1-hw-222-19-answers.pdf - ICS 222 Homework 1 Please upload your solutions to Laulima by noon on Friday Answer to Exercise 1 a b = 3 β(y = 3 y The

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Unformatted text preview: ICS 222: Homework 1 Please upload your solutions to Laulima by noon on Friday Sep 20, 2019 Answer to Exercise 1 a) @ b = 3, β(y) = 3 + y The clausal form of the induction is given by: f0 = 3 f1+n = 3 + fn Trying a few instances of f starting from 0, we get: f0 = 3, f1 = 3 + 3 = 6, f2 = 3 + 6 = 9, f3 = 3 + 9 = 12, . . . Hence, we guess that fn = 3 + 3n. To prove that this function f has the inductive specification above we apply induction: claim: L3, 3 + yMn = 3 + 3n base case: L3, 3 + yM0 = 3 = 3 + 3 · 0 step case: if L3, 3 + yMn = 3 + 3n, then (IH) L3, 3 + yM1+n = 3 + L3, 3 + yMn = 3 + 3 + 3n = 3 + 3(1 + n) b) @ b = 3, β(y) = 3y The clausal form of the induction is given by: f0 = 3 f1+n = 3fn Trying few instances of f starting from 0: f0 = 3, f1 = 3 · 3 = 9, f2 = 3 · 9 = 27, f3 = 3 · 27 = 81, . . . Hence, we guess that ∀n ∈ N. fn = 3n+1 . 1 claim: L3, 3yMn = 3n+1 base case: L3, 3yM0 = 3 = 30+1 step case: if L3, 3yMn = 3n+1 , then (IH) L3, 3yM1+n = 3 · L3, 3yMn = 3 · 3n+1 = 3(1+n)+1 c) @ b = 3, β(y) = 3y The clausal form of the induction is given by: f1+n = 3fn f0 = 3 Trying few instances of f starting from 0: 33 3 f0 = 3, f1 = 33 , f2 = 33 ·, f3 = 33 , . . . Writing the powers xy in the form x ↑ y, we guess fn = 3 ↑ · · · ↑ 3 ↑ 3. | {z } n copies To prove fn has the inductive specification above we apply induction: claim: L3, 3y M = 3 ↑ · · · ↑ 3 ↑ 3 | {z } n base case: L3, 3y M0 = 3 step case: if L3, 3y Mn = 3 ↑ · · · ↑ 3 ↑ 3, then | y {z } n 3 ↑ ··· ↑ 3 ↑3 L3,3y Mn (IH) L3, 3 M1+n = 3 | = 3 {z n } In other words, fn = 3 ↑↑ n. = 3 ↑ ··· ↑ 3 ↑3 | Answer to Exercise 2 a) γ0 = 3 · 20 = 3 suggests b = 3. Towards β, note: γ1+n = 3 · 21+n = (3 · 2n ) · 2 = 2γn 2 {z 1+n } and so β(y) = 2y. Hence we conjecture that γn = L3, 2yM. L3, 2yM5 = 2 · L3, 2yM4 = 2 · (2 · L3, 2yM3 ) = 2 · (2 · (2 · L3, 2yM2 )) = 2 · (2 · (2 · (2 · L3, 2yM1 ))) = 2 · (2 · (2 · (2 · (2 · L3, 2yM0 )))) = 2 · (2 · (2 · (2 · (2 · (3))))) = 25 · 3 b) χ0 = 0(0+1) 2 = 0 suggests b = 0. Towards β, note: χ1+n = (1 + n)(n + 2) n(1 + n) 2(1 + n) = + = χn + (1 + n) 2 2 2 Note that we cannot define β on a single parameter only since χ1+n depends on χn and 1+n. Hence, we do not have an inductive specification but a recursive specification given by L0, y + 1 + nM. The