Step by Step solution about How to resolve the algorithm Trabb Pardo–Knuth algorithm step by step in the Kotlin programming language

This code is a Kotlin program that takes 11 numbers as input from the user, processes them, and prints the results. Here is a detailed explanation of the code:

1. The program defines a function called f that takes a double as input and returns the square root of the absolute value of the input plus 5 times the input cubed.

2. The main function is the entry point of the program.

3. Inside the main function, a DoubleArray called da is created with a size of 11. This array will be used to store the 11 numbers entered by the user.

4. The program prompts the user to enter 11 numbers, and a while loop is used to read the input and store it in the da array. The loop continues until 11 numbers have been entered.

5. After the user has entered all 11 numbers, the program prints the sequence of numbers in reverse order.

6. The program then processes the sequence of numbers by iterating through the da array and calling the f function on each element. The result of the f function is printed to the console.

7. If the result of the f function is greater than 400.0, the program prints "Overflow!" to the console. Otherwise, the program prints the result of the f function.

// version 1.1.2

fun f(x: Double) = Math.sqrt(Math.abs(x)) + 5.0 * x * x * x
 
fun main(args: Array<String>) {
    val da = DoubleArray(11)
    println("Please enter 11 numbers:")
    var i = 0
    while (i < 11) {
        print("  ${"%2d".format(i + 1)}: ")
        val d = readLine()!!.toDoubleOrNull()
        if (d == null)
            println("Not a valid number, try again")
        else
            da[i++] = d
    }
    println("\nThe sequence you just entered in reverse is:")
    da.reverse()
    println(da.contentToString())
    println("\nProcessing this sequence...")
    for (j in 0..10) {
        val v = f(da[j])
        print("  ${"%2d".format(j + 1)}: ") 
        if (v > 400.0) 
            println("Overflow!")
        else
            println(v)
    }
}