1. Cleber V. G. Mira Analysis of Sorting by Transpositions based on Algebraic Formalism RECOMB 2004 João Meidanis

2. Genomes as Permutations ● Permutation ● Genome (  1  2...  n )(  n...  2  1 ) Complementary Cycles 11 22 nn  1  2 Complementary Strands  n  = (-0 0)(1 –1)...(n –n) Consider the permutation  which complements the signal of an element.

3. Working with Transpositions ● Since we are working with transpositions, we will consider only one of the strands:  = (  1  2...  n )  1  2...  k  =  ● Sorting by transpositions: ● Circular order:  (  i ) =  i+1  (  n ) =  1

4. Product of Permutations  = ( 3 2 5 1)  = (6 4 2 ) E = {1, 2, 3, 4, 5, 6}  (1) = 1  (1) = 3  (3) = 3  (3) = 2  (2) = 6  (6) = 6  (6) = 4  (4) = 4 }  (4) = 2  (2) = 5  (5) = 5  (5) = 1   = (1 3 2 6 4 5)

5. Applying a Transposition  (  i  j  k ) (  1...  i...  j-1  j...  k-1  k...  n ) = (  1...  i-1  j...  k-i  i...  j-1  k...  n ) In the Algebraic approach:  (i, j, k) = (  i  j  k )  (1, 4, 5) = (4 1 5)  = ( 4 3 2 1 5 )   = (4 1 5) ( 4 3 2 1 5 ) = ( 1 4 3 2 5 )

6. 2-cycle Decomposition ● Every permutation has a 2-cycle decomposition.  = (4 3 2 1 5) = (4 3)(3 2)(2 1)(1 5) ● Odd cycles have an even number of 2- cycles in their 2-cycle decomposition. ● The norm of  is the minimum number of cycles in the 2-cycle decomposition of .

7. 3-cycle Decompositions ● Permutations whose norm is even have a minimum decomposition on 3-cycles. ● The 3-norm is the minimum number of cycles in the 3-cycles decomposition of .  = (0 3 4 6 2 7 1 5 8) = (0 3 4)(4 6 2)(2 7 1)(1 5 8) |  | 3 = 4

8. Building a 3-cycle Decomposition ● It is possible to find a 3-cycle decomposition of  through its 2-cycle decomposition.  = (0 3 4 6 2 7 1 5 8) = (0 3)(3 4)(4 6)(6 2)(2 7)(7 1)(1 5)(5 8)  = (0 3 4 6 2 7 1 5 8) = (0 3 4)(4 6 2)(2 7 1)(1 5 8) (0 3)(3 4) = (0 3 4) (4 6)(6 2) = (4 6 2) (2 7)(7 1) = (2 7 1) (1 5)(5 8) = (1 5 8)

9. Lower Bound ● The 3-norm of a permutation  is a lower bound for the transposition distance of .  1  2...  k  =   1  2...  k =  -1 k ≥ |  -1 | 3 D t ( ,  ) ≥ |  -1 | 3

10. Splits ● A split is a transposition which is not applicable to the genome , i.e. the product of this transposition and the genome is not a genome. Ex.: (1 2 3) is not applicable to (0 3 4 6 2 7 1 5 8) since: (1 2 3)(0 3 4 6 2 7 1 5 8) = (0 1 5 8)(2 7)(3 4 6) It is not a genome!!

11. Split+Transposition Distance ● If we consider the problem of sorting genomes by splits and transpositions then the split+transposition distance of a genome  to  is: D st ( ,  ) = |    | 3

12. Bibliography ● V. Bafna and P. A. Pevzner, 1995, Sorting by Transpositions. In: “Proceedings of the Sixth Annual ACM-SIAM Symposium on Discrete Algorithms”, San Francisco, USA, pp. 614-623 ● J. Meidanis and Z. Dias, 2000, An Alternative Algebraic Formalism for Genome Rearrangements. In: “Comparative Genomics:Empirical and Analytical Approaches to Gene Order Dynamics, Map Alignment and Evolution of Gene Families” D. Sankoff and J.H. Nadeau, editors, pp. 213-223