Euler and Hamilton Paths

Transcription

1 Euler an Hamilton Paths The town of Königserg, Prussia (now know as Kaliningra an part of the Russian repuli), was ivie into four setion y ranhes of the Pregel River. These four setions C A D B Figure: Seven Briges of Koinigserg inlue the two regions on the anks of the Pregel, Kneiphof Islan an the region etween the two ranhes of the Pregel. In the eighteenth entury, seven riges onnete these regions. The following iagram shows the position of the riges on the Pregel in the town of Königserg.

2 As for an European ountry, the town folks use to walk roun the town on the Sunays, an every time they wonere if there was any means to go roun the ity without rossing these riges twie, an return to the starting point. Swiss mathematiian Leonhar Euler solve this prolem an pulishe that in In his solution, he use graphs for the first time to solve this prolem. His key points in the solutions were taking the regions as verties, an the riges as the eges etween the verties thus making a multigraph. His multigraph solution is in the following figure: C A D B

3 This iagram rephrase or showe the prolem of travelling aross every rige without rossing any rige more than one. Now, we may ask if we an raw a simple iruit from this multigraph that will ontain all the riges in the original iagram? That means, an we raw a simple graph that ontains every ege of the multigraph. Euler Ciruit: Euler Path: An Euler Ciruit in a graph G is a simple iruit ontaining every ege of G. An Euler Path in G is a simple path ontaining every ege of G. Neessary an Suffiient Conitions for Euler Ciruits an Paths What is the onition if a onnete multigraph has an Euler iruit? This is one if we an show that every vertex must have even egree. Let us suppose that one Euler iruit egins with a vertex a, an ontinues with an ege inient to a, say {a, }. The ege ontriutes 1 to eg(a). Eah time the iruit passes through a vertex it ontriutes 2 to the vertex s egree. This is eause the iruit enters from an ege inient with this vertex an leaves via another suh ege. Finally the iruit

4 terminates at the point it originate: say at vertex a, ontriuting 1 to eg(a). Therefore, eg(a) must e even, eause the iruit ontriutes 1 when it egins, an 1 when it ens, an 2 a f G1 e G2 e every time it passes through a. A vertex other than a has even egree eause the iruit ontriutes 2 to its egree eah time it passes through the vertex. We onlue that if a onnete graph has an Euler iruit, then every vertex must have even egree. Now we shall try to investigate if this neessary onitions is also suffiient for making an Euler iruit. Suppose that G 1 is a onnete graph. All its verties have even egree values.

5 In the aove graph G 1, the iretion of the onnete eges are: {a, }, {, }, {, f}, {f, a}. That is, the eges of the iruit egins from a, then passes through the verties,, f an ultimately ens in a. If we stuy this graph arefully, we see that one ege enters to a vertex an then use another ege to ome out from that vertex. That is why, for any vertex of the irete path of G 1 graph, there is an ege that is inient on that vertex (enters) an inient from that vertex (omes out). If we onsier vertex f, we see {, f} an {f, a} are two eges that are inient to this vertex. Therefore, the egree of this vertex is 2, i.e., eg(f) = 2. In the graph G 1, we i not show any iretion for the other smaller onnete paths for the verties,, e. Let us suppose that these verties {,, e} makes up another sugraph if we elete verties {a,, f}. Let us name this sugraph as G 2. Now, let us suppose a iruit in the graph G 2 as we ha rawn in G 1. Let the iruit egins from vertex, then goes to, then to e, an finally ens in. The iruit is:,, e,. An they make a path in the graph G 2. If we onsier the graph G 1, the iruit woul e: a,,,, e, f, a.

6 Theorem: A onnete multigraph has an Euler iruit if an only if eah of its verties has even egree. Königserg Brige Prolem C A D B Now let us try to solve the Königserg rige prolem. Let us see the Euler s graph of the Königserg riges here again: This graph has four verties of o egree: eg (a) = 5, eg () = 3, eg () = 3 an eg () = 3. So it oes not have an Euler iruit. There is no way to start from one point, ross eah rige one, an then ome ak to the starting point.

