The article below is an early draft of an article published here: https://www.toptal.com/algorithms/maximum-flow-linear-assignment-problem

Maximum Flow and the Linear Assignment Problem

Here’s a problem: Your business assigns contractors to fulfill contracts. It’s January first of the new year, and you’d like to make some assignments for the month ahead. You look through your rosters and decide which contractors are available for a one month engagement and you look through your available contracts to see which of them are for one month long tasks. I’ll also suppose that for each contractor-contract pair you know the profit you will receive if the contractor fulfills this contract. You’d like to maximize the profit of your business for this month, how will you do it?

This blog post describes an implementation of the Kuhn-Munkres algorithm (also known as the Hungarian algorithm) which solves the linear assignment problem. The Kuhn-Munkres algorithm takes time polynomial in the size of the input graph. My implementation will make use of a subroutine. The subroutine that I use will also execute in polynomial time with respect to its input (which will also be a graph). The subroutine is an implementation of the Edmonds-Karp algorithm. The Edmonds-Karp algorithm solves the maximum flow problem. The Edmonds-Karp algorithm is itself a modification of the Ford-Fulkerson method. I’ll describe the Ford-Fulkerson method first, then I’ll state the modification which turns it into the Edmonds-Karp algorithm and why that modification is important.

The maximum flow problem itself can be described informally as the problem of moving some fluid or gas through a network of pipes from a single source to a single terminal, supposing that the pressure in the network is sufficient to ensure that the fluid or gas cannot linger in any length of pipe or pipe fitting (those places where different lengths of pipe meet).

The codes in this post are in Python3.6. I used the namedtuple class in my examples.

Note: Due to some unfortunate oversights in the software I’m using to write this post underscores ( _ ) in $math mode$ look like this $G_{Hello!}$ , but in code mode they look like this: G_Hello!. I switch styles pretty often so please keep in mind that if I’m talking about $G_{Hello!}$ and G_Hello! I’m talking about the same thing!

Preliminaries

The ideas needed to solve these problems arise in many mathematical and engineering disciplines, often the same concepts can be known by many names and expressed in different ways (e.g. matrices or adjacency lists). Because many of the ideas are very deep, choices are made regarding how generally these concepts will be defined for any given setting. Each source makes assumptions regarding which names should be used, how the meaning behind each name will be expressed, and the level of generality in both the definitions and the discussion.

In this post I don’t want to assume any prior knowledge beyond a little introductory set theory. I want to express as little as possible and to give as few definitions as possible, making each as specific as possible (for example I don’t want to discuss infinite sets), but I still want to have what’s needed to express an graph based implementation of the Hungarian algorithm and to say something about the runtime of that implementation. I give links for sources with more information.

The implementations in this post represent the problems as directed graphs (DiGraph).

DiGraphs

A DiGraph has two attributes, setOfNodes and setOfArcs. Both these attributes are sets (unordered collections). In the code blocks on this post I’m actually using python’s frozenset but that detail isn’t particularly important.

DiGraph = namedtuple('DiGraph', ['setOfNodes','setOfArcs'])

Nodes

A Node $n$ is composed of two attributes:

$n.uid$ : A unique identifier.
This means that for any two Nodes $x$ and $y$ ,
$x.uid \ne y.uid$
$n.datum$ : This represent any data object.

Node = namedtuple('Node', ['uid','datum'])

Arcs

An Arc $a$ is composed of three attributes:

$a.fromNode$ : This is a Node, as defined above.
$a.toNode$ : This is a Node, as defined above.
$a.datum$ : This represent any data object.

Arc = namedtuple('Arc', ['fromNode','toNode','datum'])

The set of Arcs in the DiGraph represents a binary relation on the Nodes in the DiGraph. Briefly, the existence of Arc $a$ implies that a relationship exists between $a.fromNode$ and $a.toNode$ .
In a directed graph (as opposed to an undirected graph) the existence of a relationship between $a.fromNode$ and $a.toNode$ does not imply that a similar relationship between $a.toNode$ and $a.fromNode$ exists,
this is because in an undirected graph the relation being expressed is not necessarily symmetric.

For example, let the Nodes in a DiGraph represent the actions necessary for building a house, and the Arcs represent the order in which these actions can occur. The relationship that the foundation must be completed first and only afterwards can the walls be affixed to the foundation can be expressed graphically like so:

simple digraph

In the the Arc from Node A to Node B represents the relationship allows, so if the foundation to the house has been completed, this fact allows the walls to be affixed to that foundation.
The relationship is not symmetric because the previous fact does not suggest that affixing the walls to the foundation allows the foundation to be completed.

A = Node('A','Build Foundation.')
B = Node('B','Affix walls\nto foundation.')
AB = Arc(A,B,'Allows')

DiGraphs

I’ll use Nodes and Arcs to define a DiGraph but in the algorithms below I’ll represent each DiGraph as a dictionary, because that is convenient.
Here’s a method that can convert the graph representation above into a dictionary representation similar to this one:

def digraph_to_dict(G):
    G_as_dict = dict([])
    for a in G.setOfArcs:
        if(a.fromNode not in G.setOfNodes):
            err_msg = 'There is no Node {a.fromNode.uid!s} to match the Arc from {a.fromNode.uid!s} to {a.toNode.uid!s}'.format(**locals())
            pdg(G)
            raise KeyError(err_msg)
        if(a.toNode not in G.setOfNodes):
            err_msg = 'There is no Node {a.toNode.uid!s} to match the Arc from {a.fromNode.uid!s} to {a.toNode.uid!s}'.format(**locals())
            pdg(G)
            raise KeyError(err_msg)
        G_as_dict[a.fromNode] = (G_as_dict[a.fromNode].union(frozenset([a]))) if (a.fromNode in G_as_dict) else frozenset([a])
    for a in G.setOfArcs:
        if(a.fromNode not in G.setOfNodes):
            err_msg = 'There is no Node {a.fromNode.uid!s} to match the Arc from {a.fromNode.uid!s} to {a.toNode.uid!s}'.format(**locals())
            raise KeyError(err_msg)
        if a.toNode not in G_as_dict:
            G_as_dict[a.toNode] = frozenset([])
    return G_as_dict

And here’s another one that converts it into a dictionary of dictionaries, another operation that will be useful:

def digraph_to_double_dict(G):
    G_as_dict = dict([])
    for a in G.setOfArcs:
        if(a.fromNode not in G.setOfNodes):
            err_msg = 'There is no Node {a.fromNode.uid!s} to match the Arc from {a.fromNode.uid!s} to {a.toNode.uid!s}'.format(**locals())
            raise KeyError(err_msg)
        if(a.toNode not in G.setOfNodes):
            err_msg = 'There is no Node {a.toNode.uid!s} to match the Arc from {a.fromNode.uid!s} to {a.toNode.uid!s}'.format(**locals())
            raise KeyError(err_msg)
        if(a.fromNode not in G_as_dict):
             G_as_dict[a.fromNode] = dict({a.toNode : frozenset([a])})  
        else:
            if(a.toNode not in G_as_dict[a.fromNode]):
                G_as_dict[a.fromNode][a.toNode] = frozenset([a])
            else:
                G_as_dict[a.fromNode][a.toNode] = G_as_dict[a.fromNode][a.toNode].union(frozenset([a]))

    for a in G.setOfArcs:
        if(a.fromNode not in G.setOfNodes):
            err_msg = 'There is no Node {a.fromNode.uid!s} to match the Arc from {a.fromNode.uid!s} to {a.toNode.uid!s}'.format(**locals())
            raise KeyError(err_msg)
        if a.toNode not in G_as_dict:
            G_as_dict[a.toNode] = dict({})  
    return G_as_dict

When I want to talk about a DiGraph as represented by a dictionary, I’ll use G_as_dict to refer to it (I’ll make clear from the context whether it is a ‘double’ layer dictionary or just a single layer), otherwise I’ll use $G$ or $G_{something}$ .

Sometimes it’s helpful to fetch a Node from a DiGraph $G$ by it up through it’s uid (which is unique):

def find_node_by_uid(find_uid, G):
    nodes = {n for n in G.setOfNodes if n.uid == find_uid}
    if(len(nodes) != 1):
        err_msg = 'Node with uid {find_uid!s} is not unique.'.format(**locals())
        raise KeyError(err_msg)
    return nodes.pop()

In defining graphs people sometimes use the terms Node and Vertex to refer to the same concept; the same is true of the terms Arc and Edge. Two popular graph representations in python are this one which uses dictionaries and another which uses objects to represent graphs. The representation in this post is somewhere in between these two commonly used representations.

This is my DiGraph representation. There are many like it, but this one is mine.

Walks and Paths

Let $S_{Arcs}$ be a finite sequence (ordered collection) of Arcs in a DiGraph $G$ such that if $a$ is any Arc in $S_{Arcs}$ except for the last, and $b$ follows $a$ in the sequence, then there must be a Node $n = a.fromNode$ in $G.setOfNodes$ such that $a.toNode = b.fromNode$ .

Starting from the first Arc in $S_{Arcs}$ , and continuing until the last Arc in $S_{Arcs}$ collect (in the order encountered) all Nodes $n$ as defined above between each two consecutive Arcs in $S_{Arcs}$ . Label the ordered collection of Nodes collected during this operation $S_{Nodes}$ .

def path_arcs_to_nodes(s_arcs):
    s_nodes = list([])
    arc_it = iter(s_arcs)
    step_count = 0
    last = None
    try:
        at_end = False
        last = a1 = next(arc_it)
        while (not at_end):
            s_nodes += [a1.fromNode]
            last = a2 = next(arc_it)
            step_count += 1
            if(a1.toNode != a2.fromNode):
                err_msg = "Error at index {step_count!s} of Arc sequence.".format(**locals())
                raise ValueError(err_msg)
            a1 = a2
    except StopIteration as e:
        at_end = True

    if(last is not None):
        s_nodes += [last.toNode]

    return s_nodes

If any Node appears more than once in the sequence $S_{Nodes}$ then call $S_{Arcs}$ a Walk on DiGraph $G$ .
Otherwise, call $S_{Arcs}$ a Path from list(S_Nodes)[0] to list(S_Nodes)[-1] on DiGraph $G$ .

