IN SCALA PLEASE COULD YOU COMPLETE THE FUNCTIONS: is_pow_of_two, is_hard, last_odd //(1) Complete the collatz function below. It should // recursively calculate the number of steps needed // until the collatz series reaches the number 1. // If needed, you can use an auxiliary function that // performs the recursion. The function should expect // arguments in the range of 1 to 1 Million. def collatz(n: Long) : Long = { if(n == 1) 0else if(n % 2 == 0) 1 + collatz(n/2) else1 + collatz(n*3 + 1) } //(2) Complete the collatz_max function below. It should // calculate how many steps are needed for each number // from 1 up to a bound and then calculate the maximum number of // steps and the corresponding number that needs that many // steps. Again, you should expect bounds in the range of 1 // up to 1 Million. The first component of the pair is // the maximum number of steps and the second is the // corresponding number. def collatz_max(bnd: Long) : (Long, Long) = { ((1.toLong to bnd).toList.map(n => collatz(n)).max, (1.toLong to bnd).toList.map(n => collatz(n)).indexOf((1.toLong to bnd).toList.map(n => collatz(n)).max) + 1) } //(3) Implement a function that calculates the last_odd // number in a collatz series. For this implement an // is_pow_of_two function which tests whether a number // is a power of two. The function is_hard calculates // whether 3n + 1 is a power of two. Again you can // assume the input ranges between 1 and 1 Million, // and also assume that the input of last_odd will not // be a power of 2. def is_pow_of_two(n: Long) : Boolean = ??? def is_hard(n: Long) : Boolean = ??? def last_odd(n: Long) : Long = ???

C++ for Engineers and Scientists

4th Edition

ISBN:9781133187844

Author:Bronson, Gary J.

Publisher:Bronson, Gary J.

Chapter6: Modularity Using Functions

Section6.1: Function And Parameter Declarations

Problem 10E

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Question

IN SCALA PLEASE

COULD YOU COMPLETE THE FUNCTIONS: is_pow_of_two, is_hard, last_odd

//(1) Complete the collatz function below. It should

// recursively calculate the number of steps needed

// until the collatz series reaches the number 1.

// If needed, you can use an auxiliary function that

// performs the recursion. The function should expect

// arguments in the range of 1 to 1 Million.

def collatz(n: Long) : Long = {

if(n == 1) 0else

if(n % 2 == 0) 1 + collatz(n/2) else1 + collatz(n*3 + 1)

}

//(2) Complete the collatz_max function below. It should

// calculate how many steps are needed for each number

// from 1 up to a bound and then calculate the maximum number of

// steps and the corresponding number that needs that many

// steps. Again, you should expect bounds in the range of 1

// up to 1 Million. The first component of the pair is

// the maximum number of steps and the second is the

// corresponding number.

def collatz_max(bnd: Long) : (Long, Long) = {

((1.toLong to bnd).toList.map(n => collatz(n)).max, (1.toLong to bnd).toList.map(n => collatz(n)).indexOf((1.toLong to bnd).toList.map(n => collatz(n)).max) + 1)

}

//(3) Implement a function that calculates the last_odd

// number in a collatz series. For this implement an

// is_pow_of_two function which tests whether a number

// is a power of two. The function is_hard calculates

// whether 3n + 1 is a power of two. Again you can

// assume the input ranges between 1 and 1 Million,

// and also assume that the input of last_odd will not

// be a power of 2.

def is_pow_of_two(n: Long) : Boolean = ???

def is_hard(n: Long) : Boolean = ???

def last_odd(n: Long) : Long = ???

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