3. 12 15 3533 42 45 51 7362 75 86 98
Binary search: target x =
62
v
L:
Mid:
R:
1
6
12
1 2 3 4 5 6 7 8 9 10 11
12
if v(Mid) < x
else if v(Mid) > x
Else if v(Mid) == x
Break;
So throw away the left
half…
L = Left
R = Right
V = Value
4. 12 15 3533 42 45 51 7362 75 86 98v
L:
Mid:
R:
6
9
12
Binary search: target x = 62
1 2 3 4 5 6 7 8 9 10 11
12
if v(Mid) <x
else if v(Mid) > x
else if v(Mid) == x
Break;
So throw away the Right
half…
5. 12 15 3533 42 45 51 7362 75 86 98v
L:
Mid:
R:
6
7
9
Binary search: target x = 62
1 2 3 4 5 6 7 8 9 10 11
12
if v(Mid) < x
else if v(Mid) > x
else if v(Mid) == x
Break;
So throw away the left
half…
6. 12 15 3533 42 45 51 7362 75 86 98v
L:
Mid:
R:
7
8
9
Binary search: target x = 62
1 2 3 4 5 6 7 8 9 10 11
12
if v(Mid) < x
else if v(Mid) > x
else if v(Mid) == x
Break;
7. 12 15 3533 42 45 51 7362 75 86 98v
Binary search: target x = 62
1 2 3 4 5 6 7 8 9 10 11
12
So finally we got our
target value when mid
value and x both fulfill
the condition