Hashing Strings

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From the course by University of California San Diego

Data Structures

1564 ratings

University of California San Diego

Data Structures

1564 ratings

Course 2 of 6 in the 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!

From the lesson

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

Meet the Instructors

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 the previous videos,

you've learned how to quickly look up name in your phonebook given the phone number.

And we want to learn to solve the reverse problem given a name,

look up a phone number of the corresponding person.

To do that, we need to implement the Map from names to phone numbers.

And we can again use hash tables and

we can again use chaining as in the previous sections.

But we need to design a hash function that is defined names.

And more generally, we want to learn to hash arbitrary strings of characters.

And by the way in this video, you will also learn how hashing of strings and

implemented in the Java programming language.

But first, let's introduce a new notation.

Denote by lSl enclosed in vertical lines the length of string S.

For example, the length of string l"a"l is 1,

length of string l"ab"l is 2, and length of string l"abcde"l is 5.

So now how do hash strings?

Well when we're given a string, we're actually given

a sequence of characters from S[0] to S length of S- 1.

We number the characters of the strings from 0 in this lecture.

And S[i] Is an individual character that is in the i-th position in the string.

I say that we should use all the characters

when we compute our hash function of a string.

Indeed, if we don't use the first character, there will be many collisions.

For example, if the first symbol of the string is not used, then the hash value

of strings ("aa"), ("ba") and so on, up to ("za") will be the same.

Because however we compute the value of the hash function,

it doesn't use the value of the first character.

And if everything else in the strings stays the same, and

we only change the first character that doesn't influence the value of the hash

function then the value of the hash function must be the same.

And so there will be a lot of collisions and we want to avoid collisions.

So we need to use value of each of the characters.

Now, we could do a lot of things with that.

For example, sum the values of all the characters or

multiply them, but we'll do something different.

Well first, to even compute something on a string,

we need to convert each character of the string to an integer code.

For example, that can be ASCII code or Unicode,

corresponding to that symbol on your computer, that doesn't really matter.

And also we'll again need to choose a big prime number p,

the same as we used in the integer hashing.

So suppose we've chosen some big prime number p, now we introduce

a new family of hash functions called polynomial family of hash functions.

So calligraphy p is the family of hash functions which is index by small p,

which is our big prime number.

And also index by x, and x is a parameter which changes from 1 to p- 1.

So the value of a hash function index by p and

x on a string S, is the following sum.

It is a polynomial sum where we multiply the integer quote corresponding to

the ith character of S, which is noted by S of i, the same as the character itself.

We multiply it by x to the power of i.

We sum all these things up, and we take the value modular p.

So this is a family of hash functions, and

the chronology of all those hash functions is p.

So any such hash function returns value from 0 to p- 1.

And how many hash functions are there in this family?

Well of course, there are exactly p- 1 different hash functions,

because to choose to define a hash function from this family you would just

need to choose the value of x.

And x changes from 1 to p- 1, and it's an integer number of course.

So how can we implement a hash function from this family?

S, the procedure PolyHash which takes it's input string S,

prime number p and parameter x, implements the hash function from our peril.

It starts with the signing values of 0 to the result to the hash value will

return to end.

And then it will go from right to left in our string and

compute new value based on the value of the corresponding character.

And there is a formula in the code that does exactly that.

And I will show you by example that what we get in the end

by applying this formula is exactly what we want.

So basically, we start with a hash value of 0, and

then we start with i equal to 2 if the length of our string S is 3.

We start with length of S- 1 which is 2.

We have current value of hash = 0.

So we multiply the 0 by x and get 0,

then we add the value of S[i] which is S[2], and take it mod p.

And so after first iteration of the for loop, we get S[2] mod p.

What happens is the next iteration, that i is decreased and i is now 1.

And we multiply the current value S[2] by x.

And we add s[1], and take everything modular p.

And what we get is the same as of S[1] + S[2] multiply by x modular p.

And then the last iteration, i is decreased to 0.

We multiply the current value by x.

What we get is S[1] multiply by x + S[2] multiply by x squared.

And then we also add S[0] to the sum and take everything modular p.

And the result is S[0] + S[1] multiply by x + S[2] multiply by x2,

exactly as we wanted.

A polynomial hash function, with prime P and prime parameter x.

And by the way, the implementation of the built in hash code methods in

the class stream in Java, is very similar to our procedure PolyHash.

The only difference is that, it always uses x = 31.

And for some technical reasons, it avoids the modular p operator It

just computes the polynomial sum without any modular division.

So now you know how a function that is used probably trillions of times a day by

thousands and many thousands of different programs,

how this function is implemented.

So now about the efficiency of our polynomial family.

First, Lemma says that for any two different strings s1 and

s2 of length at most L + 1.

If you choose a random hash function from the polynomial family by

selecting a random value of x, parameter x from 1 to p- 1.

You can select a random hash function from the family.

So if you select a random hash from the polynomial family, then the probability

of collision on these two different strings is at most L divided by p.

So that doesn't seem like a good estimate because L can be big,

but actually it is your power to choose p.

If you choose very, very big prime number p then L over p will be very small.

And know that it won't influence the running time of the PolyHash procedure,

because the running time of this procedure is big length of S.

It only depends on the length of the string.

It doesn't depend on the length of number p more or less.

So if you select a really big number p, then the probability of collision

will be very small and the hash function will still be computed very fast.

The idea of proof of this Lemma is that the equation polynomial

equation of power L, modular prime number p has at most L different solutions x.

Basically, when we consider two strings S1 and S2.

The fact that the hash value or

some hash function from the polynomial family is the same for

these two strings means that x corresponding to our

hash function is a solution of this kind of equation.

And the fact that strings are different makes sure that at least one of

the coefficients of this equation is different from 0, and that is essential.

If the strings were the same of course,

the value of any hash function on them will be the same.

But if they're different then the probability is at most L over p.

Because there are only L or less different x for

which the hash function can give the same value on these two strings.

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