# Naming Values with Local Bindings

## Exercises

Start DrScheme.

### Exercise 1: Evaluating `let`

What are the values of the following `let`-expressions? You may use DrScheme to help you answer these questions, but be sure you can explain how it arrived at its answers.

a.

```(let ((tone "fa")
(call-me "al"))
(list call-me tone "l" tone))
```

b.

```;; Solutions to the quadratic equation x^2 - 5x + 4:
(let ((discriminant (- (* -5 -5) (* 4 1 4))))
(list (/ (+ (- -5) (sqrt discriminant)) (* 2 1))
(/ (- (- -5) (sqrt discriminant)) (* 2 1))))
```

c.

```(let ((total (+ 8 3 4 2 7)))
(let ((mean (/ total 5)))
(* mean mean)))
```

### Exercise 2: Nesting Lets

Write a nested `let`-expression that binds a total of five names, `alpha`, `beta`, `gamma`, `delta`, and `epsilon`, with `alpha` bound to 9387 and each subsequent name bound to a value twice as large as the one before it -- `beta` should be twice as large as `alpha`, `gamma` twice as large as `beta`, and so on. The body of the innermost `let`-expression should then compute the sum of the values of the five names.

### Exercise 3: Simplifying Nested Lets

Write a `let*`-expression equivalent to the `let`-expression in the previous exercise.

Consider the following procedure that squares all the values in a list.

```;;; Procedure:
;;;   square-values
;;; Parameters:
;;;   lst, a list of numbers of the form (num_1 num_2 ... num_n)
;;; Purpose:
;;;   Squares all the values in lst.
;;; Produces:
;;;   list-of-squares, a list of numbers
;;; Preconditions:
;;;   [Standard]
;;; Postconditions:
;;;   list-of-squares has the form (square_1 square_2 ... square_n)
;;;   For all i, square_i is the square of num_i (that is num_i * num_i).
(define square-values
(lambda (lst)
(let ((square (lambda (val) (* val val))))
(if (null? lst)
null
(cons (square (car lst)) (square-values (cdr lst)))))))
```

a. Verify that `square-values` works correctly.

b. Try to execute `square` outside of `square-values`. Explain what happens.

### Exercise 5: Finding the Longest Element List

Here is a procedure that takes a non-empty list of lists as an argument and returns the longest list in the list (or one of the longest lists, if there is a tie).

```;;; Procedure:
;;;   longest-list-in-list
;;; Parameters:
;;;   los, a list of lists
;;; Purpose:
;;;   Finds one of the longest lists in los.
;;; Produces:
;;;   longest, a list
;;; Preconditions:
;;;   los is a nonempty list.
;;;   every element of los is a list.
;;; Postconditions:
;;;   Does not affect los.
;;;   Returns an element of los.
;;;   No element of los is longer than longest.
(define longest-list-in-list
(lambda (los)
; If there is only one list, that list must be the longest.
(if (null? (cdr los))
(car los)
; Otherwise, take the longer of the first list and the
; longest remaining list.
(longer-list (car los) (longest-list-in-list (cdr los))))))
```

This definition of the `longest-list-in-list` procedure includes a call to the `longer-list` procedure, which returns the longer of two given lists:

```;;; Procedure:
;;;   longer-list
;;; Parameters:
;;;   left, a list
;;;   right, a list
;;; Purpose:
;;;   Find the longer of left and right.
;;; Produces:
;;;   longer, a list
;;; Preconditions:
;;;   Both left and right are lists.
;;; Postconditions:
;;;   longer is a list.
;;;   longer is either equal to left or to right.
;;;   longer is at least as long as left.
;;;   longer is at least as long as right.
(define longer-list
(lambda (left right)
(if (<= (length right) (length left))
left
right)))
```

Revise the definition of `longest-list-in-list` so that the name `longer-list` is bound to the procedure that it denotes only locally, in a `let`-expression.

### Exercise 6: Alternate Techniques

Note that there are at least two possible ways to do the previous exercise: The definiens of `longest--list-in-list` can be a `lambda`-expression with a `let`-expression as its body, or it can be a `let`-expression with a `lambda`-expression as its body. That is, it can take the form

```(define longest-list-in-list
(let (...)
(lambda (los)
...)))
```

or the form

```(define longest-list-in-list
(lambda (los)
(let (...)
...)))
```

a. Define `longest-list-in-list` in whichever way that you did not define it for the previous exercise.

b. Does the order of nesting affect what happens when the procedure is invoked? If so, which arrangement is better? Why?

c. Make a similar change to `square-values`.

### Exercise 7: Another Alternative

The two definitions you came up with in the previous exercises are not the only alternatives you have in placing the `let`. Since `longer-list` is only needed in the recursive case, you can place the `let` there.

```(define longest-list-in-list
(lambda (los)
; If there is only one list, that list must be the longest.
(if (null? (cdr los))
(car los)
; Otherwise, take the longer of the first list and the
; longest remaining list.
(let ((longer-list
(lambda (left right)
(if (<= (length right) (length left))
left
right))))
(longer-list (car los) (longest-list -in-list (cdr los))))))
```

Including the original definition, you've now seen or written four variants of `longest-list-in-list`. Which do you prefer? Why?

### Exercise 8: Checking Preconditions

Extend your favorite version of `longest-list-in-list` so that it verifies its preconditions (i.e., that `los` only contains lists and that `los` is nonempty). If the preconditions are not met, your procedure should return `#f`.

It is perfectly acceptable for you to check each list element in turn to determine whether or not it is a list, rather than to check them all at once, in advance.

## For Those With Extra Time

Using DrScheme's `(time exp)` operation, determine which of the four versions of `longest-list-in-list` is indeed the fastest. You should try a variety of lengths for the outer list. The inner lists can be mostly 0- or 1-element lists..

## History

Monday, 2 October 2000 [Samuel A. Rebelsky]

Wednesday, 28 February 2001 [Samuel A. Rebelsky]

• Removed a problem created last semester but inappropriate for this semester's class.
• Added comments in the longest string exercise and fixed some errors that I'd introduced into that exercise. (Bad Sam!)
• Added exercises 6 and 7 (further followups on the longest string exercise).
• Updated formatting.

Thursday, 1 March 2001 [Samuel A. Rebelsky]

Monday, 7 October 2002 [Samuel A. Rebelsky]

• Updated slightly.

Tuesday, 8 October 2002 [Samuel A. Rebelsky]

• Inserted exercise 4 (a simple test of local procedures).

Wednesday, 9 October 2002 [Samuel A. Rebelsky]

Sunday, 2 February 2003 [Samuel A. Rebelsky]

Disclaimer: I usually create these pages on the fly, which means that I rarely proofread them and they may contain bad grammar and incorrect details. It also means that I tend to update them regularly (see the history for more details). Feel free to contact me with any suggestions for changes.

This document was generated by Siteweaver on Tue May 6 09:19:52 2003.
The source to the document was last modified on Sun Feb 2 22:45:31 2003.
This document may be found at `http://www.cs.grinnell.edu/~rebelsky/Courses/CS153/2003S/Labs/let.html`.

Samuel A. Rebelsky, rebelsky@grinnell.edu