# Lecture 3

## Last week

• Recall that our purpose for learning a new programming language and other tools is to solve problems. And we solve those problems by creating some output from input:
• We learned about memory, which allows us to store data as bytes, and strings, which are arrays of characters.

## Searching

• Today we’ll focus on algorithms that solve problems with arrays.
• It turns out that, with arrays, a computer can’t look at all of the elements at once. Instead, a computer can only to look at them one at a time, though we can look in any order, like only being able to open one locker at a time:
• Recall that arrays are zero-indexed, meaning that the first item has an index of 0. And with $$n$$ items, the highest index would be $$n - 1$$.
• Searching is how we solve the problem of finding information. A simple problem has an input of some values, and an output of a bool, whether or not a particular value is in the array.

### Big $$O$$

• Today we’ll look at algorithms for searching. To compare their efficiency, we’ll consider running time, or how long an algorithm takes to run given some size of input.
• Computer scientists tend to describe running time with big $$O$$ notation, which we can think of as “on the order of” something, as though we want to convey an idea of running time and not an exact number of milliseconds or steps.
• In week 0, we saw a chart with different types of algorithms and the times they might take to solve a problem:
• Recall that the red line is searching linearly, one page at a time; the yellow line is searching two pages at a time; and the green line is dividing and conquering, starting in the middle and dividing the problem in half each time.
• In the chart above, if we zoomed out and changed the units on our axes, we would see the red and yellow lines end up very close together:
• So we use big $$O$$ notation to describe both the red and yellow lines, since they end up being very similar as $$n$$ becomes larger and larger. We would describe them both as having “big $$O$$ of $$n$$” or “on the order of $$n$$” running time.
• The green line, though, is fundamentally different in its shape, even as $$n$$ becomes very large, so it takes “big $$O$$ of $$\log n$$” steps. (The base of the logarithm, 2, is also removed since it’s a constant factor.)
• We’ll see some common running times:
• $$O(n^2)$$
• $$O(n \log n)$$
• $$O(n)$$
• $$O(\log n)$$
• $$O(1)$$
• Computer scientists might also use big $$Ω$$, big Omega notation, which describes the lower bound of number of steps for our algorithm, or how few steps it might take, in the best case. Big $$O$$, on the order of, is the upper bound of number of steps, or how many steps it might take, in the worst case.
• We have a similar set of the most common big $$Ω$$ running times:
• $$\Omega(n^2)$$
• $$\Omega(n \log n)$$
• $$\Omega(n)$$
• $$\Omega(\log n)$$
• $$\Omega(1)$$
• Finally, there is another notation, $$\Theta$$, big Theta, which we use to describe running times of algorithms if the upper bound and lower bound is the same.
• $$\Theta(n^2)$$
• $$\Theta(n \log n)$$
• $$\Theta(n)$$
• $$\Theta(\log n)$$
• $$\Theta(1)$$
• An algorithm with running time of $$O(1)$$ means that a constant number of steps is required, no matter how big the problem is.
• Let’s take a look at some algorithms that we can describe with these running times.
• On stage, we have seven lockers with closed doors, with numbers hidden behind them. Since a computer can only look at one element in an array at a time, we can only open one door at a time as well.
• If we want to look for the number zero, for example, we would have to open one door at a time, and if we didn’t know anything about the numbers behind the doors, the simplest algorithm would be going from left to right.
• This algorithm, linear search, would be correct but not very efficient. We might write pseudocode with:
For each door from left to right
If number is behind door
Return true
Return false

• Return false is outside the for loop, since we only want to do that after we’ve looked behind all the doors.
• We can rewrite our pseudocode to be a little closer to C:
For i from 0 to n-1
If number behind doors[i]
Return true
Return false

