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CS3024-FAZ1 Mathematical Analysis of Recursive Algorithms Design and Analysis of Algorithms (CS3024) 28/02/2006.

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Presentation on theme: "CS3024-FAZ1 Mathematical Analysis of Recursive Algorithms Design and Analysis of Algorithms (CS3024) 28/02/2006."— Presentation transcript:

1 CS3024-FAZ1 Mathematical Analysis of Recursive Algorithms Design and Analysis of Algorithms (CS3024) 28/02/2006

2 CS3024-FAZ2 Example 1 Algorithm F(n) // compute n! recursively // input: a nonnegative integer n // output: the value of n! if n = 0 return 1 else return F(n-1)*n

3 CS3024-FAZ3 Exp1: Analysis (1) Input size = n Formula: F(n) = F(n-1) * n; n>0 F(0) = 1 The number of multiplication M(n) = M(n-1) + 1; n>0 if n = 0 return 1 The call stops when n = 0 M(0) = 0 initial condition

4 CS3024-FAZ4 Exp1: Analysis (2) Solving recurrence relations: Method of backward substitution M(n) = M(n-1) + 1 = [M(n-2)+1] + 1 = M(n-2) + 2 = [M(n-3)+1] + 2 = M(n-3) + 3 General formula pattern: M(n)=M(n-i) + i Initial condition, n=0 i = n: M(n)= M(n-1)+1 = M(n-2)+2 =…= M(n-n)+n = n

5 CS3024-FAZ5 Analyzing Efficiency of Recursive Algorithms (1) 1.Decide on a parameter(s) indicating an inputs size 2.Identify the algorithms basic operation (typically, it is located in its inner most loop) 3.Check whether the number of times the basic operation is executed depends only on the size of an input. If it also depend on some additional property, the worst-case, average- case, and, if necessary, the best-case efficiencies have to be investigated separately

6 CS3024-FAZ6 Analyzing Efficiency of Recursive Algorithms (2) 4.Set up a recurrence relation, with an appropriate initial condition, for the number of times the algorithms basic operation is executed 5.Solve the recurrence or at the least ascertain(memastikan) the order of growth of its solution

7 CS3024-FAZ7 Example 2: Tower of Hanoi (1) n disks on different sizes and three pegs Initially, all disks are on the first peg in order of size. The largest on the bottom and the smallest on top The goal: move all disks to the third peg, using the second one as an auxiliary Move only one disk at a time It is forbidden to place a larger disk on top of a smaller one

8 CS3024-FAZ8 Example 2: Tower of Hanoi (2)

9 CS3024-FAZ9 Tower of Hanoi: Recursive Solution (1)

10 CS3024-FAZ10 ToH: Recursive Solution (2) To move n>1 disks from peg 1 to peg 3 (with peg 2 as an auxiliary(alat bantu)): Move recursively n-1 disk(s) from peg 1 to peg 2 (with peg 3 as an auxiluiary) Move the largest disk from peg 1 to peg 3 Move recursively n-1 disk(s) from peg 2 to peg 3 (with peg 1 as an auxiliary)

11 CS3024-FAZ11 Exp2: Analysis (1) Inputs size = the number of disks = n Basic operation = moving one disk The number of moves M(n) depends on n only: M(n) = M(n-1) M(n-1) ; for n>1 Recurrence relation: M(n) = 2M(n-1) + 1 ; for n>1 M(1) = 1 initial condition

12 CS3024-FAZ12 Exp2: Analysis (2) Backward substitution: M(n) = 2M(n-1) + 1 = 2[2M(n-2)+1]+1=2 2 M(n-2)+2+1 = 2 2 [2M(n-3)+1]+2+1=2 3 M(n-3) = 2 4 M(n-4) The pattern, after i substitution: M(n) = 2 i M(n-i) + 2 i i = 2 i M(n-i) + 2 i - 1

13 CS3024-FAZ13 Exp2: Analysis (3) Initial condition, n=1 i=n-1: M(n) = 2 i M(n-i) + 2 i - 1 = 2 (n-1) M(n-(n-1)) + 2 (n-1) -1 = 2 (n-1) M(1) + 2 (n-1) - 1 = 2 (n-1) + 2 (n-1) - 1 = 2 n - 1 Exponential algorithm! This is the most efficient algorithm It is the problems intrinsic difficulty that makes it so computationally difficult

14 CS3024-FAZ14 Example 3 Algorithm BinRec(n) //input: a positive decimal integer n //output: the number of binary digits in ns binary // representation if n = 1 return 1 else return BinRec( n/2 ) + 1

15 CS3024-FAZ15 Exp3: Analysis (1) The number of additions A(n) = A( n/2 ) + 1 ; for n>1 Recursive calls end n=1, initial condition: A(1) = 0 The presence of n/2 backward substitution stumble on values of n that are not powers of 2

16 CS3024-FAZ16 Exp3: Analysis (2) Use smoothness rule: under the very broad assumptions the order of growth for n=2 k the order of growth for all values of n n = 2 k : A( 2 k ) = A( 2 k-1 ) + 1 ; for n>1 A( 2 0 ) = 0

17 CS3024-FAZ17 Exp3: Analysis (3) Use backward substitution: A( 2 k ) = A( 2 k-1 ) + 1 = [ A( 2 k-2 ) + 1 ] + 1 = A( 2 k-2 ) + 2 = [ A( 2 k-3 ) + 1 ] + 2 = A( 2 k-3 ) + 3 … = A( 2 k-i ) + i … = A( 2 k-k ) + k = A(1) + k = k

18 CS3024-FAZ18 Exp3: Analysis (4) A( 2 k ) = A(1) + k = k n = 2 k, k = log 2 n A(n) = log 2 n (log n) The exact solution (more refined formula) A(n) = log 2 n

19 CS3024-FAZ19 Exercises Solve the following recurrence relations: x(n)=x(n-1)+5 for n>1, x(1)=0 x(n)=3x(n-1)+5 for n>1, x(1)=4 x(n)=x(n-1)+n for n>0, x(0)=0 x(n)=x(n/2)+n for n>1, x(1)=1 (solve for n=2 k ) x(n)=x(n/3)+1 for n>1, x(1)=1 (solve for n=3 k )


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