Celebration Party Problem

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Algorithmic Toolbox

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University of California San Diego

Algorithmic Toolbox

4097 个评分

课程 1（共 6 门，Specialization 数据结构与算法）

The course covers basic algorithmic techniques and ideas for computational problems arising frequently in practical applications: sorting and searching, divide and conquer, greedy algorithms, dynamic programming. We will learn a lot of theory: how to sort data and how it helps for searching; how to break a large problem into pieces and solve them recursively; when it makes sense to proceed greedily; how dynamic programming is used in genomic studies. You will practice solving computational problems, designing new algorithms, and implementing solutions efficiently (so that they run in less than a second).

从本节课中

Greedy Algorithms

In this module you will learn about seemingly naïve yet powerful class of algorithms called greedy algorithms. After you will learn the key idea behind the greedy algorithms, you may feel that they represent the algorithmic Swiss army knife that can be applied to solve nearly all programming challenges in this course. But be warned: with a few exceptions that we will cover, this intuitive idea rarely works in practice! For this reason, it is important to prove that a greedy algorithm always produces an optimal solution before using this algorithm. In the end of this module, we will test your intuition and taste for greedy algorithms by offering several programming challenges.

Celebration Party Problem6:57

Efficient Algorithm for Grouping Children5:26

Analysis and Implementation of the Efficient Algorithm5:25

与讲师见面

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

Hi, in this lesson we will discuss the problem of

organizing children into groups.

And you will learn that if you use a naive algorithm to solve this problem,

it will work very,

very slowly, because the running time of this algorithm is exponential.

But later in the next lesson, we will be able to improve the training time

significantly by coming up with a polynomial time algorithm.

Let's consider the following situation.

You've invited a lot of children to a celebration party, and

you want to entertain them and also teach them something in the process.

You are going to hire a few teachers and

divide the children into groups and assign a teacher to each of the groups

this teacher will work with this group through the whole party.

But you know that for a teacher to work with a group of children efficiently

children of that group should be of relatively the same age.

More specifically age of any two children in the same group

should differ by at most one year.

Also, you want to minimize the number of groups.

Because you want to hire fewer teachers, and

spend the money on presents and other kinds of entertainment for the children.

So, you need to divide children into the minimum possible number of groups.

Such that the age of any two children in any group differs by at most one year.

Now, let's look at the pseudo code for the naive algorithm that solves this problem.

Basically, this algorithm will consider every possible partition

of the children into groups and find the partition which both

satisfies the property that the ages of the children

in any group should differ by at most one and contains the minimum number of groups.

We start with assigning the initial value of the number of

groups to the answer m and this initial value is just the number of children.

Because we can always divide all the children into groups of one, and

then of course each group has only one child so the condition is satisfied.

Then we consider every possible partition of all children into groups.

The number of groups can be variable, so this is denoted by a number k,

and we have groups G1, G2 and up to Gk.

And then we have a partition,

we first need to check whether it's a good partition or not.

So, we have a variable good which we assigned to true initially,

because we think that maybe this partition will be good.

But then we need to check for each group whether it satisfies our condition or not.

So, we go in a for group with index i of the group from 1 to k,

and then we consider the particular group GI, and

we need to determine whether all the children in this group differ by at most

1 year, or there are two children that differ more.

To check that, it is sufficient to compare the youngest child with the oldest child.

If their ages differ more than by one, then the group is bad.

Otherwise, every two children differ by at most one year, so the group is good.

And so we go through all the groups in a for loop.

If at least one of the groups is bad,

then our variable good will contain value false by the end.

Otherwise, all the groups are good, and the variable good will contain value true.

So, after this for loop, we check the value of the variable good, and

if it's true, then we improve our answer.

At least try to improve it.

With a minimum of its current value and

the number of the groups in the current partition.

And so, by the end of the outer for loop which goes through all the partitions,

our variable m will contain the minimum possible number of groups in a partition

that satisfies all the conditions.

It is obvious that this algorithm works correctly

because it basically considers all the possible variants and

selects the best one from all the variants which satisfy our condition on the groups.

Now, let us estimate the running time of this algorithm.

And I state that the number of operations that this algorithm makes

is at least 2 to the power of n, where n is the number of children in C.

Actually, this algorithm works much slower and

makes much more operations than 2 to the power of n, but

we will just prove this lower bound to show that this algorithm is very slow.

To prove it, let's consider just partitions of the children in two groups.

Of course there are much more partitions than that.

We can divide them in two, three, four, and so on.

Much more groups.

But, let's just consider partitions in two groups and

prove that even the number of such partitions is already

at least two to the power of n.

Really, if C is a union of two groups,

G1 and G2, then basically we can make such partition for

any G1 which is a subset of such C of all children.

For any G1, just make

group G2 containing all the children which are not in the first group.

And then all the children will be divided into these two groups.

So, now the size of the set of all children is n.

And if you want to compute the number of possible groups G1 then we should

note that each item of the set, or

each child, can be either included or excluded from the group G1.

So, there can be 2 to the power of n different

groups G1. And so there are at least 2 to the power of n

partitions of the set of all children in two groups.

and it means that our algorithm will do at

least 2 to the power of n operations because this considers every partition.

Among all the partitions, there are all the partitions into two groups.

So, how long will it actually work?

We see that the Naive algorithm works in time Omega (2n),

so it makes at least 2 to the power of n operations.

And for example for just 50 children this is at least 2 to the power of 50 or

the larges number which is on the slide.

This is the number of operations that we will need to make and I estimate that

on a regular computer, this will take at least two weeks for you to compute

this if this was exactly the number of operations that you would need.

So, it works really, really slow.

But in the next lesson we will improve this significantly.

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