FV code trees with no self-synchronizing string - ITA @ UCSD

FV code trees with no self-synchronizing string Tsukasa Honda

Yusuke Tashiro

Hirosuke Yamamoto

Equity Department Nomura Securities Co., Ltd. Oota-ku, Tokyo 146-0085, Japan [email protected]

Dept. of Mathematical Informatics University of Tokyo Bunkyo-ku, Tokyo 113-8656, Japan yusuke [email protected]

Dept. of Complexity Sci. and Eng. University of Tokyo Kashiwa-shi, Chiba 277-8561, Japan [email protected]

Abstract— It is shown in this paper that all internal nodes with the same subtree can be treated as a single state in a code tree of a fixed-to-variable length code (FV code) when we use CapocelliGrango-Vaccaro algorithm and Even algorithm to check whether the FV code has a self-synchronizing string. Based on this reduced state representation, we characterize three types of code trees with no self-synchronizing string and evaluate the number of code trees included in each type. The results of exhaustive search is also given for all code trees with 21 or less leaves.

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I. I NTRODUCTION In a fixed-to-variable length code (FV code)1 , a bit error asynchronizes the parsing of codewords and causes the decoding error propagation. But, if the FV code has selfsynchronizing strings, the error propagation terminates when they appear. Hence, the self-synchronizing strings are important in the FV codes, and they are studied by many researchers [1]-[14]. Capocelli-Grango-Vaccaro (CGV) [1] proved that a FV code is statistically synchronizable if and only if it has a self-synchronizing string, and they gave an algorithm to check whether a FV code has a self-synchronizing string. Furthermore, the shortest self-synchncronizing string can be obtained by Even algorithm [2]. Bestel-Perrin [3] gave a sufficient condition for a FV code to have a synchronizing string, and several algorithms are proposed to construct an efficient FV code with a self-synchronizing string for a given source probability [4]-[8]. Furthermore, Freiling-Jungeis-Th´ebergeFeder [9] showed that almost all FV codes have a selfsynchronizing string when the code trees are sufficiently large. But, not all FV codes have a self-synchronizing string, and it has not been known except some obvious cases what kinds of FV codes have no self-synchronizing string. In CGV algorithm [1] and Even algorithm [2], all internal nodes in a code tree are treated as different states. But, we show in this paper 2 that their algorithms can be applied even if internal nodes with the same subtree are treated as a single state. This reduced state representation decreases the space and time complexities in the algorithms. Furthermore, by using the reduced sate representation, we show that three types of code trees have no self-synchronizing string, and we evaluate the number of code trees included in each type. 1 It

is assumed in this paper that FV codes are binary prefix codes. results in this paper were presented in part at the AEW4 [15] and AEW5 [16]. 2 The

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States in a code tree

II. P RELIMINARIES Since a FV code can be represented by its code tree, we mainly consider code trees rather than codes themselves. We assume that code trees are binary and complete, i.e. every internal node has two children. Definition 1 Each internal node in a code tree is called a state of the code tree. Every leaf is identified with the root state. We say a string w coalesces states s1 and s2 if w brings s1 and s2 to the same sate. If a string w brings all states including the root state to the root state, then w is called a self-synchronizing string. In the case of Fig. 1, w = 100011 is a self-synchronizing string because w brings all states “A”–“G” to the root state “A”. Definition 2 Let S0 (resp. S1 ) be the set of all states on the all zero (resp. one) path in a code tree, and its cardinality is represented by |S0 | (resp. |S1 |). For instance, we have in Fig. 1 that S0 = {A, B, D}, S1 = {A, C}, |S0 | = 3, and |S1 | = 2. Then, it is known that the following theorems hold. Theorem 1 ([3]) If |S0 | and |S1 | of a code tree are mutually prime, then the code has a self-synchronizing string.

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Reduced states in a code tree

Theorem 2 ([9]) If all states in S0 (or S1 ) coalesce by a string w, then w is a synchronizing stirng. By the following CGV algorithm, we can check whether a code tree has a self-synchronizing string. Algorithm 1 (CGV algorithm [1] ) (A1) For every state s, draw a box with s. (A2) For every pair of states s and s0 , connect s to s0 by an arrow if a codeword brings s to s0 . (A3) If every state has a path to the root state, then there exists a self-synchronizing string. We note from Theorem 2 that it suffice to check whether every state only in S0 (or S1 ) has a path to the root state in Step (A3) of Algorithm 1. We can find all self-synchronizing strings by the following Even algorithm. Algorithm 2 (Even algorithm [2] ) (B1) Let A be the set of all states in a code tree, and draw a box with A. Initialize as J ← A, U ← φ, and V ← {J }. (B2) Let K0 be the set of all states that can be reached by “0” from a state in J . If K0 6∈ (U ∪ V ), then draw a box with K0 and let U ← U ∪ {K0 }. Connect J to K0 by arrow “0”. (B3) Let K1 be the set of all states that can be reached by “1” from a state in J . If K1 6∈ (U ∪ V ), then draw a box with K1 and let U ← U ∪ {K1 }. Connect J to K1 by arrow “1”. (B4) If U = φ, then go to (B5). Otherwise, let J be a set selected arbitrarily from U , and Let U and V be U ← U \ {J } and V ← V ∪ {J }, respectively. Go back to (B2). (B5) If the box with only the root state, say the root box, has been generated, then there exists a synchronizing string. Furthermore, every path from A to the root box gives a self-synchronizing string, and hence, the shortest

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path from A to the root box gives the shortest selfsynchronizing string. In the above algorithms, every internal node is treated as a different state. But, if two states have the same subtrees as their descendants, any corresponding two nodes in the subtrees have the same movement for any string. This means that we can treat these two states as the same state. Hence, all states with the same subtree can be treated as a single state. For instance, the sates in Fig. 1 can be reduced to Fig. 2. The results obtained by applying Algorithms 1 and 2 to the reduced states are shown in Fig. 3 and 4, respectively. Since every state in Fig. 3 has a path to the root state “A”, the code tree shown in Fig. 2 has a self-synchronizing string. Furthermore, we note from Fig. 4 that “100011” is the shortest self-synchronizing string. III. C ODE TREES WITH NO SELF - SYNCHRONIZING STRING In this section, we characterize code trees with no synchronizing string based on the reduced state representation. Theorem 3 Let T1 be an arbitrary code tree, and assume that for n ≥ 2, the n-th code tree Tn is constructed recursively by connecting T1 to every leaf of the (n − 1)-th code three Tn−1 . Then, Tn has no synchronizing string.

of T˜1 . But, it contradicts the assumption that T˜1 has no selfsynchronizing string. The depth of a leaf in a code tree is the number of branches from the root to the leaf. If the initial tree T˜1 in Theorem 4 is a code tree with constant depth d ≥ 2, the great common divisor of all leaf depths in T˜n is d. Conversely, if the great common divisor of all leaf depths is d ≥ 2 in a code tree, then the tree can be constructed as T˜n from the code tree T˜1 with constant depth d. Hence, we obtain the following corollary, which was originally shown in [5].

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Corollary 1 ([5]) If the great common divisor of all leaf depths in a code tree is larger than one, then the code tree has no synchronizing string.

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wing corollary,

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Fig. 6 is an example of the code tree that has (1, 1)transition for S0 and S1 . Furthermore, we note by the next theorem that the number of such code trees is unlimited.

Proof: In Tn , let s0 be the root state, and let sm be the

6 s is larger than stateTheorem corresponding to the leaves of Tm for 1 ≤ m ≤ n − 1, T in be the a code tree with (1,representation, 1)-transition forall S0 internal and S1 . Then tring. (NoteLet that reduced state nodes the following code trees Ta and Tb also have (1, 1)-transition corresponding to the leaves of Tm are the same state in Tn . for S and S1 in their code trees. See Fig. 05 for the case of n = 2, where s1 is “C”. ) and let 1l be • T is the code tree such that both children of the root are When aastring w brings sm to the root state s0 , it also brings T . (See the left tree shown in Fig. 7.) very state in S0 s(m+1)• mod to code s1 . Hence, s0 , s1 ,by · · ·lengthening , sn−1 cannot coalesce Tb isnthe tree obtained all leaves of state in S1 to by any string. T by one bit. (See the right tree shown in Fig. 7.) tree has (k, l)a

Let S0 and S0 be the sets of all states on all zero TheoremProof: 4 path in T and Ta , respectively. Similarly let S1 and string, S1a be the ˜ for S0 and S1 , Let T1 be a code tree with no synchronizing and sets of all states on all one path in T and ˜Ta , respectively. theorem holds assume that for n ≥ 2, the n-th code tree T is constructed Then it holds that S0a = S˜0 ∪ {sa } and S1a = Sn1 ∪ {sa } where recursively connecting sa is theby root state of TaT. 1 to a leaf of the (n − 1)-th code ˜ has no synchronizing strings. tree T˜n−1 . Then, Bit “0” (resp.Tn“1”) brings the root state sa to the first state

in S0 (resp. S1 ), the last state in S0 (resp. S1 ) to sa , and other Proof: Tˆm , m = 1, 2, · · · , n be the m-th extended states in SLet 0 (resp. S1 ) to their left child states. Similarly, bit ˆm the part “1” in T˜(resp. . Then, are first the state sameintree the Sinitial n “0”) since bringsall sa Tto S1 as (resp. 0) , ˜ ˆ ˜1 Sand tree T , each state in T has the corresponding state in T tate in S0 to a the1 last state in S0 m (resp. S1 ) to sa , and other states in 0 sp. “1”) brings they(resp. have S the movement any in string in TˆmSand T˜1 . If a to their right childfor states S1 (resp. 1 ) same 0 ). Hence, p. S1 ). Hence, Tabrings has (1,a1)-transition fortoS0aa state and Ss1a0. ∈ Tˆm0 in T˜n , it brings string state s ∈ Tˆm by any string. The theorem state can be Tb in the samestate wayofassT0 avia . the corresponding ofproved s to theforcorresponding

nd S1 , then the

the root state in T˜1 . This means that if a string brings a state to the root state in T˜n , it also brings the corresponding state to the root state in T˜1 . Hence, if T˜n has a self-synchronizing string, then the string must also be a self-synchronizing string

Definition 3 Let 0k be the string 00 · · · 0 with length k and let 1l be the string 11 · · · 1 with length l. If 1k brings every state in S0 to a different state in S1 and 0l brings every state in S1 to a different state in S0 , then we say the code tree has (k, l)transition for S0 and S1 . Note that if a code tree has (k, l)-transition for S0 and S1 , it holds that |S0 | = |S1 |. The following theorem holds for such code trees. Theorem 5 If a code tree has (1, 1)-transition for S0 and S1 , then the code tree has no synchronizing string. Proof: It is obvious that bit “0” (resp. “1”) brings every state in S0 to a different state in S0 (resp. S1 ). On the other hand, since the code tree has (1, 1)-transition for S0 and S1 , bit “1” (resp. “0”) brings every state in S0 (resp. S1 ) to a different state in S1 (resp. S0 ). Hence, any two states in S0 (and S1 ) cannot coalesce by any string. Fig. 6 shows an example of the code tree that has (1, 1)transition for S0 and S1 . Furthermore, we note by the next theorem that the number of such code trees is unlimited. Theorem 6 Let T be a code tree with (1, 1)-transition for S0 and S1 . Then the following code trees Ta and Tb also have (1, 1)-transition for S0 and S1 in their code trees. • Ta is the code tree such that both children of the root are T . (See the left tree shown in Fig. 7.) • Tb is the code tree obtained by lengthening all leaves of T by one bit. (See the right tree shown in Fig. 7.) Proof: Let S0 and S0a be the sets of all states on all zero path in T and Ta , respectively. Similarly let S1 and S1a be the sets of all states on all one path in T and Ta , respectively. Then it holds that S0a = S0 ∪ {sa } and S1a = S1 ∪ {sa } where sa is the root state of Ta . Bit “0” (resp. “1”) brings the root state sa to the first state in S0 (resp. S1 ), the last state in S0 (resp. S1 ) to sa , and other states in S0 (resp. S1 ) to their left (resp. right) child states.

