PERFECT ONE-ERROR-CORRECTING CODES ON ITERATED COMPLETE GRAPHS: ENCODING AND DECODING FOR THE SF LABELING

Transcription

1 PERFECT ONE-ERROR-CORRECTING CODES ON ITERATED COMPLETE GRAPHS: ENCODING AND DECODING FOR THE SF LABELING PAMELA RUSSELL ADVISOR: PAUL CULL OREGON STATE UNIVERSITY ABSTRACT. Birchall an Teor prove that every iterate complete graph has a perfect one-error-correcting coe (PECC) an showe how to construct it []. Kleven create the SF labeling metho, a metho for assigning strings to the vertices of iterate complete graphs which has several nice properties [5]. We use these results to create a working PECC on these graphs. That is, we present a metho for encoing an ecoing: an algorithm which gives a bijection between the set of messages which can be transmitte using the PECC on an iterate complete graph, an the set of coewors in the SF labeling of that graph. In the process, we introuce a technique which shoul be useful in creating encoing an ecoing methos for any reasonable labeling of iterate complete graphs.. INTRODUCTION Many factors can cause errors in the transmission of messages. It is for this reason that error-correcting coes have been evelope. An error-correcting coe is esigne to recognize that the receive message contains an error, then fin the error an correct it. A biological example of an error-correcting coe is written communication by humans. If an English speaker reas the wor INFORMETION, he will automatically correct it to INFORMATION. This paper is concerne only with igital coes. In particular, we look at one-error-correcting coes on graphs. In this context, wors are represente by vertices in the graph, an each wor that contains an error is ajacent to the corresponing wor containing no errors. These coes come with algorithms for recognizing whether a wor contains an error an correcting the error if it oes. So in our example above, the vertex labele INFORMETION woul be ajacent to the vertex labele INFORMATION, an we woul have an algorithm which woul locate the error in INFORMETION an correct the E to an A. We look at coes on a particular family of graphs, iterate complete graphs. Much is known about coes on these graphs. In particular, every iterate complete graph has exactly one perfect one-errorcorrecting coe up to isomorphism []. Kleven gave a metho for assigning strings to the vertices of any o-imension iterate complete graph, calle the SF labeling metho, an also gave finite-state machines which recognize an correct errors in a given string [5]. Our contribution is to give an algorithm for encoing an ecoing messages. That is, we give a bijection between the set of istinct messages which can be sent using a given iterate complete graph, an the set of coewors in the SF labeling of that graph. Date: August 004. This work was one uring the Summer 004 REU program in Mathematics at Oregon State University. 6

2 Encoing an Decoing for the SF Labeling 63. BACKGROUND: ITERATED COMPLETE GRAPHS AND ERROR-CORRECTING CODES.. Iterate Complete Graphs: Definitions. Definition... A (simple) graph G = (V,E) consists of a finite set V (calle vertices) an a set E (calle eges). Elements of E are unorere pairs of elements of V. Two vertices v an v are ajacent (have an ege between them) if(v,v ) E. The ajective simple inicates that any two vertices have at most one ege between them, an that no vertex is ajacent to itself. Definition... The egree of a vertex v is the number of vertices which are ajacent to v. Definition..3. The complete graph on vertices, enote K, is the graph such that all the vertices are pairwise ajacent. Figure shows K 3, K 4, an K 5. FIGURE. K 3, K 4, an K 5. Definition..4. The secon-orer iterate complete graph, enote K, is constructe by making copies of K, then connecting each pair of copies by one ege such that no vertex ens up with egree >. The orer n iterate complete graph, enote K n, is constructe by making copies of Kn, then connecting each pair of copies by an ege between a corner vertex of one copy an a corner vertex of the other copy, such that no vertex ens up with egree >. (Note: a corner vertex of K n is a egree vertex.) Figure shows K 5, K 5, an K3 5.

3 64 Russell.. Labelings. FIGURE. K 5, K 5, an K3 5. Definition... A labeling scheme for K n is a metho of assigning strings of length n over {0,, } to the vertices of K n such that this metho gives a bijection between vertices an strings. Most labeling schemes have avantages an isavantages. This paper eals only with the SF labeling, which has several esirable properties as we will see later [5]. Examples of other labeling schemes are the C-R-E-L labeling [], the α-metho [], the Multiples of + coe [], an the C-S labeling metho [4]..3. Coes on Graphs; Perfect One-Error-Correcting Coes. Definition.3.. Let G be a graph an let V be the set of vertices of G. Then a coe on G is a subset C V. A coevertex is a vertex c C. A noncoevertex is a vertex v / C. If we have a labeling of G, then a coewor is the label of a coevertex. A noncoewor is the label of a noncoevertex. Definition.3.. A perfect one-error-correcting coe (or PECC) on a graph G is a coe such that: () No two coevertices are ajacent. () Every noncoevertex is ajacent to exactly one coevertex. Most graphs have the property that no subset of their vertices is a PECC. In fact, Cull an Nelson showe that the problem of etermining whether a given graph has a PECC is NP-complete [3]. However, Birchall an Teor showe that every iterate complete graph has a PECC an that this PECC is unique up to isomorphism []. In Section.4, we give Alspaugh et al. s construction of the PECC on K n []. Figure 3 shows three examples of PECC s on iterate complete graphs. Definition.3.3. A coewor recognizer for a PECC is an algorithm for etermining whether a given label is a coewor. An error correction machine sens a noncoewor to its corresponing coewor.

4 Encoing an Decoing for the SF Labeling 65 FIGURE 3. Perfect one-error-correcting coes on K, K 3, an K 5. Coevertices are represente by squares. The ajective perfect one-error-correcting refers to the iea that the coewors are actual messages that one might wish to transmit. An error may be mae in transmission, ue to human error, interference, noise, etc. The set of vertices which are ajacent to a particular coevertex represents the set of errors that may be mae when sening that message. The coe is one-error-correcting because every noncoevertex is at a istance of one from a coevertex. If more errors are mae, so that the actual transmitte message is no longer ajacent to the esire coevertex, then the message will be correcte to a ifferent coewor. Figure 4 shows an example where the two possible messages are LAND an SEA. An error-correction machine for this graph woul correct the strings LQND, LADN, an LLLD to the coewor LAND. No other string woul be correcte to LAND. LQND SEW LADN LAND SEA SEE LLLD ESA FIGURE 4. A PECC with two coewors (represente by rectangles). Definition.3.4. Let G be a graph. A labeling of G with the gray coe property is a labeling such that any two ajacent vertices have labels which iffer in exactly one position. One esirable property of the SF labeling is that it is a gray coe. The labeling in Figure 4 oes not have the gray coe property. Figure 5 shows a labeling with the gray coe property.

