how can we construct the automorphism of a finite graph?if you send me an axample iwill be very thankful?
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Consider K_3,3, which is the graph obtained when you draw all lines between 3 points {a,b,c} and three other points {x,y,z}. So there is a line between a and x, but not one between a and b.\n\nAn automorphism, f, will take a to one of the six points. There are two possibilities: either f(a) is in {a,b,c} or f(a) is in {x,y,z}. In the first case, {f(a), f(b), f(c)} is a permutation of {a,b,c} and {f(x),f(y),f(z)} is a permutation of {x,y,z}. In the second case, it goes the other way around. Furthermore, any two permutations in both sets will lead to an automorphism of your graph.\n\nThus, the order of the automorphism group will be 6*6*2=72.\nThere is a subgroup isomorphic to S_3 cross S_3 of index 2. Just checking commutativity tells us that the reflection between {a,b,c} and {x,y,z} is not in the center, and in fact, conjugation by the reflection reverses the order of elements in S_3xS_3. Thus, the automorphism group is the semi-direct product of S_3 cross S_3 with Z_2 where Z_2 reverses the order of the cross product.\n\nI hope this helps!
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I use GrafEq for my graphs, you could give it a try at http://www.peda.com/grafeq/
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Say X is a simple set with two elements, X={x,y}.  Then the free group on X is just all words using x,y,x^, and y^ with the rule that xx^=x^x=yy^=y^y=empty word.  The empty word is the identity.  Basically, it is the most general group generate by two elements.\n\nThis can be stated more formally by saying that Free(X), the free group on X, has a map i:X->Free(X) and the following property:\n\n   For every f:X->G, where G is a group, there exists a\n   unique group homomorphism F: Free(X)->G so that\n   f = F composed with i\n\nIn other words, Free(X) is somehow a universal case of a group containing X.\n\nThis can be generalized in lots of other areas. In Category Theory, Free is an 'adjoint functor' of the forgetful functor from Groups to Sets.  'Adjoint functors' arise in lots of interesting places, and this is one of the first ones you'll learn.
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Start here:\nhttp://www.math.pitt.edu/~sussmanm/2071Spring05/lab07/lab07.html#TheGramSchmidtMethod
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Consider a vertex v in the graph.\n\nSince it has degree at least floor(p/2), it is connected to at least floor(p/2) other vertices.  So the component of G that contains v is made up of at least floor(p/2) + 1 vertices.  \nHow many vertices could not be in this component?  Well, at most (p - floor(p/2) -1).  And this quantity is <= floor(p/2).\nBut we know that any vertex is in a component of at least size floor(p/2) + 1, so there must actually be 0 verticies not in the component that contains v.\nThus, G is connected.
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1. First find the roots (or behavior/properties of the roots) of this equation.\n2. Find the smallest field that is an extension of Q and contains the roots (the splitting field).\n3. Find the group of field automorphisms over that field. That is the Galois group.\nIdentify which of the above steps you are having trouble with and we will try to help.
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Any finite collection?\n\nsum of all subcovers is a finite cover.
576001
By the transitive property, a=c
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Isomorphisms between groups are not the same thing as isomorphisms between, for example, rings.  There is a fully general definition of isomorphism that stems from universal algebra and mathematical logic, but I risk boring you if I develop it here.  So let me tackle the specific case of normed vector spaces, as I suspect it is in this setting that you came across the term "isometric isomorphism".\n\nTwo normed vector spaces are said to be isomorphic if there is a bijective map f between them that is homogeneous and linear, and for which there is a constant C>0 such that (1/C) ||x|| < ||f(x)|| < C ||x|| for all x.  An isometric isomorphism is an isomorphism for which that constant C can be taken to be 1, provided you reinterpret the symbol < as less than or equal--it's not easy to type the appropriate symbol.  In such a case, distances are not only roughly preserved (as in the case in which one has an isomorphism with constant C>1): they are indeed actually exactly preserved.  Thus the name isometry.
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Sure. You just need a set of vertices and a set of edges (directed edges in case of a directed graph). There's no requirement for the set of edges to be non-empty. \n\nSometimes even an empty set of vertices is allowed.
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A subgroup H is generated by a set S if H is the smallest subgroup that contains S. This will mean that every element of H can be written as a finite product of elements of S and their inverses.\n\nIn your case, H will consist of {1, (12), (34), (12)(34)}. No other products are possible with (12) and (34) along with their inverses. \n\nTo show that H is normal in G, you need to show that xHx^-1=H for all x in G. Consider \n\nK={x in G:xHx^-1=H}.\n\nShow this *is* a subgroup of G (easy) and then show that K contains (12), (34), and (13)(24). If you do this, K will have to contain G and hence will equal G. \n\nTo show [G:H]=2, just notice that (12) and (34) are already in H and that the square of (13)(24) is in H. This will show there are two cosets of H in G. The rest follows.
