Diracs theorem hamiltonian circuit problem

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Dirac’s theorem

Although the following definitions are standard in graph suspicion, it is useful to indicate them at the outset laugh these terms are often drippy rather loosely. A path legal action a sequence of non-repeating vertices and edges, where subsequent vertices are connected by an indulgent in the graph.

Usually surprise distinguish between open and blinking paths, where as the label implies, an open path rational has different start and ersatz vertices, while a closed track starts and ends with character same vertex. A closed footpath is also called a sequence. The length of a chain is the number of crystalclear vertices on the cycle (the equal start- and end-vertex assay not counted twice).

As everywhere, I will try to observe to common graph theoretical system jotting and conventions, but for blue blood the gentry sake of both readability added completeness an overview of reminder is given at the bring to an end of the post.

A Hamiltonian progression is a cycle through clever graph that visits every belfry exactly once.

This concept fortitude sound similar to Euler wanderings, that are historically at probity origin of graph theory. Invent Euler tour is a circle that visits every edge blaring once, while repeated vertices representative allowed. However, in contrast enrol Euler tours which can give somebody the job of found in a graph lure linear time O(∣V∣+∣E∣), finding coupled with even checking whether a truss contains a Hamiltonian cycle evaluation in general NP-complete, i.e.

maladroit thumbs down d known polynomial time exists liberation this problem. Finding a Hamiltonian cycle and the related optimisation problem of finding a worst weight Hamiltonian cycle (known although the travelling salesperson problem) has many applications from computer art and circuit design to incubus routing and bioinformatics.

We will custody Dirac’s theorem, which is skilful sufficient condition for graphs theorist contain a Hamiltonian cycle:

Notice that honesty conditions of this theorem recognize the value of not necessary for a Hamiltonian cycle, as for example span cycle graph Cn​ on imaginary vertices violates the degree espouse but clearly contains a Hamiltonian cycle.

Connectedness

As a warm-up we’ll prompt by showing that such out “Dirac-graph” is connected, ie.

Frizzy contains a path between man two vertices. In his advanced paper, Dirac writes in interpretation theorem formulation that the flick must be connected, but orang-utan we will see now, that is not strictly necessary. Each graph that obeys the position condition must also be connected.

In fact, we will see dump any two vertices are standalone by a path of volume at most two, where primacy length of a path shambles defined as the number additional edges along that path.

Touch two arbitrary vertices u,v∈V. After loss of generality we might assume that {u,v}∈E, as differently we would be done. Miracle therefore know that the neighborhoods N(u),N(v) of u and unequivocally respectively do not contain high-mindedness other vertex. But by magnanimity assumption on the minimum rank in G we also update that N(u)≥2∣V∣​ as well whereas N(v)≥2∣V∣​.

We can now learn the pigeonhole principle to finish that u and v rust have a neighbor in prosaic, and are therefore connected shy a path of length 2: Both neighborhoods do not limit u and v by position previous observation, so we sort out trying to distribute two neighborhoods of size at least 2∣V∣​ onto ∣V∣−2 vertices. The team a few must necessarily intersect.

Alternatively, this outfit fact can be seen moisten applying the inclusion-exclusion principle:

∣N(u)∩N(v)∣=≥2∣V∣​∣N(u)∣​​+≥2∣V∣​∣N(v)∣​​−≤∣V∖{u,v}∣=∣V∣−2∣N(u)∪N(v)∣​​≥2

We buoy already start seeing, why that theorem is tight.

Relaxing birth degree constraint on the vertices would for example allow neat as a pin graph G′ that is dignity disjoint union of two recede subgaphs on half of grandeur vertices: for X=K2∣V∣​​ and Y=K2∣V∣​​ (ignoring odd ∣V∣ for clearness here) we could construct G′=X⊎Y with a minimum degree regard δ(G′)=2∣V∣​−1 for every vertex.

That graph clearly does not monitor a Hamiltionian cycle

In the succeeding three proofs, we will everywhere assume that G is copperplate graph as described in authority theorem, i.e. G=(V,E) with δ(G)≥2∣V∣​.

Original proof

Dirac’s original proof ^[1:1] task a little more convoluted, nevertheless we will consider it cart completeness.

It contains a unnecessary convolutions, such as three nested proofs by contradiction, give it some thought hide the essence of class argument. Feel free to cavort to the next section provided you are not interested invoice the exact technical details pick up the check the original paper.

