# 2.2 - Hierarchical Data and the Closure Property

## 1 Representing Sequences

### 1.1 List Operations

#### 1.1.1 Exercise 2.17

Define a procedure `last-pair' that returns the list that contains only the last element of a given (nonempty) list:

(last-pair (list 23 72 149 34)) (34)

;; ------------------------------------------------------------------- ;; Exercise 2.17 ;; ------------------------------------------------------------------- (define (last-pair list) (cond ((null? list) ()) ((null? (cdr list)) list) (else (last-pair (cdr list)))))

#### 1.1.2 Exercise 2.18

Define a procedure `reverse' that takes a list as argument and returns a list of the same elements in reverse order:

(reverse (list 1 4 9 16 25)) (25 16 9 4 1)

;; ------------------------------------------------------------------- ;; Exercise 2.18 ;; ------------------------------------------------------------------- (define (reverse list) (define (iter original new) (if (null? original) new (let ((head (car original)) (tail (cdr original))) (iter tail (cons head new))))) (iter list ()))

#### 1.1.3 Exercise 2.19

Consider the change-counting program of section *Note 1-2-2::. It would be nice to be able to easily change the currency used by the program, so that we could compute the number of ways to change a British pound, for example. As the program is written, the knowledge of the currency is distributed partly into the procedure `first-denomination' and partly into the procedure `count-change' (which knows that there are five kinds of U.S. coins). It would be nicer to be able to supply a list of coins to be used for making change.

We want to rewrite the procedure `cc' so that its second argument is a list of the values of the coins to use rather than an integer specifying which coins to use. We could then have lists that defined each kind of currency:

;; ------------------------------------------------------------------- ;; Exercise 2.19 ;; ------------------------------------------------------------------- (define us-coins (list 50 25 10 5 1)) (define uk-coins (list 100 50 20 10 5 2 1 0.5))

We could then call `cc' as follows:

(cc 100 us-coins) 292

To do this will require changing the program `cc' somewhat. It will still have the same form, but it will access its second argument differently, as follows:

(define (cc amount coin-values) (cond ((= amount 0) 1) ((or (< amount 0) (no-more? coin-values)) 0) (else (+ (cc amount (except-first-denomination coin-values)) (cc (- amount (first-denomination coin-values)) coin-values)))))

Define the procedures `first-denomination', `except-first-denomination', and `no-more?' in terms of primitive operations on list structures. Does the order of the list `coin-values' affect the answer produced by `cc'? Why or why not?

(define first-denomination car) (define except-first-denomination cdr) (define no-more? null?)

#### 1.1.4 Exercise 2.20

The procedures `+', `*', and `list' take arbitrary numbers of arguments. One way to define such procedures is to use `define' with notation "dotted-tail notation". In a procedure definition, a parameter list that has a dot before the last parameter name indicates that, when the procedure is called, the initial parameters (if any) will have as values the initial arguments, as usual, but the final parameter's value will be a "list" of any remaining arguments. For instance, given the definition

(define (f x y . z) <BODY>)

the procedure `f' can be called with two or more arguments. If we evaluate

(f 1 2 3 4 5 6)

then in the body of `f', `x' will be 1, `y' will be 2, and `z' will be the list `(3 4 5 6)'. Given the definition

(define (g . w) <BODY>)

the procedure `g' can be called with zero or more arguments. If we evaluate

(g 1 2 3 4 5 6)

then in the body of `g', `w' will be the list `(1 2 3 4 5 6)'.(4)

Use this notation to write a procedure `same-parity' that takes one or more integers and returns a list of all the arguments that have the same even-odd parity as the first argument. For example,

(same-parity 1 2 3 4 5 6 7) (1 3 5 7) (same-parity 2 3 4 5 6 7) (2 4 6)

;; ------------------------------------------------------------------- ;; Exercise 2.20 ;; ------------------------------------------------------------------- (define (same-parity n . rest) (define (iter predicate original filtered) (cond ((null? original) filtered) ((predicate (car original)) (iter predicate (cdr original) (append filtered (list (car original))))) (else (iter predicate (cdr original) filtered)))) (iter (if (even? n) even? odd?) (cons n rest) '()))

### 1.2 Mapping over lists

#### 1.2.1 Exercise 2.21:

The procedure `square-list' takes a list of numbers as argument and returns a list of the squares of those numbers.

