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lightly alter a hash programmatically

At the moment I frequently have to do something in unittests with hashes and cryptographic signatures. Sometimes they get generated, and I just need to alter one slightly and prove that something no longer works. They are strings of hex-digits 0-9 and a-f of specific length. Here is a sample 64 long:
h = '702b31faad0246cc89a5dc782cdf5235a885d0f529fb30a4e1e70e00938df91a'
I want to change just one character somewhere in there.
You can't be sure that every digit 0 - 9 and a - f will be in there, although would guess it's at least 95% certain that they all are. If you could be sure, I would just run h = h.replace('a', 'b', 1) on it.
If you do it manually, you can just look at it and see the third digit is 2 and run:
new = list(h)
new[2] = '3'
h = ''.join(new)
But if you cannot see it and it needs to happen programmatically, what is a clean and certain way to change just one character in it somewhere?

from random import randrange
h = '702b31faad0246cc89a5dc782cdf5235a885d0f529fb30a4e1e70e00938df91a'
i = randrange(len(h))
new_h = h[:i] + hex(int(h[i], 16) + randrange(1, 16))[-1:] + h[i+1:]
In words:
choose a random index i in h
split the string into the part before the index, the char at the index, and the rest
replace the char at the index with its hex value incremented by a random int between 1 and 15, modulo 16 (i.e., its rightmost hex character)
build the new string from the above pieces
Note that an increment by a value between 1 and 15 (included), followed by a modulo 16, never maps a hex digit onto itself. An increment by 0 or 16 would map it exactly onto itself.

You can just choose a random index
import random
valid_chars = '0...f'
def replace_hash(hash_digest):
idx_to_replace = random.randint(64)
char_to_replace = hash_digest[idx_to_replace]
replacements = valid_chars.replace(char_to_replace, '')
hash_digest[idx_to_replace] = replacements[random.randint(15)
return hash_digest
The most efficient way is to just replace the first char with 1 of 2 replacements. I mean, you can only collide with one char anyway so there's no need to do it randomly. But if you want a random change the function'll work.

I suggest you increment the last character of the hash (cycling to 0 after f). That way you are sure to get a different hash, only differing by one character.
You can easily extend this method to change a character at the position of your choosing, and not just the last one.
h = '702b31faad0246cc89a5dc782cdf5235a885d0f529fb30a4e1e70e00938df91a'
def change_hash(h, index=-1):
digits = list(h)
old_digit= digits[index]
v = int(old_digit, 16)
new_v = (v+1)%16
new_digit = '{:x}'.format(new_v)
digits[index] = new_digit
return ''.join(digits)
print(change_hash(h))
# 702b31faad0246cc89a5dc782cdf5235a885d0f529fb30a4e1e70e00938df91b
# ^
print(change_hash(h, 2))
# 703b31faad0246cc89a5dc782cdf5235a885d0f529fb30a4e1e70e00938df91a
# ^
EDIT:
added option to change a digit at an arbitrary position
formatting the digit using format() as it was proposed in another answer

h = chr(ord(h[0]) + ((-1) if (h[0] in "9z") else 1)) + h[1:]

Different results when return multiple values in python (Cryptopal challenges)

I'm working on problem 3(set 1) of the cryptopals challenges (https://cryptopals.com/sets/1/challenges/3)
I've already found the key ('x') and decrypted the message ('Cooking mcs like a pound of bacon')
Here is my code:
from hexToBase64 import hexToBinary
from fixedXOR import xorBuffers
def binaryToChar(binaryString):
asciiValue = 0
for i in range(int(len(binaryString))-1,-1,-1):
if(binaryString[i] == '1'):
asciiValue = asciiValue + 2**(7-i)
return chr(asciiValue)
def decimalToBinary(number):
binaryString = ""
while (number != 0):
bit = number % 2
binaryString = str(bit) + binaryString
number = int(number/2)
while(len(binaryString) < 8):
binaryString = "0" + binaryString
return binaryString
def breakSingleByteXOR(cipherString):
decryptedMess = ""
lowestError = 10000
realKey = ""
for i in range(0,128):
errorChar = 0
tempKey = decimalToBinary(i)
tempMess = ""
for j in range(0,len(cipherString),2):
#Take each byte of the cipherString
cipherChar = hexToBinary(cipherString[j:j+2])
decryptedChar = binaryToChar(xorBuffers(cipherChar,tempKey))
asciiValue = ord(decryptedChar)
if (not ((asciiValue >= 65) and (asciiValue <= 90)) \
or ((asciiValue >= 90) and (asciiValue <= 122)) \
or ( asciiValue == 32 )):
# if the character is not one of the characters ("A-Z" or "a-z"
# or " ") consider it as an "error"
errorChar += 1
tempMess = tempMess + decryptedChar
if(errorChar < lowestError):
lowestError = errorChar
decryptedMess = tempMess
realKey = chr(i)
return (realKey,decryptedMess)
if __name__ == "__main__":
print(breakSingleByteXOR("1b37373331363f78151b7f2b783431333d78397828372d363c78373e783a393b3736"))
The problem is when I use the function breakSingleByteXOR to return one value (decryptedMess), it came out okay "cOOKING mcS LIKE A POUND OF BACON"
But when I return 2 values with the function (as the code above - (key,decryptedMess)), I received a weird result ('x', 'cOOKING\x00mc\x07S\x00LIKE\x00A\x00POUND\x00OF\x00BACON'), can anyboby explain to me why this is the case?
Tbh, I'm learning python as I'm doing the challenges so hopefully I dont trigger anyone with these code.... I'd also really appreciate it if anyone could give me some advices on writing good python code
Thanks guys :D

