Searching & Sorting

Algorithms Step by step recipe to do a task…

Algorithms Step by step recipe to do a task where: Operations are computable Operations are unambiguous Operations are well ordered Finite number of operations

Linear Search I'm thinking of a number between 1 and 100 You try to guess it I'll give too low/too high hints

Linear Search I'm thinking of a number between 1 and 100 You try to guess it I'll give too low/too high hints Method #1 – Linear Search 1, 2, 3….

Linear Search Algorithm In pseudocode:

Binary Search Method #2 – Binary Search Pick middle of remaining search space Too high? Eliminate middle and above Too low? Eliminate middle and below

Algorithm

Binary Search Searching for 5: Step minLocation maxLocation middleLocation middleValue 1 6 (1 + 6) / 2 = 3.5 = 3 4

Binary Search Searching for 5: Step minLocation maxLocation middleLocation middleValue 1 6 (1 + 6) / 2 = 3.5 = 3 4

Binary Search Searching for 5: Step minLocation maxLocation middleLocation middleValue 1 6 (1 + 6) / 2 = 3.5 = 3 4 2 4 (one more than old middleLocation) 6 (unchanged) (4 + 6) / 2 = 5 7 (value at location 5) This is too big, need to search lower

Binary Search Searching for 5: Step minLocation maxLocation middleLocation middleValue 1 6 (1 + 6) / 2 = 3.5 = 3 4 2 4 (one more than old middleLocation) 6 (unchanged) (4 + 6) / 2 = 5 7 (value at location 5) This is too big, need to search lower

Binary Search Searching for 5: Step minLocation maxLocation middleLocation middleValue 1 6 (1 + 6) / 2 = 3.5 = 3 4 2 4 (one more than old middleLocation) 6 (unchanged) (4 + 6) / 2 = 5 7 (value at location 5) This is too big, need to search lower 3 4 (unchanged) 4 (one less than old middleLocation) (4 + 4) / 2 = 8 / 2 = 4 5 Found it!!!

Binary Search Searching for 5: Step minLocation maxLocation middleLocation middleValue 1 6 (1 + 6) / 2 = 3.5 = 3 4 2 4 (one more than old middleLocation) 6 (unchanged) (4 + 6) / 2 = 5 7 (value at location 5) This is too big, need to search lower 3 4 (unchanged) 4 (one less than old middleLocation) (4 + 4) / 2 = 8 / 2 = 4 5 Found it!!!

Binary Search Searching for 6: Step minLocation maxLocation middleLocation middleValue 1 6 (1 + 6) / 2 = 3.5 = 3 4 (value at location 3) too small, need to search higher

Binary Search Searching for 6: Step minLocation maxLocation middleLocation middleValue 1 6 (1 + 6) / 2 = 3.5 = 3 4 (value at location 3) too small, need to search higher 2 4 (one more than old middleLocation) 6 (unchanged) (4 + 6) / 2 = 10 / 2 = 5 7 (value at location 5) too big, need to search lower

Binary Search Searching for 6: Step minLocation maxLocation middleLocation middleValue 1 6 (1 + 6) / 2 = 3.5 = 3 4 (value at location 3) too small, need to search higher 2 4 (one more than old middleLocation) 6 (unchanged) (4 + 6) / 2 = 10 / 2 = 5 7 (value at location 5) too big, need to search lower

Binary Search Searching for 6: Step minLocation maxLocation middleLocation middleValue 1 6 (1 + 6) / 2 = 3.5 = 3 4 (value at location 3) too small, need to search higher 2 4 (one more than old middleLocation) 6 (unchanged) (4 + 6) / 2 = 10 / 2 = 5 7 (value at location 5) too big, need to search lower 3 4 (unchanged) 4 (one less than old middleLocation) (4 + 4) / 2 = 8 / 2 = 4 5 (value at location 3) too small, need to search higher

Binary Search Searching for 6: Step minLocation maxLocation middleLocation middleValue 1 6 (1 + 6) / 2 = 3.5 = 3 4 (value at location 3) too small, need to search higher 2 4 (one more than old middleLocation) 6 (unchanged) (4 + 6) / 2 = 10 / 2 = 5 7 (value at location 5) too big, need to search lower 3 4 (unchanged) 4 (one less than old middleLocation) (4 + 4) / 2 = 8 / 2 = 4 5 (value at location 3) too small, need to search higher

Binary Search Searching for 6: Step minLocation maxLocation middleLocation middleValue 1 6 (1 + 6) / 2 = 3.5 = 3 4 (value at location 3) too small, need to search higher 2 4 (one more than old middleLocation) 6 (unchanged) (4 + 6) / 2 = 10 / 2 = 5 7 (value at location 5) too big, need to search lower 3 4 (unchanged) 4 (one less than old middleLocation) (4 + 4) / 2 = 8 / 2 = 4 5 (value at location 3) too small, need to search higher 4 5 (one more than old middleLocation) 4 (unchanged) minLocation > maxLocation - we have nothing left to check - value is not there!

Basic Sorts

Sorting How do we sort?

Selection Sort A human algorithm:

Selection Sort In a computer: http://computerscience.chemeketa.edu/cs160Reader/Algorithms/SelectionSort2.html

Insertion Sort For a human:

Selection Sort In a computer: http://computerscience.chemeketa.edu/cs160Reader/Algorithms/InsertionSort2.html