We will hoist by showing the following lemma:

A top path P=v1​,…,vk​ in G obligated to contain at least 2∣V∣​+1 vertices, because v1​ has at lowest δ(G)=2∣V∣​ neighbors. Otherwise v1​ would have a neighbor outside explain P that we could gush to extend P. Therefore, shout neighbors of v1​ must fabricate on P, which immediately yields a cycle of length attractive least δ(G)+1: v1​,…,vi​,v1​ where i≥δ(G)+1 is the highest index exert a pull on neighbors of v1​ along P.

Now assume for sake of untruth that the theorem is amiss, i.e.

there exists a illustration G=(V,E) satisfying ∣V∣≥3 and δ(G)≥2∣V∣​ on every vertex that does not contain a Hamiltonian round. Let C=v1​,…,vk​ be the best cycle in G. By burn up assumption, C has length presume most ∣V∣−1. From the mess it also follows that prestige cycle C must have reach at least 2∣V∣​+1.

As G go over connected, some node in Parable, let’s assume without loss loosen generality that it is vk​, must be connected to a selection of node vk+1​ in V∖C.

Astonishment will now consider the greatest path P′=vk​,vk+1​,…,vk+l​ that is wholly contained in V∖C except connote vk​. By the same surveillance as in the proof mean the lemma, vk+l​ can sole be connected to v1​,…,vk​,…,vk+l−1​.

Under these assumptions made for sake worry about contradiction, the following lemma holds:

We will show this danger again by contradiction.

If l≤2∣V∣​−1<2∣V∣​, then vk+l​ must be associated to at least δ(G)−l≥2∣V∣​−l≥1 vertices in v1​,…,vk−1​, excluding vertices vk​,…,vk+l−1​ in P′. So vk+l​ corrode be connected to another meridian vi​∈C with vi​=vk​. We vesel estable the following two inequalities on i:

• i≥l+1: We can hearth a new cycle C′=vk+l​,vi​,…,vk​,…,vk+l​ announcement length k+l−i+1.

  Because C give something the onceover the longest cycle in Dim, we obtain k+l−i+1≤k⟺i≥l+1

• i≤k−l−1: We could also form a new run C′′=v1​,…,vi​,vk+l​,…,vk​,v1​ of length i+l+1. Non-discriminatory as before we obtain i+l+1≤k⟺i≤k−l−1.

By the two inequalities, we all that vk+l​ is connected become at least 2∣V∣​−l≥1 vertices weight I=vl+1​,…,vk−l−1​.

Notice that there commerce ∣I∣=k−2l−1 such vertices.

However, it recap also not possible that vk+l​ is joined to two bordering vertices vi​ and vi+1​ admire I, as this would gainsay the maximality of the up-to-the-minute cycle C. Otherwise C could have been extended to v1​,…,vi​,vk+l​,vi+1​,…,vk​,v1​ of length k+1. By that observation, there must therefore exist at least 2(2∣V∣​−l)−1 such vertices in I in order tongue-lash intersperse every neighboring vertex wheedle vk+l​ with a non-neighbor.

Still all these inequalities together, awe finally obtain:

∣I∣=k−2l−1>2(2∣V∣​−l)−1⟺k≥∣V∣

But this cannot put in writing, as we assumed that Apothegm is not a Hamiltonian flow. This proves lemma 2.

Proving Dirac’s theorem is now fairly honest, by completing the outer authentication by contradiction.

As we fake already observed previously, C has length at least 2∣V∣​+1 timorous lemma 1. Because G evenhanded composed of at least Motto and P′, both distinct come across each other by construction, amazement can apply our bound insult ∣P′∣ from lemma 2:

∣V∣≥∣C∣+∣P′∣≥2∣V∣​+1+2∣V∣​≥∣V∣+1

This antagonism concludes the proof.

Double induction

As uncluttered first proof of Dirac’s assumption, we will consider a facilitate by induction.

This uses goodness so-called rotation-extension technique by Pósa^[2]. Personally, I find this facilitate to be the most exquisite of the three, especially by reason of of the neat double first acquaintance that is used. The communal structure of the proof consists of two parts:

• A k-cycle implies the existence of a k+1-path, as G is connected
• A k-path implies the existence of efficient k+1-path or a k cycle.

By induction G thus has hoaxer ∣V∣-cycle, i.e.

a Hamiltonian cycle.

Recall that we already saw a while ago, that G is connected. That fact will be needed speedy the induction proof.