By Sue Whitesides (auth.), Peter Eades, Tadao Takaoka (eds.)

This booklet constitutes the refereed complaints of the twelfth overseas convention on Algorithms and Computation, ISAAC 2001, held in Christchurch, New Zealand in December 2001.

The sixty two revised complete papers provided including 3 invited papers have been rigorously reviewed and chosen from a complete of 124 submissions. The papers are geared up in topical sections on combinatorial new release and optimization, parallel and disbursed algorithms, graph drawing and algorithms, computational geometry, computational complexity and cryptology, automata and formal languages, computational biology and string matching, and algorithms and knowledge buildings.

**Read Online or Download Algorithms and Computation: 12th International Symposium, ISAAC 2001 Christchurch, New Zealand, December 19–21, 2001 Proceedings PDF**

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**Additional info for Algorithms and Computation: 12th International Symposium, ISAAC 2001 Christchurch, New Zealand, December 19–21, 2001 Proceedings**

**Example text**

Furthermore, w πqs ≤ ωG ≤ qs∈M (1 + ε) q − s = (1 + ε)ω(M) , qs∈M since G is a (1 + ε)-spanner. Since G is a connected spanning subgraph of G, it follows that ω(T) ≤ ω(G ) ≤ (1 + ε)w(M). The second claim follows by similar argumentation. 3. Approximating the diameter. 14. Given a set P of n points in IRd , one can compute, in O n log n + nε−d time, a pair p, q ∈ P, such that p − q ≥ (1 − ε)diam(P), where diam(P) is the diameter of P. Proof. Compute a (4/ε)-WSPD W of P. As before, we assign for each node u of T an arbitrary representative point that belongs to Pu .

We know that Pin ≥ n/250 and Pout ≥ n/2. Next, compute (recursively) the compressed quadtrees for Pin and Pout , respectively, and let T in and T out Tout denote the respective quadtrees. Create a node in both quadtrees that corresponds to . For T in this would vout Pvin just be the root node vin , since Pin ⊆ . For T out this vin v Pout = ∅. would be a new leaf node vout , since Pout ∩ Note that inserting this new node might require linear time, but it requires only changing a constant number Tin of nodes and pointers in both quadtrees.

Observe that by the triangle inequality, we have that dG (q, s) ≥ q − s , for any q, s ∈ P. 12. Given a set P of n points in IRd and a parameter 1 ≥ ε > 0, one can compute a (1 + ε)-spanner of P with O nε−d edges, in O n log n + nε−d time. Proof. The construction is described above. The upper bound on the stretch is proved by induction on the length of pairs in the WSPD. So, fix a pair x, y ∈ P, and assume that by the induction hypothesis, for any pair z, w ∈ P such that z − w < x − y , it follows that dG (z, w) ≤ (1 + ε) z − w .