Total unimodularity is a tool to show that a linear programming problem (LP) has integer optimal solutions. In our discussion on bipartite matchings and linear programming we saw how this is possible to do by hand in a particular case, but as the structure of the LP gets more complicated the details can get very messy. Fortunately, there is a general tool that sometimes gives what is needed.

Definition. A matrix is called totally unimodular if each of its subdeterminants equals 0, 1, or -1. (In particular, each entry of is 0, 1, or -1.)

Proposition. Let be totally unimodular and Then the polyhedron is integral (i.e. each face has an integer point.

Proof. For a minimal (w.r.t. inclusion) face of one has where resp. is a submatrix, resp. subvector, of resp. of and has full row rank. Then, possibly after permuting coordinates, we have with By total unimodularity, Then and satisfies QED.

As a corollary, the LP has integer optimum (and optimiser) whenever Moreover, the same holds for its LP dual by rewriting and applying the proposition above.

Hoffman-Kruskal Theorem. It turns out that this property characterises
total unimodularity:

1. is totally unimodular iff for any the polyhedron is integral.
2. is totally unimodular iff for any , the equality in the LP duality equation is achieved by integral vectors (assuming these min and max are finite).

The second obviously follows from the first.
The first will follow from the following:

A full rank matrix is totally unimodular iff for any the polyhedron is integral.

We omit the proof here.

Examples of totally unimodular matrices.

(Vertex-edge) incidence matrix of a graph is totally unimodular iff is bipartite.

Sketch of the proof.
1) Prove that the incidence matrix of an odd cycle is not totally unimodular, e.g. by considering the vertex numbering such that the edges are and arranging the columns of the matrix in this order, while arranging the rows in the order Then the usual decomposition of the determinant w.r.t. all the permutations of symbols will have only 2 nonzero summands, that will not cancel each other (actually, in the case of even they will cancel each other).
This shows that cannot be totally unimodular if is not bipartite.
2) Let be a square submatrix of If it has a 0 column its determinant is 0. If it has a column with just one 1, we can decompose the determinant w.r.t. this column and apply induction on the size of This leaves one with the case of having exactly two 1s per column. Same argument can be applied to argue that has, w.l.o.g., at least two 1s per row, and thus exactly two 1s, by counting 1s of in two ways. It follows that is the incidence matrix of a disjoint union of (even, as is bipartite) cycles. By reordering the rows and columns we and make to be block-diagonal, with each block the incidence matrix of an even cycle. The determinant of a block-diagonal matrix equals the product of determinants of the blocks, and so it remains to show that the incidence matrix of an even cycle is totally unimodular; this can be done as above. Q.E.D.

Excercise. Combine this with the result on the total modularity above to show that the maximum matching LP for bipartite graphs has an integer solution. Construct the LP dual for this problem and interpret its integer solutions (show that they exist first!) in graph-theoretic terms.

Excercise. The incidence matrix of a digraph is a 0,1,-1 matrix of size so that if, respectively, misses or arrives to or leaves Show that is totally unimodular.

Network flows and max-flow min-cut theorem.
Armed with the above, we can easily prove the max-flow min-cut theorem for networks. Recall that a network is a directed graph with two marked nodes and arc capacities An -flow on the network is a function satisfying for all and for any where and (The latter is called conservation law.) The value of is i.e. the “amount” of that “leaves”
And cut is a partition of into two parts and so that The capacity of the cut is the sum of the capacities of the arc leaving

Max-flow min-cut theorem. Given the network the maximum value of an -flow is equal to the minimum capacity of an -cut.

In order to formulate this as an LP, we introduce the function so that if, respectively, misses or resp. enters or resp. leaves Now we can formulate the max-flow problem as follows:

Let be the sumbatrix of the incidence matrix of obtained by removing the rows with indices and Then the latter can be rewritten as

The matrix of this LP is totally unimodular. If then we can apply the integrality result directly, and establish along the way that there exists an optimal integral flow. In general, we cannot apply the unimodularity directly. But we can take the LP dual. It is as follows:

Here the minimum is attained by integer vectors, as the polytope describing the feasible set is integral. We can extend by defining Then The set satisfies So it is a cut. To show the claim of the theorem, it suffices to show that the total capacity of the arcs leaving is bounded from above (the opposite direction is easy) by Let is such an arc. Then and Then implies so Q.E.D.

