CISC 101: Fall 2011

1. CISC 101: Fall 2011 HossainShahriar shahriar@cs.queensu.ca

2. Announcement and reminder! • Tentative date for final exam shown below, please choose a time slot! • December 19 • December 20 • December 21 • Topics to be covered in this lecture • Search

3. Search • Find a specific element from a collection of object (e.g., number, string) • We assume that the collection is sorted • Two well known types • Iterative search • Binary search • Iterative search • We compare a key with each element of a collection and • A for loop is required

4. Search (iterative) • def iterative (collection, key): • for i in range(len(collection)): • if collection[i] == key: • return True • return False • def main(): • collection = [1, 3, 6, 8, 10] #list of numbers • key = 18 #element to be searched • found = iterative(collection, key) • if found == True: • print key, " found" • else: • print key, " not found" • main()

5. Search (binary) • Binary search • We divide the collection into two equal parts • Search either left or right part depending on the key value • A while loop is required • Algorithm • Initialize low and up with lower and upper boundary of the list • Get the middle element • If the middle element equals to the key, then stop • Otherwise • If key is less than middle element, update low and up with low and mid+1, respectively • If key is higher than middle element, update low and up with mid+1 and up, respectively

6. Search (binary) • List = [1, 3, 5, 7, 9, 13, 24] , key = 3 • Key in lower half • S0: low=0, up = 7 • S1: mid=(low + up)/2=(0+7)/2=4, List[mid-1]=7, key=3, low=0, up =mid+1=5 • S2: mid = (0+3)/2 = 1, List [mid-1] = 1, key =3, low=mid=1, up = 3 • S3: mid = (1+3)/2 = 2, List [mid-1] = 3, key =3, match!!

7. Search (binary) • List = [1, 3, 5, 7, 9, 13, 24] , key = 13 • Key in upper half • S0: low=0, up = 7 • S1: mid=(low + up)/2=(0+7)/2=3, List[mid-1]=5, key=13, low=3, up=7 • S2: mid = (3+7)/2 = 5, List [mid-1] = 9, key =13, low=mid=5, up = 7 • S3: mid = (5+7)/2 = 6, List [mid-1] = 13, key =13, match!!

8. Search (binary) • List = [1, 3, 5, 7, 9, 13, 24] , key = 100 • Key not in the list • S0: low=0, up = 7 • S1: mid=(low + up)/2=(0+7)/2=3, List[mid-1]=5, key=100, low=3, up=7 • S2: mid = (3+7)/2 = 5, List [mid-1] = 9, key =100, low=mid=5, up = 7 • S3: mid = (5+7)/2 = 6, List [mid-1] = 13, key =100, low=mid=6, up =7 • S4: mid = (6+7)/2 = 6, List [mid-1] = 13, key =100, low=mid=6, up =7 • S5: mid = (6+7)/2 = 6, List [mid-1] = 13, key =100, low=mid=6, up =7 • … oops!! we will stuck in the loop forever if a key is not in list • Due to our round down operation • We better do round up operation • mid = int (math.ceil((low+up)/2.0))

9. Search (binary) • def binary (collection, key): • low=0 • up = len(collection)-1 • mid=0 • while low < = up: • mid = int(math.ceil((low+up)/2.0)) • #print mid, low, up • if collection[mid-1] == key: • return True • else: • if key < collection[mid-1]: # the element could be in lower half • up = mid +1 • elif key >= collection[mid-1]: # the element could be in upper half • low = mid +1 • return False

10. Search (binary) • Take home assignment # 4 • Write a binary search algorithm • Accept a list of string and a key string • Return True or False, if the key is found or not • Example: L = [‘have’, ‘I’, ‘just’, ‘learnt’, ‘searching’] • Key=“have’, Binary(L, key) = True • Key=“hello’, Binary(L, key) = False • Hints: see how to compare two string

11. Quiz on Wednesday (Nov 23) • String operator and function • Modular programming • Our last lecture will be on Sorting!!