The FAQ for Lab 9 is located here.

Before You Begin #

Pull the files for lab 9 from the skeleton.

The online textbook readings in for this lab are optional. The lab covers the same content as Chapter 8.2, 8.3, and 8.4 but more concisely so that there’s not as much to read. You still may want to refer back to the textbook if you prefer watching videos or reading more about the topic. Additionally there is overlap between this lab and todays’s lecture. Asymptotic analysis is both a tough topic and an important one, so this repetition is intentional and should hopefully help to reinforce the topics. However, feel free to skip around the spec if you feel like you already have a good understanding of the content.

Learning Goals #

“““““An engineer will do for a dime what any fool will do for a dollar.”

— Arthur M. Wellington” — Paul Hilfinger” — Zephyr Omaly” — Laksith Prabu” — Dominic Conricode

Efficiency comes in two flavors:

Programming cost
  • How long does it take to develop your programs?
  • How easy is it to read or modify your code?
  • How maintainable is your code?
Execution cost
  • Time complexity: How much time does your program take to execute?
  • Space complexity: How much memory does your program require?

We’ve already seen many examples of reducing programming cost. We’ve written unit tests and employed test-driven development to spend a little more time up front writing tests to save a lot of time down the line debugging programs. And we’ve seen how encapsulation can help reduce the cognitive load that a programmer needs to deal with by allowing them to think in terms of high-level abstractions like lists instead of having to deal with the nitty-gritty details of pointer manipulation. We’ve also discussed some elements of design for methods and classes, like how we prefer passing arguments around over maintaining static variables.

We’ve only just scratched the surface on methods for reducing and optimizing programming costs, but for the coming weeks, it’ll be helpful to have a working understanding of the idea of execution cost.

In this lab, we consider ways of measuring the efficiency of a given code segment. Given a function f, we want to find out how “quickly” that function runs.

Algorithms #

An algorithm is a step-by-step procedure for solving a problem, or an abstract notion that describes an approach for solving a problem. The code we write in this class, our programs, are implementations of algorithms. Indeed, we’ve already written many algorithms: the methods we’ve written for IntList, SLList, DLList, and AList are all algorithms that operate on data by storing and accessing that data in different ways.

As another example, consider the problem of sorting a list of numbers. One algorithm we might use to solve this problem is called bubble sort. Bubble sort tells us we can sort a list by repeatedly looping through the list and swapping adjacent items if they are out of order, until the entire sorting is complete.

Another algorithm we might use to solve this problem is called insertion sort. Insertion sort says to sort a list by looping through our list, taking out each item we find, and putting it into a new list in the correct order.

Several websites like VisuAlgo, Sorting Algorithms, and USF have developed some animations that can help us visualize these sorting algorithms. Spend a little time playing around with these demos to get an understanding of how much time it takes for bubble sort or insertion sort to completely sort a list. We’ll revisit sorting in more detail later in this course, but for now, try to get a feeling of how long each algorithm takes to sort a list. How many comparison does each sort need? And how many swaps?

Since each comparison and each swap takes time, we want to know which is the faster algorithm: bubble sort or insertion sort? And how fast can we make our Java programs that implement them? Much of our subsequent work in this course will involve estimating program efficiency and differentiating between fast algorithms and slow algorithms in their specific settings. This set of activities introduces an approach to making these estimates.

Measuring Execution Time #

One way to estimate the time an algorithm takes is to measure the time it takes to run the program directly. Each computer has an internal clock that keeps track of time, usually in the number of fractions of a second that have elapsed since a given base date. The Java method that accesses the clock is System.currentTimeMillis. A Timer class is provided in Timer.java.

Take some time now to find out exactly what value System.currentTimeMillis returns, and how to use the Timer class to measure the time taken by a given segment of code.

Exercise: Sorter #

The file Sorter.java contains a version of the insertion sort algorithm mentioned earlier. Its main method uses a command-line argument to determine how many items to sort. It fills an array of the specified size with randomly generated values, starts a timer, sorts the array, and prints the elapsed time when the sorting is finished.

javac Sorter.java
java Sorter 300

Compiling and running Sorter like above will tell us exactly how long it takes to sort an array of 300 randomly chosen elements.

Alternatively, you can set command line arguments in IntelliJ by clicking on the class name next to the green run button on the top left and selecting “Edit Configurations.” Next, you can enter command line arguments into the “Program arguments” field.

Visual guide: adding program arguments #

Edit run configurations

Set command line arguments

By compiling and running Sorter.java with different arguments, determine the size of the smallest array that needs 1 second (1000 milliseconds) to sort.

You may notice that other students in lab end up with different timing results and a different number of elements. What factors might contribute to these differences?

Counting Steps #

From timing the program in the previous example, we learned it isn’t very helpful in determining how good an algorithm is; different computers end up with different results! Additionally, we won’t know the runtime of our algorithm until after we’ve run the timer - we can’t make any predictions about runtime with a timer. An alternative approach is step counting. The number of times a given step, or instruction, in a program gets executed is independent of the computer on which the program is run and is similar for programs coded in related languages. With step counting, we can now formally and deterministically describe the runtime of programs.

