Cse547. Chapter 1 Problem 16 GENERALIZATION

Transcription

1 Cse547 Chapter 1 Problem 16 GENERALIZATION

2 Use the Repertoire method to solve the general fiveparameter recurrence: g(1) = α g(2n+ j) = 3g(n) + Ɛn 2 + γn + β j, for j = 0, 1 and n 1 and n є N

3 Convert recursive formula into R: g(1) = α g(2n+ 0) = 3g(n) + Ɛn 2 + γn + β 0 g(2n+ 1) = 3g(n) + Ɛn 2 + γn + β 1 (for n є N and n 1)

4 Compute few initial values for R Observe values and guess a general format for the required closed form Use Repertoire method to find exact closed formula If fails at some point, use Binary Expansion

5 g(1) = α g(2) = g(2(1) + 0) = 3g(1) + Ɛ + γ+ β 0 = 3α + Ɛ + γ+ β 0 g(3) = g(2(1) + 1) = 3g(1) + Ɛ + γ+ β 1 = 3α + Ɛ + γ+ β 1 g(4) = g(2(2) + 0) = 3g(2) + 4Ɛ + 2γ+ β 0 = 9α + 7Ɛ +5γ + 4β 0 g(5) = g(2(2) + 1) = 3g(2) + 4Ɛ + 2γ + β 1 = 9α + 7Ɛ +5γ + 3β 0 + β 1

6 g(6) = g(2(3) + 0) = 3g(3) + 9Ɛ + 3γ + β 0 = 9α + 12Ɛ +6γ + β 0 + 3β 1 g(7) = g(2(3) + 1) = 3g(3) + 9Ɛ + 3γ+ β 1 = 9α + 12Ɛ + 6γ + 4β 1 g(8) = g(2(4) + 0) = 3g(4) + 16Ɛ + 4γ + β 0 = 27α + 37Ɛ + 19γ + 13β 0 g(9) = g(2(4) + 1) = 3g(4) + 16Ɛ + 4γ + β 1 = 27α + 37Ɛ +19γ + 12β 0 + β 1 g(10) = g(2(5) + 0) = 3g(5) + 25Ɛ + 5γ + β 0 = 27α + 46Ɛ +20γ+ 10β 0 + 3β 1

7 n g(n) 1 α + 0Ɛ +0γ + 0β 0 + 0β 1 2 3α + Ɛ + γ + β 0 + 0β 1 3 3α + Ɛ + γ + 0β 0 + β 1 4 9α +7Ɛ+ 5γ+ 4β 0 + 0β 1 5 9α +7Ɛ+ 5γ+ 3β 0 + β 1 6 9α +12Ɛ + 6γ + β 0 +3β 1 7 9α +12Ɛ + 6γ + 0β 0 + 4β α +37Ɛ + 19γ + 13β 0 + 0β 1

8 n g(n) 9 27α +37Ɛ+ 19γ + 12β 0 + β α +46Ɛ + 20γ + 10β 0 + 3β 1

9 g(n) = A(n)α + B(n)Ɛ +C(n)γ + D(n)β 0 + E(n)β 1 For all n N and n 1, n can be written in the form (2 k + h), where, 2 k is the highest power of 2, not exceeding n, and (0 h < 2 k ). (Here h є N). From our table we can observe that when n = 2 k + h, then the coefficient for α = 3 k. Example, let n=10, then n = , and the coefficient for α after computing g(n) is, 3 3 = 27.

