Dr. C.V. Suresh Babu, profile picture
Uploaded byDr. C.V. Suresh Babu
9,687 views

Randomized algorithms ver 1.0

This document discusses randomized algorithms. It begins by listing different categories of algorithms, including randomized algorithms. Randomized algorithms introduce randomness into the algorithm to avoid worst-case behavior and find efficient approximate solutions. Quicksort is presented as an example randomized algorithm, where randomness improves its average runtime from quadratic to linear. The document also discusses the randomized closest pair algorithm and a randomized algorithm for primality testing. Both introduce randomness to improve efficiency compared to deterministic algorithms for the same problems.

Education◦Technology◦
In this document
Powered by AI
See More

Introduces randomized algorithms, listing categories such as recursive, backtracking, divide and conquer, dynamic programming, and emphasizing randomized algorithms.

Explains why randomness is used: to avoid worst-case behavior and achieve efficient solutions. Outlines types and operational mechanics of randomized algorithms.

Details the Quick Sort algorithm using partitioning, discusses worst, best, and average case scenarios, introduces Randomized Quick Sort with expected performance.

Contrasts deterministic algorithms (predictable output) with randomized algorithms that involve randomness, affecting execution and outcomes.

Introduces Las Vegas algorithms, ensuring correctness and polynomial stepping, along with probabilistic analysis showing success for most inputs.

Describes the closest pair problem using divide-and-conquer and randomized approaches, including cluster partitioning and distance calculations.

Details steps for randomized algorithm to find closest pairs, analyzing time complexities and distance computations across partitions.

Describes a randomized algorithm for prime testing, tests multiple random values, ensuring correctness probability, and provides examples.

Introduces number theory theorems regarding composite numbers and outlines a randomized pattern matching algorithm approach.

Explains binary representations for matching and utilizes fingerprints for matching substrings, detailing computational mechanics for comparisons.

Describes the steps for a randomized pattern matching algorithm, detailing inputs, outputs, and success probability dynamics in matching.

Analyzes mistake occurrence probability in randomized algorithms when matching patterns, including relevant number theory concepts.

Presents interactive proof methods to ensure authenticity and security in communications, introducing a randomized quadratic residue method.

Outlines step-by-step interactions between parties in a proof setup, detailing actions and algorithm mechanisms ensuring secure communication.

