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[7] K. Iwata, M. Morii, and T. Uyematsu, “An efficient universal coding algorithm for noiseless channel with symbols of unequal cost,” IEICE Trans. Fundamentals, vol. E80-A, no. 11, pp. 2232–2237, Nov. 1997. [8] R. M. Krause, “Channels which transmit letters of unequal duration,” Inform. Control, vol. 5, pp. 13–24, 1962. [9] I. Csiszár and J. Körner, Information Theory, Coding Theorems for Discrete Memoryless Systems. New York: Academic , 1981. [10] T. S. Han, Theorems on the variable-length intrinsic randomness, to be published.
A New Recursive Universal Code of the Positive Integers Hirosuke Yamamoto, Member, IEEE Abstract—A new recursive universal code of the positive integers is proposed, in which any given sequence can be used as a delimiter of codeword while bit “0” is used as a delimiter in known universal codes, e.g., Levenshtein code, Elias code, Even–Rodeh code, Stout code, Bentley–Yao code, etc. The codeword length of the proposed code is shorter than log in almost all of sufficiently large positive integers although the known codes are longer than log for any positive integer . Index Terms—Elias code, log-star function, universal code of positive integers, universal coding.
I. INTRODUCTION Many researchers have treated the universal coding of the positive integers that satisfy
P (n) P (n + 1);
(1)
where P (n) is a probability distribution on the set of positive integers = f1; 2; 3; 1 1 1g [1]–[7]. These codes can be used practically in various adaptive dictionary codes [8]. Besides the practical uses, it is an interesting coding problem to consider how efficiently we can encode the positive integers under the prefix condition. Let logk2 n be the k -fold composition of the function log2 n and let 3 log2 n be
n = log2 n + log22 n + 1 1 1 + logw2
In this correspondence, we propose a new log3 recursive structure, which satisfies that
(n)
n
(2)
where w3 (n) is the largest integer w which satisfies log2w n 0. Then, it is shown theoretically that any positive integer n can be represented with log23 n 0 w3 (n) bits if < log2 log2 e [2], [3]. On the other hand, many researchers, e.g., Levenshtein [2],1 Elias [4], Bentley–Yao [5], Even–Rodeh [6], Stout [7], etc., have proposed 3 log n-type codes with a recursive structure to attain high performance in large n. But, in their codes, codeword length l(n) cannot become shorter than log23 n although it satisfies l(n) log23 n + w3 (n) + c where c is a constant. Manuscript received June 4, 1998; revised July 23, 1999. The material in this paper was presented in part at the 1998 IEEE International Symposium on Information Theory, MIT, Cambridge, MA, August 16–21, 1998. The author is with the Department of Mathematical Engineering and Information Physics, University of Tokyo, 7-3-1 Hongo, Bunkyo-Ku, Tokyo 113-8656, Japan (e-mail:
[email protected]). Communicated by N. Merhav, Associate Editor for Source Coding. Publisher Item Identifier S 0018-9448(00)01667-9.
n-type code with a
l(n) log23 n 0 log2 (1 0 20f )wf3 (n) + cf even in the worst cases and
l(n) log23 n 0 (1 + log2 (1 0 20f ))wf3 (n) + cf in the best cases. Here, f is a parameter of the code and cf is a constant which depends on f . wf3 (n) is a similar function to w3 (n), which satisfies wf3 (n) w3 (n). Since the best and worst cases occur at infinitely many n’s, and, roughly speaking, l(n) is distributed uniformly between two extreme cases, l(n) can become shorter than log23 n in large parts of integers. In Section II, we review Elias ! code, which is a typical one of the known log3 n-type codes, and we show the reason why the codeword length cannot become shorter than log23 n in the known codes. To overcome this defect, we devise a new representation of binary numbers that never has a given sequence as a prefix. In Section III, we propose a new recursive universal code of the positive integers based on the new binary number representation and we evaluate the performance of the proposed code theoretically. It is shown that the codeword length of the proposed code is shorter than log23 n in almost all of sufficiently large positive integers. The case of r -ary universal codes are treated in Section IV. We use the following notation in this correspondence. • [n]r is the ordinary r -ary number of positive integer n such that the most significant digit of [n]r is nonzero. • [n]ir is the ordinary r -ary number of n with i digits. •
for any n 2 N ;
N
3 log2
717
btc is the largest integer not exceeding t.
