1 / 14

Simple Algorithm for Sorting the Fibonacci String Rotations

Simple Algorithm for Sorting the Fibonacci String Rotations. Manolis Christodoulakis King’s College London Joint work with Costas S. Iliopoulos Yoan Jos é Pinz ó n Ardila. Our Goal. What makes Fibonacci strings a best case input for the Burrows-Wheeler Transform (BWT)?

bryga
Download Presentation

Simple Algorithm for Sorting the Fibonacci String Rotations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Simple Algorithm for Sorting theFibonacci String Rotations Manolis Christodoulakis King’s College London Joint work withCostas S. Iliopoulos Yoan José Pinzón Ardila

  2. Our Goal • What makes Fibonacci strings a best case input for the Burrows-Wheeler Transform (BWT)? • Relationship between different rotations of a Fibonacci string • What is their lexicographic order? • Side effect: we can deduce the symbol stored at any position of any Fibonacci string in constant time (without using , provided that the fnvalues are known) SOFSEM 2006

  3. Fibonacci Strings & Numbers • The n-th Fibonacci string • Fn = Fn-1Fn-2 n≥2 F0=b, F1=a • The n-th Fibonacci number • fn = fn-1+fn-2 n≥2 f0=1, f1=1 F0 = b f0 = 1 F1 a 1 f1 = = F2 = a b f2 = 2 F3 = a b a f3 = 3 F4 = a b a a b f4 = 5 SOFSEM 2006

  4. 0 1 … i-1 i n-1 … Notation • The i-th rotation of a string • where i is taken modulo n. • rank(i,x) = the rank of Ri(x) • rot(ρ,x) = the rotation whose rank is ρ x = 0 1 … i-1 i n-1 … Ri(x) = SOFSEM 2006

  5. Burrows-Wheeler Transform (BWT) • M.Burrows and D.J.Wheeler. 1994 • Purpose: to make a string more compressible • BWT Algorithm: • Create list of all rotations • Sort them • Output last symbol of every rotation • Output the rank of the 0-th rotation SOFSEM 2006

  6. BWT on Fibonacci Strings • F5 = abaababa, f5 = 8 R0(F5) R7(F5) = = a a b a b a a a a b b a a b b a R1(F5) = b a a b a b a a R2(F5) = a a b a b a a b R2(F5) = a a b a b a a b R5(F5) = a b a a b a a b R3(F5) = a b a b a a b a R0(F5) = a b a a b a b a R4(F5) = b a b a a b a a R3(F5) = a b a b a a b a R6(F5) R5(F5) = = a b a b a a b a a b a a b a a b R6(F5) = b a a b a a b a R1(F5) = b a a b a b a a R7(F5) = a a b a a b a b R4(F5) = b a b a a b a a SOFSEM 2006

  7. Properties of Fibonacci Strings • The number of ‘b’in Fnis fn-2 • Proof: By induction. • C.S.Iliopoulos, D.W.Moore and W.F.Smyth. 1997 • Fn = Fn-2Fn-3…F1un, un = ba (n odd) • un = ab (n even) • Let’s call this the IMS formula. SOFSEM 2006

  8. Similarities in Rotations • R0(Fn) differs from Rfn-2(Fn) in 2 symbols • Proof: • R0(Fn) = Fn-2Fn-3…F1un • Rfn-2(Fn) = Fn-3…F1unFn-2 (1) • R0(Fn) =Fn-1Fn-2 • = Fn-3…F1un-1Fn-2 (2) • Ri(Fn) differs from Ri+fn-2(Fn) in 2 symbols • Proof: • Ri(Fn) = Ri(R0(Fn)) • Ri+fn-2(Fn) = Ri(Rfn-2(Fn)) SOFSEM 2006

  9. Relative Order of Rotations • Ri(Fn) < Ri+fn-2(Fn) for n odd, i  fn-1-1 • Proof: • R0(Fn) = Fn-3…F1un-1Fn-2 • Rfn-2(Fn) = Fn-3…F1unFn-2 • For i=fn-1-1: • Ri(Fn) = bFn-2Fn-3…F1a • Ri+fn-2(Fn)= aFn-2Fn-3…F1b • Similarly, • Ri(Fn) > Ri+fn-2(Fn) for n even, i  fn-1-1 = Fn-3 … F1 ab Fn-2 = Fn-3 … F1 ba Fn-2 SOFSEM 2006

  10. Sorted List of Rotations • We proved (n odd): • Ri(Fn) < Ri+fn-2(Fn) i  fn-1-1 (3) • We will now prove that there is no j s.t. • Ri(Fn) < Rj(Fn) < Ri+fn-2(Fn) • Proof: (constructive) • Start at i=fn-1 and construct the partial list • Ri Ri+fn-2Ri+2fn-2 Ri+3fn-2… Ri+kfn-2… • for as long as • i+kfn-2  fn-1-1 (mod fn)  kfn-1 • I.e. the list is complete! SOFSEM 2006

  11. Identify Rotation (i) by Rank (ρ) • Therefore, for n odd: • rot(ρ,Fn) = fn-1 • = (ρfn-2-1) mod fn • Similarly, for n even, the sorted list is constructed bottom-up giving • rot(ρ,Fn) = (-(ρ+1)fn-2-1) mod fn +ρfn-2) mod fn ( SOFSEM 2006

  12. Identify Rank (ρ) of a Rotation (i) • This is simply the inverse of the previous function • n odd • rank(i,Fn) = ((i+1)fn-2) mod fn • n even • rank(i,Fn) = ((i+1)fn-2-1) mod fn SOFSEM 2006

  13. Symbols of Fibonacci Strings • Fn[i] = ? • Observe that • Fn[i] = Ri(Fn)[0] • In the sorted list of rotations, the first fn-1rotations start with ‘a’, the rest with ‘b’ • Thus Fn[i] can be deduced from rank(i,Fn) If rank(i,Fn) ≤ fn-1then Fn[i]=a else b. SOFSEM 2006

  14. BWT & Fibonacci ― The Quick Way • The first fn-2 symbols of BWT are ‘b’ • Proof: (n odd) • We proved the first fn-2 rotations have index • (ρ·fn-2-1)modfnfor 0 ≤ ρ < fn-2 • The last symbol of these rotations is • Fn[ (ρ·fn-2-1 )modfn ] • Which for 0 ≤ ρ < fn-2is ‘b’ • The next fn-1 symbols of BWT are ‘a’ • Proof: Consequence of previous lemma +fn-1 SOFSEM 2006

More Related