User:Kichigai88/sandbox

Dénes Kőnig
 * Biography
 * Dénes Kőnig born Sept. 21, 1884 a Jew born in Budapest, Hungary. Son of an accomplished Mathematician, Gyula Kőnig.  To honor his fathers' death in 1913, Kőnig and his brother Győrgy created the Gyula Kőnig prize in 1918.  This prize was meant to be an endowment for young mathematicians, however was later devaluated. But the prize remained as a medal of high scientific recognition.  In 1899, he published his first work while still attending High School in a journal Matematikai ès Fizikai Lapok.  After his graduation in 1902, he won first place in a mathematical competition "Eőtvős Lorád".  Shortly after he wrote the first of two book collections Matematikai Mulatságok (Mathematical Entertainments).  He spent four semesters at the university in Budapest and his last five in Gottingen.  During which he studied under a famous Mathematicians József Kürschák and Hermann Minkowski.  He then received his in 1907 due to his dissertation in geometry, that same year he began working for the Technische Hochschule in Budapest and remained apart of the faculty till his death in 1944.  At first he started as an assistant in problem sessions, in 1910 he was promoted to "oberassistant", and then promoted to "Privatdocent" in 1911 teaching nomography, analysis situs (later to be known as Topology), set theory, real numbers and functions, and graph theory (the name "graph theory" didn't appear in the university catalogue until 1927).   During this time he would be a guest speaker giving mathematics lecture for architecture and chemistry students, in 1920 these lectures made their way in to book form.  He finally gained full professorship in 1935 at the Technische Hochschule.
 * From 1915 to 1942 he was on a committee to judge school contests in mathematics, collecting problems for these contests, and organizing them. Then in 1933 he was elected as secretary of the society and in 1942 he became the chairman of this committee.  He then decided to make edits in the society's journal during his time on the committee till his death.  Kőnig's activities and lectures played a vital role in the growth of graph theoritcal work of: Làszló Edyed, Pál Erdös, Tibor Gallai, Győrgy Hájos, Jószef Kraus, Tibor Szele, Pál Turán, Endre Vázsonyi, and many others.  He then went on to write the first book on graph theory Theorie de endlichen und unendlichen Graphen in 1936.  This marked the beginning of graph theory as its' own branch of mathematics.  Then in 1958, Claude Berg wrote the second book on graph theory, Thèorie des Graphes et ses applications, following Kőnig. After the occupation of Hungary by the Nazis, fortunately for persecuted mathematicians he worked to help them.  On October 15, 1944 the Nazi-affiliated Hungarian Natinal Socialist Party took over the country, days later on October 19, 1944 he committed suicide to evade persecution from the Nazis being a Hungarian Jew.


 * Accomplishments
 * 1899 - Matematikai ès Fizikai Lapok written while attending High School
 * 1902 - First place in "Eőtvős Lorád"
 * 1907 - Recieved his Doctorate's Degree
 * 1910 - promoted to "oberassistant"
 * 1911 - promoted to "Privatdocent" in 1911 teaching nomography, analysis situs (later to be known as Topology), set theory, real numbers and functions, and graph theory
 * 1935 - gain full professorship at Technische Hochschule
 * 1936 - he the wrote the first book on Graph Theory, Theorie de endlichen und unendlichen Graphen


