UNIVERSITY OF NORTH CAROLINA
Department of Statistics
Chapel Hill, N. C.
Mathematical Sciences Directorate
Air Force Office of Scientific Research
Washington 25, D. C.
-."..
AFOSR Report No. ,~...
A CODING PROBLEM ARISING IN THE
TRANSMISSION OF NUMERICAL DATA
by
R. C. Bose and 1. M. Chakravarti
University of North Carolina
and
Case Institute of Technology
April, 1961
Contract No. AF 49(638)-213
n
A biunique coding from the set of the 2 integers 1, 2,
.•• , ~ onto the set of the ~ n-place binary sequences
is shown. It has some minimal properties.
Qualified requestors may obtain copies of this report from the ASTIA
Document Service Center, Arlington Hall Station, Arlington 12, Virgin~a.
Department of Defense contractors must be established for ASTIA services
or have their need-to-know certified by the cognizant military agency
of their project or contract.
Institute of Statistics
Mimeograph Series No. 284
A CODING PROBLEM ARISING IN THE
TRANSMISSION OF NUMERICAL DATA*
R. C. Bose and I. M. Chakravarti
University of North Carolina
and
Case Institute of Technology
•
Introduction and Summary
L
Let I n denote the set of integers (i), i = 1, 2, •.. ,2n and
B the set of 2n n-place binary sequences (a). Let f be a (1,1) enn
n
coding of I into B such that the binary sequence a(i) corresponds to
n
n
the integer i and the integer a(a) corresponds to the sequence a.
Let
a.(i) denote the binary sequence which differs fro~ a(i) only in the j-th
J
position.
To transmit the integer i over a symmetric binary channel, the
n digits of the corresponding binary sequence a(i) are presented one by
one as inputs to the channel.
a'
If the corresponding output is the sequence
then the transmitted integer is interpreted as a(a').
li-a(a')1 is then the error in the interpreted value.
no error in transmission the distortion is zero.
The quantity
In case there is
If, however, there is
a single error in the j-th place the distortion is li-a.(i)l.
J
If the probability of a single digit error in a received binary
sequence is p, then the expected error in the interpreted value is pE where
provided that
(i)
(ii)
Each integer i is equally likely to occur
Errors are equally likely in any digit position
* This research was supported in part by the United States Air Force through the
Air Force Office of Scientific Research of the Air Research and Development Command,
under contract No. AF49(638)-2l3. Reproduction is permitted for any purpose of the
United States Government.
2
(iii)
Only single digit errors occur in any sequence (or we decide to
neglect cases where more than one digit is in error, since when
p is small the probability of such an occurrence is very small).
It is known from a theorem of H.H.B. Martens Ll_7,of the
Bell Telephone Laboratories, that E is minimum for the natural encoding
scheme for which the integer
n-l n-2
(l.l)
i = 2a + 2a +
l
2
+2an- l+an
corresponds to the binary sequence
(1. 2)
a = (a l , a 2 , .•. , an)' a i = 0 or 1, i = 1,2, •.. , n.
For any encoding f
(1.3)
n
let
d.{f )
~
n
= ; Ii - a{aj{i)JlI
j=l
and
n
(1. 4)
s{f )
n
= E di{f)
i=l
n
Then it follows from Marten's theorem that
There are other encodings besides the natural encoding (1.1, 1.2)
n
for which s{f ) has the minimum value 2n {2 - 1) given by (1. 5). Let t
n
J n
denote the set of all such encodings i.e.
(1.6)
The mean error for a partiCUlar integer i is p d.{f
)/n where
~
n
(1. 7)
d.
(f )
~ n
= ~ Ii
Thus for a given f
integer is
(1. 8)
- a
j=l
= min
i
n
J
'LaJ.( i )} I
the smallest expected error for a fixed
3
=
Then it is of interest to find an f n*~ In
~
such that
u(f
n
*) =
min
f
n
€
j
u( f )
n
n
= u( J )
say.
n
In this paper we give a method of construction of a set of
encodings
Jn
J n c:- i",n'
f* in J
n
n
,
=1
n
=2
n
,
s(f )
n
= 2n (2n - 1).
Hence
*)
or 2 according as n is odd or even.
u(:t n )
(1.11)
in J
such that
u(f
where 8
n
We also show by actual construction that there exists an
(1.10)
For n
such that for every f
Hence
6
or 3, one may verify by actual enumeration that the equality
holds in (1.11).
2.
