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this script prints a few lines of Pascal`s triangle ;-)

Python, 16 lines
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def pascal(n):
    """Prints n first lines of Pascal`s triangle

    author: DR#m <dnpsite.narod.ru>
    last rev 20.03.05"""
    l=[1]
    p=[]
    for i in xrange(n):
        l2=[1]
        for j in xrange(len(l)-1):
            l2.append(l[j]+l[j+1])
        l2.append(1)
        print l
        l=l2
if __name__=="__main__":
    pascal(20)

anyone going beat this? thak you

6 comments

Markus Weihs 19 years ago  # | flag

This one's a little shorter :-) (but less readable :-/)

def pascal(n):
    l = [[1]]
    for i in xrange(n-1):
        l.append([1,1])
        for j in xrange(len(l[-2])-1):
            l[-1].insert(-1,l[-2][j]+l[-2][j+1])
    return l

if __name__ == '__main__':
    for i in mypascal(20): print i
Markus Weihs 19 years ago  # | flag

In my example above it should be

if __name__ == '__main__':
    for i in pascal(20): print i
kdsfjkj klfjdsfj 19 years ago  # | flag

Shorter? Take this. def pascal2(n, t=[1], z=[[1]]): for i in xrange(1, n): z+=[[1]+[z[i-1][j]+z[i-1][j+1] for j in range(len(z[i-1])-1)]+[1]] print z

Scott David Daniels 19 years ago  # | flag

Row by Row. This highlights how the calculation is done.

def nextrow(lst):
    lag = 0
    for element in lst:
        yield lag + element
        lag = element
    yield element

row = [1]
for number in range(12):
    row = nextrow(row)
    print row
Pood Dov 19 years ago  # | flag

Another try. def pas(n=10): line=[1] for i in range(0,n): print line line=[0]+line for j in range(len(line)-1): line[j]+=line[j+1]

Eduardo gasser 8 years, 9 months ago  # | flag

Another short solution, with only one loop:

def pascal(level):
    lst = [1]
    while len(lst) < level:
        lst = [1] + map(lambda x,y: x+y if y else 1, lst, lst[1:])
        print lst

pascal(14)
Created by DR#m not on Sun, 20 Mar 2005 (PSF)
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