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Evolutionary Genome History Reconstruction through Reversal Sorting

Explore the application of reversal mutations in reconstructing evolutionary paths of genomes with reversal sequences. Understand the limitations and possibilities of k-reversal sorting.

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Evolutionary Genome History Reconstruction through Reversal Sorting

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  1. Introduction • Sorting permutations with reversals in order to reconstruct evolutionary history of genome • Reversal mutations occur often in chromosomes where each reverses the order of an interval of genes • A shortest reversal sequence sorting one genome to another corresponds to the most likely evolutionary path between them

  2. Introduction • Sorting permutations and circular permutations using as few fixed-length reversals as possible • Limiting the the transformations to reversals of length exactly k can be very restrictive

  3. Can k-reversal sort ? • Can the permutation {1,3,2,4,5} be sorted using k-reversals ? • k=1 ? well …… • k=2 ?  {1,3,2,4,5} Bubble sort  • k=3 ? • k=4 ? later on

  4. Sorting {1,3,2,4,5} with k=3 • Since 1 and 2 are separated by an odd number of items and any 3-reversal change this distance by either 0 or 2 it cannot be done !!! • {2,3,1,4,5} – distance change 0 • {1,4,2,3,5} – distance change 0 • {1,3,5,4,2} – distance change 2

  5. Sorting {1,3,2,4,5} with k=3 • {2,3,1,4,5} – distance change 0 • {1,4,2,3,5} – distance change 0 • {1,3,5,4,2} – distance change 2 • 3-rev can change position of odd elements only and even elements only • 3-rev is actually bubble sort for odd/even elements inside the permutation

  6. Notation • PG(k,n)– permutation group of size n using k-reversals • The k-reversal operation on a permutation starting the n element Rev(i): {1,…,i-1,i+k-1, i+k-2,…, i+1,i,i+k,…,n} • d– the diameter : max{ shortest path in cayley graph } or minimum reversals to get from p to q

  7. 1324 3241 3421 2134 1234 4231 2314 3124 3214 2431 1342 4312 2431 1432 3142 3124 1423 2134 3412 4132 2413 3421 1243 4213 Notation • The Cayley graphis the graph whose vertices are the elements of G, with an edge between vertices p and q iff • Cayley graph of PG(3,4):

  8. Equivalent Transformations in PG(k,n) • 4l–reversal ↔ 4–reversal ↔ ζ1,2 ,ζ2,1 • (2+4l)–reversal ↔ 2–reversal ↔ ζ1,1 • (3+4l)–reversal ↔ 3–reversal • (5+8l)–reversal ↔ 5–reversal ↔ ζ2,2 • (9+8l)–reversal ↔ 9–reversal ↔ ζ2,4 ,ζ4,2

  9. ζ1,2 ,ζ2,1  4–reversal : • ζ1,2 (1) : {1,2,3,4,…}  {2,3,1,4,…} • ζ2,1 (2) : {2,3,1,4,…}  {2,4,3,1,…} • ζ1,2 (1): {2,4,3,1,…}  {4,3,2,1,…{

  10. 4–reversal  ζ1,2 ,ζ2,1 • lemma: 4-rev  ζ2,3 , ζ3,2 • {1,2,3,4,5,6,7} → {1,5,4,3,2,6,7} → {3,4,5,1,2,6,7} • for ζ3,2 we simply reverse these operations • lemma: ζ2,3 , ζ3,2 ζ1,4 , ζ4,1 • {1,2,3,4,5,6,7} → {3,4,5,1,2,6,7} → {5, 1,2,3,4,6,7} • for ζ4,1 we simply use ζ3,2 with same operations • lemma: ζ1,4 , ζ4,1 ζ1,2 ,ζ2,1 • {1,2,3,4,5,6,7} → {1,3,4,5,6,2,7} → {1,4,5,6,2,3,7} → {2,1,4,5,6,3,7} → {2,3,1,4,5,6,7}

  11. The problem: • Given a graph PG(k,n): • How many connected components are there? • Equiv to: what is the size of any connected component? • What is the diameter of each component? • Assume n≥k+2 • If k=n there are n!/2 components • If k=n-1 there are n or 2n components, depending upon parity • n=3 {(1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,1,2) (3,2,1)} • n=4 {(1,2,3,4) (1,4,3,2) (3,2,1,4) (3,4,1,2)}

  12. The simple cases • How many connected components are in PG(2,n)? • 1component (the graph is connected) • How many connected component are in PG(3,n)? • there is only a choice of n/2 elements for odd/even places, and therefore components

  13. The number of connected components in PG(k,n)

  14. Connected components -Sign of permutation (4-rev) • The sign of permutation is • pair is disordered if i<j and ai>aj • Lemma : ζ1,2 ,ζ2,1 do not change the sign of a permutation • ζ2,1(i)= ζ1,2(i)ζ1,2 (i) • x,y,zy,z,x • sign (z-y) (z-x) (y-x) = sign (z-y) (x-z) (x-y)

  15. Connected components (4-rev) • Lemma : ζ1,2 ,ζ2,1 cannot change the sign of a permutation. The identity permutations has + sign, so permutations with – sign cannot be sorted. • Lemma: ζ2,1 can sort only half of all permutations • ζ2,1  ζ2m,1 • for i=1 to n-2 find j such that aJ = i if (j – i) is even, apply ζj - i,1) i ) else apply ζj – i - 1,1) i+1) then ζ1,2) i ) end for