Hungarian algorithm is an efficient procedure to find a maximum weight matching in a bipartite graph with parts and and a weight function Usually it is described in terms of potential functions on and looks quite mysterious, as it hides the origin of the procedure, that is a variation of the augmenting path method for finding a maximum (unweighted) matching in Here we describe this point of view.

Let be a matching of maximum weight among all matchings of size If is a maximum (unweighted) matching then we are done.

Otherwise, we construct a bipartite digraph by orienting all the edges in from to and the remaining edges in the opposite direction. We also construct the length function by setting for all and for all Observe that there no negative length (directed) cycles in Indeed, any directed cycle in contains equally many edges in from and in Note that would mean that and so is not maximum, as whereas

Let and be subsets of resp. that are unsaturated by We find (say, using the Bellman-Ford algorithm) a minimum length directed path between and By construction of is an augmenting path for So is a matching satisfying

Claim: is maximum among all matchings of size

We’ll prove this claim below. For now, we remark that it is obvious that this procedure, repeated with taking place of will find matchings of maximum weight among all matchings of size In the end it remains to select among these at most matchings one with maximum weight.

Now let us prove the claim. It certainly holds for as the procedure will select a maximum weight edge in and will consist of this edge. So we can assume Let be a matching of size of maximum weight. We need to show that As there exists an augmenting path such that But is a shortest augmenting path, so
On the other hand, is a matching of size and so by induction. Now as claimed. QED.

As described, we need to do less that loops of this procedure. In each loop, the Bellman-Ford dominates the complexity, so in total we need operations. In fact, we can stop as soon as we reached the situation when

Let a matching have maximum among matchings of size and any matching of size satisfy Then for any matching of size bigger than

Proof. Suppose be a matching contradicting the claim, minimal w.r.t. Let be the digraph constructed from and as above. The connected components of induce oriented cycles and directed paths on Each such component will have a nonnegative total length, as neither there can be a negative length cycle in (see above) nor can there be a negative length -augmenting path (otherwise we can construct ). But this implies that contradiction. QED.

In particular, the complexity above can be improved to $latex O(n|M||E|).

Bellman-Ford shortest paths algorithm. Finding minimal distances in directed graphs, when some arc lengths are negative, cannot be done by Dijkstra’s algorithm (at least not without some preparation). Moreover, it is obvious that one cannot even define the minimal distances when the digraph has a negative length cycle.
While negative length paths look artificial, they arise naturally in modeling (say, a driver has to pay to use certain stretches of the road, but is paid for driving up other stretches) and algorithms, not the least in Hungarian method for weighted bipartite matching.

Let be a directed graph and a length function on the arcs of Suppose we are interested in the shortest, w.r.t. (directed) paths from to all the vertices of (reachable from that is). We define to be the minimal length of at most -arc path from to

Certainly, (as our graph does not have a negative length cycle) and for all

Then, by induction, That’s good, but can grow arbitrarily? Well, if a walk to visits a vertex more than once, this means it contains a cycle — unless this cycle has negative length (and we don’t allow it here) it can be removed from the walk, and it will not increase walk’s length. Thus without loss in generality, a shortest path to has at most vertices, and so is the distance from to

Thus in at most operations we will find the distances from to all the vertices of

Excercise. Modify the procedure to do a simultaneous computation of the distances from each to all i.e. the distance matrix, in operations.

Potentials. In this setting, a useful concept is that of potential, a function such that

for all

When all the arc lengths are nonnegative, the zero function is a potential. When there are negative length arcs, a potential can be used to adjust the distance function so that the new one is nonnegative, and a path is shortest w.r.t. to the new function iff it is shortest w.r.t. the old one: namely, set for all arcs Indeed, when we compute the -length of the path between and we see that

Constructing a potential. Attach to an extra vertex and join it with the vertices of by arcs of length 0. Run Bellman-Ford algorithm for the distances and set As for any arc one has we indeed obtain a potential.

Again, here we assume that there are no negative length cycles in – indeed, otherwise a potential does not exist (this can be proved by summing the inequalities along a cycle.)

