Code-golf: generate pascal's triangle

Question

Generate a list of lists (or print, I don't mind) a Pascal's Triangle of size N with the least lines of code possible!

Here goes my attempt (118 characters in python 2.6 using a trick):

c,z,k=locals,[0],'_[1]'
p=lambda n:[len(c()[k])and map(sum,zip(z+c()[k][-1],c()[k][-1]+z))or[1]for _ in range(n)]

Explanation:

• the first element of the list comprehension (when the length is 0) is [1]
• the next elements are obtained the following way:
• take the previous list and make two lists, one padded with a 0 at the beginning and the other at the end.
  • e.g. for the 2nd step, we take [1] and make [0,1] and [1,0]
• sum the two new lists element by element
  • e.g. we make a new list [(0,1),(1,0)] and map with sum.
• repeat n times and that's all.

usage (with pretty printing, actually out of the code-golf xD):

result = p(10)
lines = [" ".join(map(str, x)) for x in result]
for i in lines:
    print i.center(max(map(len, lines)))

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

Answer 1

another stab (python):

def pascals_triangle(n):
    x=[[1]]
    for i in range(n-1):
        x.append(list(map(sum,zip([0]+x[-1],x[-1]+[0]))))
    return x

Answer 2

Another try, in prolog (I'm practising xD), not too short, just 164c:

s([],[],[]).
s([H|T],[J|U],[K|V]):-s(T,U,V),K is H+J.
l([1],0).
l(P,N):-M is N-1,l(A,M),append(A,[0],B),s(B,[0|A],P).
p([],-1).
p([H|T],N):-M is N-1,l(H,N),p(T,M).

explanation:

• s = sum lists element by element
• l = the Nth row of the triangle
• p = the whole triangle of size N

Answer 3

J, another language in the APL family, 9 characters:

p=:!/~@i.

This uses J's builtin "combinations" verb.

Output:

   p 10
1 1 1 1 1  1  1  1  1   1
0 1 2 3 4  5  6  7  8   9
0 0 1 3 6 10 15 21 28  36
0 0 0 1 4 10 20 35 56  84
0 0 0 0 1  5 15 35 70 126
0 0 0 0 0  1  6 21 56 126
0 0 0 0 0  0  1  7 28  84
0 0 0 0 0  0  0  1  8  36
0 0 0 0 0  0  0  0  1   9
0 0 0 0 0  0  0  0  0   1

Answer 4

Python: 75 characters

def G(n):R=[[1]];exec"R+=[map(sum,zip(R[-1]+[0],[0]+R[-1]))];"*~-n;return R

Answer 5

I wrote this C++ version a few years ago:

#include <iostream>
int main(int,char**a){for(int b=0,c=0,d=0,e=0,f=0,g=0,h=0,i=0;b<atoi(a[1]);(d|f|h)>1?e*=d>1?--d:1,g*=f>1?--f:1,i*=h>1?--h:1:((std::cout<<(i*g?e/(i*g):1)<<" "?d=b+=c++==b?c=0,std::cout<<std::endl?1:0:0,h=d-(f=c):0),e=d,g=f,i=h));}

Answer 6

F#: 81 chars

let f=bigint.Factorial
let p x=[for n in 0I..x->[for k in 0I..n->f n/f k/f(n-k)]]

Explanation: I'm too lazy to be as clever as the Haskell and K programmers, so I took the straight forward route: each element in Pascal's triangle can be uniquely identified using a row n and col k, where the value of each element is n!/(k! (n-k)!.