We define a single step as the execution of a single instruction or primitive function call. For example, the + operator which adds two numbers is considered a single step. We can see that 1 + 2 + 3 can be broken down into two steps where the first is 1 + 2 while the second takes that result and adds it to 3. From here, we can combine simple steps into larger and more complex expressions.

(3 + 3 * 8) % 3

This expression takes 3 steps to complete: one for multiplication, one for addition, and one for the modulus of the result.

From expressions, we can construct statements. An assignment statement, for instance, combines an expression with one more step to assign the result of the expression to a variable.

int a = 4 * 6;
int b = 9 * (a - 24) / (9 - 7);

In the example above, each assignment statement takes one additional step on top of however many steps it took to compute the right-hand side expressions. In this case, the first assignment to a takes one step to compute 4 * 6 and one more step to assign the result, 24, to the variable a. How many steps does it take to finish the assignment to b?

Here are some rules about what we count as taking a single step to compute:

  • Assignment and variable declaration statements
  • All unary (like negation) and binary (like addition, and, or) operators
  • Conditional if statements
  • Function calls
  • return statements

One important case to be aware of is that, while calling a function takes a single step to setup, executing the body of the function may require us to do much more than a single step of work.

Counting Conditionals #

With conditional statements like if statements, the total step count depends on the outcome of the condition we are testing.

if (a > b) {
    temp = a;
    a = b;
    b = temp;

The example above can take four steps to execute: one for evaluating the conditional expression a > b and three steps for evaluating each line in the body of the condition. But this is only the case if a > b. If the condition is not true, then it only takes one step to check the conditional expression.

That leads us to consider two quantities: the worst case count, or the maximum number of steps a program can execute, and the best case count, or the minimum number of steps a program needs to execute. The worst case count for the program segment above is 4 and the best case count is 1.

if ... else Counting #

Consider an if ... else of the form,

if (A) {
} else {

where A, B, and C are program segments. (A might be a method call, for instance.)

How many steps does it take to evaluate the entire if ... else block in terms of the number of steps it takes to evaluate A, B, and C? Think back to your practice tracing through programs to figure out which parts of the conditional will be evaluated (given the condition is true or false), and which parts won’t be evaluated.

Loop Counting #

for (int k = 0; k < N; k++) {
    sum = sum + 1;

In terms of \(N\), how many operations are executed in this loop? Remember that each of the actions in the for-loop header (the initialization of k, the exit condition, and the increment) count too!

It takes 1 step to execute the initialization, int k = 0. Then, to execute the loop, we have the following sequence of steps:

  • Check the loop condition, k < N
  • Add 1 to the sum
  • Update the value of sum by assignment
  • Increment the loop counter, k
  • Update k by assignment

This accounts for the first \(1 + 5N\) steps. In the very last iteration of the loop, after we increment k such that k now equals \(N\), we spend one more step checking the loop condition again to figure out that we need to finally exit the loop so the final number of steps is \(1 + 5N + 1\).

Example: remove #

Now consider code for the remove method, which removes the item at a given position of an array values by shifting over all the remaining elements. We notice here that things become slightly more complicated as the number of steps performed matters both on pos and len, the number of items contained in values. We now count the number of steps performed in terms of these two variables.

This example assumes that you are using an array to back some kind of list. This idea is similar to Project 1, but this implementation is not circular. Also assume that the length of the underlying array is much larger than length and will not have an effect on the runtime of this program.

The list will have the instance variables values which is the array backing the list and len which corresponds to the number of items actually contained in the list (not necessarily the same as values.length).

void remove(int pos) {
    for (int k = pos + 1; k < len; k++) {
        values[k - 1] = values[k];
    len -= 1;

Each column in the table below shows the total number of steps for computing each value of pos. These counts are written as a function of len. This way we can come up with total counts for the number of steps parameterized by len and pos.

category pos = 0 pos = 1 pos = 2 pos = len - 1
pos + 1 1 1 1   1
assignment to k 1 1 1   1
loop conditional len len - 1 len - 2   1
increment to k len - 1 len - 2 len - 3   0
update k len - 1 len - 2 len - 3   0
array access len - 1 len - 2 len - 3   0
array assignment len - 1 len - 2 len - 3   0
decrement to len 1 1 1   1
assignment to len 1 1 1   1
Total count 5 * len 5 * len - 5 5 * len - 10   5

We can summarize these results as follows: a call to remove with argument pos requires in total:

  • 1 step to calculate pos + 1
  • 1 step to make the initial assignment to k
  • len - pos loop tests
  • len - pos - 1 increments of k
  • len - pos - 1 reassignments to k
  • len - pos - 1 accesses to values elements
  • len - pos - 1 assignments to values elements
  • 1 step to decrement len
  • 1 step to reassign to len

If all these operations take roughly the same amount of time, the total is 5 * (len - pos). Notice how we write the number of statements as a function of the input arguments.

Although len is not a parameter of the remove method it is still considered to be an input as its value affects the number of steps.