Method of construction ofJ •
n
Consider the n dimensional unit hypercube Z the coordinates of
n
n
whose vertices are the n-place binary sequences 2
cube Z has n.2
n-l
n
in number.
The hyper-
edges and every vertex is connected to n different
vertices by n edges.
Let f
n
be a way of assigning the integers of I
n
to the 2 different vertices, so that each integer corresponds to a
unique vertex.
Then f
is a
n
(1,1)
encoding of (i) to
(0).
If we de-
fine the weight of an edge to be the absolute difference of the two
integers assigned to its two vertices, then it is easily seen that
1 (f )
-s
2
n
= sum
of the weights of the n.2n-l
(as deteTImined by the encoding
4
di(f )
n
= sum
of the weights of the n edges meeting
at the vertex which has the integer i assigned
to it.
Hence it is easy to recognize that
Jn
is the set of all ways of
assigning the integers (i) in a (1,1) manner to the vertices of the
unit hypercube En' such that the sum of the weights of all the edges
n l n
is 2 - (2 - 1).
Let T be any rigid motion (resultant of an orthogonal transformation
and a translation) which carries En into itself.
of all such rigid motions forms a group Sn.
tex a to T(a).
Given any encoding f
n
The set
Let T transform the ver-
in which the integer i corresponds
to the vertex a, we get a new encoding T(f ) in which i corresponds
n
T(f ) may be called equivalent to f . It is clear that th~
n
n
set of encodings is divided in this way into equivalence classes such
to T(a).
that if f
that f
I
n
n
and f' are equivalent encodings there exists a T€C such
n O n
= T(f).
n
equivalent to f
n
It is also clear that is f €
n
also belongs to J
J,
-
then any encoding
since the sum of the weights of
the edges of En is invariant under the transformations of
if f
n
and f
(2.1)
I
n
q,.
Also
are equivalent then
,
d.~ (fn )
= die f n ),
i
= 1,
n
2, •.• , 2
since the sum of the weights of the edges meeting at a given vertex
remains invariant.
We shall now describe a method of constructing an encoding
*
-J
-J
f + € n+l given an encoding f n € n
In the given encoding f n let
n l
n
a(i) be the sequence corresponding to the integer i (i = 1,2, ... ,2 ).
5
Thus i is the integer assigned to the vertex a(i) of the unit hypercube ,E •
n
In f:+
l
the sequence corresponding to i, (i
= 1,2, •.. ,2n )
is then defined to be
a*(i)
Le.
nate
= (a(i),
0)
a *(i) is the (n+l)-place sequence obtained by adding the coordi-
o to
a(i) in the (n+l)-th place.
Let f
I
n
= T(fn )
be another encoding equivalent to f .
n
sequence corresponding to the integer i in f
I
n
is a I (i)
Then the
a
= T'La(i)7.
-
n
now define the sequence corresponding to the integer i + 2 , (i
We
= 1,
2,
n
... , 2 ) as
*
n
a (i + 2 )
= (a , (i),
1)
n
i.e. a*(i + 2 ) is the sequence obtained from a'(i) by adding the coordinate 1 in the (n+l)-th place •.
Thus if we consider the unit hypercube,E
n+
1 of n+l dimensions,
* 1 establishes a correspondence between the integers 1,2, ... ,2n+l
then f n+
n
and the vertices of.E l' such that the integers 1,2, .•. ,2 correspond
n+
to the vertices of the n-dimensional hypercube,E
0 which is formed by
n,
those ver~ices of.E n+ 1 for which the (n+1)-th coordinate is zero, and
.
n
n
n+1
the ~ntegers 2 + 1, 2 + 2, .•• ,2
correspond to the vertices of the n-
dimensional hypercube .E~,l which is formed by those vertices of ,En+1
for which the (n+1)-th coordinate is unity.
The encoding f * 1 defined above is completely determined by
n+
I
the equivalent encodings f and f , and may be denoted by the notation
n
n
6
f*
+ = (f ,
, f ).
n n
n l
sion.