• Now, we’re using a variable, i, to look at each location in an array called doors.
• With n doors, we’ll need to look at all n of them. And what we do for each of the n doors, which is looking inside and possibly returning true, takes a constant number of steps each time. So the big $$O$$ running time for this algorithm would be $$O(n)$$.
• The lower bound, big Omega, would be $$\Omega(1)$$, since we might be lucky and find the number we’re looking for right away, which takes a constant number of steps.
• If we know that the numbers behind the doors are sorted, then we can start in the middle, and find our value more efficiently since we know we can go left or right, dividing the problem in half each time.
• For binary search, the pseudocode for our algorithm might look like:
If no doors
Return false
If number behind middle door
Return true
Else if number < middle door
Search left half
Else if number > middle door
Search right half

• We remember to check whether there are no doors left, since that means our number isn’t behind any of them.
• We can write this pseudocode to be more like C:
If no doors
Return false
If number behind doors[middle]
Return true
Else if number < doors[middle]
Search doors[0] through doors[middle - 1]
Else if number > doors[middle]
Search doors [middle + 1] through doors[n - 1]

• We can determine the index of the middle door with a bit of math, and then we can divide the problem into searching either the doors with indices 0 through middle - 1, or middle + 1 through n - 1.
• The upper bound for binary search is $$O(\log n)$$, since we might have to keep dividing the number of doors by two until there are no more doors left. The lower bound $$\Omega(1)$$, if the number we’re looking for is in the middle, where we happen to start.
• Even though binary search might be much faster than linear search, it requires our array to be sorted first. If we’re planning to search our data many times, it might be worth taking the time to sort it first, so we can use binary search.
• Other resources we might consider beyond the time it takes to run some code include the time it takes to write the code, or the amount of memory required for our code.

### Searching with code

• Let’s take a look at numbers.c:
#include <cs50.h>
#include <stdio.h>

int main(void)
{
int numbers[] = {4, 6, 8, 2, 7, 5, 0};

for (int i = 0; i < 7; i++)
{
if (numbers[i] == 0)
{
printf("Found\n");
return 0;
}
}
return 1;
}

• Here we initialize an array with values by using curly braces, and we check the items in the array one at a time, in order, to see if they’re equal to zero.
• If we find the value of zero, we return an exit code of 0 (to indicate success). Otherwise, after our for loop, we call return 1 (to indicate an error code).
• This is how we might implement linear search.
• We can compile our program and run it to see that it works:
$make numbers$ ./numbers
Found

• And we can change what we’re looking for to -1, and see that our program doesn’t find it:
...
if (numbers[i] == -1)
...

$make numbers$ ./numbers

• We can do the same for strings in names.c:
#include <cs50.h>
#include <stdio.h>
#include <string.h>

int main(void)
{
string names[] = {"Bill", "Charlie", "Fred", "George", "Ginny", "Percy", "Ron"};

for (int i = 0; i < 7; i++)
{
if (names[i] == "Ron")
{
printf("Found\n");
return 0;
}
}
return 1;
}

