Скачать презентацию Recursion Credits Jeff Edmonds Ping Xuan Code Скачать презентацию Recursion Credits Jeff Edmonds Ping Xuan Code

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Recursion Credits: Jeff Edmonds, Ping Xuan Recursion Credits: Jeff Edmonds, Ping Xuan

Code Representation of an Algorithm MULT(X, Y): If |X| = |Y| = 1 then Code Representation of an Algorithm MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f

Tree of Frames MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Tree of Frames MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f X = 21 Y = 23 ac = 4 bd = 3 (a+b)(c+d) = 15 XY = 483 X=2 Y=2 XY=4 X=1 Y=3 X=3 Y=5 XY=15 X = 2133 Y = 2312 ac = 483 bd = 396 (a+b)(c+d) = 1890 XY = 4931496 X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 X=3 Y=1 XY=3 X=3 Y=2 XY=6 X = 54 Y = 35 ac = 15 bd = 20 (a+b)(c+d) = 72 XY = 1890 X=6 Y=3 XY=18 X=5 Y=3 XY=15 X=4 Y=5 XY=20 X=9 Y=8 XY=72

Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f X = 21 Y = 23 ac = 4 bd = 3 (a+b)(c+d) = 15 XY = 483 X=2 Y=2 XY=4 X=1 Y=3 X=3 Y=5 XY=15 X = 2133 Y = 2312 ac = 483 bd = 396 (a+b)(c+d) = 1890 XY = 4931496 X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 X=3 Y=1 XY=3 X=3 Y=2 XY=6 One Friend for each stack frame. Each worries only about his job. X = 54 Y = 35 ac = 15 bd = 20 (a+b)(c+d) = 72 XY = 1890 X=6 Y=3 XY=18 X=5 Y=3 XY=15 X=4 Y=5 XY=20 X=9 Y=8 XY=72

Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f X = 21 Y = 23 ac = 4 bd = 3 (a+b)(c+d) = 15 XY = 483 X=2 Y=2 XY=4 X=1 Y=3 X=3 Y=5 XY=15 X = 2133 Y = 2312 ac = 483 bd = 396 (a+b)(c+d) = 1890 XY = 4931496 X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 X=3 Y=1 XY=3 X=3 Y=2 XY=6 Worry about one step at a time. Imagine that you are one specific friend. X = 54 Y = 35 ac = 15 bd = 20 (a+b)(c+d) = 72 XY = 1890 X=6 Y=3 XY=18 X=5 Y=3 XY=15 X=4 Y=5 XY=20 X=9 Y=8 XY=72

Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f X = 33 Y = 12 ac = bd = (a+b)(c+d) = XY = • Consider your input instance

Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f X = 33 Y = 12 ac = ? bd = ? (a+b)(c+d) = XY = X=3 Y=1 XY=? X=3 Y=2 XY=? X=6 Y=3 XY=? • Consider your input instance • Allocate work • Construct one or more subinstances

Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = X=3 Y=1 XY=3 X=3 Y=2 XY=6 X=6 Y=3 XY=18 • Consider your input instance • Allocate work • Construct one or more subinstances • Assume by magic your friends give you the answer for these.

Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 X=3 Y=1 XY=3 X=3 Y=2 XY=6 X=6 Y=3 XY=18 • Consider your input instance • Allocate work • Construct one or more subinstances • Assume by magic your friends give you the answer for these. • Use this help to solve your own instance.

Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 X=3 Y=1 XY=3 X=3 Y=2 XY=6 X=6 Y=3 XY=18 • Consider your input instance • Allocate work • Construct one or more subinstances • Assume by magic your friends give you the answer for these. • Use this help to solve your own instance. • Do not worry about anything else. • Who your boss is. • How your friends solve their instance.

Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 X=3 Y=1 XY=3 X=3 Y=2 XY=6 X=6 Y=3 XY=18 This technique is often referred to as Divide and Conquer

Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Consider generic Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Consider generic instances. Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f MULT(X, Y): Break X into a; b and Y into c; d If |X| = |Y| = 1 then RETURN XY e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f ac bd (a+b)(c+d) Remember! Do not worry about • Who your boss is. • How your friends solve their instance.

Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f X = 21 Y = 23 ac = 4 bd = 3 (a+b)(c+d) = 15 XY = 483 X=2 Y=2 XY=4 X=1 Y=3 X=3 Y=5 XY=15 X = 2133 Y = 2312 ac = 483 bd = 396 (a+b)(c+d) = 1890 XY = 4931496 X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 X=3 Y=1 XY=3 X=3 Y=2 XY=6 This solves the problem for every possible instance. X = 54 Y = 35 ac = 15 bd = 20 (a+b)(c+d) = 72 XY = 1890 X=6 Y=3 XY=18 X=5 Y=3 XY=15 X=4 Y=5 XY=20 X=9 Y=8 XY=72

Friends Recursive Algorithm: • Assume you have an algorithm that works. • Use it Friends Recursive Algorithm: • Assume you have an algorithm that works. • Use it to write an algorithm that works.

Friends Recursive Algorithm: • Assume you have an algorithm that works. • Use it Friends Recursive Algorithm: • Assume you have an algorithm that works. • Use it to write an algorithm that works. If I could get in, I could get the key. Then I could unlock the door so that I can get in. Circular Argument!

