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## PowerPoint Slideshow about 'CSC 221: Recursion' - vera

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Recursion: Definition

- Function that solves a problem by relying on itself to compute the correct solution for a smaller version of the problem
- Requires terminating condition: Case for which recursion is no longer needed

Recursion: Induction Basis

- Mathematical Induction:
- Prove that statement is true for first n values, given that it is true for first n-1 values
- Prove that the statement is true for a base case.

Recursion: Mathematical Induction

- Sum of first N positive integers is (N*(N+1)) / 2
- Base Case:
- 1st positive integer 1

(1 * (1+1)) / 2 =>(1*2)/2 => 2/2 => 1

- Inductive Case: Assume true for n-1
- Sum(1..N-1) = ((N-1) * (N-1+1)) / 2 => ((N-1) * (N)) / 2)

= (N2 –N)/2

- Adding N = (N2 –N)/2 + N

= (N2-N)/2 + 2N/2

= (N2 + N)/2 => (N * (N+1)) / 2

Factorial Recursion

- Factorial: n! = n * (n-1)!
- Base Case => 0! = 1
- Smaller problem => Solving (n-1)!
- Implementation:

long factorial(long inputValue)

{

if (inputValue == 0) return 1;

else return inputValue * factorial(inputValue - 1);

}

Searching

- We want to find whether or not an input value is in a sorted list:

8 in [1, 2, 8, 10, 15, 32, 63, 64]?

33 in [1, 2, 8, 10, 15, 32, 63, 64]?

Searching

int index = 0;

while (index < listSize)

{

if (list[index] == input) return index;

index++;

}

return –1;

Searching

- Better method:
- Number of operations to find input if in the list:
- Dependent on position in list
- 1 operation to size of list
- Number of operations to find input if not in the list:
- Size of list

Searching

- Better method?
- Use fact that we know the list is sorted
- Cut what we have to search in half each time
- Compare input to middle
- If input greater than middle, our value has be in the elements on the right side of the middle element
- If input less than middle, our value has to be in the elements on the left side of the middle element
- If input equals middle, we found the element.

Searching: Binary Search

for (int left = 0, right = n –1; left <= right;)

{

middle =(left + right) / 2;

if (input == list[middle]) return middle;

else if (input < list[middle]) right = middle – 1;

else left = middle + 1;

}

return – 1;

Searching: Binary Search

8 in [1, 2, 8, 10, 15, 32, 63, 64]?

1st iteration:

Left = 0, Right = 7, Middle = 3, List[Middle] = 10

Check 8 == 10 => No, 8 < 10

2nd iteration:

Left = 0, Right = 2, Middle = 1, List[Middle] = 2

Check 8 == 2 => No, 8 > 2

3rd iteration:

Left = 2, Right = 2, Middle = 2

Check 8 == 8 => Yes, Found It!

Searching: Binary Search

- Binary Search Method:
- Number of operations to find input if in the list:
- Dependent on position in list
- 1 operation if middle
- Log2 n operations maximum
- Number of operations to find input if not in the list:
- Log2 n operations maximum

Recursive Binary Search

- Two requirements for recursion:
- Same algorithm, smaller problem
- Termination condition
- Binary search?
- Search in half of previous array
- Stop when down to one element

Recursive Binary Search

int BinarySearch(int *list, const int input, const int left, const int right)

{

if (left < right)

{

middle =(left + right) / 2;

if (input == list[middle]) return middle;

else if (input < list[middle]) return BinarySearch(list, input, left, middle-1);

else return BinarySearch(list,input,middle+1,right);

}

return – 1;

}

While vs Recursion

- While and Recursion are essentially interchangeable
- Considerations:
- Efficiency
- Simplification of programming
- Readability/Understandability
- While they are equivalent, there is not always an obvious while implementation of some functions that are easily implemented with recursion

Fibonacci Computation

- Fibonacci Sequence:
- 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
- Simple definition:
- Fib[0] = 1
- Fib[1] = 1
- Fib[N] = Fib(N-1) + Fib(N-2)

Recursive Fibonacci

int fibonacci(int input)

{

if ((input == 0) || (input == 1)) return 1;

else return (fibonacci(input-1) + fibonacci(input-2));

}

Iterative Fibonacci

int fibonacci(int input)

{

int first = 1;

int second = 1;

int temp;

for (int k = 0; k < input; k++)

{

temp = first;

first = second;

second = temp + second;

}

return first;

}

Efficiency of Recursion

- Recursion can sometimes be slower than iterative code
- Two main reasons:
- Program stack usage
- Result generation

Types of Recursion

- Linear Recursion:
- 1 recursive call per function
- Factorial, Binary Search examples
- Tree Recursion:
- 2 or more recursive calls per function
- Fibonacci Example

Efficiency of Recursion

- Stack Usage:
- When a function is called by a program, that function is placed on the program call stack:

readFile()

Returns file data to be used in getData()

getData()

main()

Returns formatted data to be printed in

main()

Efficiency of Recursion

- Every stack entry maintains information about the function:
- Where to return to when the function completes
- Storage for local variables
- Pointers or copies of arguments passed in

Efficiency of Recursion

- When using recursive functions, every recursive call is added to the stack and it grows fast:
- Fibonacci (5)
- Fibonacci (5) = Fibonacci (4) +

Fibonacci(3)

- Fibonacci (1) in the computation

of Fibonacci(4) is the first time we don’t

have to call the function again.