7 Fleury s Theorem: If the originating vertex of a multigraph is aritrarily hosen, it forms a iruit y hoosing eges suessively. One an ege is hosen, that ege is remove. Eges are hosen suessively so that eah ege egins where the last ege ens, an so that this ege is not a ut ege unless there is no alternative. The theorem in logial steps is given elow: Step 1: Choose a starting vertex, say u. Step 2: Traverse any availale ege, hoosing an ege that will isonnet the remaining graph only if there is no alternative. Step 3: After traversing eah ege, remove it (together with any verties of egree 0 whih result) Step 4: If no ege remains, stop. Otherwise, hoose another availale ege an go ak to step 2. Theorem 2: A onnete multigraph has an Euler path ut not an Euler iruit if an only if has exatly two verties of o egree. Let us suppose the following graphs G an H: The graph G is an Euler Ciruit: all the verties are of egree 2.

8 a a G H The graph H, aoring to the theorem 2, has Euler path ut has no Euler iruit, as two of its verties are of egree 3. Those tow verties are a an. Hamiltonian Paths an Ciruits So far we ha evelope neessary an suffiient onitions for the existene of paths an iruits that ontain every ege of a multigraph exatly one. Here, with the help of Hamiltonian Paths an Ciruits, we shall try to see if there exists any simple paths an iruits ontaining every vertex of the graph exatly one. This terminology omes from a puzzle invente in 1857 y an Irish mathematiian Sir William Rowan Hamilton. He was evote to non-ommutative algera, an worke a lot

9 in this area. Though, he mae important ontriutions to the optis, astrat algera an ynamis. He invente Iosian Game ase on his work in non-ommutative algera. The puzzle in the example 5 is the representation of that game. Before moving to examples, let us see the efinition of Hamiltonian Path an Hamiltonian Ciruit. Hamiltonian Path: A path x 0, x 1,..., x n-1, x n in the graph G = (V, E) is alle Hamiltonian Path if V = {x 0, x 1,..., x n-1, x n } an x i x j for 0 i j n. Hamiltonian Ciruit: A iruit x 0, x 1,..., x n-1, x n (with n>1) in a graph G = (V, E) is alle a Hamiltonian iruit if x 0, x 1,..., x n-1, x n is a Hamiltonian Path. Q n is a Hamiltonian iruit. Let us suppose a simple graph G as epite in the following figures:

10 a a a e H f e M g G Figure G: The Hamiltonian iruit is present in this graph. The iruit is a,,,, e, a. Figure H: No Hamiltonian iruit is present: for a Hamiltonian iruit e present here, that must ontain the ege {a, } twie. But there is a Hamiltonian Path: a,,,. Figure M: No Hamiltonian Path or Hamiltonian Ciruit is present here. At least one vertex is left out of Hamiltonian Path, an iruit is not possile. Some neessary an suffiient onitions for Hamiltonian Ciruits an Paths There are no known simple neessary an suffiient theorems for the existene of Hamiltonian iruits.

11 Still there are some properties an e evise that may provie some give neessary onitions for Hamiltonian iruits. One of suh property says that: A graph with a vertex of egree 1 annot have a Hamiltonian Ciruit. This is only eause a Hamiltonian iruit, eah vertex is inient with two eges in the iruit. Another property says that if a vertex in the graph has egree 2, then oth eges that are inient with this vertex must e part of any Hamiltonian Ciruit. Also note that, when a Hamiltonian iruit is eing onstrute an this iruit has passe through another vertex, then all remaining ege inient with this vertex, other than the two use in the iruit, an e remove from onsieration. Furthermore a Hamiltonian iruit annot ontain a smaller iruit within it. Suffiient onitions for existene of Hamiltonian iruit: If G is a onnete simple graph with n verties where n 3, then G has a Hamilton iruit if the egree of eah vertex is at least n/2.