#
#  Copyright (C) 2004 Marco Wahl <marco.wahl@gmail.com>
#
#  No warranty.
#

"""
Call:
python trianglecount.py <no. left> <no. right>

Output:
<no. triangles>

E.g.
python trianglecount.py 2 2
returns 27.

This module has its root at the Triangle Challenge, found by accident
on the web.

Scenario:
Let there be a triangle with a base-side with the endpoints left- and
right-base-point and a left and a right side.
There can be points on the left and/or the right side.
Obtain a net of lines by drawing the lines between the points on the
right side and the left-base-point and the lines between the points
on the left side and the right-base-point.

A triangle of interest is one that can be found in the net.

Question:
How many triangles of interest are there?

Observation:
There are three kinds of triangles of interest.
1. The triangle has the base a side.
2. The triangle shares the left base-side-vertex with the base-side
   and nothing else.
3. The triangle shares the right base-side-vertex with the base-side
   and nothing else.


Algorithm: count(L, R)
----------

L: number of points on the left (excluding the top-point)
R: number of points on the right (excluding the top-point)

count(0, 0) = 1

R > 0:

count(0, R) = count(0, R-1) + no of those who use the top line
 = R + 1 + count(0, R-1)
 = (R + 1) + R + ... + 1
 = (R + 1)*(R + 2)/2

l > 0:

count(L, 0) = count(0, L)

L, R > 0:

count(L, R) = count(L, R-1)
 + #{no fo those who use any part of the left-side}
 + #{no of those including the upmost part on the right side but not use the left part}
 = count(L, R-1)
 + #{no fo those who use any part of the left-side, type 1.}
 + #{no fo those who use any part of the left-side, type 2.}
 + #{no fo those who use any part of the left-side, type 3.}
 + 0
= count(L, R-1)
 + L + 1
 + R*(L + 1)
 + count(L-1, 0)
"""

def count(le, ri):
    "See the modules documentation for documentation."
    if (0, 0) == (le, ri):
        return 1
    elif 0 == le:
        return ri + 1 + count(le, ri - 1)
    elif 0 == ri:
        return count(ri, le)
    else:
        return count(le, ri - 1) \
               + le + 1 \
               + ri * (le + 1) \
               + count(le - 1, 0)

import sys
left, right = int(sys.argv[1]), int(sys.argv[2])
print "Number of triangles: ", count(left, right)