Bubble Sort (OCR A Level Computer Science)
Revision Note
Bubble Sort
What is a Bubble Sort?
A Bubble Sort sorts items into order, smallest to largest, by comparing pairs of elements and swapping them if they are out of order
The first element is compared to the second, the second to the third, the third to the fourth and so on, until the second to last is compared to the last. Swaps occur if each comparison is out of order. This overall process is called a pass
Examiner Tips and Tricks
The highest value in the list eventually “bubbles” its way to the top like a fizzy drink, hence the name “Bubble sort”
Once the end of the list has been reached, the value at the top of the list is now in order and the sort resets back to the start of the list. The next largest value is then sorted to the top of the list
More passes are completed until all elements are in the correct order
A final pass checks all elements and if no swaps are made then the sort is complete
An example of using a bubble sort would be sorting an array of names into alphabetical order, or sorting an array of student marks from a test
Performing the Bubble sort
Figure 1: Performing the Bubble sort
Time Complexity of the Bubble Sort
To determine the algorithm's execution time as measured by the number of instructions or steps carried out Big-O notation must be used
Four statements are present in the above algorithm O(1), followed by a while loop O(n) containing two statements and a further for loop O(n). This results in the expression: 2n2 + 4
As coefficients and lower order values are not used, this give the bubble sort O(n2) worst case time complexity
The best case scenario would be an already sorted list which means a time complexity of O(n) as each item must still be compared to check if they are in order
The average case scenario would be an almost sorted list leading to O(n2/2), which if coefficients are removed still gives O(n2)
Space Complexity of the Bubble Sort
A bubble sort has a space complexity of O(1) because it operates in-place, requiring a fixed amount of memory
The fixed amount of memory is for variables such as loop counters and a temporary swap variable
Tracing a Bubble Sort
The trace table follows the algorithm through, updating values that change in the table
Each iteration of the list reduces the overall size of the list by 1 as after each pass the previously sorted final digit is in order and does not need to be rechecked. This means that 9 is in order and doesn't need to be rechecked, followed by 9 and 7, followed by 9, 7 and 6 and so on
When the if statement is checked a new row has been added to show the swap of list[j], list[j + 1] and Temp, followed by the subsequent change to Swap from FALSE to TRUE
After several iterations (that are not shown) the algorithm will eventually output a sorted list
Trace Table of the Bubble Sort
LastElement | Index | Mylist[Index] | Mylist[Index + 1] | Temp | Swap | Output |
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10 | 1 | 5 | 9 |
| FALSE |
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| 2 | 9 | 4 |
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| 4 | 9 | 4 | TRUE |
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| 3 | 9 | 2 |
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| 2 | 9 | 9 |
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| 4 | 9 | 6 |
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| 6 | 9 |
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| 5 | 9 | 7 |
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| 7 | 9 |
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| 6 | 9 | 1 |
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| 1 | 9 |
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| 7 | 9 | 2 |
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| 2 | 9 |
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| 8 | 4 | 9 |
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| 9 | 4 |
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| 9 | 9 | 3 |
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| 3 | 9 |
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9 | 1 | 5 | 4 |
| FALSE |
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| 4 | 5 | 5 | TRUE |
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| 2 | 5 | 2 |
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| 2 | 5 |
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1 |
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| FALSE | 1, 2, 2, 3, 4, 4, 5, 6, 7, 9 |
Implementing a Bubble Sort
Pseudocode
list = [5, 9, 4, 2, 6, 7, 1, 2, 4, 3]
last = len(list)
swap = True
i = 0
while i < (last - 1) and (swap = True)
swap = False
for j = 0 to last - i - 2
if list[j] > list[j+1]
temp = list[j]
list[j] = list[j+1]
list[j+1] = temp
swap = True
endif
next j
i = i + 1
endwhile
print(list)
The algorithm above implements an efficient bubble sort in that with each successive iteration, the number of comparisons decreases
This is due to the sorted end of the list not being rechecked as it is already sorted and therefore inefficient to compare them
This is managed by the j for loop where i increments after each iteration and reduces the number of compared items by 1
This is because the last iteration’s sorted element should already be in the correct position. The overall effect is reducing unnecessary swaps
Furthermore, using a while loop instead of a for loop means that if a swap does not occur then the list is in order. A for loop would continue comparing even if the items are already in order
Python
def bubble_sort(list):
n = len(list)
for i in range(n - 1):
swapped = False
for j in range(0, n - i - 1):
if list[j] > list[j + 1]:
list[j], list[j + 1] = list[j + 1], list[j]
swapped = True
if not swapped:
break # Early termination if no swaps occurred
return list
Java
public class BubbleSort {
public static void bubbleSort(int[] list) {
int n = list.length;
for (int i = 0; i < n - 1; i++) {
boolean swapped = false;
for (int j = 0; j < n - i - 1; j++) {
if (list[j] > list[j + 1]) {
int temp = list[j];
list[j] = list[j + 1];
list[j + 1] = temp;
swapped = true;
}
}
if (!swapped) {
break; // Early termination if no swaps occurred
}
}
}
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