AC MGFs Q&A 1: 0 bits in 00-free bit strings Q. What is the - - PowerPoint PPT Presentation

AC MGFs Q&A 1: 0 bits in 00-free bit strings

1

• Q. What is the average number of 0 bits in a random bit string of length n containing no 00?

a0 = 0 a1 = 1/2 1

AC MGFs Q&A 1: 0 bits in 00-free bit strings

1

• Q. What is the average number of 0 bits in a random bit string of length n containing no 00?

a0 = 0 a1 = 1/2 a2 = 2/3 0 1 1 0 1 1 1

AC MGFs Q&A 1: 0 bits in 00-free bit strings

1

• Q. What is the average number of 0 bits in a random bit string of length n containing no 00?

a0 = 0 a1 = 1/2 a3 = 1 0 1 1 0 1 0 1 0 1 1 1 0 1 1 1 a2 = 2/3 0 1 1 0 1 1 1

AC MGFs Q&A 1: 0 bits in 00-free bit strings

1

• Q. What is the average number of 0 bits in a random bit string of length n containing no 00?

a0 = 0 a1 = 1/2 a3 = 1 0 1 1 0 1 0 1 0 1 1 1 0 1 1 1 a4 = 10/8 0 1 0 1 0 1 1 0 0 1 1 1 1 0 1 1 1 0 1 0 1 1 0 1 1 1 1 0 1 1 1 1 a2 = 2/3 0 1 1 0 1 1 1

AC MGFs Q&A 1: 0 bits in 00-free bit strings

1

• Q. What is the average number of 0 bits in a random bit string of length n containing no 00?

a0 = 0 a1 = 1/2 a3 = 1 0 1 1 0 1 0 1 0 1 1 1 0 1 1 1 a4 = 10/8 0 1 0 1 0 1 1 0 0 1 1 1 1 0 1 1 1 0 1 0 1 1 0 1 1 1 1 0 1 1 1 1 a2 = 2/3 0 1 1 0 1 1 1 a5 =20/13 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 1 1 1 1 0 1 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1

AC MGFs Q&A 1: 0 bits in 00-free bit strings

2

construction OGF equation OGF enumeration

• Q. Recalling the derivation at left, fill in the boxes at right to prove that the

average number of 0 bits in a random bit string of length n containing no 00 is

BGF equation BGF cumulated cost solution

B = E + Z0 + (Z1 + Z0 × Z1) × B B(z) = 1 + z + (z + z2)B(z) B(z, u) = B(z) = 1 + z 1 − z − z2 [zn]Bu(z, 1) = [zn] f(z) (1 − z/ρ)α ∼ f(ρ) Γ(α)ρ−nnα−1 [zn]B(z) ∼ 1 + 1/φ 1 − ˆ φ/φ φn = φn+2 √ 5 ∼ n φ √ 5 ∼ n φ √ 5 [zn]Bu(z, 1) [zn]B(z)

AC MGFs Q&A 1: 0 bits in 00-free bit strings

2

construction OGF equation OGF enumeration

• Q. Recalling the derivation at left, fill in the boxes at right to prove that the

average number of 0 bits in a random bit string of length n containing no 00 is

BGF equation BGF cumulated cost solution

B = E + Z0 + (Z1 + Z0 × Z1) × B B(z) = 1 + z + (z + z2)B(z) B(z, u) = 1 + uz + (z + uz2)B(z, u) B(z) = 1 + z 1 − z − z2 [zn]Bu(z, 1) = [zn] f(z) (1 − z/ρ)α ∼ f(ρ) Γ(α)ρ−nnα−1 [zn]B(z) ∼ 1 + 1/φ 1 − ˆ φ/φ φn = φn+2 √ 5 ∼ n φ √ 5 ∼ n φ √ 5 [zn]Bu(z, 1) [zn]B(z)

AC MGFs Q&A 1: 0 bits in 00-free bit strings

2

construction OGF equation OGF enumeration

• Q. Recalling the derivation at left, fill in the boxes at right to prove that the

average number of 0 bits in a random bit string of length n containing no 00 is

BGF equation BGF cumulated cost solution

B = E + Z0 + (Z1 + Z0 × Z1) × B B(z) = 1 + z + (z + z2)B(z) B(z, u) = 1 + uz + (z + uz2)B(z, u) B(z) = 1 + z 1 − z − z2 B(z, u) = 1 + uz 1 − z − uz2 [zn]Bu(z, 1) = [zn] f(z) (1 − z/ρ)α ∼ f(ρ) Γ(α)ρ−nnα−1 [zn]B(z) ∼ 1 + 1/φ 1 − ˆ φ/φ φn = φn+2 √ 5 ∼ n φ √ 5 ∼ n φ √ 5 [zn]Bu(z, 1) [zn]B(z)

