Pascal's Triangle for Python

Question

As a learning experience for Python, I am trying to code my own version of Pascal\'s triangle. It took me a few hours (as I am just starting), but I came out with this code:

Answer 1

# combining the insights from Aaron Hall and Hai Vu,
# we get:

def pastri(n):
    rows = [[1]]
    for _ in range(1, n+1):
        rows.append([1] +
                    [sum(pair) for pair in zip(rows[-1], rows[-1][1:])] +
                    [1])
    return rows

# thanks! learnt that "shape shifting" data,
# can yield/generate elegant solutions.

Answer 2

OK code review:

import math

# pascals_tri_formula = [] # don't collect in a global variable.

def combination(n, r): # correct calculation of combinations, n choose k
    return int((math.factorial(n)) / ((math.factorial(r)) * math.factorial(n - r)))

def for_test(x, y): # don't see where this is being used...
    for y in range(x):
        return combination(x, y)

def pascals_triangle(rows):
    result = [] # need something to collect our results in
    # count = 0 # avoidable! better to use a for loop, 
    # while count <= rows: # can avoid initializing and incrementing 
    for count in range(rows): # start at 0, up to but not including rows number.
        # this is really where you went wrong:
        row = [] # need a row element to collect the row in
        for element in range(count + 1): 
            # putting this in a list doesn't do anything.
            # [pascals_tri_formula.append(combination(count, element))]
            row.append(combination(count, element))
        result.append(row)
        # count += 1 # avoidable
    return result

# now we can print a result:
for row in pascals_triangle(3):
    print(row)

prints:

[1]
[1, 1]
[1, 2, 1]

---

Explanation of Pascal's triangle:

This is the formula for "n choose k" (i.e. how many different ways (disregarding order), from an ordered list of n items, can we choose k items):

from math import factorial

def combination(n, k): 
    """n choose k, returns int"""
    return int((factorial(n)) / ((factorial(k)) * factorial(n - k)))

A commenter asked if this is related to itertools.combinations - indeed it is. "n choose k" can be calculated by taking the length of a list of elements from combinations:

from itertools import combinations

def pascals_triangle_cell(n, k):
    """n choose k, returns int"""
    result = len(list(combinations(range(n), k)))
    # our result is equal to that returned by the other combination calculation:
    assert result == combination(n, k)
    return result

Let's see this demonstrated:

from pprint import pprint

ptc = pascals_triangle_cell

>>> pprint([[ptc(0, 0),], 
            [ptc(1, 0), ptc(1, 1)], 
            [ptc(2, 0), ptc(2, 1), ptc(2, 2)],
            [ptc(3, 0), ptc(3, 1), ptc(3, 2), ptc(3, 3)],
            [ptc(4, 0), ptc(4, 1), ptc(4, 2), ptc(4, 3), ptc(4, 4)]],
           width = 20)
[[1],
 [1, 1],
 [1, 2, 1],
 [1, 3, 3, 1],
 [1, 4, 6, 4, 1]]

We can avoid repeating ourselves with a nested list comprehension:

def pascals_triangle(rows):
    return [[ptc(row, k) for k in range(row + 1)] for row in range(rows)]

>>> pprint(pascals_triangle(15))
[[1],
 [1, 1],
 [1, 2, 1],
 [1, 3, 3, 1],
 [1, 4, 6, 4, 1],
 [1, 5, 10, 10, 5, 1],
 [1, 6, 15, 20, 15, 6, 1],
 [1, 7, 21, 35, 35, 21, 7, 1],
 [1, 8, 28, 56, 70, 56, 28, 8, 1],
 [1, 9, 36, 84, 126, 126, 84, 36, 9, 1],
 [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1],
 [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1],
 [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1],
 [1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1],
 [1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1]]

Recursively defined:

We can define this recursively (a less efficient, but perhaps more mathematically elegant definition) using the relationships illustrated by the triangle:

 def choose(n, k): # note no dependencies on any of the prior code
     if k in (0, n):
         return 1
     return choose(n-1, k-1) + choose(n-1, k)

And for fun, you can see each row take progressively longer to execute, because each row has to recompute nearly each element from the prior row twice each time:

for row in range(40):
    for k in range(row + 1):
        # flush is a Python 3 only argument, you can leave it out,
        # but it lets us see each element print as it finishes calculating
        print(choose(row, k), end=' ', flush=True) 
    print()


1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1
1 17 136 680 2380 6188 12376 19448 24310 24310 19448 12376 6188 2380 680 136 17 1
1 18 153 816 3060 8568 18564 31824 43758 48620 43758 31824 18564 8568 3060 816 ...