Examples: [14]2 = 1110; [14]52 = 01110; [14]3 = 112; [14]53 = 00112; blog2 14c = 3. II. NEW BINARY NUMBER REPRESENTATION EXCLUDING A FORBIDDEN PREFIX Elias ! code CE (n) has the following recursive structure [4]:
CE (n0 ) = [nK ]2 [nK 01 ]2
1 1 1 [n1 ]2 [n0]2 0
(3)
where [n]2 is the ordinary binary number of n, the most significant bit (MSB) of which is always one. Each nk in (3) is determined recursively by nk = blog2 nk01 c. In other words, nk + 1 represents the bit length of [nk01 ]2 . The recursion in (3) stops when the length of [nK ]2 is two. Finally, bit “0” is attached as a delimiter to indicate the end of CE (n0 ).2 In the decoding, nK is obtained from the first two bits of CE (n0 ), and the length of [nk01 ]2 is recursively obtained from nk . Since the MSB of every [nk ]2 is “1,” delimiter “0” can stop the recursion and [n0 ]2 can easily be found. Levenshtein W2 code [2], Even–Rodeh code [6], and Stout code [7] have similar structures and their codes also use bit “0” as a delimiter in the same way as Elias ! code. Levenshtein W20 code [2] and Bentley–Yao search-tree code [5] have a little different structure. However, it is known that their code can be derived from Elias ! like code by gathering the MSB’s of all [nk ]2 and delimiter “0” as a prefix. 1Levenshtein code is the first log n-type code although Elias ! code is famous. 2“ n = 1” is the exception case, for which the codeword is defined as “C (1) = 0.”
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We note that the MSB of each [nk ]2 is always “1.” This means that the MSB has no information, or it is a redundant bit. But this redundant bit cannot be omitted because the MSB is used to distinguish the delimiter “0.” Since each length of [nk ]2 is given by blog2 nk c + 1 that is larger than log2 nk , the codeword length cannot become shorter than log32 n in the known recursive codes of the positive integers. The above note suggests that if we use some sequence with length f > 1, instead of “0,” as a delimiter, then some of the redundant bits may be saved from a codeword. When “0” is used as a delimiter, the prefix, i.e., the MSB “1,” of each [nk ]2 does not coincide with the delimiter “0.” Hence, if we use a sequence [a]f2 , which is the ordinary binary number of integer a with f bits, as a delimiter, then we must devise a new binary number representation of the integers, say Ba; f (n), such that the prefix of Ba; f (n) does not coincide with delimiter [a]f2 . Consider binary sequences whose length is less than j bits. Then, the total number of such binary sequences is given by 21 +22 1 1 1 +2j 01 = f j 2 0 2 while the number of the binary sequences with prefix [a]2 is 0 1 2 j 010f j 0f given by 2 + 2 + 2 + 1 1 1 + 2 =2 0 1 if j 0 1 f or 0 if j f . This means that the number of binary sequences not having prefix [a]f2 is given by 2j 0 2(j 0f ) 0 1 for any j 1 and f 1, where (t)+ is defined as (t)+ = max
ft; 0g:
TABLE I EXAMPLES OF B (n)
B~
n
AND
( )
(4)
Hence, Ba; f (n) can be represented by the following formula:
