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Suppose that $G=(V,E)$ is a simple graph and $P=(V_1,E_1)$ is a path in $G$ where $$V_1=\{v_0,v_1,\cdots,v_n\},\ E_1=\{v_0v_1,v_1v_2,\cdots,v_{n-1}v_n\}.$$ I found that if the path $P$ satisfies:

For any $v_i\in V_1$, there exist $v_j\in V_1\setminus \{v_i\}$ and $u\in V\setminus V_1$ such that $uv_i,uv_j\in E$.

Then you can always find a longer $v_0$-$v_n$ path in $G$.

I have tried to find a counterexample to this for a long time but still cannot find one. So I think maybe the above conjecture is true. Is it true or is there any known result about this? Any ideas are welcome!

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    $\begingroup$ Just take $v_i$ to be one end of the path. $\endgroup$ – Brendan McKay Jan 2 at 11:05
  • $\begingroup$ The path have the same ends with $P$. $\endgroup$ – user173856 Jan 2 at 11:23
  • $\begingroup$ Yes, sorry, I missed that aspect. $\endgroup$ – Brendan McKay Jan 2 at 11:32
  • $\begingroup$ @user173856: Is $G$ oriented or not? $\endgroup$ – Alex M. Jan 9 at 21:31
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    $\begingroup$ @WlodAA: Yes, I have already revised it. $\endgroup$ – user173856 yesterday
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Disclaimer: this is not a proof but rather some ideas which may or may not help.

If we identify the pairs $v_i, v_j$ such that there's $u$ with $v_iuv_j \in E$ with a segment $[i, j]$, we are going to have at least $\frac{n+1}{2}$ segments.

Now we can argue similarly to the Vitali covering lemma. We have a collection of segments, covering $1, \ldots, n$ (let's assume we count starting from $1$ rather then from $0$). We can find two families of segments $\{ I \}$ and $\{ J \}$ such that

  1. Segments in each family are either disjoint or nested
  2. There are disjoint subfamilies $\tilde I$ and $\tilde J$ such that each segment intersect at most two neighbouring.

Now at least one of the families has measure at least $\frac{n+1}{2}$. Assume wlog that $$ | \cup I_j | \geq | \cup J_j | $$

If we order the segments in $\tilde I$ and $\tilde J$ so that $J_k$ intersects exactly two neighbours say $I_{l-1}$ and $I_{l}$ and $|J_k \cap I_{l-1}| + |J_k \cap I_{l}| = |J_k|$, then it would suffice to walk trough the nested segments in $I_{l-1}$, then "jump" using $J_k$ to $I_l$ and walk through the nested family in $I_l$.

The scheme above assumes that all $u$'s are distinct.

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The graph $G$ below is a counterexample.

Construct $G$ as follows:

  1. Letting $n$ be quite large, $V_1 = \{v_0,v_1,\ldots, v_{n+1}\}$ and $E_1= \{v_iv_{i+1}; i=0,\ldots, n\}$ as stated in the problem.

  2. Let $k$ be an integer about $\frac{n}{3}$ and let $l=\frac{n}{2}$. So $2k=\frac{2n}{3}$ and $2k-l=\frac{n}{6}$.

  3. Then $V \setminus V_1$ $=\{u_0,\ldots, u_{l-1},u_{l+1},\ldots, u_{2k-1},u_{2k+1} \ldots, u_{n}\}$. $= \{u_j; \ j \in \{0,\ldots, n\}\setminus \{l,2k\} \}$

  4. Then the edges in $G \setminus E_1$ are:

(a) $v_0u_0$ and $u_0v_{l}$; $v_ku_k$ and $u_kv_{2k}$;

(b) $v_iu_i$ and $u_iv_{n+1}$ for $i=1,2,\ldots, l-1$;

(c) $v_mu_m$ and $u_mv_l$ for $m=n, n-1,n-2, \ldots, 2k+1,2k-1,\ldots, l+2$;

(d) $v_{2k}u_{l+1}$ and $u_{l+1}v_{l+1}$.


Here is a sketch of the proof that there is no path $P$ in $G$ starting at $v_0$ and ending at $v_{n+1}$ that is longer than $E_1$.

