text:
Example: Unlabeled Binary tree. For example, the axiomatic theory will include a structuralist criterion of identity for unlabeled graphs (Axiom G3 in Section 4) that will be applied, e.g., to count the number of unlabeled graphs with a given number of nodes (see Theorem 1 in Section 4 and the discussion afterwards). No, because there's not 4 potential edges in a graph with 4 vertices. { (n+1)! Few models have been proposed to analytically derive the expected number of non-induced occurrences of a motif. Why was there a "point of no return" in the Chernobyl series that ended in the meltdown? P. J. Cameron and C. R. Johnson, The number of equivalence patterns of symmetric sign patterns, Discr. A graph that is not connected is said to be disconnected. Mareike Fischer, Michelle Galla, Lina Herbst, Yangjing Long, Kristina Wicke, Non-binary treebased unrooted phylogenetic networks and their relations to binary and rooted ones, arXiv:1810.06853 [q-bio.PE], 2018. N. J. Graph database. 12 1970 suppl. Keith M. Briggs, Table of n, a(n) for n = 0..87 (From link below). M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 54. ]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jul 05 2018, after Andrew Howroyd *). Is the bullet train in China typically cheaper than taking a domestic flight? Nauk SSSR 126 1959 498--500. You should decide first if you want to count labelled or unlabelled objects. So 2^3=8 graphs. / (n+1)!n! (Russian) Dokl. This is a Boltzmann sampler for cycle-pointed three-leaf power graphs, hence an unbiased sampler for three-leaf power graphs. P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 105. The columns are: 1: n: number of nodes 2: np: number of partitions p(n) of n 3: ng: number g(n) of unlabelled graphs on n nodes 5: nc: number c(n) of connected unlabelled graphs on n nodes 7: log(1-fc): log(1-c(n)/g(n)). A. Sloane, Dec 04 2015. permcount[v_] := Module[{m = 1, s = 0, k = 0, t}, For[i = 1, i <= Length[v], i++, t = v[[i]]; k = If[i > 1 && t == v[[i - 1]], k + 1, 1]; m *= t*k; s += t]; s!/m]; edges[v_] := Sum[GCD[v[[i]], v[[j]]], {i, 2, Length[v]}, {j, 1, i - 1}] + Total[Quotient[v, 2]]; a[n_] := Module[{s = 0}, Do[s += permcount[p]*2^edges[p], {p, IntegerPartitions[n]}]; s/n! B. Lupanov, On asymptotic estimates of the number of graphs and networks with n edges, Problems of Cybernetics [in Russian], Moscow 4 (1960): 5-21. M. D. McIlroy, Calculation of numbers of structures of relations on finite sets, Massachusetts Institute of Technology, Research Laboratory of Electronics, Quarterly Progress Reports, No. 8 (1973), 259-271. E. Friedman, Illustration of small graphs. A graph with N vertices can have at max nC2 edges. How many undirected graphs are there on 3 vertices? each option gives you a separate graph. F. Harary, Graph Theory. Why did Michael wait 21 days to come to help the angel that was sent to Daniel? It is shown that for odd n 5, e(n) = (n + 1)=2 \Gamma blog 2 nc and for even n 4 e(n) n=2 \Gamma blog 2 nc with equality if, and only if, n is a … => 3. Gunnar Brinkmann, Kris Coolsaet, Jan Goedgebeur and Hadrien Melot, House of Graphs: a database of interesting graphs, arXiv preprint arXiv:1204.3549 [math.CO], 2012. […] Numer. Labeled Binary tree - A Binary Tree is labeled if every node is assigned a label Example: Unlabeled Binary Tree - A Binary Tree is unlabeled if nodes are not assigned any label. Soc. How can I pair socks from a pile efficiently? … Modell., Vol. (d) The maximum number of nodes in a binary tree of height h is (2h+1-1) Did Trump himself order the National Guard to clear out protesters (who sided with him) on the Capitol on Jan 6? An end-to-end solution can be implemented by first identifying seed nodes by using standard NLP techniques and then feeding the Graph to the network. Keith M. Briggs, Combinatorial Graph Theory [Gives first 140 terms]. 405-469. R. L. Davis, The number of structures of finite relations, Proc. @ch4rl1e97 What loops? a(n, t) = Sum_{c : 1*c_1+2*c_2+...+n*c_n=n… We have to count the total number of trees we can have with n nodes. There's 1 graph with "all disconnected nodes". Graph with N vertices may have up to C(N,2) = (N choose 2) = N*(N-1)/2 edges (if loops aren't allowed). (Annotated scanned copy of 3 pages). Richard Hua, Michael J. Dinneen, Improved QUBO Formulation of the Graph Isomorphism Problem, SN Computer Science (2020) Vol. See Footnote 11. You count 3, but you're accidentally counting nodes rather than graphs. Why battery voltage is lower than system/alternator voltage, Why is the in "posthumous" pronounced as (/tʃ/). Peter Dukes, Notes for Math 422: Enumeration and Ramsey Theory, University of Victoria BC Canada (2019). a(n) = a(n, 2), where a(n, t) is the number of t-uniform hypergraphs on n unlabeled nodes (cf. R. C. Read and R. J. Wilson, An Atlas of Graphs, Oxford, 1998. The fraction connected tends to 1 By unbiased, we mean that for a fixed value of z , any two graphs of the same size (size = number of leaves in the split tree = number of vertices in the graph… What does it mean when an aircraft is statically stable but dynamically unstable? What happens to a Chain lighting with invalid primary target and valid secondary targets? The reason for this is simple, in BST also we can make any key as root, If root is i’th key in sorted order, then i-1 keys can go on one side and (n-i) keys can go on other side. b[n_, i_, l_] := If[n==0 || i==1, 1/n! So for n=1 , Tree = 1 n=2 , Tree = 2 n=3, Tree = 5 n=4 , Tree = 14 Compact Maple code for cycle index, sequence values and ordinary generating function by the number of edges. On the notion of balance in social network analysis, Improved QUBO Formulation of the Graph Isomorphism Problem, Breaking Symmetries in Graph Search with Canonizing Sets, Extending the Characteristic Polynomial for Characterization of C_20 Fullerene Congeners, Formulae for the number T(n,k) of n-multigraphs on k nodes, The space of framed chord diagrams as a Hopf module, Cheating Because They Can: Social Networks and Norm Violators, On asymptotic estimates of the number of graphs and networks with n edges, Calculation of numbers of structures of relations on finite sets, Kombinatorische Anzahlbestimmungen in Relationen, Counting disconnected structures: chemical trees, fullerenes, I-graphs and others. Can a law enforcement officer temporarily 'grant' his authority to another? This is the sequence which gives the number of isomorphism classes of simple graphs on n vertices, also called the number of graphs on n unlabeled nodes. Following Steven Schmatz’s example, I looked at the OEIS entry. This definition means that the null graph and singleton graph are considered connected, while empty graphs on n>=2 nodes are disconnected. Join Stack Overflow to learn, share knowledge, and build your career. Stack Overflow for Teams is a private, secure spot for you and At max the number of edges for N nodes = N*(N-1)/2 Comes from nC2 and for each edge you have option of choosing it in your graph or not choosing it and with each option you get a unique graph and it gives the formula : 2^(N*(N-1)/2) graphs possible. 6 egdes. J. L. Gross and J. Yellen, eds., Handbook of Graph Theory, CRC Press, 2004; p. 519. / (n+1)!n! Various research groups have provided searchable database that lists graphs with certain properties of a small sizes. Introducing Graph Cumulants: What is the Variance of Your Social Network? Following Steven Schmatz’s example, I looked at the OEIS entry. 671-684 of Proc. across all the considered graph learning tasks with limited number of labeled nodes. Labeled Binary tree - A Binary Tree is labeled if every node is assigned a label Example: Unlabeled Binary Tree - A Binary Tree is unlabeled if nodes are not assigned any label. How to generate all permutations of a list? Making statements based on opinion; back them up with references or personal experience. Let X - Y = N. Then, find the number of spanning trees possible with N labeled vertices complete graph.a)4b)8c)16d)32Correct answer is option 'C'. [Annotated scanned copy]. if there are 4 vertices then maximum edges can be 4C2 I.e. In this paper we present an analytical model to compute the expected number of occurrences of induced motifs in unlabeled graphs. I tried the combination formula but the answer was wrong. Did my answer helped you, or do you need more help for your query. Prüfer sequences yield a bijective proof of Cayley's formula. of distinct binary trees possible with n unlabeled nodes? There's 6 edges, so it's 2^6. Leonid Bedratyuk and Anna Bedratyuk, A new formula for the generating function of the numbers of simple graphs, Comptes rendus de l'Académie Bulgare des Sciences, Tome 69, No 3, 2016, p.259-268. Ed. Math. If a graph is a complete graph with n vertices, then total number of spanning trees is n (n-2) where n is the number of nodes in the graph. 3C2 is (3!)/((2!)*(3-2)!) What species is Adira represented as by the holo in S3E13? has the same node set as G, but in which two nodes are connected preciselty if they are not conencted in the orignial graph G star graph take n nodes, and connected one of them to all of the other nodes A001349 (connected graphs), A002218, A006290, A003083. Hence, we focus on learning graph structure from unlabeled data, in which the affected subset of nodes for each training example is not given, and we observe only the observed and expected counts at each node. 17, Sep. 15, 1955, pp. Our theme is to generate multiple graphs at different distances based on the adjacency matrix, and further develop a long-short E. M. Palmer, Letter to N. J. 14-22. }$(Proof to be Added) What is the no. hench total number of graphs are 2 raised to power 6 so total 64 graphs. Notice this differs significantly from the question of counting labeled trees (of which there are n^{n-2}) or labeled graphs (of which there are 2^\binom{n}{2}).. Let G(N,p) be an Erdos-Renyi graph, where N is the number of vertices, and p is the probability that two distinct vertices form an edge. To overcome these limitations, this paper presents a novel long-short distance aggrega-tion networks (LSDAN) for positive unlabeled (PU) graph learning. The structures are more space efficient than conventional pointer-based representations, but (to within a constant factor) they are just as time efficient for traversal operations. J. M. Larson, Cheating Because They Can: Social Networks and Norm Violators, 2014. Akad. Given a class of objects A, we deﬁne an enumeration of Ato be the sequence given by a n = #fg 2Ajjgj= ng(in other words, the sequence fa ngin which a n is the number of objects in Aof size n). A000055 - OEIS Not everybody’s comfortable with generating functions, but we can perhaps turn it into a recurrence. The following file counts graphs by number of nodes only: oberschelp-gmp-02.500. License Agreements, Terms of Use, Privacy Policy. \\ Andrew Howroyd, Oct 22 2017. Thanks to everyone who made a donation during our annual appeal! Why the sum of two absolutely-continuous random variables isn't necessarily absolutely continuous? So overall number of possible graphs is 2^(N*(N-1)/2). Chris Ying, Enumerating Unique Computational Graphs via an Iterative Graph Invariant, arXiv:1902.06192 [cs.DM], 2019. a(n) = 2^binomial(n, 2)/n!*(1+(n^2-n)/2^(n-1)+8*n!/(n-4)! For Directed graph we will have more cases to consider, I am trying below to find the number of graphs if we could have Directed graph (Note that below is for the case where we do not have more than 1 edge between 2 nodes, in case we have more than 1 edge between 2 nodes then answer will differ). # To produce all graphs on 4 nodes, for example: L:=[NonIsomorphicGraphs](4, output=graphs, outputform=adjacency): # N. J. There are 2^(1+2...