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if n <- 1 return arr [n] else for i in 2..n arr [i] <- arr [1-1] + arr [i-2] return arr[n] d) def f(n): arr [0..n] <- [o, ..., n] if n <- 1 return arr [n] else a <- 0 for i in 0. .n a <- a + arr[i] return a

if n <- 1 return arr [n] else for i in 2..n arr [i] <- arr [1-1] + arr [i-2] return arr[n] d) def f(n): arr [0..n] <- [o, ..., n] if n <- 1 return arr [n] else a <- 0 for i in 0. .n a <- a + arr[i] return a

C++ Programming: From Problem Analysis to Program Design
C++ Programming: From Problem Analysis to Program Design
8th Edition
ISBN:9781337102087
Author:D. S. Malik
Publisher:D. S. Malik
Chapter15: Recursion
Section: Chapter Questions
Problem 8SA
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Question

The following function f uses recursion: def f(n):

if n <= 1

return n

else return f(n-1) + f(n-2)

 Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion?

a)  def f(n):

a <- 0

b <- 1

if n = 0

   return a

elsif n = 1

    return b

else for i in 1..n

    c <- a + b

    a <- b

     b <- c

     return b

b)   def f(n):

a <- 0

i <- n

while i > 0

     a <- a + i + (i-1)

     return a

c) def f(n):

  arr[0] <- 0

   arr[1] <- 1

 if n <= 1

    return arr[n]

else for i in 2..n

    arr[i] <- arr[i-1] + arr[i-2]

    return arr[n]

d) def f(n):

arr[0..n] <- [0, ..., n]

if n <= 1

return arr[n]

else a <- 0

for i in 0..n

a <- a + arr[i]

return a

Question 19. The following function f uses recursion: def f(n): if n <= 1 return n else return f(n-1) + f(n-2) Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion? a) def f(n): a <- 0 b <- 1 if n - 0 return a elsif n = 1 return b else for i in 1..n C <- a + b a <- b b <- c return b b) def f(n): a <- 0 i <- n while i > 0 a <- a + i + (i-1) return a
Transcribed Image Text:Question 19. The following function f uses recursion: def f(n): if n <= 1 return n else return f(n-1) + f(n-2) Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion? a) def f(n): a <- 0 b <- 1 if n - 0 return a elsif n = 1 return b else for i in 1..n C <- a + b a <- b b <- c return b b) def f(n): a <- 0 i <- n while i > 0 a <- a + i + (i-1) return a
c) def f(n): arr [0] <- 0 arr[1] <- 1 if n <= 1 return arr [n] else for i in 2..n arr [i] <- arr[i-1] + arr[i-2] return arr [n] d) def f(n): arr [0..n] <- [0, if n <= 1 return arr [n] n] .... else a <- 0 for i in 0..n a <- a + arr[i] return a
Transcribed Image Text:c) def f(n): arr [0] <- 0 arr[1] <- 1 if n <= 1 return arr [n] else for i in 2..n arr [i] <- arr[i-1] + arr[i-2] return arr [n] d) def f(n): arr [0..n] <- [0, if n <= 1 return arr [n] n] .... else a <- 0 for i in 0..n a <- a + arr[i] return a
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