An important tool for combinatorial optimisation problems, and graph problems among them, is linear programming, a method (and a problem) to find the optimum of a linear function on a polyhedron (i.e. on an intersection of half-spaces.) Often, such an intersection is bounded, and then one talks about a (convex) polytope rather than a polyhedron.
The optimum of a linear function on a polytope is always reached on a vertex, or, more generally, on a set of vertices of generating an optimal face. (A vertex is a point in that cannot be written as a nontrivial convex combination of two other points of )

Here we demonstrate a way to encode the problem of finding the size of a maximum matching (or, more generally, the maximum weight of a matching) in a bipartite graph as a linear programming problem.

As a warmup, we treat perfect matchings. Consider the vector of variables and the system of inequalities

Certainly, if is the characteristic vector of a perfect matching of i.e. is a 0-1 vector with 1s corresponding to the edges of it satisfies the system above. Thus maximizing the linear function on the polytope defined by gives an upper bound on the size of a maximum weight perfect matching in w.r.t. the edge weights
But in fact much more holds: namely, this bound is tight, as follows from the following theorem.

For a bipartite graph the vertices of the polytope defined by have 0-1 coordinates, and are in the one-to-one correspondence with the perfect matchings of

Proof. Let be a vertex of and be the support of i.e. Then does not contain a cycle. Indeed, a cycle in can be decomposed into the disjoint union of two (not necessarily perfect) matchings, say and as is bipartite and so each cycle in is even. Observe that for any
Thus there exists such that and so contradicting the assumption that is a vertex (and so it cannot be a convex combination of two points in ).

Thus is a forest. Let be a leaf of a subtree of Then implies that Therefore any connected component of is an edge, and so is a matching. It is a perfect matching by QED.

By this theorem, the maximum of on is reached on a perfect matching Another almost immediate corollary is the following well-known result on doubly stochastic matrices (an nonnegative real matrix is called doubly stochastic if its row and column sums are equal to 1.)

A doubly stochastic matrix is a convex combination of permutation matrices.

Indeed, -permutation matrices (the 0-1 doubly stochastic matrices) are in one-to-one correspondence to the perfect matchings of (the characteristic vector of such a matching can be viewed as an matrix, with rows corresponding to one part of the graph, and columns to the other) and so the doubly stochastic matrices are exactly the points of the perfect matching polytope of that has, by our theorem, perfect matchings as vertices (and so any point in it is a convex combination of perfect matchings).

Excercise. Formulate an analogy of system for matchings (not necessarily perfect) in bipartite graphs, and prove the corresponding result for vertices of the polytope in question. Using it, formulate the linear programming problem with optimal value being the size of maximum matching.

Excercise. Show that these constructions fail for non-bipartite graphs.

Matchings in bipartite graphs and systems of distinct representatives (SDRs) of systems of subsets are closely related. More precisely, let for a finite set. We can construct a bipartite graph where

An SDR of is such that for all and for all In particular contains exactly one element for each
Then is a matching of saturating the part In particular, it is a maximum matching.
Conversely, a matching of saturating gives rise to an SDR for

An obvious necessary condition for existing of an SDR for
is the following Hall’s condition:

In view of this, Hall’s Theorem for matchings is equivalent to the following

An SDR for exists iff it satisfies Hall’s condition.

To see that Hall’s condition is sufficient, we proceed by induction on For the claim is obvious. Assume that it is true for all collections of less than sets. We now establish that it holds for -set collections, too.

If then we take any and set for some Now by induction there is an SDR for Then is an SDR for

In the remaining case there exists such that By induction, there is an SDR for
Let Then for otherwise contradicting Hall’s condition. Now we can take an SDR for and observe that is an SDR for

A list-colouring of a graph with a colour list for is a proper colouring of by elements of so that the adjacent vertices are coloured in different colours and the colour for is in In particular, when all the s are equal to each other we have a classic graph colouring.

is called -choosable, or list-colourable if it is list-colourable for any satisfying for all The minimal for which is is choosable is called the list chromatic number of and denoted by Certainly,
It is not hard to construct examples when e.g. take and consider where vertices in the size two part of get colour sets .

Using the fact that each planar graph has a vertex of degree at most 5, as can be easily established using the Euler formula, one can prove that each planar graph is 6-choosable (employ induction by removing a minimal degree vertex).

It is less easy to show that

any planar graph is 5-choosable.