Ctrl-C to quit when you get tired of watching it, it gets very slow very fast...

Answer 3

Without using zip, but using generator:

def gen(n,r=[]):
    for x in range(n):
        l = len(r)
        r = [1 if i == 0 or i == l else r[i-1]+r[i] for i in range(l+1)]
        yield r

example:

print(list(gen(15)))

output:

[[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1], [1, 5, 10, 10, 5, 1], [1, 6, 15, 20, 15, 6, 1], [1, 7, 21, 35, 35, 21, 7, 1], [1, 8, 28, 56, 70, 56, 28, 8, 1], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1], [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1], [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1], [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1], [1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1], [1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1]]

DISPLAY AS TRIANGLE

To draw it in beautiful triangle(works only for n < 7, beyond that it gets distroted. ref draw_beautiful for n>7)

for n < 7

def draw(n):
    for p in gen(n):
        print(' '.join(map(str,p)).center(n*2)+'\n')

eg:

draw(10)

output:

      1       

     1 1      

    1 2 1     

   1 3 3 1    

  1 4 6 4 1   

1 5 10 10 5 1   

for any size

since we need to know the max width, we can't make use of generator

def draw_beautiful(n):
    ps = list(gen(n))
    max = len(' '.join(map(str,ps[-1])))
    for p in ps:
        print(' '.join(map(str,p)).center(max)+'\n')

example (2) : works for any number:

draw_beautiful(100)

Answer 4

def pascal(n):
    if n==0:
        return [1]
    else:
        N = pascal(n-1)
        return [1] + [N[i] + N[i+1] for i in range(n-1)] + [1]


def pascal_triangle(n):
    for i in range(n):
        print pascal(i)

Answer 5

# call the function ! Indent properly , everything should be inside the function
def triangle():

          matrix=[[0 for i in range(0,20)]for e in range(0,10)]         # This method assigns 0's to all Rows and Columns , the range is mentioned
          div=20/2           # it give us the most middle columns 
          matrix[0][div]=1        # assigning 1 to the middle of first row 
          for i in range(1,len(matrix)-1): # it goes column by column
               for j in range(1,20-1):  #  this loop goes row by row
                   matrix[i][j]=matrix[i-1][j-1]+matrix[i-1][j+1]               # this is the formula , first element of the matrix gets , addition of i index (which is 0 at first ) with third value on the the related row
    # replacing 0s with spaces :) 
          for i in range(0,len(matrix)):
              for j in range(0,20):
                   if matrix[i][j]==0:       #  Replacing 0's with spaces
                        matrix[i][j]=" "

          for i in range(0,len(matrix)-1):           # using spaces , the triangle will printed beautifully 
                for j in range(0,20):
                    print 1*" ",matrix[i][j],1*" ", # giving some spaces in two sides of the printing numbers
triangle() # calling the function

would print something like this

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

Answer 6

Here is my attempt:

def generate_pascal_triangle(rows):
    if rows == 1: return [[1]]

    triangle = [[1], [1, 1]] # pre-populate with the first two rows

    row = [1, 1] # Starts with the second row and calculate the next

    for i in range(2, rows):
        row = [1] + [sum(column) for column in zip(row[1:], row)] + [1]
        triangle.append(row)

    return triangle

for row in generate_pascal_triangle(6):
    print row

Discussion

• The first two rows of the triangle is hard-coded
• The zip() call basically pairs two adjacent numbers together
• We still have to add 1 to the beginning and another 1 to the end because the zip() call only generates the middle of the next row