evaluation of this function for n = 5 is given below: χ5 = 5 + χ4 = 5 + 4 + χ3 = 5 + 4 + 3 + χ2 = 5 + 4 + 3 + 2 + 1 + χ0 =5+4+3+2+1+0 Answer to Exercise 3 a) ℓn (m) = 2n + 3m ℓ0 (m) = 2 · 0 + 3m = 3m ℓ1+n (m) = 2(1 + n) + 3m = 2 + 2n + 3m = 2 + ℓ(n, m) Hence, we set g(m) = 3m and hn (y, m) = 2 + y and claim ℓ = Lg, hM. claim: L3m, 2 + yMn (m) = 2n + 3m base: L3m, 2 + yM0 (m) = 3m = 2 · 0 + 3m 3 step: L3m, 2 + yM1+n (m) = 2 + L3m, 2 + yMn (m) = 2 + 2n + 3m (by IH) = 2(1 + n) + 3m b) ρn = ( n−1 0 if n ≥ 1 otherwise Since the predecessor function ρ is defined to map ρ0 = 0 and ρ1+n = n, we set g = 0 and hn (y) = n and claim ρ = Lg, hM = L0, nM claim: L0, nMn = ρn base: L0, nM0 = 0 step: L0, nM1+n = n · n= c) fn (m) = m − ( m−n 0 if m ≥ n otherwise Since f0 (m) = m and f1+n (m) = ( ρ(fn , m)) 0 if m > n if m ≤ n, we set g(m) = m and hn (y, m) = ( ρ(y) 0 to claim f = Lg, hM. claim: base: f = Lg, hM · 0 = m = g(m) Lg, hM0 (m) = m − 4 if m > n otherwise step:  Lg, hM1+n (m) = hn Lg, hMn (m), m  = ( ρ(Lg, hMn (m)) 0 = ( ρ(Lg, hMn (m)) 0 = ( ρ(m − n) if m ≥ n + 1 0 otherwise = ( m − (1 + n) if m ≥ n + 1 0 otherwise if m > n otherwise if m ≥ n + 1 otherwise · (1 + n) =m− d) |n − m| = ( m − n if m ≥ n n − m otherwise · n) + (n − · m). We can write it as |n − m| = (m − Alternatively, |0 − m| = m, we hence put g(m) = m and since |(1 + n) − m| = ( we define hn (y, m) = ρ|n − m| if m > n 1 + |n − m| otherwise ( ρ(y) 1+y to claim f = Lg, hM. claim: base: f = Lg, hM Lg, hM0 (m) = |0 − m| = m = g(m) 5 if m > n otherwise (IH) step:  Lg, hM1+n (m) = hn Lg, hMn (m), m  = ( ρ(Lg, hMn (m)) 1 + Lg, hMn (m) = ( ρ(Lg, hMn (m)) 1 + Lg, hMn (m) = ( ρ(|m − n|) if m ≥ n + 1 1 + |m − n| otherwise = ( |m − (1 + n)| if m ≥ n + 1 |m − (1 + n)| otherwise if m > n otherwise if m ≥ n + 1 otherwise = |m − (1 + n)| Answer to Exercise 4 a) ϑ(0) = ϑ(1 + n) = 0 X i=0 1+n X i3 = 0 i3 i=0 = (1 + n)3 + n X i3 i=0 3 = (1 + n) + ϑ(n) So, we set g = 0, and h(n, y) = (1 + n)3 + y and define ϑ = Lg, hM b) claim: ϑ(n) = n X i=0 ϑ(n) = 6  !2 i n(1 + n) 2 2 (IH) base: ϑ(0) =  n(1 + n) 2 2 =  0(1 + 0) 2 2 =0 step: ϑ(1 + n) = n+1 X i3 i=0 = (1 + n)3 + n X i3 i=0 n(1 + n) 2 = (1 + n) + 2 3 2 4(1 + n) + n (1 + n)2 = 4 (1 + n)2 (4(1 + n) + n2 ) = 4 (1 + n)2 (n + 2)2 = 4   (1 + n)(1 + (1 + n)) 2 = 2 3   Since the base case and the step case are true, the claim is proved. 7 ...
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• Fall '19
• Dusko Pavlovic

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