Archive for the ‘abstraction’ Tag

Simple 2D Ray Tracer   Leave a comment

Of late I’ve been studying light, and thought it would be fun to write a program to model light striking a planar mirror. I have now got a working version out which happily traces the rays using some simple vector math and a slightly buggy ray tracing algorithm. Pictures are necessary :D

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SICP Section 2.3.4   Leave a comment

This section puts what we’ve learned about sets into practice implementing a method of encoding and decoding messages encoded with huffman code trees, and also a method of generating the trees.

;;Exercise 2.67
> (decode sample-message sample-tree)
(A D A B B C A)

;;Exercise 2.68
The encode-symbol procedure I wrote is straightforward, but takes a rather large number of steps to encode a symbol, because we have to search the set of symbols at each node for the correct branch to take. I’m not sure if there’s a better way to do this…

(define (encode-symbol symbol tree)
  (cond ((leaf? tree) '())
        ((element-of-set? symbol (symbols (left-branch tree)))
         (cons 0 (encode-symbol symbol (left-branch tree))))
        ((element-of-set? symbol (symbols (right-branch tree)))
         (cons 1 (encode-symbol symbol (right-branch tree))))
        (else (error "symbol not in tree - ENCODE-SYMBOL" symbol))))

;;Exercise 2.69
Successive-merge was actually fairly straightforward to code, however, when I first attempted to do this, I did not fully understand the actual tree generation procedure! Perhaps this is what the authors were warning about when they said it was tricky, I was certainly tricked initially :D

(define (successive-merge leaf-set)
  (if (= (length leaf-set) 1)
      (car leaf-set)
      (successive-merge
       (adjoin-set (make-code-tree (car leaf-set)
                                   (cadr leaf-set))
                   (cddr leaf-set)))))

;;Exercise 2.70
I never knew 50s rock was so repetitive ;)
Probably better than the junk that gets put out these days though…

(define freq-pairs (list (list 'A 2) (list 'BOOM 1) (list 'GET 2) (list 'JOB 2) (list 'NA 16) (list 'SHA 3) (list 'YIP 9) (list 'WAH 1)))

(define message '(GET A JOB
                  SHA NA NA NA NA NA NA NA NA
                  GET A JOB
                  SHA NA NA NA NA NA NA NA NA
                  WAH YIP YIP YIP YIP YIP YIP YIP YIP YIP
                  SHA BOOM))

(define rock50s-tree (generate-huffman-tree freq-pairs))

(define encoded-message (encode message rock50s-tree))
(define huffman-length (length encoded-message))

(define logbase2-of-8 3)
(define fixed-length (* (length message) logbase2-of-8))

fixed-length turns out to be 108 bits long, whereas the huffman-length is only 84 bits long. That’s a pretty fair length saving.

;;Exercise 2.71
I’m not really a fan of drawing large trees, so I’ll just give the code to generate it.

(define freq-pairs-5 (list (list '1 1) (list '2 2) (list '3 4) (list '4 8) (list '5 16)))
(define freq-pairs-tree-5 (generate-huffman-tree freq-pairs-5))

(define freq-pairs-10 (list (list '1 1) (list '2 2) (list '3 4) (list '4 8) (list '5 16) (list '6 32) (list '7 64) (list '8 128) (list '9 256) (list '10 512)))
(define freq-pairs-tree-10 (generate-huffman-tree freq-pairs-10))

The number of bits required for n=5 code are
Largest frequency: 1
Lowest frequency: 4
For n=10
Largest frequency: 1
Lowest frequency 9

It’s fairly obvious that the largest frequency will always have 1 bit, and the lowest frequency will have (n-1) bits.

;;Exercise 2.72
As we descend the tree we have to search the list of symbols at the node we are on, which is of the order O(n). We will need to descend n levels worst case, therefore n searches n deep is O(n2)

SICP Section 2.3.3   Leave a comment

The focus of this section is on representing mathematical sets with Scheme’s built-in data structures, and methods of combining these data structures to enable us to write faster algorithms for performing set operations. We go from a basic list based representation, to binary trees. I found this section to be quite a test of my abstract thinking skills; visualizing the operations on the representations was quite difficult for me.

;;Exercise 2.59
Implementing union-set is fairly straightforward for the unordered-list representation.

(define (union-set set1 set2)
  (cond ((null? set1) set2)
        ((null? set2) set1)
        ((element-of-set? (car set1) set2)
         (union-set (cdr set1) set2))
        (else (cons (car set1) (union-set (cdr set1) set2)))))

;;Exercise 2.60
This exercise was pretty interesting, it makes you think about situations in which what would seem to be an inefficient algorithm is actually the best for a job. Here’s my implementation of the set operations for an unordered set which allows duplicates.

