• Distance

  • distance $d(u,v)$: minimum number of edges in $(u,v)$-path
    • or: length of $(u,v)$-path
  • eccentricity of $u$: $\text{ecc}(y)$
    • $\text{ecc}(y):=\underset{v}{\max }d(u,v)$
    • i.e. maximum distance from $u$
  • center of graph: $u$ s.t.
    • $\text{ecc}(u)\le \text{ecc}(v)\forall v\in V_G$
  • radius: maximum distance from center
    • $\text{rad}(G):=\text{ecc}(u)$
  • diameter: maximum distance within the graph
    • $\text{diam}(G)=\underset{u,v}{\max }d(u,v)$

• Theorems on graphs

  • theorem: for any graph $G$:
    • $\text{rad}(G)\le \text{diam}(G)\le 2\cdot \text{rad}(G)$
  • proof:
    • left inequality: trivial
    • It cannot be the case that $\text{diam}(G)>2\cdot \text{rad}(G)$
      • center has max $\text{rad(G)}$ distance to $u,v$ (one used for diameter)
      • and must be able to connect at $2\cdot \text{rad}$ distance at least
  • theorem: if $T$ a tree w/ 3 or more vertices:
    • no leaf can be a center
  • proof:
    • suppose $l$ a leaf and $u$ its parent
    • then $d(l,v)=1+d(u,v)$ for all $v$
    • thus $l$ cannot have minimum eccentricity
  • theorem: let $T$ be tree. it has either 1 center or 2 neighboring centers
  • proof
    • base case:
      • 1 vertex: has 1 center
      • 2 vertices: has 2 neighboring center
    • (strong) induction: for nodes w/ $\ge 3$ vertices
      • we know that leaf cannot be center, thus we eliminate them
      • then, for all $v$ that's remaining, $\text{ecc}(v{)}^{\prime }=\text{ecc}(v)-1$
        • thus center does not change in any case
      • as new tree $T^{\prime }$ has less vertices, it is true
        • or: after repetition, it leads to base case

• Single-source shortest paths

  • let input: graph $G$ with vertex $v\in V_G$
  • we want output: $d(u,v)$ for all $v\in V_G$
  • solution 1: breath-first search
    • BFS: explores all vertices at distance 0 first
      • then 1, 2, and until all in CC are discovered
      • (max: $n-1$ rounds)
  • solution 2: path length calculation
    • let $\alpha (u,v,k)$: $$\alpha (u,v,k)=\{\begin{array}{ll}1& \text{if there is a }(u,b)\text{-walk w/ }k\text{ edge}\\ 0& \text{otherwise}\end{array}$$
  • distance: can be expressed as smallest $k$ w/ $\alpha =1$ $$\begin{array}{r}\alpha (u,v,k)=\underset{w\in N(v)}{\bigvee }\alpha (u,w,k-1)\\ \alpha (u,u,0)=1,\alpha (u,v,0)=0\text{ if }u\ne v\end{array}$$
    • $N(v)$: all neighbors of $v$

• Weighted graphs

  • for weighted graph w/ $w:E\to [0,\infty )$
    • length of path $\pi$:
    • $L(\pi )={\sum }_{e\in \pi }w(e)$
  • and distance given by:
    • $d(u,v)=\underset{\pi }{\min }L(\pi )$
  • initial weight:
    • $\alpha (u,u)=0$, $\forall v\ne u,\alpha (u,v,0)=\infty$
    • $\infty$: path not defined
  • for $k\ge 1$
    • $\alpha (u,v,k)=\underset{v_i\in N(v)}{\min }(\alpha (u,v_i,k-1)+w(v_i,v))$

• Adjacency matrix representation

  • adjacency matrix $A$:
    • $A[i][j]=w(\{i,j\})$
    • if no $(i,j)$-edge exists: $A[i][j]=\infty$
  • example

graph LR
    1((1))
    2((2))
    3((3))
    4((4))
    1--2---2
    1--1---3
    2--6---4
    3--0---4

$$A=\left[\begin{array}{cccc}\infty & 2& 1& \infty \\ 2& \infty & \infty & 6\\ 1& \infty & \infty & 0\\ \infty & 6& 0& \infty \end{array}\right]$$

