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Recursion Ellen Walker CPSC 201 Data Structures Hiram College

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Divide & Conquer To solve a big problem, break it down into smaller problems Examples: –Find the largest element in a list by comparing pairs of elements –Binary search: find an element in a list by deciding which half of the list it is in, and sorting the smaller half-list

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Recursion Every smaller problem (except the smallest one) is an instance of the larger problem Examples: –Binary Search –Factorial –Fibonacci number Implementation: function that calls itself

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Sequential Search as Recursion //Find (the first instance of) element in list If the list is empty, return “not found” Else if the first element of the list is the one you’re looking for, return it Else search in the smaller list starting from the second element of the list

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Sequential Search in Java //Check if value is in array of integers //Note: end is one value beyond the last in list boolean search1 (int value, int[] list, int st, int end){ if (st==end) return false; // empty list if (list[st]==value) return true; return search1(value, list, st+1, end); }

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Sequential Search as Divide & Conquer Divide the problem into two smaller problems: 1.Is the value the first element of the list? 2.Find the value in the smaller list starting at the second element. – Disadvantage: each “smaller” problem isn’t very much smaller

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Binary Search Is the element in the first or second half of the list? (For a sorted list, this is a single comparison to the middle element) Search for the element in the appropriate half of the list

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Binary Search in Java boolean search2(int val, int[] list, int st, int end){ if (st==end) return false; //empty list int middle = (end-st)/2; if (list[middle]==val) return true; if (val < list[middle]) return search2(val,list,st,middle)); else return search2(val, list, middle+1, end); }

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Binary vs. Sequential Sequential search: each list search is 1 smaller Binary search: each list search is half as big Binary search is faster, but… Binary search requires a sorted list

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Aspects of A Recursive Solution Base case (or basis or degenerate case) –List is empty, List has 1 element Recursive call –Call to search1 or search2 Recursive call must be “closer” to the base case –The list is shorter in the recursive call

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4 Questions for Recursive Algorithm Design 1. How can you define the problem in terms of a smaller problem of the same type? “single step” + “smaller problem” 2. How does each recursive call diminish the size of the problem? 3. What instance of the problem can serve as the base case? 4. As the problem size diminishes, will you reach this base case?

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4 Answers for Factorial 1. How can you define the problem… Fact(N) is N*Fact(N-1) 2. How does each … call diminish… N-1 is smaller than N 3. What … can serve as the base case? Fact(0) is 1 4. …will you reach the base case? Yes, as long as you start with N≥0 (a precondition)

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Factorial in Java Note: this is a “valued” recursive function The final result is constructed from the results of recursive calls int Fact(int N){ assert (N>=0); //make sure of precondition if (N==0) return 1; else return N*Fact(N-1); }

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Activation record Information recorded for each function call –Values of the parameters –Function’s local variables –Space for result –Where to return to (important if multiple calls!) For each function call, create a new activation record For each return, erase the activation record

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Writing a string backward Base case: Define the problem in terms of a smaller one? How does each case diminish…? Guaranteed to reach the base case?

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Multiple Calls in One Function Fibonacci sequence (counting rabbits) Rabbit(N) = Rabbit(N-1)+Rabbit(N-2) Rabbit(1) = Rabbit(2) = 1 (Rabbit is not defined for values < 1)

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A Counting Problem How many ways to visit k of n planets? –Pick some planet (p) –There are count(n-1, k) ways to visit k planets without visiting p –There are count(n-1, k-1) ways to visit k planets with visiting p –Therefore: count(n, k) = count(n-1,k)+count(n-1,k-1) –Base case?

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Impractical Recursions Fibonacci: makes as many calls as rabbits! –Solution: compute it iteratively (save intermediate steps) Counting: similar to Fibonacci –Solution: use mathematics to come up with a closed form –Count(n,k) = n! / (k! (n-k)!)

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Impractical but unavoidable Towers of Hanoi Move N disks from one tower to another, never putting a larger disk on a smaller disk Solution in 3 steps: –Recursively move N-1 disks to the extra tower –Move the Nth disk to the destination tower –Recursively move N-1 disks to the destination tower Solution time: O(2 N )

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K’th smallest element Pick a “partition” element & put smaller elements before, and larger elements after –Remember quicksort? If “partition” element is k’th, it is the answer Otherwise, recurse over the half that has the k’th element in it (change k if it’s the second half!) Impossible to predict how long it will take without knowing the exact array!

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Recursive Linked List The empty list is a linked list A list consisting of a head item followed by a list of the rest of the items is a linked list Method to –Add at the beginning of the list –Add at the end of the list Recursive method to –Count the elements –See if an element is in the list

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