Legendre chi function: Difference between revisions

From formulasearchengine
Jump to navigation Jump to search
en>Bkocsis
→‎Identities: added derivative
No edit summary
 
Line 1: Line 1:
In the [[mathematics|mathematical]] field of [[graph theory]] '''Kirchhoff's theorem''' or '''Kirchhoff's matrix tree theorem''' named after [[Gustav Kirchhoff]] is a theorem about the number of [[spanning tree (mathematics)|spanning tree]]s in a [[Graph (mathematics)|graph]], showing that this number can be computed in [[polynomial time]] as the [[determinant]] of a [[Matrix (mathematics)|matrix]] derived from the graph. It is a generalization of [[Cayley's formula]] which provides the number of spanning trees in a [[complete graph]].
I woke up the other day  and noticed - At the moment I've been solitary for some time and after much  [http://www.banburycrossonline.com vip luke bryan] bullying from pals I today find myself registered for web dating. They assured  [http://minioasis.com luke bryan today show concert]  Is the Luke bryan concert sold out [[http://lukebryantickets.neodga.com for beginners]] me that there are plenty of sweet, [http://answers.Yahoo.com/search/search_result?p=ordinary&submit-go=Search+Y!+Answers ordinary] and fun people to meet up, so here goes the toss!<br>I try and keep as toned as possible staying at the gym several-times weekly. I appreciate  [http://www.museodecarruajes.org luke bryan concert tickets] my sports and attempt to perform or view as several a potential. Being wintertime I will frequently at Hawthorn matches. Note: I have experienced the carnage of wrestling fits at stocktake revenue, Supposing that you would contemplated buying a hobby I don't brain.<br>My friends and family are amazing and spending time with them at pub gigabytes or dinners is always a must. I have never been into night clubs as I locate that you can never have a significant dialog together with the sound. Additionally, I have two quite cunning and definitely cheeky canines who are consistently enthusiastic to meet fresh individuals.<br><br>Also visit my webpage [http://lukebryantickets.pyhgy.com luke bryan concert tour 2014]
 
==Kirchhoff's theorem==
 
Kirchhoff's theorem relies on the notion of the [[Laplacian matrix]] of a graph that is equal to the difference between the graph's [[degree matrix]] (a diagonal matrix with vertex degrees on the diagonals) and its [[adjacency matrix]] (a (0,1)-matrix with 1's at places corresponding to entries where the vertices are adjacent and 0's otherwise).
 
For a given connected graph ''G'' with ''n'' labeled [[vertex (graph theory)|vertices]], let ''&lambda;''<sub>1</sub>,&nbsp;''&lambda;''<sub>2</sub>,&nbsp;...,&nbsp;''&lambda;<sub>n</sub>''<sub>&minus;1</sub> be the non-zero [[eigenvalues]] of its Laplacian matrix. Then the number of spanning trees of ''G'' is
 
:<math>t(G)=\frac{1}{n} \lambda_1\lambda_2\cdots\lambda_{n-1}\,.</math>
 
Equivalently the number of spanning trees is equal ''any'' cofactor of the Laplacian matrix of ''G''.
 
== An example using the matrix-tree theorem ==
 
[[Image:Graph_with_all_its_spanning_trees.svg|thumb|The Matrix-Tree Theorem can be used to compute the number of labeled spanning trees of this graph.]]
 
First, construct the [[Laplacian matrix]] ''Q'' for the example kite graph ''G'' (see image at right):
 
: <math>Q = \left[\begin{array}{rrrr}
2 & -1 & -1 & 0 \\
-1 & 3 & -1 & -1 \\
-1 & -1 & 3 & -1 \\
0 & -1 & -1 & 2
\end{array}\right].</math>
 
Next, construct a matrix ''Q<sup>*</sup>'' by deleting any row and any column from ''Q''. For example, deleting row 1 and column 1 yields
 
: <math>Q^\ast =
\left[\begin{array}{rrr}
3 & -1 & -1 \\
-1 & 3 & -1 \\
-1 & -1 & 2
\end{array}\right].</math>
 
Finally, take the [[determinant]] of ''Q<sup>*</sup>'' to obtain ''t(G)'', which is 8 for the kite graph. (Notice ''t(G)'' is the ''(1,1)''-cofactor of ''Q'' in this example.)
 