Embed presentation

Randomized Algorithms Dr. C.V. Suresh Babu
A short list of categories  Algorithm types we will consider include:          2 Simple recursive algorithms Backtracking algorithms Divide and conquer algorithms Dynamic programming algorithms Greedy algorithms Branch and bound algorithms Brute force algorithms Randomized algorithms Also known as Monte Carlo algorithms or stochastic methods
Why use randomness?   Avoid worst-case behavior: randomness can (probabilistically) guarantee average case behavior Efficient approximate solutions to intractable problems
Randomized algorithms   A randomized algorithm is just one that depends on random numbers for its operation These are randomized algorithms:    These are related topics:   4 Using random numbers to help find a solution to a problem Using random numbers to improve a solution to a problem Getting or generating “random” numbers Generating random data for testing (or other) purposes
Randomized algorithms   In a randomized algorithm (probabilistic algorithm), we make some random choices. 2 types of randomized algorithms: 1. 2. For an optimization problem, a randomized algorithm gives an optimal solution. The average case time-complexity is more important than the worst case time-complexity. For a decision problem, a randomized algorithm may make mistakes. The probability of producing wrong solutions is very small.
Quick Sort Select: pick an arbitrary element x in S to be the pivot. Partition: rearrange elements so that elements with value less than x go to List L to the left of x and elements with value greater than x go to the List R to the right of x. Recursion: recursively sort the lists L and R. 6
Worst Case Partitioning of Quick Sort
Best Case Partitioning of Quick Sort
Average Case of Quick Sort
Randomized Quick Sort Randomized-Partition( A, p, r ) 1. i ← Random(p, r) 2. exchange A[r] ↔ A[i] 3. return Partition(A, p, r) Randomized-Quicksort( A, p, r ) 1. if p < r 2. then q ← Randomized-Partition (A, p, r) 3. Randomized-Quicksort (A, p , q-1) 4. Randomized-Quicksort (A, q+1, r)
Randomized Quick Sort     Exchange A[r] with an element chosen at random from A[p…r] in Partition. The pivot element is equally likely to be any of input elements. For any given input, the behavior of Randomized Quick Sort is determined not only by the input but also by the random choices of the pivot. We add randomization to Quick Sort to obtain for any input the expected performance of the algorithm to be good.
Deterministic Algorithms INPUT ALGORITHM OUTPUT Goal: Prove for all input instances the algorithm solves the problem correctly and the number of steps is bounded by a polynomial in the size of the input.
Randomized Algorithms INPUT ALGORITHM OUTPUT RANDOM NUMBERS   In addition to input, algorithm takes a source of random numbers and makes random choices during execution; Behavior can vary even on a fixed input;
Las Vegas Randomized Algorithms INPUT ALGORITHM OUTPUT RANDOM NUMBERS Goal: Prove that for all input instances the algorithm solves the problem correctly and the expected number of steps is bounded by a polynomial in the input size. Note: The expectation is over the random choices made by the algorithm.
Probabilistic Analysis of Algorithms RANDOM INPUT ALGORITHM OUTPUT DISTRIBUTION Input is assumed to be from a probability distribution. Goal: Show that for all inputs the algorithm works correctly and for most inputs the number of steps is bounded by a polynomial in the size of the input.
The closest pair problem   This problem can be solved by the divide-andconquer approach in O(nlogn) time. The randomized algorithm: partition the points into several clusters:  X4 X5   3 X  1 X  2 X    6 X  7 X We only calculate distances among points within the same cluster. Similar to the divide-and-conquer strategy. There is a dividing process, but no merging process.
A randomized algorithm for closest pair finding  Input: A set S consisting of n elements x1, x2,…, xn, where S⊆ R2. Output: The closest pair in S. xi , xi ,..., xi Step 1: Randomly choose a set S1={ } where m=n2/3. Find the closest pair of S1 and let the distance between this pair of points be denoted as δ . Step 2: Construct a set of squares T with meshsize δ.  1 2 m
Step 3: Construct four sets of squares T1, T2, T3 and T4 derived from T by doubling the meshsize to 2δ . Step 4: For each Ti, find the induced decomposition S=S1(i) ∪S2(i) ∪ … ∪ Sj(i), 1≤i ≤4, where Sj(i) is a non-empty intersection of S with a square of Ti. Step 5: For each xp, xq∈Sj(i), compute d(xp, xq). Let xa and xb be the pair of points with the shortest distance among these pairs. Return xa and xb as the closest pair.
An example  27 points. S1 = {x1, x2, …, x9}, δ = d(x1, x2) 6δ 5δ X4 X3 4δ X5 X2 3δ X9 X1 2δ X7 X6 δ X8 δ 2δ 3δ 4δ 5δ 6δ
6δ 5δ X4 X3 4δ T1 X5 3δ X2 X1 X9 2δ X6 X7 δ X8 2δ δ 3δ 4δ 5δ 6δ 6δ 5δ X3 X4 4δ T2 X5 3δ X2 X1 2δ X9 X6 X7 δ X8 δ 2δ 3δ 4δ 5δ 6δ
6δ 5δ X4 X3 4δ T3 X5 3δ X2 X1 2δ X9 X6 X7 X8 δ δ 2δ 4δ 3δ 5δ 6δ 6δ 5δ X4 T4 X3 4δ X5 X2 3δ X1 2δ X9 X7 X6 δ X8 δ 2δ 3δ 4δ 5δ 6δ
Time complexity   Time complexity: O(n) in average step 1: O(n) method : Recursively apply the algorithm once, 2 i.e. randomly choose ( n ) 3 2 3 =n 2 4 9 points from the n 3 points, then solve it with a straightforward method for the n   4 9 8 points : O( n 9 ) step 2 ~ Step 4: O(n) step 5: O(n)with probability 1-2e-cn 1 6
Analysis of Step 5  How many distance computations in step 5? δ : mesh-size in step 5 T: partition in step 5 N(T): # of distance computations in partition T Fact: There exists a particular partition T0, whose mesh-size is δ0 such that (1) N(T0) ≤ c0n. (2) the probability that δ ≤ 2δ 0 is 1 - 2e 1 − cn 6 .
 Construct T1, T2, …, T16 δ 4δ0 mesh-size: 4δ0   The probability that each square in T falls into at least1 one square of Ti , i = 1, 2, …, 16 is − cn 6 1 - 2e . 1 The probability that 16 − cn 6 N (T ) ≤ ∑ N (Ti ) is 1 - 2e . i =1
 Let the square in T0 with the largest number of elements among the 16 squares have k elements. k ( k −1) 16k (16k −1) 2 = O ( k ), = O(k 2 ) 2 2  N(T0) ≤ c0n => N(Ti ) ≤ cin 16 N (T ) ≤ ∑ N (Ti ) = O( n ) with 4δ0 δ0 Ti: i =1 probability 1 - 2e 1 − cn 6 . k
A randomized algorithm to test whether a number is prime.   This problem is very difficult and no polynomial algorithm has been found to solve this problem Traditional method: use 2,3,… N to test whether N is prime. input size of N : B=log2N (binary representation) N =2B/2, exponential function of B Thus N can not be viewed as a polynomial function of the input size.