Friends & Strong Induction Recursive Algorithm: • Assume you have an algorithm that works. Friends & Strong Induction Recursive Algorithm: • Assume you have an algorithm that works. • Use it to write an algorithm that works. To get into my house I must get the key from a smaller house

MULT(X, Y): Friends & Strong Induction If |X| = |Y| = 1 then RETURN MULT(X, Y): Friends & Strong Induction If |X| = |Y| = 1 then RETURN XY Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f X = 33 Y = 12 ac = ? bd = ? (a+b)(c+d) = XY = X=3 Y=1 XY=? X=3 Y=2 XY=? • Allocate work • Construct one or more subinstances Each subinstance must be a smaller instance to the same problem. X=6 Y=3 XY=?

Friends & Strong Induction Recursive Algorithm: • Assume you have an algorithm that works. Friends & Strong Induction Recursive Algorithm: • Assume you have an algorithm that works. • Use it to write an algorithm that works. Use brute force to get into the smallest house.

Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X Friends MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Break X into a; b and Y into c; d e = MULT(a, c) and f =MULT(b, d) RETURN e 10 n + (MULT(a+b, c+d) – e - f) 10 n/2 + f MULT(X, Y): If |X| = |Y| = 1 then RETURN XY Use brute force to solve the base case instances.

Towers of Hanoi Towers of Hanoi

Towers of Hanoi Towers of Hanoi

Towers of Hanoi Get a job at the Towers of Hanoi company at level Towers of Hanoi Get a job at the Towers of Hanoi company at level n=1, 2, 3, 4, …. . Or get transferred in as the president & you decide what your vice president does.

Towers of Hanoi Time: T(1)=1, T(n) = 2 T(n-1) + 1 = (2 n) Towers of Hanoi Time: T(1)=1, T(n) = 2 T(n-1) + 1 = (2 n) (Homework 1. Assume n = 2 k)

Tower of Hanoi • http: //www. mazeworks. com/hanoi/index. htm • http: //www. cut-theknot. com/recurrence/hanoi. Tower of Hanoi • http: //www. mazeworks. com/hanoi/index. htm • http: //www. cut-theknot. com/recurrence/hanoi. shtml • http: //www. cs. yorku. ca/~andy/courses/3101/l ecture-notes/LN 0. html

Sorting Problem Specification • Precondition: The input is a list of n values with Sorting Problem Specification • Precondition: The input is a list of n values with the same value possibly repeated. • Post condition: The output is a list consisting of the same n values in non-decreasing order. 88 52 14 31 25 98 30 23 62 79 14, 23, 25, 30, 31, 52, 62, 79, 88, 98

Recursive Sorts • Given list of objects to be sorted • Split the list Recursive Sorts • Given list of objects to be sorted • Split the list into two sublists. • Recursively have a friend sort the two sublists. • Combine the two sorted sublists into one entirely sorted list.

Merge Sort 88 52 14 31 25 98 30 23 62 79 Divide and Merge Sort 88 52 14 31 25 98 30 23 62 79 Divide and Conquer

Merge Sort 88 52 14 31 25 98 30 23 62 79 Get one Merge Sort 88 52 14 31 25 98 30 23 62 79 Get one friend to sort the first half. 25, 31, 52, 88, 98 Split Set into Two (no real work) Get one friend to sort the second half. 14, 23, 30, 62, 79

Merge Sort Merge two sorted lists into one 25, 31, 52, 88, 98 14, Merge Sort Merge two sorted lists into one 25, 31, 52, 88, 98 14, 23, 25, 30, 31, 52, 62, 79, 88, 98 14, 23, 30, 62, 79

Merge Sort merge_sort ( array a[ ] ) { if (length of a > Merge Sort merge_sort ( array a[ ] ) { if (length of a > 1) then { merge_sort ( left half of a ); merge_sort ( right half of a); merge the above two sorted half arrays; } } Time: T(n) = 2 T(n/2) + n = (n log(n)) (Homework 2. Assume n = 2 k)

Recursive Images if n=0, draw if n=1 n=0 else draw And recursively Draw here Recursive Images if n=0, draw if n=1 n=0 else draw And recursively Draw here with n-1

Recursive Images if n=0, draw if n=2 n=1 else draw And recursively Draw here Recursive Images if n=0, draw if n=2 n=1 else draw And recursively Draw here with n-1

Recursive Images if n=0, draw if n=3 n=2 else draw And recursively Draw here Recursive Images if n=0, draw if n=3 n=2 else draw And recursively Draw here with n-1

Recursive Images if n=0, draw if n=30 else draw And recursively Draw here with Recursive Images if n=0, draw if n=30 else draw And recursively Draw here with n-1

Recursive Images if n=0 n=5 if n=1 n=4 n=3 n=2 Recursive Images if n=0 n=5 if n=1 n=4 n=3 n=2

Recursive Images if n=0 if n=1 n=5 n=4 n=3 n=2 Recursive Images if n=0 if n=1 n=5 n=4 n=3 n=2

Recursive Images L(n) = 4/3 L(n-1) = (4/3)n Þ ¥ Recursive Images L(n) = 4/3 L(n-1) = (4/3)n Þ ¥

Homework 3: Derive the recurrence relation for the time complexity T( N ) and Homework 3: Derive the recurrence relation for the time complexity T( N ) and solve it. Assume that N = 2 k.

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