Claim 1: (i) Any path $P$ in $G$ not starting nor ending at any vertex in $V \setminus V_1$ can take only $O(1)$ of the vertices in $V \setminus V_1$. (ii) Any path $P$ of length at least $n-2$ from $v_0$ to $v_{n+1}$ must cover almost all of $V_1$ i.e., at most $O(1)$ vertices in $V_1$ are not in $P$.

To see (i) of Claim 1, note that every $u_j$ has degree 2 and so if a path $P$ contains $u_j$ and doesn't end at $u_j$ then $P$ must contain both edges incident to $u_j$. But every $u_j$ is adjacent to at least one of $v_l, v_k,v_{2k}$, or $v_{n+1}$.

So in light of (i) of Claim 1, if there is a path $P$ from $v_0$ to $v_{n+1}$ of length $\ell$ then $P$ must contain $\ell-O(1)$ vertices in $V_1$. If $\ell \ge n-O(1)$ then almost all of $V_1$ must be in $P$.

Claim 2: The longest path in $G' \doteq G \setminus \{v_k,v_{2k},v_{n+1}\}$ has length no larger than $\max\{n-2k + (2k-l), n-2k + (l-k)\}$ $=\max\{\frac{n}{3}+\frac{n}{6}, \frac{n}{3}+\frac{n}{6}\} = \frac{n}{2} $.

Indeed, the components of $G \setminus \{v_k,v_{2k},v_l,v_{n+1}\}$ [i.e., take the graph with the vertex $v_l$ removed from $G'$] are

$C_1= G[\{v_m; 2k<m<n+1 \}+\{u_m; 2k<m<n+1 \}]$;

$C_2=G[\{v_m; l<m<2k\}+\{u_m; l<m<2k\}]$; and

$C_3= G[\{v_m; k<m<l\} +\{u_m; k<m<l\}]$.

One of these components $C_1$ has no path of length greater than $n-2k=\frac{n}{3}$; $C_2$ has no path of length greater than $2k-l=\frac{n}{6}$ and $C_3$ has no path of length greater than $l-k=\frac{n}{6}$. So this means that the length of the longest path in $G'$ is no more than the length of the longest path in $C_p$ plus the length of the longest path in $C_{p'}$ for some distinct $p, p' \in \{1,2,3\}$ where $C_1,C_2,C_3$ are as above. From this Claim 2 follows.

Remainder Of The Proof: So let $P$ be a path from $v_0$ to $v_{n+1}$, and let $e$ be the first edge in $P$ that is not in $E_1$. [Order the edges in $P$ from the starting point $v_0$ of $P$ to the finishing point $v_{n+1}$ of $P$.]

Case 1: $e=v_ku_k$. Then $e$ is the $k+1$-th edge in $P$, and the first $k$ edges of $P$ form the path $v_0v_1\ldots v_k$, and the $(k+2)$nd edge in $P$ is $u_kv_{2k}$. Thus as both $v_k$ and $v_{2k}$ are vertices that are in the first $k+2$ edges of $P$, the length of $P$ can be no more than $k+2$ plus the length of the longest path in $G' = G \setminus \{v_k,v_{2k},v_{n+1}\}$ plus 1. Thus from Claim 2 $P$ can be no longer than $k+2+\frac{n}{2}+1=\frac{5n}{6}+3 < n$.

Case 2: $e=v_{l+1}u_{l+1}$. Then $e$ is the $(l+2)$nd edge in $P$ and the first $l+1$ edges of $P$ form the path $v_0v_1\ldots v_lv_{l+1}$, and the $(l+3)$rd edge in $P$ is $u_{l+1}v_{2k}$. So $v_{2k},v_l$ and $v_{l+1}$ are vertices that appear in the first $l+3$ edges of $P$. However, note that $G \setminus \{v_{2k},v_{l+1},v_{l}\}$ has more than one connected component, and furthermore, every vertex in the set $V_{11}=\{v_{2k-1},v_{2k-2},\ldots, v_{l+2}\}$ in a different component from every vertex in the set $V_{12}=\{v_{2k+1},\ldots, v_{n}\}$. So the remaining edges of $P$ will miss at least either every vertex in $V_{11}$ or every vertex in $V_{12}$. Furthermore, neither $V_{11}$ nor $V_{12}$ intersects the first $l+3$ edges of $P$, so either $P \cap V_{11}$ or $P \cap V_{12}$ must be empty. As each of $V_{11}$ and $V_{12}$ have at least $\frac{n}{6}$ vertices, so this and Claim 1 implies that $P$ will not have length $n$.

