#### You've reached the end of your free preview.

Want to read all 7 pages?

**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 ...

View
Full Document

- Fall '19
- Dusko Pavlovic