More formally we can also see that since remove is a non-static method, we are implicitly passing in the variable this to our remove method. This means that the variables this.values and this.length are also passed in.

Comparing across a fixed value of len we can notice that for a small value of pos, the number of steps executed will be greater than if we had a larger value of pos (e.g. closer to len). And vice versa: a larger value of pos will reduce the number of steps we need to execute.

Example with Nested Loops: removeZeroes #

Counting steps in nested loops is a little more involved. As an example, we’ll consider an implementation of the method removeZeroes.

void removeZeroes() {
    int k = 0;
    while (k < len) {
        if (values[k] == 0) {
        } else {
            k += 1;

Intuitively, the code should be slowest on an input where we need to do the most work, or an array of values full of zeroes. Here, we can tell that there is a worst case (removing everything) and a best case (removing nothing). To calculate the runtime, like before, we start by creating a table to help organize information.

category best case worst case
assignment to k 1 1
loop conditional len + 1 len + 1
array accesses len len
comparisons len len
calls to remove 0 len
k + 1 len 0
update to k len 0

In the best case, we never call remove so its runtime is simply the sum of the rows in the “best case” column. Thus, the best-case count is 2 + 5 * len.

The only thing left to analyze is the worst-case scenario. Remember that the worst case makes len total calls to remove. We already approximated the cost of a call to remove for a given pos and len value earlier: 5 * (len - pos).

In our removals, pos is always 0, and only len is changing. The total cost of the removals is shown below.

\[[5 \cdot \texttt{len}] + [5 \cdot (\texttt{len} - 1)] + \cdots +[5 \cdot (1)] + [5 \cdot (0)]\] \[= 5 \cdot [(\texttt{len}) + (\texttt{len} - 1) + \cdots + (1) + (0)]\]

The challenge now is to simplify the expression. A handy summation formula to remember is the sum of the first \(k\) natural numbers.

\[1 + 2 + \cdots + k = \frac{k(k + 1)}{2}\]

This lets us simplify the cost of removals (note that above was exactly a multiple of this form with terms from 1 to len). Remembering to include the additional steps in the table, we can now express the worst-case count of removeZeroes as:

\[5 \cdot \frac{\texttt{len}(\texttt{len} + 1)}{2} + 3 \cdot \texttt{len} + 2\]

We often prefer to simplify this by substituting len for a symbolic name like \(N\).

\[5 \frac{N(N + 1)}{2} + 3N + 2\]

That took… a while.

Abbreviated Estimates #

From this section onwards, we present a set of fairly precise definitions, and we’ll be relying on the example developed in this and the previous part to build a solid definition for asymptotic notation. If you’re not fully comfortable with any of the material so far, now is the perfect time to review it with someone in your lab or your TA!

Producing step count figures even for those relatively simple program segments took a lot of work. But normally we don’t actually need an exact step count but rather just an estimate of how many steps will be executed.

In most cases, we only care about what happens for very large \(N\) as that’s where the differences between algorithms and their execution time really become limiting factors in the scalability of a program.

Why is this? Debate with someone in your lab. Then, see our thoughts below

Our take #

When our \(N\) is really small, no reasonable function will take a long time to execute! Imagine a program that takes \(N^{12}\) steps to execute. When \(N=3\), it doesn’t really matter! But when \(N \gg 3\) (\(N\) much larger than 3) we might start to get really impatient (or the universe might end)!

We want to consider what types of algorithms would best handle big amounts of data, such as in the examples listed below:

  • Simulation of billions of interacting particles
  • Social network with billions of users
  • Encoding billions of bytes of video data

Here we are using a generic value of \(N\) to represent the size of an input passed into a function. Note that we will not always use \(N\) to represent the size of a program input (though it is common).

Also notice that sometimes the program will be parameterized by more than just one input; however, for most examples in this class we will ask you to write the order of growth in terms of just one variable.

The asymptotic behavior of a function f (any one of the programs above, for example) is a description of how the execution time of f changes as the value of \(N\) grows increasingly large towards infinity. Our goal is to come up with a technique that can be used to compare and contrast two algorithms to identify which algorithm scales better for large values of \(N\).

We can then compute the order of growth of a program, a classification of how the execution time of the program changes as the size of the input grows larger and larger. For example, we say that the order of growth of \(2N + 3\) is \(N\) since, for large values of \(N\), \(2N + 3\) will be less than \(3N\) and slightly greater than \(2N\). As \(N\) tends towards infinity, the \(+ 3\) contributes less and less to the overall runtime.

This pattern holds for higher-order terms too. Applying this estimation technique to the removeZeroes method above results in the following orders of growth.

  • The order of growth for the best-case runtime of removeZeroes, \(5N + 2\), is proportional to the len, \(N\).
  • The order of growth for the worst-case runtime of removeZeroes, \(5 \frac{N(N + 1)}{2} + 3N + 2\), is \(N^2\).

The intuitive simplification being made here is that we discard all but the most significant term of the estimate and also any constant factor of that term. Later, we will see exactly why this is true with a more formal proof.

