p045.py

'''
Triangle, pentagonal, and hexagonal numbers are generated by the following
formulae:

Triangle    Tn=n(n+1)/2   1, 3, 6, 10, 15, ...
Pentagonal    Pn=n(3n-1)/2    1, 5, 12, 22, 35, ...
Hexagonal   Hn=n(2n-1)    1, 6, 15, 28, 45, ...
It can be verified that T285 = P165 = H143 = 40755.

Find the next triangle number that is also pentagonal and hexagonal.
'''

import common
import itertools

def triangles():
  for n in itertools.count(1):
    yield n*(n+1)/2
common.assertEquals([1, 3, 6, 10, 15], list(itertools.islice(triangles(), 5)))

def pentagons():
  for n in itertools.count(1):
    yield n*(3*n-1)/2
common.assertEquals([1, 5, 12, 22, 35], list(itertools.islice(pentagons(), 5)))

def hexagons():
  for n in itertools.count(1):
    yield n*(2*n-1)
common.assertEquals([1, 6, 15, 28, 45], list(itertools.islice(hexagons(), 5)))

def tripentahexas():
  ts, ps, hs = triangles(), pentagons(), hexagons()
  t, p, h = ts.next(), ps.next(), hs.next()

  while True:
    if t < p or t < h:   t = ts.next()
    elif p < t or p < h: p = ps.next()
    elif h < t or h < p: h = hs.next()
    else:
      yield t
      t, p, h = ts.next(), ps.next(), hs.next()

tphs = tripentahexas()
common.assertEquals(1, tphs.next())
common.assertEquals(40755, tphs.next())
common.submit(tphs.next(), expected=1533776805)