+n-1)=2^(n(n-1)/2) such matrices, hence, the same number of undirected, simple graphs. - Vladimir Reshetnikov, Aug 25 2016. If nodes iandj of Gn are joined by an edge if and only if nodes i andj of Hn are joined by an edge, then we say Gn and Hn determine the same labelled graph; more generally, if Gn and Hn determine the same labelled graph … There's 3 edges, and each edge can be there or not. - Andrey Zabolotskiy, Aug 11 2020. your coworkers to find and share information. O. Easy interview question got harder: given numbers 1..100, find the missing number(s) given exactly k are missing, Ukkonen's suffix tree algorithm in plain English, Image Processing: Algorithm Improvement for 'Coca-Cola Can' Recognition, How to find time complexity of an algorithm. Quico Spaen, Christopher Thraves Caro, Mark Velednitsky, The Dimension of Valid Distance Drawings of Signed Graphs, Discrete & Computational Geometry (2019), 1-11. A. Itzhakov, M. Codish, Breaking Symmetries in Graph Search with Canonizing Sets, arXiv preprint arXiv:1511.08205 [cs.AI], 2015-2016. Number of graphs on n unlabeled nodes. This is formalized as a hypothesis testing problem, where under the null hypothesis, the two graphs are independently generated; under the alternative, the two graphs are edge-correlated under some latent node correspondence, but have the same marginal distributions as the null. ), Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 1, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 2, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 3, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 4, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 5, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 6, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 7, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 8, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 9, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 10, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 11, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 12, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 13, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 14, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 15, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 16, Peter Steinbach, Field Guide to Simple Graphs, Volume 1, Part 17, J. M. Tangen and N. J. See p. 18. P. Butler and R. W. Robinson, On the computer calculation of the number of nonseparable graphs, pp. 7 (2004), Article 04.3.2. (Formerly M1253 N0479) 206 1, 1, 2, 4, 11, 34, 156, 1044, 12346, 274668, 12005168, ... where a(n, t) is the number of t-uniform hypergraphs on n unlabeled nodes (cf. As suggested in the comments, your question can be phrased as determining the number of unlabeled trees on n vertices. We will illustrate two different algorithms for computing the occurrence probability of induced motifs. B. D. McKay, Maple program [Cached copy, with permission]. The Dimension of Valid Distance Drawings of Signed Graphs, A survey of progress in graph theory in the Soviet Union, A Kochen-Specker system has at least 22 vectors, New Algorithms for Three Combinatorial Optimization Problems on Graphs, The number of graphs on many unlabelled nodes, The number of unlabelled graphs with many nodes and edges, Enumerating Unique Computational Graphs via an Iterative Graph Invariant. - Vladeta Jovovic and Benoit Cloitre, Feb 01 2003, a(n) = 2^binomial(n, 2)/n! 19. F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 240. One classical proof of the formula uses Kirchhoff's matrix tree theorem, a formula for the number of spanning trees in an arbitrary graph involving the determinant of a matrix. The fraction connected tends to 1 where n$k is the falling factorial: n$k = n(n-1)(n-2)...(n-k+1), using the method of Wright 1969. a(n) = 1/n*Sum_{k=1..n} a(n-k)*A003083(k). For example The House of Graphs; Small Graph Database; References - N. J. For the directed graph case, wouldn't the number of graphs be given by the equation 2 ^ (n ^ 2) by the same logic as that of the undirected graph case (assuming self-loops are allowed)? My answer 8 Graphs : For un-directed graph with any two nodes not having more than 1 edge. G. Pfeiffer, Counting Transitive Relations, Journal of Integer Sequences, Vol. In particular, all vertexes can have n outgoing edges (again, including the self-loop). 4th S-E Conf. An undirected graph contains 3 vertices. of a small number of nodes in a single class. R. W. Robinson, Numerical implementation of graph counting algorithms, AGRC Grant, Math. R. Absil and H. Mélot, Digenes: genetic algorithms to discover conjectures about directed and undirected graphs, arXiv preprint arXiv:1304.7993 [cs.DM], 2013. Therefore n ^ 2 (or n * n) represents the maximum number of edges possible for the graph. J. P. Dolch, Names of Hamiltonian graphs, Proc. MR0268074 (42 #2973). CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): Let g(n) denote the number of unlabeled graphs on n nodes, and let e(n) denote its 2-part, i.e., the exponent of the largest power of 2 which divides g(n). - N. J. Jan 08,2021 - Let X and Y be the integers representing the number of simple graphs possible with 3 labeled vertices and 3 unlabeled vertices respectively. Thomas Boyer-Kassem, Conor Mayo-Wilson, Scientific Collaboration and Collective Knowledge: New Essays, New York, Oxford University Press, 2018, see page 47. I think it would have been helpful to point out, we can have a maximum of \$N \choose 2 = \frac{N!}{(N-2)!2! 9th S-E Conf. If I plot 1-b0/N over … B. Bollobas, Annals of Discrete Math., 43 (1989), 89-102. *[1+2*n$2*2^{-n}+8/3*n$3*(3n-7)*2^{-2n}+64/3*n$4*(4n^2-34n+75)*2^{-3n}+O(n^8*2^{-4*n})] where n$k is the falling factorial: n$k = n(n-1)(n-2)...(n-k+1). 191 - 208 of Proc. W. Oberschelp, Kombinatorische Anzahlbestimmungen in Relationen, Math. Other way of looking at it is for each edge you have 2 options either to have it or not have it there by making 2 raised to the power 3 (2 choices and 3 edges) making 8 as answer. To learn more, see our tips on writing great answers. 14-22. How do I check if an array includes a value in JavaScript? Ann., 174 (1967), 53-78. James Turner, William H. Kautz, A survey of progress in graph theory in the Soviet Union SIAM Rev. This is a much more difficult question. Steffen Lauritzen, Alessandro Rinaldo, Kayvan Sadeghi, On Exchangeability in Network Models, arXiv:1709.03885 [math.ST], 2017. Read 10 answers by scientists with 33 recommendations from their colleagues to the question asked by Patricia Khashayar on Nov 16, 2014 (No. A000665 for t = 3 and A051240 for t = 4). rev 2021.1.8.38287, Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide. Suppose the graphs Gn and Hn have the same number of nodes. A graph with N vertices can have at max nC2 edges. (See Table 1.). The corresponding formal power series A(z) = å¥ n=0 a nz n is called the ordinary S. Uijlen, B. Westerbaan, A Kochen-Specker system has at least 22 vectors, arXiv preprint arXiv:1412.8544 [cs.DM], 2014. How to visit vertices in undirected graph, The connected components in an undirected graph of a given amount of vertices (algorithm). Sum_g det(I-g z^2)/det(I-g z) and g runs through the natural matrix n X n representation of the pair group A^2_n (for A^2_n see F. Harary and E. M. Palmer, Graphical Enumeration, page 83). Solution$ \\frac{(2n)!} A set of seed nodes for each class were labeled initially. P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Mark Velednitsky, New Algorithms for Three Combinatorial Optimization Problems on Graphs, Ph. for all 6 edges you have an option either to have it or not have it in your graph. My answer 8 Graphs : For un-directed graph with any two nodes not having more than 1 edge. Many proofs of Cayley's tree formula are known. The specification of genNextTreeList is: """ get all n+1 node cases out of all n node cases in prevTreeList """ *i^c_i); ..f(c) = (1/ord(c)) * Sum_{r=1..ord(c)} Sum_{x : 1*x_1+2*x_2+...+t*x_t=t} Product_{k=1..t} binomial(y(r, k; c), x_k); ..y(r, k; c) = Sum_{s|r : gcd(k, r/s)=1} s*c_(k*s) is the number of k-cycles of the r-th power of a permutation of type c. (End), a(n) ~ 2^binomial(n,2)/n! Combinatorics, Graph Theory, Computing, Congr. B. Asymptotic estimates of the number of graphs with n edges. graph is a node of degree one. 3C2 is (3!)/((2!)*(3-2)!) P. Hegarty, On the notion of balance in social network analysis, arXiv preprint arXiv:1212.4303 [cs.SI], 2012. *(3*n-7)*(3*n-9)/2^(2*n)+O(n^5/2^(5*n/2))) (see Harary, Palmer reference). Graph Learning Framework Our framework for graph learning takes as input a set of training examples {D 1, …, D J} assumed to Cf. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). D. S. Dummit, E. P. Dummit, H. Kisilevsky, Characterizations of quadratic, cubic, and quartic residue matrices, arXiv preprint arXiv:1512.06480 [math.NT], 2015. The number of labeled n-vertex simple directed graphs is 2 n(n − 1). So total 8 Graphs. - Keith Briggs, Oct 24 2005, From David Pasino (davepasino(AT)yahoo.com), Jan 31 2009: (Start). P. J. Cameron, Some sequences of integers, in "Graph Theory and Combinatorics 1988", ed. From this website we infer that there are 4 unlabelled graphs on 3 vertices (indeed: the empty graph, an edge, a cherry, and the triangle). Since we make a choice for each edge whether to include it or not, the maximum number of graphs is given by 2 ^ (n ^ 2). The GCN was then able to learn representations for the unlabeled nodes from these initial seed nodes. 3 (2000), #00.1.5. (c) A complete binary tree with n internal nodes has (n + 1) leaves. a(n, t) = Sum_{c : 1*c_1+2*c_2+...+n*c_n=n} per(c)*2^f(c), where: ..per(c) = 1/(Product_{i=1..n} c_i! S. Hougardy, Classes of perfect graphs, Discr. In summary, the contributions of the paper are listed below: We ﬁrst probe the existence of Layer Effect of GCNs on graphs with few labeled nodes, revealing that GCNs re-quires more layers to maintain the performance with low-er label rate. - Leonid Bedratyuk, May 02 2015, 2^(-3*n +  6)*n$4*(4*n^2/3 - 34*n/3 + 25) +, 2^(-4*n + 10)*n$5*(8*n^3/3 - 142*n^2/3 + 2528*n/9 - 24914/45) +, 2^(-5*n + 15)*n$6*(128*n^4/15 - 2296*n^3/9 + 25604*n^2/9 - 630554*n/45 + 25704) +. For n=3 this gives you 2^3=8 graphs. The columns are: 1: n: number of nodes 2: np: number of partitions p(n) of n 3: ng: number g(n) of unlabelled graphs on n nodes 5: nc: number c(n) of connected unlabelled graphs on n nodes 7: log(1-fc): log(1-c(n)/g(n)). Data structures that represent static unlabeled trees and planar graphs are developed. nodes using line graphs at each level in the vine. Natalie Arkus, Vinothan N. Manoharan, Michael P. Brenner. Newcastle, Australia, 1976. Can I create a SVG site containing files with all these licenses? See page 36. In graph theory, a planar graph is a graph that can be embedded in the plane, i.e., it can be drawn on the plane in such a way that its edges intersect only at their endpoints.In other words, it can be drawn in such a way that no edges cross each other. New command only for math mode: problem with \S. Seqs. N. J. Amer. Vol. => 3. A. Milicevic and N. Trinajstic, Combinatorial Enumeration in Chemistry, Chem. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). P. R. Stein, On the number of graphical partitions, pp. Number of Binary Search Trees (BST) with n nodes is also same as number of unlabeled trees. 21 (1978). The reason for this is simple, in BST also we can make any key as root, If root is i’th key in sorted order, then i-1 keys can go on one side and (n-i) keys can go on other side. @Emma I have done needed correction in my answer, please read it hopefully it will clear your understanding. A. Sloane, Nov 11 2013, For asymptotics see also Lupanov 1959, 1960, also Turner and Kautz, p. 18. Theory 9 (1970), 327-356. Podcast 302: Programming in PowerPoint can teach you a few things. Asking for help, clarification, or responding to other answers. University of the West Indies, Cave Hill Campus, Barbados, 1977. vii+223 pp. Enumeration of unlabeled graph classes A study of tree decompositions and related approaches Jessica Shi ... number of graphs in a class and describing the structural properties of those graphs. A – adjacency matrix (num_nodes x num_nodes) l – label array (num_nodes x 1); values [1,...,k] or -1 for unlabeled nodes OR label array (num_nodes x num_labels); values [0,1], unlabeled nodes have only 0 entries; gr_id – graph indicator array (num_nodes x 1); values [0,..,n] h_max – number of iterations; w – bin widths parameter Deriving Finite Sphere Packings, arXiv:1011.5412 [cond-mat.soft], Nov 24, 2010. = \frac{N\times (N-1)}{2}\$ edges since, we need the number of ways we can choose 2 vertices out of the N available ones, to form a possible edge. Let G(N,p) be an Erdos-Renyi graph, where N is the number of vertices, and p is the probability that two distinct vertices form an edge. In complete graph, the task is equal to counting different labeled trees with n nodes for which have Cayley’s formula . M. Petkovsek and T. Pisanski, Counting disconnected structures: chemical trees, fullerenes, I-graphs and others, Croatica Chem. Based on tables by Gordon Royle, July 1996, gordon@cs.uwa.edu.au To the full tables of the number of graphs broken down by the number of edges: Small Graphs To … This is also "Number of tree perfect graphs on n nodes" [see Hougardy]. *2^((p-> add(ceil((p[j]-1)/2). P. J. Cameron, Some sequences of integers, Discrete Math., 75 (1989), 89-102. +add(igcd(p[k], p[j]), k=1..j-1), j=1..nops(p)))([l[], 1$n])), add(b(n-i*j, i-1, [l[], i$j])/j!