Chordal bipartite graph Puns

In this proof, I do not understand why the following is true:

As 𝑎≥2, there exists some component 𝐺𝑏 such that |𝑉1∩𝑉(𝐺𝑏)|≥|𝑉2∩𝑉(𝐺𝑏)|

Does anyone see why this is true?

I suspect we can use Konig's theorem for this: I find the minimum vertex cover and show that it is larger than E(G)/D, which will tell me that the maximum matching is larger than E(G)/D. The problem is that I don't know how to practically use the number of edges in G while trying to construct a vertex cover.

I'm honestly not even sure where to begin so I don't have any work that I can show. Maybe I'm completely off the mark. Any hints would be greatly appreciated.

Is there a known algorithm that can be used to compute the minimum vertex cover if we are given a bipartite graph? I need it for some my summer project. If there is then I would request you to link it. Also if possible, if a python code exists that codes that algorithm I would request you to link that as well.

Any help would be appreciated, thank you.

Prove that a bipartite graph does not contain a cycle of odd length.

Let us assume a bipartite graph B = (V,E) with partite sets P1 and P2, let |V(B)| = n and let us satisfy the claim via contradiction.

Assume B = (V,E) contains a cycle of C_{m} as subgraph such that m = 2k+1 where k domain N and k>2. To do this, we must have at least one {v,u} domain E(B) such that v,u domain V(P1), however this is illegal because B is a bipartite graph and for all v in P1, we must have the same color, as v and u are adjacent vertices with the same color, this indicates an invalid coloring. Therefore contradiction follows, the assumption must be incorrect and the claim must be correct.

https://preview.redd.it/u54xy9xol6j61.png?width=893&format=png&auto=webp&s=aecad8d786990016ce0bd4fae2a3637410fc5b31

I'm confused as to where I should start. I know that there are two disjoint sets of vertices, and that a vertex in one set can at most connect to every vertex in the other set. However, how would I go about proving this, for the case when n is even and n is odd? Can anyone start me off? Thanks.

I'm having trouble showing that, for every bipartite graph graph with maximum degree k, there is a k-regular bipartite graph H that contains G as an induced subgraph. I can come up with examples of this, but having a hard time in actually proving it.

Any help would be greatly appreciated!

Hey guys, I am working on automating the assignments of employees to work-days in the hospital where I work. I will implement the solution in python, but atm looking for help with finding the right algorithm to solve my problem. I have first year of uni background in math, so pardon me if my terminology is not accurate :> here it goes:

We have a 'group 1' of vertices that represent employees, let's call them A, B and C. And a 'group 2' of vertices that represent days of the month, but for now let's focus on 9 days (1 through 9).

I want to attach one employee only to each day, so each day has exactly one single edge connected to it, no more, no less. The employees however, can be attached to multiple days (we will have less employees than days, so in order to satisfy the rule that all days must have one edge connected to them we will have to have multiple edges go into each employee vertex).

Not all employees can be connected to all days, however. All employees will have "possible edges" and the algorithm will have to choose as few edges as possible for each employee out of the "possible edges" in order to fill out the days. one last constraint is the days connected to a specific employee must be as evenly spread as possible and as far away from each other as possible.

Here's a visual example of the above:

Group 1 (employees A,B,C) and group 2 (days 1 through 9) and the \"possible edges\" outlined

One possible solution for the above example is this:

Each day (1-9) has a single edge connected to it. Each employee (A,B,C) gets the smallest number of edges connected to it such that the days are as evenly spread as possible.

Of course, sometimes the "possible edges" given to us as a constraint leave us with no choice to have a single employee connected to more days than their peers:

Here, employee A (blue) has to work 5 days while his peers work 2 days each.

A possible solution

If you guys have an algorithm that

There's an algorithm for iterating over the edges in a bipartite graph, to discard those edges that share a vertex.

I think it is the solution to the stable marriage problem. I know this is a CS question, but I wanted to ask it here to hear a math perspective. I heard the term in this phrasing:

>"...In this graph, you need to select two edges that do not intersect at the vertices.
>
>Let deg(x)— the number of edges included in the vertex x.
>
>Iterate over the first edge — (a,b). It will block deg(a)+deg(b)−1 of other edges (all adjacent to vertex a, to vertex b, but the edge (a,b) will be blocked twice. All non-blocked edges do not intersect with (a,b) at the vertices. So you can add (k−deg(a)−deg(b)+1) to the answer."

I want to know what it means.

so im trying to prove that if a graph is bipartite, no 2 adjacent vertices have the same distance between any other vertex. so to the contrary I've assumed there exists vets x,y,z such that x and y are adjacent they are both the same distance from ,k, from z. so my reasoning is that, if you combine the path of shortest length from x to z, and from z to y, along with the edge xy, you get a cycle of length 2k+1, which is odd, resulting in a contradiction. so I was just wondering wether my reasoning was correct, or wether I was missing something subtle, say some vertices being repeated in the cycle.

I'm getting mixed answers on this.

My professor argued that K(2) is the only bipartite value for K(n). However, when I asked my tutor, they argued K(1) is also bipartite if one set is empty of the two sets. Answers online also agree with the tutor.

I'm inclined to intuitively agree with my professor, as we defined a bipartite as adjacent vertices broken up into separate sets, and each vertex cannot have its adjacent neighbour within the same set.

So, in order to have that definition be true, would would need at least two minimum vertices for a graph to be considered to be bipartite.

So, my question is, is a K(1) graph considered bipartite? Can you argue that the vertex in one of the sets has an edge to an empty set?