Source Node

Call Node $s$ a Source Node in DiGraph $G$ if $s$ is in $G.setOfNodes$ and $G.setOfArcs$ contains no Arc $a$ such that $a.toNode = s$ .

def source_nodes(G):
    to_nodes = frozenset({a.toNode for a in G.setOfArcs})
    sources = G.setOfNodes.difference(to_nodes)
    return sources

Terminal Node

Call Node $t$ a Terminal Node in DiGraph $G$ if $t$ is in $G.setOfNodes$ and $G.setOfArcs$ contains no Arc $a$ such that $a.fromNode = t$ .

def terminal_nodes(G):
    from_nodes = frozenset({a.fromNode for a in G.setOfArcs})
    terminals = G.setOfNodes.difference(from_nodes)
    return terminals

Cuts and s-t Cuts

A Cut cut of a connected DiGraph $G$ is a subset of Arcs from $G.setOfArcs$ which partitions the set of Nodes $G.setOfNodes$ in $G$ . $G$ is connected if every Node $n$ in $G.setOfNodes$ has at least one Arc $a$ in $G.setOfArcs$ such that either $n = a.fromNode$ or $n = a.toNode$ , but $a.fromNode \ne a.toNode$ .

Cut = namedtuple('Cut', ['setOfCutArcs'])

The definition above refers to a subset of Arcs, but it can also define a partition of the Nodes of $G.setOfNodes$ . I need some functions:

For the functions predecessors_of and successors_of, $n$ is a Node in set G.setOfNodes of DiGraph $G$ , and cut is a Cut of $G$ :

def cut_predecessors_of(n, cut, G):
    allowed_arcs   = G.setOfArcs.difference(frozenset(cut.setOfCutArcs))
    predecessors   = frozenset({})
    previous_count = len(predecessors)
    reach_fringe   = frozenset({n})
    keep_going     = True
    while( keep_going ):
        reachable_from = frozenset({a.fromNode for a in allowed_arcs if (a.toNode in reach_fringe)})
        reach_fringe   = reachable_from
        predecessors   = predecessors.union(reach_fringe)
        current_count  = len(predecessors) 
        keep_going     = current_count -  previous_count > 0
        previous_count = current_count
    return predecessors

def cut_successors_of(n, cut, G):
    allowed_arcs   = G.setOfArcs.difference(frozenset(cut.setOfCutArcs))
    successors     = frozenset({})
    previous_count = len(successors)
    reach_fringe   = frozenset({n})
    keep_going     = True
    while( keep_going ):
        reachable_from = frozenset({a.toNode for a in allowed_arcs if (a.fromNode in reach_fringe)})
        reach_fringe   = reachable_from
        successors     = successors.union(reach_fringe)
        current_count  = len(successors) 
        keep_going     = current_count -  previous_count > 0
        previous_count = current_count
    return successors

Let cut be a Cut of DiGraph $G$ .

def get_first_part(cut, G):
    firstPartFringe = frozenset({a.fromNode for a in cut.setOfCutArcs})
    firstPart = firstPartFringe
    for n in firstPart:
        preds = cut_predecessors_of(n,cut,G)
        firstPart = firstPart.union(preds)
    return firstPart

def get_second_part(cut, G):
    secondPartFringe = frozenset({a.toNode for a in cut.setOfCutArcs})
    secondPart = secondPartFringe
    for n in secondPart:
        succs= cut_successors_of(n,cut,G)
        secondPart = secondPart.union(succs)
    return secondPart

cut is a Cut of DiGraph $G$ if: (get_first_part(cut, G).union(get_second_part(cut, G)) == G.setOfNodes) and (len(get_first_part(cut, G).intersect(get_second_part(cut, G))) == 0) cut is called an x-y Cut if (x in get_first_part(cut, G)) and (y in get_second_part(cut, G) ) and (x != y). When the Node $x$ in a x-y Cut cut is a Source Node and Node $y$ in the x-y Cut is a Terminal Node, then this Cut is called a s-t Cut.

STCut = namedtuple('STCut', ['s','t','cut'])

Flow Networks

I can use DiGraph $G$ to represent a flow network.

Assign each Node $n$ , where $n$ is in $G.setOfNodes$ an $n.datum$ that is a FlowNodeDatum:

FlowNodeDatum = namedtuple('FlowNodeDatum', ['flowIn','flowOut'])

Assign each Arc $a$ , where $a$ is in $G.setOfArcs$ an $a.datum$ that is a FlowArcDatum,

FlowArcDatum = namedtuple('FlowArcDatum', ['capacity','flow'])

$flowNodeDatum.flowIn$ and $flowNodeDatum.flowOut$ are positive real numbers.
$flowArcDatum.capacity$ and $flowArcDatum.flow$ are also positive real numbers.

For every node Node $n$ in $G.setOfNodes$ :

n.datum.flowIn = sum({a.datum.flow for a in G.Arcs if a.toNode == n})
n.datum.flowOut = sum({a.datum.flow for a in G.Arcs if a.fromNode == n})

DiGraph $G$ now represents a Flow Network.

The Flow of $G$ refers to the $a.flow$ for all Arcs $a$ in $G$ .

Feasible Flows

Let DiGraph $G$ represent a Flow Network.

The Flow Network represented by $G$ has Feasible Flows if:

For every Node $n$ in $G.setOfNodes$ except for Source Nodes and Terminal Nodes : $n.datum.flowIn = n.datum.flowOut$ .
For every Arc $a$ in $G.setOfNodes$ : $a.datum.capacity \le a.datum.flow$ .

Condition 1 is called a Conservation Constraint.
Condition 2 is called a Capacity Constraint.

Cut Capacity

The Cut Capacity of an s-t Cut stCut with Source Node $s$ and Terminal Node $t$ of a Flow Network represented by a DiGraph $G$ is:

def cut_capacity(stCut, G):
    part_1 = get_first_part(stCut.cut,G)
    part_2 = get_second_part(stCut.cut,G)
    s_part = part_1 if stCut.s in part_1 else part_2
    t_part = part_1 if stCut.t in part_1 else part_2
    cut_capacity = sum({a.datum.capacity for a in stCut.cut.setOfCutArcs if ( (a.fromNode in s_part) and (a.toNode in t_part) )})
    return cut_capacity

Minimum Capacity Cut

Let stCut = stCut(s,t,cut) be an s-t Cut of a Flow Network represented by a DiGraph $G$ .

stCut is the Minimum Capacity Cut of the Flow Network represented by $G$ if there is no other s-t Cut stCutCompetitor in this Flow Network such that:

    cut_capacity(stCut, G) < cut_capacity(stCutCompetitor, G)

Stripping the Flows Away

I would like to refer to the DiGraph that would be the result of taking a DiGraph $G$ and stripping away all the flow data from all the Nodes in $G.setOfNodes$ and also all the Arcs in $G.setOfArcs$ .

def strip_flows(G):
    new_nodes= frozenset( (Node(n.uid, FlowNodeDatum(0.0,0.0)) for n in G.setOfNodes) )
    new_arcs = frozenset([])
    for a in G.setOfArcs:
        new_fromNode = Node(a.fromNode.uid, FlowNodeDatum(0.0,0.0)) 
        new_toNode     = Node(a.toNode.uid, FlowNodeDatum(0.0,0.0)) 
        new_arc            = Arc(new_fromNode, new_toNode, FlowArcDatum(a.datum.capacity, 0.0))
        new_arcs          = new_arcs.union(frozenset([new_arc]))
    return DiGraph(new_nodes, new_arcs)

Maximum Flow Problem

A Flow Network represented as a DiGraph $G$ , a Source Node $s$ in $G.setOfNodes$ and a Terminal Node $t$ in $G.setOfNodes$ , $G$ can represent a Maximum Flow Problem if:

(len(list(source_nodes(G))) == 1) and (next(iter(source_nodes(G))) == s) and (len(list(terminal_nodes(G))) == 1) and (next(iter(terminal_nodes(G))) == t)

Label this representation:

MaxFlowProblemState = namedtuple('MaxFlowProblemState', ['G','sourceNodeUid','terminalNodeUid','maxFlowProblemStateUid'])

Where sourceNodeUid= $s.uid$ , terminalNodeUid= $t.uid$ , and maxFlowProblemStateUid is an identifier for the problem instance.

Maximum Flow Solution

Let maxFlowProblem represent a Maximum Flow Problem. The solution to maxFlowProblem can be represented by a Flow Network represented as a DiGraph $H$ .

DiGraph $H$ is a feasible solution to the Maximum Flow Problem on input python maxFlowProblem if:

strip_flows(maxFlowProblem.G) == strip_flows(H).
$H$ is a Flow Network and has Feasible Flows.

If in addition to 1 and 2:

There can be no other Flow Network represented by DiGraph $K$ such that strip_flows(G) == strip_flows(K) and find_node_by_uid(t.uid,G).flowIn < find_node_by_uid(t.uid,K).flowIn.

Then $H$ is also an optimal solution to maxFlowProblem.

In other words a Feasible Maximum Flow Solution can be represented by a DiGraph:

Which is identical to DiGraph $G$ of the corresponding Maximum Flow Problem with the exception that the n.datum.flowIn, n.datum.flowOut and the a.datum.flow of any of the Nodes and Arcs may be different.
Represents a Flow Network that has Feasible Flows.

and it can represent an Optimal Maximum Flow Solution if additionally:

In which the flowIn for the Node corresponding to the Terminal Node in the Maximum Flow Problem is as large as possible (when conditions 1 and 2 are still satisfied).

