Galaxy_TOC

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.

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,

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)

the

in the role of a

in the role of a

An example:

0 ≤ (bb-aa+m)

According to prerequisites

2≤m

0≤aa<m

0≤bb<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

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.

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.

and

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

Therefore, the

and the same contemplation that showed that

0≤ TXOR( TXOR(aa,bb,m) ,bb,m) < m

and

0≤ TXOR(aa, TXOR(aa,bb,m) ,m) < m

id est

If

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.

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

= ((bb-

If the formula of the

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

transforms to aa, then

and the condition

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 mod m)

According to the preconditions,

according property h4

Therefore,

Q.E.D.

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.

id est

If

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.

The example values are:

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.

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)

id est

If bb1 != bb2, then TXOR(aa,bb1,m) != TXOR(aa,bb2,m)

TXOR(aa,bb

TXOR(aa,bb

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

for the one-time pad.

I martin.vahi@softf1.com, first published the TXOR idea as a comment at minut.ee sometime before the year 2013.

This, relatively cleanly written, HTML version of the TXOR specification has been written in August 2013. The presented TXOR algorithm is probably nothing new or original, but I figured it out myself by extending the classical one-time pad by utilising the classical modulo arithmetic.

As free advertisement is always beneficial to freelancers like me and it had happened to me before that a simplistic, self-figured-out, algorithm turned out to be totally unique, I link a signed (read: timestamped) version of this document into itself.

2021_12_31 comment: In 2013_08 I received some very nice feedback on LWN.net

Thank You for reading this HTML-page. :-)