# Higher-Order Procedures

*Summary:* In this laboratory, you will use and define higher-order procedures.

## Preparation

Make sure that you understand what `map`

, `compose`

, `o`

, `left-section`

, `right-section`

, and `apply`

are intended to do.

## Exercises

### Exercise 1: Combining `map`

and `apply`

a. Use `apply`

and `map`

to sum the first elements of each list in a list of lists of numbers. The result should be a number.

```
> (apply _____ (map _____ (list (list 1 2 3) (list 4 5 6) (list 7 8 9 10) (list 11 12))))
23 ; 1 + 4 + 7 + 11
```

b. Use `apply`

and `map`

to sum the last elements of each list in a list of lists of numbers. The result should be a number.

```
> (apply _____ (map _____ (list (list 1 2 3) (list 4 5 6) (list 7 8 9 10) (list 11 12))))
31 ; 3 + 6 + 10 + 12
```

### Exercise 2: Making Lists

a. Here are four expressions to generate the successors of the squares of the first ten positive integers. Verify that each works correctly.

```
(define v1 (map increment (map square (map increment (iota 10)))))
(define v2 (map (lambda (i) (increment (square (increment i)))) (iota 10)))
(define v3 (map (compose increment (compose square increment)) (iota 10)))
(define v4 (map (o increment square increment) (iota 10)))
```

b. Which of the four definitions above you prefer? Why? Be prepared to discuss your reasons with the class.

### Exercise 3: Dot-Product

Use `apply`

and `map`

to concisely define a procedure, `(dot-product lst1 lst2)`

, that takes as arguments two lists of numbers, equal in length, and returns the sum of the products of corresponding elements of the arguments:

```
> (dot-product (list 1 2 4 8) (list 11 5 7 3))
73
> (dot-product null null)
0
```

Note that we get the first result because (1 x 11) + (2 x 5) + (4 x 7) + (8 x 3) = 11 + 10 + 28 + 24 = 73 and the second because there are no products to add.

*Note:* You should not use recursion.

### Exercise 4: Acronyms

The procedure `(acronym strings)`

is intended to produce an acronym from a list of strings. For example,

```
> (acronym (list "GNU" "Image" "Manipulation" "Program"))
"GIMP"
> (acronym (list "International" "Business" "Machinery"))
"IBM"
> (acronym (list "Grinnell" "Independent" "Musical" "Productions"))
"GIMP"
```

Write `acronym`

as concisely as possible. As a hint, you will want to use `string-ref`

, `list->string`

, `map`

, and either `l-s`

or `r-s`

.

(Recall that `(string-ref string i)`

produces the * i*th character in

*.*

`string`

`list->string`

takes a list of characters and turns it into a string.)### Exercise 5: Removing Elements

Write a procedure, `(list-remove lst predicate)`

, that takes a list and predicate as parameters and removes all elements for which * predicate* holds.

For example,

```
> (list-remove (iota 10) odd?)
'(0 2 4 6 8)
> (define remove-whitespace (r-s list-remove char-whitespace?))
> (remove-whitespace (list #\a #\space #\b #\c))
(#\a #\b #\c)
> (list->string (remove-whitespace (string->list "Hello, my name is Dr. gigls")))
"Hello,mynameisDr.gigls"
```

Note: Although this problem can be solved using `map`

and `apply`

, it is much more straightforward to solve using recursion.

### Exercise 6: Combining Predicates

It is often the case that we have situations in which we need more than one predicate to hold. For example, since the `odd?`

predicate only works with integers, it is often helpful to test whether a value is an integer before testing whether it is odd.

```
> (odd? 3)
#t
> (odd? 4)
#f
> (odd? 3.5)
odd?: expects argument of type <integer>; given 3.5
> (define odd-integer? (lambda (val) (and (integer? val) (odd? val))))
> (odd-integer? 3.5)
#f
> (odd-integer? 3)
#t
> (odd-integer? 4)
#f
```

But if itâ€™s common to combine predicates into a new predicate, we might want to write a higher-order procedure that does that. For example, we might write a procedure, `(both p1 p2)`

that takes two predicates as parameters, and returns a new predicate that holds only when both of its component predicates hold.

```
> (define odd-integer? (both integer? odd?))
> (odd-integer? 3)
#t
> (define one-element-list? (both pair? (o null? cdr)))
> (one-element-list? 2)
#f
> (one-element-list? null)
#f
> (one-element-list? (list 1))
#t
> (one-element-list? (cons 2 null))
#t
> (one-element-list? (cons 2 3))
#f
```

Write the `both`

procedure.

### Exercise 7: Manipulating Predicates, Revisited

a. Write `(either p1 p2)`

, a procedure which takes two predicates as parameters and returns a predicate that holds when either of those predicates holds.

```
> (define RGB-RED (rgb-new 255 0 0))
> (irgb? RGB-RED)
#t
> (color-name? RGB-RED)
#f
> (color-name? "red")
#t
> (irgb? "red")
#f
> (define simple-color? (either color-name? irgb?))
> (simple-color? "red")
#t
> (simple-color? RGB-RED)
#t
> (simple-color? (list 255 0 255))
#f
```

b. Write a procedure, `(negate pred)`

, that returns `#t`

on the values for which * pred* returns

`#f`

, and returns `#f`

on the values for which *holds.*

`pred`

```
> (define non-empty-list? (both list? (negate null?)))
> (non-empty-list? (list 1 2 3))
#t
> (non-empty-list? 32)
#f
> (non-empty-list? (cons 1 2))
#f
> (non-empty-list? null)
#f
```