If DiGraph $H$ represents a Feasible Maximum Flow Solution : find_node_by_uid(s.uid,H).flowOut = find_node_by_uid(t.uid,H).flowIn this follows from the Max Flow, Min Cut Theorem (discussed below). Informally, since $H$ is assumed to have Feasible Flows this means that Flow can neither be ‘created’ (except at Source Node $s$ ) nor ‘destroyed’ (except at Terminal Node $t$ ) while crossing any (other) Node (Conservation Constraints). So, since a Maximum Flow Problem contains only a single Source Node $s$ and a single Terminal Node $t$ , all flow ‘created’ at $s$ must be ‘destroyed’ at $t$ or the Flow Network does not have Feasible Flows (the Conservation Constraint would have been violated).

Let DiGraph $H$ represent a Feasible Maximum Flow Solution the value above is called the s-t Flow Value of $H$ , let mfps=MaxFlowProblemState(H,maxFlowProblem.sourceNodeUid,maxFlowProblem.terminalNodeUid,maxFlowProblem.maxFlowProblemStateUid) this means that mfps is a successor state of maxFlowProblem which just means that mfps is exacly like maxFlowProblem with the exception that the values of a.flow for arcs $a$ in maxFlowProblem.setOfArcs may be different than a.flow for arcs $a$ in mfps.setOfArcs.

def get_mfps_flow(mfps):
    flow_from_s = find_node_by_uid(mfps.sourceNodeUid,mfps.G).datum.flowOut
    flow_to_t      = find_node_by_uid(mfps.terminalNodeUid,mfps.G).datum.flowIn

    if( (flow_to_t - flow_from_s) > 0):
        raise Exception('Infeasible s-t flow')
    return flow_to_t

Here’s a visualization of a mfps along with it’s associated maxFlowProb each Arc $a$ in the image has a label, these labels are $a.datum.flowFrom / a.datum.flowTo$ , each Node $n$ in the image has a label, these labels are $n.uid\{n.datum.flowIn / a.datum.flowOut\}$ .

wiki_G_1

s-t Cut Flow

Let mfps represent a MaxFlowProblemState and let stCut represent a Cut of mfps.G. The Cut Flow of stCut is defined:

def get_stcut_flow(stCut,mfps):
    s = find_node_by_uid(mfps.sourceNodeUid, mfps.G)
    t = find_node_by_uid(mfps.terminalNodeUid, mfps.G)
    part_1 = get_first_part(stCut.cut,mfps.G)
    part_2 = get_second_part(stCut.cut,mfps.G)
    s_part = part_1 if s in part_1 else part_2
    t_part = part_1 if t in part_1 else part_2 
    s_t_sum = sum(a.datum.flow for a in mfps.G.setOfArcs if (a in stCut.cut.setOfCutArcs) and (a.fromNode in s_part) )
    t_s_sum = sum(a.datum.flow for a in mfps.G.setOfArcs if (a in stCut.cut.setOfCutArcs) and (a.fromNode in t_part) )
    cut_flow = s_t_sum - t_s_sum
    return cut_flow

s-t Cut Flow is the sum of flows from the partition containing the Source Node to the partition containing the Terminal Node minus the sum of flows from the partition containing the Terminal Node to the partition containing the Source Node.

Max Flow, Min Cut

Let maxFlowProblem represent a Maximum Flow Problem and let the solution to maxFlowProblem be represented by a Flow Network represented as DiGraph $H$ .
Let minStCut be the Minimum Capacity Cut of the Flow Network represented by maxFlowProblem.G.

Because in the Maximum Flow Problem flow originates in only a single Source Node and terminates at a single Terminal Node and because of the Capacity Constraints and the Conservation Constraints we know that the all of the flow entering maxFlowProblem.terminalNodeUid must cross any s-t Cut, in particular it must cross minStCut, this means:

get_flow_value(H, maxFlowProblem) <= cut_capacity(minStCut, maxFlowProblem.G)

Solving the Maximum Flow Problem

The basic idea for solving a Maximum Flow Problem maxFlowProblem is to start with a Maximum Flow Solution represented by DiGraph $H$ such a starting point can be H = strip_flows(maxFlowProblem.G). The task is then to use $H$ and by some greedy modification of the $a.datum.flow$ values of some Arcs $a$ in $H.setOfArcs$ to produce another Maximum Flow Solution represented by DiGraph $K$ such that $K$ doesn’t can still represent a Flow Network with Feasible Flows and get_flow_value(H, maxFlowProblem) < get_flow_value(K, maxFlowProblem). As long as this process continues the quality (get_flow_value(K, maxFlowProblem)) of the most recently encountered Maximum Flow Solution ( $K$ ) is better than any other Maximum Flow Solution that has been found. If the process reaches a point that it know that no other improvement is possible the process can terminate and it will return the optimal Maximum Flow Solution.

The description above is general and skips many proofs such as whether such a process is possible or how long it may take, I’ll give a few more details and the algorithm.

The Max Flow, Min Cut Theorem

From the book Flows in Networks by Ford and Fulkerson the statement of the Max-Flow Min-Cut Theorem (Theorem 5.1) is:

For any network the maximal flow value from $s$ to $t$ is equal to the minimum cut capacity of all cuts separating $s$ and $t$ .

Using the definitions in this post that translates to:

The solution to a maxFlowProblem represented by a Flow Network represented as DiGraph $H$ is optimal if:

get_flow_value(H, maxFlowProblem) = cut_capacity(minStCut, maxFlowProblem.G)

I like this proof of the theorem and wikipedia has another one.

The Max Flow, Min Cut Theorem is used to prove the correctness and completeness of Ford-Fulkerson Method

I’ll also give some a proof of the theorem in the section after Augmenting Paths.

The Ford-Fulkerson Method and the Edmonds-Karp Algorithm

CLRS Defines the Ford-Fulkerson Method like so (section 26.2):

FORD-FULKERSON-METHOD( $G$ , $s$ , $t$ ):
initialize flow $f$ to $0$
while there exists a augmenting path $p$ in the residual graph $G_{f}$ augment flow $f$ along $p$

Residual Graph

The Residual Graph of a Flow Network represented as the DiGraph $G$ can be represented as a DiGraph $G_{f}$ :

ResidualDatum = namedtuple('ResidualDatum', ['residualCapacity','action'])

def agg_n_to_u_cap(n,u,G_as_dict):
    arcs_out = G_as_dict[n]
    return sum([a.datum.capacity for a in arcs_out if( (a.fromNode == n) and (a.toNode == u) ) ])

def agg_n_to_u_flow(n,u,G_as_dict):
    arcs_out = G_as_dict[n]
    return sum([a.datum.flow for a in arcs_out if( (a.fromNode == n) and (a.toNode == u) ) ])

def get_residual_graph_of(G):
    G_as_dict = digraph_to_dict(G)
    residual_nodes = G.setOfNodes
    residual_arcs = frozenset([])

    for n in G_as_dict:
        arcs_from = G_as_dict[n]
        nodes_to = frozenset([find_node_by_uid(a.toNode.uid,G) for a in arcs_from])

        for u in nodes_to: 
            n_to_u_cap_sum  = agg_n_to_u_cap(n,u,G_as_dict)
            n_to_u_flow_sum = agg_n_to_u_flow(n,u,G_as_dict)
            if(n_to_u_cap_sum > n_to_u_flow_sum):
                flow = round(n_to_u_cap_sum - n_to_u_flow_sum, TOL)
                residual_arcs = residual_arcs.union( frozenset([Arc(n,u,datum=ResidualDatum(flow, 'push'))]) )
            if(n_to_u_flow_sum > 0.0):
                flow = round(n_to_u_flow_sum, TOL)
                residual_arcs = residual_arcs.union( frozenset([Arc(u,n,datum=ResidualDatum(flow, 'pull'))]) )
    return DiGraph(residual_nodes, residual_arcs)

agg_n_to_u_cap(n,u,G_as_dict) returns the sum of $a.datum.capacity$ for all Arcs in the subset of $G.setOfArcs$ where Arc $a$ is in the subset if $a.fromNode = n$ and $a.toNode = u$ .
agg_n_to_u_cap(n,u,G_as_dict) returns the sum of $a.datum.flow$ for all Arcs in the subset of $G.setOfArcs$ where Arc $a$ is in the subset if $a.fromNode = n$ and $a.toNode = u$ .

Briefly, the Residual Graph $G_{f}$ represents certain actions which can be performed on the DiGraph $G$ .

Each pair of Nodes $n,u$ in $G.setOfNodes$ of the Flow Network represented by DiGraph $G$ can generate 0, 1, or 2 Arcs in the Residual Graph $G_{f}$ of $G$ .

The pair $n,u$ does not generate any Arcs in $G_{f}$ if there is no Arc $a$ in $G.setOfArcs$ such that $a.fromNode = n$ and $a.toNode = u$ .
The pair $n,u$ generates the Arc $a$ in $G_{f}.setOfArcs$ where $a$ represents an Arc labeled a Push Flow Arc from $n$ to $u$ a = Arc(n,u,datum=ResidualNode(n_to_u_cap_sum - n_to_u_flow_sum)) if n_to_u_cap_sum > n_to_u_flow_sum.
The pair $n,u$ generates the Arc $a$ in $G_{f}.setOfArcs$ where $a$ represents an Arc labeled a Pull Flow Arc from $n$ to $u$ a = Arc(n,u,datum=ResidualNode(n_to_u_cap_sum - n_to_u_flow_sum)) if n_to_u_flow_sum > 0.0.

Each Push Flow Arc in $G_{f}.setOfArcs$ represents the action of adding a total of x <= n_to_u_cap_sum - n_to_u_flow_sum flow to Arcs in the subset of $G.setOfArcs$ where Arc $a$ is in the subset if $a.fromNode = n$ and $a.toNode = u$ .
Each Pull Flow Arc in $G_{f}.setOfArcs$ represents the action of subtracting a total of x <= n_to_u_flow_sum flow to Arcs in the subset of $G.setOfArcs$ where Arc $a$ is in the subset if $a.fromNode = n$ and $a.toNode = u$ .

