From 4d0dec9ac2fd8b2026f221c33dd5e524176bf5fe Mon Sep 17 00:00:00 2001
From: Correl Roush <correl@gmail.com>
Date: Sun, 29 Jun 2014 19:35:10 -0400
Subject: [PATCH] Sections 2.2.1-2.2.2

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+#+BEGIN_HTML
+---
+title: 2.2 - Hierarchical Data and the Closure Property
+layout: org
+---
+#+END_HTML
+
+* Representing Sequences
+** List Operations
+*** Exercise 2.17
+    Define a procedure `last-pair' that returns the
+    list that contains only the last element of a given (nonempty)
+    list:
+
+    #+begin_src scheme
+      (last-pair (list 23 72 149 34))
+      (34)
+    #+end_src
+
+    --------------------------------------------------------------------
+
+    #+begin_src scheme :tangle yes
+      ;; -------------------------------------------------------------------
+      ;; Exercise 2.17
+      ;; -------------------------------------------------------------------
+
+      (define (last-pair list)
+        (cond ((null? list) ())
+              ((null? (cdr list)) list)
+              (else (last-pair (cdr list)))))
+    #+end_src
+*** Exercise 2.18
+    Define a procedure `reverse' that takes a list as
+    argument and returns a list of the same elements in reverse order:
+
+    #+begin_src scheme
+      (reverse (list 1 4 9 16 25))
+      (25 16 9 4 1)
+    #+end_src
+
+    --------------------------------------------------------------------
+
+    #+begin_src scheme :tangle yes
+      ;; -------------------------------------------------------------------
+      ;; 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 ()))
+    #+end_src
+
+*** 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:
+
+    #+begin_src scheme :tangle yes
+      ;; -------------------------------------------------------------------
+      ;; 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))
+    #+end_src
+
+    We could then call `cc' as follows:
+
+    #+begin_src scheme
+      (cc 100 us-coins)
+      292
+    #+end_src
+
+    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:
+
+    #+begin_src scheme :tangle yes
+      (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)))))
+    #+end_src
+                          
+    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?
+
+    ----------------------------------------------------------------------
+
+    #+begin_src scheme :tangle yes
+      (define first-denomination car)
+      (define except-first-denomination cdr)
+      (define no-more? null?)
+    #+end_src
+    
+*** 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
+
+    #+begin_src scheme
+      (define (f x y . z) <BODY>)
+    #+end_src
+
+    the procedure `f' can be called with two or more arguments.  If we
+    evaluate
+
+    #+begin_src scheme
+      (f 1 2 3 4 5 6)
+    #+end_src
+    
+    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
+
+    #+begin_src scheme
+      (define (g . w) <BODY>)
+    #+end_src
+
+    the procedure `g' can be called with zero or more arguments.  If we
+    evaluate
+
+    #+begin_src scheme
+      (g 1 2 3 4 5 6)
+    #+end_src
+
+    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,
+
+    #+begin_src scheme
+      (same-parity 1 2 3 4 5 6 7)
+      (1 3 5 7)
+
+      (same-parity 2 3 4 5 6 7)
+      (2 4 6)
+    #+end_src
+
+    ----------------------------------------------------------------------
+
+    #+begin_src scheme :tangle yes
+      ;; -------------------------------------------------------------------
+      ;; 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)
+              '()))
+    #+end_src
+
+** Mapping over lists
+*** Exercise 2.21:
+    The procedure `square-list' takes a list of numbers as argument
+    and returns a list of the squares of those numbers.
+
+    #+begin_src scheme
+      (square-list (list 1 2 3 4))
+      (1 4 9 16)
+    #+end_src
+    
+    Here are two different definitions of `square-list'.  Complete
+    both of them by filling in the missing expressions:
+
+    #+begin_src scheme
+      (define (square-list items)
+        (if (null? items)
+            nil
+            (cons <??> <??>)))
+
+      (define (square-list items)
+        (map <??> <??>))
+    #+end_src
+
+    ----------------------------------------------------------------------
+
+    #+begin_src scheme :tangle yes
+      ;; -------------------------------------------------------------------
+      ;; 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))
+    #+end_src
+    
+*** 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:
+
+    #+begin_src scheme
+          (define (square-list items)
+            (define (iter things answer)
+              (if (null? things)
+                  answer
+                  (iter (cdr things)
+                        (cons (square (car things))
+                              answer))))
+            (iter items nil))
+    #+end_src
+    
+    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':
+
+    #+begin_src scheme
+          (define (square-list items)
+            (define (iter things answer)
+              (if (null? things)
+                  answer
+                  (iter (cdr things)
+                        (cons answer
+                              (square (car things))))))
+            (iter items nil))
+    #+end_src
+    
+    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.
+    
+*** 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,
+
+    #+begin_src scheme
+      (for-each (lambda (x) (newline) (display x))
+                (list 57 321 88))
+      57
+      321
+      88
+    #+end_src
+
+    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'.