It's true that the reason for the difference in the printed string is a quirk of the print function.
The deeper problem with that program is that it's not producing the correct answer. That's because the big ugly if that tries to decide whether a decrypted character is in the acceptable range is incorrect.
It's incorrect in two ways. The first is that (asciiValue >= 90) should be (asciiValue >= 97). A better way to write all of those expressions, which would have avoided this error, is to express them as (asciiValue >= ord('a')) and (asciiValue == ord(' ')) and so on, avoiding the inscrutable numbers.
The second way is that the expressions are not properly grouped. As they stand they do this:
character is not in the range 'A' to 'Z',
or character is in the range 'a' to 'z',
or character is 'space',
then count this as an error
so some of the characters that should be good (specifically 'a' through 'z' and space) are counted as bad. To fix, you need to rework the parentheses so that the condition is:
character is not in the range 'A' to 'Z',
and character is not in the range 'a' to 'z',
and character is not space,
then count this as an error
or (this is style you were trying for)
character is not (in the range 'A' to 'Z'
or in the range 'a' to 'z'
or a space)
I'm not going to give you the exact drop-in expression to fix the program, it'll be better for you to work it out for yourself. (A good way to deal with this kind of complexity is to move it into a separate function that returns True or False. That makes it easy to test that your implementation is correct, just by calling the function with different characters and seeing that the result is what you wanted.)
When you get the correct expression, you'll find that the program discovers a different "best key" and the decrypted string for that key contains no goofy out-of-range characters that behave strangely with print.

The print function is the culprit - it is translating the characters \x00 and \x07 to ASCII values when executed. Specifically, this only occurs when passing a string to the print function, not an iterable or other object (like your tuple).
This is an example:
>>> s = 'This\x00string\x00is\x00an\x00\x07Example.'
>>> s
'This\x00string\x00is\x00an\x00\x07Example.'
>>> print(s)
This string is an Example.
If you were to add the string s to an iterable (tuple, set, or list), s will not be formatted by the print function:
>>> s_list = [s]
>>> print(s_list) # List
['This\x00string\x00is\x00an\x00\x07Example.']
>>> print(set(s_list)) # Set
{'This\x00string\x00is\x00an\x00\x07Example.'}
>>> print(tuple(s_list)) # Tuple
('This\x00string\x00is\x00an\x00\x07Example.')
Edit
Because the \x00 and \x07 bytes are ASCII control characters, (\x00 being NUL and \x07 being BEL), you can't represent them in any other way. So one of the only ways you could strip these characters from the string without printing would be to use the .replace() method; but given \x00 bytes are being treated as spaces by the terminal, you would have to use s.replace('\x00', ' ') to get the same output, which has now changed the true content of the string.
Otherwise when building the string; you could try and implement some logic to check for ASCII control characters and either not add them to tempMess or add a different character like a space or similar.
References
ASCII Wiki: https://en.wikipedia.org/wiki/ASCII
Curses Module: https://docs.python.org/3.7/library/curses.ascii.html?highlight=ascii#module-curses.ascii (Might be useful if you wish to implement any logic).

Is there any easy way to reverse this hash function?