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0n−1−i 0 s0

n3s2 xsu s1s s12 s2 ssn−2 s23n−1 1 sssn−1 1 s1 sn−1 s111 snn−2 ss121 ss13n−1 1 sn nn−2 u 1s11 scn20 n−1 s11 0 s12 0 sc1n−2 0 c su1n−1 0csc1 s1n scs10c01 n−2 1 ss11n−2 s11n−1 1 s1 cn−1 c21n 1 s 1s1 1 0 1s s1ss1n−1 s2s2n 3 sssc21n−2 n1 2 3 sscn−2 s3s3x s2snn−1 sn−1 s1n−1 s13s1n s00 s1 s s s s 3 sss 3c 2 3s2n cs2n−2 c 3sf1 c n−2 c c s3n−2 c31n n n−2 21 s3 s n−1 n Vs(y|x) s f s01n2s n−2 s sc3s1ss232n−1 s s s nsn−2ss32n ssn−1 s11s31 s2n−2 s221s3n−2 susf n A code tree with (1, 1)-transition for S0 and S1 2 2 n−23 s s0 n−1 s s s s s1nn−1 n−2 n−1 1 1 nfn n−1 n n−2 2 3 n−1 n 1 1 1 2 3 n−2 c V (y|x) c2n−2 ycn 3c 10 css1n−1 cc 3 scs0c cs1 ncc2 s2 s1 c ss 1 n−1 n u1 c u cs11n ss12 cc sss21s1n−2 c1 1 1 s00 s01 s02 sc2 fns0n−2 sss0n−1 sc31 u s s s n−1 n s s s s s s n−2 nn 13snn−1 sn−2 12 s 11snu xs 1 s s s s s su n−2 n−1 n 2 3 n−2 n−1 n 1 2 n−2 n−1 n n 1 1 1 1 x 2 3 n−2 n−1 n s2 csu sn−2 s 1 1 1 u sf1 nc n 1 11 ssn−2 fn sc ssysn−1 fn sc s1n1n−2 1 111 s Vss 212 n n1 s s1 ssn−1 sc2 fnc sss1n−2 ss1n1n−2 ss1 sc2 s1 s sc12 s1n− (y|x) fn−2 sc3s sn−2 sVn−1 nc 1 ss2c snc ssc3sc1n c nn−1 sn−2 sc cs1c1 n−1 s1n c cn−1 cc 2 (y|x) sc n sc3snc2 1 n sncn−2 sc3 1 1 sscn−2 sscn−1 n−1 1 n−1 1i−1 1scn s s s u 1 1s 1n 1 x 1 s1s 2 3 n−2 n−1 n n i−1 i−1 n−1−i i−1 n−1−i x x 1 1 1 1 fn 2 3 n−2 n−1 i−1 i−1 n−1−i y fn y ns11 u 2 s12 1x s s s s s fn n u s s s s s s s s u 20fn1n 1n−2 su 1 su n−2 11 n−1 111 sn−1 ssu sn 1 20 n−2 n−1 c sufn 0 c 1 n−1 c u1 0s 210 1 nsc1n−2 xsu s c sV s13 sc 1s1cn−2 s01ccni−1 s s Ta Tbu n−1 su ss1cn−2 sci−1 12 (y|x) Vi−1 (y|x) ci−1 c i−1 1 c n−1−i n−1−i Vc (y|x) sa i−1ss V (y|x) u 2 si−1 3 i−1 n−1 n i−1 1n−1−i i−1 n−1−i n−1−i u s s s s s s s s s s 0 1 1 0 1 f 0 1 1 0 c c c c c u 1sscfcn 1 n1sccs0 2 s0c1c n−2 1 f1cc c1n−1f0n2scc0c n1 c3n11 nc cn−2 1 1 cc 0n−1c c 1cn sx2 nsu s3 s1 xf0nsu n xnsu n s V (y|x) n s1c 1 s ns0 1xnc ssn 0c 0s1y21cfy s snn−2 sn−1 scncn−2 s 2 n y s33 n0 1 sn−2 ysn−1 sn··n−1 s i−1 n−2 scn− 1 sn−1 n−1 ns n 2 ·1·n−1−i 3 n−1−i u i−1 i−1n−2 su su i−1 si−1 ·03s01n1s0un−1−i ss1n3cn10i−100si−1 s00s111s0s1c2i−1 ·i−1 scn1si−1 i−1 i−1 n−1 s10s0i−1 ·x·n−2 · s11ncs201s00i−1 s0sns01s1s1n−1−i 11··i−1 0(y|x) 0 21fn−1−i 1 s0ntree 1u·1·si−1 2u uFig. 1·l·n−1−i 0 1 1c 1 2 · ·0·1 V1 (y|x)x11n V (y|x) xn0y 1 0 0 1 c c c c c V c 9. uCode cn c cfwith c c c n (k, l)-transition for k = = n − 1(n ≥ 3). xnsu n1c V (y|x) sn−1 s1 c·c · s s sn−2xsu ssun1s·0ss·si−1 1si−1 x0n−1−i su s2 1i−1 0·1 s·11·V 0sn−1 1s1ss0ccs·1·s cssc3 i−1 11s·cn scni−1n−1−i s·sn1n·c2i−1 s12cn00i−1 ns s·c2nf(y|x) s·sn−1−i n−1−i s1n00n−1−i s1s100000si−1 s11s2u11sn−1 ·201· ·i−1 s·cs1ns03s1cni−1 s·s1n−1−i s00cnci−1 s0111sn−1−i · ·10scn sn−2 i−1 0 0n−2 11si−1 1n−1−i 1V0su(y|x) 1i−1 0i−1 0 0ysi−1 1s1130· · f· 1nss1n−1−i 0s 1si−1 nn−2 f1ynn 1Vi−1 n0u 1ns 0i−1 1f1· ·032·s1s n102i−1 2 ·n−1 n 1 0 1 0 1 1 0 1 0 1 1 0 V (y|x) y 0T n c c c 0 1 1 0 (y|x) y y f f n n f V (y|x) xsu n01 1 0nc s1n−1 sn−1−i ss1cn−1−i 1 112u1 cc ccs0n011c1 nc 1 01 12 c 0c 1 cc n c c 1 1 n−1−i 0 1 cx1n 1nnss2 V1 (y|x) i−1 s11·xsi−1 ··s·i−1 ·ssi−1 02s·s1·i+2 0s·n−1−i c n−1− i−1 ns0 si−1 i−1 ··su s1i+1 ·nss·i−1 · s·s0ns1·0sn··i−1 ss11si+1 sn−1−i s1s··n·2s···1···si−1 x 001si−1 1u·i2ss0·2i+1 ·su· · sscn 001u0su0i−1 · · · s1nsni−1 s·0sun·i+2 s3001s·110s·s·i−1 ·01·s·11·n−2 s·11ssss·00ic2100ss·x ·i−1 1··x n1s ss·n−1−i i−1 si+2 11 s0i−1 s00ss11000si−1 s0c2111i−1 ·s·c2··s·cn·11si−1 0y n 1 0 V1(y|x) 2n−1−i 1 su y1n s00fs11s1i−1 ni−1 n·n0 n−1−i n00i c0 n0 1 y 1010 1 ·u· ·00sn1 01ci−1 11·n−1−i 1 0cs 0·1s 1 u n s·11 nn−1−i n 0snc10i−1 1· 10· V 1c2n1120c10 1u c 1su 0200 1s0cs 2ssf 1s1 c0sc0f1s1·ns1·01n0· 0·01sc10·c01n1i−1 1·1n−1−i n0111 (y|x) 1 10 11 1xn 2 0 1y 1sy 2s1 f 0 1 cn 0 1Vs100 (y|x) · s s · · · s · · s s s · · · s s c 0 c 1 1 0 1 c c sb 0 1 V (y|x) s · · · s s s s s s s · · · 1sss·cn1c· n−1−i 0 1 0 1 1 c c 0 1 c c V (y|x) s 0 1 sc 0 1 c 0 1 i i+1 i+2 n s s · · · s s s s · · · s s s · · · s s s s · · · s V (y|x) s 0n 1 0 1n0 c 1 0 1ci−1 0··si·1s i+1 cnsn−1−i 1s n by0 n1 x is· ·su i+1 i+2 0s0 s 11i−1 ns· 2ss i··1·s·0i+1 i+2 ni−1 0i+2 1cs i n−1−i i+1 i+2 s · s s · s s · s s · · s s n s s · · · s · · · 01 1 2 si−1 s · · · s s · s · · s · · · s n 0 1 2 n s s s s s s s · · · scn i−1 n−1−i i−1 i−1 n−1−i i−1 su i−1 i−1 i−1 s s · · · s s s · · · s s · · s s s · · · s 0 1 n 0 1 0 n 1 n 0 1 2 n s 0 1 n 0 1 2 n i−1 n−1−i 0 1 n 0 1 0 1 n 2 n 0 1 2 n u i−1 c1 i−1 u i−1 i−1 n 01n−1−i 1 2 x mean that the code Figs. 8–10 have (k, l)-transition b 0 001 1 ns1 n 1 2 0 1 0c 0u 11 2 y n0 n 00n1n 0011cctrees n0 in 00 310y1 2c1usu n01 01 f 1 0 0 1 1 1 0 1 1 0 n T T 1 1 0 1 0 1 0 c c 0 3 c n 1 n 0 0 3 c c 2 c c 1 1 0 0 1 1 0 1 1 0 1 c c 1 2 c 0 1 1 2 c V·1ss(y|x) 0··(y|x) 1ss·00ss 1c·ss·y 2· s1·i·cs 1·s·ns·n 1si+1 2 · 0·s0·i+2 c1s10 · 1· · 1s1nnc 21 2 c00 1c1 c c 1 cc 0 cs 1 1 s· ····s· 0s 1 cxsy c ssss·s1s· ·s·V s s · s ns n 1ss 0·ss·1c0·ss·0cix ci+2 y 0 1 10 1 sb 0 11sc s ·x·0su c1b · · · s s · s 0 1 c s s · s s · · s · · · · · · s 0 i+1 i+2 i s · · · s · · c · · · s s s · · · s su 0 1 1 1 c c 0 i i+1 i+2 n s s · · · s s s s · · x 0 1 i+1 n x ·ic2n·0s·i+1 ··0s1i−1 s ···ss·i−1 ·i·su snsi+1 · · ·····s··ssn··cnssnn n−1− s0s1 · · · ssn0s1 · · s· 0ssn1s02 ·1· s· 0ssn10s1i 21 i+1 0 1 n 1ci−1 i+2 s101nn·1c·.i·2s0i+1 · si+2 1s 20 s·1n−1−i · · ·csi+2 1s s1ci+2 s00s11 ·11ss·csnx sssi+2 iss0s i+1 ··3ss00n0c1nsVnss113c(y|x) 0su 1su n s1··11sc0··s2·sc23·1sn·i·i−1 ·0sss11ns011cnss0s·c2si+2 ·01ns0··ssc1s·0211n00s··su sc·s·000i+1 nns 0ss sns001·1scyc·0s11ni+2 sc10s2s0s·i1s00·cs3s·i+1 1s··2··i−1 1·1··0·0 010ss for S0sin0000ns111and 0·c0·ss0 1 ··n··0··s 03susyS 102·csn−1−i Vsc(y|x) 111 · 1 0n 1 s0 1 · 20·0 n 01n−1−i n scsi0n1sssni+1 1si−1 s·s(y|x) 1Vs(y|x) 2·s·00·s c0n 01 ·0112s cV cs1ci+2 0s 1n 2 n· ·1 c· 0sncn1nV (y|x) c1 s · s s 0·c s c·i ··i+1 2 2 0 c c1 ns12s (y|x) i+2 ns s00s11 · · ·ss01iss12i+1 · s s · 10 2· · s0cn 1 c0· · ·s·i0sV 0 1 2 c i+2 (y|x) V i i+1 i+2 n i i+1 i+2 0 i i+1 n · · s s · · · s s s · s · s s s · · · s s · · · i+2 s s · · · s s s · · · s · · s s s · · · 0 1 i+1 i+2 n x x x s s · s s s · · · s 1 1 0 1 c c 0 1 1 0 0 1 1 c 1 c 0 1 c c y 1s0is ·120·s·0c1·si+2 s·s1isns0nc1i+1 · i·xsci+1 i+2 1·ns s 1 ssi+2 nc i+2 su 10 s10·s·y i+2 n ·0·n s1 0s··0··su ··11sscsnnc2c · ·1 ·02ss0ycns10n00s1cnsc111s0·20·1·i··3s·i+1 s0101s1i+1 si+2 ·csu · ·0i+2 s0s01s11n0s·c23·nc··i·s·1ni+1 ·10ci·s·21ci+1 s0n3ny0scnsn100c1ssc2011ss·0c3·s·n·1·is·2c1nsi+1 0cs sc i+2s00sc 11sc2n· · · scn n 0 0 c 1 · 0·i0·3si+1 0nscs 10 ·c0·30s 1cn2ncss c·ss 1n c0 s ss1scns02s0·s·scn·001·011s··1ss2s··cs122·cn2·su 0cnn1(y|x) · s0i ·s1s·i+1 ·nc· ssi ·s0·si+1 ·11s(y|x) s· ns· i+2 si·12·s··i+1 ·0s2·0(y|x) 2i s3i+1 1 sc 1· s 0· ·n1··sc·c·cs 01V s·s·i00·scs·sc·0i+1 ··0c····s·n1·s0is0·scn12ys3i+1 0s1i+2 10s··s 30csy · s1i·0sV·s02i+1 · ys·s·0xs·0111nsu c 0s ss·i+1 s·00isn··11s00·s·1·3i+1 ·1·ss··11n2·s··i+2 sssi+1 ss·ci+2 s00s11 · · · ss1i 00ss2i+1 0· ·ss i2·s i+2 ·cs·ss0ssc1is0·nV ·00·101csscn·s111scn0·s212i+2 n·(y|x) ci+2 Vi+2 · · i+2 · s0i s·32i+1 ·1 s·sc·i+2 s· · ·· ·ns· sn1s2 sc · ss0c·si+2 sci2i+1 s1ci+2 0s1·1·s ···i·1·s·3·c··i+1 ··s0scn1i+2 i+1 i+2 s · · s 2sc0s0i cis0s c1i+2 1 c 1 0 c ss01302s·0s1·cs·cn0·cssi+2 s · · · s · · s s 0 0 i i+1 i+2 0 0 0 c 1 2 c 0 2 c 0 1 i+1 n s · · · s s s s · s 1 i i+1 n 01 ci+2 s0 scn0 · · · s0n0 s0ss001s c 1 n i i+1 i+2 n 0 1 i i+1 i+2 n · ·ns·nsss0c0y0sn31s· 2·0sc·3s0·c0·3s·010ss1n02s3csc0·2s·s1·3csss·20n0·sss·33c0s·sn·s1s·0·css·1ns1·s0·s2·ns·3·ss·sc2·0·ss30nsys1ssc·n−2 · · 0·s· snc s0n−1 s30s11·sc·1c2s·ns·0nn· si cnci+1 i+2 1 s2i+2 s0 · · · siss0i+1 si+2 s n·c · ·0s2i1cs10i+1 0 ·3· · 0scn10 1 y s··1000·0i··(y|x) sV ··ss·1·s·3i+1 s · · · · · · y · s s s c· s1in 1s·11i·2ss02i+1 csci+2 0i· s 1·i+1 1 02 10 1cni i+1 c 2 s0·c·n0· s 20i+1 0 1ss 10·ci+2 ci+2 1 c ci+2 s · s s · 02c·s3·0·s i+2 n 1 i+2 n 1 i si+11si+2 2 s· ·c···0s·ns1 s n n 1 2 c 1 1 i i+1 i+2 ·c· · sn 1c 0 0·s·01·ss 0 i i+1 n s s · · s s · · s s s s s · · · · s · s s s · · · s s sscsc20··scs··i0i·sc1si+1 s0i+1 ci+2 cn 31 s 1 c 2s s·0102ss32c1ci+2 ·c· s scn10 s21i+2 s 2s·1c··1·c· sc· n1si0cs2i+1 s 0 3 · · ·csn 1s30i+2 ·c ·i ·0s i+2 0· ·0c1s nsn0cs i 0c1·s i+1 3 0i+2 c010 12n c ·n0sc0is3s01i+1 2· ·c · s s c00 s n · i+2 · si ·si+1 ss011cn0ss0s01002221sc0s013211i+2 3 0s 00s003·310ssn·2c·s 0 sc 0· ·3·ss0csc01 s12 ss300···c·i··si+1 s··2s·ss0snc3scs·31····0··ss·s0icns00s·snci+1 · ssss0cn0si+2 c·030s··0n ·s00s·s·02ns10·cnss10·n·s0i·s0·s23·cn·sc3i+1 s1s·0s0ss2s0ic1s3ssc3103i+1 ·3c·s·si+2 ·nc·· s·s··0n·s··c·s·cnsc1i+2 s00 · · · s0i ss3i+1 0 ·s 3cs·32·0c·s··3·s s100cssy·023··s·1·s3·scss00·s0cis·ns0·i+1 0·i+1 13·0n−1 s ·1si2si+1 s i+2 s s ··· · 2·s ni+2 · · · sn0 sni si+1 s s · i+2 · s s s s s · · · s s 0 3 c c 0 1 c 1 n−2 n 0 1 1 2 c 0 1 2 1 0 ·0· i+2 c c s · · · s · · · s s · · · s 0 0 i i+1 i+2 n 0 0 3 c 0 0 3 c c 0 0 3 c c 0 i i+1 i+2 n 0 1 2 n−2 n−1 n 1 s s · · · s s s · · · s 0s i· ·s i+1 i+2 nc 1i+2 0 0 s1· n·c ·is i+1 i+2 n· sni i+1 