5 66 Russell LARD SEW LEND LAND SEA SEE LANE SET.4. The G-U Construction of a PECC on K n. FIGURE 5. A labeling with the gray coe property. In this section, we give an iterative metho for constructing the PECC on K n. This metho will be a cornerstone of our encoing an ecoing scheme evelope in sections 5, 6, an 7. This G-U construction is ue to Birchall an Teor []. However, we prefer the presentation given by Alspaugh et al. an we use it here []. The G-U construction uses two types of coes on K n : G-coes an U-coes. The G-coe is the esire PECC (Theorem.4.). Let G n enote Kn with the G-coe an let U n enote Kn with the U-coe. G n an U n are constructe recursively as follows: To construct G, esignate one vertex of K as the top vertex an rotate it to the top position. Make this vertex a coevertex. Make the other vertices noncoevertices. To construct U, esignate one vertex of K Make all vertices noncoevertices. Figure 6 shows G 5 an U 5. as the top vertex an rotate it to the top position. FIGURE 6. G 5 an U 5. We now show how to construct G n an U n for arbitrary n: To construct G n when n is even: () Make copies of G n. () Connect each pair of copies by a vertex such that the top vertex of every copy remains unconnecte. (3) Designate the top vertex of some G n as the top vertex of G n.

6 Encoing an Decoing for the SF Labeling 67 To construct G n when n is o: () Create one copy of G n an copies of U n. () Connect the top vertices of the copies of U n to istinct nontop corner vertices of G n. (3) Connect each pair of copies of U n by one ege such that This ege connects a nontop corner vertex in one copy to a nontop corner vertex in the other copy. Exactly one nontop corner vertex of each U n remains unconnecte. (4) Designate the top vertex of G n as the top vertex of G n. To construct U n when n is even: () Make one copy of U n an copies of G n. () Connect the top vertices of the copies of G n to istinct nontop corner vertices of U n. (3) Connect each pair of copies of G n by one ege such that This ege connects a nontop corner vertex in one copy to a nontop corner vertex in the other copy. Exactly one nontop corner vertex of each G n remains unconnecte. (4) Designate the top vertex of U n as the top vertex of U n. To construct U n when n is o: () Make copies of U n. () Connect each pair of copies by a vertex such that the top vertex of every copy remains unconnecte. (3) Designate the top vertex of some U n as the top vertex of U n. Figure 7 shows G 5 an U 5. Figure 8 shows G3 5 an U 3 5. FIGURE 7. G 5 an U 5.

7 68 Russell FIGURE 8. G 3 5 an U 3 5. Theorem.4.. The G-coe is the unique (up to isomorphism) PECC on K n. Proof The proof is given by Birchall an Teor [] an is omitte here..5. Encoing an Decoing. Birchall an Teor [] showe that the number of coevertices c n in a PECC on K n { is n + + c n =, n even n + +, n o. One can therefore use this coe to transmit as many as c n ifferent messages. Label the possible messages by the first c n natural numbers, i.e. the set {0,,c n }. Definition.5.. An encoing an ecoing scheme for a particular labeling of K n between the set{0,,c n } an the set of coewors in the labeling. is a bijection A useful encoing an ecoing scheme works for all an n. This way, epening on the number of messages one wishes to be able to encoe, one can choose suitable an n such that c n is large enough. It is not necessary to generate the whole labeling an assign messages to coewors; one simply selects the message one wants to sen an the encoing scheme returns the corresponing coewor. After the coewor is transmitte, the receiving party runs the coewor recognizer, the error corrector if necessary, an finally the ecoer. Some labelings have simple encoing an ecoing schemes. For example, the SF labeling has a nearly trivial scheme when = 3. However, this oes not appear to be the case for the SF labeling with > 3. In Sections 5 an 6, we lay out a technique that shoul be useful in fining encoing an

8 Encoing an Decoing for the SF Labeling 69 ecoing schemes for any reasonable labeling. In Section 7, we use this technique to give an encoing an ecoing scheme for the SF labeling with arbitrary an n. 3. THE SF LABELING OF K n In this section, we escribe the construction of the SF labeling. (In Section 4, we show that there is no similar labeling of K n when is even.) We then give finite state machines for coewor recognition an error correction. 3.. The SF Labeling. The SF labeling is only efine when is o. Let 3 be an o number. The labeling of K n constructe recursively from the labeling of K n. is Label K as follows: the top vertex is labele 0, then the remaining vertices are labele,,,( ), going counterclockwise. Figure 9 shows the SF labeling of K FIGURE 9. The SF labeling of K 5. The SF labeling of K n is constructe accoring to the following algorithm: Apply the permutation α to each igit in every label of K n, where α(z)= + z (mo ). Now make copies of α(kn ). Rotate the k th copy πk raians counterclockwise, then appen k to each wor in this copy. Finally, connect the copies to form K n. Figure 0 shows the SF labeling of K 7. Figure shows the SF labeling of K3 5. All results in Section 3 are ue to Kleven [5].

9 70 Russell FIGURE 0. The SF labeling of K FIGURE. The SF labeling of K 3 5.

10 3.. Description of the Coewors. Encoing an Decoing for the SF Labeling 7 No explicit escription of the coewors has been foun. Kleven gives a recursive escription. Let G n enote the set of coewors in the SF labeling of K n, an let U n enote the set of SF labels of vertices in the U-coe on K n. Then, When n is even, G n = G n 0 T(G n ) T (G n ) T (G n ) ( ) U n = U n 0 Γ (G n ) Γ (G n ) Γ (G n ) ( ) An when n is o, G n = G n 0 Γ (U n ) Γ (U n ) Γ (U n ) ( ) U n = U n 0 T(U n ) T (U n ) T (U n ) ( ) where G n 0 means the set G n with a zero appene to every string in the set, T replaces each character x by x+ (mo ), T m means T compose with itself m times, an Γ m replaces each character x by α(t m (x)) Coewor Recognition for the SF Labeling. Kleven gives a (+ )-state machine for coewor recognition. The states are { E S, E 0, E,, E ( ), O S, O 0, O,, O ( ) }. The machine starts in state E S an reas strings from left to right. Strings of even length are accepte in the E S state an strings of o length are accepte in the O 0 state. The function δ etermines transitions among the states: δ(x y,z)=(δ (x,z) δ (y,z)) where δ an δ are given by: δ (E,z)=O δ (O,z)=E δ (x,z)=(x z) (mo ) δ (x,x)=s δ (S,z)=z Kleven proves that coewor recognition is performe correctly by this finite state machine.