We shall say that f *
n+l is derived from f n by exten-
- l' 1. e. if
We shall now show that if f n e: II n , then f *
n+ 1e: ""t
I n+
n
n
n
1
n
1
s(f ) = 2 (2 _ 1), then S(f* 1) = 2 + (2 + _ 1). If f e:J 1 then the sum
~
n
~
n
of the weights of the edges of I: n+1 (under f *
n+1 ) = sum of the weights
,
of the edges of I:n, 0+ sum of the weights of the edges of the I: n, 1
+ sum of the weights of the edges of I:n+1 for which one vertex belongs
I 1 and one vertex belongs to I: 0
to I:n,
n,
+ I:n [(2n+ i ) i=l
= 2n-l( 2n -
1 ) + 2n-l( 2n - 1 )
. shows that f *+1 e: T
17 = 2n( 2n -+
1.l )Th~s
~ +1'
Example 1.
n
If n
= 1,
the set
I
n
= 1,2
and the set Bn of the
binary sequences consists of the two sequences (0), (1). Hence there
are only two encodings
f
1 : (0)
~~---->
1,
(1)
f
1 : (0)
~------->
2,
(1) ~---> 1
which are eq'l1..ivalent and both belong to
If n = 2, the set
I
<C~--->
2
fl
J l'
= 1, 2, 3, 4
f
2
(l)B
1
1
(l)B
and the set Bn of two place binary sequences
is (0,0), (1,0), (1,1), (0,1) which we may consider
Fig. 1.
as the vertices A, B, C, D of the unit square shown in
Fig.~
Let Ml , M , N , N be the mid-points
2
2 l
AB, CD, BC and AD respectively. Let l be the
2
D (O,l)
(1,1) C
.0
line perpendicular to the plane of the square
throueh 0.
Then the group
<t
n consists of eight
rigid motions described in the following table:
A (0,0)
(1,0) B
M1
Fig. 2
7
e
Element of
•
Geometrical description
I
Analytical equation
,
- 1
,
x =x
Identity
~-x
2
R
Rotation through a right
angles about
2
R
R3
x2= xl
l
Dl
1
2
,
xl = x
2
x2
Rotation through two right
xl
angles about Ml M2
x = x
2
2
Rotation through two right
xl
angles about N N
l 2
Rotation through two right
x2
I
+ 1
= -xl
+ 1
,
I
= xl
= -x2
+ 1
I
xl = x2
I
Rotation through two right
angles about ED
Now there are
I
= -xl
I
angles about AC
D
2
= -xl +
,
x = -x + 1
l
angles about
P
2
,
xl
2
Rotation through three right
PI
I
xl = -x2+ 1
I
l
Rotation through two right
angles about
2
x2 = Xl
I
Xl = -x2 + 1
I
x2 = -xl + 1
4! or 24 different encodings, divisible into
3 distinct equivalence classes each consisting of 8 members,
From f l we obtain by extension the two non-equivalent encodings
(fl , f l ) and (f , f l ) both belonging to J 2' These with their equil
valents account for 16 of the encodings. They are diagramatically
,
exhibited in Figs. 3(a) and 3(b).
-
8
'.
By taking the weights of the edges we see that
S(fl , f l )
= S(fl ,
~~)
= 12
A third encoding fl~ not e~uivalent to either (fl , f ) or(f , f~)
l
l
is exhibited in Fig. 3(c). The encodings e~uivalent to this account
for the remaining 8 encodings.
The encodings e~uivalent to f *2= (fl , f lI ) are exhibited in the
following table, which gives in the column below each encoding the
value of i corresponding to binary
se~uence
a shown in the initial
column.
R3(f~) P1(f~)
a
(00)
P2(t;)
D1(f~) D2(f~)
2
2
413
3
1
3
4
2
(10)
2
1
(11)
3
2
1
4
4
2
3
1
(01)
4
3
2
1
3
1
2
4
Each of the above encodings is characterized by
d1 (f* ) = 4,
d (f* ) = 2,
d4 (f*
d (f* ) = 2,
2) = 4 •
2
2 2
3 2
We now define a subset $' n of J" n in the following manner.
Let
J" 1
be the set of encodings f
1
and f~ defined in Example 1.
We then
9
obtain the set of all encodings for the case n = 2 derivable from these
by extension and equivalence.
•
This is the set 16 of enCOdings}f2'
which are equivalent to (f , f ) or (f , f~). Starting from~ we
l
l
l
obtain for the case n = 3 all possible encodings derivable from any
encoding of ~2 by extension and equivalence, and denote this set of encodings by 1'3' *'::-oearly
2.
Whether
r
1
%~rn and the equality holds for n = 1 or
= n in general is not kno'WIl, but it may be conjectured
n
that the two sets are identical.
