DevonRevenge's decoding challenge #1: I scrambled a sentence can you decode it(with algorithm provided)

[devonrevenge]

So i scrambled a sentence with a function that was 16 lines of code

usigfqedtwjsvusygih



EDIT: I have no idea how easy its gonna be;if its too hard i could submit the coder but that might then be too easy

EDIT2: Heres another mulooqlrizlok

EDIT3:heres the algorithm that scrambled it: int q = (a+b)*cipher%27;
a = letter pos in sentence, b position in 27 char alphabet, cipher is the cipher
so make a program that decodes it :D

EDIT4:it is reversible, a java function decoded it

[helios]

How do we know if the function is reversible?

[devonrevenge]

it is cos i built another function to reverse it, but it might be quite hard;but i will submit an easy level too

[Peter87]

Give us some time and we will solve this.

EDIT:
When I read scrambled I thought shuffled but if it's encrypted in some more advanced way it can be very hard to solve because there could be thousands of possible solutions.

[ResidentBiscuit]

Yes knowing whether or not it is just scrambled (ie same letters as original, just different order) or encrypted would be nice to know.

[hamsterman]

I decode it as "banana". Besides it not being the word you wanted, there is no way to show that I am wrong. There does indeed exist an encoding that takes "banana" to "usigfqedtwjsvusygih". The challenge is not had, it is plain impossible.

[Catfish3]

@ devonrevenge: I want to know if your code does selective substitution.

[Dissimulation]

I figured it out with hamsterman's help

#include <iostream>
#include <string.h>

int main()
{
string EncryptedMessage;
std::cout << "Enter Encrypted Message : ";
std::cin >> EncryptedMessage;
std::cout << "Decrypted Message : Banana";
return 0;
}

[devonrevenge]

heres the algorithm that scrambled it: int q = (a+b)*cipher%27;
a = letter pos in sentence, b position in 27 char alphabet, cipher is the cipher

It was too hard to just suss out but i got the best idea for challenge 2 you will love it

@catfish, could you give me an example of that?

[helios]

int q = (a+b)*cipher%27;
a = letter pos in sentence, b position in 27 char alphabet, cipher is the cipher

This may not be reversible. If cipher is a multiple of 27, (a+b)*cipher%27 == 0 for all values of a and b.

[devonrevenge]

no it is my mathmetician freind helped me...well actually he said the same thing but then couldnt understand how i came up with code in the first place...% isnt really what mathmeticians call modulos its the coding version (it has a different name i think modulous just cought on) basically i thing the remainder is returned as an int so it is reversable (cos alphabet [30]%27 = 3 (so 'c') )

my mathmetician freind built the reverse function with the algorithm ^_^

[Catfish3]

@ devonrevenge: by "selective substitution" I meant a switch() that only replaced certain letters and left others as they were.

cipher is the cipher

Which is what? Do you mean "key"? A number given by the user?

[helios]

well actually he said the same thing but then couldnt understand how i came up with code in the first place

Just because a function is invertible in a subset of its domain doesn't mean it's completely invertible. See for example x*x and sqrt(x). You can't prove that the function you wrote is the inverse of this for all possible inputs.

% isnt really what mathmeticians call modulos its the coding version (it has a different name i think modulous just cought on)

The difference is only relevant for negative integers. For naturals, % is equivalent to the remainder function.

The remainder of the division a/b is an integer r such that a = b*q+r, where q is some integer and 0<=r<b. If cipher is a multiple of 27, then it's expressible as 27*k, where k is some integer. Then
(a+b)*cipher%27 == (a+b)*(27*k)%27 == (a*27*k+b*27*k)%27 ==
by properties of remainder:
== ((a*27*k)%27+(b*27*k)%27)%27 ==
and again:
== (((a%27)*(27%27)*(k%27))%27+((b%27)*(27%27)*(k%27))%27)%27 ==
by definition of remainder, for all integer values of c different than 0, c%c == 0. In particular, 27%27 == 0:
== (((a%27)*0*(k%27))%27+((b%27)*0*(k%27))%27)%27 ==
== (0%27+0%27)%27 == 0%27 == 0

Trust me, I've been doing proofs like this all year. The function is not fully invertible.

[devonrevenge]

cipher is just a number in this case, 4.
the reversing function has worked perfectly for all sentences so far

[helios]

cipher is just a number in this case, 4.

Well, try 27.

[Catfish3]

I'd love to see your alphabet, devon. I figured out what the messages are, but some of the letters are wrong.

[hamsterman]

even if cipher = 4, a+b could be divisible by 27. 'k' at position 1 and 'O' at position 2 (sums 108 and 81) both "encrypt" to 0. oops, assumed ascii...
Instead just 'b' at position 26 and 'a' at position 27.

[devonrevenge]

the alphabet is a=0 and so on but with 27 = ' '.

the cipher was between 2 and 9, shouldn't modulus always just keep it all together ie: 3 = 'c', 30 = 'c' 57 = 'c' etc etc rah rah rah?

my idea for the second one is sooo much cooler but i will test it a lot more

[Catfish3]

Could you please just copy paste the alphabet as it is in the code?
Because either I messed up the "decoder", or you messed up the alphabet.

[hamsterman]

oh wait, for 4 it totally is reversible. if it was 6 though, you'd have 'i' = 'r' (9 = 18) at any position.