10 CASE I: g(n) = A(n) CASE II: g(n) = 1 CASE III: g(n) = n x CASE IV: g(n) = n 2 x CASE V: g(n) = n 3 for all n є N

11 Let: Ɛ = γ = β 0 = β 1 = 0 and α =1 In this case our general formula becomes: g (n) = A(n)1 + B(n)0 + C(n)0 + D(n)0 + E(n)0 g (n) = A(n) for all n є N

12 So, now the recurrence relation becomes: A(1) = α A(2n+ j) = 3A(n) for j = 0,1 and n 1 and n є N [Recall: g(2n+ j) = 3g(n) + Ɛn 2 + γn + β j ]

13 Now we want to prove that: A(2 k + h) = 3 k α for all k 0 (Here 0 h < 2k and h є N ) It is easy to prove by induction that: A(2 k + h) = 3 k α for all k 0

14 A(1) = 1, A(2n) = 3A(n), A(2n+1) = 3A(n), A(n) = 3 k for n = 2 k + h where 0 h < 2 k Base case: k=0 => n = 1 + h, h < 1 so n = 1 g(n) = A(n) and g(1) = 1 => A(1) = 1 A(1) = 3 0 = 1 [YES] Assume A(2 k 1 + h) = 3 k 1, 0 h < 2 k 1 To prove: A(2 k + h) = 3 k, 0 h < 2 k d

15 Case n even (n = 2n 1 ): 2 k + h = 2n 1 iff h is even => n 1 = 2 k 1 + h/2 A(2n 1 ) = A(2 k + h) = 3A(n 1 ) = 3A(2 k 1 + h/2) = 3. 3 k 1 = 3 k [YES] Case n odd (n = 2n 1 + 1): 2 k + h = 2n iff h is odd => n 1 = 2 k 1 + (h 1)/2 A(2n 1 +1) = A(2 k + h) = 3A(n 1 ) = 3A(2 k 1 + (h 1)/2) = 3. 3 k 1 = 3 k [YES]

16 We have that g(n)=1, for all n є N. From g(1) = α, we find that g(1) = 1 = α. So α = 1. g(2n+0) = 3g(n) + Ɛn 2 + γn + β 0 1 = 3(1) + Ɛn 2 + γn + β 0 1= 3 + Ɛn 2 + γn + β = Ɛn 2 + γn + β 0 0 = Ɛn 2 + γn + β 0 +2, for all n є N Hence, Ɛ = γ = 0 and β 0 = 2

17 g(2n+1) = 3g(n) + Ɛn 2 + γn + β 1 1 = 3(1) + Ɛn 2 + γn + β 1 1= 3 + Ɛn 2 + γn + β = Ɛn 2 + γn + β 1 0 = Ɛn 2 + γn + β 1 +2, for all n є N Hence, Ɛ = γ = 0 and β 1 = 2

18 Plugging the values of α, Ɛ, γ, β and β 0 1 into the general form of the CF: g (n) = A(n)α + B(n)Ɛ + C(n)γ + D(n)β 0 + E(n)β 1 1 = A(n) 2D(n) 2E(n), for all n є N.

19 We have that g(n) =n, for all n є N. From g(1) = α, we find that g(1) = 1 = α. So α = 1. g(2n+0) = 3g(n) + Ɛn 2 + γn + β 0 2n = 3n + Ɛn 2 + γn + β 0 0 = 3n 2n + Ɛn 2 + γn + β 0 0 = Ɛn 2 + n + γn + β 0 0 = Ɛn 2 + (γ+1)n + β 0,, for all n є N. Hence γ = 1, Ɛ = β 0 = 0

20 We have that g(n)=n, for all n g(2n+1) = 3g(n) + Ɛn 2 + γn + β 1 2n + 1 = 3n + Ɛn 2 + γn + β 1 1 = 3n 2n + Ɛn 2 + γn + β 1 1 = Ɛn 2 + n + γn + β 1 0= Ɛn 2 + (γ+1)n 1 + β 1, for all n є N. Hence, γ = 1, Ɛ = 0, β 1 = 1

21 Plugging the values of α, Ɛ, γ, β and β 0 1 into the general form of the CF: g (n) = A(n)α + +B(n)Ɛ + C(n)γ + D(n)β 0 + E(n)β 1 n = A(n) C(n) + E(n), for all n є N