11 7 Russell 3.4. Error Correction for the SF Labeling. Kleven showe that the SF labeling is a gray coe (see Definition.3.4). Therefore, error correction involves changing exactly one igit. The error correction algorithm consists of two small algorithms. The first algorithm is a finite-state machine; it is use only when the error is in the first igit. It starts in the S state if n is even an the O state if n is o. The S state is the only non-final state. The function δ etermines transitions among the states: δ(x,z)=( x+z ) (mo ) δ(x,x)=s δ(s,z)=z The secon algorithm is use when the error is not in the first igit. To correct the wor x x n : IF x x n / U n Correct x = q, where q is the final state after running x n x through the first machine ELSE FOR i= n IF x i x IF i n Correct x i = x x i (mo ) ELSE Correct x n = 0 BREAK END FOR Kleven proves that error correction is performe correctly by these algorithms. 4. GENERALIZED TOWERS OF HANOI LABELINGS It is natural to ask whether K n amits a labeling similar to the SF labeling when is even. We will specify what we mean by similar to the SF labeling. We will then show that in fact, K n ( even) oes not amit any such labeling.

12 4.. The Towers of Hanoi Labeling of K n 3. Encoing an Decoing for the SF Labeling 73 The Towers of Hanoi is a popular puzzle. It consists of three pegs an several isks of ifferent raii which fit on the pegs. A solution to the puzzle is a configuration where all the isks are on one tower (see Figure for a picture). One is allowe to move one isk at a time, with the constraint that no larger isk may be place on top of a smaller isk. FIGURE. A solution to the Towers of Hanoi puzzle with five isks. We can escribe a configuration of the Towers of Hanoi puzzle by a string of length n over{0,, }, where n is the number of isks. The i th igit represents the position of the i th isk (first igit = smallest isk, n th igit = largest isk). The position of a isk is 0 if the isk is on the leftmost tower, if it is on the mile tower, an if it is on the rightmost tower. See Figure 3 for an example. The configuration shown is 0. FIGURE 3. A configuration of Towers of Hanoi puzzle with five isks. The associate string is 0. One can construct a graph where each vertex is labele with a configuration of the Towers of Hanoi puzzle an each ege represents a legal move in the puzzle. It turns out that this graph is actually K3 n, an the labeling is known as the Towers of Hanoi labeling of K3 n [3]. Furthermore, the resulting labeling is the SF labeling for = 3 [5].

13 74 Russell The rest of Section 4 examines the extent to which the SF labeling of K n, > 3, can be viewe as a generalization of the Towers of Hanoi labeling of K3 n. 4.. Definitions. Definition 4... The Generalize Towers of Hanoi puzzle on towers an n isks, enote GTH(,n), is the puzzle with the following rules: () Only one isk may be move at a time. () No larger isk may be place on top of a smaller isk. The towers are labele A configuration of GTH(,n) is a string of length n over {0,, }, where the i th igit gives the position (tower) of the i th isk (first igit = smallest isk; n th igit = largest isk). Definition 4... Let G be any graph with at most n vertices. A GTH(,n) labeling of G is a one-to-one map from the set of vertices of G to the set of configurations of GTH(,n), such that each ege represents a legal move in GTH(,n). Note: The SF labeling of K n, with o, is a GTH(,n) labeling. We prove this in Section 4.4. Definition The Maximal GTH(,n) graph, enote M(,n), is the labele graph on n vertices with eges corresponing to every legal move in GTH(,n). Figure 4 shows M(4,) FIGURE 4. M(4, ).

14 4.3. Nonexistence of a GTH Labeling for Even. Encoing an Decoing for the SF Labeling 75 We now show that there is no GTH(,n) labeling of K n when is even. Lemma If there is no GTH(,n ) labeling of K n, then there is no GTH(,n) labeling of K n. Proof We prove the contrapositive. Suppose we have a GTH(,n) labeling of K n. Recall that Kn consists of copies of Kn. Call them C 0,..., C. Delete all the vertices in C,...,C so that only C 0 remains. Now elete the n th character in each vertex label in C 0 (this correspons to removing the largest isk). The result is a GTH(,n ) labeling of K n. Theorem For even, there is no GTH(,n) labeling of K n. Note: We are assuming that n> since there is always a GTH(,) labeling of K. Also, we assume that > since a GTH puzzle on two towers is not interesting. Proof We prove the theorem for n = (two isks). The general result follows by inuction, using Lemma Let > be even. We first istinguish between two types of eges in M(,): Type S Eges: eges which correspon to moving the small isk Type L Eges: eges which correspon to moving the large isk We try to construct a GTH(,) labeling of K. In other wors, we must elete some of the eges in M(,) so that the resulting graph is K. We will see that this is impossible. Note that M(,) contains exactly copies of K, corresponing to the possible positions of the large isk, from which the small isk may be move to any tower. These copies of K will have to be the copies of K in K. Call them C 0...C an keep their labelings. (Note that all the eges within C i are Type S eges, an so far all the vertices in C i have egree.) Now all that is left is to connect each C i an C j, i j, by a Type L ege. This will require a total of ( ) Type L eges. If we look at only the Type L eges in M(,), we see that they take the form of istinct complete graphs on vertices. (Figure 5 illustrates this for M(4, ).) This is because there are ways to fix the position of the small isk, an then the large isk may be move freely among the remaining towers. Call these complete graphs D 0,...,D. Recall that we are trying to connect each C i an C j by a Type L ege, so we nee to use ( ) of the Type L eges which are foun in the D i s. This forces us to use at least ( ) D i. Since is even, this means that we must use at least eges from some D i. But D i has only = eges from some vertices, so some vertex must belong to two of the chosen eges. Thus, this vertex will have egree at least + in our constructe graph, which is impossible in K. Note: The reason we can fin a GTH(,n) labeling of K n when is o, is that ( ) integer which is equal to half the number of vertices in D i, so no vertex nee be use twice. = is an

15 76 Russell FIGURE 5. The type L eges in M(4,) form four triangles The SF Labeling is a GTH Labeling. In this section we prove that the SF labeling of K n is a GTH(,n) labeling. We will nee two lemmas. Lemma For all i {0,, }, we have + Proof We erive this from a tautology: (mo ) = + (mo ) = + (mo ) = + = + + (mo ) i (mo ). i i + + i i i (mo ). Lemma The SF labeling assigns the strings 00 }{{ 0},,, ( )( ) ( ) to the corner vertices of K n n times in orer, starting with the top vertex an going counterclockwise. Proof We know by efinition that the lemma is true for n=. We prove that it is also true for n=. This is sufficient to obtain the lemma for all n by inuction, since there is only one igit of information containe in the wor ii i. Notation: We say that a vertex is in position i of K if it is the ith vertex, starting with zero at the top an counting counterclockwise. We say that K contains copies of K. We call them C 0,, C, starting with C 0 at the top an counting counterclockwise. So the j th corner vertex of K is the vertex which is in position j of C j. We want to show that the j th corner vertex of K is labele j j. We go through the construction of K to show that this is in fact true. To o this, we begin with the i th vertex of K an follow it through the construction. The i th vertex of K is labele i. The first step in the construction of the SF labeling relabels this vertex + i (mo ). In the next step, we create copies of the relabele K. In the thir step, we rotate the kth

