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**Introduction to Recursion and Recursive Algorithms**

CS2110

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**Different Views of Recursion**

Recursive Definition: n! = n * (n-1)! (non-math examples are common too) Recursive Procedure: a procedure that calls itself. Recursive Data Structure: a data structure that contains a pointer to an instance of itself: public class ListNode { Object nodeItem; ListNode next, previous; … }

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**Recursion in Algorithms**

Recursion is a technique that is useful for defining relationships, and for designing algorithms that implement those relationships. Recursion is a natural way to express many algorithms. For recursive data-structures, recursive algorithms are a natural choice

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What Is Recursion? A Definition Is Recursive If It Is Defined In Terms Of Itself We use them in grammar school e.g. what is a noun phrase? a noun an adjective followed by a noun phrase Descendants the person’s children all the children’s descendants

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**What Is Recursion? Think self-referential definition**

A definition is recursive if it is defined in terms of itself Exponentiation - x raised to the y power if y is 0, then 1 otherwise it’s x * (x raised to the y-1 power)

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**Other recursive definitions in mathematics**

Factorial: n! = n (n-1)! and 0! = 1! = 1 Fibonacci numbers: F(0) = F(1) = 1 F(n) = F(n-1) + F(n-2) for n > 1 Note base case Definition can’t be completely self-referential Must eventually come down to something that’s solved “directly”

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**I know the steps needed to write a simple recursive method in Java**

Strongly Agree Agree Disagree Strongly Disagree

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**Recursive Factorial public static int factorial (int n) { if (n == 1)**

return 1; else return n * factorial(n-1); } Exercise: trace execution (show method calls) for n=5

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**Why Do Recursive Methods Work?**

Activation Records on the Run-time Stack are the key: Each time you call a function (any function) you get a new activation record. Each activation record contains a copy of all local variables and parameters for that invocation. The activation record remains on the stack until the function returns, then it is destroyed. Try yourself: use your IDE’s debugger and put a breakpoint in the recursive algorithm Look at the call-stack.

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**Broken Recursive Factorial**

public static int Brokenfactorial(int n){ int x = Brokenfactorial(n-1); if (n == 1) return 1; else return n * x; } What’s wrong here? Trace calls “by hand” BrFact(2) -> BrFact(1) -> BrFact(0) -> BrFact(-1) -> BrFact(-2) -> … Problem: we do the recursive call first before checking for the base case Never stops! Like an infinite loop!

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**Recursive Design Recursive methods/functions require:**

1) One or more (non-recursive) base cases that will cause the recursion to end. if (n == 1) return 1; 2) One or more recursive cases that operate on smaller problems and get you closer to the base case. return n * factorial(n-1); Note: The base case(s) should always be checked before the recursive call.

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**Rules for Recursive Algorithms**

Base case - must have a way to end the recursion Recursive call - must change at least one of the parameters and make progress towards the base case exponentiation (x,y) base: if y is 0 then return 1 recursive: else return (multiply x times exponentiation(x,y-1))

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**How to Think/Design with Recursion**

Many people have a hard time writing recursive algorithms The key: focus only at the current “stage” of the recursion Handle the base case, then… Decide what recursive-calls need to be made Assume they work (as if by magic) Determine how to use these calls’ results

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**Example: List Processing**

Is an item in a list? First, get a reference current to the first node (Base case) If current is null, return false (Base case #2) If the first item equals the target, return true (Recursive case – might be on the remainder of the list) current = current.next return result of recursive call on new current

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**Next: Trees and Grammars**

Lab on binary tree data structures Maybe: HW5 on grammars Lecture Later: Recursion vs. iteration Which to choose? Does it matter? Maybe Later: recursion as an design strategy

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**Next: Time Complexity and Recursion**

Recursion vs. iteration Which to choose? Does it matter? Later: recursion as an design strategy

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**Recursion vs. Iteration**

Interesting fact: Any recursive algorithm can be re-written as an iterative algorithm (loops) Not all programming languages support recursion: COBOL, early FORTRAN. Some programming languages rely on recursion heavily: LISP, Prolog, Scheme.

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**To Recurse or Not To Recurse?**

Recursive solutions often seem elegant Sometimes recursion is an efficient design strategy But sometimes it’s definitely not Important! we can define recursively and implement non-recursively Many recursive algorithms can be re-written non-recursively Use an explicit stack Remove tail-recursion (compilers often do this for you)

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**To Recurse or Not to Recurse?**

Sorting Selection sort vs. mergesort – which to choose? Factorial Could just write a loop. Any advantage to the recursive version? Binary search We’ll see two versions. Which to choose? Fibonacci Let’s consider Fibonacci carefully…

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**Recursive Fibonacci method**

This is elegant code, no? long fib(int n) { if ( n == 0 ) return 1; if ( n == 1 ) return 1; return fib(n-1) + fib(n-2); } Let’s start to trace it for fib(6)

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**Trace of fib(5) For fib(5), we call fib(4) and fib(3)**

For fib(2), we call fib(1) and fib(0). Base cases! fib(1). Base case!

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**Fibonacci: recursion is a bad choice**

Note that subproblems (like fib(2) ) repeat, and solved again and again We don’t remember that we’ve solved one of our subproblems before For this problem, better to store partial solutions instead of recalculating values repeatedly Turns out to have exponential time-complexity!