== Proof outline ==
 
First notice that the Laplacian has the property that the sum of its entries across any row and any column is 0. Thus we can transform any minor into any other minor by adding rows and columns, switching them, and multiplying a row or a column by &minus;1. Thus the cofactors are the same up to sign, and it can be verified that, in fact, they have the same sign.
 
We proceed to show that the determinant of the minor ''M<sub>11</sub>'' counts the number of spanning trees. Let ''n'' be the number of vertices of the graph, and ''m'' the number of its edges. The incidence matrix <math>E</math> is an ''n''-by-''m'' matrix. Suppose that (''i'', ''j'') is the ''k''th edge of the graph, and that ''i'' < ''j''. Then ''E<sub>ik</sub>'' = ''1'', ''E<sub>jk</sub> = &minus;1'', and all other entries in column ''k'' are 0 (see oriented [[Incidence matrix]] for understanding this modified incidence matrix E).  For the preceding example (with ''n'' = 4 and ''m'' = 5):
 
:<math>
E = \begin{bmatrix}
  1 & 1 & 0 & 0 & 0 \\
  -1 & 0 & 1 & 1 & 0 \\
  0 & -1 & -1 & 0 & 1 \\
  0 & 0 & 0 & -1 & -1 \\
\end{bmatrix}.
</math>
 
Recall that the Laplacian ''L'' can be factored into the product of the [[incidence matrix]] and its transpose, i.e., ''L'' = ''EE''<sup>T</sup>. Furthermore, let ''F'' be the matrix ''E'' with its first row deleted, so that ''FF''<sup>T</sup> = ''M<sub>11</sub>''.
 
Now the [[Cauchy-Binet formula]] allows us to write
 
:<math>\det(M_{11}) = \sum_S \det(F_S)\det(F^T_S) = \sum_S \det(F_S)^2</math>
 
where ''S'' ranges across subsets of [''m''] of size ''n'' &minus; 1, and ''F<sub>S</sub>'' denotes the (''n'' &minus; 1)-by-(''n'' &minus; 1) matrix whose columns are those of ''F'' with index in ''S''. Then every ''S'' specifies ''n'' &minus; 1 edges of the original graph, and it can be shown that those edges induce a spanning tree iff the determinant of ''F<sub>S</sub>'' is +1 or &minus;1, and that they do not induce a spanning tree iff the determinant is 0. This completes the proof.
 
== Particular cases and generalizations ==
 
=== Cayley's formula ===
{{main|Cayley's formula}}
[[Cayley's formula]] follows from Kirchhoff's theorem as a special case, since every vector with 1 in one place, &minus;1 in another place, and 0 elsewhere is an eigenvector of the Laplacian matrix of the complete graph, with the corresponding eigenvalue being ''n''.  These vectors together span a space of dimension ''n''&nbsp;&minus;&nbsp;1, so there are no other non-zero eigenvalues.
 
Alternatively, note that as Cayley's formula counts the number of distinct labeled trees of a complete graph ''K<sub>n</sub>'' we need to compute any cofactor of the Laplacian matrix of ''K<sub>n</sub>''. The Laplacian matrix in this case is
:<math>
\begin{bmatrix}
  n-1 & -1      & \cdots & -1      \\
  -1  & n-1    & \cdots & -1      \\
  \vdots & \vdots& \ddots & \vdots \\
  -1 & -1      & \cdots & n-1      \\
\end{bmatrix}.
</math>
 
Any cofactor of the above matrix is ''n<sup>n</sup>''<sup>&minus;2</sup>, which is Cayley's formula.
 
