Pascal's triangle/line

Question

I'm writing a code for homework, it needs t o output the values for pascal's triangle, but not in a triangle. for instance I input

pascalLine(2)

it outputs:

[1,2,1]

and for:

pascalLine(4)

it outputs:

[1,4,6,4,1]

I currently have this code and I believe it is far from complete, because It outputs a triangle.

def pascalLine(n): for rownum in range (n): newValue=1 PrintingList = list() for iteration in range (rownum): newValue = newValue * ( rownum-iteration ) * 1 / ( iteration + 1 ) PrintingList.append(int(newValue)) print(PrintingList) print()

"it needs t o output the values for pascal's triangle, but not in a triangle" - I'm not really sure what your desired output is here. — millimoose
– millimoose, Commented Oct 22, 2012 at 0:25
@millimoose: I think the [1,2,1] and [1,4,6,4,1] the OP gives are examples of the desired output. I.e. the function should return the nth row of Pascal's triangle as a list. — DSM
– DSM, Commented Oct 22, 2012 at 0:26

millimoose · Accepted Answer · 2015-12-26 20:30:07Z

Assuming your original code to calculate row n is correct, you can just get rid of the outer loop:

def pascalLine(n): newValue=1 row = [newValue] for i in range (n): newValue = (newValue * (n-i)) / ( i + 1 ) row.append(newValue) print(row) print()

Blender · Accepted Answer · 2012-10-22 00:45:49Z

3

You can just use this simple algorithm for computing an arbitrary row without having to compute the entire triangle:

def pascal_row(n): row = [1] for col in range(1, n): row.append(row[-1] * (n - col) / col) return row

answered Oct 22, 2012 at 0:45

Blender

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3 Comments

kmonsoor Over a year ago

the solution doesn't give n-th row of the PT, rather gives (n-1)th row. by wikipedia definition, the 1st row is considered 0-th row.

Blender Over a year ago

@kmonsoor: The rows correspond to powers of (x + y). Since (x + y)^0 = 1, the 0th row will be 1.

Blender Over a year ago

@kmonsoor: Numberings in mathematics are somewhat arbitrary, so there's no "right" answer.

Andy Richter · Accepted Answer · 2025-11-17 22:19:19Z

This gives [1] for row zero and [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1] for row 12:

 def getRow(r): row = [1] for c in range(1,r+1): row += [row[-1]*(r-c+1)//c] return row

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