Randomized prime number testing algorithm Input: A positive number N, and a parameter m.  Output: Whether N is a prime or not, with probability of being correct at least 1-ε = 1-2-m. Step 1: Randomly choose m numbers b1, b2, …, bm, 1≤ b1, b2, …, bm <N, where m≥log2(1/ε).  Step 2: For each bi, test whether W(bi) holds where W(bi) is defined as follows: N 1 (1) biN-1 ≠ 1mod N or − 2j (2) ∃ j such taht = k is an integer and the greatest common divisor of (bi)k-1 and N is not 1 or N. If any W(bi) holds, then return N as a composite number, otherwise, return N as a prime.
Examples for randomized prime number testing  Example 1: N = 12 Randomly choose 2, 3, 7 212-1 = 2048 ≠ 1 mod 12 ⇒ 12 is a composite number.
 Example 2: N = 11 Randomly choose 2, 5, 7 (1) 211-1=1024≡1 mod 11 j=1, (N-1)/2j==5 GCD(25-1, 11) = 1 W(2) does not hold . (2) 511-1=9765625≡1 mod 11 GCD(55-1, 11) = 11 W(5) does not hold . (3) 711-1=282475249≡1 mod 11 GCD(75-1, 11) = 1 W(7) does not hold .  Thus, 11 is a prime number with the probability of correctness being at least 1-2-3= 7/8.
Theorem for number theory  Theorem:  If W(b) holds for any 1≤ b<N, then N is a composite number .  If N is composite, then (N-1)/2 ≤ | { b | 1 ≤ b<N, W(b) holds } |.
Pattern matching Pattern string : X length : n Text string : Y length : m, m ≥ n To find the first occurrence of X as a consecutive substring of Y . Assume that X and Y are binary strings.  e.g. X = 01001 , Y = 1010100111  X    Straightforward method : O(mn) Knuth-Morris-Pratt’s algorithm : O(m) The randomized algorithm : O(mk) with a mistake of small probability. (k:# of testings)
Binary representation  X = x1 x2…xn∈{0,1} Y = y1 y2…ym∈{0,1} Let Y(i)=yi yi+1….yi+n-1  A match occurs if X=Y(i) for some i . Binary values of X and Y(i): B(X) = x12n-1 + x22n-2 + … + xn B(Y(i)) = yi2n-1+yi+12n-2+…+yi+n-1 , 1≤ i ≤ m-n+1
Fingerprints of binary strings    Let p be a randomly chosen prime number in {1,2,…,nt2}, where t = m - n + 1. Notation: (xi)p = xi mod p Fingerprints of X and Y(i): Bp(x) = ( ( ( (x12)p+x2)p2)p+x3)p2… Bp(Y(i)) = ( ( ( (yi2)p+yi+1)p2+yi+2)p2… ⇒ Bp(Y(i+1))= ( (Bp(Yi)-2n-1yi) 2+Yi+n)p =( ( (Bp(Yi)-( (2n-1)pyi)p )p 2)p +yi+n)p  If X=Y(i), then Bp(X) = Bp (Y(i)), but not vice versa.
Examples for using fingerprints  Example: X = 10110 , Y = 110110 n=5,m=6,t=m-n+1=2 suppose P=3. Bp(X) = (22)3 = 1 Bp(Y(1)) = (27)3 = 0 ⇒ X≠Y(1) Bp(Y(2)) = ( (0-24)3 2+0)3 = 1 ⇒ X = Y(2)
 e.g. X = 10110 , Y = 10011 , P = 3 Bp(X) = (22)3 = 1 Bp(Y(1)) = (19)3 = 1 ⇒ X= Y(1) WRONG!   If Bp(X) ≠ Bp(Y(i)), then X ≠ Y(i) . If Bp(X) = Bp(Y(i)), we may do a bit by bit checking or compute k different fingerprints by using k different prime numbers in {1,2,… nt2} .
A randomized algorithm for pattern matching   Input: A pattern X = x1 x2…xn, a text Y = y1 y2…ym and a parameter k. Output: (1) No, there is no consecutive substring in Y which matches with X. (2) Yes, Y(i) = yi yi+1.…yi+n-1 matches with X which is the first occurrence. If the answer is “No” , there is no mistake. If the answer is “Yes” , there is some probability that a mistake is made.
Step 1: Randomly choose k prime numbers p1, p2, …, pk from {1,2,…,nt2}, where t = m - n + 1. Step 2: i = 1. Step 3: j = 1. Step 4: If B(X)Pj ≠ (B(Yi))pj, then go to step 5. If j = k, return Y(i) as the answer. j = j + 1. Go to step 4. Step5: If i = t, return “No, there is no consecutive substring in Y which matches with X.” i = i + 1. Go to Step 3.
An example for the algorithm  X = 10110 , Y = 100111 , P1 = 3 , P2 = 5 B3(X) = (22)3 = 1 B5(X) = (22)5 = 2 B3(Y(2)) = (7)3 = 1 B5(y(2)) = (7)5 = 2 Choose one more prime number, P3 = 7 B7(x) = (22)7 = 1 B7(Y(2)) = (7)7 = 0 ⇒X ≠ Y(2)
How often does a mistake occur?    If a mistake occurs in X and Y(i), then B(X) - B(Y(i)) ≠ 0, and pj divides | B(X) - B(Y(i)) | for all pj’s. Let Q = π i where p j divides B ( X ) − B ( Y ( i )) B( X ) − B (Y (i )) Q<2n(m-n+1) reason: B(x)<2n, and at most (m-n+1) B(Y(i))’s 2n 2n…2n m-n-1
Theorem for number theory     Theorem: If u≥29 and a<2u, then a has fewer than π(u) diffferent prime number divisors where π(u) is the number of prime numbers smaller than u. Assume nt ≥ 29 . Q < 2n(m-n+1) = 2nt ⇒ Q has fewer than π(nt) different prime number divisors. If pj is a prime number selected from {1, 2, …, M}, π nt ) ( the probability that pj divides Q is less than π M ). ( If k different prime numbers are selected from {1, 2, … nt2} , the probability that a mistake occurs is less than k  π (nt )   provided nt ≥ 29.  2    π (nt ) 
An example for mistake probability k  π ( nt )   2   π (nt )   How do we estimate    u u Theorem: For all u ≥ 17,  ≤ π (u ) ≤ 1.25506 ln u ln u   π (nt ) nt ln(nt 2 ) ≤ 1.25506 ⋅ ⋅ 2 π (nt ) ln nt nt 2 1.25506 ln(t ) = (1 + ) t ln(nt ) Example: n = 10 , m = 100 , t = m - n + 1 = 91 π (nt ) ≤ 0.0229 2 π ( nt ) Let k=4 (0.0229)4≈2.75×10-7 // very small
Interactive proofs: method I    Two persons: A : a spy B : the boss of A When A wants to talk to B , how does B know that A is the real A, not an enemy imitating A ? Method I : a trivial method B may ask the name of A’s mother (a private secret) Disadvantage: The enemy can collect the information, and imitate A the next time.
Interactive proofs: method II    Method II: B may send a Boolean formula to A and ask A to determine its satisfiability. (an NP-complete problem). It is assumed that A is a smart person and knows how to solve this NP-complete problem. B can check the answer and know whether A is the real A or not. Disadvantage: The enemy can study methods of mechanical theorem proving and sooner or later he can imitate A. In Methods I and II, A and B have revealed too much.
A randomized algorithm for interactive proofs   Method III: B can ask A to solve a quadratic nonresidue problem in which the data can be sent back and forth without revealing much information. Definition: GCD(x, y) = 1, y is a quadratic residue mod x if z2 ≡ y mod x for some z, 0 < z < x, GCD(x, z) = 1, and y is a quadratic nonresidue mod x if otherwise. (See the example on the next page.)
An example for quadratic residue/nonresidue    Let QR = {(x, y) | y is a quadratic residue mod x} QNR = {(x, y) | y is a quadratic nonresidue mod x} Try to test x = 9, y = 7: 12 ≡ 1 mod 9 22 ≡ 4 mod 9 32 ≡ 0 mod 9 42 ≡ 7 mod 9 52 ≡ 7 mod 9 62 ≡ 0 mod 9 72 ≡ 4 mod 9 82 ≡ 1 mod 9 We have (9,1), (9,4), (9,7) ∈ QR but (9,5), (9,8) ∈ QNR
Detailed method for interactive proofs 1) 2) A and B know x and keep x confidential . B knows y. Action of B: Step 1: Randomly choose m bits: b1, b2, …, bm, where m is the length of the binary representation of x. Step 2: Find z1, z2, …, zm s.t. GCD(zi , x)=1 for all i . Step 3:Compute w1, w2, …, wm: wi ←zi2 mod x if bi=0 //(x, wi) ∈ QR wi ← (zi2y) mod x if bi=1 //(x, wi) ∈ NQR Step 4: Send w1, w2, …, wm to A.
3) Action of A: Step 1: Receive w1, w2, …, wm from B. Step 2: Compute c1, c2, …, cm: ci ←0 if (x, wi) ∈ QR ci ←1 if (x, wi) ∈ QNR Send c1, c2, …, cm to B. 4) Action of B: Step 1: Receive c1, c2, …, cm from A. Step 2: If (x, y) ∈ QNR and bi = ci for all i, then A is the real A (with probability 1-2-m).