Performing an individual Push or Pull action from $G_{f}$ on the applicable Arcs in $G$ might generate a Flow Network without Feasible Flows because the Capacity Constraints or the Conservation Constraints might be violated in the generated Flow Network.

Here’s a visualization of the Residual Graph of the previous example visualization of a Maximum Flow Solution, in the visualization each Arc $a$ represents a.residualCapacity.

wiki_R_1

Augmenting Path

Let maxFlowProblem be a MaxFlowProblem, and let G_f = get_residual_graph_of(G) be the Residual Graph of maxFlowProblem.G.

An Augmenting Path augmentingPath for maxFlowProblem is any Path from find_node_by_uid(maxFlowProblem.sourceNode,G_f) to find_node_by_uid(maxFlowProblem.terminalNode,G_f).
It turns out that an augmenting Path augmentingPath can be applied to a MaxFlowSolution represented by DiGraph $H$ generating another MaxFlowSolution represented by DiGraph $K$ where get_flow_value(H, maxFlowProblem) < get_flow_value(K, maxFlowProblem) if H is not optimal. Here’s how:

def augment(augmentingPath, H):
    augmentingPath = list(augmentingPath)
    H_as_dict = digraph_to_dict(H)
    new_nodes = frozenset({})
    new_arcs = frozenset({})
    visited_nodes  = frozenset({})
    visited_arcs = frozenset({})

    bottleneck_residualCapacity = min( augmentingPath, key=lambda a: a.datum.residualCapacity ).datum.residualCapacity
    for x in augmentingPath:
        from_node_uid = x.fromNode.uid if x.datum.action == 'push' else x.toNode.uid
        to_node_uid = x.toNode.uid   if x.datum.action == 'push' else x.fromNode.uid
        from_node = find_node_by_uid( from_node_uid, H )
        to_node = find_node_by_uid( to_node_uid, H )
        load = bottleneck_residualCapacity if x.datum.action == 'push' else -bottleneck_residualCapacity

        for a in H_as_dict[from_node]:
            if(a.toNode == to_node):
                test_sum = a.datum.flow + load
                new_flow = a.datum.flow
                new_from_node_flow_out = a.fromNode.datum.flowOut
                new_to_node_flow_in = a.toNode.datum.flowIn

                new_from_look = {n for n in new_nodes if (n.uid == a.fromNode.uid)}
                new_to_look = {n for n in new_nodes if (n.uid == a.toNode.uid)  }
                prev_from_node = new_from_look.pop() if (len(new_from_look)>0) else a.fromNode
                prev_to_node = new_to_look.pop()   if (len(new_to_look)>0)   else a.toNode
                new_nodes = new_nodes.difference(frozenset({prev_from_node}))
                new_nodes = new_nodes.difference(frozenset({prev_to_node}))

                if(test_sum > a.datum.capacity):
                    new_flow = a.datum.capacity
                    new_from_node_flow_out = prev_from_node.datum.flowOut - a.datum.flow + a.datum.capacity
                    new_to_node_flow_in    = prev_to_node.datum.flowIn - a.datum.flow + a.datum.capacity
                    load = test_sum - a.datum.capacity
                elif(test_sum < 0.0):
                    new_flow = 0.0
                    new_from_node_flow_out = prev_from_node.datum.flowOut - a.datum.flow
                    new_to_node_flow_in    = prev_to_node.datum.flowIn - a.datum.flow
                    load = test_sum
                 else:
                    new_flow = test_sum
                    new_from_node_flow_out = prev_from_node.datum.flowOut - a.datum.flow + new_flow
                    new_to_node_flow_in    = prev_to_node.datum.flowIn - a.datum.flow + new_flow
                    load = 0.0

                new_from_node_flow_out = round(new_from_node_flow_out, TOL)
                new_to_node_flow_in    = round(new_to_node_flow_in, TOL)
                new_flow               = round(new_flow, TOL)

                new_from_node = Node(prev_from_node.uid, FlowNodeDatum(prev_from_node.datum.flowIn, new_from_node_flow_out))
                new_to_node   = Node(prev_to_node.uid, FlowNodeDatum(new_to_node_flow_in, prev_to_node.datum.flowOut))
                new_arc       = Arc(new_from_node, new_to_node, FlowArcDatum(a.datum.capacity, new_flow))

                visited_nodes = visited_nodes.union(frozenset({a.fromNode,a.toNode}))
                visited_arcs  = visited_arcs.union(frozenset({a}))
                new_nodes     = new_nodes.union(frozenset({new_from_node, new_to_node}))
                new_arcs      = new_arcs.union(frozenset({new_arc}))

    not_visited_nodes = H.setOfNodes.difference(visited_nodes)
    not_visited_arcs  = H.setOfArcs.difference(visited_arcs)

    full_new_nodes = new_nodes.union(not_visited_nodes)
    full_new_arcs  = new_arcs.union(not_visited_arcs)
    G = DiGraph(full_new_nodes, full_new_arcs)

    full_new_arcs_update = frozenset([])
    for a in full_new_arcs:        
        new_from_node = a.fromNode
        new_to_node   = a.toNode
        new_from_node = find_node_by_uid( a.fromNode.uid, G )
        new_to_node   = find_node_by_uid( a.toNode.uid, G )
        full_new_arcs_update = full_new_arcs_update.union( {Arc(new_from_node, new_to_node, FlowArcDatum(a.datum.capacity, a.datum.flow))} )

    G = DiGraph(full_new_nodes, full_new_arcs_update)

    return G

In the above TOL is some tolerance value for rounding the flow values in the network, this is to avoid cascading imprecision if floating point calculations. So for example I used TOL = 10 to mean round to 10 significant digits.

Let K = augment(augmentingPath, H), then $K$ represents a Feasible MaxFlowSolution for maxFlowProblem. For the statement to be true, the Flow Network represented by $K$ must have Feasible Flows (not violate the Capacity Constraint or the Conservation Constraint.

Here’s why: In the method above, each Node added to the new Flow Network represented by DiGraph $K$ is either an exact copy of a Node from DiGraph $H$ , or it is a Node $n$ which has had the same number added to it’s n.datum.flowIn as it’s n.datum.flowOut. This means that the Conservation Constraint is satisfied in $K$ as long as it was satisfied in $H$ . The Conservation Constraint is satisfied because we explicitly check that any new Arc $a$ in the network has a.datum.flow <= a.datum.capacity thus as long as the Arcs from the set $H.setOfArcs$ which were copied unmodified into $K.setOfArcs$ do not violate the Capacity Constraint then $K$ does not violate the Capacity Constraint.

It’s also true that get_flow_value(H, maxFlowProblem) < get_flow_value(K, maxFlowProblem) if H is not optimal.

Here’s why: For an Augmenting Path augmentingPath to exist in the DiGraph representation of the Residual Graph $G_{f}$ of a MaxFlowProblem maxFlowProblem then the last Arc $a$ on augmentingPath must be a ‘push’ Arc and it must have a.toNode == maxFlowProblem.terminalNode. An Augmenting Path is defined as one which terminates at the Terminal Node of the MaxFlowProblem for which it is an Augmenting Path. From the definition of the Residual Graph it is clear that the last Arc in an Augmenting Path on that Residual Graph must be a ‘push’ Arc because any ‘pull’ Arc $b$ in the Augmenting Path will have b.toNode == maxFlowProblem.terminalNode and b.fromNode != maxFlowProblem.terminalNode from the definition of Path. Additionally, from the definition of Path it is clear that the Terminal Node is only modified once by the augment method. Thus augment modifies maxFlowProblem.terminalNode.flowIn exactly once and it increases the value of maxFlowProblem.terminalNode.flowIn because the last Arc in the augmentingPath must be the Arc which causes the modification in maxFlowProblem.terminalNode.flowIn during augment. From the definition of augment as it applies to ‘push’ Arcs the maxFlowProblem.terminalNode.flowIn can only be increased, not decreased.

Some Proofs from Sedgewick and Wayne

The book Algorithms, fourth edition by Robert Sedgewich and Kevin Wayne has some wonderful and short proofs (pages 892-894) that will be useful. I’ll recreate them here, though I’ll use language fitting in with previous definitions. My labels for the proofs are the same as in the Sedgewick book.

Proposition E: For any DiGraph $H$ representing a Feasible Maximum Flow Solution to a Maximum Flow Problem maxFlowProblem, for any stCut get_stcut_flow(stCut,H,maxFlowProblem) = get_flow_value(H, maxFlowProblem)`.

Proof: Let stCut=stCut(maxFlowProblem.sourceNode,maxFlowProblem.terminalNode,set([a for a in H.setOfArcs if a.toNode == maxFlowProblem.terminalNode])). Proposition E holds for stCut directly from the definition of s-t Flow Value. Suppose that there we wish to move some Node $n$ from the s-partition (get_first_part(stCut.cut, G)) and into the t-partition (get_second_part(stCut.cut, G)), to do so we need to change stCut.cut, which could change stcut_flow = get_stcut_flow(stCut,H,maxFlowProblem) and invalidate Proposition E. However, let’s see how the value of stcut_flow will change as we make this change. Node $n$ is at equilibrium meaning that the sum of flow into Node $n$ is equal to the sum of flow out of it (this is necessary for $H$ to represent a Feasible Solution). Notice that all flow which is part of the stcut_flow entering Node $n$ enters it from the s-partition (flow entering Node $n$ from the t-partition either directly or indirectly would not have been counted in the stcut_flow value because it is heading the wrong direction based on the definition). Additionally, all flow exiting $n$ will eventually (direly or indirectly) flow into the Terminal Node (proved earlier). When we move Node $n$ into the t-partition all the flow entering $n$ from the s-partition must be added to the new stcut_flow value, however all flow exiting $n$ must the be subtracted from the new stcut_flow value, the part of the flow heading directly into the t-partition is subtracted because this flow is now internal to the new t-partition and is not counted as stcut_flow. The part of the flow from $n$ heading into Nodes in the s-partition must also be subtracted from stcut_flow: After $n$ is moved into the t-partition these flows will be directed from the t-partition and into the s-partition and so must not be accounted for in the stcut_flow, since these flows are removed the inflow into the s-partition must be reduced by the sum of these flows, and the outflow from the s-partition into the t-partition (where all flows from s-t must end up) must be reduced by the same amount. As Node $n$ was at equilibrium prior to the process the update will have added the same value to the new stcut_flow value as it subtracted thus leaving Proposition E true after the update. The validity of Proposition E then follows from induction on the size of the t-partition.

