Preflow-push

Abstract view

Algorithmic problem: max-flow problem (standard version)

Type of algorithm: loop.

Definition:

1. For [math]\displaystyle{ a\in A }[/math], let [math]\displaystyle{ f(a) }[/math] be a real number (not necessarily foring a flow). For [math]\displaystyle{ v\in V\setminus\{s,t\} }[/math], the excess of [math]\displaystyle{ v }[/math] with respect to [math]\displaystyle{ f }[/math] is defined by [math]\displaystyle{ e_f(v):=\sum_{w:(w,v)\in A}f(w,v)-\sum_{w:(v,w)\in A}f(v,w) }[/math].
2. A node [math]\displaystyle{ v\in V\setminus\{s,t\} }[/math] is called active it its excess is positive.
3. Let [math]\displaystyle{ \Delta }[/math] denote the sum of the distance labels [math]\displaystyle{ d(v) }[/math] of all active nodes [math]\displaystyle{ v\in V\setminus\{s,t\} }[/math].

Invariant: Before and after each iteration:

1. For each arc [math]\displaystyle{ a\in A }[/math], it is [math]\displaystyle{ 0\leq f(a)\leq u(a) }[/math] .
2. For each node [math]\displaystyle{ v\in V }[/math], it is [math]\displaystyle{ e_f(v)\geq 0 }[/math].
3. The node labels [math]\displaystyle{ d }[/math] form a valid distance labeling, and it is [math]\displaystyle{ d(s)=|V| }[/math].
4. The currently active nodes are stored in a set [math]\displaystyle{ S }[/math].

Variant: At least one

1. The label [math]\displaystyle{ d(v) }[/math] of at least one node [math]\displaystyle{ v\in V\setminus\{s,t\} }[/math] is increased.
2. A saturating push is performed.
3. The value of [math]\displaystyle{ \Delta }[/math] decreases (in case of a non-saturating push).

Break condition: [math]\displaystyle{ S=\emptyset }[/math].

Induction basis

Abstract view:

1. For all arcs [math]\displaystyle{ a\in A }[/math], set [math]\displaystyle{ f(a):=0 }[/math].
2. For each arc [math]\displaystyle{ (s,v)\in A }[/math], overwrite this value by [math]\displaystyle{ f(a):=u(a) }[/math].
3. Compute a valid distance labeling [math]\displaystyle{ d }[/math], for example, the true distances from all nodes to [math]\displaystyle{ t }[/math] in the residual network.
4. Set [math]\displaystyle{ d(s):=n }[/math].

Proof: Obvious.

Induction step

Abstract view:

1. Choose an active node [math]\displaystyle{ v }[/math] from [math]\displaystyle{ S }[/math].
2. If there is at least one admissible arc leaving [math]\displaystyle{ v }[/math]:
   1. Choose one such arc [math]\displaystyle{ (v,w) }[/math], say.
   2. If [math]\displaystyle{ e_f(w)=0 }[/math], insert [math]\displaystyle{ w }[/math] in [math]\displaystyle{ S }[/math].
   3. Increase the flow over [math]\displaystyle{ f }[/math] by the minimum of [math]\displaystyle{ e_f(v) }[/math] and the residual capacity of <matrh>a</math>.
   4. If [math]\displaystyle{ e_f(v)=0 }[/math] now, extract [math]\displaystyle{ v }[/math] from [math]\displaystyle{ S }[/math].
3. Otherwise: increase [math]\displaystyle{ d(v) }[/math] to one more than the maximal label [math]\displaystyle{ d(w) }[/math] of any arc [math]\displaystyle{ (v,w) }[/math] in the residual network.

Proof: Obviously