(define (element-of-set? x set)
  (cond ((null? set) false)
        ((equal? x (car set)) true)
        (else (element-of-set? x (cdr set)))))

(define (adjoin-set x set)
  (cons x set))

(define (intersection-set set1 set2)
  (cond ((or (null? set1) (null? set2)) '())
        ((element-of-set? (car set1) set2)
         (cons (car set1)
               (intersection-set (cdr set1) set2)))
        (else (intersection-set (cdr set1) set2))))

(define (union-set set1 set2)
  (append set2 set1))

From the code we can see that, when duplicates are allowed,

  • element-of-set? is still O(n)
  • adjoin-set is O(1) from O(n)
  • intersection-set is still O(n^2)
  • union-set is O(n) from O(n^2)

By allowing duplicates, we have sped things up quite a bit, even if the algorithms are going to use up a lot more memory. In choosing which algorithm to use in a certain situation, we need to take into account the environment it will be used in. If we are, for example, on a machine with limited CPU speed, but a lot of memory (not uncommon in machines that have been upgraded), the duplicate representation would be best. On the other hand, if we were dealing with huge numbers, it would be a big drain on memory to use the duplicate representation, and the non-duplicate representation would be better for keeping overhead down.

;;Exercise 2.61
Now we’re starting to optimize the representations a bit more, it’s really neat to see how simple ideas like ordering the sets can produce fairly large performance gains :)

(define (adjoin-set x S)
  (cond ((null? S) (list x))
        ((< x (car S)) (cons x S))
        ((> x (car S)) (cons (car S) (adjoin-set x (cdr S))))
        ((= x (car S)) S)))

The ordered-list representation will take less time to adjoin, on the average, because we will not necessarily have to process the whole list in order to know if a duplicate exists.

;;Exercise 2.62
This particular exercise required a fair bit of thought. Here’s my explanation.
If the first element of set1 is less than the first element of set2, we make that element the car of a new list, and the cdr of the new list is the union of the cdr of set1 and set2…

(define (union-set set1 set2)
  (cond ((null? set1) set2)
        ((null? set2) set1)
        ((< (car set1) (car set2))
         (cons (car set1) (union-set (cdr set1) set2)))
        ((> (car set1) (car set2))
         (cons (car set2) (union-set set1 (cdr set2))))
        ((= (car set1) (car set2))
         (cons (car set1)
               (union-set (cdr set1)
                          (cdr set2))))))

;;Exercise 2.63
;;A
I think this exercise is designed to see how well we can transform a diagram into the actual representation of a tree as the most difficult part was writing down the trees :D

(define tree1 (make-tree 7
                         (make-tree 3
                                    (make-tree 1 '() '())
                                    (make-tree 5 '() '()))
                         (make-tree 9
                                    '()
                                    (make-tree 11 '() '()))))

(define tree2 (make-tree 3
                         (make-tree 1 '() '())
                         (make-tree 7
                                    (make-tree 5 '() '())
                                    (make-tree 9
                                               '()
                                               (make-tree 11 '() '())))))

(define tree3 (make-tree 5
                         (make-tree 3
                                    '()
                                    (make-tree 1 '() '()))
                         (make-tree 9
                                    (make-tree 7 '() '())
                                    (make-tree 11 '() '()))))

After running these through both functions, you’ll see that both the recursive and iterative procedures produce the same result, which is expected if they are to do the same job…

;B
Yes, they both have the same order of growth, however tree->list-2 uses less memory than tree->list-3, and thus could be faster in certain situations.

;;Exercise 2.64
Partial-tree works in a very recursive fashion :)
The best way I can explain is to work through an example.
Let’s say we have the list '(1 2 3 4 5) to turn into a tree.
Partial-tree first assigns equal, or almost equal sizes for the left and right branch of the new tree. So for our tree left-size will be 2, and right-size will be 3 (because the size of the list is not even). Then, partial tree conses together the result of partial-tree on the left and right halves of the given list. It’s quite a bit for me to wrap my head around, to be sure!

The list looks like so:

;;Exercise 2.65
Still a work in progress for me…

;;Exercise 2.66
The lookup procedure is basically element-of-set?, except instead of returning true or false, we return the item requested by key, or false if it’s not found.