More Related Content

PPT
Np cooks theorem
byNarayana Galla
 
PPTX
Randomized Algorithm
byKanishka Khandelwal
 
PPTX
Randomized Algorithm- Advanced Algorithm
byMahbubur Rahman
 
PDF
Recurrence relation solutions
bysubhashchandra197
 
PPT
0/1 knapsack
byAmin Omi
 
PPT
finding Min and max element from given array using divide & conquer
bySwati Kulkarni Jaipurkar
 
PPTX
Approximation Algorithms TSP
byP. Subathra Kishore, KAMARAJ College of Engineering and Technology, Madurai
 
PPTX
Divide and Conquer - Part 1
byAmrinder Arora
 
Np cooks theorem
byNarayana Galla
 
Randomized Algorithm
byKanishka Khandelwal
 
Randomized Algorithm- Advanced Algorithm
byMahbubur Rahman
 
Recurrence relation solutions
bysubhashchandra197
 
0/1 knapsack
byAmin Omi
 
finding Min and max element from given array using divide & conquer
bySwati Kulkarni Jaipurkar
 
Approximation Algorithms TSP
byP. Subathra Kishore, KAMARAJ College of Engineering and Technology, Madurai
 
Divide and Conquer - Part 1
byAmrinder Arora
 

What's hot

PPTX
sum of subset problem using Backtracking
byAbhishek Singh
 
PPTX
Sum of subset problem.pptx
byV.V.Vanniaperumal College for Women
 
PDF
I.BEST FIRST SEARCH IN AI
byvikas dhakane
 
PPT
Minimum spanning tree
byHinal Lunagariya
 
PPT
BackTracking Algorithm: Technique and Examples
byFahim Ferdous
 
PPTX
daa-unit-3-greedy method
byhodcsencet
 
PDF
COMPILER DESIGN Run-Time Environments
byJyothishmathi Institute of Technology and Science Karimnagar
 
PPT
Recursion tree method
byRajendran
 
PPTX
N queen problem
byRidhima Chowdhury
 
PDF
P, NP, NP-Complete, and NP-Hard
byAnimesh Chaturvedi
 
PPTX
Job sequencing with Deadlines
byYashiUpadhyay3
 
PDF
Semaphores
byMohd Arif
 
PPTX
Problem solving agents
byMegha Sharma
 
PPT
DESIGN AND ANALYSIS OF ALGORITHMS
byGayathri Gaayu
 
PDF
Time and Space Complexity
byAshutosh Satapathy
 
PPTX
Propositional logic
byRushdi Shams
 
PPTX
Breadth First Search & Depth First Search
byKevin Jadiya
 
PPTX
Code generation
byAparna Nayak
 
PPTX
Knapsack problem using greedy approach
bypadmeshagrekar
 
PPTX
Lecture 14 run time environment
byIffat Anjum
 
sum of subset problem using Backtracking
byAbhishek Singh
 
Sum of subset problem.pptx
byV.V.Vanniaperumal College for Women
 
I.BEST FIRST SEARCH IN AI
byvikas dhakane
 
Minimum spanning tree
byHinal Lunagariya
 
BackTracking Algorithm: Technique and Examples
byFahim Ferdous
 
daa-unit-3-greedy method
byhodcsencet
 
COMPILER DESIGN Run-Time Environments
byJyothishmathi Institute of Technology and Science Karimnagar
 
Recursion tree method
byRajendran
 
N queen problem
byRidhima Chowdhury
 
P, NP, NP-Complete, and NP-Hard
byAnimesh Chaturvedi
 
Job sequencing with Deadlines
byYashiUpadhyay3
 
Semaphores
byMohd Arif
 
Problem solving agents
byMegha Sharma
 
DESIGN AND ANALYSIS OF ALGORITHMS
byGayathri Gaayu
 
Time and Space Complexity
byAshutosh Satapathy
 
Propositional logic
byRushdi Shams
 
Breadth First Search & Depth First Search
byKevin Jadiya
 
Code generation
byAparna Nayak
 
Knapsack problem using greedy approach
bypadmeshagrekar
 
Lecture 14 run time environment
byIffat Anjum
 

Viewers also liked

PPT
Randomized Algorithms
byKetan Kamra
 
PPTX
Topic 1.4: Randomized Algorithms
byKM Bappi
 
PPTX
Lecture 6-cs648 Randomized Algorithms
byAnshul Yadav
 
PPTX
Dynamic programmng2
bydebolina13
 
PPTX
Introduction To Algorithms
byKM Bappi
 
PDF
An Experiment to Determine and Compare Practical Efficiency of Insertion Sort...
byTosin Amuda
 
PPT
Longest common subsequences in Algorithm Analysis
byRajendran
 
DOC
Branch and bound
byNv Thejaswini
 
PPTX
Longest common subsequence lcs
byShahariar Rabby
 
PDF
Branch and bound technique
byishmecse13
 
PPTX
The selection sort algorithm
byDavid Burks-Courses
 
PPT
Algorithm analysis
bysumitbardhan
 
PDF
Algorithm Analysis and Design Class Notes
byKumar Avinash
 
PPTX
Comparitive Analysis of Algorithm strategies
byTalha Shaikh
 
PDF
backtracking algorithms of ada
bySahil Kumar
 
PPTX
Quick sort-Data Structure
byJeanie Arnoco
 
PPT
Monte carlo
byshishirkawde
 
PPTX
Divide and conquer - Quick sort
byMadhu Bala
 
PPT
Algorithm: Quick-Sort
byTareq Hasan
 
PPT
Algorithm.ppt
byTareq Hasan
 
Randomized Algorithms
byKetan Kamra
 
Topic 1.4: Randomized Algorithms
byKM Bappi
 
Lecture 6-cs648 Randomized Algorithms
byAnshul Yadav
 
Dynamic programmng2
bydebolina13
 
Introduction To Algorithms
byKM Bappi
 
An Experiment to Determine and Compare Practical Efficiency of Insertion Sort...
byTosin Amuda
 
Longest common subsequences in Algorithm Analysis
byRajendran
 
Branch and bound
byNv Thejaswini
 
Longest common subsequence lcs
byShahariar Rabby
 
Branch and bound technique
byishmecse13
 
The selection sort algorithm
byDavid Burks-Courses
 
Algorithm analysis
bysumitbardhan
 
Algorithm Analysis and Design Class Notes
byKumar Avinash
 
Comparitive Analysis of Algorithm strategies
byTalha Shaikh
 
backtracking algorithms of ada
bySahil Kumar
 
Quick sort-Data Structure
byJeanie Arnoco
 
Monte carlo
byshishirkawde
 
Divide and conquer - Quick sort
byMadhu Bala
 
Algorithm: Quick-Sort
byTareq Hasan
 
Algorithm.ppt
byTareq Hasan
 

Similar to Randomized algorithms ver 1.0

PDF
Daa notes 2
bysmruti sarangi
 
PPTX
randomizedalgorithm and understand quick sort.pptx
byMDTahsinAmin3
 
PPTX
Programming Exam Help
byProgramming Exam Help
 
PDF
Cs6402 design and analysis of algorithms may june 2016 answer key
byappasami
 
PDF
BIN PACKING PROBLEM: A LINEAR CONSTANTSPACE -APPROXIMATION ALGORITHM
byijcsa
 
PPTX
RA-UNIT-1.pptx ( Randomized Algorithms)
bygreekathena0501
 
PDF
Randomization
bymagdakarap
 
PPT
Chap11alg
byMunhchimeg
 
PPT
Chap11alg
byMunkhchimeg
 
PPTX
Unit 2
byGunasundariSelvaraj
 
PPTX
Unit 2
byGunasundari Selvaraj
 
PPT
2010 3-24 cryptography stamatiou
byvafopoulos
 
PDF
25010001
byPremavardhan Reddy
 
PDF
Solution 3.
bysansaristic
 
PDF
Jurnal informatika
byMamaMa28
 
PDF
A New Deterministic RSA-Factoring Algorithm
byJim Jimenez
 
PDF
Csr2011 june15 11_00_sima
byCSR2011
 
PPT
A small debate of power of randomness
byAbner Chih Yi Huang
 
PDF
A Numeric Algorithm for Generating Permutations in Lexicographic Order with a...
byAfshin Tiraie
 