i+2 s s s s · · · s s · s s s s · s s s · · s s s s · · · s 11n i+2 i i+1 n c 0 1 2 c c 0 1 2 c Fig. 7. Ta and Tb f s s s s · · · s s · · · 0·cs 2 c 0 2 c c 0 1 2 c c 0 1 2 3 n s0s1s2sc3s0· s· 1·0ss2cnsc · 0·s· 03s·c ·0·csiss00i+1 0 1 i i+1 i+2 n i i+1 i+2 s0 11ss3 220ci+2 s0c31scs··1002··ss··113ss0i·ncn22s·0si+1 n 0 1 2 c c 0 1 2 3 n s · s· ·ss0s0·s0ss1·01·c1·s·0s2s0·2ns·ss31s3·0ss0··2s3s·s13c·3ssns·01nssc0· c1· sss22·n0s· 3· ·s· s··c sc·020s0sn13s110s2s01c2s3 2s13· ·0· ·s2n3·c1s1n1 cs0s012s12s3c ·0 ·2·31sn c 0·c 0···3·0s3·cnc1s002s313sn 0 0s1 0·2 ·33· cs s nsc0s · ·0· ·s1·is·s20s s s 11s· ·· ·· s 1 1 1 1 1 0 0 3 c c s · · · s s s · s s s · · 1· sc 1 i+1 i+2 0 i i+1 i+2 0 n i i+1 i+2 n c 0 c 0 0 3 c c 0 i i+1 0 i+2 i i+1 n i+2 n s s s s s 0 0 3 c c c0s1200i+2 0 s10· 02· s c2s s 0cs c 1 i+2 i+1 s scs sns0s·001·ss·0323ssnc3cs·1· s·is0hsh−1 s· 0·s· 1sscn20sn30· ·11s· 0s21sns10css021ss1f32s·2sc·c00··sss·013ns·ss02scsn−2 ·ss22snc3i+2 1u 0h−1 2 c sh nc sh+1 113css·210·ss·31s 11·n2·2s s1ns20ss·sc1n−1 s0i+1 s2ss3h+1 ssc03hs·s3·c·c·sn1cn·cni·ssssh+1 s00s11s22sc3 s· 00· s· 11sscn22sc3 · ·s· 0sscn01s2ss300i·s·111i+1 1 s2s1s 1 ·2s 1s 31 s s21s21·1s21sc3ssc30·s100n·2s·3s11·0·11··s2s1s·22cns1cns1·03c3ss122h−1 n 1sc02c3220s·1311·33·2·1·s·3cns 101 0 31s0c1ss12s2 s 01· · 3·1s c11s02ss 010n c 2 s31 · · · sn1 11s 12s 11nsss 1 03s 1ns 1c i+1s1nsn−1 2 21·s·1s 1·0s00sn2131n1s 11·0ss 1s·1·0n·sss · · 0 0 3 c c c3s·000·c·n−2 s s s s s s s s s · · · s s s s s s i i+2 s 0 0 0 3 c c 0 0 3 c c s s s s s s s s s s s s s s s s s s s s sh+1 0 3 c 0 c 0 3 c c 0h 3 i+2 ch h−1 n n3i+1 3n−1 s0s1s2s3s0· s· 0·0ss3n1sc2· ·c· 0sc 0 c1 s120sc1 20s32sc·0·fs·11s0 n2s1s303s2u i2 si+1 n h+1 i0 s n h 0h−1c20 1n−1 h+1c h+1 n−1 n0s h13·10·2· 2sc0n−2 2 n−2 00 03 3 1cn−2 h+1 1 si+2 sh−1 ·h·0h−1 ·2 s h−1 c3s 01 s12 s23 ·c· · s1n c 11121s23sc · ·1·ns1cs0s 10 s13n31ss121s32n·13·10·ss01s 0 s1 s2 s3 s · ·ns sn 01snsc01s1ss211·2sss·c31202··s1s·3ssns·10·s00ss·2s00111cn·s11ssss11002s122nssss22s1032sc3s1·c322·10·s·s·3·s31s··3s0nssc·cn1cn· 2sss1ss22nx ·1 ss1ns1s0s2s0ss3fc103n·s·cs· 1n−1 13 s s s · · · s 211 32 3 s113 s1 1n3 s0111s s 1sss 10ns 1 3 3 3 s s c 1 3 3 3 su 1 2 3 n 0 0 3 c 0 0 c3 c s0 c0 1 s2s003s0 s3s00cn−2 n−2 n−1 n n s s s s n−2 n−1 n 0 0 3 c c · · · s 1 s s s s s s s s s s s s s n−2 n−1 n−2 n n−1 n n−2 n−1 n c sh+1 sc3 s33h s 32·3·cc020 s 30 3 css·0h−1 ·h−1 On the other hand, bit “1” (resp. “0”) c02 h−10ss000s4h s0s1s2sbrings · 0· s· 1s0sn2ss3a· · 0to 0 the 10first h−1 03n 0s s31 s00s 1 00s s0c s·01·s01·s12ss2ch34·c ·1 s· 2s0 nh+1 1 2 23 · 0· · 1sn32 3 0 s0 su · 2· · s 3 3 ·cn0sccn 1ss1s300ssh1s01s0 32sss2h−1 3s ····02·s330s1s32cnss1cnhs421sc331s1·cs··sh−1 · 0ssh+1 c1s01s s00s1ss11311s32ss3ss24132sc3shsc·13··0h+1 ·s2·3·ss0n−1 s201sss1n01s0s231n2nsss3231·11·0·10s1scn2s30111 ss111s30s30200s1s313ss111ss12ss200s2s112sV330ss3·10n(y|x) 133ss001s s00 s0311· sn s0s102s3 s0 1s0n−2sss3s1001u sss0n−1 0s221s 01 02 s3s s13 n−1 s331nss33 3 n s3 s3s3 · 3103n−2 3 s 021n−1 0ss 1sns 13s 133n 111s 1s33·12·3ns 1s 3s 3n 3s 1 3 n−2 n−1 n 1 2 n−2 s s 0 3 1 2 3 n s 1 1 1 1 s s 1 2 n−2 n−1 n s s s s s n−2 s s s s x s s s s n−2 n−1 n n−2 n−1 n s s s s s s s s s s s s h+1 n−2 n−1 n s s s n−2 n 0 1) to 2a3, cand 3 0 c0 30s h−1 h 0 s200 132s h+1 s s s s s s s s h+1 n−2 n−1 n1c n−1h−1 s s s2 0s 0s 3h−1 c12h−1 0h−1 c13n−2 cs 03c · · · s1sch−1 10s n−2 n−1 n3h+1 c s3s 0ch+1 00s 31c3 c· n−1 ch s102s n s s s s s s s h+1 h+1 n−2 n−2 n−1 ns3h state in S1 (resp. S0 ) , the last state in S0 (resp. S su 0 1 n−1 n 2 0 1 h 2 h+1 3 h−1 h h+1 4 0 1 2 3 0 s 0s 1 2 1 3 h h−1 h+1 h h+1 0 0 3 c c 0 0 3 c 4 0 1 1 1 1 1 1 1 1 0 1 2 h 0 3 h−1 h h+1 s s s s s s s s · · s s s · · · s 1 · · · 0s2nss121102 3s 21ss1301s2n 2ssn01331s1s21s2s2ss3331·1s·1·3s0s03nss1 13321s2s2s11s3113s131 323·ssss3n11n s2y013 s1s131232ss131n−2 3 3 0s0 s1 s2 s s3s·sn−1 ·1c· sc ns1sn 0ss1sns1n−1 s1 ss01nss121sss1n213n−1 c 1 21s33 · 1· 2c·3ssn 0133n3 s 1s 1s1n−2 2n−1 n−1 1 s0 s31sn−1 3 s2n s 3 sx V1h−1 (y|x) 3ncn 1 s s1ss3ch+1 1s 32s3h−1 sn−2 sssnn−2 ss13h+1 ns 1s 2s 2s133 n−2 23n−2 s1shc 3ss3s3s2sscn−1 s33h+1 s3sn−2 13n−1 s00 states s01ssn−2 s3(resp. 0n 201s03ss0n−1 31n−2 h+1 ssnsnh−1 ss2n−2 sh−1 3cs1n−1 3h1n−2 1s01ssn−2 3s 3s 33n−1 3 nc13ss n−1 0s 3s s32 ssn−1 s11n−1 0 n−1 2s0n 0s h+1 s s s s s 3ss su 10 ) 1sn−2 2 s1S h−1 h1 x s1h+1 n−2 n−2 2h hcs3n ss h+1 other states in S0 (resp. S1 ) to different in S s s s s 3 c 0 c 0 3 h−1 h h+1 s s s s s s s 0 1 2 0 1 2 3 h−1 1 4 0 1 2 3 h−1 h 1 1 1 1 0 1 2 3 h−1 h h+1 s s s s · · · s 3 3 3 4 0 2 3 h−1 h s s s s 1 1 1 1 1 1 1 1 1 1 1 1 n2 313 ss1 s1n2 3s1 s3 1 s1s h−1 4n−2 h · s n ss 41 11s21s03 · ·1 · s1n11 s 1 12 1 1 3 1 h− suss 2s 2 2 2 s 3 00sss s1s12s2s1n−1 1n−1 3 3 s0 sn−2 1s 302 11 s 1s3n 1s 3 nc032ssc13s 33c3 scsn−1 3sc sn 1s3c s c1220 3c n−1 c12s02cssn c·23· s n−1 s s s 0n−2 s s V (y|x) cs cs32s s3n−2 s3n−2 s3n−1 1 1 sn−2 n−1 1n−2 1n−1 2s n−2 n−1 n2n−2 s s n−2 c0 3n−1 c0 0nyn2 s c101n−2 c c c 1 n−2 0 0 0 0 1 2 2 0 1 1 3 2 n−1 n s s s 1 1 3 3 3 3 3 3 n−2 n−1 n s s s s s s 1 1 2 2 n−2 n−2 n−1 n−1 n−1 n s s s 1 1 1 1 1 1 3 3 3 3 n s s 1 1 1 1 1 s s s s s s 1 3 3 3 3 3 s s s s s 1 1 1 1 cssn−1 c sn−2 csn−1 cs cn−1 n1shas sh−1 s sh ss1 n−2 n nsn s ss (y|x) s 2s11n−1 s1311 ss0h−1 s 1s2ssscn−2 n−2 n cs because T has (1, 1)-transition for S0 and Sss111. Hence, s1s1sh+1 sc31h+1 sn03sn−2 n−1 n−2 n−1 n−2 n−1 s21sss1sn112 sns43 h+2 1s1V 1111 s h+1 ss1ss2h+2 a sh+1 1 h+1 0n−2 h121s11s 413h+2 01h−1 1sss hs sssh+2 sscn−2 4ns 02n−2 1h−1 1n−1 sssc211ssh−1 ssshc3111s2ssh+1 s12scn−2 1s 1h−1 1 s1s 1s 11s3n 1s 12sn−1 h−1 1n 12 sn1hn−1 0s31h s3ns1s 1 n−2 3 1 s h n−1 21s 31h n−1 sss1212c0 s14 T sss1n−2 s s 1n−20 0 01sn−1 1 1 n−1 n s s s y s s s s s s s s 1 n s s s s s 1 2 n−2 n−1 n n s s s s 0 0 2 c 2 c c c c 21c c 11snn−2 nn−2 c 2s cn−2 cc2n−1 cn−1 c2 cn s cn c s s s 1n−1 n−2 n−1 ncncc 1n−2 2 ss 12 cs n−2 n3 c1c 12 n−1 1s 22 n−1 n1 11 n−1 cs sn−2 1cs ccc21 n−2 1 2 n−2 n−1 3 3 a a 1 1 1 n−2 2 2 c c c c c 1 2 n−1 2 n−1 n y s s s s s s s s c c c c c c c c c c c s s s s s s s s 1 1 1 1 1 1 1 sh+1 1 1sh+2 ssn−1 snsn−2 s0sn−1 ss n−1 sssh−1 n−2 11 1 sn−2 n−1 n n−1 nn−2 h+2 n−1 n s101 sh+2 s1ssn−2 s1n−2 s2n−2 11 s1s s301 ssn−1 (1, 1)-transition for S0 and S1 . h+1 n s1sss sh+2 sn−1 s12s112 h+1 s13sssn1n−1 s1h+1 s2 s1n−1 1 1h+1 1n 2s31n−2 3s hn−1 2n−2 2s11ns1s 3ns 111s 3s1s n−2 2ss n−1 2n−2 n−2 3 n−2 11n111nssnh−1 h n 1 11n h+1 1 1 1s 1 2s1 1 ss3s 1sn−1 111h−1 s s s 1 s s s s s 1 ss110 ss120s11 ss1n−2 s n−2 n−1 0s1 f 0ss1n−1 0 0 s s s s s 1 1 1 1 1 1 1 1 n−2 n−1 n s s 1 2 n−2 n−1 n s s s s s s s s 2 2 2 cn−2s ss11 n−1 n c ss2nc121 c sncsn−2 s1sn−1 s s12sncn cscn−2 ssccn11 c scn−1 s n−2 c sc2 sn−1 n−1 s1 ccc sc2sfn−2 sn n 2n−1 n−2 cn−2 2ns1 s s3 snssns1cn−2 c csn−2 cscn−2 1 same 2 2 way ccssnc sf cn sc c s11c c cs 1n−1 n−1 s sn−1 s2scn−1 sn−2 The theorem can be proved for Tb0 in the snc2 n−2as s3 Ta .1scn−2 s 2 sssn−1 sn−1 s1cn−2 n−1s1n3s n−2 n−1 ns s sc sc s11s21sf31cn−1 n−2 s ss3n−1 s n−1 n3 1s121 n1ss n−2 1 2n−2 n 1s1n1sn 13 s11 13 n−1 1 1cs 2sn−2 n−2 csss css21 n−1 cs1s1n c c 1 c s1 s sc1 s1 c1s21s1 1cs ss 1 13 s 1 n−2n1 n−1 1 n−1 1 s2n−1 scs1c112n−2 s s s s s s c2 cu3 s1 scn−2 cn−1 c c 2f ss cnn−1 c c ssn ccsccn−2 csncs s s s s s s s s s s s s 1 n−2 n 2 n−2 1 2 n−1 n−2 n n−1 n c f 1 2 n−1 n−2 n n−1 n f c c c c c c c c c 2 3 n−2 n−1 n c c c c c 2 n−2 f f c c c c c c c 1 2 n−2 n−1 n n c n−2 c n−1 s2 ss3c ssn−2 s s c csn−1 cn−2 sncs ss c sssss c ssss s1 ncss nc sss 1 sc sss ssc 1s 1 scs 2 s s s n s s 2 3 n−2 n−1 n s s 2 3 n−1 n s s s s s s s s 2 2 1n−2 n−2 2 3 n n−1 n−2 2n−1 32 n−2 n−2 n−1 n3 n−1 s2 21 1 n 3 n−11 n−21n n−11 12 n3 1n−1 1n 3 f n−2c n−1 3 3n−1 fn 1nn n−2 s132 cn−2 c c 1 cscn−2sc c csn−1 s1cn−1 s1n−2 s sn−1 snc f s1c cscns12 s1c c1ssc1n−2 csc1c c n sfcn s sc n ssccnc3s2 scn−2 n ssccn−1 c s2sc sscnc ss3 n−1 n suc2 sc3sc2 scn−2 ssc3nc2sfsn−1 c n−1 f ccs1f cc2scs cscsscc n n sn−2 sc3 xfnsu ssscn3c fn sss2sn−2 n−1 scn−1 sc3cn−1 scn−2 ssc3cnnc321s1n−2sfsnnscn−2 ssn−1 s n−2 2 u sn−1 s scnsn 2 2 nss3s 3nn sn−2 n−2 2 2 n−2 n−1 3 n−2 n−1 n f f 2 3 n−2 n−1 2 n fnn f f Theorem 7 1 n f f f 1 1 f 1 n f f 1 c c fnc c nnc cc c n c c c c cfnc c1 nc c n c n c sc sc cc fVn (y|x) cs s sc ss2 2 sn−2sns3 3 ssn−1 sn−2 sscnc3scn−1 s1cn−21scn scn−1 s2 ssscnc3scn−1 usn−2 sc2 ssc3 s2 scn−2 sscnsc3n−1 u s2 n−2snsn3 sn−1 ssn−2 n−2 2 n−1 2 u 3 10. sn−1 u (k, fcode fnc 1 l)-transition y n one fxnnsu exists For any positive integers k and l, there fn c tree 1cwith n n 1 fnFig. Code fn s at least f≥ 1 ffnn 1 ssc fn c1 for1skc c= n1 − 1 fn n l = cn cnfch, c −1 1(n xsu f s s s ssn−2 s3sn−1 scn−2nscn scn−1 scn n−1s2 ns3 s ≥ 2, n −1 h 2≥ 2)1 3 s n−2 2 4, n nl)-transition 1 V (y|x) tree with no self-synchronizing string and (k, for 1 1 V1 (y|x) xsu fn xn ffnn n 1 fn xsu 11 fn 1 f f 1 1 f n n n xsu su yn yn V (y|x) S0 and S1 . V (y|x) fn 1 1 fn1 1 fn 1 V (y|x) 1 1 V (y|x) 1 n In Figs. 11–13, both S and y n S1 are included in the cycle of y n 0 n y 1 y 1 1 1