(square-list (list 1 2 3 4)) (1 4 9 16)

Here are two different definitions of `square-list'. Complete both of them by filling in the missing expressions:

(define (square-list items) (if (null? items) nil (cons <??> <??>))) (define (square-list items) (map <??> <??>))

;; ------------------------------------------------------------------- ;; Exercise 2.21 ;; ------------------------------------------------------------------- (define (square-list items) (if (null? items) '() (cons (* (car items) (car items)) (square-list (cdr items))))) (define (square-list items) (map (lambda (x) (* x x)) items))

#### 1.2.2 Exercise 2.22:

Louis Reasoner tries to rewrite the first `square-list' procedure of *Note Exercise 2-21:: so that it evolves an iterative process:

(define (square-list items) (define (iter things answer) (if (null? things) answer (iter (cdr things) (cons (square (car things)) answer)))) (iter items nil))

Unfortunately, defining `square-list' this way produces the answer list in the reverse order of the one desired. Why?

Louis then tries to fix his bug by interchanging the arguments to `cons':

(define (square-list items) (define (iter things answer) (if (null? things) answer (iter (cdr things) (cons answer (square (car things)))))) (iter items nil))

This doesn't work either. Explain.

The first iterative rewrite reads the items from first to last, but builds the list last to first (cons effectively prepends the answer to the list of results).

The second version attempts to reverse the arguments of cons, however this doesn't build a proper list. Normally, a list is a value paired with a list in the second slot. This pairs a list with a value in the second slot.

#### 1.2.3 Exercise 2.23

The procedure `for-each' is similar to `map'. It takes as arguments a procedure and a list of elements. However, rather than forming a list of the results, `for-each' just applies the procedure to each of the elements in turn, from left to right. The values returned by applying the procedure to the elements are not used at all–`for-each' is used with procedures that perform an action, such as printing. For example,

(for-each (lambda (x) (newline) (display x)) (list 57 321 88)) 57 321 88

The value returned by the call to `for-each' (not illustrated above) can be something arbitrary, such as true. Give an implementation of `for-each'.

;; ------------------------------------------------------------------- ;; Exercise 2.23 ;; ------------------------------------------------------------------- (define (for-each fun list) (if (null? list) #t ))

## 2 Hierarchical Structures

### 2.1 Exercise 2.24

Suppose we evaluate the expression `(list 1 (list 2 (list 3 4)))'. Give the result printed by the interpreter, the corresponding box-and-pointer structure, and the interpretation of this as a tree (as in *Note Figure 2-6::).

;; ------------------------------------------------------------------- ;; Exercise 2.24 ;; ------------------------------------------------------------------- '(1 (2 (3 4))) ;; [ * | * ] ;; ↓ ↓ ;; 1 [ * | * ] ;; ↓ ↓ ;; 2 [ * | * ] → [ * | / ] ;; ↓ ↓ ;; 3 4 ;; * ;; / \ ;; 1 * ;; / \ ;; 2 * ;; / \ ;; 3 4

### 2.2 Exercise 2.25

Give combinations of `car's and `cdr's that will pick 7 from each of the following lists:

(1 3 (5 7) 9) ((7)) (1 (2 (3 (4 (5 (6 7))))))

;; ------------------------------------------------------------------- ;; Exercise 2.25 ;; ------------------------------------------------------------------- (car (cdr (car (cdr (cdr '(1 3 (5 7) 9)))))) (car (car '((7)))) (car (cdr (car (cdr (car (cdr (car (cdr (car (cdr (car (cdr '(1 (2 (3 (4 (5 (6 7))))))))))))))))))

### 2.3 Exercise 2.26

Suppose we define `x' and `y' to be two lists:

(define x (list 1 2 3)) (define y (list 4 5 6))

What result is printed by the interpreter in response to evaluating each of the following expressions:

(append x y) (cons x y) (list x y)

;; ------------------------------------------------------------------- ;; Exercise 2.26 ;; ------------------------------------------------------------------- ;; (append x y) '(1 2 3 4 5 6) ;; (cons x y) '((1 2 3) 4 5 6) ;; (list x y) '((1 2 3) (4 5 6))

### 2.4 Exercise 2.27

Modify your `reverse' procedure of *Note Exercise 2-18:: to produce a `deep-reverse' procedure that takes a list as argument and returns as its value the list with its elements reversed and with all sublists deep-reversed as well. For example,

