Pascal, Parentheses Balancing and Coin Change
Exercise 1: Pascal’s Triangle
The following pattern of numbers is called Pascal’s triangle. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 ...
The numbers at the edge of the triangle are all 1, and each number inside the triangle is the sum of the two numbers above it. Write a function that computes the elements of Pascal’s triangle by means of a recursive process.
Do this exercise by implementing the pascal function in Main.scala, which takes a column c and a row r, counting from 0 and returns the number at that spot in the triangle. For example, pascal(0,2)=1, pascal(1,2)=2 and pascal(1,3)=3. def pascal(c: Int, r: Int): Int
Exercise 2: Parentheses Balancing
Write a recursive function which verifies the balancing of parentheses in a string, which we represent as a List[Char] not a String. For example, the function should return true for the following strings: (if (zero? x) max (/ 1 x)) I told him (that it’s not (yet) done). (But he wasn’t listening)
The function should return false for the following strings: :-) ())(
The last example shows that it’s not enough to verify that a string contains the same number of opening and closing parentheses.
Do this exercise by implementing the balance function in Main.scala. Its signature is as follows: def balance(chars: List[Char]): Boolean
There are three methods on List[Char] that are useful for this exercise: chars.isEmpty: Boolean returns whether a list is empty chars.head: Char returns the first element of the list chars.tail: List[Char] returns the list without the first element
Hint: you can define an inner function if you need to pass extra parameters to your function.
Testing: You can use the toList method to convert from a String to a List[Char]: e.g. "(just an) example".toList.
Exercise 3: Counting Change
Write a recursive function that counts how many different ways you can make change for an amount, given a list of coin denominations. For example, there are 3 ways to give change for 4 if you have coins with denomiation 1 and 2: 1+1+1+1, 1+1+2, 2+2.
Do this exercise by implementing the countChange function in Main.scala. This function takes an amount to change, and a list of unique denominations for the coins. Its signature is as follows: def countChange(money: Int, coins: List[Int]): Int
Once again, you can make use of functions isEmpty, head and tail on the list of integers coins.
Hint: Think of the degenerate cases. How many ways can you give change for 0 CHF? How many ways can you give change for >0 CHF, if you have no coins?