+
+    ----------------------------------------------------------------------
+
+    #+begin_src scheme :tangle yes
+      ;; -------------------------------------------------------------------
+      ;; Exercise 2.23
+      ;; -------------------------------------------------------------------
+
+      (define (for-each fun list)
+        (if (null? list)
+            #t
+            ))
+
+    #+end_src
+* Hierarchical Structures
+** 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::).
+
+   -------------------------------------------------------------------
+
+   #+begin_src scheme :tangle yes
+     ;; -------------------------------------------------------------------
+     ;; Exercise 2.24
+     ;; -------------------------------------------------------------------
+
+     '(1 (2 (3 4)))
+
+     ;; [ * | * ]
+     ;;   ↓   ↓
+     ;;   1 [ * | * ]
+     ;;       ↓   ↓
+     ;;       2 [ * | * ] → [ * | / ]
+     ;;           ↓           ↓
+     ;;           3           4
+
+     ;;   *
+     ;;  / \
+     ;; 1   *
+     ;;    / \
+     ;;   2   *
+     ;;      / \
+     ;;     3   4
+   #+end_src
+   
+** Exercise 2.25
+   Give combinations of `car's and `cdr's that will
+   pick 7 from each of the following lists:
+
+   #+begin_src scheme
+     (1 3 (5 7) 9)
+
+     ((7))
+
+     (1 (2 (3 (4 (5 (6 7))))))
+   #+end_src
+
+   ----------------------------------------------------------------------
+
+   #+begin_src scheme :tangle yes
+     ;; -------------------------------------------------------------------
+     ;; 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))))))))))))))))))
+   #+end_src
+
+** Exercise 2.26
+   Suppose we define `x' and `y' to be two lists:
+
+   #+begin_src scheme
+     (define x (list 1 2 3))
+
+     (define y (list 4 5 6))
+   #+end_src
+   
+   What result is printed by the interpreter in response to
+   evaluating each of the following expressions:
+
+   #+begin_src scheme
+     (append x y)
+
+     (cons x y)
+
+     (list x y)
+   #+end_src
+
+   ----------------------------------------------------------------------
+
+   #+begin_src scheme
+     ;; -------------------------------------------------------------------
+     ;; 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))
+   #+end_src
+
+** 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,
+
+   #+begin_src scheme
+     (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))
+   #+end_src
+
+   ----------------------------------------------------------------------
+
+   #+begin_src scheme :tangle yes
+     ;; -------------------------------------------------------------------
+     ;; 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 ()))
+
+   #+end_src
+   
+** 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,
+
+   #+begin_src scheme
+     (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)
+   #+end_src
+
+   ----------------------------------------------------------------------
+
+   #+begin_src scheme
+     (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) '()))
+   #+end_src
+   
+** 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'):
+
+   #+begin_src scheme
+          (define (make-mobile left right)
+            (list left right))
+   #+end_src
+   
+   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:
+
+   #+begin_src scheme
+     (define (make-branch length structure)
+     (list length structure))
+   #+end_src
+
+      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
+
+         #+begin_src scheme
+           (define (make-mobile left right)
+             (cons left right))
+
+           (define (make-branch length structure)
+           (cons length structure))
+         #+end_src
+
+         How much do you need to change your programs to convert to
+         the new representation?
+
+   ----------------------------------------------------------------------
+
+   #+begin_src scheme :tangle yes
+     ;; -------------------------------------------------------------------
+     ;; 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)
+   #+end_src
+** Mapping over trees
+*** 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:
+
+    #+begin_src scheme
+      (square-tree
+       (list 1
+             (list 2 (list 3 4) 5)
+             (list 6 7)))
+      (1 (4 (9 16) 25) (36 49))
+    #+end_src
+
+    Define `square-tree' both directly (i.e., without using any
+    higher-order procedures) and also by using `map' and recursion.
+
+    ----------------------------------------------------------------------
+
+    #+begin_src scheme :tangle yes
+      ;; -------------------------------------------------------------------
+      ;; 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))
+    #+end_src
+*** 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
+
+    #+begin_src scheme
+      (define (square-tree tree) (tree-map square tree))
+    #+end_src
+
+    ----------------------------------------------------------------------
+
+    #+begin_src scheme :tangle yes
+      ;; -------------------------------------------------------------------
+      ;; 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))
+    #+end_src
+*** 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:
+
+    #+begin_src scheme
+      (define (subsets s)
+        (if (null? s)
+            (list nil)
+            (let ((rest (subsets (cdr s))))
+              (append rest (map <??> rest)))))
+    #+end_src
+
+    ----------------------------------------------------------------------
+    
+    #+begin_src scheme :tangle yes
+      (define (subsets s)
+        (if (null? s)
+            (list '())
+            (let ((rest (subsets (cdr s))))
+              (append rest (map (lambda (x) (cons (car s) x)) rest)))))
+    #+end_src