Fig. 6.

Proof: Code trees with (k, l)-transition and no synchronizing string are given in Figs. 6, 8–10, which correspond to the cases of (k, l) = (1, 1), (k, l) = (1, n − 1) for n ≥ 3, (k, l) = (n − 1, n − 1) for n ≥ 3, and (k, l) = (n − h, n − 1) for n ≥ 4, h ≥ 2, n−h ≥ 2, respectively. In the figures, s1d , s2d , s3d , · · · represent different states at depth d, and scd stands for the state at depth d that has the subtree with constant depth 1 n − d + 1 as its descendant. The state transition diagrams obtained by Algorithm 2 are shown in Figs. 11–13, which

the state transition diagram. In this case, any two sates in S0 (and S1 ) cannot coalesce by any string. This means that all 1 paths from all states in S0 to their leaves must be different, 1 i.e. they must be “0”, “1”,1 “1w”, where w is any binary string. 1 n, there exist n − 1 states except the root In the case of |S0 | = state. Since code trees are complete, the number of such paths must be at least 2 + 21 + 22 + · · · + 2n−2 = 2n−1 . Hence the 1 number of leaves 1in the left half of 1such a code tree must be n−1 at least 2 . The same holds for the right half of the code 1

1

s00 · · · s0h−2s2h−1s3h sch+1 · · · scn 0 s000 ·2 · 1· s0h−3s1 2h−2s3 1h−1cs3h sch+1c · · · scn s0s1s2s3 · · · i−1 sh−1sh si−1 h+1 · · · snn−i−2 0 1 1 0 1 0 0 2 3 0 0 2 c c s · · · s s s sc · · ·cscn s0 · · · sh−1shsh+1 · · · sn 0 0 0 2h−2 1 h−1 h 1 h+13 c s · · · s s s · · · s s s 0 i i+1 i+2 h−1 h h+1 · · · sn s00s11 · · · s1n 0 0 2 c c 0 0 2 c c s · · · s 2 shs1h+1 ·3· · scn s0 · · · sj sj+1sj+2 · · · sn s00 · ·0· s0h−3h−1 sh−2sh−1 sch sh+1 ·c· · scn 3 s00s01s22s13 · · · s1h−1si−1 · sn h sh+1 · ·i−1 i−1 i−1 n−i−2 1n−i−2 0 1 1 0 1 s00 · · · s10s2 s1c · · · s0cn c s00 · · · s0n−2s2nscn 0 0 0 j 2j+1 3j+2 c s · · · s s s s · · · 0 h−2 h−1 h h+1 3 scn 0 0 2 1 1 c n−h−i−1 s0 · · ·0si1si+1s1i+2 0 1 1i−1 1n−h−i−1 1 · · · sh−1 sh sh+1 · · · sn ss00s· 1· ·· ·s·0 s2n sc2n−1scn c s00s01 · · · s0n 0 0 0 n−2 0 0 0 0 1 s c0···s c 0 s1 sc s · · · 0sh−12shsh+1 · ·3· snc s0 s1 · · · sn s0s1s02 0·0· · sn n−2 n−1 0 n1 c 0 0j−h+1 1 c n−j−2 · · ·0sh−3 sh−2 0i−1 0h−i−3 s00i−1 1 0sh−1 1 sh 3 sh+1 c · · · sn c s0s1 · · · s0n s0s1s2 · · · scn s0s01s221sn−h−i−1 3 · · · sh−1 sh 1sn−h−i−1 h+1 · · · sn 1 1 n−i−2 0i−1 i−1 i−1 n−i−2 i−1 n−i−2 0 0 2 c c 1 11 0 1 0 1 s · · · s s s · · · s i−1 i−1 n−i−2 0 1 1 0 0 0 n 1j sj+1 2 1j+2 1 s00· · ·0s0h−2 s3h 1 sch+10· ·1· scn 3 1 c s0 · · · s0n−2s1n−1scn 0 1 1 3 c c 0 h−i−3 s0 · · · s0ih−1 s2i+1 si+2 · · · s0h−1 sh sh+1 · · · scn i−1 j−h+1 n−j−2 00 0 1 c c s s · · · s s s · · · s 0 0 0 00 1 c 0 c 0 0 0 0 1 h h+1 h+2 n 0 1 1 s0 s1 · · · sn ss0s1s2 ·s· · sn s0 s0 s· 0· · si si+1ss0i+2 · · · sns0 0 0 s0ss010s·1·c···s·1sn · · ·0· · 1n· sn n s0 · · · s0 s12s2ncscn s00s11 · · · s1hs3h+1s0ss1ch+2 1 n−2 n−1 n 1 2c 1n−i−2 s0s·000s· 11· s· 0· ·n−2 n · · · sc s0000· · · s0n−2 s00s01 · · · s0n 0 1 1 sn−10si n s00sn−1−i 0 h−1 s0h sh+1 1 1 0 1 c c 3 c 0i−1 csn−i−2 0s1 · · ·1s1 1si−1 · · · s s2 s1 n s3 sch+1 · · · scn s s · · · · · · s s s s · · · s 0 1 0 001 0t−1 h+i−1 n00c c 0h−3 h−2 h−1 h 21 1 ch+i 1 h+i+1 3c c c s 0 1 n 0 1 2 n 2 1 3 0 1 3 0 i n−1−i n−h−t−1 s s s s · · · s s s · · · s 0 1 1 3 c c s01h1s12h+1 s2·3··3··ss·h+2 s1h−1 ·ss·00·sss10sn· 2·ns·1s·n· · s1 s3 sc · · · sc s11· ·ss·0h+i+1 · 1·s·hshnsh+1 0 ss0h+i 0s0n−2 1 s0n−1 0 00s0n1 s0s11 · 3· · sh+i−1 0h−1 1 h+1 s00 s01 s02 ns h−1 n c c 0 s1 · · · sn s00 ·0 · 1· s20j s32j+1 scj+2 · · ·hsc h+1 1 3 c 0 0 01 s0c 0 1 c 0 c s001·c · · s0h−2 s2h−1s3h ni−1 sch+1 · · · sn−i−2 01 c 1 c 0i 0n−1−is0s1 · · · shsh+1 sh+2 · s· s00· s00ss11n11·0··0·· ·ss1n−2 i−1 n 0sn−1 scn1 n 0 2 1 · s s s · · · s 1 · ssnn−11s 1 n−2 n s00 s01 s02ss10s·00·s·01ss·0n−2 s 0 1 1 3 0 2 1 1 3cs3 n−j−2 cs 0 1 1 3 c j−h+1 ·h−i−3 ·isssii+1 si+1 s0si+2 ·2i−1 ·s·1h−1 s·h−1 ·c2··1s·cns1cn c 0 1 3 1c 0 s1 s2s · n· · sn s s s n−1 0 1 1 3s 0si−1 c· ·s·0ss·0··0 · s s · · · s s · · s s h h+1 · · s · · · s i−1 n−i−2 0 0 0 0 0 1 3 i+2 h h+1 s s · · · s s s 00 1n 1 2 n−2 0 1 n−1 h h+1n i+2 0 0 n 1h+t−1 1 h+t s0sh+t+1 02s3 · · ·ss0snh−1 1·0 ·· ··ss·h0si ssh+1 · · · s s s · · · scn 0 11 c n−2 n−1 2 sc·i+2 0 0 1 c s · · s i+1 h−1 h h+1 0 1 3 c c 1 n n−i−2 1 0 0· ·21· s s s · · · sn s0ss10ss21 ·· ····ssh+t−1 sh+t ss00h+t+1 13n 13c 3· · 3 cc 0sc 0 n−2 0n sn 23 c c 0 1 1i 1n−1−i 1 s · · · s · 1 1 3 3 1 n i n−1−i 0 1 s s s s · · · s s s · · · 1 0 1 2hs 3h+1 sh−1 h+2 h s h+1 n 0 1n s0s· · 1· sh−1 ssh sh+1 · · · sn s s 0 1 1 1 0 0 0 0 1 c c s s · · · s 4 0 1 2 3 h−1 h 0 0 2 1 3 c c 1 · · ·cs n 1s 33h−3 s0s1s12 · · · sn s1 s1 sss300css011s·1·ss··003s·s·0·1n−2 3s0h+1 ···c1·s·s0h−3 ss2scnsh−2 s3sh scch+1 c0s3h−1 h−1 n 2 i−1 s00ss0011······ss0i1ns1i+1sci+2 s00s·s11·s11·c20s·cn1· · sscn12 1 ss1n−2 3 n−h−i−1 0 s31 · · c· ss n1 100·· ·· ··1s c s1 c · · · sc 1n−h−i−1 ssn−1 sn0scn01h−1 s1h−2 4 0 2 ssh+2 3 h−2 h−1 ssn−2 · · · s0hs2ssn−1 s3 ss1h−1 · · ·ssch+1 n−1 h n c 1h+1 n n−h−t−1 s≤n−1−i · 0· · 1scn scn 0transition 1 1 sh0sc00si+2 sc01 ·0· ·s·sh−3 10c01 1 1 10n Fig.1 11. State diagram 8. i s≤h+i+1 s00s11cs·c· · s01n−2ss3n−1 c1 ssn0 s101s·t−1 2cs3 1 s0c c 1 i· ·i+1 3· s nsc0 2 h ch+1 c 1· ·1sh+i−1 0i (1 s00s11s1c2 ·s·00·0for sscn11 ·Fig. ·sn−2 ··0·s··n·s·h+i−1 0c n − 2) 0 10h+i c s s s s s s · · · s s s · · · 1 1 c s0s1 · ·c· 1sn s · · · s s s · · · s 0 1 0 1 h+i h+i+1 n 0 0 2 1 1 3 c c h+1 h+2 n−1 n i i+1 i+2 h−1 h h+1 n s s · · · s s s s · · · s 0 j j+1 j+2 n s0n · · · si si+1si+2 · · · sn 0 s1 s2 sn−2 sn 0 s1 c n 1css32cs3 · · · sh−1 3 s21s2h−133s3h scs1ch+1 3 ·s··0·s 3 sh sh+1 · · · sn ss0h−2 s20 1 s3sn−1 scn−2 s0 c1 2 i−1 1 03 ni−1 s14ss01s0sss002s10s·0s1··1···s··0h−2 s·01·h−1 1 sh−130s2h c h+1 c·0s·n 2s 3 c c j−h+1 0 1 n−i−2 c c n−1 0s 2ss c0hh−i−3 1 3 i−1n0 s s 1 3 c c · s s s · · · · s s s s · · · s 0 n 1n−i−2 0n−j−2 sch0 · ·sh+1 ·3sn−j−2 s n00 sh−1 ·h· · h+1 n−2 n−1 nc s1 0 0si+2 · · · sn s1 1 1 s001 ·c · ·0sic ss h−2 h−1 1 3 3 01 11 2 3c13 21 3 j−h+1 h−3 h h+1 n 0i−1 i−1 1i−1 s0ss00s100s· 11···· s· ssh1i01ss2i+1 si+2 0c s 33h−1 1 s 0s c h−2 s·02· ··ss·cnss0·00s0ss11s0n··s0n−2 h+1 ·2· s11n1 2 scs001sn−1 · 3·c · scn i+1 sci+2 c ss0101n−2 s202211h−i−3 sscnn−1 4 scnscn−2 ss110c 0ss121 scs0h+1 s1 1n s· 3h·h0·i−1 ss·003······scs··3h+2 ss0h−3 ss2h−2 1· s· 2·s n ss·100sscnn−i−2 ·0s· 2·0s0n−2 s12n·s·cn· s1 1i−1 h−1 hh−1 h+1 n−2 n−1 0· · · i−1 1n−i−2 3 c n−h−i−1 c n−1 h+1 · · s ·2·ss3i s·i+1 si+2 shn sh+1 n· · · sn c c212h−1 s · · s s s · · · s 0 0 c c 1 1n−h−i−1 0 1 0 1 0 1 0 1 h−1 n−i−2 0 i i+1 i+2 h−1 h h+1 n 0 0 c c 1 0 0 sh−1 c c c sh+2 1 ssn−2 s s·0s s sh+1 ···s 0 s0n−1 0···2·s·h−1 1 snhsshh+1 s00 scn−1s01 scns02 sh+1 s0n−2 0· s1· · ·0ss3n ns2cs00· 3·1· ·ss0cc 1s2 sc c · · · sc 0 3 sn−2 c 0 0 0 0 s 1 2c s0 · c· s s · · · s s s · · n−1 n 3 s · · · s s s · · · s · s s s s s · · · s s s s · · · s 0 i i+1 i+2 h−1 h h+1 n 0 h−1 h h+1 n f 0 h−2 h−1 h h+1 n 1 1 n−1 n0 s10 · 0· ·1s0s13 nsc ·3· 0· s1c c2s0s01 ·n· · sn−2 1 3 3 3cs1 cc 3 c ss01s c01c·0· · sn 03 11 1c cc sci+2 s00 s·s14s ·11·s0sc 0i00s·s2i+1 ·s312·11n−2 · scn−2 sh+1 ·s0n·0·ss1ns3h−1 s s n−1 s·n−1 ·s0·2 ss10h−2 sh−1 0 j−1 ic i+1 i+2 sn0· · ·cscn s101 3 sc 02 s0n−2 s1 0n−1 n−j−1 i+2ss 3· · s 1 3 c c j−h+1 h−i−3 ·s0·sn0·1 s·1n·s·2ssn0hsi−1 hsc12shh+1 s0 ·1· · s·0i ·s0i3i+1 sh+i ·ssi+1 1s·2i−1 0si−1 1nn−i−2 sc2 sc3 0 scn−2 scnsn h+i+1 0 3c s00ss10h−1 0n−j−2 h+i−1 0 1 1 1c c n20 2sni+2 ·1·0· sn c s1 c· ·cs h−2 h−1 sh 0sh+1 · · · sn 0 s1 ss0sch+2 1ss cn 3· ····ss us10s1 0·1· ·ssn−1 · ·s·js2snj2j+1 ·ccnscns200·c0·1c·sscnh−3 1 1 1i−1 · ssnh+1 s0000i−1 ·0·s1· ·s·1i·0ssn−i−2 j+1 0 n−1 1ss c c sc2s2s0s1 s· c3s· n−2 sn−2 n−2 0nj+2 0cn−1 c sn−i−2 0s·00·s·n−1 j+2 1i+1 s 2 i+2c1· · · sh−1 sh sh+1 · · · sn 1 s · · · s s s · · · s · 0 1 · ·3· sn c c 10 1 2 c 1 c 1 s · · · s s s ·2·0jss12j+1 s s s 00 h+1 ch sh+1 n·0·ss 0 s 0 ch c h−1 n· · ·1sj+2 0 0 1s 0 3 n−2 c n−1 snc s s · · · s 0 2h−3 1h−2 3h−1 s s s f 0 0 2 c 0 1 2 3 h−1 h h+1 n s · · · s s s · · · s s s s s · · · s n s0 · · ·fsnh−3 s sh−1 s sh0ssh+1 s · · sn· · · s3ns00sc0s2s0 1 · · 2·c s1 3 s3 c sc · · c· sc s 1 2 3 02s0 n3 0j−h+1 n−2 0 