Here are some example Flow Networks to help visualize the not obvious cases where Proposition E holds, in the image the red areas indicate the s-partition, the blue areas represent the t-partition, and the green Arcs indicate an s-t Cut. In the second image flow between Node A and Node B increases while the flow into Terminal Node t doesn’t change.:

e2_1
e2_2

Corollary: No s-t Cut Flow value can exceed the capacity of any s-t Cut.

Proposition F. (Maxflow-mincut theorem): Let $f$ be an s-t Flow the following 3 conditions are equivalent:

There exists an st Cut whose capacity equals the value of the flow $f$ .
$f$ is a maxflow.
There is no augmenting path with respect to $f$ .

Condition 1 implies condition 2 by the corollary. Condition 2 implies condition 3 because the existence of an augmenting path implies the existence of a flow with larger values, contradicting the maximality of $f$ . Condition 3 implies condition 1: Let $C_{s}$ be the set of all Nodes that can be reached from $s$ with an Augmenting Path in the Residual Graph. Let $C_{t}$ be the remaining Arcs, then $t$ must be in $C_{t}$ (by our assumption). The Arcs crossing from $C_{s}$ to $C_{t}$ then form an stCut which contains only Arcs $a$ where either a.datum.capacity = a.datum.flow or a.datum.flow = 0.0. If this were otherwise then the Nodes connected by an Arc with remaining residual capacity to $C_{s}$ would be in the set $C_{s}$ since there would then be an Augmenting Path from $s$ to such a Node. The flow across the stCut is equal to the stCut’s capacity (since Arcs from $C_{s}$ to $C_{t}$ have flow equal to capacity) and also to the value of the stFlow (by Proposition E).

This statement of the Maxflow Mincut Theorem implies the earlier statement from Flows in Networks.

Corollary. (Integrallity property) When capacities are integers, there exists an integer-valued maxflow, and the Ford-Fulkerson algorithm finds it.
Proof: Each Augmenting Path increases the flow by a positive integer (the minimum of the unused capacities in the ‘push’ Arcs and the flows in the ‘pull’ Arcs**, all of which are always positive integers.

This justifies the Ford Fulkerson Method description from CLRS. The method is to keep finding Augmenting Paths and applying augment to the latest maxFlowSolution coming up with better solutions, until no more Augmenting Path meaning that the latest Maximum Flow Solution is optimal.

From Ford-Fulkerson to Edmonds-Karp

The remaining questions regarding solving Maximal Flow Problems are:

How should Augmenting Paths be constructed?
Will the method terminate if we use real numbers and not integers?
How long will it take to terminate (if it does)?

The Edmonds-Karp Algorithm specifies that each Augmenting Path is constructed by a Bread First Search (BFS) of the Residual Graph, it turns out that this decision of point 1 above will also force the algorithm to terminate (point 2) and allows the asymptotic time and space complexity to be determined.

First, here’s a BFS implementation:

def bfs(sourceNode, terminalNode, G_f):
    G_f_as_dict = digraph_to_dict(G_f)
    parent_arcs = dict([])
    visited = frozenset([])
    deq = deque([sourceNode])
    while len(deq) > 0:
        curr = deq.popleft()
        if curr == terminalNode:
            break
        for a in G_f_as_dict.get(curr):
            if (a.toNode not in visited):
                visited = visited.union(frozenset([a.toNode]))
                parent_arcs[a.toNode] = a
                deq.extend([a.toNode])
    path = list([])
    curr = terminalNode
    while(curr != sourceNode):
        if (curr not in parent_arcs):
            err_msg = 'No augmenting path from {} to {}.'.format(sourceNode.uid, terminalNode.uid)
            raise StopIteration(err_msg)
        path.extend([parent_arcs[curr]])
        curr = parent_arcs[curr].fromNode
    path.reverse()

    test = deque([path[0].fromNode])
    for a in path:
        if(test[-1] != a.fromNode):
            err_msg = 'Bad path: {}'.format(path)
            raise ValueError(err_msg)
        test.extend([a.toNode])

    return path

I used a deque from the python collections module.

To answer question 2 above I’ll paraphrase another proof from Sedgewick and Wayne: Proposition G. The number of Augmenting Paths needed in the Edmonds-Karp algorithm with $N$ Nodes and $A$ Arcs is at most $\frac{NA}{2}$ . Proof: Every Augmenting Path has a bottleneck Arc- an Arc that is deleted from the Residual Graph because it corresponds either to a ‘push’ Arc that becomes filled to capacity or a ‘pull’ Arc through which the flow becomes 0. Each time an Arc becomes a bottleneck Arc the length of any Augmenting Path through it must increase by a factor or 2. This is because each Node in a Path may appear only once or not at all (from the definition of Path) since the Paths are being explored from shortest Path to longest that means that at least one more Node must be visited by the next path that goes through the particular bottleneck Node that means an additional 2 Arcs on the path before we arrive at the Node. Since the Augmenting Path is of length at most $N$ each Arc can be on at most $\frac{N}{2}$ Augmenting Paths, and the total number of Augmenting Paths is at most $\frac{NA}{2}$ .

The Edmonds-Karp Algorimth executes in $O(NA^{2})$ . If at most $\frac{NA}{2}$ Paths will be explored during the algorithm and exploring each Path with BFS is $N+A$ then the most significant term of the product and hence the asymptotic complexity is $O(NA^{2})$ .

Let mfp be a MaxFlowProblemState.

def edmonds_karp(mfp):
    sid, tid = mfp.sourceNodeUid, mfp.terminalNodeUid
    no_more_paths = False
    loop_count = 0
    while(not no_more_paths):
       residual_digraph = get_residual_graph_of(mfp.G)
       try:
           asource = find_node_by_uid(mfp.sourceNodeUid, residual_digraph)
           aterm   = find_node_by_uid(mfp.terminalNodeUid, residual_digraph)
           apath   = bfs(asource, aterm, residual_digraph)
           paint_mfp_path(mfp, loop_count, apath)
           G = augment(apath, mfp.G)
           s = find_node_by_uid(sid, G)
           t = find_node_by_uid(tid, G)
           mfp = MaxFlowProblemState(G, s.uid, t.uid, mfp.maxFlowProblemStateUid)
           loop_count += 1
       except StopIteration as e:
           no_more_paths = True
    return mfp

The version above is inefficient and has worse complexity than $O(NA^{2})$ since it constructs a new Maximum Flow Solution and new a Residual Graph each time (rather than modifying existing DiGraphs as the algorithm advances). To get to a true $O(NA^{2})$ solution the algorithm must maintain both the DiGraph representing the Maximum Flow Problem State and it’s associated Residual Graph. So the algorithm must avoid iterating over Arcs and Nodes unnecessarily and update their values and associated values in the Residual Graph only as necessary.

To write a faster Edmonds Karp algorithm I re-wrote several pieces of code from the above. I hope that going through the code which generated a new DiGraph was helpful in understanding what’s going on. In the fast algorithm I use some new tricks and python data structures that I don’t want to go over in detail. I will say that a.fromNode and a.toNode are now treated as strings and uids to Nodes. For this code let mfps be a MaxFlowProblemState

import uuid 

def fast_get_mfps_flow(mfps):
    flow_from_s = {n for n in mfps.G.setOfNodes if n.uid == mfps.sourceNodeUid}.pop().datum.flowOut
    flow_to_t   = {n for n in mfps.G.setOfNodes if n.uid == mfps.terminalNodeUid}.pop().datum.flowIn

    if( (flow_to_t - flow_from_s) > 0.00):
        raise Exception('Infeasible s-t flow')
    return flow_to_t

def fast_e_k_preprocess(G):
    G = strip_flows(G)
    get = dict({})
    get['nodes'] = dict({})
    get['node_to_node_capacity'] = dict({})
    get['node_to_node_flow'] = dict({})
    get['arcs'] = dict({})
    get['residual_arcs'] = dict({})

    for a in G.setOfArcs:
        if(a.fromNode not in G.setOfNodes):
            err_msg = 'There is no Node {a.fromNode.uid!s} to match the Arc from {a.fromNode.uid!s} to {a.toNode.uid!s}'.format(**locals())
            raise KeyError(err_msg)
        if(a.toNode not in G.setOfNodes):
            err_msg = 'There is no Node {a.toNode.uid!s} to match the Arc from {a.fromNode.uid!s} to {a.toNode.uid!s}'.format(**locals())
            raise KeyError(err_msg)
        get['nodes'][a.fromNode.uid] = a.fromNode  
        get['nodes'][a.toNode.uid]   = a.toNode  
        lark = Arc(a.fromNode.uid, a.toNode.uid, FlowArcDatumWithUid(a.datum.capacity, a.datum.flow, uuid.uuid4()))
        if(a.fromNode.uid not in get['arcs']):
            get['arcs'][a.fromNode.uid] = dict({a.toNode.uid : dict({lark.datum.uid : lark})})  
        else:
            if(a.toNode.uid not in get['arcs'][a.fromNode.uid]):
                get['arcs'][a.fromNode.uid][a.toNode.uid] = dict({lark.datum.uid : lark})
            else:
                get['arcs'][a.fromNode.uid][a.toNode.uid][lark.datum.uid] = lark

    for a in G.setOfArcs:
        if a.toNode.uid not in get['arcs']:
            get['arcs'][a.toNode.uid] = dict({})  