• But when we try to compile our program, we get:
$make names names.c:11:22: error: result of comparison against a string literal is unspecified (use an explicit string comparison function instead) [-Werror,-Wstring-compare] if (names[i] == "Ron") ^ ~~~~~ 1 error generated. make: *** [<builtin>: names] Error 1  • It turns out that we can’t compare strings directly in C, since they’re not a simple data type built into the language, but rather an array of many characters. Luckily, the string library has function, strcmp, string compare, which compares strings for us. strcmp returns a negative value if the first string comes before the second string, 0 if the strings are the same, and a positive value if the first string comes after the second string. • We’ll change our conditional to if (strcmp(names[i], "Ron") == 0), so we can check whether our two strings are actually equal. • And if we wrote if (strcmp(names[i], "Ron")), then any non-zero value, positive or negative, would be considered true, which would be the opposite of what we want. • We could actually write if (!strcmp(names[i], "Ron")) to invert the value, which would work in this case, but it would be arguably worse design since it doesn’t explicitly check for the value of 0 as the documentation indicates. ## Structs • We might want to implement a phone book, with names and phone numbers, in phonebook0.c: #include <cs50.h> #include <stdio.h> #include <string.h> int main(void) { string names[] = {"Carter", "David"}; string numbers[] = {"+1-617-495-1000", "+1-949-468-2750"}; for (int i = 0; i < 2; i++) { if (strcmp(names[i], "David") == 0) { printf("Found %s\n", numbers[i]); return 0; } } printf("Not found\n"); return 1; }  • We have two arrays, names and numbers, and we’ll make sure that each person’s phone number has the same index as their name in each array. • We’ll search our names array for someone’s name, and then return their corresponding phone number at the same index. • This program is correct, but not well-designed since we’ll have to maintain both arrays carefully to make sure that the names and numbers line up. • And feel free to call or text David for a surprise! • It turns out in C that we can define our own data type, or data structure. It would be a better design for our program to have some array with a data type person that includes both their name and phone number, so we can just say person people[];. • In C, we can create a struct with the following syntax: typedef struct { string name; string number; } person;  • typedef struct tells the compiler that we’re defining our own data structure. And person at the end of the curly braces will be the name of this data structure. • Inside our struct, we’ll have two strings, name and number. We’ll use strings for phone numbers since they might include punctuation, and other types of numbers, like zip codes, might have a leading 0 that would disappear if treated as a number. • We’ll use this in phonebook1.c: #include <cs50.h> #include <stdio.h> #include <string.h> typedef struct { string name; string number; } person; int main(void) { person people[2]; people[0].name = "Carter"; people[0].number = "+1-617-495-1000"; people[1].name = "David"; people[1].number = "+1-949-468-2750"; for (int i = 0; i < 2; i++) { if (strcmp(people[i].name, "David") == 0) { printf("Found %s\n", people[i].number); return 0; } } printf("Not found\n"); return 1; }  • We see new syntax to set the values for each variable inside each person struct by using the dot operator, .. • In our loop, we can also use .name or .number to access variables in our structs, and be certain that they are from the same person. • In computer science, encapsulation is the idea that related data is grouped together, and here we’ve encapsulated two pieces of information, name and number into the same data structure. The color of a pixel, with red, green, and blue values, might also be well-represented by a data structure as well. • We can also imagine that a struct can be used to store precise decimal values or large integer values, perhaps with arrays that we can use to store large numbers of digits. ## Sorting • Sorting is solving the problem which takes in an unsorted list of numbers as input, and producing an output of a sorted list of numbers: • For