(define (lookup key records)
  (cond ((null? records) false)
        ((= key (entry records)) (record-value (entry records)))
        ((< key (entry records))
         (lookup key (left-branch records)))
        ((> key (entry records))
         (lookup key (right-branch records)))))

;;Where a record structure looks like
(define (make-record key value)
  (cons key value))

Posted 23/09/2010 by Emmanuel Jacyna in Scheme, SICP

Tagged with , , , , , , , , ,

SICP – Section 2.3.1-2.3.2   Leave a comment

This section introduces the idea of “symbolic data”, and demonstrates how we can abstract our code with it. I found it to be really enjoyable, the questions were very challenging at the end. I notice I seem to say that I find each section enjoyable, and I do! I cannot recommend this book enough, not only for the knowledge and ideas you gain from it, but for the fun you’ll have solving the problems.

;;Exercise 2.53

As usual, just a check to make sure we understand how to manipulate symbolic data.
>(list '(a b c))
(a b c)
>(list (list 'george))
((george))
>(cdr '((x1 x2) (y1 y2)))
(y1 y2)
>(cadr '((x1 x2) (y1 y2)))
y1
>(pair? (car '(a short list)))
#f
>(memq 'red '((red shoes) (blue socks)))
#f
>(memq 'red '(red shoes blue socks))
(red shoes blue socks)

;;Exercise 2.54

A fairly straightforward implementation of the recursive description they gave.

(define (my-equal? a b)
  (cond ((and (eq? a '()) (eq? b '())) true)
        ((and (symbol? a) (symbol? b))
         (eq? a b))
        ((and (pair? a) (pair? b))
         (and (my-equal? (car a) (car b))
              (my-equal? (cdr a) (cdr b))))
        (else false)))

;;Exercise 2.55

The first challenging question. Had me stumped for a while, even when I typed the code in at my interpreter I didn’t get it, however, if you notice the footnote no. 34, it explains that the interpreter replaces each use of ' with a quote. So ''abracadabra is '(quote abracadabra), and the car of that list is the symbol quote!

;;Exercise 2.56

We’ve been introduced to the differentiation program, a very nice example of abstraction. The key to this exercise is pretty much to copy what happens for the other rules, just supplying different selectors and constructors, and a bit more code in the deriv function. I’m a bit of a rebel ;) I chose to use the ‘^ symbol instead of ‘**, as ‘^ seems more intuitive to me :)

Here are the selectors, predicate, and constructors:

;;'(^ b a)
(define (exponentiation? x)
  (and (pair? x) (eq? (car x) '^)))

(define (base x)
  (cadr x))

(define (exponent x)
  (caddr x))

(define (make-exponentiation base exponent)
  (cond ((and (number? base) (number? exponent))
         (expt base exponent))
        ((=number? exponent 0) 1)
        ((=number? exponent 1) base)
        (else (list '^ base exponent))))

And here is the modified deriv function:

(define (deriv expr var)
  (cond ((number? expr) 0)
        ((variable? expr)
         (if (same-variable? expr var) 1 0))
        ((exponentiation? expr) ;;EXPONENTIATION RULE
         (make-product
           (make-product (exponent expr)
                         (make-exponentiation (base expr)
                                              (- (exponent expr) 1)))
           (deriv (base expr) var)))
        ((sum? expr)
         (make-sum (deriv (addend expr) var)
                   (deriv (augend expr) var)))
        ((product? expr)
         (make-sum
           (make-product (multiplier expr)
                         (deriv (multiplicand expr) var))
           (make-product (deriv (multiplier expr) var)
                         (multiplicand expr))))
        (else
         (error "unknown expression type -- DERIV" expr))))

And a test,
>(deriv '(^ x 4) 'x)
(* 4 (^ x 3))

;;Exercise 2.57

I had some initial difficulties with this question, I understood what I was attempting to do in the selectors, but the constructors were a little trickier. With the selectors, the only interesting bit is in the augend selector. When we get the augend, we must remember if there is a list, we need to return it as the sum of the rest of the items in the list, not a a list of two or more numbers. eg. (augend ‘(+ 1 2 3)) is (+ 2 3) not (2 3).