(define x (list (list 1 2) (list 3 4))) x ((1 2) (3 4)) (reverse x) ((3 4) (1 2)) (deep-reverse x) ((4 3) (2 1))

;; ------------------------------------------------------------------- ;; Exercise 2.27 ;; ------------------------------------------------------------------- (define (deep-reverse list) (define (iter original new) (if (null? original) new (let ((head (car original)) (tail (cdr original))) (iter tail (cons (if (pair? head) (deep-reverse head) head) new))))) (iter list ()))

### 2.5 Exercise 2.28

Write a procedure `fringe' that takes as argument a tree (represented as a list) and returns a list whose elements are all the leaves of the tree arranged in left-to-right order. For example,

(define x (list (list 1 2) (list 3 4))) (fringe x) (1 2 3 4) (fringe (list x x)) (1 2 3 4 1 2 3 4)

(define (fringe tree) (define (iter original new) (if (null? original) new (let ((head (car original)) (tail (cdr original))) (if (pair? head) (iter (append head tail) new) (iter tail (cons head new)))))) (iter (deep-reverse tree) '()))

### 2.6 Exercise 2.29

A binary mobile consists of two branches, a left branch and a right branch. Each branch is a rod of a certain length, from which hangs either a weight or another binary mobile. We can represent a binary mobile using compound data by constructing it from two branches (for example, using `list'):

(define (make-mobile left right) (list left right))

A branch is constructed from a `length' (which must be a number) together with a `structure', which may be either a number (representing a simple weight) or another mobile:

(define (make-branch length structure) (list length structure))

a. Write the corresponding selectors `left-branch' and `right-branch', which return the branches of a mobile, and `branch-length' and `branch-structure', which return the components of a branch.

b. Using your selectors, define a procedure `total-weight' that returns the total weight of a mobile.

c. A mobile is said to be "balanced" if the torque applied by its top-left branch is equal to that applied by its top-right branch (that is, if the length of the left rod multiplied by the weight hanging from that rod is equal to the corresponding product for the right side) and if each of the submobiles hanging off its branches is balanced. Design a predicate that tests whether a binary mobile is balanced.

d. Suppose we change the representation of mobiles so that the constructors are

(define (make-mobile left right) (cons left right)) (define (make-branch length structure) (cons length structure))

How much do you need to change your programs to convert to the new representation?

;; ------------------------------------------------------------------- ;; Exercise 2.29 ;; ------------------------------------------------------------------- (define (make-mobile left right) (list left right)) (define (make-branch length structure) (list length structure)) (define left-branch car) (define right-branch cadr) (define branch-length car) (define branch-structure cadr) ;; Test Data ;; ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (define (make-test-mobile) (make-mobile (make-branch 3 4) (make-branch 1 (make-mobile (make-branch 1 10) (make-branch 5 2))))) ;; Calculations ;; ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (define (branch-weight branch) (let ((structure (branch-structure branch))) (if (number? structure) structure (+ (branch-weight (left-branch structure)) (branch-weight (right-branch structure)))))) (define (total-weight mobile) (+ (branch-weight (left-branch mobile)) (branch-weight (right-branch mobile)))) (define (torque branch) (* (branch-length branch) (branch-weight branch))) (define (balanced? mobile) (define (balanced-branch? branch) (let ((structure (branch-structure branch))) (if (number? structure) #t (balanced? structure)))) (let ((left (left-branch mobile)) (right (right-branch mobile))) (and (= (torque left) (torque right)) (and (balanced-branch? left) (balanced-branch? right))))) ;; New representation ;; ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (define (make-mobile left right) (cons left right)) (define (make-branch length structure) (cons length structure)) (define right-branch cdr) (define branch-structure cdr)

### 2.7 Mapping over trees

#### 2.7.1 Exercise 2.30

Define a procedure `square-tree' analogous to the `square-list' procedure of *Note Exercise 2-21::. That is, `square-list' should behave as follows:

(square-tree (list 1 (list 2 (list 3 4) 5) (list 6 7))) (1 (4 (9 16) 25) (36 49))

Define `square-tree' both directly (i.e., without using any higher-order procedures) and also by using `map' and recursion.

