A step-by-step illustration of Binary Search to help you walk through a series of operations. Illustration is accompanied by actual code with bold line indicating the current operation. https://github.com/yoshiwatanabe/Algorithms/blob/master/Finding/BinarySearch.cs

1. Binary Search
Illustrated walk through

2. Iterative binary search
int begin = 0;
int last = array.Length - 1;
int mid = 0;
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
Part #1 Initialize pointers
Part #2 Search

3. begin
last
mid
x
4
This is what we
search for
[0]
[1]
[2]
[3]
2
4
5
6
Let’s look for 4

4. begin
0
x
last
mid
3
0
[0]
[1]
[2]
[3]
2
4
5
6
int begin = 0;
int last = array.Length - 1;
int mid = 0;
4

5. begin
0
x
0 <= 3
is true
mid
last
3
0
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
4

6. begin
0
x
last
(0+3)/2 = 1
mid
3
1
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
4

7. begin
0
x
last
mid
4
3
1
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
4<4
is false

8. begin
0
x
last
mid
4
3
1
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if ( array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
4>4
is false

9. begin
0
x
last
mid
4
3
1
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
We found x=4 at
index 1!
return -1;

10. begin
last
mid
x
5
This is what we
search for
[0]
[1]
[2]
[3]
2
4
5
6
Let’s look for 5

11. begin
0
x
last
mid
3
0
[0]
[1]
[2]
[3]
2
4
5
6
int begin = 0;
int last = array.Length - 1;
int mid = 0;
5

12. begin
0
x
0 <= 3
is true
mid
last
3
0
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
5

13. begin
0
x
last
(0+3)/2 = 1
mid
3
1
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
5

14. begin
0
x
last
mid
5
3
1
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
4<5
is true

15. begin
2
x
last
mid
3
1
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
5

16. begin
2 <= 3
is true
2
x
last
mid
3
1
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
5

17. begin
2
x
last
(2+3)/2 =2
mid
3
2
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
5

18. begin
2
x
last
mid
3
2
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
5
5<5
is false

19. begin
2
x
last
mid
3
2
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if ( array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
5
5>5
is false

20. begin
2
x
last
mid
5
3
2
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
We found x=5 at
index 2!
return -1;

21. begin
last
mid
x
6
This is what we
search for
[0]
[1]
[2]
[3]
2
4
5
6
Let’s look for 6

22. begin
0
x
last
mid
3
0
[0]
[1]
[2]
[3]
2
4
5
6
int begin = 0;
int last = array.Length - 1;
int mid = 0;
6

23. begin
0
x
0 <= 3
is true
mid
last
3
0
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
6

24. begin
0
x
last
(0+3)/2 = 1
mid
3
1
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
6

25. begin
0
x
last
mid
6
3
1
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
4<6
is true

26. begin
2
x
last
mid
3
1
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
6

27. begin
2 <= 3
is true
2
x
last
mid
3
1
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
6

28. begin
2
x
last
(2+3)/2 =2
mid
3
2
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
6

29. begin
2
x
last
mid
3
2
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
6
5<6
is true

30. begin
last
mid
3
3
2
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
x
6

31. begin
last
3 <= 3
is true
mid
3
3
2
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
x
6

32. begin
last
3
mid
(3+3)/2 = 3
3
3
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
x
6

33. begin
3
last
3
mid
3
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
x
6
6<6
is false

34. begin
3
last
3
mid
3
[0]
[1]
[2]
[3]
2
4
5
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if ( array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
return -1;
x
6
6>6
is false

35. begin
3
last
3
mid
3
[0]
[1]
[2]
4
5
6
[3]
2
x
6
while (begin <= last) {
mid = (begin + last) / 2;
if (array[mid] < x) {
begin = mid + 1;
}
else if (array[mid] > x) {
last = mid - 1;
}
else {
return mid;
}
}
We found x=6 at
index 3!
return -1;