• $A[i][j]$: shortest walk with one edge from $i$ to $j$
  • for $t$ steps: $$A^t[i][j]=\underset{k}{\min }(A^{t-1}[i][k]+A[k][j])$$
    • 👨‍🏫 a 'modified' matrix multiplication
      • multiplication into addition
      • sum into $\min$
  • for there to be $t$ step walk from $i$ to $j$:
    • must be $t-1$ step walk from $i$ to $k$, and edge between $k$ to $j$
• new matrix multiplication
  • $C=A\times B⟹C[i][j]=\underset{k}{\min }(A[i][k]+B[k][j])$
• logically, we obtain
  • $A^{100}=A^{50}\times A^{50}$
    • as any 100 step path: can be divided into two 50 step paths
  • $A^{25}=A^{12}\times A^{12}\times A$
• finally, $\text{pow}(A,t)=A^t$ can be computed by:

  if t = 1:
      return A
  if t = 0:
      return I // only valid path: to itself
  if t is even:
      B = pow(A, t/2)
      return B × B
  if t is odd:
      B = pow(A, ⌊t/2⌋)
      return B × B × A
  

• then distance can be computed by
  • $d(u,v)=\underset{k}{\min }{A}^k[u,v]$
  • w/ shortest walk on $k$ edges
• to support shortest walk w/ at most $k$ edges
  • self-loops can be added to each vertex $A[i][i]=0$
  • in this case, we have:
  • $d(u,v)=A^{n-1}[u][v]$
    • any CC: must be connected in $n-1$ edges at least
    • $n$: no. of vertices

• Dijkstra's Algorithm

  • Input: weighted graph $G$ and vertex $y\in V_G$
    • w/ non-negative edge weights $E_G\to [0,\infty )$
  • Output: distance $d(u,v)$ for every vertex $v$
  • intuition
    • check vertex $u$
    • for all of its neighbors: update its weight
    • for a walk with minimum length (say: $(u,v)$):
      • shortest walk for two vertices are finalized
      • as all other alternatives: longer than $u,v$
    • then, update weights for all of $v$'s neighbors, repeat
  • rough pseudocode
    1. initialize distances: $d(u,v)=\infty$ for all $u\ne v$
       • $d(u,v)=0$
    2. create a priority queue $Q$ containing all vertices in $V$
    3. while $Q$ is not empty:
       1. extract vertex $v$ w/ minimum distance from $Q$
       2. for each neighbor $w$ of $v$:
          1. if $d(u,v)+w(v,w)<d(u,w)$:
             1. $d(u,w)=d(u,v)+w(v,w)$
             2. update the priority of $w$ in $Q$
  • Dijkstra's algorithm: results in a spanning tree
    • prove that it is indeed shorted path

• Matchings

  • matching $M\subseteq E$
    • s.t. no two edges in $M$ sharing an endpoint
    • not necessarily covering all vertices
  • example

graph LR
    1(( ))
    2(( ))
    3(( ))
    4(( ))
    5(( ))
    1-->2
    2--M-->1
    1-->4
    2-->3
    3--M-->4
    4-->3
    4-->5

• Types of matchings

  • $M$: maximal matching if $\nexists M^{\prime }\supset M$
    • cannot add any more match
  • example

graph LR
    1(( ))
    2(( ))
    3(( ))
    4(( ))
    1-->2
    2--M-->1
    2-->3
    3--M-->4
    4-->3

• $M$: maximum matching is $\forall M^{\prime }|M^{\prime }|\le |M|$
  • all maximum matching: maximal
    • not the other way around
  • example

graph LR
      1(( ))
      2(( ))
      3(( ))
      4(( ))
      1-->2
      2--M-->3
      3-->4

    - maximal (cannot add anymore), but not maximum

• Alternating walks and augmenting paths

  • suppose: $M$ being a matching in $G$
  • alternating $(u,v)$-walk: defined at
    • walk that alternates between edges in the matching $M$ and edges in the set E</span>M
    • 👨‍🎓 ~=changing group every turn?
  • augmenting walk/path: alternating path starting & ending in E</span>M
  • augmenting edge: cant be used for expanding the matching
    • simply: swap the colors, within the alternating path

• Theorem on maximum matchings [Berge]

  • theorem: a matching $M$ is maximum iff it does not contain an augmented path
  • proof:
    • if there is augmented -> $M$ not maximum: trivial
      • as stated above
    • $M$ not maximum -> $\exists$ augmented
      • let $M$ be a non-maximum matching
      • let $M^{\ast }$ be a maximum matching
      • let $F\subset E$
        • s.t. $F=M\Delta M^{\ast }$
          • i.e. $M\cup M^{\ast }-M\cap M^{\ast }$
        • and let $H=(V,F)$: a graph
        • degree of $v\in H$: at most 2
          • i.e. one from $M$, another from $M^{\ast }$
      • previously: a graph w/ degree at most 2: made of cycles / paths
        • for a cycle (w/ every edge having degree 2)
          • it must be an even cycle
          • as $M$ and $M^{\ast }$ within it must alternate all the time
        • however, as $|M|<|M^{\ast }|$, there must be a path
          • starting with $M^{\ast }$ and ending with $M^{\ast }$
          • which is an augmented path for $M$
        • if the matching was maximum: there won't be an augmented path