  16. example for i=1 to n-2 find j such that aJ = i if (j – i) is even, apply ζj - i,1) i ) else apply ζj – i - 1,1) i+1) then ζ1,2) i ) end for 1 2 3 4 5 i = 1,j = 4, j – i=3  ζ2,1) 2)

  17. example for i=1 to n-2 find j such that aJ = i if (j – i) is even, apply ζj - i,1) i ) else apply ζj – i - 1,1) i+1) then ζ1,2) i ) end for 1 2 3 4 5 ζ2,1) 2)  ζ1,2) 1)

  18. example for i=1 to n-2 find j such that aJ = i if (j – i) is even, apply ζj - i,1) i ) else apply ζj – i - 1,1) i+1) then ζ1,2) i ) end for 1 2 3 4 5

  19. example for i=1 to n-2 find j such that aJ = i if (j – i) is even, apply ζj - i,1) i ) else apply ζj – i - 1,1) i+1) then ζ1,2) i ) end for 1 2 3 4 5 i = 2, j = 5, j – i=3  ζ2,1) 3 )

  20. example for i=1 to n-2 find j such that aJ = i if (j – i) is even, apply ζj - i,1) i ) else apply ζj – i - 1,1) i+1) then ζ1,2) i ) end for 1 2 3 4 5 ζ2,1) 3 )  ζ1,2) 2 )

  21. example for i=1 to n-2 find j such that aJ = i if (j – i) is even, apply ζj - i,1) i ) else apply ζj – i - 1,1) i+1) then ζ1,2) i ) end for 1 2 3 4 5

  22. Connected components (4-rev) • the i th iteration places aJ into i th position, where aJ = i • at termination, either • because they have different signs, using prev lemma we know cannot be transformed into using ζ1,2 • thus, ζ1,2 divides the permutation group into 2 equal size sub-groups, and ζ1,2 sorts just half of all permutations

  23. The number of connected components in PG(k,n)

  24. Circular permutations -Notation • CPG(k,n)– circular permutation group of size n using k-reversals • Each permutation in CPG(n)represents a set of n permutations on PG(n) equivalent under the shift operation • {1,2,3,4} = { (1,2,3,4) , (2,3,4,1) , (3,4,1,2) , (4,1,2,3) } • Any permutation can be rearranged to exactly n arrangements by shift • PG(n) has n! permutations  CPG(n) has n!/n = (n-1)! permutations

  25. 1234 1324 3421 2134 2431 2314 3124 3214 1342 4312 1423 3124 2431 1432 2134 3421 2413 3412 1243 4213 3241 4231 3142 4132 Notation • The Cayley graphis the graph whose vertices are the elements of CPG, with an edge between vertices p and q iff • Cayley graph of CPG(3,4):

  26. Notation • The Cayley graphis the graph whose vertices are the elements of CPG, with an edge between vertices p and q iff • Cayley graph of CPG(3,4): 1234 1342 1324 1432 1243 1423

  27. Equivalent Transformations in CPG(k,n) • All PG(k,n) transformations hold for n > k+2 • 4l–reversal ↔ 4–reversal ↔ ζ1,2 ,ζ2,1 • (2+4l)–reversal ↔ 2–reversal ↔ ζ1,1 • (3+4l)–reversal ↔ 3–reversal • (5+8l)–reversal ↔ 5–reversal ↔ ζ2,2 • (9+8l)–reversal ↔ 9–reversal ↔ ζ2,4 ,ζ4,2

  28. The problem: • Given a graph CPG(k,n): • How many connected components are there? • Equiv to: what is the size of any connected component? • What is the diameter of each component? • Assume n≥k+2 • If k=n or k=n-1 there are (n-1)!/2 components • Since all PG(k,n) transformations hold: • # Components in CPG(n) ≤ # Components in PG(n)

  29. Connected comp. of CPG(k,n) for even k • Recall: How many connected components are in PG(2,n)? • 1component (the graph is connected) • same in CPG(2,n) –holds for all n

  30. Connected comp. of CPG(k,n) for even k & even n CPG(4l,2m) is connected (a single component) Proof: • Recall that: • 4 – reversals → ς1,2, ς2,1 • ς1,2, ς2,1 sort all permutations to {1,...,2m-1,2m} or {1,…,2m,2m-1} • ς1,2 can sort circular permutation {1,…,2m,2m-1} to {1,...,2m-1,2m} : • 1,2,3,4,6,5 → 5,1,2,3,4,6 (shift) • 5,1,2,3,4,6 → 1,2,5,3,4,6 → 1,2,3,4,5,6

  31. Connected comp. of CPG(k,n) for even k & odd n • Recall: 4–reversals do not change the sign of permutations. • If n is odd a shift operation doesn’t change the sign • x1,…,x2m,x2m+1 → x2m+1,x1,…,x2m • 2m = even #(disorders) • We can use the algorithm • 4l-reversals sorts half of CPG(k,n)

  32. Connected comp. of CPG(k,n) for even k • So far:

  33. Diameter of CPG(k,n) bounds • Upper bound =O(n2/k +nk) • Lower bound =Ω(n2/k2+n)

  34. Conclusions & Open problems • A complete answer to the connectedness question of the Cayley Graphs for permutations and circular permutations • Bounds to the diameter of CPG(k,n) • Can we tighten these bounds ? • What is the diameter of PG(k,n) ? • What happens with signed permutations where each element has 2 possible orientations ? • What happens if we allow numerous reversals ?

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