* < 2n+l +3n - 8n where
Existence of f *€
such that u(f)
n
n
n =
6
5 = 1 or 2 according as n is odd or even.
n
1:
Let us define a sequence {ril by the recurrence formulae
r 2m
m
r 2m+l= r 2m+ 2 , r = 2; m = 1,2,3, ...
l
= r 2m_l ,
It follows that
r
2m
= r 2m_l = 32
( 1 + 2 2m-I) ,
m = 1,2,3, ..•
In particular
r l = r 2= 2,
(3.3)
r
3
= r 4= 6,
r = r6= 22.
5
Suppose there exists an n-place binary encoding f n
€
:t:,
such that
(f )
d
rn
n
as n is odd or even.
=
2n+l + 3n - 8
~n
6
Since f
We shall derive from f
n
n
€
~n'
,
8 = 1 or 2 according
n
it also bas the minimal property
by extension an encoding f *
n+l such that
* Proceeding in this manner we can always derive
n
from
n-l'
10
We shall first prove some useful Lemmas.
= (c l ,
c 2 ' ••. ' cn ) is a binary sequence, an
addition of binary sequences is defined as addition of corresponding coLemma I.
If r
ordinates (mod 2), then there exists a rigid motion T which carries
r
the n-dimensional unit hypercube
~
n
into itself and any vertex a into
the vertex a + r.
The required transformation is
= -+
i = 1, 2, •.. , n
xi + c i
where the upper or the lower sign is taken in ± xi according as c i is
o or 1.
x Ii
Lemma II.
be the encoding in which the integer i corresponds
n
to the binary sequence a(i), i = 1, 2, ..• , 2 . Let f' = T (f ) where
, n r n
T is defined as in Lemma I, and let f *+ = (f , f ) be derived from f
n
n
n
n l
r
n
by extension, then for 1 < i < 2
= =
n
(i) di(f:+ l ) = di(fn ) + /2 _ i + a {r + aCi)} I
(ii)
Let f
n
d i +2n (f:+ l )
= d/fn ) + /2 n+ i - a { r + a(i)} /
The encoding f n is equivalent to assigning the integer i to the
corresponding vertex
encoding f'
n
a(i) of the unit hypercube
= Tr (fn )
The equivalent
is equivalent to assigning the integer i to the
vertex a( i) + r of ~.
n
~n.
If t3 is any arbitrary vertex of ~
the integer corresponding to t3 is a(r + t3).
n
then in f
The encoding f *
n+l is
I
n
11
equivalent to assigning the integer i to the vertex a(i), 0 or ~ n+1
n
and the integer 2 + a { a( i) + 'Y } to the vertex a( i), 1 of ~n+1"
d i (f*
n+1 )
~n+1 meeting
= sum of the weights of the edges of
at a(i), 0
= sum of the weights (under f n ) of the edges of
~
n
meeting a a(i) + weight of the edge joining
a(i), 0 and a(i), 1.
The integer corresponding to a(i), 0 in f * 1 is i, and the integer
n+
n
corresponding to a(i), 1 is 2 + a [a(i) + 'I}' Hence
d i (f:+ 1 )
= di(fn ) + f2 n+ a [a(i) +
'I} - ii, 1
~
~
i
2
n
Again
*
di+2n(fn+l)
= sum
of the weights of the edges of ~n+1
meeting at (a(i) + 'I, 1)
= sum
of
of the weights (under fl) of the edges
n
~
n
meeting at (a(i) + 'I) + weight
of the
edge joining (a(i) + 'I, 1) and (a(i) + 'I, 0).
* 1 is 2n+ i
The integer corresponding to (a(i) + 'I, 1) in f n+
and the integer corresponding to (a( i) + r, 0) is a { a( i) + 'I }.
n
d i +2n(f:+ 1 ) = di(f'n) + /2 + i - a [a(i) + 'I} I
1
<
i
<2
Hence
n
=
This completes the proof of Lemma II.
Case I.
Let n be Odd, say n
= 2m
- 1.
exists an encoding f 2m _1 £ ~2m-1 such that
22m + 3(2m-1) - 0
2m-1
( f2m_1 ) = --------=~
d
r 2m=l
6
By
hypothesis there
02m_1
=1
.
12
Let the rigid motion T be defined as in Lemma I, and let
r
f*
1 = (fn , f n ) be derived by extension from f n , where f I n
n+
Putting i = r
= r 2m_l in Lemma II, part (i) we have
2m
I
r
Let us choose
= Tr (fn ).
such that
i.e. in f' the integer 1 is assigned to the vertex r + a(r
2m
from
l ).
_
Then
(;.9)
2m
= 2 + ;(2m-l) 6
22m+l+ 6m - 8
=
i.e.
