Complexity and Big-O Notation [interview question]

An important question is: How efficient is an algorithm or piece of code? Efficiency covers lots of resources, including:

CPU (time) usage
memory usage
disk usage
network usage

All are important but we will mostly talk about CPU time in 367. Other classes will discuss other resources (e.g., disk usage may be an important topic in a database class).

Be careful to differentiate between:

Performance: how much time/memory/disk/... is actually used when a program is run. This depends on the machine, compiler, etc. as well as the code.
Complexity: how do the resource requirements of a program or algorithm scale, i.e., what happens as the size of the problem being solved gets larger.
Complexity affects performance but not the other way around.
The time required by a method is proportional to the number of "basic operations" that it performs. Here are some examples of basic operations:

one arithmetic operation (e.g., +, *).
one assignment
one test (e.g., x == 0)
one read
one write (of a primitive type)
Some methods perform the same number of operations every time they are called. For example, the size method of the Sequence class always performs just one operation: return numItems; the number of operations is independent of the size of the sequence. We say that methods like this (that always perform a fixed number of basic operations) require constant time.

Other methods may perform different numbers of operations, depending on the value of a parameter or a field. For example, for many of the Sequence methods, the number of operations depends on the size of the sequence. We call the important factor (the parameters and/or fields whose values affect the number of operations performed) the problem size or the input size.

When we consider the complexity of a method, we don't really care about the exact number of operations that are performed; instead, we care about how the number of operations relates to the problem size. If the problem size doubles, does the number of operations stay the same? double? increase in some other way? For constant-time methods like the size method, doubling the problem size does not affect the number of operations (which stays the same).

Furthermore, we are usually interested in the worst case: what is the most operations that might be performed for a given problem size (other cases -- best case and average case -- are discussed below). For example, the addBefore method (for a sequence implemented using an array) has to move the current item and all of the items that come after the current item one place to the right in the array. In the worst case, all of the items in the array must be moved. Therefore, in the worst case, the time for addBefore is proportional to the number of items in the sequence, and we say that the worst-case time for addBefore is linear in the number of items in the sequence. For a linear-time method, if the problem size doubles, the number of operations also doubles.

Constant and linear times are not the only possibilities. For example, consider method createSeq:

Sequence createSeq( int N ) {
Sequence s = new Sequence();
for (int k=1; k<=N; k++) s.addBefore(new Integer(k));
return s;
}
Note that, for a given N, the for-loop above is equivalent to:

s.addBefore( new Integer(1) );
s.addBefore( new Integer(2) );
s.addBefore( new Integer(3) );
...
s.addBefore( new Integer(N) );
As discussed above, the number of operations for addBefore is proportional to the number of items in the sequence when addBefore is called. For the N calls shown above, the sequence lengths are: 0, 1, 2, ..., N-1. So what is the total time for all N calls? It is proportional to 0 + 1 + 2 + ... + N-1.

Recall that we don't care about the exact time, just how the time depends on the problem size. For method createSeq, the "problem size" is the value of N (because the number of operations will be different for different values of N). It is clear that the time for the N calls (and therefore the time for method createSeq) is not independent of N (so createSeq is not a constant-time method). Is it proportional to N (linear in N)? That would mean that doubling N would double the number of operations performed by createSeq. Here's a table showing the value of 0+1+2+...+(N-1) for some different values of N:


N 0+1+2+...+(N-1)
4 6
8 28
16 120

Clearly, the value of the sum does more than double when the value of N doubles, so createSeq is not linear in N. In the following graph, the bars represent the lengths of the sequence (0, 1, 2, ..., N-1) for each of the N calls.



The value of the sum (0+1+2+...+(N-1)) is the sum of the areas of the individual bars. You can see that the bars fill about half of the square. The whole square is an N-by-N square, so its area is N2; therefore, the sum of the areas of the bars is about N2/2. In other words, the time for method createSequence is proportional to the square of the problem size; if the problem size doubles, the number of operations will quadruple. We say that the worst-case time for createSeq is quadratic in the problem size.

