Chapter 6 Quotes:

Section D.4 Chapter 6 Quotes:

Hill's Cryptography in an Algebraic Alphabet:

Cryptography in an Algebraic Alphabet.

By Lester S. Hill, Hunter College

1. The Bi-Operational Alphabet

Let \(a_0,\ a_1,\ \ldots,\ a_{25}\) denote any permutation of the letters of the English alphabet; and let us associate the letter \(a_i\) with the integer \(i\text{.}\) We define operations of modular addition and multiplication (modulo 26) over the alphabet as follows:

\begin{gather*} a_i+a_j=a_r,\\ a_i\, a_j=a_t, \end{gather*}

where \(r\) is the remainder obtained upon dividing the integer \(i+j\) by the integer 26 and \(t\) is the reaminder obtained on dividing \(ij\) by 26. The integers \(i\) and \(j\) may be the same or different.

It is easy to verify the following salient propositions concerning the bi-operational alphabet thus set up:

(1) If \(\alpha,\ \beta,\ \gamma\) are letters of the alphabet,

\(\alpha+\beta=\beta+\alpha\) and \(\alpha\beta=\beta\alpha\) [commutative law]
\(\alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma\) and \(\alpha(\beta\gamma)=(\alpha\beta)\gamma\) [associative law]
\(\alpha(\beta+\gamma)=\alpha\beta+\alpha\gamma\) [distributive law]

(2) There is exactly one “zero” letter, namely \(a_0\text{,}\) characterized by the fact that the equation \(\alpha+a_0=\alpha\) is satisfied whatever the letter denoted by \(alpha\text{.}\)

(3) Given any letter \(\alpha\text{,}\) we can find exactly one letter \(\beta\text{,}\) dependent on \(\alpha\text{,}\) such that \(\alpha+\beta=a_0\text{.}\) We call \(\beta\) the “negative” of \(\alpha\text{,}\) and we write: \(\beta=-\alpha\text{.}\)

(4) Given any letters \(\alpha,\ \beta\) we can find exactly on letter \(\gamma\) such that \(\alpha+\gamma=\beta\) [i.e. \(\gamma=\beta-\alpha\) is unique].

(5) Distinguishing the twelve letters,

\begin{gather*} a_1,\ a_3,\ a_5,\ a_7,\ a_9,\ a_{11},\ a_{15},\ a_{17},\ a_{19},\ a_{21},\ a_{23},\ a_{25}, \end{gather*}

with subscripts prime to 26, as “primary” letters, we make the assertion, easily proved: If \(\alpha\) is any primary letter and \(\beta\) is any letter, there is exactly one letter \(\gamma\) for which \(\alpha\gamma=\beta\text{.}\)

(6) In any algebraic sum of terms, we may clearly omit terms of which the letter \(a_0\) is a factor; and we need not write the letter \(a_1\) explicitly as a factor in any product.

2. An Illustration

Let the letters of the alphabet be associated with the integers as follows:

Table D.4.1.

a	b	c	d	e	f	g	h	i	j	k	l	m
5	23	2	20	10	1	8	4	18	25	0	16	13
n	o	p	q	r	s	t	u	v	w	x	y	z
7	3	1	19	6	12	24	21	17	14	22	11	9

or, in another convenient formulation:

0	1	2	3	4	5	6	7	8	9	10	11	12
k	p	c	o	h	a	r	n	g	z	e	y	s
13	14	15	16	17	18	19	20	21	22	23	24	25
m	w	f	l	v	i	q	d	u	x	b	t	j

It will be seen that

\begin{gather*} c+x=t,\ j+w=m,\ f+y=k,\ -f=y,\ -y=f,\ etc.\\ an=z,\ hm=k,\ cr=s,\ etc. \end{gather*}

The zero letter is \(k\text{,}\) and the unit letter is \(p\text{.}\) The primary letters are: \(a\) \(b\) \(f\) \(j\) \(n\) \(o\) \(p\) \(q\) \(u\) \(v\) \(y\) \(z\text{.}\)

Since this particular alphabet will be used several times, in illustration of further developments, we append the following table of negatives and reciprocals:

Letter
:
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
t
u
v
w
x
y
z

Neg.
:
u
o
t
r
l
y
i
x
g
p
k
e
m
q
b
j
n
d
w
c
a
z
s
h
f
v

Rec.
:
u
v

n

j

f
z
p
y

a
b

q
o

The solution to the equation \(z+\alpha=t\) is \(\alpha=t-z\) or \(\alpha=t+(-z)=t+v=f\text{.}\)

The system of linear equations: \(o\, \alpha+u\, \beta = x\text{,}\) \(n\, \alpha+i\, \beta = q\) has solution \(\alpha = u\text{,}\) \(\beta=o\text{,}\) which may be obtained by the familiar method of elimination or by formula. [5, pp.306-308]