16 Encoing an Decoing for the SF Labeling 77 copy, C k, by k positions clockwise. This means that when k is equal to + i (mo ), the vertex labele k (or equivalently, the vertex labele + i (mo )) is in position i k. But by Lemma 4.4., we have that k i k. So the vertex labele k is in the k th position of C k (this is the k th corner vertex of K.) In the fourth step, we appen k to each vertex in C k, so that kk is the k th corner vertex of K. We can now show that the SF labeling is a GTH labeling. Theorem The SF labeling of K n is a GTH(,n) labeling for all 3. Proof We fix an prove the theorem by inuction on n. The result is obvious for n=. Now assume that the SF labeling of K n is a GTH(,n ) labeling. We examine the construction of K n. Note that after the first step in the construction (permuting the igits in the labeling of K n ), the resulting labeling is still a GTH(,n ) labeling. This is because permuting the igits in a configuration of the GTH puzzle is analogous to gluing all the isks to the towers they are currently on, then shuffling the towers aroun, isks an all. Legal moves are sent to legal moves. So we nee only worry about the eges that will be introuce between istinct copies of K n. Call these copies D 0,, D. Now, the clockwise rotation scheme guarantees that the ege between D i an D j (i j) will connect two vertices with ientical labels. Furthermore, these two vertices are corner vertices of D i an D j respectively, but they are not corner vertices of K, so by Lemma 4.4., they are labele aa a for some a i j. The final step in the construction appens i to one vertex an j to the other, leaving an ege between aa ai an aa a j. This represents a legal move since a i j. 5. AN INDEXING SYSTEM FOR THE CODEVERTICES IN K n In this section we give a metho for representing a coevertex in K n by an (n )-tuple over {0,, }. The avantage of this technique is that the(n )-tuple contains explicit information, in a simple form, about the position of the coevertex insie K n. This representation will later serve as an intermeiate step in our encoing an ecoing scheme. 5.. A Scheme for Representing Coevertices by Vectors. Let C n be a PECC on Kn, let c n be the number of coevertices in C n, an pick v Cn. We give a recursive algorithm for assigning an(n )-tuple(w,w,,w n ) to v.

17 78 Russell Algorithm. First of all, K n contains copies of Kn. Label them 0,, with 0 at the top. If v is containe in copy i, then let w = i. This K n contains copies of K n. Label them 0,, with 0 being the top copy, where top is taken in the sense of the G-U construction. If v is containe in copy j, then let w = j. In general, suppose w k (k < n ) is given, so that v is containe in the w th k copy of K n k insie a copy of K n k+. If v is containe in the l th copy of K n k insie this K n k (where the 0 th copy is the top copy, with top taken in the sense of the G-U construction), then let w k+ = l. Suppose w n is given. There is a small change at this step. If v is containe in the p th copy of G insie a copy of K, then let w n = p. Note that if this copy of K is a U, then the pth copy of G is the(p+) st K subgraph. con- Note: No two istinct coevertices have the same corresponing(n )-tuple since a copy of K tains at most one coevertex. Definition 5... We say that an (n )-tuple η over {0,, } represents a coevertex in C n if there is some v C n whose corresponing(n )-tuple is η. 5.. A Map Between Natural Numbers an Coevertices. Let Γ enote the set of(n )-tuples over{0,, }. We efine a function by Φ :{0,,, n } Γ. m (m in base, but written backwars). We give an explicit algorithm for Φ below. (Φ will later be pare own to a bijection between the set {0,,c n } an the set of(n )-tuples which represent coewors in K n. This bijection will provie a big piece of our encoing an ecoing scheme.) Let m {0,,, n }. The(n )-tuple is prouce accoring to the following algorithm: ALGORITHM t 0 = m v = t 0 (mo ) k= WHILE k<n t k = t k v k v k+ = t k (mo ) k=k+ END WHILE Φ(m)=(v,v,,v n )

18 Encoing an Decoing for the SF Labeling 79 Notation. Let M enote the set {0,,,c n }. Let C Γ enote the set of (n )-tuples over {0,, } which represent coevertices in K n. Now that Φ is efine, we will procee by the following steps: () We first show that Φ is a bijection (Proposition 5.3.3). () Next we show that if m M, then Φ(m) C (Lemma 5.3.6). (3) These results give the theorem that Φ M is a bijection between M an C (Theorem 5.3.7). (4) Finally, we give an explicit expression for the inverse map Φ : C M (Proposition 5.3.8) Properties of Φ. Our first goal here is to show that Φ is a bijection. This requires two lemmas. Lemma Let 0 m m n. Then Φ(m m )=Φ(m ) Φ(m ) (where subtraction is componentwise). Proof First of all, (m m ) (mo )=m (mo ) m (mo ). = v (m m )=v (m ) v (m ). Furthermore, t (m m )= m m v (m m ) = m m v (m )+v (m ) (by the above) = m v (m ) m v (m ) = t (m ) t (m ). The rest of the algorithm is completely etermine by v an t, so the lemma is prove. Lemma Let m {0,, n } an Φ(m) = (v,v,,v n ), an suppose that v = v = =v k = 0 for some k n. Then m 0 (mo k ). Proof We first show that for all 0 j k, we have t j = m. This is clearly true for j = 0 since j t 0 = m. Now suppose t l = m for all 0 l j. Then t l l+ = t l v l+. But v l+ = 0 since j k, so t l+ = t l = m. l+ We can now prove the lemma. Suppose that v ==v k = 0. Then by the above, we have t k = m. k We also have v k = 0, i.e., t k 0 (mo ). So m 0 (mo ). Therefore m is ivisible by k. k Proposition Φ is a bijection between {0,,, n } an Γ.