AC MGFs Q&A 1: 0 bits in 00-free bit strings

2

construction OGF equation OGF enumeration

• Q. Recalling the derivation at left, fill in the boxes at right to prove that the

average number of 0 bits in a random bit string of length n containing no 00 is

BGF equation BGF cumulated cost solution

B = E + Z0 + (Z1 + Z0 × Z1) × B B(z) = 1 + z + (z + z2)B(z) B(z, u) = 1 + uz + (z + uz2)B(z, u) B(z) = 1 + z 1 − z − z2 B(z, u) = 1 + uz 1 − z − uz2 [zn]Bu(z, 1) = [zn] f(z) (1 − z/ρ)α ∼ f(ρ) Γ(α)ρ−nnα−1 [zn]B(z) ∼ 1 + 1/φ 1 − ˆ φ/φ φn = φn+2 √ 5 [zn] z (1 − z − z2)2 ∼ n φ √ 5 ∼ n φ √ 5 [zn]Bu(z, 1) [zn]B(z)

AC MGFs Q&A 1: 0 bits in 00-free bit strings

2

construction OGF equation OGF enumeration

• Q. Recalling the derivation at left, fill in the boxes at right to prove that the

average number of 0 bits in a random bit string of length n containing no 00 is

BGF equation BGF cumulated cost solution

B = E + Z0 + (Z1 + Z0 × Z1) × B B(z) = 1 + z + (z + z2)B(z) B(z, u) = 1 + uz + (z + uz2)B(z, u) B(z) = 1 + z 1 − z − z2 B(z, u) = 1 + uz 1 − z − uz2 [zn]Bu(z, 1) = [zn] f(z) (1 − z/ρ)α ∼ f(ρ) Γ(α)ρ−nnα−1 [zn]B(z) ∼ 1 + 1/φ 1 − ˆ φ/φ φn = φn+2 √ 5 ∼ 1/φ (1 − ˆ φ/φ)2 nφn = nφn+1 5 [zn] z (1 − z − z2)2 ∼ n φ √ 5 ∼ n φ √ 5 [zn]Bu(z, 1) [zn]B(z)

AC MGFs Q&A 2

3

• Q. Which of the following is true of the average number of 0s in a random bit string
• f length n excluding any fixed pattern p, as n grows?

Approaches 0 Approaches 1 Approaches cn for some fixed constant c Has periodic behavior Asymptotic order of growth depends on the pattern

AC MGFs Q&A 2

3

• Q. Which of the following is true of the average number of 0s in a random bit string
• f length n excluding any fixed pattern p, as n grows?

Approaches 0 Approaches 1 Approaches cn for some fixed constant c Has periodic behavior Asymptotic order of growth depends on the pattern

F

AC MGFs Q&A 2

3

• Q. Which of the following is true of the average number of 0s in a random bit string
• f length n excluding any fixed pattern p, as n grows?

Approaches 0 Approaches 1 Approaches cn for some fixed constant c Has periodic behavior Asymptotic order of growth depends on the pattern

F F

AC MGFs Q&A 2

3

• Q. Which of the following is true of the average number of 0s in a random bit string
• f length n excluding any fixed pattern p, as n grows?

Approaches 0 Approaches 1 Approaches cn for some fixed constant c Has periodic behavior Asymptotic order of growth depends on the pattern

F F F

AC MGFs Q&A 2

3

• Q. Which of the following is true of the average number of 0s in a random bit string
• f length n excluding any fixed pattern p, as n grows?

Approaches 0 Approaches 1 Approaches cn for some fixed constant c Has periodic behavior Asymptotic order of growth depends on the pattern

T F F F

AC MGFs Q&A 2

3

• Q. Which of the following is true of the average number of 0s in a random bit string
• f length n excluding any fixed pattern p, as n grows?

Approaches 0 Approaches 1 Approaches cn for some fixed constant c Has periodic behavior Asymptotic order of growth depends on the pattern

T F F F F