This was established in 1994 by Thomassen.
In fact cannot be improved — there are examples of planar for which The smallest known (in 2003) example has 63 vertices.

The classical analogy of this, the Heawood theorem that says that each planar graph is 5-colourable, is known since 1890, and the 4-colourability is a recent and quite difficult result by Appel and Haken.

To show the 5-choosability, one first observes that it suffices to prove it for the planar graphs for which every face, except possibly the exterior face of the plane embedding, is a triangle. We call such graphs almost triangulated. Then, one proceeds by induction on Assume that we know that for any almost triangulated planar with exterior face and such that for all resp. for all we can extend a proper -list-colouring of an edge on to a full proper -list-colouring (as is obviously true provided giving us a basis of induction).

Let us now show this for with Let be an edge of that we colour using and
If has a chord, i.e. there is an edge that joins two non-adjacent vertices of the cycle induced on (and so is not cutting through the fact bounded by the cycle on ), we spilt into two planar parts and so that and have in common. W.l.o.g. by induction, we can extend this colouring to a list-colouring of This colouring will fix a proper list-colouring of the edge that is on the exterior face of Again, by induction, we can extend this colouring of to a list-colouring of Combining and we obtain a list-colouring of as required.

It remains to consider the case when does not have a chord.
Then let be the vertex adjacent to Take any two colours in (this is possible as ) and remove from for all Then remove and all the edges on it from We obtain a planar graph with one vertex less, and with the valid exterior face conditions. By induction, we can extend the colouring of to a list-colouring of It remains to see that can be properly coloured, too. Indeed, we can colour it with either or depending on the colour of the neighbour of on distinct from Q.E.D.

The concept of prime ideal in a commutative ring with is one of several natural generalisations of the concept of prime integer number.

An ideal is called prime if for any the following holds:
implies that at least one of these elements, and is in

E.g. the principal ideal is prime if and only if is prime.

A maximal ideal is prime.

Indeed, let Assume that We need to show that then If this were not the case then and are two non-zero elements in the ring such that But this is not possible, as is a field. Thus as claimed.

Analysing this proof, one can easily see that

If is prime then has no zero-divisors, i.e. it is an integral domain.

Further important property of prime ideals is that they are radical, i.e.
where the radical of the ideal ideal is Indeed, implies that either or is in and we derive by applying this reduction.

Yet another interesting observation is that is multiplicatively closed.

Rings of fractions

A subset is called multiplicatively closed (or multiplicative) if and for any

Given a multiplicatively closed set one defines
a relation on as follows:

It is not hard to show that is an equivalence relation.
To simplify the notation, write its equivalence class with representative as We define

is the ring of fractions of w.r.t. with addition and multiplication given by the rules

It is easy to check that this is well-defined. We also have

so that is a ring homomorphism.

Note that need not be injective, i.e. need not hold. Indeed, if is a zero-divisor such that for then and so

The most well-known example is the case being an integral domain, and Then is a field, called the field of fractions of

Examples:

• is the field of fractions of
• for the ring of polynomials over a field the field is the field of rational functions over
• Let be non-nilpotent. Then isa multiplicative set. Moreover, then (it is not completely trivial to prove this, though). Intuitively, we make the variable behave like the inverse of as in this ring.

Now let us look at the case for a nonzero prime ideal. In this case is denoted by and called the localisation of at

Example: Let be the ring of polynomials over a field and Then is prime, and is equal to

The ring has unique maximal ideal

It suffices to show that is invertible in iff Indeed, if then On the other hand, if then there exists such that Thus and so (if it was, would be in as is an ideal).

There is one glaring omission in our Linear Algebra curriculum – it avoids talking about the dual space of a vector space. This makes talking about relationship between subspaces and equations that define them exceedingly difficult. Better late than never, so here it comes.

Let be a vector space over a field Denote by the set of linear functions

Examples
Let the space of continuous functions on Then the function given by is linear on

Let be the vector space of polynomials with real coefficients.
Then the function given by is linear on

Note that as is linear, one has for any Thus we have defined by so that To simplify notation, we will write instead As well, we can define for any and more generally, And there is the zero function for any Thus we have all the ingredients of a vector space, as can be easily checked.

is a vector space over It is called the dual space of

So far, we haven’t used the linearity of our functions at all (we actually did not need the fact that ). Indeed, any closed under addition and multiplication set of functions would form a vector space.
What makes the dual space so special is that to define it suffices to define on a basis of Indeed, so we can compute for any once we know the ’s.

