Solution: 2013 Equations, Jan. 1, 2013

Here’s my solution to the problem for Jan. 1, 2013. The solution is written in Python.

The task:

As we begin the new year, we note that 109-8*7+654*3-2/1 = 2013. There are three other combinations of the numbers 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, in order, combined with the five operators NIL, +, -, * and / that also evaluate to 2013.

Your task is to write a program that finds all four expressions that evaluate to 2013.

My solution:

# This function generates equations based on the positions of the
# operations in a list. After producing an equation, it sends the
# equation to another function to be evaluated. It would be faster
# if I only generated and checked solutions that didn't start with
# '+', '*', or '/' (as solutions of this sort are not valid in my
# scheme) but this presumably doesn't add a significant amount of
# computation time, so it's unnecessary to worry about.
def generateEqsAndSolve():
	Ops = ['-', '', '+', '*', '/']
	for x in (itertools.product(Ops, repeat=10)):
		processEq(list(x))

# Create an equation from the ordered list of operations.
# Then solve for the value of the equation.
def processEq(L):
	if L[0] == '*' or L[0] == '/' or L[0] == '+':
		return

	# Get the equation as a string.
	equation = ''
	for x in range(len(L)):
		if (L[x] == '-' or L[x] == '+') and x != 0:
				equation += ' '
		equation += L[x]
		equation += str(10-x)

	equationLen = len(equation)

	# Evaluate the equation...p
	finalVal = 0.0
	num = 0.0
	i = 0
	while i < equationLen:
		eqPart = ''
		if equation[i] == '-':
			i += 1
			while equation[i] != ' ' and i < equationLen:
				eqPart += equation[i]
				i += 1
				if not i < equationLen:
					break

			num = processNum(eqPart)
			num *= -1
		elif equation[i] == '+' or equation[i] == '':
			i += 1
			while equation[i] != ' ' and i < equationLen:
				eqPart += equation[i]
				i += 1
				if not i < equationLen:
					break

			num = processNum(eqPart)

		finalVal += num
		i += 1

	# If the equation has the value of 2013, print it!
	if math.fabs(finalVal - 2013) < .004:
		print equation

	return

# Helper function that takes string representing
# part of an eqaution produced in processEq and
# returns the value of the part of the equation
def processNum(eqPart):
	partLen = len(eqPart)
	i = 0
	count = 0
	strNum = ''
	num = 0
	storedNum = 1.0
	op = ''
	while i < partLen:
		if not eqPart[i].isdigit():
			break
		strNum += eqPart[i]
		i += 1

	storedNum = float(strNum)
	strNum = ''

	while i < partLen:
		op = eqPart[i]
		i += 1

		while eqPart[i].isdigit() and i < partLen:
			strNum += eqPart[i]
			i += 1
			if not i < partLen:
				break
		num = int(strNum)

		if op == '*':
			storedNum *= num
		else:
			storedNum /= num

		strNum = ''

	return storedNum