Proof: Universal Family for Integers (Optional)

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Do curso por University of California San Diego

Data Structures

1713 classificações

University of California San Diego

Data Structures

1713 classificações

Curso 2 de 6 em Specialization Data Structures and Algorithms

A good algorithm usually comes together with a set of good data structures that allow the algorithm to manipulate the data efficiently. In this course, we consider the common data structures that are used in various computational problems. You will learn how these data structures are implemented in different programming languages and will practice implementing them in our programming assignments. This will help you to understand what is going on inside a particular built-in implementation of a data structure and what to expect from it. You will also learn typical use cases for these data structures. A few examples of questions that we are going to cover in this class are the following: 1. What is a good strategy of resizing a dynamic array? 2. How priority queues are implemented in C++, Java, and Python? 3. How to implement a hash table so that the amortized running time of all operations is O(1) on average? 4. What are good strategies to keep a binary tree balanced? You will also learn how services like Dropbox manage to upload some large files instantly and to save a lot of storage space!

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Hash Tables

In this module you will learn about very powerful and widely used technique called hashing. Its applications include implementation of programming languages, file systems, pattern search, distributed key-value storage and many more. You will learn how to implement data structures to store and modify sets of objects and mappings from one type of objects to another one. You will see that naive implementations either consume huge amount of memory or are slow, and then you will learn to implement hash tables that use linear memory and work in O(1) on average! In the end, you will learn how hash functions are used in modern disrtibuted systems and how they are used to optimize storage of services like Dropbox, Google Drive and Yandex Disk!

Phone Book Problem4:16

Phone Book Problem - Continued6:49

Universal Family9:42

Hashing Integers9:17

Proof: Upper Bound for Chain Length (Optional)8:43

Proof: Universal Family for Integers (Optional)11:40

Hashing Strings9:33

Hashing Strings - Cardinality Fix7:00

Conheça os instrutores

Alexander S. Kulikov
Visiting Professor
Department of Computer Science and Engineering
Michael Levin
Lecturer
Computer Science
Daniel M Kane
Assistant Professor
Department of Computer Science and Engineering / Department of Mathematics
Neil Rhodes
Adjunct Faculty
Computer Science and Engineering

Hi, in this video we will prove that the set of functions we suggested to you

with integers is really a universal family of hash functions.

This video will be heavy on math but it is optional.

We will use properties of prime numbers of modular arithmetics.

We will use the concept of one to one correspondence.

Properties of upper integral parts of real number and probabilities of course.

So the theorem is that the set of functions was suggested polygraphic H with

index B is the set of all hash functions indexed by B,

A, and B where B is a prime number and A and B are parameters.

And those parameters generate hash functions.

So the formula for the hash function is on the side, and parameters a and

b are from one to p-1 and from zero to p-1, respectively.

So this is a family of hash functions which contains p multiplied by p-1,

different hash functions for different bayers ab.

So these functions is the universal family for

the universe of keys, which are integers from 0 to p- 1.

And of course, it will be also universal family for any sub-set of this universe.

Now, let's prove this theorem.

To do that, we first need a Lemma.

So, select some hash function with parameters a and b from our set, a b.

And fix some keys, x and y which are different, then, the intermediate values,

which would compute while computing the hash value of keys x and y.

So ax + b modular p and ay + b modular p,

those are the intermediate values before we take the value modular m and

get the actual value of the hash function from our family.

So those values are always different for different keys in the universe.

We will prove this Lemma by contradiction.

So suppose is equal to s or (ax + b) is equal to (ay +b) mod p.

We subtract the right part from the left part and

we see that a(x-y) is equal to 0 mod p.

And that means that b divides product of a and x minus y.

Now recall that b is a prime number and so

it means that either b divides a or that b divides x minus y.

But a is between 1 and p minus 1 so a is a positive integer less than p and

p is prime so p cannot divide a and thus p must divide x minus y.

But x and y are between 0 and p minus 1.

So their difference is, by absolute value, less than p.

And p divides x minus y.

So, the only value for x minus y, for which it is possible is 0.

x minus y is 0.

And so x is equal to y.

But this contradicts our initial assumption that x and

y are different keys.

And so we prove by contradiction that r is not equal to s.

There is a corollary from that that for hash function h

which is without taking modular m, there are no collisions.

And an important corollary is that for

a hash function, which is before taking modulo m,

there are no collisions for the universe of integer keys from 0 to p minus 1.

Now we need another lemma.

That for any pair of different keys x and y,

there is a 1 to 1 correspondence between pairs of numbers a and

b which generally different hash functions in our family.

