python recursive pascal triangle

Question

After completing an assignment to create pascal's triangle using an iterative function, I have attempted to recreate it using a recursive function. I have gotten to the point where I can get it to produce the individual row corresponding to the number passed in as an argument. But several attempts to have it produce the entire triangle up to and including that row have failed. I even tried writing a separate function which iterates over the range of the input number and calls the recursive function with the iterated digit while appending the individual lines to list before returning that list. The desired output should be a list of lists where each internal list contains one row of the triangle. Like so:

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

Instead it returns a jumbled mess of a nested list completely filled with 1's.

Here is the recursive function in question, without the second function to append the rows (I really wanted 1 all inclusive function anyway):

def triangle(n):
    if n == 0:
        return []
    elif n == 1:
        return [1]
    else:
        new_row = [1]
        last_row = triangle(n-1)
        for i in range(len(last_row)-1):
            new_row.append(last_row[i] + last_row[i+1])
        new_row += [1]
    return new_row

To be clear, I have already completed the assigned task, this is just to provide a deeper understanding of recursion...

Iterative solution:

def triangle(n):
    result = []
    for row in range(n):
        newrow = [1]
        for col in range(1, row+1):
            newcell = newrow[col-1] * float(row+1-col)/col
            newrow.append(int(newcell))
        result.append(newrow)
    return result

It looks like it only returns the `n`th list. Is this intended? What exactly is your current output? — robert, May 17 '12 at 01:28
Why don't you pass a list of lists as a parameter rather than a number? Especially since the results of each row depends on the results of the previous one. — Joel Cornett, May 17 '12 at 01:28
@robert the intention is to output a list of lists, where each list is a row of the triangle. I got to this point, but any attempt I have made to collect the rows into a list have resulted in arbitrarily nested lists containing 1's. — Verbal_Kint, May 17 '12 at 01:37

score 13 · Accepted Answer · edited May 17 '12 at 01:44

13

You just need to pass a list of lists through the recursion, and pick off the last element of the list (i.e. the last row of the triangle) to build your new row. Like so:

def triangle(n):
    if n == 0:
        return []
    elif n == 1:
        return [[1]]
    else:
        new_row = [1]
        result = triangle(n-1)
        last_row = result[-1]
        for i in range(len(last_row)-1):
            new_row.append(last_row[i] + last_row[i+1])
        new_row += [1]
        result.append(new_row)
    return result

edited May 17 '12 at 01:44

robert

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answered May 17 '12 at 01:41

happydave

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No problem. Recursion can be confusing. I'd recommend using a lot of print statements, or a good debugger. – happydave May 17 '12 at 01:48
This still print out [1 ,1]? – Timothy Leung Jul 09 '13 at 06:18
how about add one more base case for n = 2 return [1, 2, 1] – Timothy Leung Jul 09 '13 at 06:18

score 4 · Answer 2 · answered May 17 '12 at 01:48

4

An alternative to happydave's solution, using tail recursion:

def triangle(n, lol=None):
    if lol is None: lol = [[1]]
    if n == 1:
        return lol
    else:
        prev_row = lol[-1]
        new_row = [1] + [sum(i) for i in zip(prev_row, prev_row[1:])] + [1]
        return triangle(n - 1, lol + [new_row])

answered May 17 '12 at 01:48

Joel Cornett

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I think it's a little clearer as `[x + y for x, y in zip([0] + prev_row, prev_row + [0])]`. – Karl Knechtel May 17 '12 at 04:45
1

@KarlKnechtel: I think you're right. I like the `zip(a,a[1:])` trick though :) – Joel Cornett May 17 '12 at 07:47

score 0 · Answer 3 · answered Nov 05 '22 at 08:41

I think its shod be helpful, this code draw triangle and do it recursively:

def traingle(n):
    if n == 1:
        print(1)
        return [1]
    else:
        answer = [1]
        print_able = '1 '
        previos = traingle(n-1)
        for index in range(len(previos)-1):
            eleman = previos[index]+previos[index+1]
            answer.append(eleman)
            print_able += str(eleman)+' '
        answer.append(1)
        print_able += '1'
        print(print_able)
        return answer


end = int(input())
traingle(end)

score -1 · Answer 4 · answered Jun 03 '18 at 17:06

-1

Yes, as Karl Knechtel also showed, recursive Pascal Triangle can go this way :

P=lambda h:(lambda x:x+[[x+y for x,y in zip(x[-1]+[0],[0]+x[-1])]])(P(h-1))if h>1 else[[1]]
print(P(10))

answered Jun 03 '18 at 17:06

Cépagrave

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python recursive pascal triangle

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