x3c-solver-complete

from itertools import combinations
import re

X =[1,7,3,4,5,6]
C = (1,2,3),(3,4,5),(1,2,5),(1,5,6),(1,7,6)

if len(X) % 3 > 0:
  print('no')
  quit()

# I will check each possible set of "len(X)//3-combinations" from
# the list C.
# This algorithim belongs to Hopee1995x on Reddit

trip_wire_for_no = 0
for jj in combinations(C, len(X)//3):
    if len(jj) > 1:
        one_by_one = str(jj)
        # Below, I then convert each combination
        # of jj into an indidvudual list.
        loop_one_by_one = (re.sub("[^0-9,]", "", one_by_one))
        loop_one_by_one = loop_one_by_one.split(',')
        # Below, I convert "strings-of-integers"
        # into int's.
        # (eg. '2' is now converted into 2 in the list)
        convert = list(map(int, loop_one_by_one))
        # Below, I make sure that ALL OF JJ's elements exist in X.
        make_sure_all_elements_exist_in_X =  all(elem in convert for elem in X)
        # if make_sure_all_elements_exist_in_X == true, I then loop through the new list. (convert)
        if make_sure_all_elements_exist_in_X:
            for yy in range(0, len(convert)):
                # Below, this statement makes sure that
                #every element in "convert" occurs less than 2-times.
                # So that our X3C solution only contains
                # ONE-ELEMENT-OF-EACH.
                if convert.count(int(convert[yy])) < 2:
                    # This statement makes sure that all elements
                    #in convert occur once in the list X.
                    if X.count(int(convert[yy])) == 1:
                        # This statement makes sure that
                        # the len(convert) == len(X).
                        if len(convert) == len(X):
                            print(convert, 'yes')
                            trip_wire_for_no = 1
                            break


if trip_wire_for_no == 0:
  print('no')