- Stack entries take up space and

require extra processing

Fibonacci(1)

Fibonacci(2)

Fibonacci(3)

Fibonacci(4)

Fibonacci(5)

main()

Efficiency of Recursion

- Another Reason for Slowdowns [Tree Recursion]
- Traditional Recursion doesn’t save answers as it executes
- Fib(5)

= Fib(4) + Fib(3)

= Fib(3) + Fib(2) + Fib(3)

= Fib(2) + Fib(1) + Fib(2) + Fib(3)

= Fib(1) + Fib(0) + Fib(1) + Fib(2) + Fib(3)

= Fib(1) + Fib(0) + Fib(1) + Fib(1) + Fib(0) + Fib(3)

= Fib(1) + Fib(0) + Fib(1) + Fib(1) + Fib(0) + Fib(2) + Fib(1)

= Fib(1) + Fib(0) + Fib(1) + Fib(1) + Fib(0) + Fib(1) + Fib(0) + Fib(1)

- Solution: Dynamic programming – saving answers as you go and reusing them

Dynamic Programming

- Fibonacci Problem
- We know the upper bound we are solving for
- Ie Fibonacci (60) = 60 different answers
- Generate an array 60 long and initalize to –1
- Everytime we find a solution, fill it in in the array
- Next time we look for a solution, if the value in the array for the factorial we need is not –1, use the value present.

Fibonacci Examples

- Three implementations of fibonacci:
- Naive recursion implementation

(worst performance)

- Dynamic Programming recursion implementation (better)
- Iterative implementation

(best)

Recursive Datastructures

- Recursion is useful when datastructure is inherently recursive
- Unix directory hierarchy is a tree datastructure

/

/home

/var

/usr

/home/turketwh

/home/turketwh/CS112

/home/turketwh/CS221

Directory Traversal

- ls –R in Unix => recursive list
- Potential implementation?
- listDirectory(directory baseDirectory)

{

file[] files = getFiles();

int fileCount = getFileCount();

for (int i = 0; i < fileCount; i++)

{

if (files[I].type == “dir”) listDirectory(file);

else listFile(file);

}

}

Recursive Datastructures

- Will see a lot of datastructures that are recursive
- Lists [Atom + Smaller List]
- Trees [ Root + Subtrees]
- Have simple implementations because their functionality can be defined recursively.

Towers of Hanoi

Not allowed

General Problem: For any number of boxes, move boxes from start peg to

destination peg. Can never place a bigger box on top of a smaller box.

Towers Of Hanoi

- Recursive?
- 1 box from peg 1 to peg 3

1 _ _ _ _ 1

- 2 boxes from peg 1 to peg 3

1 1

2 _ _ 2 1 _ _ 1 2 _ _ 2

Towers Of Hanoi

- 3 boxes from peg 1 to peg 3

1

2 2 1 1

3 _ _ 3 _ 1 3 2 1 3 2 _ _ 2 3

1

2 2

1 2 3 1 _ 3 _ _ 3

- For n boxes,
- Solve the n-1 problem from start to the temp peg
- Move the nth box to the destination peg
- Solve the n-1 problem from the temp peg to the destination peg

Space Efficiency

- Factorial(N)
- Factorial(4) =>

= 4*Factorial(3)

= 4 * 3 * Factorial(2)

= 4 * 3 * 2 * Factorial(1)

= 4 * 3 * 2 * 1

- Has to make a maximum of N calls before base case is reached and function returns.
- N activation records will be placed on the stack.
- Since the size of the input is N, this function requires memory that is linearly related to the size of the input

Space Efficiency

- Fibonacci(N) – Standard recursive implementation
- Fib(5) = Fib(4) + Fib(3)
- For Fib(4), also puts Fib(3),Fib(2), Fib(1) on stack
- For Fib(3), also puts Fib(2), Fib(1) on stack
- Fib(4), Fib(3) compute separately
- Space Efficiency considers maximum amount required at one time – Fib(4) branch in this case
- Requires memory linearly related to input size

Tail Recursion

Tail Recursion: When the results of a recursive call are not used after the recursive call within the calling function.

long factorialHelper(long startValue, long inputValue)

{

if (startValue == 0) return 1;

else if (inputValue == 1) return startValue;

else return factorialHelper(startValue * (inputValue-1), inputValue - 1);

}

long factorial(long inputValue)

{

return factorialHelp(inputValue, inputValue);

}

Space Efficiency: Tail Recursion

- Since the results of the recursive call aren’t needed for other computations, the stack frames aren’t needed to hold partial results.
- With each call, new frame replaces old frame.
- The compiler handles these optimizations.
- Requires constant amount of memory and is not dependent on the size of the input.

Space Efficiency: Tail Recursion

factorialHelper(3,3)

main()

factorialHelper(6,1)

Return 6

factorialHelper(6,2)

Call

factorialHelper(6,2)

Return 6

main()

Call

factorialHelper(3,3)

Return 6

Call

main()

factorialHelper(6,1)

main()

Without Optimization

With Optimization looks like this:

(each call of fh is done when it calls

the next, so take it off the stack)

Recursion

- Key Ideas
- Decomposition: Solve smaller problem(s) and combine answers
- Tends to allow for very simple writing of code
- Used naively, may lead to
- Significant stack usage
- Repeated computations
- Have touched on methodologies to help work around those issues:
- Tail Recursion
- Storing answers (‘dynamic programming’)
- Suggests a new technique may be needed for computing number of operations for algorithm to complete

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