I've got the following python method, it gets a string and returns an integer. I'm looking for the correct input that will print "Great Success!"
input = "XXX"
def enc(pwd):
inc = 0
for i in range(1, len(pwd) + 1):
_1337 = pwd[i - 1]
_move = ord(_1337) - 47
if i == 1:
inc += _move
else:
inc += _move * (42 ** (i - 1))
return inc
if hex(enc(input)) == 0xEA9D1ED352B8:
print "Great Success!"

It's encoding the input in base 42 (starting from chr(47) which is '/'), and easy to decode:
def dec(x):
while x:
yield chr(47 + x % 42)
x //= 42
print ''.join(dec(0xEA9D1ED352B8))
The output is: ?O95PIVII

Yes it is easy to crack. The answer is "?O95PIVII".
>>> enc("?O95PIVII")==0xEA9D1ED352B8
True
The check in your example is invalid by the way, you need to drop the hex from it because the literal 0xEA9D1ED352B8 is parsed as an int.
edit See Rawing's comment for a hint of how I got the answer. A bit more detail: because its summing powers of 42 based on the index of the char in the string, it's easy to figure out the required length of the input string: math.ceil(math.log(target)/math.log(42)) which gives us 9. I then worked my way from the end to the start of the string, because the last character has the most effect on the sum.

You can try to bruteforce it.
Just make a while loop where the input is encrypted by the function until you have the same hash.
But with no informations about the length of the input and so on it could take a while.

how to make an imputed string to a list, change it to a palindrome(if it isn't already) and reverse it as a string back

A string is palindrome if it reads the same forward and backward. Given a string that contains only lower case English alphabets, you are required to create a new palindrome string from the given string following the rules gives below:
1. You can reduce (but not increase) any character in a string by one; for example you can reduce the character h to g but not from g to h
2. In order to achieve your goal, if you have to then you can reduce a character of a string repeatedly until it becomes the letter a; but once it becomes a, you cannot reduce it any further.
Each reduction operation is counted as one. So you need to count as well how many reductions you make. Write a Python program that reads a string from a user input (using raw_input statement), creates a palindrome string from the given string with the minimum possible number of operations and then prints the palindrome string created and the number of operations needed to create the new palindrome string.
I tried to convert the string to a list first, then modify the list so that should any string be given, if its not a palindrome, it automatically edits it to a palindrome and then prints the result.after modifying the list, convert it back to a string.
c=raw_input("enter a string ")
x=list(c)
y = ""
i = 0
j = len(x)-1
a = 0
while i < j:
if x[i] < x[j]:
a += ord(x[j]) - ord(x[i])
x[j] = x[i]
print x
else:
a += ord(x[i]) - ord(x[j])
x [i] = x[j]
print x
i = i + 1
j = (len(x)-1)-1
print "The number of operations is ",a print "The palindrome created is",( ''.join(x) )
Am i approaching it the right way or is there something I'm not adding up?

Since only reduction is allowed, it is clear that the number of reductions for each pair will be the difference between them. For example, consider the string 'abcd'.
Here the pairs to check are (a,d) and (b,c).
Now difference between 'a' and 'd' is 3, which is obtained by (ord('d')-ord('a')).
I am using absolute value to avoid checking which alphabet has higher ASCII value.
I hope this approach will help.
s=input()
l=len(s)
count=0
m=0
n=l-1
while m<n:
count+=abs(ord(s[m])-ord(s[n]))
m+=1
n-=1
print(count)

This is a common "homework" or competition question. The basic concept here is that you have to find a way to get to minimum values with as few reduction operations as possible. The trick here is to utilize string manipulation to keep that number low. For this particular problem, there are two very simple things to remember: 1) you have to split the string, and 2) you have to apply a bit of symmetry.
First, split the string in half. The following function should do it.
def split_string_to_halves(string):
half, rem = divmod(len(string), 2)
a, b, c = '', '', ''
a, b = string[:half], string[half:]
if rem > 0:
b, c = string[half + 1:], string[rem + 1]
return (a, b, c)
The above should recreate the string if you do a + c + b. Next is you have to convert a and b to lists and map the ord function on each half. Leave the remainder alone, if any.
def convert_to_ord_list(string):
return map(ord, list(string))
Since you just have to do a one-way operation (only reduction, no need for addition), you can assume that for each pair of elements in the two converted lists, the higher value less the lower value is the number of operations needed. Easier shown than said:
def convert_to_palindrome(string):
halfone, halftwo, rem = split_string_to_halves(string)
if halfone == halftwo[::-1]:
return halfone + halftwo + rem, 0
halftwo = halftwo[::-1]
zipped = zip(convert_to_ord_list(halfone), convert_to_ord_list(halftwo))
counter = sum([max(x) - min(x) for x in zipped])
floors = [min(x) for x in zipped]
res = "".join(map(chr, floors))
res += rem + res[::-1]
return res, counter
Finally, some tests:
target = 'ideal'
print convert_to_palindrome(target) # ('iaeai', 6)
target = 'euler'
print convert_to_palindrome(target) # ('eelee', 29)
target = 'ohmygodthisisinsane'
print convert_to_palindrome(target) # ('ehasgidihmhidigsahe', 84)
I'm not sure if this is optimized nor if I covered all bases. But I think this pretty much covers the general concept of the approach needed. Compared to your code, this is clearer and actually works (yours does not). Good luck and let us know how this works for you.

