Homework 1_Validating Credit Card Numbers

When solving the homework, strive to create not just code that works, but code that is stylish and concise. See the style guide on the website for some general guidelines. Try to write small functions which perform just a single task, and then combine those smaller pieces to create more complex functions. Don’t repeat yourself: write one function for each logical task, and reuse functions as necessary. Be sure to write functions with exactly the specified name and type signature for each exercise (to help us test your code). You may create additional helper functions with whatever names and type signatures you wish.
Validating Credit Card Numbers1
Have you ever wondered how websites validate your credit card number when you shop online? They don’t check a massive database of numbers, and they don’t use magic. In fact, most credit providers rely on a checksum formula for distinguishing valid numbers from random collections of digits (or typing mistakes).
In this section, you will implement the validation algorithm for credit cards. It follows these steps:
• Double the value of every second digit beginning from the right. That is, the last digit is unchanged; the second-to-last digit is doubled; the third-to-last digit is unchanged; and so on. For example, [1,3,8,6] becomes [2,3,16,6].
• Add the digits of the doubled values and the undoubled digits from the original number. For example, [2,3,16,6] becomes 2+3+1+6+6 = 18. 1Adapted from the first practicum assigned in the University of Utrecht functional programming course taught by Doaitse Swierstra, 2008-2009.
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• Calculate the remainder when the sum is divided by 10. For the above example, the remainder would be 8. If the result equals 0, then the number is valid.
Exercise 1 We need to first find the digits of a number. Define the functions
toDigits :: Integer - [Integer] toDigitsRev :: Integer - [Integer] toDigits should convert positive Integers to a list of digits. (For 0 or negative inputs, toDigits should return the empty list.) toDigitsRev should do the same, but with the digits reversed. Example: toDigits 1234 == [1,2,3,4] Example: toDigitsRev 1234 == [4,3,2,1] Example: toDigits 0 == [] Example: toDigits (-17) == []
Exercise 2 Once we have the digits in the proper order, we need to double every other one. Define a function
doubleEveryOther :: [Integer] - [Integer] Remember that doubleEveryOther should double every other number beginning from the right, that is, the second-to-last, fourth-to-last, ...numbers are doubled. Example: doubleEveryOther [8,7,6,5] == [16,7,12,5] Example: doubleEveryOther [1,2,3] == [1,4,3]
Exercise 3 The output of doubleEveryOther has a mix of one-digit and two-digit numbers. Define the function
sumDigits :: [Integer] - Integer to calculate the sum of all digits. Example: sumDigits [16,7,12,5] = 1 + 6 + 7 + 1 + 2 + 5 = 22
Exercise 4 Define the function
validate :: Integer - Bool that indicates whether an Integer could be a valid credit card number. This will use all functions defined in the previous exercises. Example: validate 4012888888881881 = True Example: validate 4012888888881882 = False
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The Towers of Hanoi2
Exercise 5 The Towers of Hanoi is a classic puzzle with a solution that can be described recursively. Disks of different sizes are stacked on three pegs; the goal is to get from a starting configuration with all disks stacked on the first peg to an ending configuration with all disks stacked on the last peg, as shown in Figure 1.
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Figure 1: The Towers of Hanoi
The only rules are • you may only move one disk at a time, and
• a larger disk may never be stacked on top of a smaller one. For example, as the first move all you can do is move the topmost, smallest disk onto a different peg, since only one disk may be moved at a time. Figure 2: A valid first move.From this point, it is illegal to move to the configuration shown in Figure 3, because you are not allowed to put the green disk on top of the smaller blue one.
Figure 3: An illegal configuration.
To move n discs (stacked in increasing size) from peg a to peg b using peg c as temporary storage, 1. move n−1 discs from a to c using b as temporary storage 2. move the top disc from a to b 3. move n−1 discs from c to b using a as temporary storage. For this exercise, define a function hanoi with the following type:
type Peg = String type Move = (Peg, Peg) hanoi :: Integer - Peg - Peg - Peg - [Move] 2Adapted from an assignment given in UPenn CIS 552, taught by Benjamin Pierce
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Given the number of discs and names for the three pegs, hanoi should return a list of moves to be performed to move the stack of discs from the first peg to the second. Note that a type declaration, like type Peg = String above, makes a type synonym. In this case Peg is declared as a synonym for String, and the two names Peg and String can now be used interchangeably. Giving more descriptive names to types in this way can be used to give shorter names to complicated types, or (as here) simply to help with documentation. Example: hanoi 2 "a" "b" "c" == [("a","c"), ("a","b"), ("c","b")]
Exercise 6 (Optional) What if there are four pegs instead of three? That is, the goal is still to move a stack of discs from the first peg to the last peg, without ever placing a larger disc on top of a smaller one, but now there are two extra pegs that can be used as “temporary” storage instead of only one. Write a function similar to hanoi which solves this problem in as few moves as possible. It should be possible to do it in far fewer moves than with three pegs. For example, with three pegs it takes 215 −1 = 32767 moves to transfer 15 discs. With four pegs it can be done in 129 moves. (See Exercise 1.17 in Graham, Knuth, and Patashnik, Concrete Mathematics, second ed., Addison-Wesley, 1994.)