PPTX
Algorithms Design
byWajahatAli602286
 
Daa notes 2
bysmruti sarangi
 
randomizedalgorithm and understand quick sort.pptx
byMDTahsinAmin3
 
Programming Exam Help
byProgramming Exam Help
 
Cs6402 design and analysis of algorithms may june 2016 answer key
byappasami
 
BIN PACKING PROBLEM: A LINEAR CONSTANTSPACE -APPROXIMATION ALGORITHM
byijcsa
 
RA-UNIT-1.pptx ( Randomized Algorithms)
bygreekathena0501
 
Randomization
bymagdakarap
 
Chap11alg
byMunhchimeg
 
Chap11alg
byMunkhchimeg
 
Unit 2
byGunasundariSelvaraj
 
Unit 2
byGunasundari Selvaraj
 
2010 3-24 cryptography stamatiou
byvafopoulos
 
25010001
byPremavardhan Reddy
 
Solution 3.
bysansaristic
 
Jurnal informatika
byMamaMa28
 
A New Deterministic RSA-Factoring Algorithm
byJim Jimenez
 
Csr2011 june15 11_00_sima
byCSR2011
 
A small debate of power of randomness
byAbner Chih Yi Huang
 
A Numeric Algorithm for Generating Permutations in Lexicographic Order with a...
byAfshin Tiraie
 
Algorithms Design
byWajahatAli602286
 

More from Dr. C.V. Suresh Babu

PPTX
Data analytics with R
byDr. C.V. Suresh Babu
 
PPTX
Association rules
byDr. C.V. Suresh Babu
 
PPTX
Clustering
byDr. C.V. Suresh Babu
 
PPTX
Classification
byDr. C.V. Suresh Babu
 
PPTX
Blue property assumptions.
byDr. C.V. Suresh Babu
 
PPTX
Introduction to regression
byDr. C.V. Suresh Babu
 
PPTX
DART
byDr. C.V. Suresh Babu
 
PPTX
Mycin
byDr. C.V. Suresh Babu
 
PPTX
Expert systems
byDr. C.V. Suresh Babu
 
PPTX
Dempster shafer theory
byDr. C.V. Suresh Babu
 
PPTX
Bayes network
byDr. C.V. Suresh Babu
 
PPTX
Bayes' theorem
byDr. C.V. Suresh Babu
 
PPTX
Knowledge based agents
byDr. C.V. Suresh Babu
 
PPTX
Rule based system
byDr. C.V. Suresh Babu
 
PPTX
Formal Logic in AI
byDr. C.V. Suresh Babu
 
PPTX
Production based system
byDr. C.V. Suresh Babu
 
PPTX
Game playing in AI
byDr. C.V. Suresh Babu
 
PPTX
Diagnosis test of diabetics and hypertension by AI
byDr. C.V. Suresh Babu
 
PPTX
A study on “impact of artificial intelligence in covid19 diagnosis”
byDr. C.V. Suresh Babu
 
PDF
A study on “impact of artificial intelligence in covid19 diagnosis”
byDr. C.V. Suresh Babu
 
Data analytics with R
byDr. C.V. Suresh Babu
 
Association rules
byDr. C.V. Suresh Babu
 
Clustering
byDr. C.V. Suresh Babu
 
Classification
byDr. C.V. Suresh Babu
 
Blue property assumptions.
byDr. C.V. Suresh Babu
 
Introduction to regression
byDr. C.V. Suresh Babu
 
DART
byDr. C.V. Suresh Babu
 
Mycin
byDr. C.V. Suresh Babu
 
Expert systems
byDr. C.V. Suresh Babu
 
Dempster shafer theory
byDr. C.V. Suresh Babu
 
Bayes network
byDr. C.V. Suresh Babu
 
Bayes' theorem
byDr. C.V. Suresh Babu
 
Knowledge based agents
byDr. C.V. Suresh Babu
 
Rule based system
byDr. C.V. Suresh Babu
 
Formal Logic in AI
byDr. C.V. Suresh Babu
 
Production based system
byDr. C.V. Suresh Babu
 
Game playing in AI
byDr. C.V. Suresh Babu
 
Diagnosis test of diabetics and hypertension by AI
byDr. C.V. Suresh Babu
 
A study on “impact of artificial intelligence in covid19 diagnosis”
byDr. C.V. Suresh Babu
 
A study on “impact of artificial intelligence in covid19 diagnosis”
byDr. C.V. Suresh Babu
 

Recently uploaded

PDF
Multiple Myeloma , definition, etiology, PP, CM , DE and management
byNisha Borsa
 
PPTX
Industrial production, estimation and utilization of the following phytoconst...
byD. Y. Patil College of Pharmacy, Kolhapur.
 
PDF
M.Sc. Nonchordates Complete Syllabus PPT | All Important Topics Covered
byKNIPSS SULTANPUR
 
PDF
Advanced Process Flow and Micromachining Techniques for MEMS Fabrication: A D...
byGS Virdi
 
PPTX
PARENTAL ROUTES OF DRUGS ADMINISTRATION .pptx
byAneetaSharma15
 
PPTX
Multiple Correlation - Introduction, Meaning, Examples, Options, Formulas and...
bySundar B N
 
PDF
Madhumathi Endowment Quiz Finals by Sreejesh P S and Shyam Krishnan P - Keral...
bySreejesh P S
 
PDF
SIHMA Scalabrini Institute for Human Mobility in Africa(SIHMA) - Annual Repor...
byScalabrini Institute for Human Mobility in Africa
 
PPTX
GROWTH & DEVELOPMET MILESTONES (TODDLER).pptx
byAneetaSharma15
 
PPTX
Partial Correlation of Coefficient - Values of r₁₂.₃, r₂₃.₁ & r₁₃.₂
bySundar B N
 
PDF
IMANI Africa files RTI request seeking full disclosure on 2026 SIM registrati...
bynservice241
 