Manipulating string to repeat based on the length of another string

I am working on a python project, where I am required to include an input, and another value (which will be manipulated).
For example,
If I enter the string 'StackOverflow', and a value to be manipulated of 'test', the program will make the manipulatable variable equal to the number of characters, by repeating and trimming the string. This means that 'StackOverflow' and 'test' would output 'testtesttestt'.
This is the code I have so far:
originalinput = input("Please enter an input: ")
manipulateinput = input("Please enter an input to be manipulated: ")
while len(manipulateinput) < len(originalinput):
And I was thinking of including a for loop to continue the rest, but am not sure how I would use this to effectively manipulate the string. Any help would be appreciated, Thanks.

An itertools.cycle approach:
from itertools import cycle
s1 = 'Test'
s2 = 'StackOverflow'
result = ''.join(a for a, b in zip(cycle(s1), s2))
Given you mention plaintext - a is your key and b will be the character in the plaintext - so you can use this to also handily manipuate the pairing...
I'm taking a guess you're going to end up with something like:
result = ''.join(chr(ord(a) ^ ord(b)) for a, b in zip(cycle(s1), s2))
# '\x07\x11\x12\x17?*\x05\x11&\x03\x1f\x1b#'
original = ''.join(chr(ord(a) ^ ord(b)) for a,b in zip(cycle(s1), result))
# StackOverflow

There are some good, Pythonic solutions here... but if your goal is to understand while loops rather than the itertools module, they won't help. In that case, perhaps you just need to consider how to grow a string with the + operator and trim it with a slice:
originalinput = input("Please enter an input: ")
manipulateinput = input("Please enter an input to be manipulated: ")
output = ''
while len(output) < len(originalinput):
output += manipulateinput
output = output[:len(originalinput)]
(Note that this sort of string manipulation is generally frowned upon in real Python code, and you should probably use one of the others (for example, Reut Sharabani's answer).

Try something like this:
def trim_to_fit(to_trim, to_fit):
# calculate how many times the string needs
# to be self - concatenated
times_to_concatenate = len(to_fit) // len(to_trim) + 1
# slice the string to fit the target
return (to_trim * times_to_concatenate)[:len(to_fit)]
It uses slicing, and the fact that a multiplication of a X and a string in python concatenates the string X times.
Output:
>>> trim_to_fit('test', 'stackoverflow')
'testtesttestt'
You can also create an endless circular generator over the string:
# improved by Rick Teachey
def circular_gen(txt):
while True:
for c in txt:
yield c
And to use it:
>>> gen = circular_gen('test')
>>> gen_it = [next(gen) for _ in range(len('stackoverflow'))]
>>> ''.join(gen_it)
'testtesttestt'

What you need is a way to get each character out of your manipulateinput string over and over again, and so that you don't run out of characters.
You can do this by multiplying the string so it is repeated as many times as you need:
mystring = 'string'
assert 2 * mystring == 'stringstring'
But how many times to repeat it? Well, you get the length of a string using len:
assert len(mystring) == 6
So to make sure your string is at least as long as the other string, you can do this:
import math.ceil # the ceiling function
timestorepeat = ceil(len(originalinput)/len(manipulateinput))
newmanipulateinput = timestorepeat * manipulateinput
Another way to do it would be using int division, or //:
timestorepeat = len(originalinput)//len(manipulateinput) + 1
newmanipulateinput = timestorepeat * manipulateinput
Now you can use a for loop without running out of characters:
result = '' # start your result with an empty string
for character in newmanipulateinput:
# test to see if you've reached the target length yet
if len(result) == len(originalinput):
break
# update your result with the next character
result += character
# note you can concatenate strings in python with a + operator
print(result)