trbarron/abs_brackets.py

## abs_brackets.py


def eval_equation(braces, input):
    # Takes in input as a list of numbers
    # Takes in braces as a list of "-1"s (closed |'s) and 1's (opened |'s)
    # Returns the evaluation

    eq = ""
    for i in range(num_braces):

        # map our abs logic to something the computer can interpret
        if braces[i] == 1: eq += "abs("
        if braces[i] == -1: eq += ")"

        if i != num_braces - 1:
            # add the number
            eq += str(input[i])

            # after every number, if its not followed by a ')' then you want a '*'
            if braces[i + 1] != -1: eq += "*"
    return eval(eq)

def find_results(input):
    answers = set()  # use a set to avoid repeated entries

    # This takes all the possibilities and only evaluates those that will yield answers
    for i in range(2 ** num_braces):
        braces = list(str(bin(i))[2:])
        while len(braces) < num_braces: braces.insert(0, "-1")  # prepend "0"s since it doesn't have leading 0's
        for m in range(num_braces): braces[m] = int(
            braces[m]) * 2 - 1  # turn the binary into -1's and 1's for our logic later

        running_sum = 0
        failed_attempt = False
        for m in range(num_braces):
            running_sum += braces[m]
            if running_sum < 0: failed_attempt = True  # if we ever have the sum be negative that means we had more closed braces than open
        if sum(braces) == 0 and not failed_attempt: answers.add(
            eval_equation(braces, input))  # sum must be 0 since opens must equal closeds
    return(answers,len(answers))

for i in range(40):
    input = [-1 * j -1 for j in range(2*i-1)]
    num_braces = len(input) + 1
    a,b = find_results(input)
    print("input: ",input)
    print("results: ",b)



	def eval_equation(braces, input):
	# Takes in input as a list of numbers
	# Takes in braces as a list of "-1"s (closed \|'s) and 1's (opened \|'s)
	# Returns the evaluation

	eq = ""
	for i in range(num_braces):

	# map our abs logic to something the computer can interpret
	if braces[i] == 1: eq += "abs("
	if braces[i] == -1: eq += ")"

	if i != num_braces - 1:
	# add the number
	eq += str(input[i])

	# after every number, if its not followed by a ')' then you want a '*'
	if braces[i + 1] != -1: eq += "*"
	return eval(eq)

	def find_results(input):
	answers = set() # use a set to avoid repeated entries

	# This takes all the possibilities and only evaluates those that will yield answers
	for i in range(2 ** num_braces):
	braces = list(str(bin(i))[2:])
	while len(braces) < num_braces: braces.insert(0, "-1") # prepend "0"s since it doesn't have leading 0's
	for m in range(num_braces): braces[m] = int(
	braces[m]) * 2 - 1 # turn the binary into -1's and 1's for our logic later

	running_sum = 0
	failed_attempt = False
	for m in range(num_braces):
	running_sum += braces[m]
	if running_sum < 0: failed_attempt = True # if we ever have the sum be negative that means we had more closed braces than open
	if sum(braces) == 0 and not failed_attempt: answers.add(
	eval_equation(braces, input)) # sum must be 0 since opens must equal closeds
	return(answers,len(answers))

	for i in range(40):
	input = [-1 * j -1 for j in range(2*i-1)]
	num_braces = len(input) + 1
	a,b = find_results(input)
	print("input: ",input)
	print("results: ",b)