0 M (j; f )] ; if M (j; f ) n < M (j; f ) + N (j; f; a); [n 0 M (j; f 0 1)] ; if M (j; f ) + N (j; f; a) n < M (j + 1; f ) [n
Ba; f (n) =
j
2
2
2
2
2
j
2
2
2
2
2
(5)
where
M2 (j; f ) = 2j 0 2(j 0f )
N2 (j; f; a) = b2 0f ca =
j
2
j
0;
0f a;
if j f if j < f:
(6) (7)
Especially, if a = 0, i.e., [a]f2 = 00 1 1 1 0, then (5) can be simplified as follows:
B0; f (n) = [n 0 M2 (j; f 0 1)]2j ; if M2 (j; f ) n < M2 (j + 1; f ): (8) We note that letting f (n) be the length of Ba; f (n), then f (n) is given by if M2 (j; f ) n < M2 (j + 1; f ):
f (n) = j;
(9)
Some examples of Ba; f (n) are shown in Table I. Any Ba; f (n) does not have [a]f2 as a prefix. But, [a]f2 may appear as a prefix when Ba; f (n) with length f (n) < f is concatenated by another Ba; f (n). For instance, when f = 3 and [a]f2 = 100, “Ba; f (5) = 10” and “Ba; f (7) = 000” makes “Ba; f (5) Ba; f (7) = 10000.” In order to prevent such cases, we remove the sequences that coincide with a prefix of [a]f2 from fBa; f (n)g. Since one sequence is removed for ~a; f (n) is given each length j if j < f , the obtained binary number B by
0 M (j; f ) + L(j; f )] ; if M (j; f ) 0 L(j; f ) n < M (j; f ) 0L(j; f ) + N~ (j; f; a) [n 0 M (j; f 0 1) + L(j + 1; f )] ; ~ (j; f; a) n if M (j; f ) 0 L(j; f ) + N < M (j + 1; f ) 0 L(j + 1; f ) [n
j
2
B~a; f (n) =
2
2
j
2
2
2
2
2
L(j; f ) = (f 0 1) 0 (f 0 j )+ ;
N~2 (j; f; a) = b2j 0f ac:
(11) (12)
~a; f (n) becomes If [a]f2 = 00 1 1 1 0, then B
B~0; f (n) = [n 0 M2 (j; f 0 1) + L(j + 1; f )]j2 ; if M2 (j; f ) 0 L(j; f ) n < M2 (j + 1; f ) 0 L(j + 1; f ):
(13)
~ f (n) of B ~a; f (n) is given by The length
~f (n) = j; if M2 (j; f ) 0 L(j; f ) n < M2 (j + 1; f ) 0 L(j + 1; f ): (14) ~a; f (n) are also shown in Table I. Some examples of B
III. NEW RECURSIVE UNIVERSAL CODE OF POSITIVE INTEGERS
2
2
where
(10)
~a; f (n) defined by (10), a new recursive universal code of Using B the positive integers can be defined similarly to (3) as follows:
Ca; f (n0 ) ~a; f (nK 01 ) 1 1 1 B ~a; f (n1 )B ~a; f (n0 )[a]f ~a; f (nK )B =B 2
(15)
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where each nk is given by ~ f (nk01 ) nk =
TABLE II EXAMPLES OF C
01
n
( )
(16)
and K is the integer k that satisfies nk = 1. Ca; f (n) can also be represented recursively as follows: ~a; f (n)[a]f Ca; f (n) = C 2
(17)
~a; f (n); B ~ f (n) ~a; f ( C
~a; f (n) = C
0 1)B~
if n = 1 if n > 2.
n);
a; f (
(18)
~a; f (1) is always equal to “0” or “1,” the first We note that since B ~ segment Ba; f (1) can be omitted in the binary case. Some examples of ~a; f (nk ) does Ca; f (n) are shown in Table II. Since a prefix of any B not coincide with [a]f2 , [a]f2 can delimit the codeword. We now derive upper bounds on the codeword length la; f (n) of code Ca; f (n) defined by (15) (or (17) and (18)). Let wf3 (n) be K + 1 or the integer k that satisfies nk = 0 for nk recursively defined by (16). Note that wf3 (n) is a monotonically increasing function of n and wf3 (n) w3 (n). Then, the following theorem holds.