Asymptotic Analysis #

Recap: Simplified Analysis Process #

Rather than building the entire table of all the exact step counts, we can instead follow a simplified analysis process.

Choose a cost function
These are the underlying assumptions about the costs of each step or instruction for our machine. In this course, we’ll assume all of the basic operations (Java operators, assignment statements, return statements, array access) each take the same amount of time (1 unit to execute). In CS 61C, we’ll see how this fundamental assumption often isn’t true.
Compute the order of growth
Given the cost function, we can then compute the order of growth for a program. In the removeZeroes example, we saw how we could compute an exact count and then find the correct order of growth runtime classification for it by simplifying the expression.

Later, we’ll learn a few shortcuts and practice building intuition / inspection to determine orders of growth, but it’s helpful to remember that we’re always solving the same fundamental problem of measuring how long it takes to run a program, and how that runtime changes as we increase the value of \(N\).

Big-Theta Notation #

Computer scientists often use special notation to describe runtime. The first one we’ll learn is called big-theta, represented by the symbol \(\Theta\).

Suppose we have a function, \(R(N)\), with order of growth \(f(N)\). We could say,

\(R(N) \in \Theta(f(N))\), or “\(R(N)\) is in \(\Theta(f(N))\)”

Why do we say “in” \(\Theta\)? Formally, \(\Theta(f(N))\) is a family of functions that all grow proportional to \(f\). Here, proportional can be thought of as roughly equal. Thinking back to our working definition of order of growth as a method for classification, \(\Theta(f(N))\) refers to the entire set of all functions that share the same order of growth.

The advantage of using notation like big-theta is that it provides a common definition for asymptotic analysis which reduces the amount of explaining we need to do when we want to share our ideas with others. It also makes sure we’re all on the same page with the claims we make, so long as we use them carefully and precisely.

Learning new notation can be a little daunting, but we’ve actually already been making statements in big-theta terms. The first claim about removeZeroes that we made earlier,

The order of growth for the best-case runtime of removeZeroes, \(5N + 2\), is proportional to the length of the array, \(N\).

is essentially equivalent to the claim: In the best-case, removeZeroes is in \(\Theta(N)\).

And, likewise, the second claim that we made earlier,

The order of growth for the worst-case runtime of removeZeroes, \(5 \frac{N(N + 1)}{2} + 3N + 2\), is \(N^2\).

has its own equivalent in big-theta notation: In the worst-case, removeZeroes is in \(\Theta(N^2)\).

When we use removeZeroes here, we mean the runtime of the function rather than the function itself. In practice, we’ll often use this English shortcut as long as the meaning is clearly communicated, though it would be more accurate to say the runtime of the function.

Asymptotic Variables #

You may have observed, in our analysis of removeZeroes, that we were careful to make clear what the running time estimate depended on, namely the value of len and pos.

Unfortunately, people are sometimes careless about specifying the quantity on which an estimate depends. Don’t just use \(N\) without making clear what \(N\) means. This distinction is important especially when we begin to touch on sorting later in the course. It may not always be clear what \(N\) means.

We’ll often qualify our runtimes by stating, “where \(N\) is the length of the list”, but we often also say things like, “where \(N\) is the value of the largest number in the list”.

Asymptotic Bounds #

Formally, we say that \(R(N) \in \Theta(f(N))\) if and only if there exist positive constants \(k_1, k_2\) such that \(k_1 \cdot f(N) \leq R(N) \leq k_2 \cdot f(N)\) for all \(N\) greater than some \(N_0\) (very large \(N\)).

In other words, \(R(N)\) must be bounded above and below by \(f(N)\) asymptotically. But we’ve already seen something like this too.

We say that the order of growth of \(2N + 3\) is \(N\) since, for large values of \(N\), \(2N + 3\) will be less than \(3N\) and slightly greater than \(2N\). As \(N\) tends towards infinity, the \(+ 3\) contributes less and less to the overall runtime.

In this example, we chose \(k_1 = 2\) and \(k_2 = 3\). These two choices of \(k\) constitute a tight-bound for \(2N + 3\) for all values of \(N \geq 3\).

This idea of big-theta notation as a tight-bound is very useful as it allows us to, very precisely, state exactly how scalable a function’s runtime grows as the size of its input (\(N\)) grows. When a \(\Theta(N)\) function’s input size increases, we’d expect the runtime to also increase linearly.

Big-O #

But, there are many scenarios where we can’t actually give a tight bound: sometimes, it just doesn’t exist. And, practically-speaking, one of the common use scenarios for runtime in the real world is to help choose between several different algorithms with different orders of growth. For these purposes, it’s often sufficient just to give an upper-bound on the runtime of a program.

There exists a very common asymptotic notation, big-O, represented by the symbol, \(O\).

If we could think of big-theta as an \(=\) (equals) sign, then big-O is like a \(\leq\) sign. Likewise, the formal definition for big-O follows, \(R(N) \in O(f(N))\) if and only if there exists a positive constant \(k_2\) such that \(R(N) \leq k_2 \cdot f(N)\) for all \(N\) greater than some \(N_0\) (very large \(N\)).