Thus for a finite-dimensional vector space one sees a (dependent upon the choice of a basis in ) bijection between and This bijection, that is even an isomorphism of vector spaces, is defined by the dual basis of given by coordinate functions where ’s are the coefficients of is the decomposition of in the basis

Finite-dimensionality is crucial here. E.g. let us consider the vector space of polynomials It is a countable space: one can view it as the set of infinite 0-1 strings, with only finitely many 1’s occurring in each string. On the other hand, can be viewed as the set of all the infinite 0-1 strings, which is uncountable, so there cannot be a bijection between and

Given one can define a function as follows: It is linear, as Here we do not see any dependence on the choice of a basis in and we have

The vector space of linear functions on is (canonically) isomorphic to via the mapping

Indeed, we see immediately that and so we need only to check that this mapping is bijective. Let be a basis in and its dual basis in Then if and 0 otherwise. Thus is the basis of which is dual to the basis of and the mapping sends the vector with coordinates to the vector with the same coordinates in
the basis of Hence the latter is bijective.

In view of the latter, we can identify with and write instead The set of such that is a subspace, called annihilator of of dimension More generally, the following holds.

Let be a subspace of and Then the annihilator of is a subspace of of dimension

Indeed, we can choose a basis in so that is a basis of where Then we have the dual basis of and is the subspace with the basis

In view of this, each can be obtained as the set of solutions of a system of homogeneous linear equations for of rank

Dual spaces and annihilators under a basis change
Let be a linear transformation of and a subspace of Then is a subspace. How can one look at By writing out in a basis for any in the dual basis we get equation Thus, considering as a matrix, we get where denotes the action of on It follows that i.e. We have, considering that acts on by right multiplication, and not by left ones, to take the transpose, too.

acts on as

An example.
Let We work in the standard basis of Then the dual basis of is , so that
Let be the group of matrices It fixes, in its left action on by multiplication, the vector . Let be the 1-dimensional subspace of generated by Then is generated by and The group preserves in its action on As is 2-dimensional, there should be a nontrivial kernel in this action, and indeed, it consists of the elements of the form

A particularly simple case is Then is isomorphic to the symmetric group on 4 letters, as can be seen in its action on the 4 elements of outside On the other hand, it acts on the 3 nonzero elements of as

Excercises in algebra

A classical example are -necklaces and -bracelets. These are words of length in a -letter alphabet, with beginnings and ends of the words glued together and made invisible (so one can visualise them as regular with letters from the alphabet placed at the vertices). Two necklaces are identical when one is obtained from another by rotations; two bracelets are identical when one can be obtained from another by rotation and a “flip”. One can view counting of necklaces and bracelets as counting of orbits of certain groups on the sets of “labelled” necklaces (resp. bracelets). For the necklace case, the group is and for the bracelet case it is the dihedral group of order
The set of labelled necklaces/bracelets is just the set of words of length in a -letter alphabet; so one has

Orbit counting is often simplified by the following result, often attrubuted to Burnside, although it was certainly known to Frobenius.

Let a finite group act on a finite set Denote by for the set of elements of fixed by Then the number of orbits of on is the average, over of i.e.

The proof involves counting of edges in the bipartite graph, with one part being and the other being a pair for and is an edge iff Thus the total number of edges in this graph is
On the other hand, a pair is an edge iff the stabiliser of in Thus, for a -orbit one has the total number of edges leaving equal to Summing over all the -orbits, one obtains the total number of edges completing the proof.

Often it is easier to sum over distinctive representatives of the conjugacy classes as So one has

where denoting the centraliser of in

Let us see how this can be applied to counting bracelets. For simplicity, let us assume that is prime. We consider natural the action of the dihedral group on the set of words of a -letter alphabet.

We know that is generated by the cyclic permutation and an order two “flip” permutation The group can be decomposed into the union of cosets
The only words fixed by a nonidentity element of the subgroup are the “constant” ones, i.e. words consisting of a single letter, and so The coset equals (an unusual coincidence), and so it suffices to find
For a word to be in the latter set, it must have … so there are possibilites.
Thus we count
E.g. for one obtains