=== Kirchhoff's theorem for multigraphs ===
 
Kirchhoff's theorem holds for [[multigraph]]s as well; the matrix ''Q'' is modified as follows:
* if vertex ''i'' is adjacent to vertex ''j'' in ''G'', ''q<sub>i,j</sub>'' equals &minus;''m'', where ''m'' is the number of edges between ''i'' and ''j'';
* when counting the degree of a vertex, all [[loop (graph theory)|loops]] are excluded.
 
=== Explicit enumeration of spanning trees ===
Kirchhoff's theorem can be strengthened by altering the definition of the Laplacian matrix. Rather than merely counting edges emanating from each vertex or connecting a pair of vertices, label each edge with an [[indeterminant]] and let the (''i'', ''j'')-th entry of the modified Laplacian matrix be the sum over the indeterminants corresponding to edges between the ''i''-th and ''j''-th vertices when ''i'' does not equal ''j'', and the negative sum over all indeterminants corresponding to edges emanating from the ''i''-th vertex when ''i'' equals ''j''.
 
The determinant above is then a [[homogeneous polynomial]] (the Kirchhoff polynomial) in the indeterminants corresponding to the edges of the graph. After collecting terms and performing all possible cancellations, each [[monomial]] in the resulting expression represents a spanning tree consisting of the edges corresponding to the indeterminants appearing in that monomial. In this way, one can obtain explicit enumeration of all the spanning trees of the graph simply by computing the determinant.
 
===Matroids===
The spanning trees of a graph form the bases of a [[graphic matroid]], so Kirchhoff's theorem provides a formula to count the number of bases in a graphic matroid. The same method may also be used to count the number of bases in [[regular matroid]]s, a generalization of the graphic matroids {{harv|Maurer|1976}}.
 
==See also==
*[[Prüfer sequences]]
*[[Minimum spanning tree]]
*[[List_of_graph_theory_topics#Trees|List of topics related to trees]]
 
==References==
*{{citation
| last1 = Harris | first1 = John M.
| last2 = Hirst | first2 = Jeffry L.
| last3 = Mossinghoff | first3 = Michael J.
| edition = 2nd
| publisher = Springer
| series = Undergraduate Texts in Mathematics
| title = Combinatorics and Graph Theory
| year = 2008}}.
*{{citation
| last = Maurer | first = Stephen B.
| issue = 1
| journal = SIAM Journal on Applied Mathematics
| mr = 0392635
| pages = 143–148
| title = Matrix generalizations of some theorems on trees, cycles and cocycles in graphs
| volume = 30
| year = 1976}}.
*{{citation
| last = Tutte | first = W. T.
| isbn = 978-0-521-79489-3
| page = 138
| publisher = Cambridge University Press
| title = Graph Theory
| year = 2001}}.
 
==External links==
*[http://www.math.fau.edu/locke/Graphmat.htm A proof of Kirchhoff's theorem]
*[http://www.math.ku.edu/~jmartin/mc2004/graph1.pdf A discussion on the theorem and similar results]
 
[[Category:Algebraic graph theory]]
[[Category:Spanning tree]]
[[Category:Theorems in graph theory]]

Latest revision as of 06:41, 21 December 2014

I woke up the other day and noticed - At the moment I've been solitary for some time and after much vip luke bryan bullying from pals I today find myself registered for web dating. They assured luke bryan today show concert Is the Luke bryan concert sold out [for beginners] me that there are plenty of sweet, ordinary and fun people to meet up, so here goes the toss!
I try and keep as toned as possible staying at the gym several-times weekly. I appreciate luke bryan concert tickets my sports and attempt to perform or view as several a potential. Being wintertime I will frequently at Hawthorn matches. Note: I have experienced the carnage of wrestling fits at stocktake revenue, Supposing that you would contemplated buying a hobby I don't brain.
My friends and family are amazing and spending time with them at pub gigabytes or dinners is always a must. I have never been into night clubs as I locate that you can never have a significant dialog together with the sound. Additionally, I have two quite cunning and definitely cheeky canines who are consistently enthusiastic to meet fresh individuals.

Also visit my webpage luke bryan concert tour 2014