Finite Playground

Crossing number of Kn and Gioan’s Theorem

The Harary-Hill Conjecture states that the crossing number of complete graph Kn is

$\displaystyle \frac{1}{4} \left\lfloor\frac{n\vphantom{-0}}{2}\right\rfloor \left\lfloor\frac{n-1}{2}\right\rfloor \left\lfloor\frac{n-2}{2}\right\rfloor \left\lfloor\frac{n-3}{2}\right\rfloor$.

Instead of looking at all possible drawings (which is infinite), we can partition the drawings into classes, depending on the rotation system of the drawing.

The following theorem guarantees us that the crossing number is still well-defined.

Gioan’s Theorem.  Let D1 and D2 be two drawing of graph Kn in the plane.  Then one can transform D1 into Dusing only $\Delta$-moves.  As a corollary, the crossing number of D1 and Dis the same.

Noted that in general this is not true when the graph is not complete.

Perfect matching equivalence

There are so many things can be written…

Anyway, the following observation is really interesting; details can be found in the book “matching theory” by Lovasz and Plummer.

Consider a general graph $G$.  An equivalence relationship can be defined on the nodes of the graph as follows: two nodes $x$ and $y$ are equivalent if $G-x-y$ has no perfect matching(!).

Dilworth Theorem and equivalence to Konig-Egervary Theorem

Dilworth’s Theorem. If $P$ is a finite poset, then the maximum size of an antichain in $P$ equals the minimum number of chains needed to cover the elements of $P$.

Proof. Induction on $|P|$. We separate into two cases.

(a) Some maximum antichain $A$ omits both a maximal element and a minimal element of $P$. Then neither $D[A]$ or $U[A]$ contains all of $P$; the maximality of $A$ implies $P = D[A] \cup U[A]$. Both $D[A]$ and $U[A]$ has width $w(P)$ because they contain $A$, and they only intersect in $A$. Apply induction hypothesis on both sets, we obtain $k$ chains covering $D[A]$ and $k$ chains covering $U[A]$; combining these chains form $k$ chains covering $P$.

(b) Every maximum antichain of $P$ consists of all maximal elements or all minimal elements. Thus $w(P-\{x,y\}) \le k-1$ if $x$ is a minimal element and $y$ is a maximal element below $x$. Apply induction hypothesis on $P-\{x,y\}$ gives us $k-1$ chains; add chain $\{x,y\}$ to get a covering of $P$.

Theorem.  Dilworth’s Theorem is equivalent to Konig-Egervary Theorem.

Proof.  Dilworth to Konig-Egervary: View an $n$-node bipartite graph $G$ as a bipartite poset. The nodes of one part are maximal elements, and nodes of the other part are minimal. Every covering of the poset of size $n-k$ uses $k$ chains of size 2, which is actually a matching. Each antichain of size $n-k$ corresponds to an independent set in $G$, and the rest of $k$ nodes forms a vertex cover.

Konig-Egervary to Dilworth: Consider an arbitrary poset $P$ of size $n$. Define a bipartite graph $G$ as follow: For each element $x \in P$, create two nodes $x^-$ and $x^+$. The two parts of the graph are $\{x^- : x \in P\}$ and $\{x^+ : x \in P\}$. The edge set is $\{x^- y^+ : x <_P y\}$.

Every matching in $G$ yields a chain-covering in $P$ as follow: We start we $n$ chain, each contains a unique element. If $x^- y^+$ is in the matching then $x$ and $y$ are combined in the same chain. Therefore because each node in $G$ appears in at most on edge of the matching, a matching of $k$ edges forms a covering of $n-k$ chains.

Given a vertex cover $R$ in $G$ of size $k$, define a corresponding antichain $A = \{x \in P : x^-, x^+ \notin R\}$; this is indeed an antichain because if there is a relation $x <_P y$ in $A$ then edge $x^- y^+$ will be presented, and vertex cover $R$ needs to contain at least one of $x^-, y^+$.

Claim. No minimal vertex cover of $G$ uses both $\{x^-,x^+\}$, because by transitivity the sets $\{z^- : z \in D(x)\}$ and $\{y^+ : y \in U(x)\}$ induce a complete bipartite subgraph in $G$. A vertex cover of a complete bipartite graph must use all of one part. Since $x^-,x^+$ have all the neighbors in these two sets, we can remove one of $x^-,x^+$ that is not in the right part and decrease the size of the vertex cover.

Thus a minimum vertex cover of size $k$ yields an antichain of size $n-k$.

Graph coloring theorems revisited II – Vizing’s Theorem

Alexandr Kostochka found the following new proof to the classical theorem.

Vizing’s Theorem.  For every simple graph $G$, $\chi'(G) \le \Delta(G) + 1$.

Proof.  We prove by contradiction; assume that $G$ is a minimal counter-example (in terms of number of edges) to the statement; that is, $G$ has no edge-($\Delta+1$)-coloring.

Let $\phi$ be an edge-($\Delta+1$)-coloring of $G-xy$. For every $v \in V(G)$, let $O(v)=O_\phi(v)$ denote the set of colors in $[\Delta+1]$ not used to color the edges incident to $v$; one can see that $O(x) \cap O(y) = \emptyset$.