And pairs of numbers r and s, which are the intermediate values in

the computation of hash function of keys x and y.

Let's prove that.

First know that different pairs a, b generate different pairs r, s.

And that is because if we have some pair r, s we can

uniquely solve for a and b module b with these formulas.

So if there is any other pair, a prime b prime which leads to the same r and

s, then actually a prime and b prime have to be equal to a and

b respectively because of these equations.

So now we know that different pairs a and b generate different pairs r, s.

And we know that the total number of pairs a,b is p multiplied by p-1 because

there are p-1 values for a and independently p values for b.

So total number of pairs is p multiplied by p-1.

And also, it turns out that the total number of pairs r,s is also p multiplied

by p-1 Because the total number of pairs r,

s with values from 0 to p- 1 is p squared.

But r must be different from s so we must subtract p from that,

and that will be p squared- p which is the same as p multiplied by p- 1.

So the number of pairs a, b is the same as the number of pairs r, s.

And different pairs a,b led to different pairs (r,s) and that means that

there is a one to one correspondence between these pairs and these pairs.

And the corollary from that is that if we select some pair of different keys,

x and y and we select any hash function at random.

For all our family with equal probability, which is one of p multiplied

by p minus one, because they're all, p multiplied by p minus 1,

different hash functions in our family, then each pair of values are S.

Which are intermediate values, happen with equal probability,

again, one over P multiplied by p- 1 so why is that?

Well we know that there's a one to one correspondence between pairs a, b and

pairs r, s and selecting the hash function from our family is the same

selecting random pair a, b we know that the probability of any pair a,

b is one over p by p -1.

And so the probability of any pair r,s is the same as the probability of

the corresponding pair a,b.

So it is always 1/p(p- 1).

So we proved our corrollary.

Now let's prove the initial theorem.

The probability of collision for a fixed pair of keys, x and y, which are different

is the same as the probability that r module m is the same as s module m.

We know that r is different from s, but

they still can be equal module m because m can be less than Pm.

In most cases it is actually less than b.

So we, again, know that each pair r, s has probability 1 over p by p minus 1.

And we'll estimate the probability of collision using this fact.

Know that for each fixed r from 0 to p minus 1, there are at most.

Upper integral part of p over m,

minus 1 such value of s that s is different from r and

r modular m is the same as s modular m.

So why is that?

Well, let's assume, without loss of generality, that r modular m is 0.

And look at the roll of numbers from 0-P-1

than the numbers which have the same remainder

module m as r are 0 M, 2M, 3M, 4M and so

on up to a paranted real part of P over M-1 multiplied by M.

We cannot multiply the next integer by m because that is already

bigger than p minus 1, so in total there are part of pure m,

different integers between 0 and

p minus 1 which have the same remainder modular m as r.

But, s also needs to be different from r.

So, the number of such s', less by 1.

And now, the probability of collision is equal to probability r module m,

equal to s module m, and that is, at most, some of probabilities

of all such pairs are s for which this equality holds.

So we sum over all different values of r from 0 to p-1 and

then we sum over all pairs which contain this r.

And for which s module m is the same as r module m.

We estimated that their at most integral part of p over m-1

such as for any fixed r.

And the probability of any final pair of this r and

any s with the same remainder module m is 1 over p

by p-1 so the total probability is at most this sum.

And the sum has exactly p and it has p in the denominator so

we can just remove the sum and the denominator and

we get upper integral part of p over m-1 divided by p-1.

Then we use the property of that upper integral part of p over m at most

p plus m minus one over m.

And then, when we make the common denominator, and remove

all the things that cancel out, we see that it is equal to 1 over m.

So we just proved that the probability of

r modular m being the same as s modular m is, at most, 1 over m.

And that proves that the probability of collision for

some fixed pair of different keys, x and y is, at most, 1 over m.

And that is the property of universal family that we needed to prove.

So we finally proved our theorem.

In conclusion, we've just proved that a suggested family is

really a universal family for integers, but now the important thing

that integers are not unbounded on the integers from 0 to p- 1.

So if you need to hash really, really big integers,

you will need to select a prime number that is bigger, even bigger than

the largest of those numbers and then you will get universal family for

that prime number and it will work for your integers.

We also proved in the previous video the upper bound for

expected chain length using universal family and constant amortized expected

running time of operations with hash table, also provided in the last lecture.

So now we have the whole picture.

We have a universal family for integers.

And we have proofs that if you use a universal family,

that your operations work fast.

So now we are confident that if you use the suggested family for

integers, then the operations with the hash tables storing

those integers will run fast in practice, at least on average.

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