Encryption/Decryption in Python; decryption not working for few random bytes

Question

first of all I'm reasonably new to Python. I'll try to explain this as best I can! I'm currently making an Encryption/Decryption program which encrypts and decrpyts a file, everything works fine, no errors given. Only when I check the newly decrypted file to the original in a Hex editor, very few but some bytes have not been decrpyted correctly. The only places I can see this happening is with any 'FF' byte is now a '00' byte in the decrypted file. I can't work out why this is happening, code is below:

def shift_byte(x):
    if(ord(x) == 0):
        return (ord(x))
    elif(ord(x) == shiftrange):
        v1 = (ord(x) + shift)
        return v1
    else:       
        return ((ord(x) + shift) % 255)

That is the encryption, just a simple shift cipher, this is the decryption from another file:

def shift_byte(x):
    if(x == 0):
        return (x)
    elif(x == shiftrange):
        v1 = x - shift
        return v1
    else:       
        return ((x - shift) % 255)

My only thoughts are that it is something to do with this line:

return ((x - shift) % 255)

Is there something wrong here? Thanks.

How can ord(x) be a range? Shouldn't ord(x) be within a range? Why not always use the modulus operator? — Maarten Bodewes
– Maarten Bodewes, Commented Jan 11, 2015 at 16:15
That's just bad variable naming by me, shiftrange isn't actually a range, its just an integer. — mattybaines
– mattybaines, Commented Jan 11, 2015 at 16:28
I have included the modulus on each statement now and again it has improved slightly, but still some random bits are not as they should be in the decrypted file — mattybaines
– mattybaines, Commented Jan 11, 2015 at 18:04
I've just found that the byte that is affected by this issue depends on the shift no. being used, the problem may lie somewhere else — mattybaines
– mattybaines, Commented Jan 11, 2015 at 18:07

Maarten Bodewes · Accepted Answer · 2015-01-11 16:16:25Z

1

It has to be:

return ((x + shift) % 256)

and

return ((x - shift) % 256)

With % 255 your values only range from 0 to 254(0x00 to 0xFE) and not from 0 to 255(0x00 to 0xFF).

edited Jan 11, 2015 at 16:16

Maarten Bodewes

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answered Jan 11, 2015 at 15:25

nablahero

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5 Comments

mattybaines Over a year ago

Thank you! Not sure how I missed that, works fine now!

mattybaines Over a year ago

Actually hasn't fully worked! I'm now getting the same issue but with different bytes, the decrypted image can now be open, but the image is a bit messed up!

President James K. Polk Over a year ago

This is incorrect. The decryption has to 'undo' the encryption, no matter if %256 or %255 is used. If %255 is performed on encryption then that's what is also needed on decryption.

mattybaines Over a year ago

Yeah I have changed %255 to %256 on both and I am now able to open the decrypted image, only the image is a bit scattered

nablahero Over a year ago

Thaats's not correct Greg. If you use %255 the input can only be between 0 and 254. Here the input ranges from 0x00 to 0xFF, so from 0 to 255 which means it has to be % 256.

Artjom B. · Accepted Answer · 2015-01-12 23:07:24Z

0

I really dont get why you use this structure:

if(x == 0):
    return (x)
elif(x == shiftrange):
    v1 = x - shift
    return v1
else:       
    return ((x - shift) % 256)

why don't you only return ((x-shift) % 256) ? If i understand this correctly you do not encrypt or decrypt 00 right? because if x == 0 you just return x. In this case it wont be "encrypted".

For me this is a problem, because is you encrypt F with a shift of 1, it will be 0. When calling the decryption function 0 will not be shifted back to F, it will just return 0.

Just use:

def shift_byte(x):       
    return ((x - shift) % 256)

Does this help?

edited Jan 12, 2015 at 23:07

Artjom B.

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answered Jan 12, 2015 at 23:04

nablahero

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