example, 6 3 8 5 2 7 4 1 might be input, and output would be 1 2 3 4 5 6 7 8. ### Selection sort demonstration • We’ll have eight volunteers come up to the stage, in unsorted order: 5 2 7 4 1 6 3 0  • We ask our volunteers to sort themselves, and everyone moves to the correct position: 0 1 2 3 4 5 6 7  • Unfortunately, computers can only look at one number and move one of them at a time, at least with programs we have written so far. • We’ll start with unsorted numbers again, and going step-by-step, we’ll look for the smallest number first, looking at each number in order: 5 2 7 4 1 6 3 0 ^  • Now, we’ll swap the smallest number with the number at the beginning, since it’s easier than shifting all of the other numbers down: 0 | 2 7 4 1 6 3 5  • Now, our problem has gotten smaller, since we know at least the beginning of our list is sorted. So we can repeat what we did, starting from the second number in the list: 0 | 2 7 4 1 6 3 5 ^ 0 | 1 7 4 2 6 3 5  • 1 is the smallest number now, so we’ll swap it with the second number. • We’ll repeat this again … 0 1 | 7 4 2 6 3 5 ^ 0 1 | 2 4 7 6 3 5  • … and again … 0 1 2 | 4 7 6 3 5 ^ 0 1 2 | 3 7 6 4 5  • … and again … 0 1 2 3 | 7 6 4 5 ^ 0 1 2 3 | 4 6 7 5  • … and again … 0 1 2 3 4 | 6 7 5 ^ 0 1 2 3 4 | 5 7 6  • … and again, until we’ve swapped the last number in our list: 0 1 2 3 4 5 | 7 6 ^ 0 1 2 3 4 5 | 6 7  ### Bubble sort demonstration • We’ll start with our unsorted list, but this time we’ll look at pairs of numbers and swap them if they are out of order: 5 2 7 4 1 6 3 0 ^ ^ 2 5 7 4 1 6 3 0 ^ ^ 2 5 7 4 1 6 3 0 ^ ^ 2 5 4 7 1 6 3 0 ^ ^ 2 5 4 1 7 6 3 0 ^ ^ 2 5 4 1 6 7 3 0 ^ ^ 2 5 4 1 6 3 7 0 ^ ^ 2 5 4 1 6 3 0 7  • Now, the highest number is all the way to the right, so we’ve improved our problem. • We’ll repeat this again: 2 5 4 1 6 3 0 | 7 ^ ^ 2 5 4 1 6 3 0 | 7 ^ ^ 2 4 5 1 6 3 0 | 7 ^ ^ 2 4 1 5 6 3 0 | 7 ^ ^ 2 4 1 5 6 3 0 | 7 ^ ^ 2 4 1 5 3 6 0 | 7 ^ ^ 2 4 1 5 3 0 6 | 7  • Now the two biggest values are on the right. • We’ll repeat again … 2 4 1 5 3 0 | 6 7 ^ ^ 2 4 1 5 3 0 | 6 7 ^ ^ 2 1 4 5 3 0 | 6 7 ^ ^ 2 1 4 5 3 0 | 6 7 ^ ^ 2 1 4 3 5 0 | 6 7 ^ ^ 2 1 4 3 0 5 | 6 7  • … and again … 2 1 4 3 0 | 5 6 7 ^ ^ 1 2 4 3 0 | 5 6 7 ^ ^ 1 2 3 4 0 | 5 6 7 ^ ^ 1 2 3 4 0 | 5 6 7 ^ ^ 1 2 3 0 4 | 5 6 7  • … and again … 1 2 3 0 | 4 5 6 7 ^ ^ 1 2 3 0 | 4 5 6 7 ^ ^ 1 2 3 0 | 4 5 6 7 ^ ^ 1 2 0 3 | 4 5 6 7  • … and again … 1 2 0 | 3 4 5 6 7 ^ ^ 1 2 0 | 3 4 5 6 7 ^ ^ 1 0 2 | 3 4 5 6 7  • … and finally: 1 0 | 2 3 4 5 6 7 ^ ^ 0 1 | 2 3 4 5 6 7  • Notice that, as we go through our list, we know more and more of it becomes sorted, so we only need to look at the pairs of numbers that haven’t been sorted yet. ### Selection sort • The first algorithm we saw is called selection sort, and we might write pseudocode like: For i from 0 to n–1 Find smallest number between numbers[i] and numbers[n-1] Swap smallest number with numbers[i]  • We need to sort all n numbers in our list, so we’ll have a loop with i to keep track of how many numbers we’ve sorted so far. • The first step in the loop is to look for the smallest number in the unsorted part of the list, from index i to n-1. • Then, we swap the smallest number we found with the number at index i, which makes everything up to i sorted. • And we’ll repeat this until the entire list is sorted, from left to right, as i goes from 0 to n-1. • For this algorithm, we started with looking at all $$n$$ elements, then only $$n - 1$$, then $$n - 2$$, and so on: $$n + (n – 1) + (n – 2) + ... + 1$$ $$n(n + 1)/2$$ $$(n^2 + n)/2$$ $$n^2/2 + n/2$$ $$O(n^2)$$ • We can use some math formulas to get to $$n^2/2 + n/2$$ steps. Since $$n^2$$ is the biggest, or dominant, factor as $$n$$ gets really large, we can say that the algorithm has an upper bound for running time of $$O(n^2)$$. • In the best case, where the list is already sorted, our selection sort algorithm will still look at all the numbers and repeat the loop, so it has a lower bound for running time of $$\Omega(n^2)$$. ### Bubble sort • The pseudocode for bubble sort might look like: Repeat n-1 times For i from 0 to n–2 If numbers[i] and numbers[i+1] out of order Swap them  • Since we are comparing each pair of numbers at i and i+1, we only need to go up to n–2 for i. • Then, we swap the two numbers if they’re out of order. • We