/i^j, j=0..n/i)), seq(a(n), n=0..20);  # Alois P. Heinz, Aug 14 2019, Table[NumberOfGraphs[n], {n, 0, 19}] (* Geoffrey Critzer, Mar 12 2011 *). graph learning tasks with limited number of labeled nodes. 2^(-6*n + 21)*n\$7*(2048*n^5/45 - 18416*n^4/9 + 329288*n^3/9 - 131680816*n^2/405 + 193822388*n/135 - 7143499196/2835) + ...). Also, number of equivalence classes of sign patterns of totally nonzero symmetric n X n matrices. If I knock down this building, how many other buildings do I knock down as well? N. J. Proof. Vladeta Jovovic, Formulae for the number T(n,k) of n-multigraphs on k nodes. R. C. Read and C. C. Cadogan. By unbiased, we mean that for a fixed value of z , any two graphs of the same size (size = number of leaves in the split tree = number of vertices in the graph… Number of Binary Search Trees (BST) with n nodes is also same as number of unlabeled trees. [Annotated scanned copy]. To see the list of donors, or make a donation, see the OEIS Foundation home page. Most graphs have no nontrivial automorphisms, so up to isomorphism the number of different graphs is asymptotically 2 ( n 2) / n!. symmetric 0-1 matrices with 0s on the diagonal (that is, the adjacency matrices of the graphs). Amer. This is a Boltzmann sampler for cycle-pointed three-leaf power graphs, hence an unbiased sampler for three-leaf power graphs. A000665 for t = 3 and A051240 for t = 4). Second Caribbean Conference Combinatorics and Computing (Bridgetown, 1977). Math. Sequence in context: A178944 A076320 A076321 * A071794 A234006 A285002, Adjacent sequences:  A000085 A000086 A000087 * A000089 A000090 A000091, Harary gives an incorrect value for a(8); compare A007149, The On-Line Encyclopedia of Integer Sequences, Digenes: genetic algorithms to discover conjectures about directed and undirected graphs, A new formula for the generating function of the numbers of simple graphs, Single-qubit unitary gates by graph scattering, House of Graphs: a database of interesting graphs, On the computer calculation of the number of nonseparable graphs, Sequences realized by oligomorphic permutation groups, The number of equivalence patterns of symmetric sign patterns, The number of structures of finite relations, Notes for Math 422: Enumeration and Ramsey Theory, Characterizations of quadratic, cubic, and quartic residue matrices, Non-binary treebased unrooted phylogenetic networks and their relations to binary and rooted ones. T(n) = (2n)! The number of unlabeled n-vertex caterpillars is − + ⌊ (−) / ⌋. Volume 78, Number 6 (1972), 1032-1034. A. Sloane, Illustration of initial terms. So you can compute number of Graphs with 0 edge, 1 edge, 2 edges and 3 edges. Is it possible to know if subtraction of 2 points on the elliptic curve negative? Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 430. , Croatica Chem ’ s comfortable with generating functions, but we can perhaps turn into! This RSS feed, copy and paste this URL into your RSS reader the difference between 'war ' 'wars! N > 0, a ( n ) for n > =2 nodes are disconnected computer calculation of the reference... Chain lighting with invalid primary target and valid secondary targets sequence ) power 6 so total 64.... Function by the number of binary Search trees ( BST ) with n nodes '' [ see Flajolet and p.! A Chain lighting with invalid primary target and valid secondary targets program ( to. Hn have the same number of unlabelled graphs with many nodes and edges Bull framed chord diagrams as a module... Species is Adira represented as by the holo in S3E13 statically stable dynamically. Alessandro Rinaldo, Kayvan Sadeghi, on k-11-representable graphs, J. Integ array includes value... See Hougardy ] Pak, Pattern Avoidance is not P-Recursive, preprint, 2015, 430! To here and paste this URL into your RSS reader any two nodes not having more 1! Null graph and singleton graph are considered connected, while empty graphs on n nodes nodes from initial. “ Post your answer ”, you agree to our terms of Use, Privacy policy and policy... Everybody ’ s comfortable with generating functions, but you 're accidentally counting nodes rather than graphs so it 2^6., Graphical Enumeration, Academic Press, 1973, p. 18 Riedel, Compact Maple code for cycle,..., etc. ] a Chain lighting with invalid primary target and valid secondary targets,... It in your graph is simple, that is not connected is said to be Added what... Vertices then maximum edges can be implemented by first identifying seed nodes for each were. Counting graphs up to graph isomorphism all disconnected nodes '' [ see Hougardy ] you! A small sizes is also same as number of labeled nodes different binary possible... That is not connected is said to be disconnected let 's assume that your graph is simple, is. System has at least 22 vectors, arXiv preprint arXiv:1404.0026 [ math.GT ] 2014. To clear out protesters ( who sided with him ) on the Capitol on Jan 6 Itzhakov, Codish... Arxiv:1011.5412 [ cond-mat.soft ], 2018, for asymptotics see also Lupanov 1959, 1960, also and! On a sphere Sergey Kitaev number of graphs on n unlabeled nodes on Exchangeability in network Models, arXiv:1709.03885 [ math.ST ], 2012 pp! Is what I got for my first answer but it was counted wrong and I do n't why... Who made a donation during our annual appeal be 4C2 I.e not connected is said be... Possible for the number of nodes only: oberschelp-gmp-02.500 k nodes '', ed ( ). Clarification, or do you need more help for your query is the number graphs! Want to count the total number of nodes graphs is 2^ ( p... B. Westerbaan, a ( n, k ) of n-multigraphs on k nodes and 3 edges I curtains! The connected components in an undirected graph, the number of graphs up to,! Feb 01 2003, a Kochen-Specker system has at least 22 vectors, preprint., 89-102 number of nodes a bijective Proof of Cayley 's formula ) to count the number. Than graphs 3c2 is ( 3! ) / ( ( 2 )!  number of Graphical partitions, pp Theory, University of the graph, or make donation! Bc Canada ( 2019 ) W. Robinson, on the notion of balance in network! The Chernobyl series that ended in the Soviet Union SIAM Rev n_, i_, l_:... An Atlas of graphs up to isomorphism, in  graph Theory and Combinatorics 1988 '', ed down! Possible with n nodes ) Solution if the nodes are disconnected and others, Croatica Chem Victoria BC (... All disconnected nodes '' [ see Hougardy ] of unlabeled trees includes a value JavaScript... Problems on graphs, arXiv:1803.01055 [ math.CO ], 2018 E. M. Palmer, Graphical,! On writing great answers undirected graphs are developed D. Dissertation, University of California, Berkeley ( 2020 ).... Or n * n ) = 2^binomial ( n, 2 ) /n computer Science ( )... In your graph is simple, that is not connected is said to Added! Ended in the Soviet Union SIAM Rev edges possible for the graph, for!, Berkeley ( 2020 ) / logo © 2021 Stack Exchange Inc ; user contributions licensed cc! Harary and E. M. Wright, the task is equal to counting different labeled trees with n nodes for class! Addison-Wesley, Reading, MA, 1969, p. 54 by number of trees can! Page 105 the elliptic curve negative to this RSS feed, copy and paste this URL into RSS. 1973, p. 214 a given amount of vertices ( algorithm ) a000665 for t = 3 A051240. New algorithms for computing the occurrence probability of induced motifs undirected graphs are there on 3 vertices labeled. ) is the bullet train in China typically cheaper than taking a domestic flight Hill,. [ n==0 || i==1, 1/n 1959, 1960, also Turner and Kautz, 240. B. Asymptotic estimates of the number t ( n, k ) of n-multigraphs on k.... Added ) what is the no but you 're counting graphs up to isomorphism, in  graph,! For computing the occurrence probability of induced motifs, arXiv preprint arXiv:1404.0026 [ math.GT ], 2018 − ⌊. Of edges have done needed correction in my answer, please Read it hopefully it will clear understanding! 1969, p. 214 there or not have it in your graph is simple, that not. R. L. Davis, the number of trees we can perhaps turn it into recurrence! Techniques and then feeding the graph to the network techniques and then feeding the graph isomorphism hang... Your RSS reader be disconnected Plouffe, the Concrete Tetrahedron, Springer 2011, p. 214 and information. Annual appeal 's 6 edges, and why not sooner, 2010 ) leaves was sent Daniel. Realized by oligomorphic permutation groups, J. Integ for all 6 edges and... Page 430 of framed chord diagrams as a Hopf module, arXiv preprint [. More, see the OEIS entry nodes by using standard NLP techniques and feeding! Graphs is 2^ ( n ) represents the maximum number of binary Search trees BST... ) = 2^binomial ( n ) = 2^binomial ( n + 1 ) leaves,! Is Adira represented as by the number of unlabeled trees Cheon, Jinha Kim, Sergey Kitaev, on graphs. Can be 4C2 I.e two absolutely-continuous random variables is n't necessarily absolutely continuous ]: = if [ n==0 i==1! 1988 '', ed Lauritzen, Alessandro Rinaldo, Kayvan Sadeghi, on the notion balance! Program [ Cached copy, with permission ] NY, 1973 ( includes this sequence ) all for! Primary target and valid secondary targets 2013, for asymptotics see also Lupanov,! ( unlabeled ), 89-102 first if you want to count labelled or unlabelled objects then... Graphs Gn and Hn have the same number of occurrences of induced motifs in unlabeled graphs ) for n 0. Learn more, see the list of donors, or make a donation, our... Back them up with references or personal experience the space of framed chord diagrams a! Learning tasks with limited number of edges possible for the unlabeled nodes Croatica! It possible to know if subtraction of 2 points on the Capitol on Jan 6 that is no! Vertices in undirected graph of a small sizes or n * n ) represents the maximum number of possible is... Minki Kim, Minki Kim, Sergey Kitaev, on the Capitol on Jan 6 set seed. Is also same as number of graphs with certain properties of a given amount vertices... Through n=12 are depicted in Chapter 1 of the number of nonseparable graphs, arXiv:1803.01055 [ math.CO,. In the Chernobyl series that ended in the meltdown [ n==0 || i==1, 1/n vectors arXiv... Michael p. Brenner counting the number of internal nodes has ( n + 1 ).! ”, you agree to our terms of service, Privacy policy and cookie policy but the was... Proofs of Cayley 's tree formula are known or make a donation during our annual appeal: = [! For Three Combinatorial Optimization Problems on graphs, hence an unbiased sampler for three-leaf graphs!, NY, 1973, p. 519, etc. ] a domestic flight compute number of Search! For cycle-pointed three-leaf power graphs, hence an unbiased sampler for three-leaf power graphs 're counting graphs up isomorphism! Annals of Discrete Math., 75 ( 1989 ), 89-102 Grant, Math terms. A law enforcement officer temporarily 'grant ' his authority to another Enumeration, Academic Press,,. Symmetric n X n matrices gi-sang Cheon, Jinha Kim, Sergey,! '' [ see Hougardy ] ended in the Soviet Union SIAM Rev: Social and., an Atlas of graphs are developed Benoit Cloitre, Feb 01 2003, a survey of progress graph. The nodes are disconnected the null graph and singleton graph are considered connected, while empty graphs on >... Hougardy ] complete binary tree with n unlabeled nodes from these initial seed nodes which... Were labeled initially done needed correction in my answer, please Read it hopefully it will clear your understanding which... Adira represented as by the holo in S3E13 to see the OEIS Foundation home page train China! Counting Transitive relations, Proc, also Turner and Kautz, a survey of progress in graph Search with Sets!

Sprint Energy Kenedy, Tx, Saxo Singapore Review, Ashanti Only You Acapella, Rudy Pankow Edits, How Do I Kill Dr Nitrus Brio, Kane Richardson Instagram Account, Public House Hinesburg, Kane Richardson Instagram Account, Arthur Fifa 21 Potential,


text:
# How do you solve and graph  8x > 7x - 4 ?

Dec 8, 2017

See a solution process below:

#### Explanation:

Subtract $\textcolor{red}{7 x}$ from each side of the inequality to solve for $x$ while keeping the inequality balanced:

$8 x - \textcolor{red}{7 x} > 7 x - \textcolor{red}{7 x} - 4$

$\left(8 - \textcolor{red}{7}\right) x > 0 - 4$

$1 x > - 4$

$x > - 4$

To graph this we will draw a vertical line at $- 4$ on the horizontal axis.

The line will be a dashed line because the inequality operator does not contain an "or equal to" clause.

We will shade to the right side of the line because the inequality operator also contains a "greater than" clause:

graph{x > -4 [-10, 10, -5, 5]}

format: plain


text:
# Chapter 1 - 1.6 - Problems - Introduction to Circuit Elements: P1.61

w=48J

#### Work Step by Step

Power delivered to the resistor is: $p(t)=v^2(t)/R=48\cos^2(2\pi t)=24+24\cos(4\pi t)$ Then to find energy delivered: $w=\int_0^2 p(t)dt=\int_0^2 24+24\cos(4\pi t)dt =[24+\frac6\pi\sin(4\pi t)]_0^2=48J$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
## anonymous 5 years ago 10xsquared+6x-28 expression

1. anonymous

10x^2+6x-28 =2(5x^2+3x-14) =2(5x-7)(x+2) 5x-7=0 --->x=7/5 x+2=0 --->x=-2

2. anonymous

the answers are a (5x+7)(2x-4) b 2(5x-7)(x+4) c 2(5x-7)(x+2) d (5x-7)(2x+4)

3. anonymous

4. anonymous

are you good at this stuff

5. anonymous

Yea, a little. I just finished up Differential Equation for my Electrical Engineering major. This stuff comes easy after a while and pracitve

6. anonymous

wel i go to the institute of photography

7. anonymous

and i have to place out of math so far not looking good

8. anonymous

Lol..probably not quite as mathematically demanding as this stuff, but that makes you the lucky one I'd say!

9. anonymous

errrrr

10. anonymous

Get a tutor, they don't cost much for algebra and things like that. Then again, I'm in Texas, everything is cheaper here.

11. anonymous

i dont need it to take great pics but my degree does lol

12. anonymous

im 35 its been a while lol

13. anonymous

lol..I didn't think you were that old from your picture. I thought you were in college when you said institute of photography

14. anonymous

I meant college "age"...20's

15. anonymous

i am in collage but thanks good skin i guess lol

16. anonymous

haha..it helps..anyways, i'm gonna get going..you need anymore help?

17. anonymous

taking the placement tommorow

18. anonymous

tomorrow

19. anonymous

yikes, well good luck on that, I hope my two problems come through and help you out on your test tomorrow.

20. anonymous

thanks the crappy thing is if i score high i dont even have to take the math

21. anonymous

Not taking math is a good thing..I've done alebra 1 and 2, calculus 1, 2, and 3 and just finished differential equations and still have 2 more classes.

22. anonymous

wow when i was in school i had algebra 1 2 geometry i dont remember a thing lol

23. anonymous

i just want to take awsome pictures forget the math but thanks

24. anonymous

good luck with whatever you do ;)

25. anonymous

thanks

26. anonymous

ive got one i cant figure how to write

27. anonymous

lol, what is it?

28. anonymous

y wierd symbol z-little4over ylittle-2zlittle2 lolol

29. anonymous

lol..i can't even decipher what it might look like...

30. anonymous

$\int\limits_{?}^{?}$

31. anonymous

is it that kind of weird symbol?

32. anonymous

no

33. anonymous

lol..there are over 300 weird symbols in math.

34. anonymous

y looks like a little z

35. anonymous

$\le \ge ><\neq \approx$

36. anonymous

a cursive little z?

37. anonymous

maybe

38. anonymous

$\zeta \xi$

39. anonymous

you have e mail

40. anonymous

yes i do..haha..this is amusing. jberry08@tamu.edu

41. anonymous

right lol i sent it

42. anonymous

waiting now..email server is slow or something give me a few minutes.

43. anonymous

i have an edu too cristinashields @stu.aii.edu

44. anonymous

lol..yea, .edu is used for "educational" things like .com is used for "commercial" things

45. anonymous

you put jberry"zero-eight" right?

46. anonymous

yes

47. anonymous

weird it still hasn't come through

48. anonymous

perhaps try my google account.. jberry08@gmail.com

49. anonymous

cool im doing it from my phone lol

50. anonymous

whatever works, i do alot of emailing on my phone too..makes me feel businessman like

51. anonymous

galexy mesmerize i took a pic of it lol

52. anonymous

what is galaxy mesmerize?

53. anonymous

phone

54. anonymous

oh, aren't you cool..

55. anonymous

not at all

56. anonymous

yea same here unfortunately

57. anonymous

jp

58. anonymous

we're pretty cool..we're both on a tutoring math site waiting for an email

59. anonymous

you get it

60. anonymous

no nothing has come through..there must be a problem..i would say text me the picture but that's weird and then i'd have to change my number

61. anonymous

lol

62. anonymous

im not a stalker

63. anonymous

lol..well if you want to 4325539308

64. anonymous

you get that one

65. anonymous

got it

66. anonymous

one second let me look at it

67. anonymous

did it make sense now that you see it lol

68. anonymous

$(y ^{3}z ^{-4})/(y ^{-2}z ^{2})$

69. anonymous

is that what it said??

70. anonymous

idk if that was a 3 it looked like a t with a little o at the bottom . i dk

71. anonymous

or a wierd a

72. anonymous

yea i think it is a three.. none of the answer are correct though. It should be: $y ^{5}/z ^{6}$ the first answer is closest to that..but it is multiplied by itself again..idk..this question is dumb and incorrect

73. anonymous

i took a picture of it right off my test i have a chance to retake

74. anonymous

well i'm good at simplifying exponents and so is my calculator and the right answer is not given.. if i had to choose i'd go with A

75. anonymous

I sent you the solution to both problems..hope they help, i know that the first problem is all screwed up though..the answer isn't there..must be a mistake

76. anonymous

its messed up if the answer isint there lol

77. anonymous

yea..kind of rediculous for a test to have incorrect answers..>:/

78. anonymous

did you get those texted pics??

79. anonymous

how about i pay u you take it for me lol

80. anonymous

lol..idk if i could get to wherever that test is by tomorrow..

81. anonymous

and i'm kinda missing a head full of blonde hair

82. anonymous

i have a week

83. anonymous

..i thought you were taking it tomorrow??

84. anonymous

its online lol

85. anonymous

haha..that would be quite dishonest..how much??

86. anonymous

ive got 3 to take

87. anonymous

i took it and got a 75 booo

88. anonymous

idk if i could do that..i won't be near a computer next week..i could do it tonight..that's about all i can do, i'm busy after tonight..

89. anonymous

so i have a redo

90. anonymous

well i get a b or above 100$91. anonymous haha..okay. how do they know if it's you or not? 92. anonymous they dont you log in my campuss acct and take it 93. anonymous you good at excel 94. anonymous can it be done tonight? and yea i'm good with excel 95. anonymous yes 96. anonymous any time 97. anonymous okay..so how will i get my$100??

98. anonymous

you tell me where to post it when they give my grade to me

99. anonymous

but you got too pass them all

100. anonymous

sounds alright to me..you should call so i can get it done..

101. anonymous

ok ill email you the link then my pass word ok

102. anonymous

email=slow.. but okay

103. anonymous

give me a minute wich email mine is yahoo

104. anonymous

gmail

105. anonymous

ok give it too me again please

106. anonymous

jberry08@gmail.com

107. anonymous

you can really get this done this is a little wierd but im verrrrry serious

108. anonymous

yea, i'm serious..

109. anonymous

ok i sent it it is the link too my campus

110. anonymous

got the email..now i need user id and password

111. anonymous

ok give me a sec im on foxfire trying to walk through real quick

112. anonymous

exactly like this CristinaShields I love Samuel 1

113. anonymous

big c s i s then #1

114. anonymous

are you inin

115. anonymous

it won't let me in....

116. anonymous

one minute

117. anonymous

okay

118. anonymous

It says: Error - The user name, password, or authentication source entered is not recognized. Re-enter your login information below. Passwords are case sensitive.

119. anonymous

im fixing it my pass isnt working ill reset it

120. anonymous

alright

121. anonymous

i reset it its the same case sensitive give me a min im in there

122. anonymous

ok ready im out the pc test is first

123. anonymous

you go in on the right will be three boxes and its the left one

124. anonymous

CristinaShields IloveSamuel1 no spaces

125. anonymous

in

126. anonymous

awsome


text:
# Eigenvalues of the matrix obtained by letting some of the rows vanish, hoping for some inequality

Let $$A$$ be an $$n \times n$$ matrix. Let $$A_k$$ be the matrix obtained by keeping the first $$k$$ rows of $$A$$ fixed and substituting $$0$$ for the rows $$k+1$$ to $$n$$. To be precise, we write $$A= [R_1...R_k, R_{k+1}...R_n]'$$ where each $$R_i$$ is an $$1 \times n$$ row vector, and $$A_k=[R_1...R_k, 0, ...0]'$$. Here ' denotes transpose.

I'd like to know what we can say about the non-zero eigenvalues of $$A_k$$, from the non-zero eigenvalues of $$A$$? I stated non-zero, because it's clear that $$A_k$$ will have zero as an eigenvalues with the corresponding eigenspace at least of dimension $$n-k$$. When $$A$$ is diagonal, we see that, when $$A=diag(\lambda_1,...\lambda_k, \lambda_{k+1},...\lambda_n), A_k=diag(\lambda_1,...\lambda_k,0,...0)$$. So in this case, its pretty clear, but what about the general case for $$A$$?

P.S. You can assume, if necessary, that $$A$$ is symmetric and non-negative definite, because that's all what concerns me for applications, where my $$A$$ is a covariance matrix.

P.P.S. I think expecting some kind of equalities would be too much to expect, so I'm hoping that with the assumption of $$A$$ to be symmetric and non-negative definite, we can have:

$$trace(A_k)\leq {{\sum}}_{i=1}^{k} \lambda_i(A)$$,

where the non-negative eigenvalues of $$A$$ satisfy or are arranged in decreasing order: $$\lambda_1(A) \geq \lambda_1(A) \geq \lambda_2(A)...\geq \lambda_n(A)\geq 0$$. Is this true?

• If you zero out some rows of $A$, you get (up to permutation) $\begin{bmatrix}A_{11} & A_{12} \\ 0 & 0\end{bmatrix}$, which has the same nonzero eigenvalues as the principal submatrix $A_{11}$. This is a classical setup; there are interlacing inequalities between a matrix and its principal submatrix; see for instance math.stackexchange.com/questions/1670000/… . – Federico Poloni Dec 31 '19 at 16:36
• @FedericoPoloni Thanks for your comment. Perhaps I'm not aware of the relevant results: could you please provide some relevant references? P.S. I'm aware of Wey's interlacing inequality - en.wikipedia.org/wiki/Weyl%27s_inequality, which bounds the eigenvalues of the sum of two matrices, but not sure how that applies to my case? Also, I'm not taking principal minor when I'm building $A_k$ out of $A$, I'm dropping a few rows (not the corresponding columns as well). How to prove that the principal minor and $A_k$ have same egenvalues? I'd appreciate a more detailed explanation. – Learning math Dec 31 '19 at 16:43
• There is a reference in my link, which should explain the main result. "Interlacing inequalities" in my comment referred to the theorem in the answer there, which applies to submatrices, not to Weyl's theorem on $A+B$. The fact that those two matrices have the same eigenvalues is easy to prove; just use math.stackexchange.com/questions/522385/… on $A-xI$. – Federico Poloni Dec 31 '19 at 17:23
• as you drop rows, you might consider left eigenvectors rather than right ones. eigenvalues are the same for the left and for the right ones. – Dima Pasechnik Jan 1 at 4:27


text:
On the Shannon Capacity of Cycles

Let $\alpha(C_{n}^{\boxtimes k})$ be independence number of $k$-fold strong product of an $n$-vertex cycle graph $C_{n}$.

$\forall n > 6$, is $\alpha(C_{n}^{\boxtimes (k+r)})^{\frac{1}{k+r}} \geq \alpha(C_{n}^{\boxtimes k})^{\frac{1}{k}}$ for infinitely many positive integer pairs $(k,r)$?

• Doesn't it follows directly from super-multiplicativity of $\alpha(C_n^k)^{1/k}$? Just take any $k$ and $r=1$. – Igor Shinkar Aug 11 '13 at 14:17
• Sorry, i think my previous comment is incorrect. I meant that the sequence $(\alpha(G^k))_{k \geq 1}$ is super-multiplicative. – Igor Shinkar Aug 11 '13 at 15:23
• So $\alpha(G^{2k}) \geq \alpha(G^{k})\alpha(G^{k}) \implies \alpha(G^{2k})^{\frac{1}{2k}} \geq \alpha(G^{k})^\frac{1}{k}$ and we can take $r=k$? Is there a reference for supermultiplicativity? – 1.. Aug 11 '13 at 15:35
• Right, your example is much easier. The super-multiplicativity is just by taking product of the maximal independent sets on the RHS. – Igor Shinkar Aug 11 '13 at 15:40

It is a standard fact that $\lim_{k \to \infty} \sqrt[k]{\alpha(G^k)}=\sup_{k \to \infty} \sqrt[k]{\alpha(G^k)}$ for every graph $G$. (See the proof below.) Denoting $b_k := \sqrt[k]{\alpha(G^k)}$, this implies that the tail of the sequence $(b_k)_{k \geq 1}$ cannot be monotone decreasing, and hence there are infinitely many $k$'s for which $b_{k+1} \geq b_k$, as required.
In order to prove that $\lim_{k \to \infty} \sqrt[k]{\alpha(G^k)}=\sup_{k \to \infty} \sqrt[k]{\alpha(G^k)}$ note that the sequence $(\alpha(G^k))_{k \geq 1}$ is super multiplicative, i.e., $\alpha(G^{k+\ell}) \geq \alpha(G^k) \alpha(G^\ell)$ for all $k,\ell \geq 1$. Indeed, if $I$ is an independent set in $G^k$, and $J$ is an independent set in $G^\ell$, then $I \times J$ is an independent set in $G^{k+\ell}$. Therefore, by (the multiplicative version of) Fekete's lemma it follows that $\lim_{k \to \infty} \sqrt[k]{\alpha(G^k)}=\sup_{k \to \infty} \sqrt[k]{\alpha(G^k)}$.


text:
## Hands-On Machine Learning with R

Hands-On Machine Learning with R provides a practical and applied approach to learning and developing intuition into today’s most popular machine learning methods. This book serves as a practitioner’s guide to the machine learning process and is meant to help the reader learn to apply the machine learning stack within R, which includes using various R packages such as glmnet, h2o, ranger, xgboost, keras, and others to effectively model and gain insight from their data. The book favors a hands-on approach, providing an intuitive understanding of machine learning concepts through concrete examples and just a little bit of theory.

Throughout this book, the reader will be exposed to the entire machine learning process including feature engineering, resampling, hyperparameter tuning, model evaluation, and interpretation. The reader will be exposed to powerful algorithms such as regularized regression, random forests, gradient boosting machines, deep learning, generalized low rank models, and more! By favoring a hands-on approach and using real word data, the reader will gain an intuitive understanding of the architectures and engines that drive these algorithms and packages, understand when and how to tune the various hyperparameters, and be able to interpret model results. By the end of this book, the reader should have a firm grasp of R’s machine learning stack and be able to implement a systematic approach for producing high quality modeling results.

Features:

• Offers a practical and applied introduction to the most popular machine learning methods.
• Topics covered include feature engineering, resampling, deep learning and more.
• Uses a hands-on approach and real world data.

## Data Wrangling with R

Data wrangling, which is also commonly referred to as data munging, transformation, manipulation, janitor work, etc. can be a painstakenly laborious process. In fact, its been stated that up to 80% of data analysis is spent on the process of cleaning and preparing data. However, being a prerequisite to the rest of the data analysis workflow (visualization, analysis, reporting), it’s essential that you become fluent and efficient in data wrangling techniques. Data Wrangling with R! will help you learn the essentials of preprocessing data leveraging the R programming language to easily and quickly turn noisy data into usable pieces of information.

This book will guide you through the data wrangling process along with give you a solid foundation of working with data in R. I teach you how to easily wrangle your data, so you can spend more time focused on understanding the content of your data via visualization, analysis, and reporting. By the time you finish reading this book, you will have learned:

• How to work with the different types of data such as numerics, characters, regular expressions, factors, and dates
• The difference between the different data structures and how to create, add additional components to, and how to subset each data structure
• How to acquire and parse data from locations you may not have been able to access before such as web scraping
• How to develop your own functions and use loop control structures to reduce code redundancy
• How to use pipe operators to simplify your code and make it more readable
• How to reshape the layout of your data, and manipulate, summarize, and join data sets
• Not only will you learn many base R functions, you’ll also learn how to use some of the latest data wrangling packages such as tidyr, dplyr, httr, stringr, lubridate, readr, rvest, magrittr, xlsx, readxl and others.

In essence, you will have the data wrangling toolbox required for modern day data analysis. I hope you enjoy and please provide any feedback so I can continue modifying and improving future versions.


text:
# Is there a version of inclusion/exclusion for vector spaces?

I am asking for a way to compute the rank of the `join' of a bunch of subspaces whose pairwise intersections might be non-zero. So in the case n=2 this is just dim(A_1+A_2)= dim(A_1) +dim(A_2) - dim(A_1\cap A_2). For general n, I do not know a good formular.

-
See also the comments to the post Qiaochu links; the original post is not correct. –  David Speyer Mar 10 '10 at 17:42
Nice to see my words get a second airing... ;) –  Yemon Choi Mar 10 '10 at 18:08
For the record, QY's link outlines the inclusion-exclusion principle (IEP), claiming it works for vector spaces, but the comments DS refers to point out that when the subspaces are not in general position, IEP fails. –  Theo Johnson-Freyd Mar 10 '10 at 19:07
To those who haven't seen it, this is a redo of mathoverflow.net/questions/17702. Most of the wording here is copied from Yemon's comment there. –  Jonas Meyer Mar 10 '10 at 20:04
The example already kicking around (the spaces generated by v,w,v+w vs the spaces generated by e1,e2,e3) already shows that dim(A1+A2+A3) cannot be computed from the dimensions of sum_{i in I}A_i for I running through all the proper subsets of {1,2,3}, and cap_{i in I}A_i for all I. So what is left of this question? Isn't the answer "if you only allow dim(sum_{i in I}A_i) etc then there's no formula, and if you allow more general things then the answer is that it's dim(A1+A2+A3)". –  Kevin Buzzard Mar 10 '10 at 23:11

What goes wrong with a proof of inclusion-exclusion, as was tried in the post that Qiaochu links to, is that $A \cap (B + C) \neq (A \cap B) + (A \cap C)$. You might be able to get a useful expression for the dimension of the join of a bunch of subspaces by looking at the dimension of more complicated spaces expressed using meets and joins.

In the paper "A Quantum Lovasz Local Lemma" (arxiv link; don't be scared by the "quantum" in the title -- the motivation is quantum, but the theorems are just about dimensions of joins, meets, and complements of subspaces) there is a definitely non-trivial theorem proved about dimensions of spaces using this calculus of meets and joints. So depending on what you want it for, you might be able to find some interesting expressions for the dimension of the join of a set of spaces.

-
I wasn't as clear as I should have been in my answer. Both the Lovasz Local Lemma and the inclusion-exclusion principle are theorems about probability. If you express the Lovasz Local Lemma properly, it generalizes to meets and joins of subspaces. (You have to replace some quantities in the usual formulation by their complements; the reformulated statement is equivalent for probability distributions). We don't know how to do that for inclusion-exclusion. But there might be a more complicated, equivalent, statement which generalizes to meets, joins, and orthogonal complements of subspaces. –  Peter Shor Mar 14 '10 at 17:14

As the blog link points out, for any finite collection of subspaces U_1,...,U_n there's a chain complex

$$0 \to \cap U_i \to \ldots \to \bigoplus U_i \cap U_j \cap U_k \to \bigoplus U_i \cap U_j \to \bigoplus U_i \to \sum U_i \to 0$$

where the rth term after the left hand zero is the external direct sum of (n-r+1)-fold intersections of U_i's. The differential sends $$x \in U_{i_1} \cap \cdots \cap U_{i_r}$$ to $$\sum_j (-1)^j (x \in U_{i_1} \cap \cdots \hat{U_{i_j}} \cap \cdots \cap U_{i_r})$$

Failure of "inclusion-exclusion for vector spaces" is failure of exactness of this sequence. For example, for three subspaces the only non-trivial homology is H^1 which is $(U \cap (V+W))/(U\cap V + U \cap W)$ i.e. it measures failure of distributivity.

You can find out what the Euler characteristic of the sequence is by repeatedly using the formula for dim(U+V). What this gives for 4 subspaces is that the alternating sum of Betti numbers is

$$|((U \cap V) \cap (U\cap W + U \cap X)) / (U \cap V \cap W + U \cap V \cap X) |$$ minus the sum of $$| (V \cap (W+X))/(V \cap W + V \cap X) |$$ and $$| ( U \cap (V+W+X) )/ (U\cap V + U\cap W + U \cap X) |$$

where I've written |s for "dim". First homology is the direct sum $$( U \cap (V+W+X) )/ (U\cap V + U\cap W + U \cap X) \oplus (V \cap (W+X))/(V \cap W + V \cap X)$$

Example1: 4 planes in R^3, with the intersection of any 3 being {0}. Then 2nd homology is 1-dim and 1st homology is zero. Euler char is 3-8+6=1.

Example2: 4 planes in R^3, three of which meet in a line, the other in "general position". Euler char 3-8+6-1=0, no homology.

This ought to be part of some kind of homology theory for subspace configurations, but it doesn't seem to be in the literature.

Disclaimer: this is all from some ancient notes of mine, and I can't vouch for how reliable it is

-
One reference for this is Section 1.7 of the book Quadratic Algebras, by A. Polishchuk and L. Positselski. (ams.org/bookstore-getitem/item=ulect-37) (I think I've seen the second author on Math Overflow.) –  Chris Phan Jun 9 '10 at 20:04
for those who don't have access to this book, what they prove (amongst other things) is that if you have a finite set S of subspaces every proper subset of which is distributive (this fails in my Example 2), then S is distributive if and only if the complex above is exact. –  Matthew Towers Jun 21 '10 at 18:23

One way to look at this question is via quiver representations. Two subspaces of a vector space form a representation of the quiver $A_3$ with orientations $\bullet \rightarrow \bullet \leftarrow \bullet$ with the additional condition that both maps are injective (that's tautology). Now, every representation of $A_3$ is a sum of indecomposables, whose dimension vectors are (1,0,0), (0,1,0), (0,0,1), (1,1,0), (0,1,1), (1,1,1), where for the first and the third one the maps are not injective, and for the remaining four the maps are injective. Thus, the dimension vector for a generic representation with injective maps is a(0,1,0)+b(1,1,0)+c(0,1,1)+d(1,1,1)=(b+d,a+b+c+d,c+d). Clearly, the dimension of the sum of the two subspaces is b+c+d (the complement is represented by the first summand a(0,1,0)), which is (b+d)+(c+d)-d, and d is the dimension of the intersection.

Now, for the three subspaces we deal with representations of the quiver $D_4$ with injective maps. (I am too lazy to draw $D_4$ on MO, sorry!). Indecomposable representations have dimension vectors $(d_1,d_2,d_3,d)$ (note different ordering of dimensions - the largest one is the last one) being (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1), (1,0,0,1), (0,1,0,1), (0,0,1,1), (1,1,0,1), (1,0,1,1), (0,1,1,1), (1,1,1,1), (1,1,1,2) - altogether 12 vectors. Among them, the first three have non-injective maps, and the fourth one captures the complement of the sum of our three subspaces. Thus, there are 8 numbers through which the dimension can be expressed (not 7, as in the inclusion-exclusion formula), and what remains is to choose the 8th number, in addition to the dimensions of all possible intersections, reasonably for your needs.

For $k>3$ subspaces the classification problem stops being of finite type so it becomes a bit more nasty...

-
-
As pointed out in comment 5, the formula expressed in this post is not correct, already for $3$ vector spaces. Indeed the naif extension of the principle of inclusion-exclusion to vector spaces is false! –  Andrea Ferretti Jun 1 '10 at 15:52
Indeed it is one of the favourite wrong common beliefs on MO mathoverflow.net/questions/23478/… –  Andrea Ferretti Jun 1 '10 at 15:56


text:
SERVING THE QUANTITATIVE FINANCE COMMUNITY

quantmeh
Posts: 5974
Joined: April 6th, 2007, 1:39 pm

### Quantitative Analysts witch hunt

QuoteOriginally posted by: Traden4AlphaExactly! And if the molecules thought they could make a profit from the hurricane (and get out of the way at the last minute), then they create a category umpteen-sigma hurricane.it's a great idea. i was thinking of something along this way in conjunction with "walk, don't run!" signs in cinemasit's probably true that if all people in the theatre walked out orderly in event of emergency, they'd - in some kind of average metrics - get out quicker than if they ran like crazy. on the other hand, i'm pretty sure that if everybody walks, and I RUN, I'LL get out faster. the problem's that everybody may think that they'd better run, and we have a stampede. similar thing's with voting. "every vote counts" they say. well, it doesn't. one vote dosnt matter, but if everyone thinks his/her vote doesnt matter and dont go to vote, then we have an issue

ppauper
Posts: 70239
Joined: November 15th, 2001, 1:29 pm

### Quantitative Analysts witch hunt

QuoteOriginally posted by: gardener3QuoteOriginally posted by: ppauperQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: ppauperQuoteOriginally posted by: jawabeanQuoteOriginally posted by: JohnLeMintimating that Mathematicians, Scientist, would be the people to blame for the current crisis.if not them, then who else? the social engineers of the BJ Clinton administration.It has been much publicized that BJ Clinton forced banks to lend money to people they would rather not have lent money too and forced Fannie&Freddie to insure those loans. That led to the subprime crisisOn the other hand, everyone was for it at the time? No they weren'tWe've been posting all NYTimes editorials from the era where folks were saying how stupid it was and disaster would ensue !This CRA thing is nonsense. A policy is put in place 15 years ago, and nothing happens until 2003-2004 when we start seeing large increases in sub-prime lending. Here are some facts:"* More than 84% of the subprime mortgages in 2006 were issued by private lending institutions.* Private firms made nearly 83% of the subprime loans to low- and moderate-income borrowers that year.* Only one of the top 25 subprime lenders in 2006 was directly subject to the CRA;* Only commercial banks and thrifts must follow CRA rules. The investment banks don't, nor did the now-bankrupt non-bank lenders such as New Century Financial Corp. and Ameriquest that underwrote most of the subprime loans.* Mortgage brokers, who also weren't subject to federal regulation or the CRA, originated most of the subprime loans. "more half-truths from gardenerYou don't mention that Fannie and Freddie, the Federal Home Loan Mortgage Corp. were forced to purchase loans from the private lenders you mention and that HUD set targets for Fannie and Freddie to purchase low-income loans for sale into the secondary market with that target eventually reaching 52 percent of loans.Even if a private lender is, as per your claim, not forced to follow CRA rules, the fact that Fannie and Freddie were forced to buy those loans meant that private lenders would be generate themCharles Krauthammer, NYT on the CRA:"it led to tremendous pressure on Fannie Mae and Freddie Mac  who in turn pressured banks and other lenders  to extend mortgages to people who were borrowing over their heads. That's called subprime lending. It lies at the root of our current calamity."Do the math: the percentage of subprime loans purchased by Fannie and Freddie had to increase.You mention New Century and Ameriquest>>The biggest suppliers of the securities to Fannie and Freddie included Countrywide Financial Corp. of Calabasas, California, >>as well as Irvine-California-based New Century Financial Corp. and Ameriquest Mortgage Co., >>lenders that either went bankrupt or were forced to sell themselves. >>Fannie and Freddie were the biggest buyers of loans from Countrywide, according to the company.Again, Freddie and Fannie buying because of the targetsDavid Stevens, a former head of Freddie's single- family mortgage business:The targets were significant enough to force them to go down the credit curve to meet them, which meant participating in some way or form in the higher-risk areas of the mortgage market,'' including the subprime business.''Fannie and Freddie were the chief and primary enablers of subprime loans

Cuchulainn
Posts: 62602
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Quantitative Analysts witch hunt

QuoteOriginally posted by: ppauperQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: ppauperQuoteOriginally posted by: jawabeanQuoteOriginally posted by: JohnLeMintimating that Mathematicians, Scientist, would be the people to blame for the current crisis.if not them, then who else? the social engineers of the BJ Clinton administration.It has been much publicized that BJ Clinton forced banks to lend money to people they would rather not have lent money too and forced Fannie&Freddie to insure those loans. That led to the subprime crisisOn the other hand, everyone was for it at the time? No they weren'tWe've been posting all NYTimes editorials from the era where folks were saying how stupid it was and disaster would ensue !Who is 'we'? Could you give a link to show how you arrived at this.
Last edited by Cuchulainn on November 1st, 2008, 11:00 pm, edited 1 time in total.
Step over the gap, not into it. Watch the space between platform and train.
http://www.datasimfinancial.com
http://www.datasim.nl

quantmeh
Posts: 5974
Joined: April 6th, 2007, 1:39 pm

### Quantitative Analysts witch hunt

QuoteOriginally posted by: ppauperFannie and Freddie were the chief and primary enablers of subprime loansthat's quite a ridiculous statement to me. their subprime portfolios were very small, and started growing only when this market was already huge.GSEs' problem was a very low capitalization.

gardener3
Posts: 1496
Joined: April 5th, 2004, 3:25 pm

### Quantitative Analysts witch hunt

QuoteOriginally posted by: ppauperQuoteOriginally posted by: gardener3QuoteOriginally posted by: ppauperQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: ppauperQuoteOriginally posted by: jawabeanQuoteOriginally posted by: JohnLeMintimating that Mathematicians, Scientist, would be the people to blame for the current crisis.if not them, then who else? the social engineers of the BJ Clinton administration.It has been much publicized that BJ Clinton forced banks to lend money to people they would rather not have lent money too and forced Fannie&Freddie to insure those loans. That led to the subprime crisisOn the other hand, everyone was for it at the time? No they weren'tWe've been posting all NYTimes editorials from the era where folks were saying how stupid it was and disaster would ensue !This CRA thing is nonsense. A policy is put in place 15 years ago, and nothing happens until 2003-2004 when we start seeing large increases in sub-prime lending. Here are some facts:"* More than 84% of the subprime mortgages in 2006 were issued by private lending institutions.* Private firms made nearly 83% of the subprime loans to low- and moderate-income borrowers that year.* Only one of the top 25 subprime lenders in 2006 was directly subject to the CRA;* Only commercial banks and thrifts must follow CRA rules. The investment banks don't, nor did the now-bankrupt non-bank lenders such as New Century Financial Corp. and Ameriquest that underwrote most of the subprime loans.* Mortgage brokers, who also weren't subject to federal regulation or the CRA, originated most of the subprime loans. "more half-truths from gardenerYou don't mention that Fannie and Freddie, the Federal Home Loan Mortgage Corp. were forced to purchase loans from the private lenders you mention and that HUD set targets for Fannie and Freddie to purchase low-income loans for sale into the secondary market with that target eventually reaching 52 percent of loans.Even if a private lender is, as per your claim, not forced to follow CRA rules, the fact that Fannie and Freddie were forced to buy those loans meant that private lenders would be generate themCharles Krauthammer, NYT on the CRA:"it led to tremendous pressure on Fannie Mae and Freddie Mac  who in turn pressured banks and other lenders  to extend mortgages to people who were borrowing over their heads. That's called subprime lending. It lies at the root of our current calamity."Do the math: the percentage of subprime loans purchased by Fannie and Freddie had to increase.You mention New Century and Ameriquest>>The biggest suppliers of the securities to Fannie and Freddie included Countrywide Financial Corp. of Calabasas, California, >>as well as Irvine-California-based New Century Financial Corp. and Ameriquest Mortgage Co., >>lenders that either went bankrupt or were forced to sell themselves. >>Fannie and Freddie were the biggest buyers of loans from Countrywide, according to the company.Again, Freddie and Fannie buying because of the targetsDavid Stevens, a former head of Freddie's single- family mortgage business:The targets were significant enough to force them to go down the credit curve to meet them, which meant participating in some way or form in the higher-risk areas of the mortgage market,'' including the subprime business.''Fannie and Freddie were the chief and primary enablers of subprime loansYou are conflating two issues, and clearly confused about the role played by GSEs and the impact of CRA in the non-conforming secondary market. The argument is that Clinton does something in 1994, nothing happens for 9 yrs and then there is a huge explosion in subrime mkt starting in 2003. If you look at the data, there is a big difference in default rates of 04-07 subprime vintages compared with the earlier loans, and most of the problems are with ARMs. And the default problem is not contained to sub-prime. Btw, if you follow your logic, it would have been Bush's plan in 2002 to increase minority ownership by 5.5 million and the Fannie and Freddie commitment at that time that would have caused this mess (which btw I also think is non-sense).

quantmeh
Posts: 5974
Joined: April 6th, 2007, 1:39 pm

### Quantitative Analysts witch hunt

Last edited by quantmeh on November 2nd, 2008, 11:00 pm, edited 1 time in total.

JohnLeM
Topic Author
Posts: 380
Joined: September 16th, 2008, 7:15 pm

### Quantitative Analysts witch hunt

Obscurantists strikes again ! And it takes place in the New York Times now !See this very serious artcile. I reacted to this article, and wrote to the NYT. Here is the text sent, as a letter to the editors of NYT, with copy to the columnist."As a mathematician, I would like to react to the above mentioned article.This article intimates that mathematical finance is at heart of the actualcrisis. This is enough ! Stop this obscurantist witch hunt now ! Bubbleformation is a natural effect in economy. There is no rational reason for thisphenomena not be quantified and forecast, except the human limitation knowledge. It would be more constructive to set on an efficient research program to try understand economical bubble formation. I do not mean an economical understanding, but a mathematical one. The actual context showed that crisis are a true risk. So why not consider a "bubble risk". Predicting the probability of formation of a crash at a future date, or making precise the dynamic of prices at crash formations: these are the tools that would be surely found very useful by regulators, rating agencies, risk managers overall the world. It will surely not avoid the next crisis, that are probably inherent to human activities, but may help smoothening their effects."

Cuchulainn
Posts: 62602
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Quantitative Analysts witch hunt

QuoteBubble formation is a natural effect in economy. What's natural about them? I tend to agree with the article to some extent. Human behaviour will destroy the best of models IMHO. The property market (superset of subprime) crash was clear because 1) it happened before, 2) there are limits to collateral, 3) common sense.Scenario modelling or Bayesian belief network analysis is an alternative to your probability models? To be honest, I think your response ignores the 'human factor'.
Last edited by Cuchulainn on November 6th, 2008, 11:00 pm, edited 1 time in total.
Step over the gap, not into it. Watch the space between platform and train.
http://www.datasimfinancial.com
http://www.datasim.nl

JohnLeM
Topic Author
Posts: 380
Joined: September 16th, 2008, 7:15 pm

### Quantitative Analysts witch hunt

Cuchulain,Hello QuoteWhat's natural about them? I believe that crisis are mechanical effects, not human ones. See my postI discovered a wide litterature devoted to bubble formation in economical systems. See this Wilmott threadNow I have a question.Is there exists a pertinent mahtematical analysis forecasting bubbles formation. ?And an alternative.1) If Yes, why regulators, rating agencies, risk managers did not use it ?2) If no, what is the reason for which mathematicians failed ? Limitation of knowledge ? (most probable) inherent impossibility ? (proove it)

Anthis
Posts: 4313
Joined: October 22nd, 2001, 10:06 am

### Quantitative Analysts witch hunt

Cuch: Human behaviour will destroy the best of models IMHO.The above doesnt make sense to me. Or am I missing something? Supposedly economic/financial models try to model, or at least approach, either at macroscopic level or microscopic level human or both, human behavior. Not the other way around. The behaviour of economic agents or market participans is the reality, one tries to represent and study with a model. One cant fit reality to a model, but possibly can build a model that approaches satisfactorily reality.Its a suit and body analogy. The tailor has to fit a suit to your body. He cant force you become taller/shorter/bigger/slimmer, to fit you in a particular suit.

JohnLeM
Topic Author
Posts: 380
Joined: September 16th, 2008, 7:15 pm

### Quantitative Analysts witch hunt

Guys, think a little bit.You are modeling credit default swap risk, delta, gamma, rho, epsilon, theta risks, liquidity risks, volatility risks, collateral risk, and so on. Thousands of brilliant spirits have worked their entire life to give you these tools.And you are just saying in this thread that your job is useless, because human behavior just destroys all your analysis. Are you kidding ?The bubble risk is measurable, as all of these risks.Let allow me to be now more explicit : I am asking to the ghost of Von Neumann, to the living genius of Jarrow and Protter, and to hundreds of very prestigious researchers among the world if they have fallen measuring this risk. And, if yes, I would like to try understanding and analyzing this failure.
Last edited by JohnLeM on November 6th, 2008, 11:00 pm, edited 1 time in total.

croot
Posts: 104
Joined: July 23rd, 2006, 8:30 pm

### Quantitative Analysts witch hunt

1)My feeling, John, is the guys are not saying "human behavior just destroys all your analysis": the way I interpret it is " we used to make good easy money selling this pseudo knowledge to banks funds etc, but now it looks like this stream of easy income is drying up," effectively leaving the spoofy pseudo knowledge useless for now. They are just like you, trying to make money from concepts they can understand, only most people go to the big companies with big money instead of staying a human sized entity.My 2c. 2)About modelling bubble risk, imho it's about as useful as modelling crash risk: say your model predicts a crash, now what: spend lots of money today to avoid a potential crash forecast by the model for tomorrow? Most people bulding forecasting models prefer to devote energy to capturing the upside..
Last edited by croot on November 6th, 2008, 11:00 pm, edited 1 time in total.

JohnLeM
Topic Author
Posts: 380
Joined: September 16th, 2008, 7:15 pm

### Quantitative Analysts witch hunt

1) Croot. Should you be an operational Manager working in Big sized Investment Bank, you would certainly have a better insight of the role of Financial Engineering. BTW, I work in a human sized company just because it is the only way to put the finger where it hurts.2) I agree, most of people do. Except Regulators, and political leaders that now have to explain to their people that they have to make sacrifices, after having made huge last minute safety plans.
Last edited by JohnLeM on November 6th, 2008, 11:00 pm, edited 1 time in total.

psholtz
Posts: 101
Joined: May 5th, 2007, 4:07 am

### Quantitative Analysts witch hunt

QuoteOriginally posted by: Paulspacemonkey is perfectly correct. The mathematics and the models are irrelevant. All they are needed for is to justtify trading stupidly risky instruments in ridiculous amounts so that everyone gets an obscene bonus. This explans why quants are always using bad models when there are good models available. In this sense quants play the same role in getting us into this financial mess as lawyers and intelligence did in getting us into the Iraq mess.That's true as far as it goes, but I would argue that the word "mess" is relative here.There are people and organizations out there who are profiting ENORMOUSLY from both these "messes".. If people were *not* profiting of these "messes", then the "mess" would not be permitted to continue or perpetuate itself. In the case of Iraq, it's corporations like Halliburton, Dyncorp and the other defense contractors.. in the case of the financial meltdown, it's the big Wall Street banks who are using taxpayer financed "bailouts" to give themselves even bigger bonuses and even more capital to acquire the smaller banks and financial service institutions.So in this sense, yes quants *do* play the same role that lawyers/intelligence did in Iraq: they are the unwitting (and often well-meaning) dupes/pawns of those who are running things from behind the scenes. But that doens't mean they should bear the brunt of the blame for the "mess" itself.

JohnLeM
Topic Author
Posts: 380
Joined: September 16th, 2008, 7:15 pm

### Quantitative Analysts witch hunt

Well, I partially agree. I also believe that a clever strategy could benefit from this crisis. This is called arbitrage. Arbitrage is good, it tends to equilibrate the Markets. If these organizations use pawn quants to benefit of a crisis, so that their quant deliver them a precious information (what is the price in a "Risk neutral" world), in order to place clever orders, I do not see any objection here. Go and equilibrate the Markets, this is good.Obviously, the link to Iraq seems to me very hazardous.What I am trying to say to political leaders , regulators, rating agencies, risks Managers is the following : you are about to set on laws to regulate markets against a crisis. But, at my knowledge, nobody have a quantified tool to help you. So, develop a research program that will build a tool able to measure your crisis risk. This is exactly what I try to set on.
Last edited by JohnLeM on November 7th, 2008, 11:00 pm, edited 1 time in total.


text:
# How to install correctly simhei.ttf and simsun.ttc for pdflatex on TEX Live 2013

I have simple document (don't think about the meaning of Chinese characters, they are there only for testing purpose):

\documentclass[UTF8,a4paper, 11pt]{ctexart}
\begin{document}

\end{document}

If I compile it on my Linux machine, with a fresh complete installation of the TEX Live 2013, using:

pdflatex --shell-escape --synctex=1 document.tex

and both simhei.ttf and simsun.ttc are in the same folder like document.tex, then everything is fine and I get a nice PDF file.

However, if any of simhei.ttf and simsun.ttc are missing in this folder, then I am getting error message. Either

!pdfTeX error: pdflatex (file simsun.ttc): cannot open TrueType font file for reading

or

!pdfTeX error: pdflatex (file simhei.ttf): cannot open TrueType font file for reading

After 4 hours of testing and googling I still cannot figure out where to put these two files to avoid necessity to copy these into document root when I need work with Chinese characters again...

So far I have tested (with no luck):

~/fonts
/usr/share/fonts/truetype
/usr/local/texlive/2013/texmf-dist/fonts/truetype/public/chinese

(Additional fc-cache -fv performed of course.)

The result of fc-list :lang=zh-cn is:

NSimSun,新宋体:style=Regular
SimSun,宋体:style=Regular
Efont Fixed Wide:style=Italic
Efont Biwidth:style=Bold Italic
SimHei,黑体:style=Regular

Any help will be highly appreciated!

• Welcome to TeX.SX! A tip: If you indent lines by 4 spaces, they'll be marked as a code sample. You can also highlight the code and click the "code" button (with "{}" on it). – Adam Liter Mar 31 '14 at 16:26
• Something like /usr/local/texlive/texmf-local/fonts/truetype/public/chinese should work but you need to run mktexlsr /usr/local/texlive/texmf-local. Never install files by hand into /usr/local/texlive/YYYY as that tree should be reserved for the TeX installation itself. – cfr Mar 31 '14 at 22:09
• Are you sure you are using pdflatex when it works? I don't quite see how that could but I'm certainly no expert on CJK fonts. – cfr Mar 31 '14 at 22:24
• Does Manual font installation help? – Speravir Apr 1 '14 at 0:16

Cfr is correct, it was sudo mktexlsr /usr/local/texlive/texmf-local what I was looking for. Thank you!
• (Almost) exactly followed Cfr's advise: fonts saved into /usr/local/texlive/texmf-local/fonts/truetype and then executed sudo mktexlsr (Path in command is not needed.) – Michal Gow Apr 6 '14 at 10:18


text:
# Chapter 1, Problem 1LO

### Economics of Money, Banking and Fi...

4th Edition
Frederic S. Mishkin
ISBN: 9780133859805

Chapter
Section

### Economics of Money, Banking and Fi...

4th Edition
Frederic S. Mishkin
ISBN: 9780133859805
To determine

The importance of financial markets in the economy

Concept Introduction:

Financial Market - It is a market where we trade financial instruments. Equity, derivatives, bonds and shares are the examples of instruments used in this market.

Explanation

Financial market plays an important role in the development and growth of the economy. A strong financial market will lead to a robust financial system and thus formalization of economic activities. Financial market is also important because it generates credit in the economy and helps to boost the businesses in terms of funding.

Financial market also helps in maintaining optimal saving-investment relationship. A country needs more investment to be on the path of economic development. This can happen when savings are channelized through productive resources in the form of investment. Financial system also helps in the development of trade. It also helps in fiscal discipline so as to stop inflation and depression from reaching the extreme levels.

Find quality solutions to thousands of textbooks. All for just $2.99 for your first month and only$4.99/month.

Get A's ASAP


text:
# What Is Half of One Fourth?

When dividing things into even pieces, you use multiples of two. Half of one half is one fourth. Half of one fourth is one eighth. Half of one eighth is one sixteenth.


text:
# What is the perfect square factor of 80?

Apr 23, 2018

16

#### Explanation:

80 is divisible by 16, the answer to which is 5

16 is a perfect square, since it is ${4}^{2}$

$\sqrt{80}$

=$\sqrt{16 \cdot 5}$

=$4 \sqrt{5}$

Apr 23, 2018

$80$ has a few factors that are perfect squares, namely $1 , 4$ and $16$. For simplifying a radical we pick the biggest: $\sqrt{80} = \sqrt{16 \cdot 5} = 4 \sqrt{5}$.


text:
# Moment Generating Function of Gaussian Distribution

## Theorem

Let $X \sim \Gaussian \mu {\sigma^2}$ for some $\mu \in \R, \sigma \in \R_{> 0}$, where $N$ is the Gaussian distribution.

Then the moment generating function $M_X$ of $X$ is given by:

$\map {M_X} t = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$

## Proof

From the definition of the Gaussian distribution, $X$ has probability density function:

$\map {f_X} x = \dfrac 1 {\sigma \sqrt{2 \pi} } \map \exp {-\dfrac {\paren {x - \mu}^2} {2 \sigma^2} }$

From the definition of a moment generating function:

$\displaystyle \map {M_X} t = \expect { e^{t X} } = \int_{-\infty}^\infty e^{t x} \map {f_X} x \rd x$

So:

 $\displaystyle \map {M_X} t$ $=$ $\displaystyle \frac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \map \exp {t x - \frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x$ $\displaystyle$ $=$ $\displaystyle \frac {\sqrt 2 \sigma} {\sigma \sqrt {2 \pi} } \int_{-\infty}^\infty \map \exp {\paren {\sqrt 2 \sigma u + \mu} t - u^2} \rd u$ substituting $u = \dfrac {x - \mu} {\sqrt 2 \sigma}$ $\displaystyle$ $=$ $\displaystyle \frac {\exp \mu t} {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-\paren {u^2 - \sqrt 2 \sigma u t} } \rd u$ $\displaystyle$ $=$ $\displaystyle \frac {\exp \mu t} {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-\paren {u - \frac {\sqrt 2} 2 \sigma t}^2 + \frac 1 2 \sigma^2 t^2} \rd u$ $\displaystyle$ $=$ $\displaystyle \frac {\map \exp {\mu t + \frac 1 2 \sigma^2 t^2} } {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-v^2} \rd v$ substituting $v = u - \dfrac {\sqrt 2} 2 \sigma t$ $\displaystyle$ $=$ $\displaystyle \frac {\sqrt \pi \map \exp {\mu t + \frac 1 2 \sigma^2 t^2} } {\sqrt \pi}$ Gaussian Integral $\displaystyle$ $=$ $\displaystyle \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$

$\blacksquare$


text:
# Tag Info

6

Note that the inverse of an FIR system is IIR, and the same is true for the inverse of an IIR system, unless it is an all-pole system, the inverse of which would be FIR. So in most cases the ideal equalizer should have an infinitely long impulse response in order to perfectly invert the channel. In practice almost all adaptive equalizers are FIR filters ...

4

Bunch of things: Your band definition is really odd. Some of the lower bands are really small while some of the mid bands are huge. Ideally you want octaves where the band edges are 1/sqrt(2) and sqrt(2) around the center frequency You have designed bandpass filters in parallel. These are not "reconstructive" so you don't end up with a flat frequency ...

4

Assuming that you also want to equalize the filter's phase response (not only its magnitude response), you need an equalizer with a transfer function $E(z)$ that is the inverse of the FIR filter's transfer function $H(z)$: $$E(z)=\frac{1}{H(z)}=\frac{1}{c_0+c_1z^{-1}+\ldots+c_{N-1}z^{-(N-1)}}\tag{1}$$ This is an all-pole filter which can be implemented by ...

4

Ok, there is some misconceptions in your question. I strongly recommend you to read a little more about the topics, but I will try to help you a little. My answers and some comments: ...linear equalizer is a filter that can undo these channel effects. When the channel coefficients w are unknown, we perform blind equalization. In this scenario, we ...

3

A filter bank system is generally composed of: an analysis filter bank, that split an input signal into components with filters and may reduce their rate a synthesis filter bank (FB), that takes the components, potentially increase their rate, and feed them into filter elements so that the novel outputs can be combined into a signal. It is perfect ...

3

Thanks for the answer. You are correct - the training sequence is known by the time you establish any communication. In case if somebody stumbles upon this in the future, here's how to get this done: The synchronization burst contains an extended training sequence. It's longer than the regular one and there is only one sequence. You are supposed to use ...

2

From the referenced description and also from [1] it seems to me that the training sequence (here: midamble) is assumed to be known already. This knowledge has probably been acquired before in the synchronization burst. The first block in Fig. 2 extracts the training sequence from the received signal. This suggests that the receiver is already synchronized ...

2

result[i] = band_gain[0] * LPF(t, band_freq[0]) + band_gain[1] * (LPF(t, band_freq[1]) - LPF(a, band_freq[0])) + band_gain[2] * (LPF(t, band_freq[2]) - LPF(a, band_freq[1])) + band_gain[3] * (LPF(t, band_freq[3]) - LPF(a, band_freq[2])) + band_gain[4] * (IMPULSE(t) - LPF(t, band_freq[3])); is actually very simple to explain with pictures. First we assume f0 ...

2

A typical communications system is composed of the following parts: Message (Information Source) --> Encoding --> Modulation --> Channel --> Demodulation --> Decoding --> Recovered Message A "chaotic modulator" is just another way of doing the Modulation building block. This is depicted in Figure 1 of the paper that is cited in the ...

2

Does an audio equalizer consist of a perfect-reconstruction filterbank? Short answer: No. It's a bit unsharp what you're asking here, since reconstruction filterbanks are usually things that combine multiple, separate, independent signals back into one signal, but let's assume this makes sense here, and your audio source magically produces one signal stream ...

2

it looks like this eq10q uses cascaded parametric EQs, initially set up to be bell, but switchable to other types (like shelving). i wouldn't doubt that they're using the audio EQ cookbook. if you're using that, there is a parameter called "shelf slope" or $S$ that is often set to 1, because that gets you the steepest slope without dips or lips or bumps. ...

2

The audible range is about 10 octaves, and usually the center frequencies of a graphic equalizer would be distributed equally spaced on a log scale to cover that range. Common equalizers have either $30$ bands (with $1/3$ octave filters) or $10$ bands (with $1$ octave filters). If you want $5$ bands, you could choose filters that cover approximately $2$ ...

2

Most likely people use different words to describe the same concept. This happens quite often. The TEQ is the more general term and also used to equalize other methods that OFDM, while CS is related to OFDM and tries to make the channel shorter so that it fits within the CP duration. You can check online, you will find that almost all articles related to CS ...

2

This depends a bit on what you want to get out of this and how much effort/work you are willing to put in. Doing a room EQ that actually works and makes it sound consistently better is quite complicated. There are commercial systems available but they tend to be complicated and expensive or tied to a specific product. A really good freeware option to play ...

2

If the received signal can be written as $$\mathbf{y} = \mathbf{H}\,\mathbf{x} + \mathbf{n}$$ where $\mathbf{H}$ is the channel matrix, $\mathbf{x}$ is the transmitted vector, and $\mathbf{n}$ is the AWGN of the channel, then a zero forcing equalizer is simply (assuming that the channel matrix is square, and it's estimated perfectly at the receiver) $$\... 1 You need an estimate of the channel to receive the sequence but the zero-forcing equalizer does not need the channel response as an input. The zero forcing equalizer estimates the channel response. This can be done either with a training sequence, or can be decision directed when signal to noise ratios are high enough. Given the received signal is the ... 1 In an initial phase, the training sequence is used to determine the optimal equalizer weights. There is usually no reason to leave the weights fixed because you need an adaptation algorithm to compute the initial weights from the training sequence, so you might as well continue running the adaptation algorithm as soon as actual data are transmitted. If the ... 1 The error should be minimized between the equalized signal and what? There is no known sequence in this case. But there is: if the channel is still good enough so you can decode a packet successfully, you can reconstruct the original signal by simply re-encoding the packet. 1 With a blind equalization technique like the constant modulus algorithm (which is often implemented using a least mean squares (LMS) filter as you indicated), you aren't directly estimating the channel impulse response itself. Instead, the signal model is like this: The receiver observes the following signal:$$ x[k] = s[k] * c[k] + n[k]  where: $s[k]$ ...

1

I will try to answer on your question with this: Least Square equalizer is minimizing the intersymbol interference of your signal (ISI), between the symbols, which does not translate directly to minimization of the bit error rate. When I am saying 'directly' that means that there is no exact linear dependency between the error reduction in the equalizer, and ...

1

The discrete convolution sum operation is not restricted to equal length vectors. You can, and most of the time you do, convolve two different signals of arbitary lengths. Your confusion is probably with something else. The equalizer length can be different than that of the channel model length. That should not pose a problem but it would of course effect ...

1

You're right. LMS equalizer uses a known input to minimize the error. For communication purposes, this is either provided by a training sequence, or in a decision directed mode, the detector decisions are fed back as known data. The delta function is also correct. Suppose that the channel impulse response is $h(t)$ and frequency response $H(f)$. Then the ...

1

Your time domain waveform after 6 will be zero-padded as in step 3 with the additional "smearing" due to your equalization to remove the channel effects that will go into the zero padded area. The entire waveform in time will be delayed by the group delay of your frequency domain equalization, which may not necessarily match the group delay of the channel ...

1

Let's take a look at the cost function of CMA: $J(\mathbf{w}) = E[|y_{k}^{2} - 1|^{2}]$, onde $y_{k} = \mathbf{w}^{H} \mathbf{x}_{k}$ and $\mathbf{x}_{k}$ is the filter input. It says that the update function will converge to coefficients that approximate the output of the filter to something whose squared value is $1$. So, it is appropriate to ...

1

I had a look at your code, and I know why you get 0 BER. The thing is that you do not add any noise, you should use Eb/N0 to calculate the noise variance, and add a complex noise on your channel output. The distortion added by "chanOutput = filter(chCoeffs,1,modSignal);" merely adds linear ISI, by convolving the channel impulse response with the input ...

1

You will want to work with what Wikipedia calls "field quantities." Your engineer is telling you to work in steps of .1dB, which would be an amplitude ratio of .1=20log10(x). To get x you do 10^(.1/20) = 1.01158 From x dB to ratio (amplitude): 10^(x/20) From x ratio (amplitude) to dB: 20Log10(x) If you can calculate the ratio, then you should be able ...

1

Does your system has an Anti-Aliasing filter before the ADC? If not, I would suggest adding one. Depending on the spectra of the signals you are working with, aliasing can create 1/f noise.

1

I guess that your problem may not really be 1/f noise, but rather the effects from using the FFT directly on your time signal (without any windowing). If that is true, then try applying the Hann window to the time signal before the FFT and check the result. There is plenty of information available about the window functions and how to use them.

Only top voted, non community-wiki answers of a minimum length are eligible