CS683: calculating the effective resistances

Similar documents
2.4 Linear Inequalities and Interval Notation

I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3

The graphs of Rational Functions

Special Relativity solved examples using an Electrical Analog Circuit

Continuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom

Chapter E - Problems

Section 6.1 Definite Integral

Suppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = -2.

7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus

Regular expressions, Finite Automata, transition graphs are all the same!!

1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

CS415 Compilers. Lexical Analysis and. These slides are based on slides copyrighted by Keith Cooper, Ken Kennedy & Linda Torczon at Rice University

Designing Information Devices and Systems I Fall 2016 Babak Ayazifar, Vladimir Stojanovic Homework 6. This homework is due October 11, 2016, at Noon.

Bridging the gap: GCSE AS Level

Scientific notation is a way of expressing really big numbers or really small numbers.

1 Nondeterministic Finite Automata

Chapters Five Notes SN AA U1C5

We know that if f is a continuous nonnegative function on the interval [a, b], then b

Data Structures and Algorithms CMPSC 465

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved.

Designing Information Devices and Systems I Spring 2018 Homework 7

Lecture 2: January 27

4.4 Areas, Integrals and Antiderivatives

DIRECT CURRENT CIRCUITS

4.1. Probability Density Functions

PHYS Summer Professor Caillault Homework Solutions. Chapter 2

Vectors , (0,0). 5. A vector is commonly denoted by putting an arrow above its symbol, as in the picture above. Here are some 3-dimensional vectors:

EECS 141 Due 04/19/02, 5pm, in 558 Cory

5: The Definite Integral

The Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.

Chapter 9 Definite Integrals

5.2 Exponent Properties Involving Quotients

We partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

Potential Changes Around a Circuit. You must be able to calculate potential changes around a closed loop.

Lecture 20: Numerical Integration III

Designing finite automata II

Things to Memorize: A Partial List. January 27, 2017

Review of Gaussian Quadrature method

Polynomials and Division Theory

Hints for Exercise 1 on: Current and Resistance

AUTOMATA AND LANGUAGES. Definition 1.5: Finite Automaton

6.5 Numerical Approximations of Definite Integrals

ES.182A Topic 32 Notes Jeremy Orloff

Section 4: Integration ECO4112F 2011

Physics 2135 Exam 1 February 14, 2017

Section 6: Area, Volume, and Average Value

Physics 1402: Lecture 7 Today s Agenda

M344 - ADVANCED ENGINEERING MATHEMATICS

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

Chapter 0. What is the Lebesgue integral about?

KEY CONCEPTS. satisfies the differential equation da. = 0. Note : If F (x) is any integral of f (x) then, x a

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?

Homework 3 Solutions

y = f(x) This means that there must be a point, c, where the Figure 1

Resources. Introduction: Binding. Resource Types. Resource Sharing. The type of a resource denotes its ability to perform different operations

Algorithm Design and Analysis

Lecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations.

Matrix Eigenvalues and Eigenvectors September 13, 2017

Linear Systems with Constant Coefficients

Bases for Vector Spaces

Homework Assignment 6 Solution Set

Reading from Young & Freedman: For this topic, read the introduction to chapter 24 and sections 24.1 to 24.5.

Discrete Mathematics and Probability Theory Spring 2013 Anant Sahai Lecture 17

1B40 Practical Skills

University of. d Class. 3 st Lecture. 2 nd

Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University

Connected-components. Summary of lecture 9. Algorithms and Data Structures Disjoint sets. Example: connected components in graphs

PROPERTIES OF AREAS In general, and for an irregular shape, the definition of the centroid at position ( x, y) is given by

Partial Derivatives. Limits. For a single variable function f (x), the limit lim

Lecture 2e Orthogonal Complement (pages )

Exam 1 Solutions (1) C, D, A, B (2) C, A, D, B (3) C, B, D, A (4) A, C, D, B (5) D, C, A, B

Surface maps into free groups

SUMMER KNOWHOW STUDY AND LEARNING CENTRE

1. Extend QR downwards to meet the x-axis at U(6, 0). y

Homework 4. 0 ε 0. (00) ε 0 ε 0 (00) (11) CS 341: Foundations of Computer Science II Prof. Marvin Nakayama

Nondeterminism and Nodeterministic Automata

Motion. Acceleration. Part 2: Constant Acceleration. October Lab Phyiscs. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.

Lecture Solution of a System of Linear Equation

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1

7.1 Integral as Net Change Calculus. What is the total distance traveled? What is the total displacement?

Chapter 1: Logarithmic functions and indices

Minnesota State University, Mankato 44 th Annual High School Mathematics Contest April 12, 2017

September 13 Homework Solutions

CS 275 Automata and Formal Language Theory

Triangles The following examples explore aspects of triangles:

Coalgebra, Lecture 15: Equations for Deterministic Automata

Section 7.1 Area of a Region Between Two Curves

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014

Generalized Fano and non-fano networks

Designing Information Devices and Systems I Anant Sahai, Ali Niknejad. This homework is due October 19, 2015, at Noon.