    for n in get['nodes']:
        get['residual_arcs'][n] = dict()
        get['node_to_node_capacity'][n] = dict()
        get['node_to_node_flow'][n] = dict()
        for u in get['nodes']:
            n_to_u_cap_sum  = sum(a.datum.capacity for a in G.setOfArcs if (a.fromNode.uid == n) and (a.toNode.uid == u) )
            n_to_u_flow_sum = sum(a.datum.flow     for a in G.setOfArcs if (a.fromNode.uid == n) and (a.toNode.uid == u) )

            if(n_to_u_cap_sum > n_to_u_flow_sum):
                flow = round(n_to_u_cap_sum - n_to_u_flow_sum, TOL)
                get['residual_arcs'][n][u] = Arc(n,u,ResidualDatum(flow, 'push'))
            if(n_to_u_flow_sum > 0.0):
                flow = round(n_to_u_flow_sum, TOL)
                get['residual_arcs'][u][n] = Arc(u,n,ResidualDatum(flow, 'pull'))

            get['node_to_node_capacity'][n][u] = n_to_u_cap_sum
            get['node_to_node_flow'][n][u]     = n_to_u_flow_sum

    return get

def fast_bfs(sid, tid, get):
    parent_of   = dict([])
    visited     = frozenset([])
    deq         = deque([sid])
    while len(deq) > 0:
        n = deq.popleft()
        if n == tid:
            break
        for u in get['residual_arcs'][n]:
            if (u not in visited):
                visited = visited.union(frozenset({u}))
                parent_of[u] = get['residual_arcs'][n][u]
                deq.extend([u])
    path = list([])
    curr = tid
    while(curr != sid):
        if (curr not in parent_of):
            err_msg = 'No augmenting path from {} to {}.'.format(sid, curr)
            raise StopIteration(err_msg)
        path.extend([parent_of[curr]])
        curr = parent_of[curr].fromNode
    path.reverse()
    return path

def fast_edmonds_karp(mfps):
    sid, tid = mfps.sourceNodeUid, mfps.terminalNodeUid
    get = fast_e_k_preprocess(mfps.G)
    no_more_paths, loop_count = False, 0
    while(not no_more_paths):
       try:
           apath   = fast_bfs(sid, tid, get)
           get = fast_augment(apath, get)
           loop_count += 1
       except StopIteration as e:
           no_more_paths = True
    nodes = frozenset(get['nodes'].values())
    arcs = frozenset({})
    for from_node in get['arcs']:
        for to_node in get['arcs'][from_node]:
            for arc in get['arcs'][from_node][to_node]:
                arcs |= frozenset({get['arcs'][from_node][to_node][arc]})

    G = DiGraph(nodes, arcs)
    mfps = MaxFlowProblemState(G, sourceNodeUid=sid, terminalNodeUid=tid, maxFlowProblemStateUid=mfps.maxFlowProblemStateUid)
    return mfps

Here’s a visualization of how this algorithm solves the example Flow Network from above. The visualization shows the steps as they are reflected in the DiGraph $G$ representing the most up-to-date Flow Network and as they are reflected in the Residual Graph of that flow network. Augmenting Paths in the Residual Graph are shown as red paths, in the DiGraph representing the problem the set of Nodes and Arcs affected by a given Augmenting Path is highlighted in green. In each case I’ll highlight the parts of the graph that will be changed (in red or green) and then show the graph after the changes (just in black).

Flow Network DiGraph

Residual DiGraph

Here’s another visualization of how this algorithm solving a different example Flow Network. Notice that this one use real numbers and contains multiple Arcs with the same fromNode and toNode values.
Also notice that because Arcs with a ‘pull’ ResidualDatum may be part of the Augmenting Path, the nodes affected in the DiGraph of the Flown Network may not be on a path in $G$ ! (see the picture labeled G:15).

Flow Network DiGraph

Residual DiGraph





[mathx_R_6

Bipartite Graphs

Suppose we have a DiGraph $G$ , $G$ is Bipartite if it’s possible to partition the Nodes in $G.setOfNodes$ into two sets ( $part_{1}$ and $part_{2}$ ) such that for any Arc $a$ in $G.setOfArcs$ it cannot be true that $a.fromNode$ in $part_{1}$ and $a.toNode$ in $part_{1}$ . It also cannot be true that $a.fromNode$ in $part_{2}$ and $a.toNode$ in $part_{2}$ . In other words $G$ is Bipartite if it can be partitioned into two sets of Nodes such that every Arc must connect a Node in one set to a Node in the other set.

Testing Bipartite

Suppose we have a DiGraph $G$ , we want to test if it is Bipartite. We can do this in $O(|G.setOfNodes|+|G.setOfArcs|)$ by greedy coloring the graph into two colors. First, we need to generate a new DiGraph $H$ , this graph will have will have the same set of Nodes as $G$ but it will have more Arcs than $G$ . Every Arc $a$ in $G$ will create 2 Arcs in $H$ , the first Arc will be identical to $a$ , the second Arc reverses the director of $a$ ( b = Arc(a.toNode,a.fromNode,a.datum) ).

Bipartition = namedtuple('Bipartition',['firstPart', 'secondPart', 'G'])
def bipartition(G):
    nodes = frozenset(G.setOfNodes
    arcs  = frozenset(G.setOfArcs)
    arcs = arcs.union( frozenset( {Arc(a.toNode,a.fromNode,a.datum) for a in G.setOfArcs} ) )
    H = DiGraph(nodes, arcs)
    H_as_dict = digraph_to_dict(H)
    color = dict([])
    some_node = next(iter(H.setOfNodes))
    deq = deque([some_node])
    color[some_node] = -1
    while len(deq) > 0:
        curr = deq.popleft()
        for a in H_as_dict.get(curr):
            if (a.toNode not in color):
                color[a.toNode] = -1*color[curr]
                deq.extend([a.toNode])
            elif(color[curr] == color[a.toNode]):
                print(curr,a.toNode)
                raise Exception('Not Bipartite.')

    firstPart  = frozenset( {n for n in color if color[n] == -1 } )
    secondPart = frozenset( {n for n in color if color[n] == 1 } )

    if( firstPart.union(secondPart) != G.setOfNodes ):
        raise Exception('Not Bipartite.')

    return Bipartition(firstPart, secondPart, G)

Matchings and Maximum Matchings

Suppose we have a DiGraph $G$ and matching is a subset of Arcs from $G.setOfArcs$ matching is a Matching if for any two Arcs $a$ and $b$ in matching: len(frozenset( {a.fromNode} ).union( {a.toNode} ).union( {b.fromNode} ).union( {b.toNode} )) == 4. In other words, no two Arcs in a Matching share a Node. Matching matching, is a maximum matching if there is no other Matching alt_matching in $G$ such that len(matching) < len(alt_matching), in other words matching is a Maximum Matching if it is the largest set of Arcs from $G.setOfArcs$ that still satisfies the definition of Matching (the addition of any Arc not already in the matching will break the Matching definition). A Maximum Matching matching is a Perfect Matching if every for Node $n$ in $G.setOfArcs$ there exists an Arc $a$ in matching where a.fromNode == n or a.toNode == n.

Maximum Bipartite Matching

A Maximum Bipartite Matching is a Maximum Matching on a DiGraph $G$ which is Bipartite. Given that $G$ is Bipartite the problem of finding a Maximum Bipartite Matching can be transformed into a Maximum Flow Problem solvable with the Edmonds-Karp algorithm and then the Maximum Bipartite Matching can be recovered from the solution to the Maximum Flow Problem. Let bipartition be a Bipartition of $G$ .

To do this I need to generate a new DiGraph ( $H$ ) with some new Nodes( $H.setOfNodes$ ) and some new Arcs( $H.setOfArcs$ ). $H.setOfNodes$ contains all the Nodes in $G.setOfNodes$ and two more Nodes $s$ (a Source Node) and $t$ (a Terminal Node). $H.setOfArcs$ will contain one Arc for each $G.setOfArcs$ , if an Arc $a$ is in $G.setOfArcs$ and $a.fromNode$ is in bipartition.firstPart and $a.toNode$ is in bipartition.secondPart then include $a$ in $H$ (adding a FlowArcDatum(1,0)). If $a.fromNode$ is in bipartition.secondPart and $a.toNode$ is in bipartition.firstPart then include Arc(a.toNode,a.fromNode,FlowArcDatum(1,0)) in $H.setOfArcs$ . The definition of a Bipartite graph ensures that no Arc connects any Nodes where both Nodes are in the same partition. $H.setOfArcs$ also contains an Arc from Node $s$ to each Node in bipartition.firstPart. Finally, $H.setOfArcs$ contains an Arc each Node in bipartition.secondPart to Node $t$ . $a.datum.capacity = 1$ for all $a$ in $H.setOfArcs$ .

