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The Breakpoint Graph. 1 5- 2- 4 3 . The Breakpoint Graph. 6 1 5- 2- 4 3 0. Augment with 0 = n+1. The Breakpoint Graph.

The Breakpoint Graph

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1 5- 2- 4 3

6 1 5- 2- 4 3 0

- Augment with 0 = n+1

11 2 1 9 10 3 4 8 7 6 5 0

6 1 5- 2- 4 3 0

- Augment with 0 = n+1
- Vertices 2i, 2i-1 for each i

11 2 1 9 10 3 4 8 7 6 5 0

6 1 5- 2- 4 3 0

- Augment with 0 = n+1
- Vertices 2i, 2i-1 for each i
- Blue edges between adjacent vertices

11 2 1 9 10 3 4 8 7 6 5 0

6 1 5- 2- 4 3 0

- Augment with 0 = n+1
- Vertices 2i, 2i-1 for each i
- Blue edges between adjacent vertices
- Red edges between consecutive labels 2i,2i+1

Sort a given breakpoint graph

11 2 1 9 10 3 4 8 7 6 5 0

into n+1 trivial cycles

11 10 9 8 7 6 5 4 3 2 1 0

Sort a given breakpoint graph

11 2 1 9 10 3 4 8 7 6 5 0

into n+1 trivial cycles

11 10 9 8 7 6 5 4 3 2 1 0

Conclusion:We want to increase number of cycles

Def:A reversal acts on two blue edges

11 2 1 9 10 3 4 8 7 6 5 0

cutting them and re-connecting them

11 2 1 9 10 3 4 7 8 6 5 0

A reversal can either

11 2 1 9 10 3 4 8 7 6 5 0

Act on two cycles, joining them (bad!!)

11 2 1 9 10 3 4 7 8 6 5 0

A reversal can either

11 2 1 9 10 3 4 8 7 6 5 0

Act on one cycle, changing it (profitless)

11 2 1 5 6 7 8 4 3 10 9 0

A reversal can either

11 2 1 9 10 3 4 8 7 6 5 0

Act on one cycle, splitting it (good move)

11 10 9 1 2 3 4 8 7 6 5 0

(Bafna, Pevzner 93)

Where d=#reversals needed (reversal distance),

and c=#cycles.

Proof: Every reversal changes c by at most 1.

(Bafna, Pevzner 93)

Where d=#reversals needed (reversal distance),

and c=#cycles.

Proof: Every reversal changes c by at most 1.

Alternative formulation:

where b=#breakpoints, and c ignores short cycles

Red edges can be :

Oriented{

Right-to-Right

Left-to-Left

Unoriented{

Left-to-Right

Right-to-Left

Red edges can be :

Oriented{

Right-to-Right

Left-to-Left

Unoriented{

Left-to-Right

Right-to-Left

Def:This reversal acts on the red edge

Red edges can be :

Oriented{

Right-to-Right

Left-to-Left

Unoriented{

Left-to-Right

Right-to-Left

Def:This reversal acts on the red edge

Thm: A reversal acting on a red edge is good

the edge is oriented

Def: Two red edges are said to be overlapping if they span intersecting intervals which do not contain one another.

Def: Two red edges are said to be overlapping if they span intersecting intervals which do not contain one another The lines intersect

Def: Two red edges are said to be overlapping if they span intersecting intervals which do not contain one another The lines intersect

Thm:A reversal acting on a red edge flips the orientation of all edges overlapping it, leaving other orientations unchanged

Def: Two red edges are said to be overlapping if they span intersecting intervals which do not contain one another The lines intersect

Thm:if e,f,g overlap each other, then after applying a reversal that acts on e,f and g do not overlap

Nodes correspond to red edges.

Two nodes are connected by an arc if they overlap

Nodes correspond to red edges.

Two nodes are connected by an arc if they overlap

Def:Unoriented connected components in the overlap graph - all nodes correspond to oriented edges.

Nodes correspond to red edges.

Two nodes are connected by an arc if they overlap

Def:Unoriented connected components in the overlap graph - all nodes correspond to oriented edges.

Cannot be solved in only good moves

- A profitless move on an oriented edge, making its component to oriented

- A profitless move on an oriented edge, making its component to oriented
or:

- A bad move (reversal) joining cycles from different unoriented components, thus merging them flipping the orientation of many components on the way

- Def:Hurdle - an unoriented connected component which is consecutive along the cycle

- Def:Hurdle - an unoriented connected component which is consecutive along the cycle
- Thm: (Hannenhalli, Pevzner 95)
Proof: A hurdle is destroyed by a profitless move, or

at most two are destroyed (merged) by a bad move.

- Def:Hurdle - an unoriented connected component which is consecutive along the cycle
- Thm: (Hannenhalli, Pevzner 95)
Proof: A hurdle is destroyed by a profitless move, or

at most two are destroyed (merged) by a bad move.

- Thm: