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Generating Bracelets in Constant Amortized Time
 SIAM JOURNAL ON COMPUTING
, 2001
"... A bracelet is the lexicographically smallest element in an equivalence class of strings under string rotation and reversal. We present a fast, simple, recursive algorithm for generating (i.e., listing) kary bracelets. Using simple bounding techniques, we prove that the algorithm is optimal in the s ..."
Abstract

Cited by 10 (3 self)
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A bracelet is the lexicographically smallest element in an equivalence class of strings under string rotation and reversal. We present a fast, simple, recursive algorithm for generating (i.e., listing) kary bracelets. Using simple bounding techniques, we prove that the algorithm is optimal
Generating Bracelets in Constant Amortized Time
, 2001
"... Abstract A bracelet is the lexicographically smallest element in an equivalence class of strings under string rotation and reversal. We present a fast, simple, recursive algorithm for generating (ie., listing) kary bracelets. Using simple bounding techniques, we prove that the algorithm is optimal ..."
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Abstract A bracelet is the lexicographically smallest element in an equivalence class of strings under string rotation and reversal. We present a fast, simple, recursive algorithm for generating (ie., listing) kary bracelets. Using simple bounding techniques, we prove that the algorithm is optimal
Fast Algorithms to Generate Necklaces, Unlabeled Necklaces, and Irreducible Polynomials over GF(2)
, 2000
"... this paper ## Sawada 23 developed an algorithm to generate kary bracelets in constant ## amortized time. Proskurowski et al. 17 show that the orbits of the ' Z. Z. Z . composition of b and d can be generated in amortized Oktime, which is CAT if k is fixed. It remains an interesting challenge ..."
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Cited by 23 (9 self)
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this paper ## Sawada 23 developed an algorithm to generate kary bracelets in constant ## amortized time. Proskurowski et al. 17 show that the orbits of the ' Z. Z. Z . composition of b and d can be generated in amortized Oktime, which is CAT if k is fixed. It remains an interesting challenge