s1sn−j−2 n 2 icn−1 c i+1 0· si+2 2h 21 ·ch+1 s000s·311·01··c0h−2 · · ss1n00h−1 s0 3 csn−i−2 s0nn 0i−10 00 ss21000·h−i−3 h−1 ·11c· s2 h−2 sh hsh+1h+1 · · · sn n 0 0 1 2 s3 ·s·n−1 n s · · · sc1h−1 shs00s·ssh+1 · 3· · ssh−1 1 s 1 i−1 11·cs n s2ss·s0n−2 · ·sn−2 i−1· · · sci−1 n−1−is1 s00s01s22s13 · · s s ss s··ch+1 0 xs 1 1 s120 s1·13i−1 0 s21cnsn−j−1 010n−2 s·s11s1nc21h−1 ··ss·n−2 s· 2s23sscnn3csnc ss1n−1 ·0sn−1 s1 s1 c c fsnc1n−1 c n−2 fn n−2 0 1 00n12 0n 1·j−1 0s h−1 hush+1 su00s·c·shs·02s3·00h−2 c c snh sn00s·h+1 · · n−1 s·0·s·2sn sscs000n·n·······s·s0n−2 s0sscn002ss112n·ss·cn1· s1n· · · s1 1 s3 sc i · · · scn−1−i s1· ·s· ·3sn s31s1001s11·sc2· ··3·s·11s1cn s3 3 s2c 1 · 3· · n−2 h−1 su 3s·0· 0 0 3 c c s s s s s s 1 1 0 0 2 3 c 0n−1s 0 s2 n sc ccj j+1 0j+2 0 i2 i+11 i+2 0 1 h−131 h c h+1 n 2 3 s n−2 s s s s s s 0 i i+1 i+2 h−1 h h+1 n s02s1s2cs3 · · · snc u s · · · s · · · s 0 0 3 c c s s s s s 0 1 2 3 n−2 n−1 0 0 2 1 1 3 c c · · · s s s · · · s 0 1 1 0 h−2 h−1 h h+1 n 0 0 2 c c 0· · sj+20 s2 sn1 ·n····ss3i si+1 · · ·c s sh i sh+10· n−1−i · · sc 0 n−2 n−1 n u s00s1s21sj3s·j+1 ··s0·s0s·1h−1 s0s10 V · · ·(y|x) s sj+1 s1j+2 · 1· sns2 scs 3 s0s1s2s3 · · · sn i 1 n−1−i 2 i+2 schssh+1 · · · sc i si+1 h si+2 c ·1··s h−1 n h sh+1 ss001s3n−1 · c··s1h−i−3 0c 0 0· · · s· n·1sh−1 03 c 0 n c · c· sh+1 s0 · · 0· sj0n−2 s s00 · · · s0i s2i+1fsn1i+2 · ·s·s02s3hs0nc sh−1 1 ·s n−2 x0nsu···s··00·sss001h−1 n 10 0s 1001s1131 h+1 s00s111 n−1 ·s·1·csn1n0sj−h+1 s·cc2·0cs·n−j−2 n 0· · s 0n 2 s0c0s0 · · · 0 00 0 2 2 1 1 1· s· 13sn c cc c 1 0i−1 0 23ncs0n−j−1 cj−1 s · s s n 3 c c s · · · s s s s · s · · · s s s s s · · · s s s · · · s x 0 h−1 h h+1 n 1 2 n−2 n−1 n 0 1 n s n c c c c c 3 3 3 3 3 0 n−2 n n 0 1 h h+1 i+2 n 0 s 1 h−2 2 3sh−10sh h−1s h 0 · h+1 n · · su s 33 sshsn−1 s0n−2 ssnn 3 c 3 ss0 · · ·0sh−3 ··s fn xsu 0 1 ys0s1·2s· · si ss 13 si+2 00 s 2 0c s2 c c si+1 s2s·n−2 3 ·s·3cs· n−1 s0 s·u2· ·Vs(y|x) s sn−2 · ·s1·ns0n s2 n−1 3h+1 1 s3c0s01s·c ·h+1 · s· n· · sc ns00s11sc2 · · · scn c s · · ·ss ··n··s·scs·0h−3 ssss00·h−1 s2i+2 · ·s· s0cnc sn−2 00 ·· ··3··hs 1 0i s3n−10cn−1−i s0h+1 s2c00s·nn−2 · ·c s· h−2 s1h−1s1h−1 sss3hs0000s3h·ch+1 · ·2·ss·cn2h−2 s00s110 · 1· · s31n 01s0s1 2· · 1· sc1 0 V 10·0·· ·s1c· ss1i1s cs0 · · · s0 0ss2 c ysnsh−1 ·s·i+2 sh−3 ·sscj+2 0 10 j sj+1 h+1 s1sc2(y|x) s1n−2 c 1n· ·s·2 s1nci+1 xnsu s300s110scc2 ·s·31n−1 sn0sh−1 · · h·ns10n h+1 s100s11nsc23 · · ·cscn s03h i2ssi+1 s0s1 0·s·n1· s·n−2 2 snns c 0 s1 · ·s·0s s·12s·n· · sc0si 1nsc3 0n−1−i sn−2u Vs(y|x) 1ss 00s c · · · sn c 0 h−3 s h−2 h+1 1 · · · s1n n−1 nns00 ·h−1 ·0s1js1sn−1 scn1i1s3n2i+1 0 2 0 01 32 c 0 1· i+2 · · s s s · · · s j+1 sj+2 0 0 c c i−1 s0 s1 ·h−i−3 j−h+1 n−j−2 s s · · · s s s n s · · · s s s · · · s s s 10 s j+2 n sc c h+i+1 c n c h+i i+2 · ·0· s h−1c h h+1 · · · sn s0 · · ·0sh−2 s0h+1 0 0 fxsu 00 20 j cj+1 y 0 · · · sc h−13ishi+1 n 01 0 sc1 0 32sc 0c 1 sc cy n01 s0c 1 3 c h+i−1 s03 0 0scn−2s0s1sc ·s·Fig. 1 · n· · sj00sj+1 sj+2 010. V (y|x) · ·· ·h sc1i+1 s2 ≤ · · ·iscn≤ cs0(1c ≤ sdiagram ct s≤ nsc 13. sState n−1 n s0transition sc s 2 c 0− 01 for 23 Fig. s22h−1 snsch+1 · s· cis− ·s··1·1s·s02h−1 ss1i s·ss·h1i+1 ss0c3i+2 · · s ssc1c0sn−1 n0s·2· · ssn s00s010s22s ·n−2 s30 ·1···c·s·0h−3 s3hch+1sss00·ch+1 0 ·····s··s h−2 n 2, i+2 c· s s s 0 1 2 x· su n 1· s 1sh sch+1 11 1n · ·c · sn c s0 s1 · · · sn s · s 0h 0 11 c h−2 h−1ssh−1 h s n·n−2 1s03f·n 0s3s1h+1 c n3· · ·sh−1 0 0 2 (y|x) c h−2 V n n 0 n · 0 s s u · · · s s s · · · s 0 0 0 0 1 c c s0 · · · si si+11si+23· · · csn s0ss1 0·s·1· ·s1·n·0s11 s2 0 h1s 2h+1 ·n−2 · · i+2 s s0 · · · sh−2sh−10sh sh+1 ···s 1n−1 n 0 1· · ·2s1n1 in i+1 i+2 3s cs sn·1· · s c s1s s s · · · s s1c1n−2 − 3, h ≤ j ≤0 ns− 1 cn−1 yn · s0n−2s2nscnn 2 1s00 s01 ss shs0h+1 s00 ·s·2·ssc0h−3·s·2h−2 sc 0sh sh+1 · · · scn0 0 · ·2) u1 scj+2n· · · scn 0 · · · si nsi+1ssc2i+2 f· n·0· sn n 0scnn1 0 1c 2 nc hV (y|x) s0 · · ·0s0h−1 s00s11 · · · s1j s02j+1 2 c s2sc32 3 · · · ssh−1 sn−1 h h+1 s0·csn h−1 n−2 s0 s· · · s0n−2 0 0 s1 s02 sn−2 s0n−1 1 3 c c 0 s 0 s0n 2 0 s2 sc 03 2 s2c c · · ·ss0c0s·0c·· ·· s· h−1 · · · s0 0n s01· · · si si+1si+2 · · · sn s00 · 0· · 1s0i0s2i+11sfs0n1i+2 s s s s n c s0 · ·2 ·n−1 10 · ·3c· s1h−1 3 sh+1 c ·c· ·0 sn 0c 3 h h+1 s c c h s · · · s s s · · · s 3 c 0 0 0 h−2 h−1 h h+1 n 0 0 2 c c f 0 1 n x s1 s ·s·i+2 ·1sh+i−1 s2h+i 0 h−1 h h+1 n ·· ·· ssin0si+1 si+2 ·n· · s0 n·1· ·us03s1c sc · ·c· sc s0ss ·0··ss·2hu s0 s1 s2 sn−2 s0n−1 s0n 3·1ssn c·c··s··h+i+1 s0 · · · sh−1shsu s2 h+1 ·s01sh+1 0 0· · · sn0 1 0·s s·ss00·si+2 ·sfor s·nscnscs0cn0···Fig. 2 (2 1 ≤si, 1 i+1 2s 3 0s j ≤ i sn i+1− i+2 n Fig. 12. 9. 1) s s · · · s 2 s3 ·i· · s h−1diagram h1c s h+1 0 1 State 1s00s1transition 0 0s002· · · sc 0h−2 s2h−1cs3h sch+1 · · · scn · · s s s · · · s · · · s c c c 0 1 n 0 i i+1 i+2 h−1 h h+1 n V (y|x) s0s1 · · · sn sn2 s0 · · ·0sj2sj+1csj+2 · · ·csn f s s s s n c c c c c u 3 n−2 n−1 n 0 0 0 0 0 0 0 1 x s s 3 c cs c s sn−2 sn−1 s ss1n s2 3 sn−2 sn−1 sn n s0 · · · s0 s0s·2h·s·ch+1 s ·c · · sj sj+1sj+2 · · · sn s00 · 0· · 1s0h−3su s21s0nss0s1h−1 23 1sh c sh+1 1 · ·3· sccnu2 3 c3 3 s0s·2 · · sscj+2 n · · · 0sn 0 0 s13 ·ss·h+i+1 0 0· s0h−1 s3 0 s33 ns030 c s3n−2 s0ss1 0· ·· ·· ·sh−2 s0j s2j+1y scj+20· · · scnh−1 1 s2sh+i 0·0· · ·s·n· s n−1 x0i0su s211 c leaves. s12 c s1n−2For s1n−1 s1n s s31s·sc3·h·s·s·3hsh+1 ·n13ssscnsch+1 0s·n· · C(u s02 c ss0n−2 s0n−1 sby 0s1c 2 1 s2 s0 1 x sh+i−1 s32i+1 0 1 (y|x) given − 1). Let Tj j+1 be a code ci+2 ·0 · ·1 c2h−1 ssu s00 ·s·2s·with 0 · · · si sni+1 si+2 · · · sh−1 h sh+1 · · · sn s s00s01s22s13 · ·0·Vsf1h−1 s s · · · s h+1 · ·s·1sn s0 · · ·0tree s0n−2 s1sch−1shs1 s12m s s1n−1 s1n n h h+1 n n 1 n n−2 0 0 2 nc 1n 3 c c f 0 0 20 c 0 cc 12 ns2V0(y|x) n 2 s · · · s s s 1 s00 · 0· · 0s0h−2 s s · · · s x s · · · s s of sm · ·s= ·nssnn 3 0, T n−2 n nbe generated from h· s0 s h+1 2 1 h−1 32 3c s1 n c 1xsus33sc V 1(y|x) sn−2 instance, can sy·00n· ·s·1h−1 · ·s·3 scnc su 3 3 0 · · ·j+2 s0ss1s0 y 3· · ·s·s3c·n1sh−1 ss001· · · in s0n−2the s2n−1scn0casej sj+1 i+2 1 1 2·s·3· ·s h+1s 2 ss1h−1 s21 ishsi+1 ·0ssh·2cn· h+1 0 0 2 s1n−1 1h+1V·c1·s 3 s 1 ·s·n−2 tree. Hence 30 theorem ch · ss0i1s3i+1 ·11scn ss1n1 1sn−1s2s11 sn−2 1 n 1 holds. (y|x) sc · · c· scn 4c i+2 V (y|x) s00 · · ·the s0s2 0following s1 h−3 · · · sh−2 s00s01 2 · · ·ss00n· · · sj sj+1 3sh sh+1 · · · s sn−2 1 yn−1 h−1 ns10 2 n scn−1 scn s s s s s s 0 i i+1 0 i+22 c c initial code trees with u = 3, u = 3 , u = 3s3c21,cj+2u s= 3sn−2 , and 0 0 2 c 3 n 1 2 n−2 n−1 n 0 s0 s 0 s0 · 0· · sh−10sh2s1s0h+1 · · s · · · s 0· s 2n 1 1 3 c 1 3cc c c n 1 s s · · · s 0 n−2 n n ·2si· s· i+1 si+2 · sh−1 s2 s3 scn−2 scn−12 scn 0 1 n s0 ·s·0· ·s·i ·ssi+1 ·3sh−1 · shsnh+1 · · · sn y 1 00 ·s·i+2 c sh·s· h+1 c· · s 1 6 s s s · · · s 0 0 c h−2 3h−1 c h h+1 c n uc = 3 c. But, the code trees generated sfrom 3, u = 3 , s2ns= s0 2h−2s1h−1 s sh+1 0 · · · sn−2u Theorems00 8· ·s·0 s· 0h−3 fn n sc11 · · · ssn12 s11n−2 s1n−1 c s1n sc2 0 2 c h sc3 c scn−2 sn−1 sn s00s01 · · · s0n 6 0 0· · sj sj+1 0 2ss0j+2 1··s· 0·ss3n 1sc 3· · c· sc sc c sc sn−2sc 1 scn−1 scn c c c 33 are included 1from u = 3 . f s · · · s s 3 0· si si+1 00 ·s·i+2 2 i·c· i+1 ch sh+1n· ·2· sn u = in the code trees generated n i+2 u s s s s 0 h−1 1n · · · s no shsh+1 ···s 2 3 n−2 S0 n−1 If a code trees has self-synchronizing string and both 1 1 s00 · · · s0h−2s0 2h−1s3hh−1sch+1 · · · scn n u 4 s fn included in 1 n c |S1 |c c = n and S1 with s00|S = cycle of 1 But, the code trees generated from u = 3 sare xnot fsnc included · ·0·0s|0i s3i+1 ·c · s· sn c sc are 0 s2i+2 scn−1 fin scn the su n u n−2 6 n 0 0 s0 2· · c· sj sj+1 scj+21 2· · · sn 3 1 u = 3 . u x the code trees generated from s · · · s s s · · · s 1 V (y|x) su 0 h−1 h h+1 n u the state transition diagram obtained by Algorithm s 2, then the s (y|x)y n s fn xnsu Hence, the total number of code trees Vysatisfying Theorem n n n number of must n s00 · ·leaves · s0j s2j+1scj+2 · · · scn be 1at leastxsu2 . xsu V (y|x) 1 1 V (y|x) 3 is given by the following theorem. n V (y|x)y 1 y n 1 NO IV. N UMBER OF CODE TREES WITH 1

