Python function to find every possible permutation of a input hexademical key where one bit of one byte in the key is flipped, extremely quick and dirty...

bitFlip.py

def bitFlip(key):
	flippedKeys = []
	for i in range(0, len(key), 2):
		bits = bin(int(key[i:i+2], base=16))
		for j, b in enumerate(bits[2:len(bits)]):
			flip = list(copy.deepcopy(bits[2:len(bits)]))
			flip[j] = (1)^int(b)
			#if j < len(flip)-1:
			#	flip[j+1] = (1)^int(b)
			bitStr = ""
			for f in flip:
				bitStr += str(f)
			flippedKey = copy.deepcopy(key)
			if not flippedKey[0:i-2]+hex(int(bitStr, base=2))[2:4] in flippedKeys and not flippedKey[0:i-2]+hex(int(bitStr, base=2))[2:4]+flippedKey[i:len(flippedKey)] in flippedKeys:
				if len(flippedKey[0:i-2]+hex(int(bitStr, base=2))[2:4]) == len(key):
					flippedKeys.append(flippedKey[0:i-2]+hex(int(bitStr, base=2))[2:4])#+flippedKey[i+2:len(flippedKey)])
				elif len(flippedKey[0:i-2]+hex(int(bitStr, base=2))[2:4]+flippedKey[i:len(flippedKey)]) == len(key) :
					flippedKeys.append(flippedKey[0:i-2]+hex(int(bitStr, base=2))[2:4]+flippedKey[i:len(flippedKey)])
	return flippedKeys