(define (addend s) (cadr s))

(define (augend s)
  (let ((aug (cddr s)))
    (if (eq? (cdr aug) '())
        (car aug)
        (make-sum (car aug)
                  (cdr aug)))))

The constructor also held me up for a little while, as I had a bit of a hiccup in my code.
I was not combining the augend a2 into the list (‘+ a1 a2) correctly. This is best shown with some code…
>(make-sum x (2 (* 5 y)))
(+ x (2 (* 5 y)))

As you can see, instead of combining the 2 at the same level as x, it is a separate list, (2 (* 5 y)) which the program does not know how to parse. It should of course be (+ x 2 (* 5 y)). How do we combine the lists? Using append! Here is the constructor then:

(define (make-sum a1 a2)
  (cond ((=number? a1 0) a2)
        ((=number? a2 0) a1)
        ((and (number? a1) (number? a2)) (+ a1 a2))
        (else (append (list '+ a1) a2))))

Note that we can leave all the simplification code in, it will still work even in the new format. The changes to product are almost exactly the same as the changes to sum, I feel as though there’s an underlying abstraction somewhere ;)

Here’s the full code, including the product as well as sum, remember that this is all available on my sicp github repo.

;;(+ x y z)
(define (addend s) (cadr s))

(define (augend s)
  (let ((aug (cddr s)))
    (if (eq? (cdr aug) '())
        (car aug)
        (make-sum (car aug)
                  (cdr aug)))))

(define (make-sum a1 a2)
  (cond ((=number? a1 0) a2)
        ((=number? a2 0) a1)
        ((and (number? a1) (number? a2)) (+ a1 a2))
        (else (append (list '+ a1) a2))))

;;'(* x y z)
(define (multiplier p) (cadr p))

(define (multiplicand p)
  (let ((mult (cddr p)))
    (if (eq? (cdr mult) '())
        (car mult)
        (make-product (car mult)
                      (cdr mult)))))

(define (make-product m1 m2)
  (cond ((or (=number? m1 0) (=number? m2 0)) 0)
        ((=number? m1 1) m2)
        ((=number? m2 1) m1)
        ((and (number? m1) (number? m2)) (* m1 m2))
        (else (append (list '* m1) m2))))

;;Exercise 2.58
;;Part A

Things are getting interesting, we are now getting out of the safe world of polish notation, and into the rather complex one of parsing infix expressions. However, part a is very easy to implement, it simply requires swapping the parsing of the operator and the first operand, the change is exactly the same for multiplication.

(define (sum? x)
  (and (pair? x) (eq? (cadr x) '+)))

(define (addend s) (car s))

(define (augend s) (caddr s))

(define (make-sum a1 a2)
  (cond ((=number? a1 0) a2)
        ((=number? a2 0) a1)
        ((and (number? a1) (number? a2)) (+ a1 a2))
        (else (list a1 '+ a2))))

;;Part B

Part b is simply put, a beast of a question. It most certainly is quite a difficult problem, but it’s definitely possible! In fact, I suggest you stop reading now, so that you can have the same exhiliration that I got upon solving it! We are now required to parse infix expressions with any or no parenthization. I mulled over this one for a couple days, and spent more than a few nights writing attempts to solve it in my note book. I finally hit epiphany at badminton on Thursday (6/5/10) night.

Before we do anything, we should state what we are attempting to accomplish with our code. This is a bit trickier than it sounds, so let’s go through a couple examples and see if we can find anything interesting.

(x * (6 + y) + 5) is equivalent to
((x * (6 + y)) + 5) (note the extra parenthesis before x)

So one thing we need to do is replace parentheses to make it easier to parse.

(x * (6 + y) + 5)
can be reduced like so:

addend: x * (6 + y)
      multiplier: x
      multiplicand: (6 + y)
                 addend: 6
                 augend: y
augend:

This looks like a job for recursion to me, and it also looks like we’ll need to worry about putting sums in the right place before the products. For example, if we tried to grab the multiplicand of (3 * 6 + 5) we would obtain 6 + 5 instead of 6.

We can come up with a simple recursive definition for this:

If the augend of expr is a sum, then the augend of expr is the sum of everything after the sum operator in expr (y + 3* z). If the augend of expr is a product, then the augend of expr is the product of everything after the product operator in expr. If the augend of expr is a single variable then the augend of expr is the first item after the sum operator. Finally if the augend of expr is a single number, then the augend of expr is the first item after the sum operator.

So we can see that there are four rules to apply here which will eventually filter down to the base case of the single variable or number. So, the first thing we will implement is the definition of a sum and product. An expression is a sum if it has the ‘+ operator in it, and an expression is a product if it has only the ‘* operator in it. This gives us the proper operator precedence. The in? predicate here tells whether a given symbol is in a list.