;; ------------------------------------------------------------------- ;; Exercise 2.30 ;; ------------------------------------------------------------------- (define (square-tree tree) (cond ((null? tree) '()) ((not (pair? tree)) (square tree)) (else (cons (square-tree (car tree)) (square-tree (cdr tree)))))) (define (square-tree tree) (map (lambda (sub-tree) (if (pair? sub-tree) (square-tree sub-tree) (square sub-tree))) tree))

#### 2.7.2 Exercise 2.31

Abstract your answer to *Note Exercise 2-30:: to produce a procedure `tree-map' with the property that `square-tree' could be defined as

(define (square-tree tree) (tree-map square tree))

;; ------------------------------------------------------------------- ;; Exercise 2.31 ;; ------------------------------------------------------------------- (define (tree-map f tree) (map (lambda (sub-tree) (if (pair? sub-tree) (tree-map f sub-tree) (f sub-tree))) tree)) (define (square-tree tree) (tree-map square tree))

#### 2.7.3 Exercise 2.32

We can represent a set as a list of distinct elements, and we can represent the set of all subsets of the set as a list of lists. For example, if the set is `(1 2 3)', then the set of all subsets is `(() (3) (2) (2 3) (1) (1 3) (1 2) (1 2 3))'. Complete the following definition of a procedure that generates the set of subsets of a set and give a clear explanation of why it works:

(define (subsets s) (if (null? s) (list nil) (let ((rest (subsets (cdr s)))) (append rest (map <??> rest)))))

(define (subsets s) (if (null? s) (list '()) (let ((rest (subsets (cdr s)))) (append rest (map (lambda (x) (cons (car s) x)) rest)))))

## 3 Sequences as Conventional Interfaces

### 3.1 Sequence Operations

;; =================================================================== ;; Section 2.2.3: Sequences as Conventional Interfaces ;; =================================================================== (define (filter predicate sequence) (cond ((null? sequence) nil) ((predicate (car sequence)) (cons (car sequence) (filter predicate (cdr sequence)))) (else (filter predicate (cdr sequence))))) (define (accumulate op initial sequence) (if (null? sequence) initial (op (car sequence) (accumulate op initial (cdr sequence)))))

#### 3.1.1 Exercise 2.33

Fill in the missing expressions to complete the following definitions of some basic list-manipulation operations as accumulations:

(define (map p sequence) (accumulate (lambda (x y) <??>) nil sequence)) (define (append seq1 seq2) (accumulate cons <??> <??>)) (define (length sequence) (accumulate <??> 0 sequence))

;; ------------------------------------------------------------------- ;; Exercise 2.33 ;; ------------------------------------------------------------------- (define (map p sequence) (accumulate (lambda (x y) (cons (p x) y)) '() sequence)) (define (append seq1 seq2) (accumulate cons seq2 seq1)) (define (length sequence) (accumulate (lambda (e acc) (+ 1 acc)) 0 sequence))

#### 3.1.2 Exercise 2.34

Evaluating a polynomial in x at a given value of x can be formulated as an accumulation. We evaluate the polynomial

a_n r^n | a_(n-1) r^(n-1) + ... + a_1 r + a_0

using a well-known algorithm called "Horner's rule", which structures the computation as

(... (a_n r + a_(n-1)) r + ... + a_1) r + a_0

In other words, we start with a_n, multiply by x, add a_(n-1), multiply by x, and so on, until we reach a_0.(3)

Fill in the following template to produce a procedure that evaluates a polynomial using Horner's rule. Assume that the coefficients of the polynomial are arranged in a sequence, from a_0 through a_n.

(define (horner-eval x coefficient-sequence) (accumulate (lambda (this-coeff higher-terms) <??>) 0 coefficient-sequence))

For example, to compute 1 + 3x + 5x^3 + x^(5) at x = 2 you would evaluate

(horner-eval 2 (list 1 3 0 5 0 1))

;; ------------------------------------------------------------------- ;; Exercise 2.34 ;; ------------------------------------------------------------------- (define (horner-eval x coefficient-sequence) (accumulate (lambda (this-coeff higher-terms) (+ this-coeff (* higher-terms x))) 0 coefficient-sequence))

#### 3.1.3 Exercise 2.35

Redefine `count-leaves' from section *Note 2-2-2:: as an accumulation:

(define (count-leaves t) (accumulate <??> <??> (map <??> <??>)))

;; ------------------------------------------------------------------- ;; Exercise 2.35 ;; ------------------------------------------------------------------- (define (count-leaves t) (accumulate + 0 (map (lambda (node) (if (pair? node) (count-leaves node) 1)) t)))