need to repeat this n-1 times since each time we go over the list, only one number moves all the way to the right. • To determine the running time for bubble sort, we have $$n – 1$$ comparisons in the loop, and $$n – 1$$ loops: $$(n - 1) \times (n - 1)$$ $$n^2 - 1n - 1n + 1$$ $$n^2 - 2n + 1$$ $$O(n^2)$$ • The largest factor is again $$n^2$$ as $$n$$ gets larger and larger, so we can say that bubble sort has an upper bound for running time of $$O(n^2)$$. • If we add a little logic to our algorithm, we can optimize the running time in the best case: Repeat n-1 times For i from 0 to n–2 If numbers[i] and numbers[i+1] out of order Swap them If no swaps Quit  • If we looked at all the numbers, and didn’t make any swaps, then we know that our list has been sorted. • The lower bound for running time of bubble sort would be $$\Omega(n)$$. • We look at a visualization with animations for how numbers are moved for selection sort and bubble sort. ## Recursion • Recursion is the ability for a function to call itself. We haven’t seen this in code yet, but we’ve seen this in pseudocode for binary search: If no doors Return false If number behind middle door Return true Else if number < middle door Search left half Else if number > middle door Search right half  • We’re using the same “search” algorithm for each half. This seems like a cyclical process that will never end, but we’re actually changing the input to the function and dividing the problem in half each time, stopping once there are no more doors left. • Our pseudocode from week 0 used “go back” to repeat some steps, like a loop: 1 Pick up phone book 2 Open to middle of phone book 3 Look at page 4 If person is on page 5 Call person 6 Else if person is earlier in book 7 Open to middle of left half of book 8 Go back to line 3 9 Else if person is later in book 10 Open to middle of right half of book 11 Go back to line 3 12 Else 13 Quit  • But we can similarly have our algorithm use itself after it divides the problem: 1 Pick up phone book 2 Open to middle of phone book 3 Look at page 4 If person is on page 5 Call person 6 Else if person is earlier in book 7 Search left half of book 8 Else if person is later in book 9 Search right half of book 10 Else 11 Quit  • In week 1, too, we implemented a “pyramid” of blocks in the following shape: # ## ### ####  • Our code to print this out might look like: #include <cs50.h> #include <stdio.h> void draw(int n); int main(void) { int height = get_int("Height: "); draw(height); } void draw(int n) { for (int i = 0; i < n; i++) { for (int j = 0; j < i + 1; j++) { printf("#"); } printf("\n"); } }  • We have a draw function which takes in an argument, n, and uses a loop to print n rows with more and more bricks in each row. • We can run our code and see that it works: $ make iteration
$./iteration Height: 4 # ## ### ####  • But notice that a pyramid of height 4 is actually a pyramid of height 3 with an extra row of 4 blocks added on. And a pyramid of height 3 is a pyramid of height 2 with an extra row of 3 blocks. A pyramid of height 2 is a pyramid of height 1 with an extra row of 2 blocks. And finally, a pyramid of height 1 is a pyramid of height 0 (no blocks) with a row of 1 block added. • Since a pyramid is a recursive structure, we can write a recursive function to draw a pyramid, a function that calls itself to draw a smaller pyramid before adding another row: #include <cs50.h> #include <stdio.h> void draw(int n); int main(void) { int height = get_int("Height: "); draw(height); } void draw(int n) { if (n <= 0) { return; } draw(n - 1); for (int i = 0; i < n; i++) { printf("#"); } printf("\n"); }  • We can rewrite our draw function to use itself. • If n is 0 (or negative somehow) we’ll stop without printing anything. And we need to make sure we stop for some base case, so our function doesn’t call itself over and over forever. • Otherwise, we’ll call draw again, to print a pyramid of size n - 1 first. • Then, we’ll print the row of blocks we need for our pyramid, of size n. • We can run our code and see that it works: $ make recursion
$./recursion Height: 4 # ## ### ####  • We can change our conditional to if (n == 0), and type in a negative number to see what happens: $ make recursion
\$ ./recursion
Height: -100
Segmentation fault (core dumped)

