Given an undirected, connected and unweighted simple graph $G(V,E)$.

We call $j$ a neighbor of $i$ if $(i,j)\in E$. $N_i$ is the set of neighbors of $i$. Each vertex only knows its neighbors and has no idea of any other vertex in $G$. We want to find a walk $L$ in $G$ such that $L$ covers all vertices in $V$, but the walk need not be a trail, i.e., one can travel each edge in $E$ more than once.

Given any undirected, connected and simple graph $G(V,E)$,each node of which is considered as a city. We call $j$ a neighbor of $i$ if $(i,j)\in E$. $N_i$ is the set of neighbors of $i$. $|V|=N$

There is a traveler who starts travelling from some city and wants to visit all cities. But he does not have a map,it means he only knows about the local information,i.e. he only knows the neighbors of a city when staying at that city. Each time, he moves to a neighbor of currently staying city.

Suppose he traveled from city a to b 3 times and b to a 2 times, then we call link (a,b) (or (b,a)) has been used totally 5 times.

I am considering an algorithm: each time, suppose he stays in some city $i$, and he selects a neighbor city of $i$ whose link between $i$ has been used least times in the past, except the city he visited at last time, and move to that city. If more than 1, any of them is OK.

I want to know for any graph $G$, after how many times, he will must visit all the cities.

An undirected  , connected and unweighted simple graph G(V,E)

We call j is a neighbor of i if (i,j) is in E. N_i is the neighbor set of i. each vertex only know its neighbor and have no idea on any other vertex in G. We want to find a walk L in G such that L covers all nodes in V, but the walk does not need to be a trail, i.e., one can travel each edge in E for many times.

Given an undirected, connected and unweighted simple graph $G(V,E)$.

We call $j$ a neighbor of $i$ if $(i,j)\in E$. $N_i$ is the set of neighbors of $i$. Each vertex only knows its neighbors and has no idea of any other vertex in $G$. We want to find a walk $L$ in $G$ such that $L$ covers all vertices in $V$, but the walk need not be a trail, i.e., one can travel each edge in $E$ more than once.