FractionalGeometry-Chap2

# A moments reection reveals the diculty one repair

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Unformatted text preview: an N -string not already in L, we are ﬁnished. How does the prefer 1 algorithm generate an Eulerian cycle in a de Bruijn graph? Consider the right side of Fig. 2.32. The 3-string 000 is generated by 0 by the edge 00 −→ 00. This edge is labeled 1 on the graph on the right of Fig. 1 2.32. The prefer 1 algorithm next applies 1 by traversing the edge 00 −→ 10. This edge is labeled 2 on the graph. In the table below we show the Eulerian path generated by the prefer 1 algorithm. edge label 1 2 3 4 5 6 7 8 current 3 string 000 100 110 111 011 101 010 001 growing string 000 1000 11000 111000 0111000 10111000 010111000 0010111000 The entries of the middle column are all the binary 3-strings. Reading the last entry of the right column from right to left we ﬁnd all these 3-strings. There are a LOT of de Bruijn sequences B (k, N ) of length N over an alphabet of size k , speciﬁcally, N −1 (k !)k kN For example, for binary sequences, N 1 2 3 4 5 6 7 ... 10 2.6 number of length N de Bru...
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