recursion3

College of Computing & Information Technology
King Abdulaziz University
CPCS-204 – Data Structures I
And More
Recursion

And More Recursion page 2
© Dr. Jonathan (Yahya) Cazalas
Binary Search – A reminder
 Array Search
 We are given the following sorted array:
 We are searching for the value, 19 (for example)
 Remember, we said that you search the middle
element
 If found, you are done
 If the element in the middle is greater than 19
 Search to the LEFT (cuz 19 MUST be to the left)
 If the element in the middle is less than 19
 Search to the RIGHT (cuz 19 MUST then be to the right)
index 0 1 2 3 4 5 6 7 8
value 2 6 19 27 33 37 38 41 118

Binary Search – A reminder
 Array Search
 We are given the following sorted array:
 We are searching for the value, 19
 So, we MUST start the search in the middle
INDEX of the array.
 In this case:
 The lowest index is 0
 The highest index is 8
 So the middle index is 4
index 0 1 2 3 4 5 6 7 8
value 2 6 19 27 33 37 38 41 118

Binary Search
 Array Search
 Correct Strategy
 We would ask, “is the number I am searching for, 19,
greater or less than the number stored in index 4?
 Index 4 stores 33
 The answer would be “less than”
 So we would modify our search range to in between
index 0 and index 3
 Note that index 4 is no longer in the search space
 We then continue this process
 The second index we’d look at is index 1, since (0+3)/2=1
 Then we’d finally get to index 2, since (2+3)/2 = 2
 And at index 2, we would find the value, 19, in the array

Binary Search
 Binary Search code:
public static int binSearch(int[] a, int value) {
int low = 0;
int high = a.length;
while (low <= high) {
int mid = (low + high)/2;
if (a[mid] == value)
return 1; // return 1 if found
else if (a[mid] < value)
low = mid + 1;
else
high = mid - 1;
}
return 0; // this only happens if not found

Binary Search
 Binary Search code:
 At the end of each array iteration, all we do is
update either low or high
 This modifies our search region
 Essentially halving it
 As we saw previously, this runs in log n time
 But this iterative code isn’t the easiest to read
 We now look at the recursive code
 MUCH easier to read and understand

Binary Search – Recursive
 Binary Search using recursion:
 We need a stopping case:
 We need to STOP the recursion at some point
 So when do we stop:
1) When the number is found!
2) Or when the search range is nothing
 huh?
 The search range is empty when (low > high)
 So how let us look at the code…

 Binary Search Code (using recursion):
 We see how this code follows from the
explanation of binary search quite easily
public static int binSearch(int[] a, int low, int high, int searchVal) {
int mid;
if (low <= high) {
mid = (low+high)/2;
if (searchVal < a[mid])
return binSearch(a, low, mid-1, searchVal);
else if (searchVal > a[mid])
return binSearch(a, mid+1, high, searchVal);
else
return 1;
}
return 0;
}

 Binary Search Code (using recursion):
 So if the value is found
 We return 1
 Otherwise,
 if (searchVal < a[mid])
 Then recursively call binSearch to the LEFT
 else if (searchVal > a[mid])
 Then recursively call binSearch to the RIGHT
 If low ever becomes greater than high
 This means that searchVal is NOT in the array

Brief Interlude: Human Stupidity

Recursive Exponentiation
 Example from Previous lecture
 Our function:
 Calculates be
 Some base raised to a power, e
 The input is the base, b, and the exponent, e
 So if the input was 2 for the base and 4 for the exponent
 The answer would be 24 = 16
 How do we do this recursively?
 We need to solve this in such a way that part of the
solution is a sub-problem of the EXACT same nature of
the original problem.

 Our function:
 Using b and e as input, here is our function
 f(b,e) = be
 So to make this recursive, can we say:
 f(b,e) = be = b*b(e-1)
 Does that “look” recursive?
 YES it does!
 Why?
 Cuz the right side is indeed a sub-problem of the original
 We want to evaluate be
 And our right side evaluates b(e-1)

 Our function:
 f(b,e) = b*b(e-1)
 So we need to determine the terminating condition!
 We know that f(b,0) = b0 = 1
 So our terminating condition can be when e = 1
 Additionally, our recursive calls need to return an
expression for f(b,e) in terms of f(b,k)
 for some k < e
 We just found that f(b,e) = b*b(e-1)
 So now we can write our actual function…

 Code:
// Pre-conditions: e is greater than or equal to 0.
// Post-conditions: returns be.
public static int Power(int base, int exponent) {
if ( exponent == 0 )
return 1;
else
return (base*Power(base, exponent-1));
}

 Say we initially call the function with 2 as our base
and 8 as the exponent
 The final return will be
 return 2*2*2*2*2*2*2*2
 Which equals 256
 You notice we have 7 multiplications (exp was 8)
 The number of multiplications needed is one less
than the exponent value
 So if n was the exponent
 The # of multiplications needed would be n-1

Fast Exponentiation
 This works just fine
 BUT, it becomes VERY slow for large exponents
 If the exponent was 10,000, that would be 9,999 mults!
 How can we do better?
 One key idea:
 Remembering the laws of exponents
 Yeah, algebra…the thing you forgot about two years ago
 So using the laws of exponents
 We remember that 28 = 24*24

Fast Exponentiation
 One key idea:
 Remembering the laws of exponents
 28 = 24*24
 Now, if we know 24
 we can calculate 28 with one multiplication
 What is 24?
 24 = 22*22
 and 22 = 2*(2)
 So… 2*(2) = 4, 4*(4) = 16, 16*(16) = 256 = 28
 So we’ve calculated 28 using only three multiplications
 MUCH better than 7 multiplications

Fast Exponentiation
 Example of Fast Exponentiation
 So, in general, we can say:
 bn = bn/2*bn/2
 So to find bn, we find bn/2
 HALF of the original amount
 And to find bn/2, we find bn/4
 Again, HALF of bn/2
 This smells like a log n running time
 log n number of multiplications
 Much better than n multiplications
 But as of now, this only works if n is even

Fast Exponentiation
 bn = bn/2*bn/2
 This works when n is even
 But what if n is odd?
 Notice that 29 = 24*24*2
/ 2 / 2
/ 2 / 2
( ) if n is even
( )( ) if n is odd
n n
n
n n
a a
a
a a a

= 


Fast Exponentiation
 Also, this method relies on “integer division”
 We’ve briefly discussed this
 Basically if n is 9, then n/2 = 4
 Integer division
 Think of it as dividing
 AND then rounding down, if needed, to the nearest integer
 So if n is 121, then n/2 = 60
 Finally, if n is 57, then n/2 = 28
 Using the same base case as the previous
power function, here is the code…

Fast Exponentiation
 Code:
Public static int powerB(int base, int exp) {
if (exp == 0)
return 1;
else if (exp == 1)
return base;
else if (exp%2 == 0)
return powerB(base*base, exp/2);
else
return base*powerB(base, exp-1);
}

Recursion
WASN’T
THAT
BODACIOUS!

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