First partition the Nodes in $G.setOfNodes$ the two disjoint sets (part1 and part2) such that no Arc in $G.setOfArcs$ is directed from one set to the same set (this partition is possible because $G$ is Bipartite). Next, add all Arcs in $G.setOfArcs$ which are directed from part1 to part2 into $H.setOfArcs$ . Then create a single Source Node $s$ and a single Terminal Node $t$ and create some more Arcs

construct a MaxFlowProblemState

def solve_mbm( bipartition ):
    s = Node(uuid.uuid4(), FlowNodeDatum(0,0))
    t = Node(uuid.uuid4(), FlowNodeDatum(0,0))

    translate = {}
    arcs = frozenset([])
    for a in bipartition.G.setOfArcs:
        if ( (a.fromNode in bipartition.firstPart) and (a.toNode in bipartition.secondPart) ):
            fark = Arc(a.fromNode,a.toNode,FlowArcDatum(1,0))
            arcs = arcs.union({fark})
            translate[frozenset({a.fromNode.uid,a.toNode.uid})] = a
        elif ( (a.toNode in bipartition.firstPart) and (a.fromNode in bipartition.secondPart) ):
            bark = Arc(a.toNode,a.fromNode,FlowArcDatum(1,0)) 
            arcs = arcs.union({bark})
            translate[frozenset({a.fromNode.uid,a.toNode.uid})] = a
    arcs1 = frozenset( {Arc(s,n,FlowArcDatum(1,0)) for n in bipartition.firstPart } )
    arcs2 = frozenset( {Arc(n,t,FlowArcDatum(1,0)) for n in bipartition.secondPart } )
    arcs = arcs.union(arcs1.union(arcs2))

    nodes = frozenset( {Node(n.uid,FlowNodeDatum(0,0)) for n in bipartition.G.setOfNodes} ).union({s}).union({t})
    G = DiGraph(nodes, arcs)
    mfp = MaxFlowProblemState(G, s.uid, t.uid, 'bipartite') 
    result = edmonds_karp(mfp)

    lookup_set = {a for a in result.G.setOfArcs if (a.datum.flow > 0) and (a.fromNode.uid != s.uid) and (a.toNode.uid != t.uid)}
    matching = {translate[frozenset({a.fromNode.uid,a.toNode.uid})] for a in lookup_set}

    return matching

Minimal Node Cover

A [Node Cover] in a DiGraph $G$ is a set of Nodes (cover) from $G.setOfNodes$ such that for any Arc $a$ of $G.setOfArcs$ this must be true: (a.fromNode in cover) or (a.toNode in cover). A Minimal Node Cover is the smallest possible set of Nodes in the graph that is still a Node Cover. Kőnig’s theorem states that in a Bipartite graph the size of the Maximum Matching on that graph is equal to the size of the Minimal Node Cover, and it suggests how the Node Cover can recovered from a Maximum Matching:

Suppose we have the Bipartition bipartition and the Maximum Matching matching. Define a new DiGraph $H$ , $H.setOfNodes=G.setOfNodes$ , the Arcs in $H.setOfArcs$ are the union of two sets.

The first set is Arcs $a$ in matching, with the change that if a.fromNode in bipartition.firstPart and a.toNode in bipartition.secondPart then $a.fromNode$ and $a.toNode$ are swapped in the created Arc give such Arcs a a.datum.inMatching=True attribute to indicate that they were derived from Arcs in a Matching.

The second set is Arcs $a$ NOT in matching, with the change that if a.fromNode in bipartition.secondPart and a.toNode in bipartition.firstPart then $a.fromNode$ and $a.toNode$ are swapped in the created Arc (give such Arcs a a.datum.inMatching=False attribute).

Next, run a Depth First Search (DFS) starting from each Node $n$ in bipartition.firstPart which is neither n == a.fromNode nor n == a.toNodes for any Arc $a$ in matching. During the DFS some Nodes are visited and some are not (store this information in a n.datum.visited field). The Minimum Node Cover is the union of the Nodes {a.fromNode for a in H.setOfArcs if ( (a.datum.inMatching) and (a.fromNode.datum.visited) ) } and the Nodes {a.fromNode for a in H.setOfArcs if (a.datum.inMatching) and (not a.toNode.datum.visited)}.

This can be shown to lead from a Maximum Matching to a Minimal Node Cover by a proof by contradiction, take some Arc $a$ that was supposedly not covered and consider all four cases regarding whether a.fromNode and a.toNode belong (whether as toNode or fromNode) to any Arc in Matching matching. Each case leads to a contradiction due to the order that DFS visits Nodes and the fact that matching is a Maximum Matching.

Suppose we have a function to execute these steps an return the set of Nodes comprising the Minimal Node Cover when given the DiGraph $G$ , and the Maximum Matching matching:

ArcMatchingDatum = namedtuple('ArcMatchingDatum', ['inMatching' ])
NodeMatchingDatum = namedtuple('NodeMatchingDatum', ['visited'])

def dfs_helper(snodes, G):
    sniter, do_stop = iter(snodes), False
    visited, visited_uids = set(), set()
    while(not do_stop):
        try:
            stack = [ next(sniter) ]
            while len(stack) > 0:
                curr = stack.pop()
                if curr.uid not in visited_uids:
                    visited = visited.union( frozenset( {Node(curr.uid, NodeMatchingDatum(False))} ) )
                    visited_uids = visited.union(frozenset({curr.uid}))
                    succ = frozenset({a.toNode for a in G.setOfArcs if a.fromNode == curr})
                    stack.extend( succ.difference( frozenset(visited) ) )
        except StopIteration as e:
            stack, do_stop = [], True
    return visited

def get_min_node_cover(matching, bipartition):
    nodes = frozenset( { Node(n.uid, NodeMatchingDatum(False)) for n in bipartition.G.setOfNodes} ) 
    G = DiGraph(nodes, None)
    charcs = frozenset( {a for a in matching if ( (a.fromNode in bipartition.firstPart) and (a.toNode in bipartition.secondPart) )} )
    arcs0 = frozenset( { Arc(find_node_by_uid(a.toNode.uid,G), find_node_by_uid(a.fromNode.uid,G), ArcMatchingDatum(True) ) for a in charcs } )
    arcs1 = frozenset( { Arc(find_node_by_uid(a.fromNode.uid,G), find_node_by_uid(a.toNode.uid,G), ArcMatchingDatum(True) ) for a in matching.difference(charcs) } )
    not_matching = bipartition.G.setOfArcs.difference( matching )
    charcs = frozenset( {a for a in not_matching if ( (a.fromNode in bipartition.secondPart) and (a.toNode in bipartition.firstPart) )} )
    arcs2 = frozenset( { Arc(find_node_by_uid(a.toNode.uid,G), find_node_by_uid(a.fromNode.uid,G), ArcMatchingDatum(False) ) for a in charcs } )
    arcs3 = frozenset( { Arc(find_node_by_uid(a.fromNode.uid,G), find_node_by_uid(a.toNode.uid,G), ArcMatchingDatum(False) ) for a in not_matching.difference(charcs) } )
    arcs = arcs0.union(arcs1.union(arcs2.union(arcs3)))

    G = DiGraph(nodes, arcs)
    bip = Bipartition({find_node_by_uid(n.uid,G) for n in bipartition.firstPart},{find_node_by_uid(n.uid,G) for n in bipartition.secondPart},G)
    match_from_nodes = frozenset({find_node_by_uid(a.fromNode.uid,G) for a in matching}) 
    match_to_nodes = frozenset({find_node_by_uid(a.toNode.uid,G) for a in matching})
    snodes = bip.firstPart.difference(match_from_nodes).difference(match_to_nodes)
    visited_nodes = dfs_helper(snodes, bip.G)
    not_visited_nodes = bip.G.setOfNodes.difference(visited_nodes)

    H = DiGraph(visited_nodes.union(not_visited_nodes), arcs)
    cover1 = frozenset( {a.fromNode for a in H.setOfArcs if ( (a.datum.inMatching) and (a.fromNode.datum.visited) ) } ) 
    cover2 = frozenset( {a.fromNode for a in H.setOfArcs if ( (a.datum.inMatching) and (not a.toNode.datum.visited) ) } )
    min_cover_nodes = cover1.union(cover2)
    true_min_cover_nodes = frozenset({find_node_by_uid(n.uid, bipartition.G) for n in min_cover_nodes})

    return min_cover_nodes

The Linear Assignment Problem

The linear assignment problem consists of finding a maximum weight matching in a weighted bipartite graph. Problems like the one at the very start of this post can be expressed as a Linear Assignment Problem. Given a set of workers, a set of tasks, and a function indicating the profitability of an assignment of one worker to one task, we want to maximize the sum of all assignments that we make, this is a Linear Assignment Problem. Assume that the number of tasks and workers are equal, though I will show that this assumption is easy to remove. In the implementation I represent Arc Weights with an attribute a.datum.weight for an Arc $a$ .

WeightArcDatum = namedtuple('WeightArcDatum', ['weight'])

Kuhn-Munkres Algorithm

The Kuhn-Munkres Algorithm solves the The Linear Assignment Problem. A good implementation can take $O(N^{4})$ time, (where $N$ is the number of Nodes in the DiGraph representing the problem). An implementation that is easier to explain takes $O(N^{5})$ (for a version which regenerates DiGraphs) and $O(N^{4})$ for (for a version which maintains DiGraphs). This is similar to the two different implementations of the Edmonds-Karp algorithm.

For this description I’m only working with Complete Bipartite Graphs (those where half the Nodes are in one part of the Bipartition and the other half in the second part). In the worker, task motivation this means that there are as many workers as tasks. This seems like a significant condition (what if these sets are not equal!) but it is easy to fix this issue, I talk about how to do that in the last section.

The version of the algorithm described here uses the useful concept of zero weight arcs. Unfortunately, this concept only makes sense when we are solving a Minimization (if rather than maximizing the profits of our worker-task assignments we were instead minimizing the cost of such assignments). Fortunately, it is easy to turn a Maximum Linear Assignment Problem into a Minimum Linear Assignment Problem by setting each the Arc $a$ weights to $M-a.datum.weight$ where M=max({a.datum.weight for a in G.setOfArcs}). The solution to the original Maximizing Problem will be identical to the solution Minimizing Problem after the Arc weights are changed. So for the remainder assume that we make this change.

The Kuhn-Munkres Algorithm solves minimum weight matching in a weighted bipartite graph by a sequence of Maximum Matchings in unweighted Bipartite graphs. If a we find a Perfect Matching on the DiGraph representation of the Linear Assignment Problem, and if the weight of every Arc in the Matching is zero then we have found the minimum weight matching since this matching suggests that all Nodes in the DiGraph have been Matched by an Arc with the lowest possible cost (no cost can be lower than 0, based on prior definitions). No other Arcs can be added to the Matching (because all Nodes are already matched) and no Arcs should be removed from the Matching because any possible replacement Arc will have at least as great a weight value.