#### 3.1.4 Exercise 2.36

The procedure `accumulate-n' is similar to `accumulate' except that it takes as its third argument a sequence of sequences, which are all assumed to have the same number of elements. It applies the designated accumulation procedure to combine all the first elements of the sequences, all the second elements of the sequences, and so on, and returns a sequence of the results. For instance, if `s' is a sequence containing four sequences, `((1 2 3) (4 5 6) (7 8 9) (10 11 12)),' then the value of `(accumulate-n + 0 s)' should be the sequence `(22 26 30)'. Fill in the missing expressions in the following definition of `accumulate-n':

(define (accumulate-n op init seqs) (if (null? (car seqs)) nil (cons (accumulate op init <??>) (accumulate-n op init <??>))))

;; ------------------------------------------------------------------- ;; Exercise 2.36 ;; ------------------------------------------------------------------- (define (accumulate-n op init seqs) (if (null? (car seqs)) '() (cons (accumulate op init (map car seqs)) (accumulate-n op init (map cdr seqs)))))

### 3.2 Exercise 2.37

Suppose we represent vectors v = (v_i) as sequences of numbers, and matrices m = (m_(ij)) as sequences of vectors (the rows of the matrix). For example, the matrix

+- -+ | 1 2 3 4 | | 4 5 6 6 | | 6 7 8 9 | +- -+

is represented as the sequence `((1 2 3 4) (4 5 6 6) (6 7 8 9))'. With this representation, we can use sequence operations to concisely express the basic matrix and vector operations. These operations (which are described in any book on matrix algebra) are the following:

__ (dot-product v w) returns the sum >_i v_i w_i (matrix-*-vector m v) returns the vector t, __ where t_i = >_j m_(ij) v_j (matrix-*-matrix m n) returns the matrix p, __ where p_(ij) = >_k m_(ik) n_(kj) (transpose m) returns the matrix n, where n_(ij) = m_(ji)

We can define the dot product as(4)

(define (dot-product v w) (accumulate + 0 (map * v w)))

Fill in the missing expressions in the following procedures for computing the other matrix operations. (The procedure `accumulate-n' is defined in *Note Exercise 2-36::.)

(define (matrix-*-vector m v) (map <??> m)) (define (transpose mat) (accumulate-n <??> <??> mat)) (define (matrix-*-matrix m n) (let ((cols (transpose n))) (map <??> m)))

#### 3.2.1 Exercise 2.38

The `accumulate' procedure is also known as `fold-right', because it combines the first element of the sequence with the result of combining all the elements to the right. There is also a `fold-left', which is similar to `fold-right', except that it combines elements working in the opposite direction:

(define (fold-left op initial sequence) (define (iter result rest) (if (null? rest) result (iter (op result (car rest)) (cdr rest)))) (iter initial sequence))

What are the values of

(fold-right / 1 (list 1 2 3))

(fold-left / 1 (list 1 2 3))

(fold-right list nil (list 1 2 3))

(fold-left list nil (list 1 2 3))

Give a property that `op' should satisfy to guarantee that `fold-right' and `fold-left' will produce the same values for any sequence.

;; ------------------------------------------------------------------- ;; Exercise 2.38 ;; ------------------------------------------------------------------- (define fold-right accumulate) (define (fold-left op initial sequence) (define (iter result rest) (if (null? rest) result (iter (op result (car rest)) (cdr rest)))) (iter initial sequence)) ;; (fold-right / 1 (list 1 2 3)) ;; 3/2 ;; (fold-left / 1 (list 1 2 3)) ;; 1/6 ;; (fold-right list nil (list 1 2 3)) ;; (((() 1) 2) 3) ;; (fold-left list nil (list 1 2 3)) ;; (1 (2 (3 ())))

#### 3.2.2 Exercise 2.39

Complete the following definitions of `reverse' (*Note Exercise 2-18::) in terms of `fold-right' and `fold-left' from *Note Exercise 2-38:::

(define (reverse sequence) (fold-right (lambda (x y) <??>) nil sequence)) (define (reverse sequence) (fold-left (lambda (x y) <??>) nil sequence))

;; ------------------------------------------------------------------- ;; Exercise 2.39 ;; ------------------------------------------------------------------- (define (reverse sequence) (fold-right (lambda (x y) (append y (list x))) '() sequence)) (define (reverse sequence) (fold-left (lambda (x y) (cons y x)) '() sequence))