7. Indefinite Integrals

r 0 ( ) cos( ) r( )sin( ). 1. Last time, we calculated that for the cardioid r( ) =1+sin( ),

CS103B Handout 18 Winter 2007 February 28, 2007 Finite Automata

What else can you do?

Definite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)

List all of the possible rational roots of each equation. Then find all solutions (both real and imaginary) of the equation. 1.

Transcription:

CS683: clculting the effective resistnces Lecturer: John Hopcroft Note tkers: June Andrews nd Jen-Bptiste Jennin Mrch 7th, 2008 On Ferury 29th we sw tht, given grph in which ech edge is lelled with resistnce (nlogy with n electricl circuit), nd given two vertices of this grph nd, the proility of reching efore returning to (considering pth tht strts from ) is P escpe = C eff C = R R eff. The purpose of this lecture will e to clculte C eff, nd thus P escpe, in some specil cses. We re minly interested in clculting the proility of escpe from the center of volume, first in one dimension, then in 2 dimension nd 3 dimensions. In ll the lecture we will keep the nottion for our strting node. Throughout, C = x C x, where C x = R x nd x is neighor of in the grph. In one dimension In one dimension our volume looks like: 4 3 2 0 2 3 4 Bending in hlf, this cn lso e viewed s: 0 v = 2 3 4 2 3 4 Therefore R eff = n+ 2, so P escpe = C eff n +. 2 In two dimensions n n v = C = C R eff = 2R eff = n+ 0 when Here lso, we wnt to clculte the resistnce. Here we wnt lower ound on the resistnce, so we put some resistnces to zero (the ones tht re old in the

scheme nd correspond to seril edges ); the effect of doing this is to reduce R eff nd give us n upper ound on P escpe : This cn lso e viewed s: 4 2 20 Therefore R eff 4 + 2 + 20 + = ( 4 + 3 + 5 + ) = inf 4 i= Θ(log n). Therefore the proility of escping 0 P escpe 4Θ(log n) P escpe 0 when n. 3 In three dimensions 3. Lower ound on P escpe 2 i = nd so Let us present the technique in 2 dimensions, then pply it to the 3-dimensionl cse. As efore, to get lower ound on P escpe, we need n upper ound on R eff, therefore we will put some resistnces to infinity, i.e., just cut the wires(on prllel edges ). We will do tht in the following wy: first drw in dshes the lines of eqution x + y = 2 n (this will e x + y + z = 2 n in 3 dimensions); then from the origin, follow pth up nd pth right until hitting dotted line; when hitting dotted line, split ech pth reching the dotted line into pth up nd pth right gin, etc. Doing tht, ll the resistnces tht re not 2

on pth re considered to e +, nd the others re considered to e. Here is picture illustrting this process: Now, this cn lso e viewed s(p, r), where p is the numer of pths nd r is the resistnce of ech pth: 2 4 2 Now, since the resistnce of ech pth etween dotted lines grows t rte of 2 i nd the numer of prllel pths etween dotted lines lso grows t rte of 2 i, these fctors lnce ech other nd therefore we would hve R eff 2 + 2 4 + 4 8 +. But we re relly in 3 dimensions, therefore we get tht the numer of pths grows fster thn 2 i, infct it grows t the rte of 3 i, while the length of ech pth grows t the sme rte: 3

R eff 3 + 2 9 + 4 27 + + ( ) k 2 3 3 k=0 3 ( 2 ) 3 R eff Now for P escpe P escpe = C R eff P escpe C P escpe 3 3.2 Upper ound on P escpe Now to get P escpe s upper ound we find lower ound on R eff using the sme principles s with the 2D lower ound, ie using cues insted of squres: R eff 6 + 6 9 + 6 25 + + 6 numerofnodesonsideofcue + R eff ( + 6 9 + ) 25 + (( + 6 4 + 9 + 6 + 25 + ) ( 36 4 + 6 + )) 36 + (( ) ( + 4 6 4 + 9 )) + R eff π2 6 Hence getting to the upper ound on P escpe : 4 Pge Rnk R P escpe R eff P escpe 8 3π 2 We now riefly discuss specil grphs with respect to pge rnk nd leve detiled discussion for the next lecture. 4

When crwling the we it is possile to lnd node, or group of nodes, such tht there re no out edges. This presents prolem when ending crwl or in unduly lnding t certin nodes too often for pge rnk to e ccurte. An effective wy round this is to rndomly restrt we crwl, for some reson restrting with proility 0.5 works well. Consider the cse in which we crwl from pge restrts with proility 0.5, leves the pge ltogether with proility 0.425 nd follows self-loop with proility 0.425. Allowing tht the pge s rnk is x, the pge rnk will eventully converge to the following formul: x = + 0.425xx =.74 Even if the person seeks to rtificlly increse their pge rnk y dding n infinite numer of self loops the pge rnk will still converge to every out possiility going to the pge nd with proility 0.5 the we crwl will leve giving: x = + 0.85xx = 6.6666 5