Note that this is a looser condition than big-theta since big-O doesn’t include the lower bound.

To see why we would prefer a theta bound consider the following simplified example. Would you know more about a person’s age if they told you either 1. “I am between 30 and 40 years old” OR 2. “I am less than 40 years old”?

Big-Omega #

Sometimes it will also be useful to describe function runtimes using an \(\Omega\) (Omega) bound, which you can think of as a lower bound. For example, if a tight \(\Theta\) bound does not exist you could supply an \(O\) and \(\Omega\) bound. Practically speaking, an \(\Omega\) bound by itself might be less useful than a \(\Theta\) or \(O\) bound, but in certain cases it will provide useful information.

Similar to above, if we could think of big-theta as an \(=\) (equals) sign, then big-\(\Omega\) is like a \(\geq\) sign. Likewise, the formal definition for big-Omega follows, \(R(N) \in \Omega(f(N))\) if and only if there exists a positive constant \(k_1\) such that \(R(N) \geq k_1 \cdot f(N)\) for all \(N\) greater than some \(N_0\) (very large \(N\)).

Note that this too is a looser condition than big-theta since big-Omega doesn’t include the upper bound.

We can return to our simplified example to again show why we prefer a theta bound. Would you know more about a person’s age if they told you either 1. “I am between 30 and 40 years old” OR 2. “I am more than 30 years old”?

Back to Big Theta #

Now that we have learned big-Omega and big-O notation, where does big theta fall into this? We already said that big theta means roughly equal, but what does that mean? Given we have defined big-O as an upper bound and big-Omega as a lower bound, what happens if the tightest big-O bound is in the same family as the tightest big-Omega bound? This means the code is both upper bounded AND lower bounded by the same family of functions! This is what we mean by equal.

A theta bound only exists if the tightest big-O bound is equal to the tightest big-Omega bound.

Discussion: Case Analysis #

Read the following three sections of Chapter 8.4 from Runtime Analysis Subtleties all the way through Big Omega. You don’t need to read the last section on Amortized Analysis, yet. We will see that in a few labs!

Discuss with someone why each of the following claims are true.

  • removeZeroes is in \(\Omega(1)\).
  • removeZeroes is in \(\Omega(N)\).
  • removeZeroes is in \(O(N^2)\).
  • A \(\Theta(\cdot)\) bound does not exist for removeZeroes.
  • In the best case, removeZeroes is in \(O(N^2)\).
  • In the best case, removeZeroes is in \(\Theta(N)\).
  • In the worst case, removeZeroes is in \(\Omega(\log N)\).
  • In the worst case, removeZeroes is in \(\Theta(N^2)\).

Limit Definition #

An alternative, the calculus-based limit definition is also sometimes useful, as you can apply L’Hopital’s Rule to derive asymptotic simplifications like dropping multiplicative constants and additive lower-order terms.

We generally won’t use this too often though, as the first definition provides a more useful and intuitive visualization of the lower and upper bounds.

Common Orders of Growth #

Here are some commonly-occurring estimates listed from no growth at all to fastest growth.

  • Constant time, often indicated with \(1\).
  • Logarithmic time or proportional to \(\log N\).
  • Linear time or proportional to \(N\).
  • Linearithmic time or proportional to \(N \log N\).
  • Polynomial time or proportional to \(N^{k}\) for some constant \(k\).
  • Exponential time or proportional to \(k^{N}\) for some constant \(k\).
  • Factorial time or proportional to \(N!\) (\(N\) factorial).

Orders of Growth

Logarithmic Algorithms #

First, if you are shaky on the properties of logarithms, I suggest looking through this Khan academy section on log properties!

We will shortly encounter algorithms that run in time proportional to \(\log N\) for some suitably defined \(N\). Recall from algebra that the base-10 logarithm of a value is the exponent to which 10 must be raised to produce the value. It is usually abbreviated as \(\log_{10}\). Thus

  • \(\log_{10} 1000\) is 3 because \(10^{3} = 1000\).
  • \(\log_{10} 90\) is slightly less than 2 because \(10^{2} = 100\).
  • \(\log_{10} 1\) is 0 because \(10^{0} = 1\).

In algorithms, we commonly deal with the base-2 logarithm, written as \(\lg\), defined similarly.

  • \(\lg 1024\) is 10 because \(2^{10} = 1024\).
  • \(\lg 9\) is slightly more than 3 because \(2^{3} = 8\).
  • \(\lg 1\) is 0 because \(2^{0} = 1\).

Another way to think of log is the following: \(\log_{\text{base}} N\) is the number of times \(N\) must be divided by the base before it hits 1. For the purposes of determining orders of growth, however, the log base actually doesn’t make a difference because, by the change of base formula, we know that any logarithm of \(N\) is within a constant factor of any other logarithm of \(N\). We usually express a logarithmic algorithm as simply \(\log N\) as a result.