SELF - SYNCHRONIZING STRING 1

1 1

The number of code trees with m leaves and no selfsynchronizing string is given in Table I by the exhaustive search for m ≤ 21. It is well known that the total number of code trees with m leaves is given by Catalan number C(m − 1), where C(t) is defined as µ ¶ 1 2t C(t) = . (1) t+1 t In Table-I, columns “th. 3”, “th. 4”, and “th. 7” corresponds to the code trees that satisfy Theorems 3, 4, and 7, respectively, and “oth.” represents the code trees that cannot be classified into any one of them. Since some code trees satisfy more than one theorem, the total shown in the most right column may not be equal to the sum of other columns for some m. We note that for m ≤ 21, all code trees with no synchronizing string satisfy at least one of Theorems 3, 4, or 7. Since the code trees satisfying Theorems 3 or 4 are constructed recursively, the number of such codes can be evaluated theoretically. First we consider the case of Theorem 3. If the initial tree T1 has u leaves, then the code tree Tn has un leaves. Since different initial code trees with u leaves generate different code trees with un leaves, the number of such generated codes is

Theorem 9 There exist code trees satisfying Theorem 3 if and only if the number of leaves m can be represented as m = pq for some integers p ≥ 2 and q ≥ 2. Assume that p cannot be represented as p = uv for any u ≥ 2 and v ≥ 2, and q has divisors a1 , a2 , · · · , ak , which are less than q. Then the number of the code trees satisfying Theorem 3, say N (m), is given by X N (m) = C(pai − 1) (2) i:ai is not a divisor of other aj

For example, we have that N (4) = N (22 ) = C(2 − 1) = 1 N (8) = N (23 ) = C(2 − 1) = 1 N (9) = N (32 ) = C(3 − 1) = 2 N (16) = N (24 ) = C(22 − 1) = 5 which coincide with Table I. Furthermore, we can predict that N (25) = N (52 ) = C(5 − 1) = 14 N (27) = N (33 ) = C(32 − 1) = 1, 430 .. .

s0n

TABLE I E XHAUSTIVE SEARCH

FOR CODE TREES WITH NO SYNCHRONIZING STRING .