• A segmentation fault means that we’ve touched memory in our computer that we shouldn’t have, and this happened since our function has called itself over and over so many times and ended up using too much memory.

## Merge sort

• We can take the idea of recusion to sorting, with another algorithm called merge sort. The pseudocode might look like:
If only one number
Quit
Else
Sort left half of number
Sort right half of number
Merge sorted halves

• Our strategy will be sorting each half, and then merge them together. Let’s start with two sorted halves that look like:
2 4 5 7 | 0 1 3 6

• We’ll look at just the first number in each list, and move the smaller, 0, into the start of a new list. Then we can look at the next number in the list:
2 4 5 7 | 0 1 3 6
^         ^

2 4 5 7 |   1 3 6
^           ^
0

• We’ll move the next smallest number, 1, down, and look at the next number in that list:
2 4 5 7 |     3 6
^             ^
0 1

• This time, the 2 is the next smallest, so we’ll move that down, and repeat this process:
  4 5 7 |     3 6
^           ^
0 1 2

4 5 7 |       6
^             ^
0 1 2 3

5 7 |       6
^           ^
0 1 2 3 4

7 |       6
^         ^
0 1 2 3 4 5

7 |
^
0 1 2 3 4 5 6

|
0 1 2 3 4 5 6 7

• So this is how we would merge two sorted halves together.
• We’ll start at the beginning now with an entirely unsorted list of numbers:
5 2 7 4 1 6 3 0

• We start by looking at the first half, which is a list of size 4:
5 2 7 4

• We’ll have to sort the left half, which is a list of size 2:
5 2

• The left half is size 1, so we’re done, and the right half is also size 1, so we can merge both halves together:
2 5

• Now we’ll need to sort the right half:
7 4

• Again, we’ll merge both of these halves together, by taking the smallest element from each list
4 7

• Now we’ve sorted both halves, each of size 2, and can merge them together:
2 5 | 4 7
^     ^

5 | 4 7
^   ^
2

5 |   7
^     ^
2 4

|   7
^
2 4 5

2 4 5 7

• We’ll repeat this for the right half, another list of size 4:
1 6 3 0

• Sorting the left half gives us 1 6, and the right half merges to 0 3.
• We’ll merge those with:
1 6 | 0 3
^     ^

1 6 |   3
^       ^
0

6 |   3
^     ^
0 1

6 |
^
0 1 3

0 1 3 6

• Now, we have two sorted halves, 2 4 5 7 and 0 1 3 6, which we can merge together as we saw above.
• Every time we merged two halves, we only needed to look at each number once. And we divided our list of 8 numbers three times, or $$\log n$$ times. We needed more memory to merge our new lists into, but the upper bound for running time for merge sort is only $$O(n \log n)$$. Since $$\log n$$ is less than $$n$$, $$O(n \log n)$$ is less than $$O(n^2)$$.
• The lower bound of our merge sort is still $$\Omega(n \log n)$$, since we have to do all the work even if the list is sorted. So merge sort also has $$\Theta(n \log n)$$.
• We look at a final visualization of sorting algorithms with a larger number of inputs, running at the same time.