Theorem 1: la; f (n) satisfies for any n that la; f (n)
log3 n + F (f )w3 (n) + c 2
2
f
f
+ 2 (n)
(19)
+ 22 (n)
(20)
and la; f (n) satisfies for infinitely many n that
log3 n 0 (1 0 F (f ))w3 (n) + c
la; f (n)
2
2
f
f
where
0 log
F2 (f ) =
0 2)
cf = 5(f
2 (n)
+
logn
2
e
(1
2
0 20
f
)
(21)
+ f + 5F2 (f )
1+
3
(w (n)
(22)
0 1)(log
e)w
2
01
(n)
log2 n
4n:7 :
(23)
~ (n) = j , the Proof: We note from (14) that in integers n with smallest and largest ones are given by n = M2 (j; f ) 0 L(j; f ) and n = M2 (j + 1; f ) 0 L(j + 1; f ) 0 1, respectively. Hence, from (16), la; f (n) 0 log3 n has a local maximum at the following n:
nK = 1
0 L(n
k
+ 1; f )
n = n0
0 L(n
k
+1
02
n
+ 2; f )
0 1:
0f 0 (f 0 1):
+1
f 0 2, n satisfies n 0 =2 0 1 0 (f 0 1) + (f 0 n 0 1) 2 0 2 0 0 (f 0 1) 0 (1 0 2 ) 0 (f 0 1): =2
Furthermore, for nk
+ (f
1
k
2
n1 + 1
+1
n
+1
(27)
(28)
n
+1
+1 f
(30)
+ F2 (f )
log
2
2
2 2
2
n+
2
n
log2 e
n
(f
0 1) + F (f ):
(31)
n2 + 1
log (n + (f 0 1)) + F (f ) log log n + logn e (f 0 1) + F (f ) + (f 0 2) 2
1
2
2
2
2
2
+ F2 (f ) log
2
= log2 n + log2
1+
e
n
(f
0 1) + F (f ) + (f 0 2) 2
log2 n
+ F2 (f )
log
k
n
2
log (n + (f 0 1)) + F (f ) f 01 = log n + log 1+
k
n
0 1)) + F (f ):
From (30) and (31), n2 + 1 is bounded by
In the following, we derive an upper bound of la; f (n) for these two extreme cases. We first consider the former case, i.e., the worst case. For nk f 0 1, (25) becomes n
log (n 0
(26)
nk01 = M2 (nk + 2; f )
nk01 = 2
1 1 1, we have
From (30), n1 is upper-bounded by
(25)
while it has a local minimum at the following n:
1
nk + 1
(24)
nk01 = M2 (nk + 1; f )
k
Hence, for any k = 1; 2;
f
(29)
2 2
n+
+ F2 (f ):
log2 e
log2 e
log2 n
n
(f
0 1) + F (f ) + (f 0 2) 2
(32)
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holds for b = log2 e. Furthermore, Aw be bounded above as follows:
Repeating such procedure, we obtain
nk +1 log2k n +
log2
e
(log2
e)2
01 n + (logk01 n)(logk02 n) log2 2 2 k01 (log2 e) +111 + k01 k02 (log2 n)(log2 n) 1 1 1 (log2 n) k
Aw
k
+
n +(1+ Ak (n))F2 (f )+ Ak+1 (2n )(f 0 2)+Bk (n) (33)
2
A1 (n) = 0
Ak (n) =
02
j =0
Bk (n) =
(log2
e)j +1
01 n)(logk02 n) 1 1 1 (logk010j n) (log2 2 2 k
e)k : k01 k02 (log2 n)(log2 n) 1 1 1 (log2 n)n (log2
la; f (n) = w
01
w
b
b2
+
4
42
03 (2) (n)03 (n)03
1 1 1 exp b +111 + 4
(n)
w
2
w w
i
0b
(40)
b2
+
2
1+
=
b
1+
b 2 4
b3 2 exp2 (2)
+ 2b
+
b2
2 exp2 (2)
0b
=
111
+
111
b(4 + b) 40b
(41)
