The Berge Mystery Story Presented by Supriya Janjanam.

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Transcript of The Berge Mystery Story Presented by Supriya Janjanam.

The Berge Mystery Story

Presented bySupriya Janjanam

Outline

• A Mystery in the Library & Facts

• Interval Graph

• Properties

• Solving the Mystery

• References

A Mystery in the Library

Six professors has been to the library on the day that the rare tractate was stolen.

Each has entered once, stayed for sometime, and then left.

If two were in the library in the same time , then at least one of them saw the other.

Detectives questioned the professors and gathered the following testimony :

The Facts: Abe said that he saw Burt and Eddie

Burt said that he saw Abe and Ida

Charlotte claimed to see Desmond and Ida

Desmond said that he saw Abe and Ida

Eddie testified to seeing Burt and Charlotte

Ida said that she saw Charlotte and Eddie.

ONE OF THE PROFESSOR LIED!! WHO WAS IT?

Interval Graph

A graph is an interval graph if we can find intervals on the line so that two vertices are joined by an edge if and only if their corresponding intervals overlap.

Phase 1

Phase 2

Phase 3Task 4

Task 5

1 2 3

4 5

The interval graph G

Interval Graph

v1 v2

v3v4

v1 v2

v3

v1v2

v3v1

v2v4

V3??

An interval graph NOT an interval graph

Applications of Interval Graphs

• Genetics

• Scheduling

• Artificial intelligence

• VLSI design

• Computer storage

• Frequency assignment

• More …

Properties of Interval Graph

• Two properties characterize interval graphs:

- The Chordal Graph Property

- The co-TRO Property

The Chordal Graph Property

Every cycle of length > 4 has a chord

i.e., they may not contain chordless cycles!

Interval Graphs are Chordal

Interval graphs may not contain chordless cycles!

- i.e., they are chordal. Why?

Interval Graphs are Chordal

Interval graphs may not contain chordless cycles!

- i.e., they are chordal. Why?

The co-TRO PropertyThe transitive orientation (TRO) of the

complement

i.e., the complement must have a TRO

Not transitive ! Transitive !

Solving the Mystery

Abe said that he saw Burt and EddieBurt said that he saw Abe and IdaCharlotte claimed to see Desmond and IdaDesmond said that he saw Abe and IdaEddie testified to seeing Burt and CharlotteIda said that she saw Charlotte and Eddie.

The Testimony Graph

Solving the Mystery

Clue 1:

Double arrows imply TRUTH

The Testimony Graph

Solving the MysteryUndirected Testimony Graph

cycle

We know there is a lie, since {A, B, I, D} is a chordless 4-cycle.

Intersecting Intervals cannot form Chordless Cycles

Burt Desmond

Abe

Here No place for Ida’s interval:It must hit both B and D but cannot hit A.

which is Impossible!

Solving the Mystery

There are three chordless 4-cycles:{A, B, I, D}{A, E, I, D}

{A, E, C, D}

One professor from the chordless 4-cycle must be a liar.

Solving the Mystery

There are three chordless 4-cycles:{A, B, I, D}{A, E, I, D}

{A, E, C, D}

Burt is NOT a liar: He is missing from the second cycle.

One professor from the chordless 4-cycle must be a liar.

Solving the Mystery

There are three chordless 4-cycles:{A, B, I, D}{A, E, I, D}

{A, E, C, D}

Ida is NOT a liar: She is missing from the third cycle.

One professor from the chordless 4-cycle must be a liar.

Solving the Mystery

There are three chordless 4-cycles:{A, B, I, D}{A, E, I, D}

{A, E, C, D}

Charlotte is NOT a liar: She is missing from the second.

One professor from the chordless 4-cycle must be a liar.

Solving the Mystery

There are three chordless 4-cycles:{A, B, I, D}{A, E, I, D}

{A, E, C, D}

Eddie is NOT a liar: He is missing from the first cycle.

One professor from the chordless 4-cycle must be a liar.

Solving the Mystery

Burt is NOT a liar: He is missing from the second cycle. Ida is NOT a liar: She is missing from the third cycle. Charlotte is NOT a liar: She is missing from the second. Eddie is NOT a liar: He is missing from the first cycle.

WHO IS THE LIAR? Abe or Desmond ?

Solving the Mystery

WHO IS THE LIAR? Abe or Desmond ?

If Abe were the liar and Desmond truthful, then {A, B, I, D} would remain a chordless 4-cycle, since B and I are truthful.

Therefore:

Desmond is the liar.

References

[1] http://www.cs.haifa.ac.il/~golumbic/courses/algorithmic-graph-theory/slides_and_notes_of_lectures/Lecture%201%20-%20Introduction/intro-agt-v1b-part1.pdf

[2] https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CB4QFjAA&url=http%3A%2F%2Fliawww.epfl.ch%2FTeaching%2FCourses%2FAlgorithmicGraphTheory%2Fdoc%2FRAT-thailand.ppt&ei=RGNFVOzKNdSwyATMs4K4BQ&usg=AFQjCNFtXDmvuMXAftqXMkqltQ8AGkLqpA&sig2=muUpMdF2FPCr9pKAxgKrQQ&bvm=bv.77880786,d.aWw&cad=rja

[3] https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=4&ved=0CDcQFjAD&url=http%3A%2F%2Fwww.dis.uniroma1.it%2F~seminf%2Fseminars%2Fgolumbic.ppt&ei=RGNFVOzKNdSwyATMs4K4BQ&usg=AFQjCNGS79daY-KflIW7ErRF1jaM_vIR-w&sig2=oPFk-4Si8FwLYESjmtQaYQ&bvm=bv.77880786,d.aWw&cad=rja

ANY

QUERIES….???

Thank you…