Change of Base Formula
\[\log_b x = \frac{\log_a x}{\log_a b}\]

Algorithms for which the running time is logarithmic are those where processing discards a large proportion of values in each iterations. The binary search algorithm is an example. We can use binary search in order to guess a number that a person is thinking of. In each iteration, we guess a number and are told whether the number they’re thinking of is higher or lower than our guess. The algorithm then discards half the possible values for the searched-for number, repeating the process in the other half. Thus, we continually divide the size of the problem by 2 until there is only one value left.

For example, say you started with a range of 1024 numbers in the number guessing game. Each time you would discard half of the numbers so that each round would have the following numbers under consideration:

Round # Numbers left
1 1024
2 512
3 256
4 128
5 64
6 32
7 16
8 8
9 4
10 2
11 1

We know from above that \(\lg 1024 = 10\) which gives us an approximation of how many rounds it will take. We will see further applications of logarithmic behavior when we work with trees in subsequent activities.

Analyzing Iteration #

We’ve thus far defined the language of asymptotic analysis and developed some simple methods based on counting the total number of steps. However, the kind of problems we want to solve are often too complex to think of just in terms of number iterations times however much work is done per iteration.

Consider the following function, repeatedSum.

long repeatedSum(int[] values) {
    int N = values.length;
    long sum = 0;
    for (int i = 0; i < N; i += 1) {
        for (int j = i; j < N; j += 1) {
            sum += values[j];
    return sum;

In repeatedSum, we’re given an array of values of length N. We want to take the repeated sum over the array as defined by the following sequences of values j takes on:

  • \[0, 1, 2, 3, \cdots, N - 1\]
  • \[1, 2, 3, 4, \cdots, N - 1\]
  • \[2, 3, 4, 5, \cdots, N - 1\]

Notice that each time, the number of elements, or the iterations of j, being added is reduced by 1. While in the first iteration, we sum over all \(N\) elements, in the second iteration, we only sum over \(N - 1\) elements. On the next iteration, even fewer: just \(N - 2\) elements. This pattern continues until the outer loop, i, has incremented all the way to \(N\).

One possible approach to this problem is to draw a bar chart to visualize how much work is being done for each iteration of i. We can represent this by plotting the values of i across the X-axis of the chart and the number of steps for each corresponding value of i across the Y-axis of the chart.

Empty plot

Now, let’s plot the amount of work being done on the first iteration of i where i = 0. If we examine this iteration alone, we just need to measure the amount of work done by the j loop. In this case, the j loop does work proportional to \(N\) steps as the loop starts at 0, increments by 1, and only terminates when j = N.

How about the next iteration of i? The loop starts at 1 now instead of 0 but still terminates at \(N\). In this case, the j loop is proportional to \(N - 1\) steps. The next loop, then, is proportional to \(N - 2\) steps.

Partial linear plot

We can start to see a pattern forming. As i increases by 1, the amount of work done on each corresponding j loop decreases by 1. As i approaches \(N\), the number of steps in the j loop approaches 0. In the final iteration, when i = N - 1, the j loop performs work proportional to 1 step.

Extrapolated linear plot

We’ve now roughly measured each loop proportional to some number of steps. Each independent bar represents the amount of work any one iteration of i will perform. The runtime of the entire function repeatedSum then is the sum of all the bars, or simply the area underneath the line.

Complete linear plot

The problem is now reduced to finding the area of a triangle with a base of \(N\) and height of also \(N\). Thus, the runtime of repeatedSum is in \(\Theta(N^{2})\).

We can verify this result mathematically by noticing that the sequence can be described by the following summation:

\(1 + 2 + 3 + ... + N = \frac{N(N + 1)}{2}\), which after dropping lower order terms and multiplicative constants we see is in \(\Theta(N^{2})\). It’s useful to know both the formula as well as its derivation through the chart above.

Multivariate Analysis #

Sometimes we care about how the runtime of an algorithm will grow with respect to multiple variables. As an example, consider the rectangle function below that computes the area of a rectangle with side lengths N and M.

int rectangle(int N, int M) {
    int area = 0;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            area += 1;
    return area;

Analyzing this algorithm we see that the outer loop runs \(N\) times and the inner loop runs \(M\) times per each iteration of the outer loop. Since each iteration of the inner loop only does a constant amount of work, we find the resulting runtime to be in \(\Theta(MN)\).

Analyzing Recursion #

Now that we’ve learned how to use a bar chart to represent the runtime of an iterative function, let’s try the technique out on a recursive function, mitosis.

int mitosis(int N) {
    if (N == 1) {
        return 1;
    return mitosis(N / 2) + mitosis(N / 2);

Let’s start by trying to map each \(N\) over the x-axis like we did before and try to see how much work is done for each call to the function. The conditional contributes a constant amount of work to each call. But notice that in our return statement, we make two recursive calls to mitosis. How do we represent the runtime for these calls? We know that each call to mitosis does a constant amount of work evaluating the conditional base case but it’s much more difficult to model exactly how much work each recursive call will do. While a bar chart is a very useful way of representing the runtime of iterative functions, it’s not always the right tool for recursive functions.

