COMPUTING Euler Tours

Neeldhara Misra

DSA1 Week 5

Recall

A walk is a sequence of edges
\(e_1, \ldots, e_{n-1}\)
 

such that there exists a sequence of vertices
\(v_1, \ldots, v_n\)

for which \(e_i = (v_i, v_{i+1})\)
for all \(1 \leqslant i \leqslant n-1\).

An (closed/open) Euler Tour of a graph \(G\)

is a (closed/open) walk

that contains every edge exactly once

(i.e, no repeats and no omissions).

If a graph \(G\) has:

(a) more than two vertices of odd degree, or

(b) exactly one vertex of odd degree, then

it does not have an Euler Tour of any kind

(closed or open).

What we discovered

If a graph \(G\) has:

(a) exactly two vertices of odd degree, or

(b) no vertex of odd degree, then

does it have an Euler Tour of some kind

(closed or open)?

Food for thought

Closed Euler Tours

Getting back to where we started.

If a graph \(G\) has any vertex of odd degree,

it does not have a closed Euler Tour.

What we know

If a graph \(G\) has any vertex of odd degree,

it does not have a closed Euler Tour.

What we know

If a graph \(G\) has any vertex of odd degree,

it does not have a closed Euler Tour.

What we know

If a graph \(G\) has any vertex of odd degree,

it does not have a closed Euler Tour.

What we know

If a graph \(G\) has any vertex of odd degree,

it does not have a closed Euler Tour.

What we know

If a graph \(G\) has any vertex of odd degree,

it does not have a closed Euler Tour.

What we know

If a graph \(G\) has any vertex of odd degree,

it does not have a closed Euler Tour.

What we know

If a graph \(G\) has any vertex of odd degree,

it does not have a closed Euler Tour.

What we know

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

Will you get stuck at this point?

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

Will you get stuck at this point?

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

Can we conclude that we have an Euler Tour?

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

Can we conclude that we have an Euler Tour?

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

Now can we conclude that we have an Euler Tour?

If a graph \(G\) has no vertex of odd degree,

it does have a closed Euler Tour.

What we don't know

Now can we conclude that we have an Euler Tour?

If a graph \(G\) has no vertex of odd degree,

AND and all of its vertices with nonzero degree belong to a single connected component,

then it does have a closed Euler Tour.

What we now know

find_tour(G,v,marked):

    set curr := v

    set S := empty

    while there is an outgoing edge e = (curr,u)     which is not in S or marked:

        append e to S

        res_deg[curr] = res_deg[curr]-1

        set curr := u

return S

init marked = empty

init fragments = empty

while there is a vertex v such that res_deg(v) > 0:

    F = find_tour(G,v,marked)

    Add F to fragments

patchup():
    Let S[i] for i in 1, 2, ..., f
    denote the set of all fragments

    while f > 1:
        look for a fragment S[i] that intersects S[0]
        if S[i] does not exist:
            return false
        else:
            expand S[0] along S[i]
            remove S[i] from the set of all fragments