where inequality 3 holds because of
01 n 1 02 n 2
log
w
(n)
log
w
(n)
w
(n)
log
(nk+1 + 1) + f
for k (nk + 1) + f
(37)
k=1
02
(n)
0k n expk02 (2) 2
3. Hence, we have
(n)
=
w
4
and
k=0
b
2
2 exp2 (2) exp22 (2)
2 exp2 (2)
b
b
(n)
=
+
b3
4
=
~(nk ) + f
k=0
2
w
i=0
n) 3 b +
(n)
(W )
(n)
2b
(n) (
(36)
The codeword length in the local maximum (or worst) cases, say
01
Aw
(35)
(W ) la; (n), has the following upper-bound from (16) and (33) f w
=
(34)
k
1+
1 b
where Ak (n) and BK (n) are defined as
02 )(logw (n)03 ) 2 bw (n)02 (n)02 w (n)03 )(log ) 1 1 1 log n
2 exp2 (2)
111 +
2b
w
(log2
b2
2
e) (f 0 1)+ F2 (f ) k01 (log2 n)(log2k02 n) 1 1 1 (log2 n)n (log2
k
111 +
3 b +
2
= log2
log2
n) can
(n) (
b2
02 n + (logw 2
(n)
w
+
1 (F (f )+(f 0 2)) +
b
01 (n) =
(n)
01 (n) and Aw
(n)
w
(n)
Ak (n)
k=1
k=1
+ Ak+1 (2 )(f n
0 2) + B
log3 n + w3 (n)F (f ) +
On the other hand,
n)) + f
w
(n)
Ak (n)F2 (f )
2
f
(n)
w
Ak+1 (2n ) =
k=1
(n)
Ak+1 (2
n
+
)(f
0 2) +
w
w
k=1
(n)
Bk (n) + f:
02
k=1
Ak (n)
4b 3(4
0 b)
Ak (2n )
Ak (2n )
(43)
k=1
(38)
because w3 (2n ) = w3 (n) + 1 and A1 (2n ) = 0 by the definition. Therefore, (43) is also bounded by
We note from [3, eq. (A-12)] that inequality (n)
(2 )
=
k=1
w
(n)+1
k=2
k=1 w
(42)
k(
w
2
b(4 + b) 2
n + (1 + Ak (n))F2 (f )
k
2b
0 b) + 4 0 b + 4 0 b b 22 = +b G : 40b 3
(n)
(log2
4b
3(4
w
(n)
(39) k=1
Ak+1 (2n ) G2 :
(44)
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Furthermore, the sum of Bk (n), say 2 (n), can be bounded by3 Bk (n)
(W )
la; f
k=1
=
n (n)
w
1+
k
(log2
k01 (log2 e) 01 n)(logk02 n) 1 1 1 (log
2
(B )
2
la; f (n)
n)
log2 e
1+
n
3
(w (n)
0 1)(log
2
e)w
2
01
(W )
(47)
log3 n + w3 (n)F (f ) + G F (f ) + G (f 0 2) 2
2
f
2
2
2
+
+ f + 2 (n):
(48)
Next we treat the local minimum (or best) case given by (27). In this case, we have for nk f 0 2 nk01 = 2
n
+2
02
n
0f 0 (f 0 1) 0 1
+2
n
2
+2 +2
n
+2
k
+2
nk + 2
f
f
log
2
(50)
(51)
0 2)+2B
n):
k(
(52)
R(n
k)
f
R R
(W )
(B )
(nk + 1) + f
(nk + 2)
=
R
(B )
0 w3 (n) + f
f
2
+ G2 (f
0 2)
+
(57)
0 L(n + 1; f ) n 0 0 L(n + 1; f )g
(58)
0 L(n + 1; f ) 0 L(n + 2; f )g:
(59)
k
1
k
k
1
k
< M2 (nk + 2; f )
(B )
k
k
k
(W )
1
k
(60)
k
k
2
1
k
k
log
2
(nk01 + (f
0 1)) + F (f ):
2 R (n ), it satisfies n 0 1:5M (n + 1; f ) 0 L(n
0 2) + 2B
k
(B )
(61)
1
(53)
tight bound (n) 4=n can be obtained by directly calculating (45), where the second term in the bracket has the maximum at n = 4.