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milkyway: \nI think your answer is incomplete. Unfortunately, I don't have my own proof, but let me criticize yours. \nSorry.:-)\n\nWhat is definitely proven by induction in your post is that you are able to construct an acceptable set of N points. However, in my view, your reasoning does not prove the statement for a given set.\n\n1. From the very beginning, you have N points. How can one realize the induction scheme in this case? Perhaps, you mean that you can temporarily erase N-3 points (keeping in mind their positions) and start with 3 points forming a triangle, or "3 couples" in your terms. Then you restore the erased points one by one at their positions.  I believe this should be explained in your proof. \n\n2. Note that if you start with 3 aligned points, by "restoring" the 4th point, you can generate an allowed set of N+1 points out of the prohibited set of N points. This is missing in your proof, where you deal only with allowed N-sets.  \n\nBut this is not the most serious flaw there.\n\n3. You write: "now, this way there always are the old two couples, plus two new couples (new point-non aligned point of each of such two couples). So going from N to N+1, the number of acceptable couples is always >=3."  \n\nI disagree with this statement. \n\nIndeed, imagine that the three couples, which work at the stage N, form a triangle. Aligning the newly "restored" point with one side of the triangle, you kill one working couple. Still you have the two remaining old couples and get a possibly working new couple (new point -- non aligned point of the triangle). However, it is not at all guaranteed that there is no other point lying on the line going through this new possibly working couple. If such point exists, the number of working couples in this case gets reduced from 3 to 2 when going from N to N+1. At the next step, this way I will reduce the number of working couples from 2 to 1 and so on.\n\nThe transition from N to N+1 is therefore not justified.\n\n4. A general question. For N=infinity the statement is false for points located at nodes of a regular \n(say, square) lattice. Doesn't it contradict to the very idea of proving this statement by induction?\n\n\n------------------------------\n\n\n\n\nHere is my proof.\n\nGiven a set of N points, the number of all possible couples is M=N(N-1)/2.\n\nFor each couple draw a straight line going through it. We get M lines.\n\nLet us assume that the statement we are going to prove is false: i.e., none of those M lines goes through ONLY TWO points.\n\nThen, of course, each of those M lines coincides with at least the two other lines, so the actual number of distinguishable lines K < M.\n\nWe have N points and K different lines.\n\nEach of the NK pairs (point + line) is characterised by a distance S(n,k) between  the point n and the line k. \n\nThis distance S(n,k) is defined as a length of a segment perpendicular to the line k with one of the ends lying on the line k and the other end coinciding with the point n.\n\nLet us list all the NK distances S(n,k). Since NK is a finite integer number, one can find the shortest value (or several equal shortest values) of S(n,k). \n\nTake a closer look at the point P and the line L with the shortest S(n=P,k=L).\n\nBy assumption, there are at least three points (L1,L2, and L3) lying on the line L. Therefore, at least two of these three points (say, L1 and L2) can be found to the one side (say, to the right) of the point A where the perpendicular to the line L starting at the point P hits the line L. The case when one of the three points (say, L1) coincides with the point A is treated in exactly the same way.\n\nNow we have a triangle (P-L1-L2), with two vertices L1 and L2 lying on the line L to the right of A in the order A,L1,L2 from left to right. \n\nSince L1 lies to the right of A (or coincides with A), the angle between P-L1 and L1-L2 is greater than (or equal to) pi/2.\n\nTherefore t
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Not true. It is true for continuous homomorphisms. In that case a homomorphism is of the form f(x)=ax for some constant a. THen a^n=1, so a=+-1.
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Karnaugh map:\n\nCheck out the following link which will help.
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This sounds like an excellent application of random graph theory (each person is a random vertex, and each edge has a weight giving each path a weight as well). I read a few paragraphs on this a while back, but its been too long to fashion an application for you. I think you will be interested in it.\nCheck out Random Graphs, by Bollobas or The Strange Logic of Random Graphs, by Spencer.
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there is a problem in maths that is named "graphes" and the funtion of these routers is to use this solution, u can find it in math books!!!
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I don't believe it can.  This is an example of a unicursal problem and you can find some more details at the puzzle link in the reference below, you'll need to scroll down to puzzle number 51.\n\nFrom my days of O Level Maths I remember that we touched on these and I seem to recall that there was an equation to do with the number of vertices and nodes that a shape had which would determine if it was unicursal or not.\n\nBasically only unicursal shapes (those with a specific ratio of nodes and vertices) can be drawn without removing a pencil from the page and without tracing over an existing line.\n\nI'll leave further research to you.\n\nHave fun