Instead, let’s try another strategy: drawing call trees. Like the charting approach we used for iteration earlier, the call tree will reduce the complexity of the problem and allow us to find the overall runtime of the program on large values of \(N\) by taking the tree recursion out of the problem. We will draw each call to the function as a node with its input size within the node, and recursive calls are drawn as children of a node. Consider the call tree for fib below.

int fib(int N) {
    if (N <= 1) {
        return 1;
    return fib(n - 1) + fib(n - 2);

Fibonacci tree

At the root of the tree, we make our first call to fib(n). The recursive calls to fib(n - 1) and fib(n - 2) are modeled as the two children of the root node. We say that this tree has a branching factor of two as each node contains two children. It takes a constant number of instructions to evaluate the conditional, addition operator, and the return statement as denoted by the 1 to the upper-right of each node. (Note that you may see call trees with nodes’ input size and work performed in opposite positions. Be sure to understand what the numbers in your call tree represent.)

We can see this pattern occurs for all nodes in the tree: each node performs the same constant number of operations if we don’t consider recursive calls. If we can come up with a scheme to count all the nodes, then we can simply multiply by the constant number of operations to find the overall runtime of fib.

For a tree with branching factor \(b\) and height \(h\) we can compute the number of nodes as \(b^{h+1}-1\) (if we think of the root as height 0). For the sake of asymptotic analysis, it is sufficient to use \(b^{h}\) or \(b^{h+1}\) as we will below (as an exercise, can you justify why?).

Spend a little time thinking about the maximum height of this tree: when does the base case tell us the tree recursion will stop?

Note that the counting the number of nodes in this recursive call tree for fib ends up being a little tricky because the tree is not symmetric i.e. each of the two recursive calls made from one call to fib are not the same.

If you are interested, you can check out this article which explains how to compute a tight bound for fib.

Returning to the original problem of mitosis, the call tree is setup just like fib except instead of decrementing \(N\) by 1 or 2, we now divide \(N\) in half each time. Each node performs a constant amount of work but also makes two recursive calls to mitosis(N / 2).

N-time tree

Like before, we want to identify both the branching factor and the height of the tree. In this case, the branching factor is 2 like before. Recall that the series \(N, N/2, \cdots , 4, 2, 1\) contains \(\log_{2} N\) elements since, if we start at \(N\) and break the problem down in half each time, it will take us approximately \(\log_{2} N\) steps to completely reduce down to 1, so the height of the tree will be \(\log_{2} N\).

Plugging into the formula, we get \(2^{\log_{2} N}\) nodes which simplifies to \(N\). Therefore, \(N\) nodes performing a constant amount of work each will give us an overall runtime in \(\Theta(N)\).

There is another way to approach this analysis: going level by level. We can see that on the first level of the recursive call tree there is a single node doing a constant amount of work (1). On the second level we double the number of nodes, but each node still is doing a constant amount of work, so the total work on this layer is 1+1=2. Similarly, we can see that the work on level 3 is 4, on level 4 is 8, and so on following the pattern that the work on level \(i\) is \(2^{i}\). From the analysis above, we know that this tree has \(\log_{2} N\) levels. Now, to figure out the total work done by the function we just need to sum up all the work done by all of the levels:

\[1 + 2 + 4 + \cdots + 2^{(log_{2} N) - 2} + 2^{(log_{2} N) - 1}\] \[= 1 + 2 + 4 + \cdots + \frac{N}{4} + \frac{N}{2}\]

This is a geometric sum which is dominated by its last element. Here, the last element is larger than all of the elements that came before it combined (Exercise: prove this to yourself. As a hint, try out the first few powers of 2). This allows us to say the overall runtime is proportional to the last term giving us an overall runtime in \(\Theta(N)\)

It will not always be the case that each layer in a recursive call tree will have the same amount of summed work, nor will it always be the case that each node in a recursive call tree will do the same amount of work. You will have to use the techniques at your disposal to solve these problems, which is a skill that takes time and practice to develop.

In general, for a recursion tree, we can think of the total work as:

\[\sum_{\text{layers}} \frac{\text{nodes}}{\text{layer} }\frac{\text{work}}{\text{node}}\]

For mitosis, we have \(\log N\) levels, \(2^(i-1)\) nodes in layer \(i\), with \(1\) work per node. Thus we see the summation will be as follows, which matches the quantity we just calculated:

\[\sum_{i = 0}^{\log N} 2^i (1)\]