 * The Labyrinth Problem
 * The notion of a labyrinth is synonymous with a maze, however here it is looked at from a mathematical point of view. A point of view in which Dénes Kőnig sought out to find a solution to.  Although he postulated three solutions to a labyrinth problem, he used the works of Wiener, Tremaux, and Tarry to formulate the solutions.  Each solution involves looking a labyrinth as a finite connected graph, where the width of the graph does not matter.  The passages of the labyrinth as Kőnig relates them to the edges of a graph and the dead ends as the vertices of the graph.  A series of "walks" and "reverse walks" on an edge of the graph would determine how to proceed throughout the labyrinth in reaching a definite spot in the labyrinth, usually designated as the center, and back to the starting point.
 * Since the spot in the graph is indefinite to be considered as the center of the graph, the solutions Kőnig addresses is a method by which every point in the labyrinth is reached. Of course, if the layout of the labyrinth is known, then in fact, there are a finite number of possibilities, since there is no difficulty.   Instead, we assume that the layout of the labyrinth is unknown and by walking from a certain place X (or an endpoint) is reached, the labyrinth (the graph), known as the “neighborhood of X”, has passages (edges) going to X are known as well as the property whether the passage has already been traversed, therefore relating to the neighborhood X.
 * Wiener’s Solution
 * First let us assume that walking in a labyrinth a vertex P is reached, for every edge going to P is known whether or not it has been traversed earlier, at the same time we can assume that it is possible to traverse the same path in the opposite direction with repetition. This problem was first solved by Wiener who was also the first to view the labyrinth problem as a mathematical problem.  Wiener’s method is as follows:
 * "Beginning at an arbitrary vertex A as starting point, the edges AB, BC, CD, … are chosen quite arbitrarily until an endpoint Q of the graph is reached or such a vertex Q, which has already been passed. Having reached Q, we turn around and retrace in the opposite direction the path already traversed, until a vertex R is reached which is an edge going to it which has not yet been traversed.  Having reached R, we turn off and follow any edge RS not yet traversed, and then we continue walking over as yet untraversed edges in an arbitrary manner until we reach an endpoint or such a vertex which was already traversed; here we turn back again, etc.  we now prove that the sequence of edges ABC… Q… RS…
 * 1. Leads back to A and 
 * 2. Then contains every edge of the graph.
 * If there is no turning off, then we, of course, come back to A. But if there is a turning off, then, since with every turning off a new edge is added, there must be a last turning off because of the finite number of edges.  Since the vertices are also finite in number, after this turning off we must reach an endpoint or a vertex Z already traversed.  By the directions we must, since no more turnings off are possible, traverse this whole sequence of edges F = AB… Z in the opposite direction, by means of which we reach A.  We see that every edge of the original graph G was traversed in the following way.  We suppose that this is not the case; then the edges not traversed form a proper subgraph G’’ of G, while the edges traversed form the (connected) subgraph G’ of G.  G’ and G’’ have no common edge; but they have at least one common vertex P, since otherwise G = G’ + G’’ would not be connected.  Since P belongs to G’, P is a vertex of the sequence of edges F = AB… Z.  Since P also belongs to G’’, there is an untraversed edge ending in P.  in the above mentioned travering of G in the direction of A we must therefore turn off at P, and this contradicts the fact that the last turning off already took place.” 
 * Tremaux' Solution
 * Here we assume that we are not always able to retrace the same path in both directions. That is in our walk to a vertex P, then every edge to P we know whether it has been traversed. This is the elegant solution given by Tremaux:
 * "If, beginning at an arbitrary starting vertex A1, we reach by an edge PQ a vertex Q which is not an endpoint of the graph and which is reached for the first time, then we continue the walk with an arbitrary other edge QR. But if Q is either a vertex traversed earlier or is an endpoint, and PQ was traversed for the first time, then we turn back, that is, we continue the walk with the same edge PQ (in the opposite direction QP).  If both Q and PQ were traversed earlier, then we continue on some other edge QR not yet traversed or, if there is no such edge, continue on an edge QR, which was traversed only once.  If these rules are followed, then each edge is tarversed at most twice; because of the finiteness of the number of edges the walk must come to an end, that is, we must finally reach a vertex K, such that all edges going to K have already been traversed twice.  Now K must be A1, since we otherwise would have a contradiction."
 * Here is a link to images of proof of Tremaux' theorem by Kőnig: Tremaux Proof part 1, Tremaux Proof part 2, Tremaux Proof part 3, Tremaux Proof part 4
 * Tarry's Solution
 * A simpler solution to the labyrinth problem was introduced by Tarry in the following assumptions unlike Tremaux. If upon arrival at some vertex P, there exist an edge PK going to P, whether the edge has been traversed in either direction is irrelevant, it is always possible to determine whether an edge has reached P for the first time.  Tarry's solution is as follows:
 * "When we come on an edge PQ to a vertex Q, we should continue the walk with any edge QR which was still not traversed in the direction towards R; but we should choose the entrance edge of Q for QR only if all other edges QK going to Q were already traversed in the direction towards K."
 * Here is a link to images of proof of Tarry's theorem by Kőnig: Tarry Proof part 1, Tarry Proof part 2
 * Connection between Tremaux' and Tarry's Solutions
 * Kőnig noticed that there was a connection between Tremaux' and Tarry's solution to the Labyrinth problem and expressed this by the following theorem: "Any Tremaux walk is at the same time a Tarry walk." That is, in a Tremaux walk there exists a vertex P, which a distinct initial vertex of the walk.  Then the walk is initiated by the first edge out of P if and only if all other edges say PK, which end at the vertex P, had traversed in the same direction to K.
 * First let us assume that P is not reached for the first time and not an endpoint of the graph. Let q be the number of edges going to P after P has been passed n-times (n=1, 2, 3, ...) and traversed only once.  Thus, q=2. However, if P has been arrived at an nth time (n>1) on an edge traversed for the first time, then we must turn back and consequently q[n]=q[n-1].  But if the edge has been traversed before, then it is now an edge that has been traversed more than once. Then, q[n]=q[n-1]-2 or q[n]=q[n-1] whether an edge that we continue our walk on has been traversed once or more after P has been reached for an nth time.  So, q[1]=2, which implies q[2]= 0 or 2, q[3]=0 or 2, etc.: then q[n]=0 or 2 (of course if some q[i]=0, then q[i+1]=q[i+2]=...=0).  Now, let us assume that at an nth arrival at the vertex P and edge Q[n]P (n>1), Q[1]P is the chosen entrance edge continuing the walk to P (Q[n]P does not equal Q[1]P because n>1, then Q[1]P would have been traversed three times).  After a certain passing, n-1, Q[1]P and Q[n]P both would have been already traversed.  For Q[1]P this is clear since n>1, however for Q[n]P it follows that we would have to turn around back onto Q[n]P, instead of PQ[1], on the other hand these edges were traversed only once before arriving at P for the nth time.  Otherwise traversing these edges passing P n-times would violate Tremaux rulse.  So, q[n-1] must be less than or equal to 2, we let q[n-1]=2.  Before arriving at P for the nth time, all edges KP, only Q[1]P and Q[n]P were traversed once and all other edges going to P were traversed twice.  If one of the edges were not traversed, then an edge PQ[1] had already been traversed once after the arrival at P n-times would not be according to the rules.  So, according to Tremaux' theorem, no edge can be traversed more than twice in the same direction; so, PQ[1] as an exception, all edges PK have been traversed in the same direction to a vertex K.  After arriving at P for the nth time, the edge PQ[1] is the only edge PK which still hasn't been traversed in the direction to K.
 * Here, adding a remark about Tarry's solution, if Tarry's rulse were modified in that if no edge be traversed twice in the same direction, it is required that no edge be traversed three times. Then the labyrinth problem would not be solved, so as proved above Tremaux rules derive from Tarry's rules through specialization. This broad result is obtained using Tremaux' solution to the labyrinth problem and Tarry's about the requirement of opposite directions, which is satisfied by Tremaux walks.  The "direction in which an edge was traversed does not have to be assumed as known at the return to an endpoint of the edge."