Big-O Notation
We express complexity using big-O notation. For a problem of size N:

a constant-time method is "order 1": O(1)
a linear-time method is "order N": O(N)
a quadratic-time method is "order N squared": O(N2)
Note that the big-O expressions do not have constants or low-order terms. This is because, when N gets large enough, constants and low-order terms don't matter (a constant-time method will be faster than a linear-time method, which will be faster than a quadratic-time method). See below for an example.

Formal definition:

A function T(N) is O(F(N)) if for some constant c and for values of N greater than some value n0:
T(N) <= c * F(N)
The idea is that T(N) is the exact complexity of a method or algorithm as a function of the problem size N, and that F(N) is an upper-bound on that complexity (i.e., the actual time/space or whatever for a problem of size N will be no worse than F(N)). In practice, we want the smallest F(N) -- the least upper bound on the actual complexity.

For example, consider T(N) = 3 * N2 + 5. We can show that T(N) is O(N2) by choosing c = 4 and n0 = 2. This is because for all values of N greater than 2:

3 * N2 + 5 <= 4 * N2
T(N) is not O(N), because whatever constant c and value n0 you choose, I can always find a value of N greater than n0 so that 3 * N2 + 5 is greater than c * N.

How to Determine Complexities
In general, how can you determine the running time of a piece of code? The answer is that it depends on what kinds of statements are used.

Sequence of statements
statement 1;
statement 2;
...
statement k;
(Note: this is code that really is exactly k statements; this is not an unrolled loop like the N calls to addBefore shown above.) The total time is found by adding the times for all statements:
total time = time(statement 1) + time(statement 2) + ... + time(statement k)

If each statement is "simple" (only involves basic operations) then the time for each statement is constant and the total time is also constant: O(1). In the following examples, assume the statements are simple unless noted otherwise.

if-then-else statements
if (cond) {
sequence of statements 1
}
else {
sequence of statements 2
}
Here, either sequence 1 will execute, or sequence 2 will execute. Therefore, the worst-case time is the slowest of the two possibilities: max(time(sequence 1), time(sequence 2)). For example, if sequence 1 is O(N) and sequence 2 is O(1) the worst-case time for the whole if-then-else statement would be O(N).

for loops
for (i = 0; i < N; i++) {
sequence of statements
}
The loop executes N times, so the sequence of statements also executes N times. Since we assume the statements are O(1), the total time for the for loop is N * O(1), which is O(N) overall.

Nested loops
for (i = 0; i < N; i++) {
for (j = 0; j < M; j++) {
sequence of statements
}
}
The outer loop executes N times. Every time the outer loop executes, the inner loop executes M times. As a result, the statements in the inner loop execute a total of N * M times. Thus, the complexity is O(N * M). In a common special case where the stopping condition of the inner loop is j < N instead of j < M (i.e., the inner loop also executes N times), the total complexity for the two loops is O(N2).
Statements with method calls:
When a statement involves a method call, the complexity of the statement includes the complexity of the method call. Assume that you know that method f takes constant time, and that method g takes time proportional to (linear in) the value of its parameter k. Then the statements below have the time complexities indicated.


f(k); // O(1)
g(k); // O(k)
When a loop is involved, the same rule applies. For example:
for (j = 0; j < N; j++) g(N);
has complexity (N2). The loop executes N times and each method call g(N) is complexity O(N).

Some methods may require different amounts of time on different calls, even when the problem size is the same for both calls. For example, we know that if addBefore is called with a sequence of length N, it may require time proportional to N (to move all of the items and/or to expand the array). This is what happens in the worst case. However, when the current item is the last item in the sequence, and the array is not full, addBefore will only have to move one item, so in that case its time is independent of the length of the sequence; i.e., constant time.

In general, we may want to consider the best and average time requirements of a method as well as its worst-case time requirements. Which is considered the most important will depend on several factors. For example, if a method is part of a time-critical system like one that controls an airplane, the worst-case times are probably the most important (if the plane is flying towards a mountain and the controlling program can't make the next course correction until it has performed a computation, then the best-case and average-case times for that computation are not relevant -- the computation needs to be guaranteed to be fast enough to finish before the plane hits the mountain).

On the other hand, if occasionally waiting a long time for an answer is merely inconvenient (as opposed to life-threatening), it may be better to use an algorithm with a slow worst-case time and a fast average-case time, rather than one with so-so times in both the average and worst cases.