Case 3: $e=v_lu_m$ for some $m < 2k$. Then $e$ is the $(l+1)$st edge in $P$ and the first $l+1$ edges of $P$ form the path $v_0v_1\ldots v_lu_m$ and the $(l+2)$nd edge in $P$ is $u_mv_{m}$. However, note that the only way $P$ can be extended to reach $v_{n+1}$ is if the rest of $P$ is $v_mv_{m+1}\ldots v_{n+1}$. [Indeed, note that $v_k$ and $v_{l}$, and $v_m$ are in the first $l+2$ edges in $P$, and so the remaining edges of $P$ are all (a) in a single component of $G \setminus \{v_k,v_l,v_m\}$ that has a vertex $v'$ adjacent to $v_m$ that isn't in the first $l+2$ edges of $P$, and (b) form a path starting at $v'$ and ending at $v_{n+1}$. The only such vertex $v'$ is $v_{m+1}$, and furthermore, one can see that the component of $G \setminus \{v_k,v_l,v_m\}$ containing $v_{m+1}$ is a tree, and so the path $v_{m+1} \ldots v_{n+1}$ is the only path from $v'=v_m$ to $v_{n+1}$ and so these must be the remaining edges of $P$.] As $m > l+2$ this implies that $P$ will not have length longer than that of $v_0v_1\ldots v_{n+1}$.

Case 4: $e=v_lu_m$ for some $m > 2k$. Can be finished in a similar fashion as Case 3.

Case 5: $e=v_0u_0$ which is the first edge in $P$. Then $v_0u_0$ is the first edge in $P$, and the 2nd edge in $P$ is $u_0v_l$. We divide into two subcases.

The first subcase of Case 5 is that the next vertex in $P \cap V_1$ after $v_l$ is in the set $\{v_m: m > l\}$. Then, by Claim 1 it follows that $P$ must contain almost every vertex in $\{v_i: i<l\}$ as well, and so as every path in $G \setminus \{v_{n+1}\}$ from $\{v_m: m > l\}$ and $\{v_i: i< l\}$ must pass through $\{v_l,v_{2k}\}$, it follows that $P$ must enter back into $\{v_i: i< l\}$ via the edges $v_{2k}u_kv_k$, and then $P$ must cover almost every vertex in $\{v_i: i< l\}$. But this is impossible as every vertex in $\{v_i: k<i<l\}$ is in a different component of $G \setminus \{v_l,v_k,v_{n+1}\}$ from every vertex $\{v_i: 0<i<k\}$, so $P$ cannot be extended to cover vertices in both $\{v_i: k<i<l\}$ and $\{v_i: 0<i<k\}$. So $P$ cannot be extended from $v_k$ to cover vertices in both the sets $\{v_i: 0<i<k\}$. $\{v_i: k<i<l\}$. Each of these sets has $\frac{n}{6}$ vertices. So by Claim 1 $P$ cannot have length $n$.

The 2nd subcase of Case 5 is that the next vertex in $P \cap V_1$ is in the set $\{v_i: i< l\}$. One can finish using the same line of reasoning as in the above subcase.

*Case 6: $e=v_iu_i$; $i=1,\ldots, l-1$; $i \not = k$. Then $e$ is the $i+1$-th edge in $P$, and the $i+2$th edge in $P$ is $u_iv_{n+1}$, which implies that $P$ has length $i+2 \le l+1 < n$.

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  • $\begingroup$ I see $v_{l+2}$ adjacent to $u_{l+2}$ in (c), but there is no $u_{l+2}$. $\endgroup$ – Brendan McKay yesterday
  • $\begingroup$ Sorry was a typo on my part I editted that $\endgroup$ – Mike yesterday
  • $\begingroup$ @BrendanMcKay made some more edits $\endgroup$ – Mike 20 hours ago

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