19 80 Russell Proof The preimage set an the target set have the same carinality, so it is sufficient to show that Φ is one-to-one. Suppose we have m m such that Φ(m )=Φ(m ). Φ(m ) Φ(m )=(0,0,,0) = Φ(m m )=(0,0,,0) by Lemma = m m is ivisible by n, by Lemma = m m must be zero since m m is nonnegative an the next multiple of n is not in {0,, n }. = m = m. We now use Proposition to show that by restricting the omain of Φ to M, we obtain a bijection between M an C. This uses three lemmas. Lemma Proof First note that Φ(c n )= c n = ( 0,, 0,,, 0, n (, 0,,, 0, n ), n o ), n even. { n + = n n + n 3 +, n even n + = n n + n 3 +, n o. When we write c n in this form, the result is clear by inspection. Lemma The(n )-tuple in C which maximizes Φ is { (0,, 0,,, 0, ), n o (, 0,,, 0, ), n even. Proof We prove the result by inuction. When n=, we have c =. So M ={0,, } an Φ(m)=(m). Therefore, ( ) C is the coevertex which maximizes Φ. This(n )-tuple is of the esire form. Now suppose the lemma is true in K q. Case : q is o. Then we are given that the coevertex in K q which maximizes Φ is ( 0,,, 0, ). q Since q is even, k q+ consists of copies of G q, so there are coevertieces in Kq+ which are of the form(i, 0,,, 0, q (, 0,,, 0, ). ), i=0. So the coevertex in K q+ which maximizes Φ is

20 Encoing an Decoing for the SF Labeling 8 Case : q is even. Then we are given that the coevertex in K q which maximizes Φ is(, 0,,, 0, ). q Now, K q+ consists of one copy of G q an copies of U q. We look for i such that ( i,, 0,,, 0, ) maximizes Φ over the coevertices in K q+. But by Lemma 5.3.4, } {{ } q Φ ((, 0,,, 0, ))=c q. So ( i,, 0,,, 0, ) cannot be q in a K q subgraph with fewer than c q coevertices. Since U q has fewer than c q coevertices, ( i,, 0,,, 0, ) must be in the copy of G q, which is the top subgraph of Kq+, so we are force to pick i=0, an the coevertex in K q+ which maximizes Φ is of the esire form. Lemma If m M, then Φ(m) C. Proof Since Φ is a bijection, we prove the equivalent statement: if(w,w,,w n ) C, then Φ ((w,w,,w n )) M. Fix(w,w,,w n ) C. Case : w n. Then w w n can be anything an this still represents a coevertex, so the result hols automatically. Case : w n =. Lemmas an tell us that the (n )-tuple which maximizes Φ is exactly Φ(c n ). So there is no c C with Φ (c) / M. Theorem The restricte map Φ : M C is a bijection. Proof Lemma tells us that Φ(M) C. Furthermore, by Proposition 5.3.3, Φ must be a bijection between M an Φ(M). Therefore, since M an C have the same carinality, Φ(M)= C an Φ : M C is a bijection. Finally, we give an explicit expression for Φ. Proposition Let (w,w,,w n ) C. Then Φ ((w,w,,w n ))= ( ( ( ( ( w n +w n )+w n 3 )+w n 4 )+ )+w )+w. n 7 times n 7 (Call this number t 0.) Proof By Proposition 5.3.3, Φ ((w,w,,w n )) exists. So it is sufficient to compute Φ(t 0 ) = (v,v,,v n ) an see that v i = w i for all i n. First note that t 0 w (mo ), so v = w. So we have t = t 0 w Now let <k<n an suppose = ( ( ( ( ( w n + w n )+w n 3 )+w n 4 )+ )+w 3 )+w n 8 times n 8 t k = ( ( n k 6 times ( ( ( w n + w n )+w n 3 )+w n 4 )+ n k 6 )+w k+ )+w k

21 8 Russell Then t k = t k w k = ( ( n k 7 times an v k+ = t k (mo )=w k+. ( ( ( w n + w n )+w n 3 )+w n 4 )+ n k 7 )+w k+ )+w k+ 6. USING Φ TO DESCRIBE THE POSITION OF THE m th CODEVERTEX IN K n In this section we give an algorithm which uses Φ(m) to escribe explicitely the position of the corresponing coevertex in K n. We will arrive at a bijection between {0,,c n } an the set of positions of coevertices. This is an extremely useful technique. In particular, it provies the basis for our encoing an ecoing scheme for the SF labeling. Perhaps more importantly, though, it coul be use to create an encoing an ecoing scheme for any reasonable labeling of K n. 6.. Describing the Position of any Vertex in K n : Right-Sie-Up Coorinates. Since there are n vertices in K n, we can escribe the position of each vertex by an n-tuple(u,u,,u n ) over{0,, }. Obviously, there are numerous ways to o this. We introuce a system calle Right- Sie-Up (RSU) coorinates. In RSU coorinates, every K m subgraph of K n is viewe as being oriente the same way as Kn. If our vertex is containe in the i th K m subgraph of a K m+ subgraph (where 0 points the same way as the top vertex of K n an we count counterclockwise), then we simply let u n m = i. (Note: the vertex itself is a K 0 subgraph.) As an example, Figure 6 shows the RSU coorinates of each vertex in K5. (0,0) (0,) (0,4) (,0) (0,) (0,3) (4,0) (,) (,4) (4,) (4,4) (,) (,3) (4,) (4,3) (,0) (3,0) (,) (,4) (3,) (3,4) (,) (,3) (3,) (3,3) FIGURE 6. RSU coorinates on K 5.

22 Encoing an Decoing for the SF Labeling 83 In Section 6. we give the forwar algorithm which takes in Φ(m) an returns the RSU coorinates of the m th coewor. Section 6.3 contains a proof of the algorithm. In Section 6.4, we give the inverse algorithm, an in Section 6.5 we prove it. These steps give us the complete bijection between {0,,c n } an the RSU coorinates of coevertices. 6.. Forwar Algorithm. We break the forwar algorithm into two parts. The first part uses Φ(m) to prouce the vector S = (s,,s n ), whose i th component is either (G, n i+) or (U, n i+), epening on whether the coevertex escribe by Φ(m) is containe in a G n i+ or a U n i+. In the process, we also translate Φ(m) into what we refer to as the Relative coorinates (efine later in Definition 6.3.) of the coevertex. The secon part of the algorithm translates Relative coorinates into RSU coorinates. FORWARD ALGORITHM: PART (Prouces the vector S. Figure 7 is a visual representation of this part of the algorithm.) (v,v,,v n )=Φ(m) k= s =(G,n) WHILE k< n IF s k =(G, n k+ ) IF n k+ is even ELSE s k+ =(G, n k) IF v k = 0 ELSE s k+ =(G, n k) s k+ =(U, n k) ELSE (i.e. if s k =(U, n k+ )) IF n k+ is even ELSE v k = v k + IF v k = 0 ELSE s k+ =(U, n k) s k+ =(G, n k) s k+ =(U, n k)