PPTX
cis.jackson.alaysia.ResearchPresentation 1.Microsoft Teams vs Zoom vs Webex v...
byMerritt College
 
PPTX
How to Access Revenue Report in Odoo 18 Events
byCeline George
 
PPTX
GROWTH & DEVELOPMENT MILESTONES (INFANTS).pptx
byAneetaSharma15
 
PDF
Introduction to Descriptive Statistics.pdf
byThelma Villaflores
 
PPTX
Partial Correlation - Values of r₁₂.₃, r₂₃.₁ & r₁₃.₂ r₁₂, r₁₃ and r₂₃
bySundar B N
 
PPTX
Access partner report in Odoo 18.2 _ Odoo 19
byCeline George
 
PDF
Environmental Studies Module 2 BBA,BCCA Sem 1 NEP.pdf
byJayanti Pande
 
PPTX
Expected Revenue Report In Odoo 18 CRM
byCeline George
 
PDF
Cultivating Greatness Pune's Best Preschools and Schools.pdf
byWellington College
 
Multiple Myeloma , definition, etiology, PP, CM , DE and management
byNisha Borsa
 
Industrial production, estimation and utilization of the following phytoconst...
byD. Y. Patil College of Pharmacy, Kolhapur.
 
M.Sc. Nonchordates Complete Syllabus PPT | All Important Topics Covered
byKNIPSS SULTANPUR
 
Advanced Process Flow and Micromachining Techniques for MEMS Fabrication: A D...
byGS Virdi
 
PARENTAL ROUTES OF DRUGS ADMINISTRATION .pptx
byAneetaSharma15
 
Multiple Correlation - Introduction, Meaning, Examples, Options, Formulas and...
bySundar B N
 
Madhumathi Endowment Quiz Finals by Sreejesh P S and Shyam Krishnan P - Keral...
bySreejesh P S
 
SIHMA Scalabrini Institute for Human Mobility in Africa(SIHMA) - Annual Repor...
byScalabrini Institute for Human Mobility in Africa
 
GROWTH & DEVELOPMET MILESTONES (TODDLER).pptx
byAneetaSharma15
 
Partial Correlation of Coefficient - Values of r₁₂.₃, r₂₃.₁ & r₁₃.₂
bySundar B N
 
IMANI Africa files RTI request seeking full disclosure on 2026 SIM registrati...
bynservice241
 
cis.jackson.alaysia.ResearchPresentation 1.Microsoft Teams vs Zoom vs Webex v...
byMerritt College
 
How to Access Revenue Report in Odoo 18 Events
byCeline George
 
GROWTH & DEVELOPMENT MILESTONES (INFANTS).pptx
byAneetaSharma15
 
Introduction to Descriptive Statistics.pdf
byThelma Villaflores
 
Partial Correlation - Values of r₁₂.₃, r₂₃.₁ & r₁₃.₂ r₁₂, r₁₃ and r₂₃
bySundar B N
 
Access partner report in Odoo 18.2 _ Odoo 19
byCeline George
 
Environmental Studies Module 2 BBA,BCCA Sem 1 NEP.pdf
byJayanti Pande
 
Expected Revenue Report In Odoo 18 CRM
byCeline George
 
Cultivating Greatness Pune's Best Preschools and Schools.pdf
byWellington College
 

