Pascal's Pyramid and Higher Dimensions

Question

Pascal's Triangle is a familiar mathematical construct with many interesting properties. It is constructed by starting with a 1 on top, and generating the numbers in the next row from the sum of the number to the left and right in the row above (with implied 0's around the boundary):

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

The Triangle can be expanded into N dimensions by using the same rules. For example, the 3-D Pascal's Triangle would be a pyramid with each row being a cross-section of the pyramid, i.e., a triangle. Each triangle's entry is generated by adding the three numbers in the "row" above forming a triangle directly above that entry. An example of the first four "rows" is shown below:

1

 1
1 1

  1
 2 2
1 2 1

   1
  3 3
 3 6 3
1 3 3 1

The 4-D Pascal's Triangle would have a 3-D Pascal's Triangle for the cross section, and so forth.

Your task is to write a program that takes as input the dimension of the Pascal's Triange, and a row index to display. The row index starts at 0, so row 0 is 1 for every dimension. Your program should abide by the following specifications:

• Take input N = dimension number, R = row number (0-based). The input can be either command-line arguments or from stdin. N >= 2, R >= 0
• Output the R'th row in the N-dimensional Pascal's Triangle, by displaying each row or triangle on a separate line as in the above example.
• The format can vary as long as it is 'triangle-like' and obvious where each row and entry in the row belongs with respect to other rows.

Here is some sample output (I believe these are correct):

>./pascal 2 5
1 5 10 10 5 1
>./pascal 3 4
        1
      4   4
    6  12   6
  4  12  12   4
1   4   6   4   1
>./pascal 4 3
1

 3
3 3

  3
 6 6
3 6 3

   1
  3 3
 3 6 3
1 3 3 1
>./pascal 5 2
1


2

 2
2 2


1

 2
2 2

  1
 2 2
1 2 1

Bonus: What is the significance of each "row" in an N-dimensional Pascal's triangle? I can think of at least three interesting facts.

Answer 1

Ruby, 181

def P(a)a.min<0?0:a.max<1?1:(b=a*1;t=k=0;b.map{b[k]-=1;k+=1;t+=P(b)};t)end
def d(n,a)print n>1?((0..a[-1]).map{|i|d(n-1,a+[i])};"\n"):"#{P(a)} "end
d(ARGV[0].to_i,[ARGV[1].to_i])

The first approach using the following recursive formula

P(...,-1,...) = 0
P(0,0,...,0) = 1
P(a,b,c,....) = P(a-1,b,c,...) + P(a-1,b-1,c,...) + P(a-1,b-1,c-1,...) + ...

and then printing row P(R,...).

The output is triangle-like - for pascal 4 2 it looks like

1 

3 
3 3 

3 
6 6 
3 6 3 

1 
3 3 
3 6 3 
1 3 3 1