23 84 Russell k=k+ END WHILE RETURN S=(s,,s n ) SK = ( G, n - k + ) SK = ( U, n - k + ) n - k + even n - k + o n - k + o n - k + even S K + = ( G, n - k ) k = k + S K + = ( U, n - k ) k = k + Vk = Vk + Vk = 0... Vk = Vk = - Vk = 0... Vk = Vk = - S K + = ( G, n - k ) S K + = ( U, n - k ) S K + = ( U, n - k ) k = k + k = k + k = k + S K + = ( G, n - k ) k = k + FIGURE 7. A visual representation of Part of the forwar algorithm. FORWARD ALGORITHM: PART (Translates the new(v,v,,v n ), which is in Relative coorinates, into RSU coorinates) R =(r,, r n )=(v,, v n, 0 ) k= WHILE k<n IF s k =(G, n k+ ) where n k+ is even R k+ = R k +( 0,, 0, r k, r k,, r k ) k times (n k) times IF s k =(U, n k+ ) where n k+ is o Note: the superscript on ri is an inex, not a power.

24 ELSE R k+ = R k +( 0,, 0, r k, r k,, r k ) k times (n k) times R k+ = R k k=k+ END WHILE RETURN R n Encoing an Decoing for the SF Labeling A Proof of the Forwar Algorithm. We make a few concepts precise before beginning the proof of the algorithm. The proof will consist of three lemmas which together imply that the algorithm returns a coevertex in RSU coorinates. Definition The top K m subgraph in a copy of G m or U m is the copy of K m containing the top vertex of G m or U m, where top vertex is taken in the sense of the G-U construction of the SF labeling. We will nee to consier four types of subgraphs in K n : G m, where m is o. The top K m subgraph (labele 0) is G m. Subgraphs through are copies of U m which are all oriente the same way as G m. G m, where m is even. Subgraphs 0 through are copies of G m. Copy i is rotate πi with respect to the orientation of G m. U m, where m is o. Subgraphs 0 through are copies of U m. Copy i is rotate πi with respect to the orientation of U m. U m, where m is even. raians counterclockwise raians counterclockwise The top K m subgraph (subgraph 0) is U m. Subgraphs through are copies of G m which are all oriente the same way as U m. Definition The Relative coorinates of a vertex in K n are obtaine as follows. Suppose the vertex is containe in a certain G m or U m subgraph of K n. Label the Km subgraphs of this G m or U m with the numbers 0 through, where 0 is the top copy (see Definition 6.3.) an we count counterclockwise. If our vertex is containe in copy i, then the(n m+) st Relative coorinate is i.

25 86 Russell Remark The ifference between RSU coorinates an Relative coorinates is that Relative coorinates implicitely contain the subgraph rotations inherent in the G-U construction, while RSU coorinates are not aware of the G-U construction. Lemma If we apply Part of the forwar algorithm to every element in the image set{φ(m) 0 m c n }, then we obtain the correct number of coevertices in each type of subgraph. For example, if we let = 5 an n = 3, then we will obtain ( (G,3), (G,), (G,) ) five times an ((G,3), (U,), (G,))sixteen times. Proof Part is simply a concrete representation of the four cases liste after Definition 6.3., except for a small ajustment when s k =(U,n k+ ) where n k+ is even. We have to make this ajustment because the number of coevertices in U n k+ (n k+ even) is congruent to (mo ), so for i {0,, }, Φ returns v k = i once more than it returns v k =. But subgraphs through of U n k+ (all copies of G n k ) each contain one more coevertex than subgraph 0 (a copy of U n k ). The ajustment (aing to v k ) compensates for this before the algorithm is allowe to move on, so that we en up with the correct number of vertices in all the copies of G n k an U n k. There is no such problem with G n k+ (n k+ o) since in this case, the top K n k subgraph is the one with one more coevertex than the others. There is no such problem with G n k+ (n k+ even) an U n k+ subgraphs have the same number of coevertices. (n k+ o) because all K n k Lemma The vector R at the beginning of Part gives the Relative coorinates of a coevertex. Proof By Lemma an the efinition of Relative coorinates, the vector (v,v,,v n ) returne by Part gives the first n components of the Relative coorinates of a coevertex. Since each coevertex is the top vertex of its respective copy of G, we let the nth coorinate equal zero, so that R n =(v,,v n,0) gives the Relative coorinates of a coevertex. Lemma Part of the forwar algorithm takes in the Relative coorinates of a vertex an rotates each K m subgraph the correct number of times so as to return the RSU coorinates of the same vertex. Proof (See Remark for an explanation of why rotations are the issue here.) Suppose we have a K m graph in RSU coorinates. Then when the graph is rotate counterclockwise by π raians, a vertex whose coorinates were(u,,u m ) now has coorinates(u +,,u m + ). Now, a K m subgraph in K n π has been rotate counterclockwise by raians a total of γ= ri i B times, where B is the set {i n m s i {(G, even), (U, o)} }. Relative coorinates o not show this rotation. So in orer to uno the rotation of this subgraph, we nee to ajust the Relative coorinates of each vertex in our K m subgraph (i.e., ajust w m+ through w n ) by aing γ to each w m+ w n. Part performs this ajustment one rotation at a time.

26 Encoing an Decoing for the SF Labeling 87 Theorem The forwar algorithm takes in Φ(m) an returns the corresponing coevertex in RSU coorinates. Proof The theorem is a irect consequence of Lemmas 6.3.4, 6.3.5, an Inverse Algorithm. In this section we provie an inverse to the algorithm escribe in Section 6.. The inverse algorithm takes in the RSU coorinates of a coevertex an returns Φ(m). Like the forwar algorithm, we break the inverse algorithm into two parts. Part of the inverse algorithm inverts Part of the forwar algorithm (this is Lemma 6.5.). INVERSE ALGORITHM: PART (Changes from RSU coorinates to Relative coorinates. Figure 8 is a visual representation of this part of the algorithm.) Y =(y,,y 3 n )= the given RSU coorinates. k= g =(G,n) WHILE k<n IF g k =(G, n k+ ) IF n k+ is even g k+ =(G, n k) Y k+ = Y k ( 0,, 0, y k k,, yk k ) k times (n k) times ELSE IF y k k = 0 g k+ =(G, n k) Y k+ = Y k ELSE g k+ =(U, n k) Y k+ = Y k ELSE (g k =(U, n k+ )) IF n k+ is even IF y k k = 0 g k+ =(U, n k) Y k+ = Y k ELSE 3 Note: the superscript on yi is an inex, not a power.

