def getSum(x, y):
	#Returns the sum of x and y
	return x + y

def getProduct(x, y):
	#Returns the product of x and y
	return x * y

def getFactors(z):
	#Returns the factors of a positive integer z, but not itself and 1, and not if z is 2
	i = 2
	factorList = []
	while (i < z):
		if (z % i == 0):
			other = z / i
			if (i not in factorList and other not in factorList):
				factorList.append(i)
				factorList.append(z/i)
		i = i + 1
	return factorList

def getAddends(z):
	#Returns the addends of a positive integer z, but not itself minus 1 and 1
	i = 2
	addendList = []
	while (i < (z-1)):
		if (i not in addendList):
			addendList.append(i)
			addendList.append(z-i)
		i = i + 1
	return addendList

def hasOneFactorSumList(addendListSum):
	#Returns True if any addend pair's product has one factor, and false if they all have more than one
	i = 0
	oneFactor = False
	while (i < len(addendListSum)):
		firstNum = addendListSum[i]
		secondNum = addendListSum[i+1]
		hypotheticalProduct = getProduct(firstNum, secondNum)
		hypotheticalFactors = getFactors(hypotheticalProduct)
		if (len(hypotheticalFactors) <= 2):
			return True
		i = i + 2
	return False

def findPossibleCombos(factorsListProduct):
	#Returns a possible combination for any factor pairs that don't sum to a number, that being a product has only one pair of factors
	possibleCombos = []
	i = 0
	while (i < len(factorsListProduct)):
		firstNum = factorsListProduct[i]
		secondNum = factorsListProduct[i+1]
		hypotheticalSum = getSum(firstNum, secondNum)
		#P is now mentally at what S would've seen before calculating to tell P that he knows P doesn't know
		hypotheticalAddends = getAddends(hypotheticalSum)
		oneFactor = hasOneFactorSumList(hypotheticalAddends)
		if (not oneFactor):
			possibleCombos.append(firstNum)
			possibleCombos.append(secondNum)
		i = i + 2
	return possibleCombos

def riddleSolver():
	print("STARTING THE RIDDLE")
	for x in range(2,100):
		#reduce redundant checks where x and y are flipped
		for y in range(x,100):
			#x and y are different variables, can't be the same number
			if (x == y):
				continue
			print("COMPUTING FOR X: " + str(x) + " AND Y: " + str(y))
			productVar = getProduct(x,y)
			factorsListProduct = getFactors(productVar)
			#the product of x and y must have multiple factors or P guy would know right away
			if (len(factorsListProduct) >= 4 and len(factorsListProduct) %  2 == 0):
				sumVar = getSum(x,y)
				addendListSum = getAddends(sumVar)
				#the sum of x and y must have multiple addends or S guy would know right away
				if (len(addendListSum) >= 4 and len(addendListSum) %  2 == 0):
					#none of those addend pairs must produce a number that being a product has only one pair of factors or else S can't tell P he doesn't know
					oneFactor = hasOneFactorSumList(addendListSum)
					if (not oneFactor):
						#P can now sum his original factors list and find which pair(s) do(es)n't sum to a number that would let S tell him that, namely, which pair doesn't sum to a number that being a product has only one pair of factors
						possiblePCombos = findPossibleCombos(factorsListProduct)
						if (len(possiblePCombos) == 2):
							#P now knows what x and y are and tells S that he does, which spurs S to also know, this means that P had eliminated all other factor pairs because they sum to a number that being a product has only one pair of factors which wouldnt've allowed S to make his statement
							print(str(possiblePCombos) + " IS A POSSIBLE COMBINATION")
							#S can do the same math with all of his addend pairs, getting a product for each, and then getting factors pairs for those, and finding the one which doesn't sum to a number that being a product has only one pair of factors
							#Only one of S's addend pairs must meet this criteria
							criteriaCounter = 0
							solution = []
							i = 0
							while (i < len(addendListSum)):
								firstNum = addendListSum[i]
								secondNum = addendListSum[i+1]
								sumProduct = getProduct(firstNum,secondNum)
								sumProductFactors = getFactors(sumProduct)
								possibleSCombos = findPossibleCombos(sumProductFactors)
								if (len(possibleSCombos) == 2):
									criteriaCounter = criteriaCounter + 1
									solution = possibleSCombos
								if (criteriaCounter > 1):
									break
								i = i + 2
							if (criteriaCounter == 1):
								print("SOLVED! " + str(solution) + " IS THE ANSWER")
								return
						else:
							continue
					else:
						continue
				else:
					continue
			else:
				continue
	print("RIDDLE COMPLETED WITHOUT FINDING AN ANSWER")

riddleSolver()