascal(n):
"""
Yield up to row ``n`` of Pascal's triangle, one row at a time.

The first row is row 0.

>>> list (pascal (0))
[ [1] ]
>>> list (pascal (1))
[ [1], [1, 1]]
>>> list (pascal (4))
[ [1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1] ]
>>> list (pascal (-1))
Traceback (most recent call last):
ValueError: n must be an integer >= 0
"""
if not isinstance(n, int):
raise TypeError('n must be an integer')
if n < 0:
raise ValueError('n must be an integer >= 0')

def newrow(row):

"Calculate a row of Pascal's triangle given the previous one."
prev = 0
for x in row:
yield prev + x
prev = x
yield 1

prevrow = [1]
yield prevrow
for x in xrange(n):
prevrow = list(newrow(prevrow))
yield prevrow
```

In all other aspects, this is identical to the first implementation.

## Python 3

Python 3 removed list-based iterators. This is a good thing, but it needs one minor change to make the Pascal's Triangle generator work: replace the `xrange(n)` with `range(n)`. In Pythons before 3.x, `range(n)` would allocate a list of `n` elements of the arithmetic sequence `[0,1,...,n-1]`, with an associated O(n) storage requirement. Because of this, `xrange` was introduced which used a generator instead, for an O(1) (constant) storage requirement. In Python 3, list-based iterators are gone entirely, and all the generator-based iterators have been renamed. Hence the issue.