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**Non-recursive Fibonacci**

Two bottom-up iterative solutions: Create an array of size n, and fill with values starting from 1 and going up to n Or, have a loop from small values going up, but only remember two previous Fibonacci values use them to compute the next one (See next slide)

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**Iterative Fibonacci long fib(int n) {**

if ( n < 2 ) return 1; long answer; long prevFib=1, prev2Fib=1; // fib(0) & fib(1) for (int k = 2; k <= n; ++k) { answer = prevFib + prev2Fib; prev2Fib = prevFib; prevFib = answer; } return answer;

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**Next: Putting Recursion to Work**

Divide and Conquer design strategy Its form Examples: Binary Search Merge Sort Time complexity issues

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Divide and Conquer It is often easier to solve several small instances of a problem than one large one. divide the problem into smaller instances of the same problem solve (conquer) the smaller instances recursively combine the solutions to obtain the solution for original input Must be able to solve one or more small inputs directly This is an algorithm design strategy Computer scientists learn many more of these

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**General Strategy for Div. & Conq.**

Solve (an input I) n = size(I) if (n <= smallsize) solution = directlySolve(I); else divide I into I1, …, Ik. for each i in {1, …, k} Si = solve(Ii); solution = combine(S1, …, Sk); return solution;

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**Why Divide and Conquer? Sometimes it’s the simplest approach**

Divide and Conquer is often more efficient than “obvious” approaches E.g. BinarySearch vs. Sequential Search E.g. Mergesort, Quicksort vs. SelectionSort But, not necessarily efficient Might be the same or worse than another approach We must analyze each algorithm's time complexity Note: divide and conquer may or may not be implemented recursively

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Top-Down Strategy Divide and Conquer algorithms illustrate a top-down strategy Given a large problem, identify and break into smaller subproblems Solve those, and combine results Most recursive algorithms work this way The alternative? Bottom-up Identify and solve smallest subproblems first Combine to get larger solutions until solve entire problem

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**Binary Search of a Sorted Array**

first mid last Strategy Find the midpoint of the array Is target equal to the item at midpoint? If smaller, look in the first half If larger, look in second half

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**Binary Search (non-recursive)**

int binSearch ( array[], target) { int first = 0; int last = array.length-1; while ( first <= last ) { mid = (first + last) / 2; if ( target == array[mid] ) return mid; // found it else if ( target < array[mid] ) // must be in 1st half last = mid -1; else // must be in 2nd half first = mid + 1 } return -1; // only got here if not found above

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**Binary Search (recursive)**

int binSearch ( array[], first, last, target) { if ( first <= last ) { mid = (first + last) / 2; if ( target == array[mid] ) // found it! return mid; else if ( target < array[mid] ) // must be in 1st half return binSearch( array, first, mid-1, target); else // must be in 2nd half return binSearch(array, mid+1, last, target); } return -1; No loop! Recursive calls takes its place But don’t think about that if it confuses you! Base cases checked first? (Why? Zero items? One item?)

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**Mergesort is Classic Divide & Conquer**

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**Algorithm: Mergesort Specification: Algorithm:**

Input: Array E and indexes first, and Last, such that the elements E[i] are defined for first <= i <= last. Output: E[first], …, E[last] is sorted rearrangement of the same elements Algorithm: void mergeSort(Element[] E, int first, int last){ if (first < last) { int mid = (first+last)/2; mergeSort(E, first, mid); mergeSort(E, mid+1, last); merge(E, first, mid, last); // merge 2 sorted halves }

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**Exercise: Trace Mergesort Execution**

Can you trace MergeSort() on this list? A = {8, 3, 2, 9, 7, 1, 5, 4};

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**Efficiency of Mergesort**

Wait for CS2150 and CS4102 to study efficiency of this and other recursive algorithms But… It is more efficient that other sorts like Selection Sort, Bubble Sort, Insertion Sort It’s O(n lg n) which is the same order-class as the most efficient sorts (also quicksort and heapsort) The point is that the D&C approach matters here, and a recursive definition and implementation is a “win”

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**Merging Sorted Sequences**

Problem: Given two sequences A and B sorted in non-decreasing order, merge them to create one sorted sequence C Input size: C has n items, and A and B have n/2 Strategy: Determine the first item in C: it should be the smaller of the first item in A and the first in B. Suppose it is the first item of A. Copy that to C. Then continue merging B with “rest of A” (without the item copied to C). Repeat!

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**Algorithm: Merge merge(A, B, C) if (A is empty) else if (B is empty)**

append what’s left in B to C else if (B is empty) append what’s left in A to C else if (first item in A <= first item in B) append first item in A to C merge (rest of A, B, C) else // first item in B is smaller append first item in B to C merge (A, rest of B, C) return

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**Summary of Recursion Concepts**

Recursion is a natural way to solve many problems Sometimes it’s a clever way to solve a problem that is not clearly recursive Sometimes recursion produces an efficient solution (e.g. mergesort) Sometimes it doesn’t (e.g. fibonacci) To use recursion or not is a design decision for your “toolbox”

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**Recursion: Design and Implementation**

“The Rules” Identify one or more base (simple) cases that can be solved without recursion In your code, handle these first!!! Determine what recursive call(s) are needed for which subproblems Also, how to use results to solve the larger problem Hint: At this step, don’t think about how recursive calls process smaller inputs! Just assume they work!

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**Exercise: Find Max and Min**

Given a list of elements, find both the maximum element and the minimum element Obvious solution: Consider first element to be max Consider first element to be min Scan linearly from 2nd to last, and update if something larger then max or if something smaller than min Class exercise: Write a recursive function that solves this using divide and conquer. Prototype: void maxmin (list, first, last, max, min); Base case(s)? Subproblems? How to combine results?

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**What’s Next? Recursive Data Structures**

Binary trees Representation Recursive algorithms Binary Search Trees Binary Trees with constraints Parallel Programming using Threads

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224 3/30/98 CSE 143 Recursion [Sections 6.1,6.3-6.7]

224 3/30/98 CSE 143 Recursion [Sections 6.1,6.3-6.7]

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