Each vertex in set A is connected to every other vertex in set B with a weight.

I don't care about the no. of edges that are picked, as long as the total sum of weights of edges pick is the largest possible. Note that there are only positive weights.

1:1 in this case means a vertex from set A can only be match to a maximum of one vertex from set B and vice versa.

What is the algorithm I can employ for this problem?

Hi,

I have been reading about weighted bipartite graphs. I was wondering if there is a formula to find the number of perfect matchings in a weighted bipartite graph given the number of matching for the same graph.

I was looking through the internet but couldn't find any proof or resources.

If the number of vertices was even it would be pretty easy to find a Eulerian Circuit(since you would divided the vertices to two independent sets of even vertices)

But what about this? I think it's impossible but I am not really sure why

Hey /r/learnmath.

So, I've written a proof of the fact that a graph is bipartite if and only if it contains no odd-length cycles (i.e. cycles where the number of edges is odd). However, I'm not sure if it's "formal" enough. I'd love some feedback:

First, we prove that if a graph is bipartite, it cannot contain an odd cycle. We do this by contradiction. Assume that there exists a graph [; G = (E, V) ;] that is bipartite and contains an odd cycle. Now, we define [; G' = (E', V') ;] as the subgraph of the cycle. Picking an arbitrary vertex [; v_1 \in V' ;], and note its colour as red. We then traverse the path [; P ;] from [; v_1 ;] to [; v_n \in V' ;], where [; |V'| = n ;].

We make the observation that every edge traversed performs a colour flip (i.e. if we step from a red vertex, the next one will be blue, and vice versa). Thus, if the number of edges between any two vertices in the cycle is even, then they will be the same colour (and if it is odd, they will be of different colours). We also note that the number of edges in [; P ;] is even, and thus [; v_1 ;] and [; v_n ;] have the same colour.

However, since there is an edge between [; v_1 ;] and [; v_n ;], they cannot have the same colour, as this violates the fact that [; G ;] is bipartite.

QED.

Next, we prove that if a graph contains no odd cycles, then it is bipartite.

We do this in two parts.

First we consider a graph that contains no cycles at all. This is simple -- we can provide a 2-colouring of the graph by picking a random vertex [; v ;] in it, colouring it red, and then performing breadth-first search from that vertex. Every "layer" of the search we colour the opposite colour from the last. Note that since there is a unique path from [; v ;] to any other vertex in the graph, this is guaranteed to work.

Secondly, we consider a graph [; G = (E, V) ;] with an even cycle. [; u, v \in V ;] are two vertices within this graph such that there are two paths between them. Let us define, then, [; x ;] as the vertex where these paths diverge, and [; y ;] as the vertex where they converge once more. Thus there are two paths between [; x ;] and [; y ;]. Note that the length of the cycle is the sum of the lengths of these two paths.

Since the cycle length is even, we know that the difference in the lengths of the two paths is even. Given that the two paths start on the same vertex, an even length difference means that their shared terminatin

Hello everyone!

So, I'm kind of stumped in this question...

By definition, "a bipartite is one where its vertices are in two disjoint and independent sets where every edge connects one vertex from the first set to the second".

If every set is a different colour, this means a bipartite graph has 2 sets and 2 colours. Hence, by definition, the chromatic number of a bipartite graph will always be 2.

This means that a graph with more than 2 colours is not bipartite.

It almost feels like the answer is trivial since that's the definition of a bipartite graph.

Is there maybe a more formal proof that I can use? I get that feeling that my answer feels a bit incomplete, as if I'm missing something, but not sure what.

I've been looking around and found some properties of bipartite graphs that indicate the existence of a perfect matching, but not really any concrete algorithms I can use for implementation.

As K3,3 is non-planar and any graph with a homeomorphism of K3,3 is non-planar can we say that K6,6 is also non planar?

But should this status quo continue? What do you think, /r/math?

My thoughts on this is, but first, yes, arrays of numbers are notationally more convenient, and in research papers I wouldn't recommend writing matrices as figures of bipartite graphs instead of rectangular arrays of numbers. But textbooks that are intended to teach matrices to those who don't know matrices, should have figures of bipartite graphs and rectangular arrays of numbers so that readers learn both forms.

Any help would be appreciated, thank you

Explain why a graph with n vertices, where each vertex has degree at least n - 2 cannot be bipartite when n ≥ 5.

Let us prove the claim via contradiction . Assume ∃ bipartite graph B = (V,E) with partite sets P_{1} and P_{2} such that |V(P_{1})| = m, |V(P_{2})| = k, therefore |V(B)| = m + k. Also assume that ∀ v ∈ V(B), deg(v) ≥ m+k-2, and m+k ≥ 5.

Let us try to satisfy condition deg(v) ≥ m+k-2 ∀ v ∈ V(P_{2}), to do this, we must have that m+k-2 ≤ m, that is k - 2 ≤ 0, and therefore k ≤ 2. Let's say we arrange cardinality of P_{2} and P_{1} regarding these inequalities and the claim holds ∀ v ∈ V(P_{2}). To satisfy the claim for P_{2} vertices, we need to have that ∀ u ∈ V(P_{1}), deg(u) ≥ m+k-2. Since m+k-2 ≥ k (holds because m+k ≥ 5 and k ≤ 2) and we cannot have {u,z} ∈ E(P_{2}) such that u,z ∈ V(P_{2}), the claim does not hold ∀ u ∈ V(P_{1}). Therefore, contradiction follows, the claim must be true.