Fully generalized COUNTLESS algorithm with a (relatively) memory efficient dynamic programming solution

countlessND.py

from itertools import combinations
from functools import reduce
import numpy as np

def countlessND(data, factor):
  assert len(data.shape) == len(factor)

  sections = []

  mode_of = reduce(lambda x,y: x * y, factor)
  majority = int(math.ceil(float(mode_of) / 2))

  data += 1 # offset from zero
  
  # This loop splits the 2D array apart into four arrays that are
  # all the result of striding by 2 and offset by (0,0), (0,1), (1,0), 
  # and (1,1) representing the A, B, C, and D positions from Figure 1.
  for offset in np.ndindex(factor):
    part = data[tuple(np.s_[o::f] for o, f in zip(offset, factor))]
    sections.append(part)

  pick = lambda a,b: a * (a == b)
  lor = lambda x,y: x + (x == 0) * y # logical or

  subproblems = [ {}, {} ]
  results2 = None
  for x,y in combinations(range(len(sections) - 1), 2):
    res = pick(sections[x], sections[y])
    subproblems[0][(x,y)] = res
    if results2 is not None:
      results2 = lor(results2, res)
    else:
      results2 = res

  results = [ results2 ]
  for r in range(3, majority+1):
    r_results = None
    for combo in combinations(range(len(sections)), r):
      res = pick(subproblems[0][combo[:-1]], sections[combo[-1]])
      
      if combo[-1] != len(sections) - 1:
        subproblems[1][combo] = res

      if r_results is not None:
        r_results = lor(r_results, res)
      else:
        r_results = res
    results.append(r_results)
    subproblems[0] = subproblems[1]
    subproblems[1] = {}
    
  results.reverse()
  final_result = lor(reduce(lor, results), sections[-1]) - 1
  data -= 1 
  return final_result