If we find a Maximum Matching of the subgraph of $G$ which contains only zero weight Arcs, and it is not a Perfect Matching we don’t have a full solution (since the Matching is not Perfect). However, we can produce a new DiGraph $H$ by changing the weights of Arcs in $G.setOfArcs$ in a way that new 0-weight Arcs appear and the optimal solution of $H$ is the same as the optimal solution of $G$ . Since we guarantee that at least one zero weight Arc is produced at each iteration we guarantee that we will arrive at a Perfect Matching in no more than |G.setOfNodes|^{2}=N^{2} such iterations.

Suppose that in Bipartition bipartition, bipartition.firstPart contains Nodes representing workers, and bipartition.secondPart represents Nodes representing tasks.

The algorithm starts by generating a new DiGraph $H$ . $H.setOfNodes = G.setOfNodes$ . Some Arcs in $H$ are generated from Nodes $n$ in bipartition.firstPart. Each such Node $n$ generates an Arc $b$ in $H.setOfArcs$ for each Arc $a$ in $bipartition.G.setOfArcs$ where $a.fromNode = n$ or $a.toNode = n$ , b=Arc(a.fromNode, a.toNode, a.datum.weight - z) where z=min(x.datum.weight for x in G.setOfArcs if ( (x.fromNode == n) or (x.toNode == n) )).

More Arcs in $H$ are generated from Nodes $n$ in bipartition.secondPart. Each such Node $n$ generates an Arc $b$ in $H.setOfArcs$ for each Arc $a$ in $bipartition.G.setOfArcs$ where $a.fromNode = n$ or $a.toNode = n$ , b=Arc(a.fromNode, a.toNode, ArcWeightDatum(a.datum.weight - z)) where z=min(x.datum.weight for x in G.setOfArcs if ( (x.fromNode == n) or (x.toNode == n) )).

KMA: Next, form a new DiGraph $K$ composed of only the zero weight Arcs from $H$ , and the Nodes incident on those Arcs. Form a bipartition on the Nodes in $K$ , then use solve_mbm( bipartition ) to get a Maximum Matching (matching) on $K$ . If matching is a Perfect Matching in $H$ (the Arcs in matching are incident on all Nodes in $H.setOfNodes$ ) then the matching is an optimal solution to the Linear Assignment Problem.

Otherwise, if matching is not Perfect generate the Minimal Node Cover of $K$ using node_cover = get_min_node_cover(matching, bipartition(K)). Next, define z=min({a.datum.weight for a in H.setOfArcs if a not in node_cover}). Define nodes = H.setOfNodes, arcs1 = {Arc(a.fromNode,a.toNode,ArcWeightDatum(a.datum.weigh-z)) for a in H.setOfArcs if ( (a.fromNode not in node_cover) and (a.toNode not in node_cover)}, arcs2 = {Arc(a.fromNode,a.toNode,ArcWeightDatum(a.datum.weigh)) for a in H.setOfArcs if ( (a.fromNode not in node_cover) != (a.toNode not in node_cover)}, arcs3 = {Arc(a.fromNode,a.toNode,ArcWeightDatum(a.datum.weigh+z)) for a in H.setOfArcs if ( (a.fromNode in node_cover) and (a.toNode in node_cover)}. The != symbol in the previous expression acts as an XOR operator. Then arcs = arcs1.union(arcs2.union(arcs3)). Next, H=DiGraph(nodes,arcs). Go back to the label KMA. The algorithm continues until a Perfect Matching is produced, this Matching is also the solution to the Linear Assignment Problem.

def kuhn_munkres( bip ):
    nodes = bip.G.setOfNodes
    arcs = frozenset({})
    original_weights = dict({})

    for a in bip.G.setOfArcs:
        original_weights[a.fromNode.uid,a.toNode.uid] = a.datum.weight

    for n in bip.firstPart:
        z = min( {x.datum.weight for x in bip.digraph.setOfArcs if ( (x.fromNode == n) or (x.toNode == n) )} )
        arcs = arcs.union( frozenset({Arc(a.fromNode, a.toNode, WeightArcDatum(a.datum.weight - z))
                                      for a in bip.digraph.setOfArcs 
                                      if ( (a.fromNode == n) or (a.toNode == n) )}) )
    for n in bip.secondPart:
        z = min( {x.datum.weight for x in bip.digraph.setOfArcs if ( (x.fromNode == n) or (x.toNode == n) )} )
        arcs = arcs.union( frozenset({Arc(a.fromNode, a.toNode, WeightArcDatum(a.datum.weight - z))
                                      for a in bip.digraph.setOfArcs 
                                      if ( (a.fromNode == n) or (a.toNode == n) )}) )

    H = DiGraph(nodes, arcs)
    assignment, value, not_done = dict({}), 0, True

    while( not_done ):
        zwarcs = frozenset( {a for a in H.setOfArcs if a.datum.weight == 0} )
        znodes = frozenset( {n.fromNode for n in zwarcs} ).union( frozenset( {n.toNode for n in zwarcs} ) )
        K = DiGraph(znodes, zwarcs)    
        k_bipartition = bipartition(K)
        matching = solve_mbm( k_bipartition )
        mnodes = frozenset({a.fromNode for a in matching}).union(frozenset({a.toNode for a in matching}))
        if( len(mnodes) == len(H.setOfNodes) ):
            for a in matching:
                assignment[ a.fromNode.uid ] = a.toNode.uid
                value += original_weights[a.fromNode.uid,a.toNode.uid]
            not_done = False
        else:
            node_cover = get_min_node_cover(matching, bipartition(K))
            z = min( frozenset( {a.datum.weight for a in H.setOfArcs if a not in node_cover} ) )
            arcs1 = frozenset( {Arc(a.fromNode,a.toNode,WeightArcDatum(a.datum.weight-z))
                                for a in H.setOfArcs 
                                if ( (a.fromNode not in node_cover) and (a.toNode not in node_cover))} )
            arcs2 = frozenset( {Arc(a.fromNode,a.toNode,WeightArcDatum(a.datum.weight)) 
                                for a in H.setOfArcs 
                                if ( (a.fromNode not in node_cover) != (a.toNode not in node_cover))} ) 
            arcs3 = frozenset( {Arc(a.fromNode,a.toNode,ArcWeightDatum(a.datum.weigh+z)) 
                                for a in H.setOfArcs 
                                if ( (a.fromNode in node_cover) and (a.toNode in node_cover))} ) 
            nodes = H.setOfNodes
            arcs = arcs1.union(arcs2.union(arcs3))
            H = DiGraph(nodes,arcs) 
    return value, assignment

This implementation is $O(N^{5})$ because it generates a new Maximum Matching matching at each iteration, similar to the previous two implementations of Edmonds-Karp this algorithm can be modified so that it keeps track of the matching and adapts it intelligently to each iteration. When this is done the complexity becomes $O(N^{4})$ . A more advanced and more recent version of this algorithm (requiring some more advanced data structures can run in $O(N^{3})$ . Details of both the simpler implementation above and about the more advanced implementation can be found at this post which motivated this blog post.

None of the operations on Arc weights modify the final assignment returned by the algorithm. Here’s why: Since our input graphs are always Complete Bipartite Graphs a solution must map each Node in one partition to another Node in the second partition, via the Arc between these two Nodes. Notice that the operations performed on the Arc weights never changes the order (ordered by weight) of the Arcs incident on any particular Node. Thus when the algorithm terminates at a Perfect Complete Bipartite Matching each Node is assigned a zero weight Arc, since the relative order of the Arcs from that Node hasn’t changed during the algorithm, and since a zero weight Arc is the cheapest possible Arc and the Perfect Complete Bipartite Matching guarantees that one such Arc exists for each Node. This means that the solution generated is indeed the same as the solution from the original Linear Assignment Problem without any modification of Arc weights.

Unbalanced Assignments

It seems like the algorithm is quite limited since as described it operates only on Complete Bipartite Graphs (those where half the Nodes are in one part of the Bipartition and the other half in the second part). In the worker, task motivation this means that there are as many workers as tasks (seems quite limiting). However, there is an easy transformation that removes this restriction. Suppose that there are fewer workers than tasks, we add some dummy workers (enough to make the resulting graph a Complete Bipartite Graph). Each dummy worker has an Arc directed from the worker to each of the tasks. Each such Arc has weight 0 (placing it in a matching gives no added profit). After this change the graph is a Complete Bipartite Graph which we can solve for. Any task assigned a dummy worker is not initiated.

Suppose that there are tasks than worker, we add some dummy tasks (enough to make the resulting graph a Complete Bipartite Graph). Each dummy task has an Arc directed from each worker to the dummy task. Each such Arc has weight 0 (placing it in a matching gives no added profit). After this change the graph is a Complete Bipartite Graph which we can solve for. Any worker assigned to dummy task is not employed during the period.

Here’s another example of this type of modification.

Linear Assignment Example

Ok, finally let’s do an example with the code I’ve been using. I’m going to modify the example problem from [here][kunmonk]. We have 3 tasks: we need to Clean The Bathroom, Sweep The Floor, and Wash The Windows. The workers available to use are Alice, Bob, Charlie, and Diane. Each of the workers gives us the wage they require per task. Here are the wages per worker:

	Clean The Bathroom	Sweep The Floor	Wash The Windows
Alice	$2	$3	$3
Bob	$3	$2	$3
Charlie	$3	$3	$2
Diane	$9	$9	$1

If we want to pay the least amount of money, but still get all the tasks done, who should do what task? Start my introducing a dummy task to make the DiGraph representing the problem Bipartite.

	Clean The Bathroom	Sweep The Floor	Wash The Windows	Do Nothing
Alice	$2	$3	$3	$0
Bob	$3	$2	$3	$0
Charlie	$3	$3	$2	$0
Diane	$9	$9	$1	$0

Supposing that the problem is encoded in a DiGraph $G$ , then kuhn_munkres( bipartition(G) ) will solve the problem and return the assignment. It’s easy to verify that the optimal (lowest cost) assignment will cost $5. Here’s a visualization of the solution the code above generates:

assign_diag

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The means of production for dubious induction

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