exact cover by 3sets

import sympy
import random
from itertools import permutations

# Only input lists with 3-elements.
# Lists should be treated as sets.
s = [1,2,3,4,5,6]
c = [[1,4,3],[4,6,5],[1,3,2]]


miss = []
delete = []
remove = []
remove_t=[]
no = 0
def input_validation():
    # checking for missing elements.
    no = 0
    for a in range(0, len(c)):
        for b in range(0, len(c[a])):
            miss.append(c[a][b])
    for d in s:
        if d not in miss:
            print('no', d)
            no = 1
            break
    if no == 1:
        quit()
    # Throw out unwanted orderings of sub-lists
    for a in range(0, len(c)):
        for i in permutations(c[a], 3):
            if list(i) in c[a:]:
                if list(i) != c[a]:
                    delete.append(list(i))
    for a in range(0, len(delete)):
        if delete[a] in c:
            del c[c.index(delete[a])]
    # Lists are treated as sets, so lists
    # with repeating elements is thrown out
    for rem in range(0, len(c)):
        if len(c[rem]) != len(set(c[rem])):
            remove.append(c[rem])
    for rem_two in range(0, len(remove)):
        if remove[rem_two] in c:
            del c[c.index(remove[rem_two])]
    # Throw out lists that have elements that do
    # not exist in s.
    for j in range(0, len(c)):
        for jj in range(0, len(c[j])):
            if any(elem not in s for elem in c[j]):
                remove_t.append(c[j])
    for rem_two in range(0, len(remove_t)):
        if remove_t[rem_two] in c:
            del c[c.index(remove_t[rem_two])]





input_validation()

# This reduction aims at reducing the running time
# within the length of the input.

# This is a bijective mapping of values in both C & S to their
# index-location values.

def reduction_t():
    s_copy = s.copy()
    for a in range(0, len(s)):
        s[a] = a
    for b in range(0, len(c)):
        for bb in range(0, len(c[b])):
            c[b][bb] = s_copy.index(c[b][bb])

# Use primes to help spread values apart.
# So that one value shouldn't replace another.
# Causing a false positive.

# (eg. C = [3], [3,4] == 10, and so
# does S = 1,2,3,4. But 1 & 2 are missing.)
def reduction_r():
    s_copy = s.copy()
    prime = 1
    for a in range(0, len(s)):
        while True:
            prime = prime + 1
            if sympy.isprime(prime) == True:
                if prime > a+1:
                    if prime not in s:
                        m = random.randint(2, len(c))
                        if prime*m not in s:
                            s[a] = prime*m
                            break
    for b in range(0, len(c)):
        for bb in range(0, len(c[b])):
            c[b][bb] = s[c[b][bb]]


reduction_t()
reduction_r()



r = []
for a in range(0, len(c)):
    r.append(sum(c[a]))

# Remove lists with repeating values
r = [r[x] for x in range(len(r)) if not(r[x] in r[:x])]
# k is our target-sum for SubsetSum function
k = sum(s)

# Courtesy to google search.
# I did not write this code
# below.
def isSubsetSum(set, n, sum):

    # The value of subset[i][j] will be
    # true if there is a
    # subset of set[0..j-1] with sum equal to i
    subset =([[False for i in range(sum + 1)]
            for i in range(n + 1)])

    # If sum is 0, then answer is true
    for i in range(n + 1):
        subset[i][0] = True

    # If sum is not 0 and set is empty,
    # then answer is false
    for i in range(1, sum + 1):
         subset[0][i]= False

    # Fill the subset table in botton up manner
    for i in range(1, n + 1):
        for j in range(1, sum + 1):
            if j<set[i-1]:
                subset[i][j] = subset[i-1][j]
            if j>= set[i-1]:
                subset[i][j] = (subset[i-1][j] or
                                subset[i - 1][j-set[i-1]])

    # uncomment this code to print table
    # for i in range(n + 1):
    # for j in range(sum + 1):
    # print (subset[i][j], end =" ")
    # print()
    return subset[n][sum]

# Driver code
if __name__=='__main__':
    set = r
    sum = k
    n = len(set)
    if (isSubsetSum(set, n, sum) == True):
        print("There should be an Exact Cover with sets of 3.")
    else:
        print("No")

# This code is contributed by
# sahil shelangia.