1
2m
6
2n+2 + ;(n+l) - 8 +
n 1
=
6
Let n be even, say n = 2m. By hypothesis there exists
*
dn+l(fn+l )
Case II.
an encoding f 2m
~2m such that
£
22m+l + 6m - 02m
(f
d
r 2m
2m
)
= ------.;=
6
Let the rigid motion T be defined as in Lemma I, and let
r
Putting i = r 2m in Lemma II, part (ii)
f*n+l =(rn , r')
n where r'n = Tr (fn ).
m
and noting that r 2m+ 2 = r2m+l we have
d
r 2m+l
(r*2m+l) = d
r 2m
(r2m ) + 12
2m
+ r2 - a
m
{r
+ a( r 2 ) }
m
I
13
Let us choose
1. e. in f
I
r
such that
2m
the integer 2
is assigned to the vertex r + a( r
2m
).
Then
from (3. 11) ,
=
22m+l + 6m - 2
6
+
2
3
m1
(1 + 2 - )
2m
2 +2 + 3(2m+l) - 82m+l
=
6
i.e.
To complete the induction we note that for the trivial encoding
f
l
in Example 1
f
since n
l
(0)
:
= 1,
r
l
=2
Example 2.
<i~-->
and 81
1,
(1) <i---> 2
= 1.
(i) Starting from f l , we get f 1
= Tr(fl )
choosing r to satisfy (3.10) i.e.
r
Hence
'e
r = (1)
+
a(2) = a(l)
and
f 1:
(0) - - > 2,
(1)
---> 1
by
14
Hence
f; = (f1 , f '1):
(0,0) <0,.--> 1,
(1,0) <0,.....--> 2,
> 4,
(1,1) <0,.....----> 3.
(0,1) <0
This encoding is geometrically exhibited in Fig. 3(b).
d
n 1
2 + + 3n -
*
r2
Clearly
l::.
v
=- - - - -n
(f2 )
6
0=2
2
(ii)
Changing terminology we can now define f 2 by
f
2
: (0,0) <0,.....----> 1,
(1,1)
~
> 3.
We get f~ = T (f ) by choosing
r 2
r
+
(1,0) <0------> 2, (0,1) ~~---> 4,
r to satisfy (3.12),i.e.
a(2) = a(4)
Hence
r = 0(4) - a(2) = (0,1) - (1,0) = (1,1)
I
f 2:
(1,1) «-..---> 5,
(0,1) ~,.....--> 6,
(1,0) ~--> 8,
(0,0) <-> 7
f * = (f2 , f I2): (0,0,0) = 1, (1,0,0)
3
(1,1,0) = 3, (0,0,1)
(0,1,1) = 6,
f * is geometrically exhibited in Fig. 4.
3
2n+1+ 3n - 8
n
d (f* ) = - - - - - - r3 3
6
Since d (f* )
6 3
= 4,
n
= 3, 03
= 1
= 4,
= 2,
(0,1,0)
= 7,
(1,0,1) = 8,
15
(iii)
Again changing the terminology and taking the encoding
exhibited in Fig. 4 to be f , we get f/2
3
to satisfy (3.10),i.e .
•
= 0:(1)
f*
4
= (0,0,0)
- 0:(6)
(0,0,0,0)
by choosing r
= 0:(1)
r + 0:(6)
Hence r
= T (f2 )
-(0,1,1)
~>1,
= (0,1,1), and
>2
(1,0,0,0) <:
(0,1,0,0) <:--> 4 ,
(1,1,0,0) <-----> 3 ,
(0,0,1,0) <
> 7 ,
(1,0,1,0) <
(0,1,1,0) <
>6
(0,0,0,1)
,
~>14,
(1,1,1,0)
>8 ,
~>
5 ,
(1,0,0,1) <
> 13,
(0,1,0,1) <
> 15,
(1,1,0,1) <
> 16,
(0,0,1,1) <
> 12,
(1,0,1,1) <
> 11,
(0,1,1,1) <
> 9 ,
(1,1,1,1) <
> 10 .
2n+1 + 3n- 0
n
d
(f* ) =
r
4
6
4
since 06 (f* )
4
*
= 7,
*
n
= 4,
*
04
*
=2
*
*
*
*
Reference
1
£:1_7 H.H.B. Martens,lI A remark on error minimizing codes:
Unpublished manuscript, with the Bell Telephone Laboratories, Murray
Hill, New Jersey.
)