27 88 Russell ELSE k=k+ END WHILE g k+ =(G, n k) Y k+ = Y k g k+ =(U, n k) Y k+ = Y k ( 0,, 0, y k k,, yk k ) k times (n k) times RETURN Y n =(y n,,yn n) gk = (G, n - k + ) gk = (U, n - k + ) n - k + even n - k + o gk+ = ( G, n - k ) Yk+ = Yk - ( 0 0, ) k = k+ { k y k k { n - k y k k n - k + o gk+ = ( U, n - k ) { y k k Yk+ = Yk - ( 0 0, ) k = k+ k { n - k y k k n - k + even y k k = 0... y k k = y k k = - y k k = 0... y k k = y k k = - gk+ = ( G, n - k ) gk+ = ( U, n - k ) gk+ = ( U, n - k ) gk+ = ( G, n - k ) Yk+ = Yk k = k+ Yk+ = Yk k = k+ Yk+ = Yk k = k+ Yk+ = Yk k = k+ FIGURE 8. A visual representation of Part of the inverse algorithm. Part of the inverse algorithm takes in Y n, which is in relative coorinates, an returns the corresponing vector of the form Φ(m). Therefore this part inverts Part of the forwar algorithm (this is Lemma 6.5.). In fact, it is essentially the same as Part of the forwar algorithm, with the exception that when s k =(U,n k+), with n k+ even, we subtract from v k instea of aing ; also, this algorithm returns(v,v,,v n ) instea of S.

28 Encoing an Decoing for the SF Labeling 89 INVERSE ALGORITHM: PART (Changes from Relative coorinates to a vector of the form Φ(m)) (v,v,,v n )=(y n,,yn n ) k= s =(G,n) WHILE k< n IF s k =(G, n k+ ) IF n k+ is even ELSE s k+ =(G, n k) IF v k = 0 ELSE s k+ =(G, n k) s k+ =(U, n k) ELSE (i.e. s k =(U, n k+ )) IF n k+ is even ELSE v k = v k IF v k = 0 ELSE k=k+ END WHILE s k+ =(U, n k) s k+ =(G, n k) s k+ =(U, n k) RETURN (v,v,,v n ) Φ(m)=(v,v,,v n )

29 90 Russell 6.5. A Proof of the Inverse Algorithm. Lemma Part of the inverse algorithm inverts Part of the forwar algorigthm. Proof Let p p u n be all the numbers such that s pi {(G, even),(u, o)}. Then, accoring to Part of the forwar algorithm, R n = R + We want to show that Y n = R, i.e., that Y n = R n u j= u j= ( 0, 0,, 0 p j ( 0, 0,, 0 p j, r p j p j, r p j p j,, r p j p j p, r j n p j ) p j, r p j p j,, r p j p j n p j ). Note that R ==R p an Y = = Y p. Furthermore, r p i = ri = y p i, since we have that Y = R n an r p i = ri n for i= p. So s i = g i for all i= p +. Thus, the first vector that gets subtracte from Y in Part of the inverse algorithm is ( 0, 0,, 0, y p, y p,, y p )=(0, 0,, 0, r p, r p,, r p p n p p n p ) Now Y p + an R have the first p entries ientical, an S an G have the first p + entries ientical, so the next vector which gets subtracte from Y p + is an so on. ( 0, 0,, 0, r p, r p,, r p p n p ) Lemma Part of the inverse algorithm inverts Part of the forwar algorithm. Proof This is clear, since the two algorithms are ientical except that one contains v k = v k + while the other contains v k = v k Theorem The inverse algorithm inverts the forwar algorithm. Proof Direct consequence of Lemmas 6.5. an 6.5..

30 Encoing an Decoing for the SF Labeling 9 7. ENCODING AND DECODING FOR THE SF LABELING In this section, we present an encoing an ecoing scheme for the SF labeling. Definition An encoing an ecoing scheme for a particular labeling of K n is a bijection between the set of natural numbers { 0,, c n }, where c n is the number of coevertices in a PECC on K n, an the set of coewors in the labeling. Note that, unlike the results from Sections 5 an 6, an encoing an ecoing scheme is esigne for one particular labeling metho. 7.. Towar an Encoing Scheme. Cull an Nelson present an easy encoing an ecoing scheme for the case when = 3. Their scheme uses the fact that in the = 3 case, every istance neighborhoo of a coewor contains exactly one wor which is a multiple of 4 (when viewe as a number in base 3). They enote by G n the set of coewors in the SF labeling of K3 n. To encoe the number m < G n, one simply runs the number 4m through the finite state machine for error correction. [3] We aske if Cull an Nelson s technique coul be easily generalize for > 3 (using multiples of + instea of multiples of 4). Unfortunately, the answer is no. This is because in the general case, some coewors are not ajacent to any multiple of +, while others are ajacent to two multiples of +. An easy example is K5 3, where the coewor 44 is ajacent to 34 an 44, both of which are multiples of 6. After trying some similar ieas an failing, we ecie to esign an encoing an ecoing scheme which woul use few (if any) theoretical properties of the SF labeling. In particular, we wrote two algorithms, iscusse in etail in Sections 5 an 6, which together give a bijection between { 0,, c n } an the set of RSU coorinates of coevertices. Now all that is left is to fin a simple bijection between RSU coorinates of coevertices an the coewors of the SF labeling. Fortunately, this turns out to be fairly straightforwar. Note: The avantage of this approach is that our bijection between natural numbers an RSU coorinates has nothing to o with any labeling of K n. Furthermore, any reasonable labeling has an easy map between the position of a vertex an its label. Therefore, our technique coul be imitate to create encoing an ecoing schemes for most labelings.

31 9 Russell 7.. The SF Labeling as a Tree. In this section we give an intuitive motivation for our encoing an ecoing scheme. We show how one can view the construction of the SF labeling as a -ary tree, where each layer of the tree represents the SF labeling of K m for a particular m (this makes sense because the SF labeling is constructe recursively). Each noe in the m th layer represents a vertex in K m an is labele with the SF label of that vertex as well as its RSU coorinates. The encoing an ecoing scheme passes between these two pieces of information by traveling up the tree through one type of information, turning aroun at the root, an travelling back own to the same vertex through the other type of information. Suppose we have the SF labeling of K m. Then the SF labeling of Km+ is constructe by applying α to the labels in K m, creating copies of the graph, rotating each an appropriate number of times, an appening the appropriate igit to all the labels in each (see Section 3.). So if b b m is the label of a vertex in K m, then b b m gives rise to aughter labels in K m+. These labels are α(b b m ) i where i {0,, }. So we can think of the construction of the SF labeling as a -ary tree where each vertex in K m is the root of a subtree of the form in Figure 9. b b k Appen 0 Appen ( -) ( b b k) 0 ( b b ) ( -) k FIGURE 9. A vertex label in K m an its aughter labels. As we sai above, the SF labeling makes copies of α(k m) an rotates the ith copy πi raians clockwise before connecting all the copies by eges. So the i th augter of the vertex whose RSU coorinates in K m were ( a,, a m ), is the vertex in K m+ with RSU coorinates ( i, a i,, a m ). Figure 0 is the same as Figure 9, with the aition that each noe is also labele with the RSU coorinates of the corresponing vertex. Our encoing an ecoing scheme is the composition of the algorithm from Sections 5 an 6, an the algorithm we give in Section 7.3, which is a bijection between SF labels of coevertices an RSU coorinates of coevertices. Note that these last two types of ata are shown in Figure 0. The technique will be to specify one type of ata for a noe in the tree, then follow a path up to the root by computing this same ata for all the noes along the path. One then specifies the other type of ata for the last noe in the path, then follows the reverse path back own by computing the new ata for each noe, finally arriving at the new ata for the original noe.