Randomized algorithms ver 1.0

  • 1.
  • 2.
    A short listof categories  Algorithm types we will consider include:          2 Simple recursive algorithms Backtracking algorithms Divide and conquer algorithms Dynamic programming algorithms Greedy algorithms Branch and bound algorithms Brute force algorithms Randomized algorithms Also known as Monte Carlo algorithms or stochastic methods
  • 3.
    Why use randomness?   Avoidworst-case behavior: randomness can (probabilistically) guarantee average case behavior Efficient approximate solutions to intractable problems
  • 4.
    Randomized algorithms   A randomizedalgorithm is just one that depends on random numbers for its operation These are randomized algorithms:    These are related topics:   4 Using random numbers to help find a solution to a problem Using random numbers to improve a solution to a problem Getting or generating “random” numbers Generating random data for testing (or other) purposes
  • 5.
    Randomized algorithms   In arandomized algorithm (probabilistic algorithm), we make some random choices. 2 types of randomized algorithms: 1. 2. For an optimization problem, a randomized algorithm gives an optimal solution. The average case time-complexity is more important than the worst case time-complexity. For a decision problem, a randomized algorithm may make mistakes. The probability of producing wrong solutions is very small.
  • 6.
    Quick Sort Select: pickan arbitrary element x in S to be the pivot. Partition: rearrange elements so that elements with value less than x go to List L to the left of x and elements with value greater than x go to the List R to the right of x. Recursion: recursively sort the lists L and R. 6
  • 7.
  • 8.
  • 9.
    Average Case ofQuick Sort
  • 10.
    Randomized Quick Sort Randomized-Partition(A, p, r ) 1. i ← Random(p, r) 2. exchange A[r] ↔ A[i] 3. return Partition(A, p, r) Randomized-Quicksort( A, p, r ) 1. if p < r 2. then q ← Randomized-Partition (A, p, r) 3. Randomized-Quicksort (A, p , q-1) 4. Randomized-Quicksort (A, q+1, r)
  • 11.
    Randomized Quick Sort     ExchangeA[r] with an element chosen at random from A[p…r] in Partition. The pivot element is equally likely to be any of input elements. For any given input, the behavior of Randomized Quick Sort is determined not only by the input but also by the random choices of the pivot. We add randomization to Quick Sort to obtain for any input the expected performance of the algorithm to be good.
  • 12.
    Deterministic Algorithms INPUT ALGORITHM OUTPUT Goal: Provefor all input instances the algorithm solves the problem correctly and the number of steps is bounded by a polynomial in the size of the input.
  • 13.
    Randomized Algorithms INPUT ALGORITHM OUTPUT RANDOM NUMBERS   Inaddition to input, algorithm takes a source of random numbers and makes random choices during execution; Behavior can vary even on a fixed input;
  • 14.
    Las Vegas Randomized Algorithms INPUT ALGORITHM OUTPUT RANDOMNUMBERS Goal: Prove that for all input instances the algorithm solves the problem correctly and the expected number of steps is bounded by a polynomial in the input size. Note: The expectation is over the random choices made by the algorithm.
  • 15.
    Probabilistic Analysis of Algorithms RANDOM INPUT ALGORITHM OUTPUT DISTRIBUTION Inputis assumed to be from a probability distribution. Goal: Show that for all inputs the algorithm works correctly and for most inputs the number of steps is bounded by a polynomial in the size of the input.
  • 16.
    The closest pairproblem   This problem can be solved by the divide-andconquer approach in O(nlogn) time. The randomized algorithm: partition the points into several clusters:  X4 X5   3 X  1 X  2 X    6 X  7 X We only calculate distances among points within the same cluster. Similar to the divide-and-conquer strategy. There is a dividing process, but no merging process.
  • 17.
    A randomized algorithmfor closest pair finding  Input: A set S consisting of n elements x1, x2,…, xn, where S⊆ R2. Output: The closest pair in S. xi , xi ,..., xi Step 1: Randomly choose a set S1={ } where m=n2/3. Find the closest pair of S1 and let the distance between this pair of points be denoted as δ . Step 2: Construct a set of squares T with meshsize δ.  1 2 m
  • 18.
    Step 3: Constructfour sets of squares T1, T2, T3 and T4 derived from T by doubling the meshsize to 2δ . Step 4: For each Ti, find the induced decomposition S=S1(i) ∪S2(i) ∪ … ∪ Sj(i), 1≤i ≤4, where Sj(i) is a non-empty intersection of S with a square of Ti. Step 5: For each xp, xq∈Sj(i), compute d(xp, xq). Let xa and xb be the pair of points with the shortest distance among these pairs. Return xa and xb as the closest pair.
  • 19.
    An example  27 points. S1= {x1, x2, …, x9}, δ = d(x1, x2) 6δ 5δ X4 X3 4δ X5 X2 3δ X9 X1 2δ X7 X6 δ X8 δ 2δ 3δ 4δ 5δ 6δ
  • 20.
  • 21.
  • 22.
    Time complexity   Time complexity:O(n) in average step 1: O(n) method : Recursively apply the algorithm once, 2 i.e. randomly choose ( n ) 3 2 3 =n 2 4 9 points from the n 3 points, then solve it with a straightforward method for the n   4 9 8 points : O( n 9 ) step 2 ~ Step 4: O(n) step 5: O(n)with probability 1-2e-cn 1 6
  • 23.
    Analysis of Step5  How many distance computations in step 5? δ : mesh-size in step 5 T: partition in step 5 N(T): # of distance computations in partition T Fact: There exists a particular partition T0, whose mesh-size is δ0 such that (1) N(T0) ≤ c0n. (2) the probability that δ ≤ 2δ 0 is 1 - 2e 1 − cn 6 .
  • 24.
     Construct T1, T2,…, T16 δ 4δ0 mesh-size: 4δ0   The probability that each square in T falls into at least1 one square of Ti , i = 1, 2, …, 16 is − cn 6 1 - 2e . 1 The probability that 16 − cn 6 N (T ) ≤ ∑ N (Ti ) is 1 - 2e . i =1
  • 25.
     Let the squarein T0 with the largest number of elements among the 16 squares have k elements. k ( k −1) 16k (16k −1) 2 = O ( k ), = O(k 2 ) 2 2  N(T0) ≤ c0n => N(Ti ) ≤ cin 16 N (T ) ≤ ∑ N (Ti ) = O( n ) with 4δ0 δ0 Ti: i =1 probability 1 - 2e 1 − cn 6 . k
  • 26.
    A randomized algorithmto test whether a number is prime.   This problem is very difficult and no polynomial algorithm has been found to solve this problem Traditional method: use 2,3,… N to test whether N is prime. input size of N : B=log2N (binary representation) N =2B/2, exponential function of B Thus N can not be viewed as a polynomial function of the input size.
  • 27.
    Randomized prime number testingalgorithm Input: A positive number N, and a parameter m.  Output: Whether N is a prime or not, with probability of being correct at least 1-ε = 1-2-m. Step 1: Randomly choose m numbers b1, b2, …, bm, 1≤ b1, b2, …, bm <N, where m≥log2(1/ε).  Step 2: For each bi, test whether W(bi) holds where W(bi) is defined as follows: N 1 (1) biN-1 ≠ 1mod N or − 2j (2) ∃ j such taht = k is an integer and the greatest common divisor of (bi)k-1 and N is not 1 or N. If any W(bi) holds, then return N as a composite number, otherwise, return N as a prime.