22 We have that g(n) = n 2, for all n. From g(1) = α, we deduce that g(1) = 1 2 = 1 = α. So, α =1. g(2n+1) = 3g(n) + Ɛn 2 + γn + β 1 (2n+1 = 3n 2 + Ɛn 2 + γn + β 1 4n n = 3n 2 + Ɛn 2 + γn + β 1 n n = Ɛn 2 + γn + β 1 0 = (Ɛ 1)n 2 + (γ 4)n + β 1, for all n є N. Hence Ɛ = 1, γ = 4, β 1 = 1

23 g(2n+0) = 3g(n) + Ɛn 2 + γn + β 0 (2n+ 0 = 3n 2 + Ɛn 2 + γn + β 0 4n 2 = 3n 2 + Ɛn 2 + γn + β 0 0 = (Ɛ 1)n 2 + γn + β 0 => Ɛ = 1, γ = 0, β 0 = 0 Ɛ = 1, γ = 4, β 1 = 1 But for g(2n+1) we just obtained: Contradiction! Method fails!

24 We get a similar contradiction here too (No need to proceed beyond first contradiction)

25 Let us consider the case when the input value to R is of the form 2n+0 = 2n. Let the binary representation of 2n be: (b m,., b 1, b 0 This means 2n = 2 m b m + + 2b 1 + b 0 We observe that b 0 = 0, because 2n is even Dividing by 2: n = 2 m 1 b m + + b 1 or n = (b m,., b 1... Fact [1]

26 Now, let us consider the case when the input value to R is of the form (2n+1). Let the binary representation of (2n+1) be: (b m,., b 1, b 0 So, (2n+1) = 2 m b m + 2 m 1 b m b 1 + b 0 Since 2n+1 is an odd number b 0 = 1

27 2n+1 = 2 m b m + + 2b 1 + b 0 2n+1 = 2 m b m + + 2b n = 2 m b m + + 2b 1 n = 2 m 1 b m + + b 1 This means that n = (b m,,b 1... Fact [2] From [1] & [2], we observe same representation for both even/odd n. So, the CF need not have 2 cases

28 g(2n+0) = g(2n+1) = g( (b m,., b 1, b 0 ) = 3g( (b m,., b 1 ) + Ɛ[(b m,., b 1 ] 2 + γ(b m, b m 1,., b 1 + β b0 = 3{3g( (b m,., b 2 ) + Ɛ[(b m,., b 2 ] 2 + γ(b m, b m 1,., b 2 + β b1 } + Ɛ[(b m,., b 1 ] 2 + γ(b m,., b 1 + β b0

29 g ( (b m,., b 1, b 0 ) = 3 2 g( (b m,., b 2 ) + 3Ɛ[(b m,., b 2 ] 2 +3γ(b m, b m 1,., b 2 + 3β b1 + Ɛ[(b m,., b 1 ] 2 + γ(b m,., b 1 + β b0 = 3 2 {3g( (b m,., b 3 ) + Ɛ[(b m,., b 3 ] 2 + γ(b m, b m 1,., b 3 + β b2 } + 3Ɛ[(b m,., b 2 ] 2 + 3γ(b m, b m 1,., b 2 + 3β b1 + Ɛ[(b m,., b 1 ]2 + γ(b m,., b 1 + β b0

30 (repeated) = 3 2 {3g( (b m,., b 3 ) + Ɛ[(b m,., b 3 ] 2 + γ(b m, b m 1,., b 3 + β b2 } + 3Ɛ[(b m,., b 2 ] 2 + 3γ(b m, b m 1,., b 2 + 3β b1 + Ɛ[(b m,., b 1 ]2 + γ(b m,., b 1 + β b0 = 3 3 g( (b m,., b 3 ) Ɛ[(b m,., b 3 ] γ(b m,., b β b2 + 3Ɛ[(b m,., b 2 ] 2 + 3γ(b m,., b 2 + 3β b1 + Ɛ[(b m,., b 1 ] 2 + γ(b m, b m 1,., b 1 + β b0