Construct the auxiliary multidigraph $H$ as follows: $V(H) = N_G(y)$ and there is a directed edge from $u$ to $v$ in $E(H)$ if $\phi(vy) \in O(u)$. Intuitively an edge from $u$ to $v$ means that if we can recolor $vy$, then we can recolor $uy$ as well. Let $X$ be the set of vertices reachable in $H$ from $x$ ($x \in X$ as well); any recoloring of an edge $wy$ for $w\in X$ will imply that we can color $xy$. Since the out-neighbors of a vertex reachable from $x$ also are reachable from $x$, $N^+_H(v) \subseteq X$ for every $v \in X$.  Therefore

$\sum_{v \in X} d^+_{H[X]}(v) = |{E(H[X])}| = \sum_{v \in X} d^-_{H[X]}(v) \text,$

and the theorem will be proven if we can show that

1. $d^+_{H[X]}(v) \ge \Delta - d(v) + 1 + [v = x]$ and
2. $d^-_{H[X]}(v) \le 1-[v=x]$ for all $v \in X$;

this implies

$0 = \sum_{v \in X} \{d^+_{H[X]}(v) - d^-_{H[X]}(v)\} \ge 2 + \sum_{v \in X} (\Delta - d(v)) \ge 2$,

a contradiction.  Items (1) and (2) follow from the following claims. Let $\alpha \in O(y)$.

Claim 1 (Exchange argument).  $\alpha\notin O(v)$ for every $v\in X$.
Proof sketch.  Recolor the edges of the form $wy$ for each $w$ on the path from $x$ to $v$.

Claim 1 yields that for every $v \in X$ and $\beta \in O(v)$, there is some $w \in N(y)$ with $\phi(yw)=\beta$. Then by the definition of $H$, $vw\in H[X]$. Because $|{O(v)}| \ge \Delta+1-d(v)$ for each $v\in V(G)$ and $|{O(u)}| \ge \Delta+1-d_{G-xy}(u) = \Delta-d(u)+2$ for $u \in\{x,y\}$; so

$d^+_{H[X]}(v) \ge \Delta+1-d(v)$ for every $v \in X$ and $d^+_{H[X]}(x) \ge \Delta-d(x)+2$.

Claim 2 (Distinct $O$-sets).  For all distinct $v,w \in X$, $O(v) \cap O(w) = \emptyset$.
Proof sketch.  A $[\beta,\gamma]$-path in $G$ is a path whose edges are alternately colored with $\beta$ and $\gamma$. A $[\beta,\gamma](a,b)$-path is a $[\beta,\gamma]$-path from $a$ to $b$ in $G$.

Claim: If $v \in X$ and $\beta \in O(v)$, then $G$ contains an $[\alpha,\beta](v,y)$-path.  The claim follows from a Kempe-chain-like argument to recolor $vy$ with $\alpha$.

Then suppose there are $v,w\in X$ with $\beta\in O(v)\cap O(w)$, then the $[\beta,\alpha]$-path starting at $y$ must end at both $v$ and $w$, a contradiction.

Claim 2 yields that $d^-_{H[X]}(v) \le 1$ for every $v \in X$ and $d^-_{H[X]}(x) = 0$, because the color on edge $vy$ corresponds to a unique in-neighbor of $v$ (and there are no color on edge $xy$).

Graph coloring theorems revisited I – Brook’s Theorem and Rubin’s Block Theorem

Rubin’s Block Theorem.  令 G 為一雙連通圖; 若 G 不是完全圖或奇圈, 則 G 中有一個含有最多一條弦 (chord) 的偶圈.

Lemma 1.  給定連通圖 G 與一色組函數 L, (i) 若有某個點 v 的顏色選擇多於 degree, 則圖 G 可順利著色; (ii) 若圖 G 為雙連通, 且有兩個點上的顏色組不同, 則圖 G 可順利著色.

Lemma 2.  連通圖 G 若存在一 degree-choosable 生成子圖 H, 則圖 G 也是 degree-choosable.
Proof.  By Lemma 1(a).

Theorem.  圖 G 不為 degree-choosable 若且唯若每個雙連通集 (block) 皆為完全圖或奇圈.
Proof.  由 Lemma 2 與 Block Theorem 我們只需要證明含有最多一條弦的偶圈是 degree-choosable; 但利用 Lemma 1(b) 這很明顯.

Brooks’ Theorem (extended).  若圖 G 不是完全圖或奇圈, 則 $\chi_{\ell}(G) \le \Delta(G)$.
Proof.  利用上面的 Theorem 我們只需要處理當所有雙連通集都是完全圖或奇圈的情形.  如果有個點的 degree 比 $\Delta(G)$ 小, 利用 Lemma 1(a) 就結束了; 所以 G 一定是 $\Delta(G)$-正則 (regular); 但在這種情況下表示我們只有一個 block (不然斷點的 degree 會有問題), 證明完成.

A lower bound to Ramsey numbers and an application

Ramsey’s Theorem.  $R(p_1,\ldots,p_k;r)$ is finite.

$R(p_1,\ldots,p_k;r) \le R(p_1',\ldots,p_k';r-1)+1$, where $p_i' = R(p_1,\ldots,p_{i}-1,\ldots,p_k;r)$.

Theorem. $R(p,p) = \Omega(p 2^{p/2})$.
Proof.  每個特定的$p$-clique 或$p$-independent set 出現的機會都是$2^{-{p \choose 2}}$; 因此若$2{n \choose p} 2^{{n \choose 2}-{p\choose 2}} < 2^{n \choose 2}$的話就表示某個大小為$n$的圖同時沒有大小為$p$的 cliques 或 independent sets, 因此$R(p,p) > n$.  整理不等式便得出我們的結果.

Shortest paths in graphs with integral weights

Thorup ’03, directed graphs, $O(m + n \log\log n)$, by integral priority queue
Thorup ’99, undirected graphs, $O(m)$, by hierarchical bucketing.

van Emde Boas ’77, $O(m \log\log C)$,
Thorup ’03, $O(m + n \log\log C)$.

Survey: Zwick ’01.