Galaxy_TOC

# Text-XOR (TXOR)

TXOR is an encryption algorithm that differs from the one-time pad cipher only by a fact that in stead of using only 2 different values, the "true" and the "false", 1 and 0, {0,1}, the number of values that the TXOR uses can be any whole, finite, number that is greater than or equal to 2.

## Motivation

The main motivation for developing the TXOR has been to use an encryption algorithm that has the security properties of the one-time pad, but that is computationally efficient to use within text processing oriented scripting languages like PHP, JavaScript, Ruby.

The targeted use case of the TXOR is to use the maximum Unicode code value in stead of the classical "1" of the XOR based one-time pad.

As the range, [0,max(Unicode codes)], contains numbers that have not been assigned to any existing character, cipher-texts can not be saved in text form without using some special encoding. For example, the cipher-text might be encoded as a comma-separated list of integers.

## The Algorithm

TXOR(aa,bb,m)=((bb-aa+m) mod m)

Prerequisites:
2≤m
0≤aa<m
0≤bb<m

The aa,bb,m are whole numbers.

The bb must always be in the role of a key and
the aa must always be either
in the role of a clear-text or
in the role of a cipher-text.

An example:

ciphertext = TXOR(cleartext,key,m)
cleartext = TXOR(ciphertext,key,m)

## Various Checks

### Property h1

0 ≤ (bb-aa+m)

According to prerequisites
2≤m
0≤aa<m
0≤bb<m

Therefore the (bb-aa) has its smallest value when bb=0 and aa=(m-1).
0≤(bb-aa+m) because 0≤((0-(m-1))+m)=1

### Property h2

0 ≤ ((bb-aa+m) mod m)
id est
0 ≤ TXOR(aa,bb,m)

According to the property h1 0 ≤ (bb-aa+m). According to the prerequisites (0 < 2 ≤ m).
A remainder of a positive dividend and a positive divisor is a positive number.

### Property h3

((bb-aa+m) mod m) < m
id est
TXOR(aa,bb,m) < m

According to the property h2 and the prerequisites both, m and the ((bb-aa+m) mod m), are positive numbers.
((bb-aa+m) mod m) is a remainder of a division ((bb-aa+m) mod m)/m.
If the dividend and the divisor are both positive numbers and the divisor is greater than zero, then a remainder is always smaller than the divisor.

### Property h4

0≤ ((bb- ((bb-aa+m) mod m) +m) mod m) < m
and
0≤ ((((bb-aa+m) mod m) -aa+m) mod m) < m
id est
0≤ TXOR( TXOR(aa,bb,m) ,bb,m) < m
and
0≤ TXOR(aa, TXOR(aa,bb,m) ,m) < m

According to properties h2 and h3 0≤ TXOR(aa,bb,m) < m
Therefore, the TXOR(aa,bb,m) is within the same bounds as the aa and the bb.
and the same contemplation that showed that 0≤ TXOR(aa,bb,m) < m applies for showing that
0≤ TXOR( TXOR(aa,bb,m) ,bb,m) < m
and
0≤ TXOR(aa, TXOR(aa,bb,m) ,m) < m

### Property #1aa

aa = TXOR( TXOR(aa,bb,m) ,bb,m)
id est
If bb is a key and aa is the cleartext, then the decryption result always matches with the encrypted cleartext.

ciphertext = TXOR(aa,bb,m) = ((bb-aa+m) mod m)

TXOR( TXOR(aa,bb,m),bb,m) = TXOR(ciphertext,bb,m) = ((bb-ciphertext+m) mod m) =

= ((bb-((bb-aa+m) mod m)+m) mod m) = g

If aa = g, then the Property #1aa holds.
If the formula of the g, the
g = ((bb-{ ((bb-aa+m) mod m) }+m) mod m)
transforms to aa, then aa = g can be transformed to aa = aa
and the condition aa = g is met.

bb-((bb-aa+m) mod m)+m-g=c1*m

-((bb-aa+m) mod m)=c1*m-m+g-bb
-((bb-aa+m) mod m)=(c1-1)*m+g-bb
-((bb-aa+m) mod m)=c2*m+g-bb

((bb-aa+m) mod m)=bb-g-c2*m

(bb-aa+m)-(bb-g-c2*m)=c3*m
bb-aa+m-bb+g+c2*m=c3*m
-aa+g+(c2+1)*m=c3*m
-aa+g=c3*m-(c2+1)*m
-aa+g=c3*m-c4*m
-aa+g=c5*m

aa=(g mod m)

According to the preconditions, 0 ≤ aa < m and
according property h4 0 ≤ g < m.
Therefore, g = aa.

Q.E.D.

### Property #1bb

Generally bb = TXOR(aa, TXOR(aa,bb,m) ,m) does not hold.
id est
If aa is a key and bb is the cleartext, then the decryption result might not match with the encrypted cleartext.

bb = TXOR(aa, TXOR(aa,bb,m) ,m) can be shown to be false by finding one example, where it is false.

The example values are:

aa = key = 2

bb = clear-text = 6

m = 10

ciphertext = TXOR(aa,bb,m) = ((bb-aa+m) mod m) = ((6-2+10) mod 10) = 14 mod 10 = 4

TXOR(aa, TXOR(aa,bb,m) ,m) = TXOR(aa, ciphertext, m) = decryption-result =

= ((ciphertext-aa+m) mod m) = ((4-2+10) mod 10) = (12 mod 10) = 2

2 != 4

Q.E.D.

### Property #2com

In respect to aa and bb the TXOR(aa,bb,m) is not commutative.
id est
There exist cases, where TXOR(x,y,m)!=TXOR(y,x,m)

Sample values:

x=2, y=6, m=10

TXOR(x,y,m) = (y-x+m) mod m) = ((6-2+10) mod 10) = 4
TXOR(y,x,m) = (x-y+m) mod m) = ((2-6+10) mod 10) = 6
4 != 6

Q.E.D.

That is to say that

XOR(x,y) = XOR(y,x)
but
TXOR(x,y,m) != TXOR(y,x,m)

### Property #3stat

If cleartext is a constant, then ciphertexts of different keys never collide.
id est
If bb1 != bb2, then TXOR(aa,bb1,m) != TXOR(aa,bb2,m)

TXOR(aa,bb1,m) = ((bb1-aa+m) mod m) = ((bb1+c1) mod m)
TXOR(aa,bb2,m) = ((bb2-aa+m) mod m) = ((bb2+c1) mod m)

Due to the "mod m" part of the equations the
((bb1+c1) mod m) equals with the ((bb2+c1) mod m) only,
if bb1=bb2+c2*m. Due to one of the prerequisites, the
(0 ≤ bb < m ),
the c2 = 0, because otherwise m ≤ bb2, which would violate the prerequisite.

If the TXOR(aa,bb1,m) = TXOR(aa,bb2,m) only, when the bb1 = bb2, then
TXOR(aa,bb1,m) != TXOR(aa,bb2,m), whenever bb1 != bb2.

Q.E.D.

Consequently, for any clear-text value there exists as many possible ciphertext values as there are values in the key-space,
(0 ≤ bb < m), number of possible keys is (m-1),
and the same kind of unbreakability proof applies as is used