2
(62)
+ 1; f )
(63)
k
2
k
k
= M2 (nk + 1 + log2 1:5; f )
f
+ f + 22 (n):
1
k
jR (n )j = jR (n )j = 1 : jR(n )j jR(n )j 2 When n 0 2 R (n ), it satisfies n 0 M (n + 1; f ) 0 L(n + 1; f ):
When nk01
3 k (n)) 0 wf (n) + f 3 n 0 (1 0 F2 (f ))w (n) + G2 F2 (f ) n
k
Since the cardinality of R(nk ), jR(nk )j, is equal to M2 (nk + 1; f ) for nk f , the following relation holds:
nk + 1
k
log3
(56)
Hence, in this case, we have from (30) that
(log2 n + (1 + Ak (n))F2 (f ) + Ak+1 (2 )(f
n 0
f
(n)
k=1
k
(nk ) = nk01 : 1:5M2 (nk + 1; f )
k=1
+ 1; f )
0 L(n + 1; f ) n 0 0 L(n + 2; f )g:
f
k
k=1
(n)
k
(nk ) = nk01 : M2 (nk + 1; f )
k
(n)
la; f (n) =
0 L(n
= nk01 : M2 (nk + 1; f )
(W )
(B )
Furthermore, from (37), (42), (44), and (52), la; f (n) can be bounded by
w
+ 1; f )
We note from (21), (54), and (55) that F2 (f ) can be approximated as (log2 e)=2f , and by setting f large, the coefficient of wf3 (n) in the worst case becomes very small while the one in the best case becomes almost 01. Hence, we can conjecture that la; f (n) is shorter than log3 n in large parts of the positive integers. In the remainder of this section, we show that this conjecture is true by considering a general case instead of the best and worst cases. For a given nk , nk01 must be included in a region R(nk ) defined as
k
n
w
k
< 1:5M2 (nk + 1; f )
(nk01 + f ) + F2 (f ):
n +(1+ Ak (n))F2 (f )+ Ak+1 (2 )(f
w
2
F2 (f )
Hence, in the same way as (33), we have k
(55)
We divide this region R(nk ) into two regions, the worse region (W ) (B ) (nk ) and the better region R (nk ), which are defined as
0 1 0 (f 0 1) + (f 0 n 0 2) 0 1 0 2 0 0 (f 0 1) 0 1 0 (1 0 2 ) 0 f: n
+
f
instead of (25). But (56) also induces the same inequality (30) as (25). Hence (54) holds for any n. On the other hand, (55) holds for infinite many n’s that satisfy (27). Q.E.D.
This means that, for any k , nk + 2 can be bounded by
2
M (n
< M2 (nk + 2; f )
n
=2
nk01
(49)
f 03
nk01 = 2
2
Finally, note that any n satisfies
(46)
log2 n
where the last inequality holds because the second term in the bracket of (46) has the maximum at n = 16. (W ) From (38), (42), and (44), la; f (n) can be bounded by la; f (n)
log3 n 0 (1 0 F (f ))w3 (n) + 5(f 0 2)
(n)
4n:7
3A
2
+ f + 5F2 (f ) + 22 (n):
(45)
log
+
f
(54)
k=2
nk +2
2
2
2
log2 e
1
and for nk
1 1 1 < 5 because b = log e = 111 (48) and (53) that 3 3 (n) log n + F (f )w (n)+5(f 0 2) + f +5F (f )+ (n)
G2 is bounded by G2 = 4:95 1:4427 . Hence, we have from
(n)
w
2 (n) =
721
0 L(n
k
+ 1; f ):
(64)
Therefore, we can obtain similarly that nk + 1 + log2 1:5
log
2
(nk01 + (f
0 1)) + F (f ): 2
(65)
722
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 46, NO. 2, MARCH 2000
For a given n, letting K(B ) (n) be the set of k such that nk 2 R(B) (nk+1 ) and letting (n) be the ratio defined as
(n) =
jK
(n)j wf3 (n) (B )
also holds for any k . Hence, a sequence xK 01 xK 02 1 1 1 x1 x0 with m = occurs with probability Pr (x) bounded above by
(66) Pr(x)
0, i.e.,
(n) >
0 log (1 0 20 ) log e 20 2:5 1 20 log 1:5 log 1:5 f
2
2
2
f
f
Xk =
0;
if nk if nk
1;
2R 2R
(W ) (B )
(69)
(nk+1 )
for such probability distribution. Then, from the definition of (W ) (B ) (nk+1 ) and R (nk+1 ), we have
fX
Pr
K
g
=0 =1
(70)
where K = w 0 1. Furthermore, since (W )
(B )
j
(nk+1 )
holds for any nk+1 f and any integer included in R(B ) (nk+1 ) is larger than integers in R(W ) (nk+1 ), we can easily show that for w 3 and 1 k K 0 1
fX
Pr
jX
k =0
K
= 0;
XK 01 = xk01 ; 1 1 1 ; Xk+1 = xk+1 g