  • 28.
    Examples for randomizedprime number testing  Example 1: N = 12 Randomly choose 2, 3, 7 212-1 = 2048 ≠ 1 mod 12 ⇒ 12 is a composite number.
  • 29.
     Example 2: N= 11 Randomly choose 2, 5, 7 (1) 211-1=1024≡1 mod 11 j=1, (N-1)/2j==5 GCD(25-1, 11) = 1 W(2) does not hold . (2) 511-1=9765625≡1 mod 11 GCD(55-1, 11) = 11 W(5) does not hold . (3) 711-1=282475249≡1 mod 11 GCD(75-1, 11) = 1 W(7) does not hold .  Thus, 11 is a prime number with the probability of correctness being at least 1-2-3= 7/8.
  • 30.
    Theorem for numbertheory  Theorem:  If W(b) holds for any 1≤ b<N, then N is a composite number .  If N is composite, then (N-1)/2 ≤ | { b | 1 ≤ b<N, W(b) holds } |.
  • 31.
    Pattern matching Pattern string: X length : n Text string : Y length : m, m ≥ n To find the first occurrence of X as a consecutive substring of Y . Assume that X and Y are binary strings.  e.g. X = 01001 , Y = 1010100111  X    Straightforward method : O(mn) Knuth-Morris-Pratt’s algorithm : O(m) The randomized algorithm : O(mk) with a mistake of small probability. (k:# of testings)
  • 32.
    Binary representation  X =x1 x2…xn∈{0,1} Y = y1 y2…ym∈{0,1} Let Y(i)=yi yi+1….yi+n-1  A match occurs if X=Y(i) for some i . Binary values of X and Y(i): B(X) = x12n-1 + x22n-2 + … + xn B(Y(i)) = yi2n-1+yi+12n-2+…+yi+n-1 , 1≤ i ≤ m-n+1
  • 33.
    Fingerprints of binarystrings    Let p be a randomly chosen prime number in {1,2,…,nt2}, where t = m - n + 1. Notation: (xi)p = xi mod p Fingerprints of X and Y(i): Bp(x) = ( ( ( (x12)p+x2)p2)p+x3)p2… Bp(Y(i)) = ( ( ( (yi2)p+yi+1)p2+yi+2)p2… ⇒ Bp(Y(i+1))= ( (Bp(Yi)-2n-1yi) 2+Yi+n)p =( ( (Bp(Yi)-( (2n-1)pyi)p )p 2)p +yi+n)p  If X=Y(i), then Bp(X) = Bp (Y(i)), but not vice versa.
  • 34.
    Examples for usingfingerprints  Example: X = 10110 , Y = 110110 n=5,m=6,t=m-n+1=2 suppose P=3. Bp(X) = (22)3 = 1 Bp(Y(1)) = (27)3 = 0 ⇒ X≠Y(1) Bp(Y(2)) = ( (0-24)3 2+0)3 = 1 ⇒ X = Y(2)
  • 35.
     e.g. X =10110 , Y = 10011 , P = 3 Bp(X) = (22)3 = 1 Bp(Y(1)) = (19)3 = 1 ⇒ X= Y(1) WRONG!   If Bp(X) ≠ Bp(Y(i)), then X ≠ Y(i) . If Bp(X) = Bp(Y(i)), we may do a bit by bit checking or compute k different fingerprints by using k different prime numbers in {1,2,… nt2} .
  • 36.
    A randomized algorithmfor pattern matching   Input: A pattern X = x1 x2…xn, a text Y = y1 y2…ym and a parameter k. Output: (1) No, there is no consecutive substring in Y which matches with X. (2) Yes, Y(i) = yi yi+1.…yi+n-1 matches with X which is the first occurrence. If the answer is “No” , there is no mistake. If the answer is “Yes” , there is some probability that a mistake is made.
  • 37.
    Step 1: Randomlychoose k prime numbers p1, p2, …, pk from {1,2,…,nt2}, where t = m - n + 1. Step 2: i = 1. Step 3: j = 1. Step 4: If B(X)Pj ≠ (B(Yi))pj, then go to step 5. If j = k, return Y(i) as the answer. j = j + 1. Go to step 4. Step5: If i = t, return “No, there is no consecutive substring in Y which matches with X.” i = i + 1. Go to Step 3.
  • 38.
    An example forthe algorithm  X = 10110 , Y = 100111 , P1 = 3 , P2 = 5 B3(X) = (22)3 = 1 B5(X) = (22)5 = 2 B3(Y(2)) = (7)3 = 1 B5(y(2)) = (7)5 = 2 Choose one more prime number, P3 = 7 B7(x) = (22)7 = 1 B7(Y(2)) = (7)7 = 0 ⇒X ≠ Y(2)
  • 39.
    How often doesa mistake occur?    If a mistake occurs in X and Y(i), then B(X) - B(Y(i)) ≠ 0, and pj divides | B(X) - B(Y(i)) | for all pj’s. Let Q = π i where p j divides B ( X ) − B ( Y ( i )) B( X ) − B (Y (i )) Q<2n(m-n+1) reason: B(x)<2n, and at most (m-n+1) B(Y(i))’s 2n 2n…2n m-n-1
  • 40.
    Theorem for numbertheory     Theorem: If u≥29 and a<2u, then a has fewer than π(u) diffferent prime number divisors where π(u) is the number of prime numbers smaller than u. Assume nt ≥ 29 . Q < 2n(m-n+1) = 2nt ⇒ Q has fewer than π(nt) different prime number divisors. If pj is a prime number selected from {1, 2, …, M}, π nt ) ( the probability that pj divides Q is less than π M ). ( If k different prime numbers are selected from {1, 2, … nt2} , the probability that a mistake occurs is less than k  π (nt )   provided nt ≥ 29.  2    π (nt ) 
  • 41.
    An example formistake probability k  π ( nt )   2   π (nt )   How do we estimate    u u Theorem: For all u ≥ 17,  ≤ π (u ) ≤ 1.25506 ln u ln u   π (nt ) nt ln(nt 2 ) ≤ 1.25506 ⋅ ⋅ 2 π (nt ) ln nt nt 2 1.25506 ln(t ) = (1 + ) t ln(nt ) Example: n = 10 , m = 100 , t = m - n + 1 = 91 π (nt ) ≤ 0.0229 2 π ( nt ) Let k=4 (0.0229)4≈2.75×10-7 // very small
  • 42.
    Interactive proofs: methodI    Two persons: A : a spy B : the boss of A When A wants to talk to B , how does B know that A is the real A, not an enemy imitating A ? Method I : a trivial method B may ask the name of A’s mother (a private secret) Disadvantage: The enemy can collect the information, and imitate A the next time.
  • 43.
    Interactive proofs: methodII    Method II: B may send a Boolean formula to A and ask A to determine its satisfiability. (an NP-complete problem). It is assumed that A is a smart person and knows how to solve this NP-complete problem. B can check the answer and know whether A is the real A or not. Disadvantage: The enemy can study methods of mechanical theorem proving and sooner or later he can imitate A. In Methods I and II, A and B have revealed too much.
  • 44.
    A randomized algorithmfor interactive proofs   Method III: B can ask A to solve a quadratic nonresidue problem in which the data can be sent back and forth without revealing much information. Definition: GCD(x, y) = 1, y is a quadratic residue mod x if z2 ≡ y mod x for some z, 0 < z < x, GCD(x, z) = 1, and y is a quadratic nonresidue mod x if otherwise. (See the example on the next page.)
  • 45.
    An example forquadratic residue/nonresidue    Let QR = {(x, y) | y is a quadratic residue mod x} QNR = {(x, y) | y is a quadratic nonresidue mod x} Try to test x = 9, y = 7: 12 ≡ 1 mod 9 22 ≡ 4 mod 9 32 ≡ 0 mod 9 42 ≡ 7 mod 9 52 ≡ 7 mod 9 62 ≡ 0 mod 9 72 ≡ 4 mod 9 82 ≡ 1 mod 9 We have (9,1), (9,4), (9,7) ∈ QR but (9,5), (9,8) ∈ QNR
  • 46.
    Detailed method for interactiveproofs 1) 2) A and B know x and keep x confidential . B knows y. Action of B: Step 1: Randomly choose m bits: b1, b2, …, bm, where m is the length of the binary representation of x. Step 2: Find z1, z2, …, zm s.t. GCD(zi , x)=1 for all i . Step 3:Compute w1, w2, …, wm: wi ←zi2 mod x if bi=0 //(x, wi) ∈ QR wi ← (zi2y) mod x if bi=1 //(x, wi) ∈ NQR Step 4: Send w1, w2, …, wm to A.
  • 47.
    3) Action of A: Step1: Receive w1, w2, …, wm from B. Step 2: Compute c1, c2, …, cm: ci ←0 if (x, wi) ∈ QR ci ←1 if (x, wi) ∈ QNR Send c1, c2, …, cm to B. 4) Action of B: Step 1: Receive c1, c2, …, cm from A. Step 2: If (x, y) ∈ QNR and bi = ci for all i, then A is the real A (with probability 1-2-m).

Editor's Notes