32 Encoing an Decoing for the SF Labeling 93 Label { RSU { b b k a,, a k ( ) Appen 0 Appen ( -) Label { RSU { ( b b k) 0 ( b b k) ( -) ( 0, a -0,, - 0 ) ( -, a -( -),, a k - ( -)) a k 7.3. Encoing an Decoing Algorithms. FIGURE 0. Figure 9 with RSU coorinates ae. Recall that the forwar algorithm (Section 6) actually gives a bijection between the set of vectors of the form Φ(m) an the set of RSU coorinates of coevertices. Call this map F. So the map Λ=F Φ is a bijection between the first c n natural numbers an the RSU coorinates of coevertices. Our encoing scheme is the map ENCODE from RSU coorinates of coevertices to coewors. The ecoing scheme, DECODE, is the inverse of ENCODE. ALGORITHM: ENCODE: PART 4 Q =(q,, q n )=Λ(m) k= WHILE k<n Q k+ =( q k+,, q k+ n ) =( q k,, qk n k+ )+( qk,, qk n ) n k k=k+ END WHILE ENCODE: PART y = q n k= 4 Note: In both ENCODE an DECODE, the superscript on qi is an inex, not a power. Also, in ENCODE: PART, the Q i s are vectors of ifferent lengths; in particular, Q i has one more component than Q i+.

33 94 Russell WHILE k<n y k+ = α(y k ) q n k k=k+ END WHILE RETURN y n ENCODE(Λ(m))=y n (where α(y k ) means apply the permutation α to the igits of the string y k in orer, proucing a new string, an string string means appen string to string. ALGORITHM: DECODE: PART Given coewor c c n q = c n k= WHILE k<n q k+ = α k (c n k ) k=k+ END WHILE DECODE: PART FOR i=n q i = q i i j= q j END FOR RETURN Q

34 7.4. Proofs of ENCODE an DECODE. Encoing an Decoing for the SF Labeling 95 To ai in the proof of our encoing an ecoing algorithm, Figure gives a more global view of the SF labeling tree. This iagram uses the same notation as ENCODE an DECODE. (0) (k) ( -) 0 k - Appen 0 Appen k Appen - Appen 0 Appen k Appen - Appen 0 Appen k Appen - (0) 0 (0) k (0) ( -) (k) 0 (k) k (k) ( -) (0,0-0) (k,0-k) ( -,0-( -)) (0,k-0) (k,k-k) ( -,k-( -)) ( -) 0 ( -) k ( -) ( -) (0,( -)-0) (k,( -)-k) ( -,( -)-( -)) n q n- q ( ( ( ) ) ), ( q, ) q 3 n n- q, q q q q Appen 0 Appen k Appen - n n- q ) q ) ) q ) n 0 ( n- (( ( q ) q ))) q ) k q n n- (, q -,, k ) ( ( ( ( ) ( 0, q -0,, q ) n n- - q n k k ( n- -, q - ( -),, q - q - -( - )) q n n- q ) q ) ) q ) ( -) ( ( ( ( ) FIGURE. Tree representation of the SF labeling, using notation from ENCODE an DECODE. At each noe, the top label gives the SF label an the bottom label gives the RSU coorinates.

35 96 Russell Proposition ENCODE takes in the RSU coorinates of a vertex in K n of the same vertex. an returns the SF label Proof Suppose we have the RSU coorinates of a vertex v in K n. Part of ENCODE computes the RSU coorinates of its parent vertex in K n, then computes the RSU coorinates of the parent vertex of that vertex, an so on, up to its ancestor in K. By inspection of the general subtree in Figure, ENCODE performs the necessary operation correctly at each step. Part of ENCODE follows the reverse path back own the tree to the original noe. Note that any path through the tree is completely etermine by which igits are appene to successive labels. The algorithm follows the path which is etermine by information gathere in Part. It computes α an then appens the appropriate igit. By inspection of the general subtree in Figure, this is performe correctly at each step. Part, therefore, finishes by returning the label of the original vertex. Proposition DECODE takes in the SF label of a vertex in K n of the same vertex. Proof The proposition can be prove in a similar fashion to Proposition an returns the RSU coorinates 8. CONCLUSION We have presente an encoing an ecoing metho for the SF labeling of o-imension iterate complete graphs. The metho uses general properties of PECC s on iterate complete graphs; only in the last step oes it use any information which is particular to the SF labeling. Further work coul inclue testing this technique on other labeling schemes to etermine if it will be useful as preicte. Work coul also be one to simplify the algorithms themselves as well as the proofs. REFERENCES [] Shawn Alspaugh, Nathan Knight, an Kathleen Meloney. Perfect One Error Correcting Coes on Iterate Complete Graphs. Proceeings of the REU Program in Mathematics. NSF an Oregon State University. Corvallis, Oregon. 00. [] Be Birchall an Jason Teor. Perfect One Error Correcting Coes on Iterate Complete Graphs. Proceeings of the REU Program in Mathematics. NSF an Oregon State University. Corvallis, Oregon [3] Paul Cull an Ingri Nelson. Perfect Coes, NP-Completeness, an Towers of Hanoi Graphs. Bulletin of the ICA 6:3-38, 999. [4] Christopher Frayer an Shalini Rey. Perfect One Error Correcting Coes an Iterate Complete Graphs. Proceeings of the REU Program in Mathematics. NSF an Oregon State University. Corvallis, Oregon. 00. [5] Stephanie Kleven. Perfect Coes on O Dimension Serpinski Graphs. Proceeings of the REU Program in Mathematics. NSF an Oregon State University. Corvallis, Oregon UNIVERSITY OF PENNSYLVANIA aress: prussell@math.upenn.eu