Recap #

Runtime Minimization
One of the most important properties of a program is the time it takes to execute. One goal as a programmer is to minimize the time (in seconds) that a program takes to complete.
Runtime Measurement
  • Measure the number of seconds that a program takes to complete using a stopwatch (either physical or in software). This tells you the actual runtime, but is dependent on the machine and inputs.
  • Count the number of operations needed for inputs of a given size. This is a machine independent analysis, but still depends on the input, and also doesn’t actually tell you how long the code takes to run.
  • Derive an algebraic expression relating the number of operations to the size of an input. This tells you how the algorithm scales, but does not tell you how long the code takes to run.
Algorithm Scaling
While we ultimately care about the runtime of an algorithm in seconds, we’ll often say that one algorithm is better than another simply because of how it scales. By scaling, we mean how the runtime of a piece of code grows as a function of its input size. For example, inserting at the beginning of ArrayList on an old computer might take \(R(N) = 0.0001N\) seconds, where \(N\) is the size of the list. For example, if the runtime of two algorithms is \(R_1(N) = N^2\), and \(R_2(N) = 5000 + N\), we’d say algorithm \(R_2\) is better, even though \(R_1\) is much faster for small \(N\).
Order of Growth
The result of applying our last 3 simplifications gives us the order of growth of a function. So for example, suppose \(R(N) = 4N^2 + 3N + 6\), we’d say that the order of growth of \(R(N)\) is \(N^2\).

The terms “constant”, “linear”, and “quadratic” are often used for algorithms with order of growth \(1\), \(N\), and \(N^2\), respectively. For example, we might say that an algorithm with runtime \(4N^2 + 3N + 6\) is quadratic.

Simplified Analysis
Once we’ve chosen a cost function, we can either:
  • Compute the exact expression that counts the number of operations.
  • Use intuition and inspection to find the order of growth of the number of operations.

This latter approach is generally preferable, but requires a lot of practice. One common intuitive/inspection-based approach is use geometric intuition. For example, if we have nested for loops where i goes from 0 to N, and j goes from i + 1 to N, we observe that the runtime is effectively given by a right triangle of side length N. Since the area of a such a triangle grows quadratically, the order of growth of the runtime is quadratic.

To formalize our intuitive simplifications, we introduce big-theta notation. We say that a function \(R(N) \in \Theta(f(N))\) if there exists positive constants \(k_1\) and \(k_2\) such that \(k_1 \cdot f_1(N) \leq R(N) \leq k_2 \cdot f_2(N)\).

When using \(\Theta\) to capture a function’s asymptotic scaling, we avoid unnecessary terms in our \(\Theta\) expression. For example, while \(4N^2 + 3N + 6 \in \Theta(4N^2 + 3N)\), we will usually make the simpler claim that is \(4N^2 + 3N + 6 \in \Theta(N^2)\).

Big-theta is exactly equivalent to order of growth. That is, if a function \(R(N)\) has order of growth \(N^2\), then we also have that \(R(N) \in \Theta(f(N))\).

In the final section, we applied what we learned about counting steps, estimation, and orders of growth to model algorithmic analysis for larger problems. Two techniques, charting and drawing call trees, helped us break down challenging problems into smaller pieces that we could analyze individually and recombine to form the final solution.

Below, we will list some tips and formulas that will be handy when you start finding the asymptotic runtimes of functions.

Practical Tips #

  1. Before attempting to calculate a function’s runtime, first try to understand what the function does.
  2. Try some small sample inputs to get a better intuition of what the function’s runtime is like. What is the function doing with the input? How does the runtime change as the input size increases? Can you spot any ‘gotchas’ in the code that might invalidate our findings for larger inputs?
  3. Try to lower bound and upper bound the function runtime given what you know about how the function works. This is a good sanity check for your later observations.
  4. If the function is recursive, draw a call tree to map out the recursive calls. Within each node, denote how much work that specific node does. Then, note the total (sum) work done on each level. You should also find the height of the recursive call tree. This breaks the problem down into smaller parts that we can analyze individually. Once each part of the tree has been analyzed, we can then reassemble all the parts to determine the overall runtime of the function.
  5. If the function has a complicated loop, draw a bar chart to map out how much work the body of the loop executes for each iteration.
  6. Only consider what happens for very large N. If you see a statement like if (N < 1) {return 0} at the top of an algorithm this doesn’t mean that we can immediately say the algorithm has a best case runtime in \(\Theta(1)\). This is a very common mistake.

Useful Formulas #

  • \(1 + 2 + 3 + 4 + \cdots + N\) is in \(\Theta(N^2)\).
  • There are \(N\) terms in the sequence \(1, 2, 3, 4, \cdots, N\).
  • \(1 + 2 + 4 + 8 + \cdots + N\) is in \(\Theta(N)\).
  • There are \(\log N\) terms in the sequence \(1, 2, 4, 8, \cdots, N\).
  • The number of nodes in a complete tree, \(N\), is approximately \(b^h\) where \(b\) is the branching factor and \(h\) is the height of the tree.
  • All logarithms are proportional to each other by the Change of Base formula so we can express them generally as just \(\log\).

It’s worth spending a little time proving each of these to yourself with a visual model! I personally recommend Desmos or WolframAlpha.

Deliverables #

A quick recap of what you need to do to finish today’s lab.

  • Look through the Timer class and try timing the algorithm in Sorter.java for different inputs. Discuss with your neighbors what you came up with.
  • Read through the content on asymptotic analysis (big-theta, O, and omega) focusing on how to handle logarithmic, iterative, and recursive algorithms.
  • Complete the online assessment on Gradescope. There is no coding submission.