The Fibonacci sequence is defined as 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,... The first two terms, f[0) and f(1), are equal to 1. Every subsequent term is the sum of the two previous terms. Thus for each n≥2n≥2, we have f(n)=f(n-1)+f(n-2). Here are two Python programs to calculate f(n)f(n). One of these two methods uses bottom-up dynamic programming and one of these two methods uses recursion. But which one is which? def fibi(n): if n<=1: return 1 else: return fib1(n-1)+fib1(n-2) def fib2(n): C- table = [0 for i in range(n+1)] for i in range(n+1): if i<=1: table[i]=1 else: table[i]=table[i-1]+table[i-2] return table[n] For each of our two Fibonacci-calculating algorithms above, we can determine the running time. Here are six statements Here are six statements The following multiple-choice options contain math elements, so you may need to read them in your screen reader's "reading" or "browse" mode instead of "forms" or "focus" mode. Choice 1 of 6: 1. fib1(n) is 2. fib1(n) is 3. fib1(n) is 4. fib1(n) is 5. fib1(n) is 6. fib1(n) is (n2)(n2) and fib2(n) is 0 (n) (n2)(n2) and fib2(n) is (n2)0 (n2) (2n) (2n) and fib2(n) is 0 (n) (2n)e(2n) and fib2(n) is (n2) 0 (n2) (n!) and fib2(n) is 0 (n) (n!) and fib2(n) is 0(n2)
The Fibonacci sequence is defined as 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,... The first two terms, f[0) and f(1), are equal to 1. Every subsequent term is the sum of the two previous terms. Thus for each n≥2n≥2, we have f(n)=f(n-1)+f(n-2). Here are two Python programs to calculate f(n)f(n). One of these two methods uses bottom-up dynamic programming and one of these two methods uses recursion. But which one is which? def fibi(n): if n<=1: return 1 else: return fib1(n-1)+fib1(n-2) def fib2(n): C- table = [0 for i in range(n+1)] for i in range(n+1): if i<=1: table[i]=1 else: table[i]=table[i-1]+table[i-2] return table[n] For each of our two Fibonacci-calculating algorithms above, we can determine the running time. Here are six statements Here are six statements The following multiple-choice options contain math elements, so you may need to read them in your screen reader's "reading" or "browse" mode instead of "forms" or "focus" mode. Choice 1 of 6: 1. fib1(n) is 2. fib1(n) is 3. fib1(n) is 4. fib1(n) is 5. fib1(n) is 6. fib1(n) is (n2)(n2) and fib2(n) is 0 (n) (n2)(n2) and fib2(n) is (n2)0 (n2) (2n) (2n) and fib2(n) is 0 (n) (2n)e(2n) and fib2(n) is (n2) 0 (n2) (n!) and fib2(n) is 0 (n) (n!) and fib2(n) is 0(n2)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Which is the correct choice?
![The Fibonacci sequence is defined as 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
The first two terms, f(0) and f(1), are equal to 1. Every subsequent term is the sum of the
two
previous terms. Thus for each n≥2n22, we have f(n)=f(n-1)+f(n-2).
Here are two Python programs to calculate f(n)f(n). One of these two methods uses
bottom-up dynamic programming and one of these two methods uses recursion. But which
one is which?
def fibi(n):
if n<=1: return 1
else: return fib1(n-1)+fib1(n-2)
def fib2(n):
table
C-
[ for i in range(n+1)]
for i in range(n+1):
if i=1: table[i]=1
else: table[i]=table[i-1]+table[i-2]
return table[n]
For each of our two Fibonacci-calculating algorithms above, we can
determine the running time.
Here are six statements
Here are six statements
The following multiple-choice options contain math elements, so you may need to read
them in your screen reader's "reading" or "browse" mode instead of "forms" or "focus"
mode.
Choice 1 of 6:
1. fib1(n) is
2. fib1(n) is
3. fib1(n) is
4. fib1(n) is
5. fib1(n) is
6. fib1(n) is
(n2)(n2) and fib2(n) is 0 (n)
(n2)(n2) and fib2 (n) is (n2)0 (n2)
(2n) (2n) and fib2(n) is 0 (n)
(2n)Ⓒ(2n) and fib2(n) is 0 (n2) Ⓒ (n2)
(n!) and fib2(n) is 0 (n)
(n!) and fib2(n) is 0 (n2)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9ec53c5c-d1c4-42dd-957e-ce8596851c7e%2F94dd39d8-d0c0-4fe2-bdaf-35aa0a27fcc6%2Fibbplmo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The Fibonacci sequence is defined as 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
The first two terms, f(0) and f(1), are equal to 1. Every subsequent term is the sum of the
two
previous terms. Thus for each n≥2n22, we have f(n)=f(n-1)+f(n-2).
Here are two Python programs to calculate f(n)f(n). One of these two methods uses
bottom-up dynamic programming and one of these two methods uses recursion. But which
one is which?
def fibi(n):
if n<=1: return 1
else: return fib1(n-1)+fib1(n-2)
def fib2(n):
table
C-
[ for i in range(n+1)]
for i in range(n+1):
if i=1: table[i]=1
else: table[i]=table[i-1]+table[i-2]
return table[n]
For each of our two Fibonacci-calculating algorithms above, we can
determine the running time.
Here are six statements
Here are six statements
The following multiple-choice options contain math elements, so you may need to read
them in your screen reader's "reading" or "browse" mode instead of "forms" or "focus"
mode.
Choice 1 of 6:
1. fib1(n) is
2. fib1(n) is
3. fib1(n) is
4. fib1(n) is
5. fib1(n) is
6. fib1(n) is
(n2)(n2) and fib2(n) is 0 (n)
(n2)(n2) and fib2 (n) is (n2)0 (n2)
(2n) (2n) and fib2(n) is 0 (n)
(2n)Ⓒ(2n) and fib2(n) is 0 (n2) Ⓒ (n2)
(n!) and fib2(n) is 0 (n)
(n!) and fib2(n) is 0 (n2)
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