Leaves

Trees

m 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Total

1 2 5 14 42 132 429 1430 4,862 16,796 58,786 208,012 742,900 2,674,440 9,694,845 35,357,670 129,644,790 477,638,700 1,767,263,190 6,564,120,420 8,987,427,466

Trees with no synchronizing string Th3 Th4 Th7 Oth. Total

1

4 1 2

2 22

140

5

8 969 36 7084

9

8263

6 6 8 6 4 35

0

0 0 1 0 0 4 1 4 22 0 0 140 0 8 973 42 6 7092 6 4 8303

n 2 3 4 5 6

Similarly, the case of p = 8 is given by n 2 3

Correspondence between T˜n and a p-ary tree

m C8 (n) 15 8 22 92

We can use four initial code trees in the case of p = 9. Hence, we have n m C9 (n) C9 (n) × 4 2 17 9 36 3 25 117 468 These numbers coincide with the numbers shown in Table I. From Theorem 6, we can construct code trees with (1, 1)transition for S0 and S1 by applying Ta and Tb recursively. If T has p leaves, then both Ta and Tb have 2p leaves. Let Mn be the number of code trees that can be constructed from the code tree shown in Fig. 6 by applying Ta or Tb n times. Then, the n-th constructed code tree has m = p × 2n−1 leaves and Mn is given by Mn = M0 × 2n−1 . If we use the code tree of Fig. 6 and its reversed tree in both sides as the initial code tree T , the following table is obtained. n 0 1 2

Fig. 14.

m C4 (n) 7 4 10 22 13 140 16 969 19 7, 084

m Mn 9 2 18 4 36 8

We note from Table I that besides Figs. 6, 8-10, there are many structures of code trees that satisfy Theorem 7. V. C ONCLUDING REMARKS

Next we consider the case of Theorem 4. Let the initial code tree T˜1 in Theorem 4 has p leaves. Then T˜1 and generated code trees T˜n have correspondence to p-ary trees as shown in Fig. 14. Since the total number of p-ary trees is given by Fuss-Ccatalan number Cp (n) defined by µ ¶ 1 pn Cp (n) = , (3) (p − 1)n + 1 n

In this paper, we showed that there exist three types of code trees wtih no self-synchronizing string. From Table I, all code trees with no self-synchronizing string can be classified into one of three categories. But, there exist code trees that have no self-synchronizing string and don’t belong to any one of the three categories. Such an example is shown in Fig. 15.

the number of code trees satisfying Theorem 4 is given by the following theorem.

This work is partially supported by the JSPS Grant-in-Aid for Scientific Research (B), No. 17360174.

Theorem 10 If the initial code tree T˜1 in Theorem 4 has p leaves, then the number of different T˜n and the number of their leaves are given by Cp (n) and m = (p − 1)n + 1, respectively.

R EFERENCES

Since we have code trees with no self-synchronizing string when the number of leaves is 4, 8, 9, · · · in Table I, we can use these code trees as an initial code tree T˜1 . Therefore, in the case of p = 4, we have the following table.

ACKNOWLEDGMENT

[1] R. M. Capocelli, L. Gargano, and U. Vaccaro, “On the characterization of statistically synchronizable variable-length codes,” IEEE Trans. Inform. Theory, vol 34, no. 4, pp. 817–825, July 1988. [2] S. Even, “Test for synchronizability of finite automata and variable length codes,” IEEE Trans. on Inform. Theory, vol. IT-10, pp. 185–189, July 1964 [3] J. Berstel and D. Perrin, Theory of Codes, New York: Academic, 1985. [4] B. Rudner, “Construction of minimum-redundancy codes with an optimum synchronizing property,” IEEE Trans. Inform. Theory, vol. 17, pp. 478–487, no. 4, Sep. 1971.