For addBefore, for a sequence of length N, the worst-case time is O(N), the best-case time is O(1), and the average-case time (assuming that each item is equally likely to be the current item) is O(N), because on average, N/2 items will need to be moved.

Note that calculating the average-case time for a method can be tricky. You need to consider all possible values for the important factors, and whether they will be distributed evenly.


When do Constants Matter?
Recall that when we use big-O notation, we drop constants and low-order terms. This is because when the problem size gets sufficiently large, those terms don't matter. However, this means that two algorithms can have the same big-O time complexity, even though one is always faster than the other. For example, suppose algorithm 1 requires N2 time, and algorithm 2 requires 10 * N2 + N time. For both algorithms, the time is O(N2), but algorithm 1 will always be faster than algorithm 2. In this case, the constants and low-order terms do matter in terms of which algorithm is actually faster.

However, it is important to note that constants do not matter in terms of the question of how an algorithm "scales" (i.e., how does the algorithm's time change when the problem size doubles). Although an algorithm that requires N2 time will always be faster than an algorithm that requires 10*N2 time, for both algorithms, if the problem size doubles, the actual time will quadruple.

When two algorithms have different big-O time complexity, the constants and low-order terms only matter when the problem size is small. For example, even if there are large constants involved, a linear-time algorithm will always eventually be faster than a quadratic-time algorithm. This is illustrated in the following table, which shows the value of 100*N (a time that is linear in N) and the value of N2/100 (a time that is quadratic in N) for some values of N. For values of N less than 104, the quadratic time is smaller than the linear time. However, for all values of N greater than 104, the linear time is smaller.


N 100*N N2/100
102 104 102
103 105 104
104 106 106
105 107 108
106 108 1010
107 109 1012


Best-case and Average-case Complexity

Class not registered in IE7

Every time when i open my IE7, and browsed certain pages, the browser will showed "Class not registered" javascript error. Open a command window, and try re-registering the IE dlls with a batch file like this:

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Something we do need to know what hardware specification for our PC without open the CPU casing. There is a simple way to do it:

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2. Unzip the file and run the cpuz.exe

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Thunderbird Certificate Error

Everytime when you send out an email using Thunderbird, and there is an error message telling that there is a Certificate error. To remove this message , please do following:

Step 1: Down the thunderbird "remember_mismatched_domains-1.4.6-fx+tb+sm.xpi" addon
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Backup Thunderbird email

(1) Find the profile Thunderbird folderThey are located in different places for different versions of Windows and they may be assigned random file names that make them difficult to recognize.
On Windows 2000/XP/2003/Vista machines the location for your Thunderbird profile is:

C:\Documents and Settings\\Application Data\Thunderbird\Profiles\\
On Windows 9x/Me PCs it can usually be found at:C:\Windows\Application Data\Thunderbird\Profiles\\

If you can't locate your profiles then check out this document for more information:

http://kb.mozillazine.org/Profile_Folder

(2) Copy the Thunderbird profile folder - Done!To back these up just copy the profiles to another location, drive, CD or USB stick. It's as simple as that! You can do it manually but you cab also use Windows Scheduler to do the job automatically with a batch file (for experts).

Remove "welcome to Thunderbird" page

1. In Thunderbird, Go to Tools> Options> General.

2. Uncheck the option for Thunderbird Start Page

Force Thunderbird to send HTML

Force Thunderbird to send HTML

Step 1:go to “Tools”->”Options”

Step 2Make sure you are in “Composition” tab and click on the “Send Options”

Step 3Change the default setting from “Ask me what to do” to “Send the message in both plain text and HTML”

Thunderbird : Potentially executable content

Error Message:
An error occurred while sending mail. The mail server responded: Potentially executable content. If you meant to send this file then please package it up as a zip and try again.
Forward Inline or As Attachment by Default
To change whether Mozilla Thunderbird inserts the forwarded message as an attachment or inline in the new email:

1. Select Tools Options... (or Thunderbird Preferences...) from the menu.
2. Go to the Composition category.
3. Make sure you're on the General tab.
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