• Matching (cont.)

  • design an algorithm giving a maximum matching
    • pick any trivial matching: empty or maximal matching
    • if it's not a maximum: search for an augmented path
      • while such path exists: expand the matching
    • finally, if no augmented path can be found
    • it's a maximum matching
  • similar technique: starting with suboptimal solution
    • and keep improving the method
      • 👨‍🏫 still "augmented"
    • until it can't be further improved
    • theorem: if it can't be further improved => it's optimal

• Network flow

  • problem of interest: network flow
  • directed graph $G=(V,E)$
    • w/ capacity function $c:E\to [0,\infty )$
    • claim: there are two special vertices
      • source $s$ and sink $t$ $\in V$
    • example

graph LR
      a((a))
      b((b))
      c((c))
      d((d))
      a--4-->b
      a--5-->c
      b--6-->d
      b--1-->c
      c--5-->d

- source: <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal">a</span></span></span></span>
- sink: <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6944em;"></span><span class="mord mathnormal">d</span></span></span></span>

• 👨‍🏫 consider: edge as unidirectional water pipe
  • and capacity: of water flow
• question: what's the max possible flow from source to sink, s.t. all water reaches sink?
  • 👨‍🏫 name: maximum flow algorithm / max flow
• from above example: $a\to b$ has capacity $4$, thus we can't send flow larger than 4
• instead of edge: assign capacity to every pair of vertices
  • ~= Dijkstra?
  • now: $c:V\times V\to [0,\infty )$
• flow: function $f:V\times V\to \mathbb{R}$
  • $f(u,v)$: flow from $u$ to $v$
• condition / constraints
  • $\forall u,v\in V,f(u,v)\le c(u,v)$
  • $\forall u,v\in V,f(u,v)=-f(v,u)$
    • skew symmetry
    • sending 2 units $u\to v$: same as sending -2 units $v\to u$
  • ∀u,∈V</span>{s,t}, ∑v​f(u,v)=0=∑v​f(v,u)
    • all except source & sink: indegree = outdegree
    • preservation of flow: water coming out from source = water reaching sink
• for source: ${\sum }_{v}f(s,v)>0$
  • there may be edge to source as well
  • and for sink: ${\sum }_{v}f(v,t)<0$
• solution $|f|$, flow s.t. it satisfies 3 conditions above:
  • $|f|={\sum }_{v}f(s,v)={\sum }_{v}f(v,t)$
• intuition
  • for a walk: total flow that can be pass through it = minimum capacity of an edge in walk
    • e.g. 1 unit across $a\to b\to c\to d$
      • capacity: 4,1,5
  • interestingly: one can send flow in reverse direction, too
    • e.g. 1 unit across $a\to c\to b\to d$
      • capacity: 5,-1,6 respectively
      • as it "cancels" my previous $b\to c$ flow
  • formally speaking:
    • if there is flow $f(a,b)=1$
      • then $f(b,a)=-1$
      • and $c(b,a)=0\ge f(b,a)$, satisfying condition
  • thus: find a path $s\to t$
    • using only the edges: $f(u,v)<c(u,v)$
• residual graph of flow $f$: $G_f$
  • w/ same vertices as $G$
  • $r(u,v)=c(u,v)-f(u,v)$
  • any non-zero edges on this graph: can be utilized for more flow
  • upon move $f(u,v)$: $r(u,v)-=f(u,v)$
    • and $r(v,u)+=f(v,u)$
• is there a case: our flow is maximal, but not maximum?

• Ford-Fulkerson's algorithm

  • basically the idea above
  • start with initial flow as 0
  • while there exists an augmented path from the source to the sink
    1. find an augmenting path using any path-finding algorithms
       • BFS or DFS
    2. determine: amount of flow that can be sent along the path
       • i.e. minimum residual capacity along the edges of the path
    3. increase the flow along the augmented path by determined amount
  • return the maximum flow

• Proof of correctness

  • first: extend $f$
    • s.t. for a set $A,B$: $f(A,B)={\sum }_{a\in A}{\sum }_{b\in B}f(a,b)$
  • $f(s,V)=|f|={\sum }_{v\in V}f(s,v)$
  • and $f(V,t)={\sum }_{v\in V}f(v,t)=|f|$
  • suppose: $A\cup B=\varnothing$
    • then, $f(A\cup B,C)=f(A,C)+f(B,C)$
  • define: cut in $G$: division $V=A\bigsqcup B$
    • s.t. $s\in A\wedge t\in B$
    • what is $f(A,B)$ is $(A,B)$: a cut?
      • $f(A,B)=|f|$
        • as for any non-$s,t$ $u,v$: $f(u,v)=0$
      • $f(A,B)=f(A,V)-f(A,A)$
        • $f(A,A)=0$ (due to condition)
      • =f(s,V)+f(A</span>{s},V)
        • f(A</span>{s},V)=0 as sum of flow from any non-s,t vertex: 0
      • $=|f|$, almost by definition
  • similarly, define $c(A,B)$ for vertex group $A,B$
    • $c(A,B)={\sum }_{a\in A}{\sum }_{b\in B}c(a,b)$
    • then: $f(A,B)\le c(A,B)$ (as inequality holds for every elements)
  • a flow is optimal if $f(A,B)=c(A,B)$ (can't be further improved)
    • thus: find a cut satisfying the condition
  • let $A$: all set of vertices that an be reached from $s$ in residual graph
    • and the rest: $B$
    • $t\in B$ as algorithm continued until you can't reach $s\to t$ in residual anymore
  • when the algorithm is stuck: there is no edge going $A\to B$
    • if there was: it's reachable, and thus marked $B$
    • as original graph and residual graph: shares vertices, but not edges
      • there "might" be an edge $u\in A\to v\in B$
      • but its non-existence in residual: shows that all its capacity has been used up
  • how about $\alpha \in B\to \beta \in A$?
    • in original graph, it might have $c(\alpha ,\beta )\ne 0$
    • however: in can't have positive flow as that would add positive edge $\beta \to \alpha$ in residual graph
      • which is $A\to B$ edge, impossible
  • finally: $f(A,B)={\sum }_{u\in A}{\sum }_{v\in B}f(u,v)={\sum }_u{\sum }_{v}c(u,v)=c(A,B)$
    • $=|f|$
    • thus: the algorithm outputs optimal solution
  • related idea: it also gives minimum capacity with $|f|$
  • does this algorithm runs forever?
    • if flows doesn't start with 0 / set is of real numbers: yes
      • each step: increases flow at least by one
    • new range of $f:V\times V\to \mathbb{N}$
    • furthermore, upper bound for flow exists: capacity
      • thus it can't run forever
  • 👨‍🏫 in 3711, you make it run for $\mathbb{R}$ w/ slight modification