31 (repeated) = 3 3 g( (b m,., b 3 ) Ɛ[(b m,., b 3 ] γ(b m, b m 1,., b β b2 + 3Ɛ[(b m,., b 2 ] 2 + 3γ(b m,., b 2 + 3β b1 + Ɛ[(b m,., b 1 ] 2 + γ(b m,., b 1 + β b0 Generalizing: g ((b m,., b 1, b 0 ) = 3 m g((b m ) + Ɛ{3 m 1 [(b m ] m 2 [(b m b m 1 ] [(b m b 1 ] 2 } + γ{3 m 1 (b m + 3 m 2 (b m b m (b m,.,b 1 } + 3 m 1 β bm m 2 β bm β b β b0

32 Input value 1 is represented by only one bit b m. In this case b m = 1. Also, the only time when b m is 0, is when the decimal number itself is 0. Any natural number is going to have b m = 1. Since the input to R are all natural numbers, therefore, in our computation b m = 1. So, g(b m ) = g(1) = α

33 g ((b m,., b 1, b 0 ) = 3 m α + Ɛ{3 m 1 [(b m ] m 2 [(b m b m 1 ] [(b m b 1 ] 2 } + γ{3 m 1 (b m + 3 m 2 (b m b m (b m,.,b 1 } + 3 m 1 β bm 1 + 3m 2 β bm β b β b0 = 3 m α + Ɛ{Σ3 m i [(b m b m i+1 ] 2 } + γ{σ3 m i (b m b m i+1 } + Σ3 m i β bm i (i = 1 to m)

34 n = 8: Binary representation: (1000 m = 3, b 0 = 0, b 1 =0, b 2 = 0 and b 3 = b m = 1 CF: g ((b m,., b 1, b 0 ) = 3 m α + Ɛ{Σ3 m i [(b m b m i+1 ] 2 } + γ{σ3 m i (b m b m i+1 } + Σ3 m i β bm i g((1000 ) = 3 3 α + ƐΣ3 3 i (b 3 b 3 i+1 + γσ3 3 i (b 3 b 3 i+1 ) + Σ3 3 i β b3 i

35 g((1000 ) = 3 3 α + ƐΣ3 3 i (b 3 b 3 i+1 + γσ3 3 i (b 3 b 3 i+1 ) + Σ3 3 i β b3 i = 3 3 α + Ɛ{3 3 1 [(b 3 b ] [(b 3...b ] [(b 3 b ] 2 } + γ{3 3 1 (b 3 b (b 3...b (b 3 b } β b β b β b3 3 = 3 3 α + Ɛ{3 2 [(1 ] [(10 ] [(100 ] 2 } + γ{3 2 ( ( (100 }+ 3 2 β β β 0

36 (repeated) = 3 3 α + Ɛ{3 2 [(1 ] [(10 ] [(100 ] 2 } + γ{3 2 ( ( (100 }+ 3 2 β β β 0 = 27α + Ɛ{9[1] + 3[4] + [16]} + γ{9(1) + 3(2) + 1(4)}+ 9β 0 + 3β 0 + β 0 = 27α +37Ɛ + 19γ + 13β 0 D b b b

37 n g(n) 1 α + 0Ɛ +0γ + 0β 0 + 0β 1 2 3α + Ɛ + γ + β 0 + 0β 1 3 3α + Ɛ + γ + 0β 0 + β 1 4 9α +7Ɛ+ 5γ+ 4β 0 + 0β 1 5 9α +7Ɛ+ 5γ+ 3β 0 + β 1 6 9α +12Ɛ + 6γ + β 0 +3β 1 7 9α +12Ɛ + 6γ + 0β 0 + 4β α +37Ɛ + 19γ + 13β 0 + 0β 1