# The Algorithm itself

A project log for Attempt @ BPP algorithim for an NP-hard problem

Exact Three Cover.

```from itertools import groupby
import random
from itertools import combinations
from itertools import permutations
import sympy

s = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30
c = [[1,2,3],[1,2,4],[1,2,5],[1,2,6],[3,4,5],[3,4,6],[3,4,7],[4,5,6],[4,5,7],[6,3,7],[7,10,11],[22,1,11],[22,27,29],[30,1,14],[30,13,14],[14,20,11],[25,27,29],[14,1,2],[14,2,3],[14,5,1],[26,4,13],[26,7,17],[27,29,18],[19,27,13],[3,4,13],[19,2,1],[19,20,9],[18,3,2],[29,26,1],[5,3,11],[13,16,19],[13,20,21],[13,1,9],[17,19,23],[1,3,5],[5,3,2],[6,8,1],[7,30,23],[11,19,7],[7,4,28],[7,4,28],[18,13,1],[7,12,15],[15,2,3],[29,1,2],[29,2,3],[29,2,4],[29,5,4],[29,23,22],[3,26,5],[3,26,4],[3,25,3],[10,11,2],[30,1,2],[30,3,4],[30,4,5],[30,8,9],[30,18,19],[30,20,13],[6,22,23],[6,24,15],[18,1,2],[18,2,3],[18,4,5],[18,16,1],[24,26,27],[10, 2, 7], [10, 13, 14], [10, 5, 9], [6, 3, 8], [10, 26, 27], [10, 19, 16], [10, 25, 23], [10, 12, 15], [10, 29, 30], [10, 20, 21],[27,28,30],[28,29,30],[10,2,1],[10,7,8],[10,13,14],[10,11,8],[15,1,2],[15,8,4],[13,14,15],[11,12,15],[22,1,2],[22,21,3],[22,25,23],[10,4,8],[17,2,3],[17,2,4],[17,20,21],[18,2,3],[18,2,4],[18,19,16],[15,4,9],[4,5,1],[4,5,10],[5,7,9],[4,5,9]]

# Need a prime
# 3sets apart.

count = len(s)//3
while True:
count = count + 1
if sympy.isprime(count) == True:
prime = count
break

# I'm going to need this for
# my formula.
# Here, I am counting
# all possible Three Element
# combinations.
# But, its 241 times larger
# 241 is prime. Theortically,
# this should space them out
# better.

combinations_count = len(s)*241*((len(s)*241)-1)*((len(s)*241)-2)//6
combinations_three = len(s)*1*((len(s)*1)-1)*((len(s)*1)-2)//6

# The formula I need
# to use to calculate
# odds of finding
# a len(s)//3 length
# solution.

p = (len(s)//3)/combinations_count * 100

if len(s) % 3 != 0:
print('invalid input')
quit()

# Sort list to remove
# sets like (1,2,3) and (1,3,2)
# but leave (1,2,3)

delete = []
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])]

# remove sets
# that have
# repeating
# elements

remove = []
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])]

# remove sets
# that have
# elements
# that don't
# exist in S.

remove=[]
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.append(c[j])

for rem_two in range(0, len(remove)):
if remove[rem_two] in c:
del c[c.index(remove[rem_two])]

# Remove repeating sets

solutions =[c[x] for x in range(len(c)) if not(c[x] in c[:x])]

# check left and right for solutions

if len(c) >= len(s)//3*2:
for jj in combinations(c[0:len(s)//3*2], len(s)//3):
jj_flat = [item for sub in jj for item in sub]
jj_set = set(jj_flat)
if set(s) == jj_set and len(jj_set) == len(jj_flat):
print('yes', jj)
quit()

if len(c) >= len(s)//3*2:
X = list(reversed(c))
for jj in combinations(X[0:len(s)//3*2], len(s)//3):
jj_flat = [item for sub in jj for item in sub]
jj_set = set(jj_flat)
if set(s) == jj_set and len(jj_set) == len(jj_flat):
print('yes', jj)
quit()

# Well if length(c) is small
# use brute force with polynomial constant

if len(c) <= len(s)//3*2:
for jj in combinations(c, len(s)//3):
jj_flat = [item for sub in jj for item in sub]
jj_set = set(jj_flat)
if set(s) == jj_set and len(jj_set) == len(jj_flat):
print('yes', jj)
quit()

if len(c) <= len(s)//3*2:
quit()

# will need these Three (what a prime!)
# just in case my algorithim
# needs to reverse in loop.

length = len(solutions)
ss = s
c = solutions

# Shuffle these
# annoying
# counter-examples
# that might ruin
# my day

# Using prime number to help
# non-sequitir. Primes
# have been observed
# in nature to help
# avoid conflict.
# I'm applying the concept
# from nature onto my algorithim.
# It can't hurt!

print('Initalizing', prime,'shuffles')
for a in range(0, prime):
random.shuffle(c)

# while loop to see
# if we can find
# an Exact Three Cover
# in poly-time.

stop = 0
Potential_solution = []

while True:

if len(Potential_solution) == len(ss) // 3:
# break if Exact
# three cover is
# found.
print(Potential_solution, 'YES')
print('took',stop,'steps in while loop')
break

# Keep c
# and append to
# end of list
# del c
# to push >>
# in list.

c.append(c)
del [c]
Potential_solution = []
s = set()
stop = stop + 1
for l in c:
if not any(v in s for v in l):
Potential_solution.append(l)
s.update(l)

if len(Potential_solution) != len(ss)//3:
if stop == length:
# Reverse list just
# to reduce false -/+
for cc in range(0, len(c)):
c = list(reversed(c))

if stop >= combinations_three:
break

# Repeating while loop a
# second time. The first
# while loop was to address
# a bug that would get stuck
# in an infinite loop.

stop = 0
Potential_solution = []

while stop <= int(100/p*combinations_count):

# Shuffling c randomly
# to reduce chance of
# error.

random.shuffle(c)

if len(Potential_solution) == len(ss) // 3:
# break if Exact
# three cover is
# found.
print(Potential_solution, 'yes')
print('took',stop,'steps in while loop')
break

# Keep c
# and append to
# end of list
# del c
# to push >>
# in list.

c.append(c)
del [c]
Potential_solution = []
s = set()
stop = stop + 1
for l in c:
if not any(v in s for v in l):
Potential_solution.append(l)
s.update(l)

if len(Potential_solution) != len(ss)//3:
if stop == length:
# Reverse list just
# to reduce false -/+ 