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This e-book constitutes the refereed lawsuits of the fifth Asian Computing technological know-how convention, ASIAN'99, held in Phuket, Thailand, in December 1999.

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**Extra info for Advances in Computing Science — ASIAN’99: 5th Asian Computing Science Conference Phuket, Thailand, December 10–12,1999 Proceedings**

**Example text**

Namely, we start from any vertex v, we assign to v color 1 and to its neighbors color 2; in the general step i of the pointer jumping loop, we assign the color of vertex j to not yet colored vertices at distance 2i from j. First observe that after log n − 1 iterations each vertex has a color, as guaranteed by the pointer jumping technique. The found coloring is trivially a 2-coloring and it is valid. Indeed, vertex v assigns its color (that is, 1) to all vertices with even distance from it, while v’s adjacent vertices assign their color (that is, 2) to all vertices with even distance from them, and so with odd distance from v.

Then B1 , . . , Bb has multiple output variables (B1 . . Bb ). , R1 : If . . Then B1 , . . , Rb : If . . , If X1 and X2 and . . and Xm Then Y . The disjunctive clauses A and B in the form: If A or B Then C is broken down into two disjunctive rules: R1 : If A Then C, and R2 If B Then C. Further details of this conversion can be found in [5]. step 3 Construct a CSG using the following procedure. Algorithm 1 Construct CSG Input: Rule base R = {ri }, a current starting state sc Output: an update the current state graph A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 begin l max = Find maxLabel(R) /* find the input var.

Definition 8 (sort measure). We define a measure d on sorts as follows: d(0) = 0 d(s1 , . . , sn ) = 1 + d(s1 ) + . . + d(sn ) . Lemma 6. Suppose r1 →let 2e · · · →let 2e rn+1 →in rn+2 where the name received in input in the last reduction has sort s and n ≥ d(s). e. r1 ≡ r1 →let 2e · · · →let 2e rd(s)+1 →in rd(s)+2 →let 2e · · · →let 2e rn+2 and rn+2 ∼ = rn+2 . Proofhint. The construction of a name a of sort s taken in input needs at most d(s) (let 2e ) reductions. All the other (let 2e ) reductions can be moved to the right of the input by iterated application of the lemma 5(1-2).