A B C D E F G H I J K A B C D E F G A B C D E F G B I AA C JDF AB BKD BC C E AC BDH B CE EG FCHDD F GIDE E G A J EF FK FGG G A DC D F GE F G CKH BFABD GJK BA G BED CB CFE D EG CFE F G I JH A KIH B JI H I IJBAJE JKC K DCG CEH GF B ED HID KAA J A K A B C D E F G A B C D E F G AH B C I HECD JGJK F E G F G H I H J I K J K I A ABD JCAB KHDBCEH I J K A B C D E F G AFI D E F G A B C D E F G IIC JJ D KK E A KH BH ABD GJK F J GK j−1 J1 II K K1CEH HCEH I H GJK J H I K JABD K JJ K 0HGJK I ABD J CEH K H JGJK KH1n−j−1I HH CEH IIGJK D CEH ABD ABD ABD CEH AFI GJK CEH GJK AFI A E F G B CEHC GJK D A E BGJK F C GD ABD CEH ABD GJK ABD j−1A ABD n−j−1 ABD CEH GJK B C DI E F G ABD1 GJK CEH GJK AFI AFI 0CEH 1 ABD 1j−1AFI 1n−j−1CEH AFI CEH GJK AFI AFI 0 1 1 H J K I GGJABD K CEHn−j−2 GJK i−1 B EH 0CEH FH AACEH B DD1Dn−j−1 Fh−i−3 ABD CEH n−j−1 j−1 n−j−1 j−1 FGJK n−j−1 ABD AFI AFI AFI GJK 0CEH 0j−h+1 ABD GJK ABD CEH GJK AFI ABD GJK IG J K0 AFI 1 10j−1 AFI 1CEH 1 GJK 1C 1ABD 0AGJK 1CEH 1CC AFI 0B 1j−1 1EE 10ABD 11j−1 1n−j−1 10j−1 11n−j−1 j−1 n−j−1 j−1 n−j−1 j−1 j−1 n−j−1 Hi−1 K j−1 n−j−1 II0 I1An−j−1 JJ1J0 B KK j−1 n−j−1 AFI j−1 n−j−1 AFI 0 1 0 1 1 0AFI 1 AFI 1HH 1 1 1 C D E F G AFI h−i−3 j−h+1 n−j−2 AFI AFI 11 11h−i−3 11 j−h+11 1 n−j−2 1 1 0AFI 001 0i−1 0 0 0 0 0 0 j−1 n−j−1 ABD CEH GJK j−1 n−j−1 j−1 j−1 n−j−1 j−1n−j−1 j−1 n−j−1 ABD CEH GJK n−j−1 j−1 n−j−1 n−j−1 0Jh−i−3 1j−1 2 0c 11 CEH 1 1 GJK 1 1 0 10i−1 10 11Hh−i−3 j−h+1 n−j−2 i−1 n−j−2 1c011n−j−2 i−1 1 0I 1i−10h−i−3 100n−j−2 h−i−3 1·j−h+1 s1100Ks01100·0j−h+1 ·0n−j−2 s1j101sj−h+1 · · ·0sn−j−2 0i−1 0 0h−i−3 0ABD j+1 sj+2 n 0i−1 0 0h−i−3 0 0000j−h+1 000j−h+1 AFI AFI i−1 h−i−3 i−1 j−h+1 h−i−3 n−j−2 j−h+1 i−1 h−i−3 j−h+1 i−1 n−j−2 h−i−3 j−h+1 n−j−2 ABD2 00CEH i−1 h−i−3 j−h+1 ABD GJK i−1 h−i−3 j−h+1 00 0n−j−2 ABD CEH GJK n−j−2 0j−1 0 j−1 0 0n−j−2 0 0GJK00 1 00j−h+1 0 0 c CEH 00ch−i−3 0n−j−2 1 2 0 00 c 0 AFI c0 i000 n−j−2 n−1−i n−j−1 si−1 s11 · · · s1j sh−i−3 si−1 · · ·j−h+1 snn−j−2 0h−i−3 1 1 1n−j−1 0h−i−3 j+1 j+2 s s · · · s s s · · · s i−1 h−i−3 j−h+1 0 1 0 0 0 1 1 1 i−1 j−h+1 n−j−2 i−1 j−h+1 n−j−2 i−1 h−i−3 j−h+1 n−j−2 0 1 j j+1 j+2 n i−1 h−i−3 j−h+1 n−j−2 i−1 h−i−3 j−h+1 n−j−2 AFI AFI i−1 h−i−3 j−h+1 n−j−2 j−1 n−j−1 0 0 0 0 0 1 1 AFI 0 0 0 0 01 · · ·0s1 c1 20 c1 s020ss001s·c10ABD 0 CEH 0 c··s·cs·cGJK 0· ·00sc 0 00 00 i 1 n−1−i · · · s1j s2j+1ss00scj+2 s·c·0··ss··11·0ss··1j2j+1 ·ssj−1 ·c2j−1 ss1jcj+2 ss2cj+2·0s·s·0n−j−1 0sn−j−1 1 0 02 10c 00 01 1 2c 0c01 · · ·ssj+2 n 0· ·01·ss n1·n· ·0snn j+2 n−j−1 1 · ss js 0s 1 j+1 101nj+1 c 0jcs0j+1 10j+2 11sj−1 12 j+1 11ns110c 1jAFI c c· · ·s· ·ss cs11s12 ss0s0sccs11· ·······s·s001s1scs1122· ·s·scsc1js s · s s s · · · s s00s11 · · · s1j ss2j+1 s · · · · · 0 1s0j+2 ·1s · · · i−1 h−i−3 · · · s 0 1 j+1 j+2 j j+1 j+2 n n 0 1 j j+1 j+2 n n s · · · s s s · · · s 0 1 1 0 i−1 c0 n−j−2 s ···s 00n11· · s j j0j+1 j+2 nn 0 1 c · ·0·j−h+1 j+1 j+20 0 01h−i−3 j j+1 j+2 s 1s 0 0j−h+1 0n−j−2 0 1 10 21 0 c 10 11 2 n c0c10 12 c·c· scnsi−1 n11s c· ·0· s00100s2122111·1n−j−1 n · · j−h+1 2·s·10s·3ccn n−j−2 s··c·0cj+2 ·02j+1 1·s s·1ss1·cncnc··ss·0c·ss101jsh−i−3 sc22j+1sscj+2 · scn 0 · j+2 ·0s·ss0·1ssns·j+2 ··11·ss·11cj+2 scn 1s·s00·00s·s111s11j−1 1j s0j+1 · · s s · · · s· 0ss1j 1s·sj+1 js j+1 s s · · s s · · · 0 0 0 0 · · · s s s · · 0 0 0 1 ·s s s s · · s s s 0 1 j j+2 n 0 1 j j+1 j+2 · · · s1n s0ss001s·11····s·ssjs00sss1nj+1 · · · · s j j+1 j+2 n s0s1n·n· ·h−i−3 sh−i−3 0 1j−h+1 2j−h+1 h h+1 s30s1 · ·h−1 · sn−j−2 s0s1s2 · n· · sn 1i−1 n i−1 h−i−3 j−h+1 n−j−2 nn−j−2 1 · · · sn 0 01i−1 s00s11 · · · s1ns0s0s2s1 0·0s·00·ss11 · · ·sss0300010ss11c· · · ·s·1n· s00cs00s0000s0s11112· ··1·· ·ss001n01ns00011110· · · s031ns11002sc 11 c· c· · s1nc c c 0 1 1h−1 10 121 315. 01 A h n tree h+1 n010s0self-synchronizing s s s · · · s sstring · ··c·2···s·s··1ns·cnsnsc0s1 ·c· · s1j s2j+1scj+2 · · · scn 0 1 i−1 h−i−3 j−h+1 s · · · s s code with no s s · · · s sn−j−2 0s3i+2 1sj+2 0 1 10 1 0 Fig. 1 1 h+1 j1i+1 0cih10 sj+1 s01 ·0s·3·2sssn10cs01 ··2··1s··10sssnc1s3·0s·s10·0scss0n2ss0321s·010··2c··sc·11ss1s00s0ss311s011c·3·s··0c··s·3sc1n·sn0·0s1s·1·c110··2·s0·s··2·c03·s·scssn1n1000c1h−1 1· · s 0s 1s1j·css s · · · · s s · j+2 · · s·n· · sn s22s13 · · ·ss0s1h−1 0 1 j+1ns s s s s · · · s s s · · · s 0 i i+1 i+2 h−1 h h+1 s · s · · · s s s · · · s ·s·1·ss2sh−1 sn·00sh· 031s1sh−1 s01h+1 s2 ns03h−1 c· · hsh−1 c h2 ss30h+1 12s0h h ch+1 h+1 ccs·h−1 h+1 2s2h+1 ch 0 n n i ni+1 i+2 0 s01 s 3 ·ss ·213·1s1·31j2j+1 c3 c c 0 j+2 21···1·n c330 10c c 2n 31 cc c 1 0s322sc1s1·c0c·s·0 s·21·1s·1ss 00s0s011·21··s0··c·s0s c00ss0s j+1 j+2 s10h·h0· 1ssh−1 s23s0s·h·3· ···c·ss·sh+1 · 0s·h−1 s0s01s22s13 · ·s·0s·1h−1 ··1··sj+1 s0s·scnsj+2 ss33·s··ss3··nsn·scsns01h−1 1jsj 2s 0· · 1css 1 h+1 · · ·0 sn s s s s 0n·0 0·11·1 s2h−1 3 ·ns 1h+1 n·h· s0s hs s0h+1 11s 2h−1 h2·0s h+1 s s · · s · · s 2 · · s · · 0 1 3 n 0 1 2 3 h−1 h+1 n 1 2 3 h h+1 s s · hc c2 s h+1 n1i+1 · · · s s · · · s s s · · · 0 0 1 1 3 c c 0i+2 11 cc 31h−1 2 c s s s s s s s · · s 0 0 2 1 1 3 c c 0 1 n 0 0 2 1 0 0 120 10 3 0 i 2c1i+1 3 0 0 2 1 1 3 c c 0 i i+2 h−1 h h+1 n 1 c c 0 1 2 n−2 n−1 n 0 0 2 1 3 c c s·3·h·0··c·s0s1s1ch−1 ·n2·h+1 ·0s1shj32·1ss·0j+1 0s1·13s s s 0· · · sn s s s s·s01·0 s2· 1sssh−1 ·s0ss1·h−1 ·s·h+1 s01·c·1sn·ssc22·sj+2 3 sss·s c2·1 s 013 ·s·s0c0·0hs3·s102s 3s1s csscs0s 21s·s ns n2c2sss ·3··s·sscn·nn· sh−1 0h−1 1ss1n·2··0s s32h−2 ·s·3·sh·snh·s11·0·c·sss·s0ch+1 ·h+1 c1s ·····s·s··0·s·0·s3hss·sns·1h−1 1h−3 h−1 · · s0i s2i+1ss01i+2 ·1····s·s·sn0·i1sh1·sand ss1h−1 ·h+1 sJ. sh−1 ·1·h−1 sh+1 0·0sss h· s·s·h+1 nh+1 ·si+1 sch+1 s3h3ch+1 0··3h−1 s 0·Huffman s0 ch h+1 sn s001 0··2····3s[5] s0 i+1 ·Ferguson, ·s·s2s·sh+1 ·H. si+2 sih−1 ·T. · ·sh2s1J.0i13ss0shh+1 sshi+1 1·h−1 i+2 n·n·0h ii+2 Rabinowitz, “Self-synchronizing nh−1 i+1 h+1 nhs0h+1 ·nsnsi3si+1 21c c 1 3 n−1 c ·0h−1 2 i+2 311 0 cc2 20n 111 c3n−2 s0000·s0s·001111s·i+2 00s1c022· s c c· · ·s·1· sc s3 sc 3·sss 1·s··10·2·ss n10ss02c ·3· ·1·s·c0·ss2c ss11s00c· ·· ··3·s · · s s s · · · · s · s s s ·h+1 s0 · · · s0i scodes,” s · s s · · · s 0 s · · · s s · · · 0 2c 1 h−1 c · · · snc s s s · · · s s s · · · 0i+1 i+1 03 h−1 i+2 h−1 i+2 h s0ss h s3 h+1 i+1 i+2 hsi s0i+1 h+1 i ·i+1 nTheory, i+2 hi+2 h+1 ssh−1 ·Trans. · · s · · s s · · · 0h−1 0 isih−1 230, 1i s c nhih3 i+1 c 4, 0 0 2 1 1 c s · · · s s s s · · · s IEEE Inform. vol. pp. 687–693, no. 0 h+1 n 0 i+1 i+2 h−1 h+1 n 0 i+2 h h+1 n s s s · · · 3sn1h−1 0 h−3 h−2 h−1 h h+1 n s s · 0s02200·c ·c·11sh−3 11s1s2h−1 3·s· h·c10s c1h−1 s s s s 0·· s 2cc·scs 1 ·0· ·1ccsc2n 31 c h sh+1c · · · sn 0 0 20 1 00 2 1 10 2s30 s11·c1· · s13s001·cc· · s 0 2 3 h−2 h+1 n 3 c 0 2 3 0 1 3 h h+1 0 3 c 0 0 2 1 1 3 c s s · · · s s s · · 0 0 2 3 c c ·nn· s·h−1 s0 ·0s· 1· si ss03i+1 s·ci+2 s·····sss2·0sh−1 ·0·0h+1 s·s·sn···i+2 csi+2 0s 2·s 1n·h+1 ·i+2 ··11··ss·h+1 nh+1 0·0s·s1s 2·s cs·0h·hs shh−1 sni+1 ···1··s·ssh+1 0h+1 ·s3hsssh10cci3h·is·s·i+1 ··ncn1s·ss·h−1 h−1 ss·hi+2 · · ss1 nh sh+1 · · · sn 2· ss 30 s 0 ·Jul. 0i··s 3h+1 csn·h ·sci+1 ·s··3·s·ssn·1hsh·issh−1 i+2 h+1 0·1i+1 h−1 · · s0h−3s2h−2 ··s1i+1 ·2s1·sh10·ch+1 s1s··0h−3 ·s·scni+1 si+1 s1·s2h−2 scs·ss10s30sh−1 2h−1 ·i+2 s3hch+1 s··s·ccs··is·i·ssh−1 si+2 h−2 h+1 s0 0·h−1 · · ssh−3 ·h+1 · 2s1984. sh−2 ·h−1 1·sh−3 1h+1 3· s·cs h0 ·ss n0·h0i02·s 00s csh−1 h−3 h·s n2··h+1 2 02c33 s ss12322s0sh−1 s0·h−3 s0·h−2 ·ccn1cnsh+1 0s12cc2 n0 1s 1cn−2 0 3binary 2ccs 1 c 3sn c sn2·ss3hh−1 chsh−2 0ss s00100sMontgomery, B.h−2 L.sh−1 and Abrahams, 1·s cof c 0n−1 3·c···3·s h−1 hJ. h+1 ns030s“Synchronization h−1 0·s 1h+1 3s·s0··0s··s1·s h−1 h+1 1 · ss 1c0s 1h·· s 1· s0n · · s · · s · s s s s · · s sh+1 · · ·1 scn 3 c s0 · · [6] · s0h−3 s s · · · s s · · s 0h−2 0s011s 223·3s cs cs · · s s s · · s · · s s s s · · s 0 h−3 h−2 h−1 h−3 h−2 h+1 h−1 h−1 h h+1 0 n h−3 h−2 h−1 h h+1 n s · · s s s · · · s 0 0 2 3 c ss2 n−2 s sno. 1 nsn3 · s s0 ·01· ·codes,” sh−2 s 1IEEE s1ch+1 ·1h−1 · snc0s3h00Theory, h−3 h−2 hsh1· ·ch+1 h−3 h−1 h+1 nh−2 ·ch+1 s3i s·h2i+1 scn1i+2 · · · s s s · · · scn 1csh−2 ·s02·2cs·h+1 s00sh−1 ssshc10ich−1 0·Inform. 0 2s 0h 2 Trans. · s s · · 0 ·i+2 2 c6, ccs 0s11·sn 0 0 0 2source 23 0 0ch−1 3h−3 c· h−2 pp. 0·101· s·vol. 2·c3223· 32, 3849–854, c2·c··ss·0n· 1n−1 0 h+1 0 1 3 c 0 1 3 c i+1 h−1 h 0 0 c 0 0 3 s · · · s s s · · s s s s s · · s s s · · s s s ······sscnc h−1 h h+1 s · · · s s s · · · s s s s · s · · s s · · · s c0 0 h−2 c0 3· · 2·h−2 ·s··c·ns··scs·h−3 s·s·h·h0s· isssnh+1 ·i+2 ····sh−2 2h−1 3h ch+1 s s s s s 02c·3·h−1 h−3 h−2 h−1 h ssh+1 h+1 nh+1 03n2h+1 3cn c··h−2 cssh−2 0·00h−2 10chs h·s·s·0·s h+1 n · · s h−3 h−1 hh sh+1 s · s 0 0s3 h−3 h−3 h−1 h0s h+1 h−1 · · s · · · s s 2· s c·0h−3 c2ss 0 3 h−1 n 0 h−3 h−2 h−1 n · · s0h−2s2h−1 s · · · · s s s · · s · · s s s s 0 h−3 h−1 h 0 i+1 h−1 h+1 n s · · · s s s · s s0 0·h· · sh+1 s· h−1 sn1986. s0·h+1 sh−2 · 2sh−2 ·1h−1 · ··hs·h−1 2sn1h−1 1h 0 s0h0· 0·h−2 h+1 h+1 n 2s 1s0h−2 1s cscn ·c · 0·ns·ccsc 0c 1 02 0 ·Nov. h+1 nhs ·iiss0i·2i+1 s·h−2 ·1h−1 c shs003s00s·h−1 ss3ch+1 sc3hs3hss03hh+1 0c·0s· ···s2·n·2n··nss 33ss c·c ·s·3s0scc c s2· · · ssc3 sc 0s 0· ··c·s0s i+1 i+2 h−1 h+1 3h−1 i+1 h+1 · · · s s0 · · [7] · s0h−2 ·s0ssi+2 scni+2 ··2··D. ·· s·ss0h−2 sss2h−1 · · · s s s s · · · s 0 0 2 1 c c c c c · · · s s s s · · · s 0 h−2 0 h−1 h h−2 h+1 h−1 n h h+1 n h ss h+1 0 h h+1 n · · · s s · · · s R.sh−1 M. sCapocelli, A. A. Santis, L. Gargano, and U. Vaccaro, “On 0 0 2 1 3 c c 0 h−2 h−1 h h+1 n 012 h+1 n h−2 s0n−1 sh3 cscccch−2 ss ccsh−1s3h sch+1 · · · scn 0 2h−1 3303nh−2 ss0sc2c2·cc2·h−1 · ·ss··3h−3 0 ssh−1 3· ·· ·· ·scs h−3 030 c2 0 0s c·1·h· s·00·h+1 cscchs12sh+1 30···s n nh−2 0s00s 2·h−1 ·of ·statistically 1 0 s·h+1 · ····n−2 sn s00 · · · s0h−2 s00s·2h−1 · · sconstruction ·s·3h·s00 s2h−1 scn·ch+1 ·s··c·sss0h−2 s·cnsnh+1 ·s·2h−2 s0h−2 the codes,” IEEE Trans. In-ss3hhscssch+1 sh−2 ·h+1 ·h−1 sh−2 ss3hi·s·i+1 s1·sch+1 · ·ssynchronizable ·h−2 ss·scn00ss00s·h−1 h−2 h s h+1 n hhh−2 h+1 s s s 0sh−3 h−1 h+1 ·· ·· ·· s · · s · · · s h0 ·ss ni+2 · s s h−1 h n s s s h−1 h n 0 h−1 h+1 0 h−1 n n−2 n−1 n 2 11 1 33 31 c1c c 1 2 cc c 3 0s0 s238, 2s 1 1 h−2 sh−1 sh+1 ss0000s·00·· ··1·ss·0h−3 spp. sshhshssh+1 ·no. form. Theory, vol. 407–414, 1992. sh−2 sh−1 ·· ·· ·s2, s·nnsMay. h−3 n h−3 h−2 h−1 h+1 0 c 0 12 3 c c 0 0 2 3 c 1 1 0 1 s · · · s s f 1 1 scodes 1 1 3 Huffman c s n· · ·cequivalent [8] A. E. Escott, sand· · S. Perkins, with sh−2 sh+1 · · 0· sn h−2c h−1sh sh+1 · · · sn 0 sh−1 ·1s0h−3 s2h−2s“Binary 0 h−1 sh fsh+1 ·0· · s · s110h−2hs2h−1 s3h44,sch+1 · · ·1 sn 1 n s0 · · Theory, 1 1 n a short 0synchronizing codeword,” IEEE Trans. Inform. vol. u 1 0 0 2 3 c c 0 2 3 c c 0s s 2s s 3s s cs · · · s· cs ss000s·0·····ss·h−2 sh−1 shh1998. h−2 h−1 h sh+1 h+1· · · snnun h−2 h−1 h+1 no. 1, pp. 346–351, Jan. s 0 0 2 3 1 s · · · s s sch+1s · · · scnand 1 all [9] C. F. Freiling, D. F. Theberge, 0 S. Jungreis, h−2 h−1 sh xnsu K. Zeger, “Almost 1 complete binary prefix codes have a self-synchronizing string,” IEEE n xsu V (y|x) 11 1 Trans. Inform. Theory, vol. 49, no. 9, pp. 2219–2225, Sep. 2003. yn 1 V (y|x) [10] V. K. W. Wei, R. A. Scholtz, ”On the characterization of statistically n y synchronizable codes,” IEEE Trans. Inform. Theory, vol. 26, no. 6, pp. 733–735, Sep. 1980. 1 [11] Y. Takishima, M. Wada, and H. Murakami, “Error states and synchro1 nization recovery for variable length codes,” IEEE Trans. Communications, vol. 42, pp. 783–792, no. 2, Feb. 1994. [12] W.-M, Lam, and S. R. Kulkarni, “Extend synchronizing codewords for binary prefix codes,” IEEE Trans. Inform. Theory, Vol. 42, pp. 984–987, No. 3, May. 1996. [13] M. R. Titchener, “The synchronization of variable-length codes,” IEEE Trans. Inform. Theory, vol. 43, pp. 683–691, no. 2, Mar. 1997. [14] G. Zhou, and Z. Zhang, “Synchronization recovery of variable-length codes,” IEEE Trans. Inform. Theory, Vol. 48, pp. 219–227, No. 1, Nov. 2002. [15] T. Honda and H. Yamamoto, “Synchronizing Strings of FV Codes and Identification of FV Code Trees,” AEW4, pp. 31–34, Viareggio Italy, Oct. 6-8, 2004. [16] Y. Tashiro and H. Yamamoto, “FV Code Trees with no Synchronizing String,” AEW5, Jeju, South Korea, October 25-27, 2006.