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:Location: 1 2 3 4 5 6 Value:

Step minLocation maxLocation middleLocation middleValue1 1 6 (1 + 6) / 2

= 3.5 = 3

4

Binary Search

Searching for 5:Location: 1 2 3 4 5 6 Value:

Step minLocation maxLocation middleLocation middleValue1 1 6 (1 + 6) / 2

= 3.5 = 3

4

Binary Search

Searching for 5:Location: 1 2 3 4 5 6 Value:

Step minLocation maxLocation middleLocation middleValue1 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:Location: 1 2 3 4 5 6 Value:

Step minLocation maxLocation middleLocation middleValue1 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:Location: 1 2 3 4 5 6 Value:

Step minLocation maxLocation middleLocation middleValue1 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:Location: 1 2 3 4 5 6 Value:

Step minLocation maxLocation middleLocation middleValue1 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:Location: 1 2 3 4 5 6 Value:

Step minLocation maxLocation middleLocation middleValue1 1 6 (1 + 6) / 2

= 3.5 = 3

4 (value at location 3)too small, need to search higher

Binary Search

Searching for 6:Location: 1 2 3 4 5 6 Value:

Step minLocation maxLocation middleLocation middleValue1 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:Location: 1 2 3 4 5 6 Value:

Step minLocation maxLocation middleLocation middleValue1 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:Location: 1 2 3 4 5 6 Value:

Step minLocation maxLocation middleLocation middleValue1 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:Location: 1 2 3 4 5 6 Value:

Step minLocation maxLocation middleLocation middleValue1 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:Location: 1 2 3 4 5 6 Value:

Step minLocation maxLocation middleLocation middleValue1 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