(define (sum? x)
  (and (pair? x) (in? '+ x)))

(define (product? x)
  (and (pair? x) (not (in? '+ x)) (in? '* x)))

Now that we have our definitions of what constitutes a sum, we need to give the definitions of addend and augend. The addend is everything before the first ‘+ operator, but we also need to process the addend (remember the recursive definition) to correctly resolve operator precedence in it. We will do this with the recursive process stated above as a procedure named process-expr which I will leave unspecified. Remember that black box abstraction, and wishful thinking are very important. Anyway, here’s the definition of addend and augend, split-on is a function which gives a cons where the car is everything before the symbol, and the cdr is everything after the symbol.

>(split-on '+ (1* x + 2 * y))
((1 * x) 2 y)

(define (addend s)
  (process-expr (car (split-on '+ s))))

(define (augend s)
  (process-expr (cdr (split-on '+ s))))

Now it’s almost all in place, there’s one more definition left, how do we construct a sum? This is very easy, and stays the same as what we implemented for part A, (list a1 '+ a2).

Now, to implement the process-expr procedure. Since we have already defined all the selectors and constructors, this will be very easy.

(define (process-expr expr)
  (cond ((if (= (length expr) 1) (variable? (car expr)) false) (car expr))
        ((if (= (length expr) 1) (number? (car expr)) false) (car expr))
        ((sum? expr) (make-sum (addend expr)
                               (augend expr)))
        ((product? expr) (make-product (multiplier expr)
                                       (multiplicand expr))))))

This is a beautiful example of mutual recursion, and the great thing is, that it works! I first wrote this while waiting to play at badminton, and when I tranferred it from paper, it worked first try, with just a couple hiccups. One of those hiccups was the parsing of a parenthized expression such as (3 * (5 + 4)). The augend of this expression is ((5 + 4)), which is (list (list 5 + 4)). When process-expr encounters this, it checks if it is a variable, but while the length of (list (list 5 ‘+ 4)) is 1, it will not satisfy either the variable? or the number? predicate. It will then check if it is a sum, but there is no ‘+ in (list (list 5 + 4)), and it will also check if it is a product, but it won’t satisfy that either, so the cond expression will return #void! This is a bit of a bug, but it is fairly easy to fix. Let’s think of a parenthized-expression as another type of expression. It must satisfy these requirements to be considered parethized:

  • Must have a length of 1
  • Must not be a variable
  • Must not be a number
  • It’s car must be either a sum or a product

This is readily expressed as a procedure.

(define (parenthized-expression? expr)
  (and (not (variable? expr))
       (not (number? expr))
       (= (length expr) 1)
       (or (sum? (car expr))
           (product? (car expr)))))

;;Selector
(define (deparenthize expr) (car expr))

Now, we just need to add a rule for handling parenthized-expressions to process-expr. This is accomplished by adding a let expression, which will bind expr correctly:

(define (process-expr expr)
  (let ((expr (if (parenthized-expression? expr) (deparenthize expr) expr)))
    (cond ((if (= (length expr) 1) (variable? (car expr)) false) (car expr))
          ((if (= (length expr) 1) (number? (car expr)) false) (car expr))
          ((sum? expr) (make-sum (addend expr)
                                 (augend expr)))
          ((product? expr) (make-product (multiplier expr)
                                         (multiplicand expr))))))

We’re done! We have now created a fairly useful symbolic differentiator, which can handle (almost) arbitrary expressions. There is nothing as exhilirating as solving a really tough problem, and this can definitely be ranked as one of them! Here is all the code, and remember, this code is on my sicp github repo.

Happy Hacking!

;;Split an expr on symb
(define (split-on symb expr)
  (define (iter r l)
    (cond ((eq? l '()) r)
          ((eq? symb (car l)) (cons r (cdr l)))
          (else (iter (append r (list (car l))) (cdr l)))))
  (iter '() expr))

;;Is symb in expr?
(define (in? symb expr)
  (cond ((eq? expr '()) false)
        ((eq? symb (car expr)) true)
        (else (in? symb (cdr expr)))))

;; representing algebraic expressions
(define (variable? x) (symbol? x))

(define (same-variable? v1 v2)
  (and (variable? v1) (variable? v2) (eq? v1 v2)))

(define (=number? exp num)
  (and (number? exp) (= exp num)))

;;Is expr a parenthized expression? eg. ((1 + 2))
(define (parenthized-expression? expr)
  (and (not (variable? expr))
       (not (number? expr))
       (= (length expr) 1)
       (or (sum? (car expr))
           (product? (car expr)))))

(define (process-expr expr)
  (let ((expr (if (parenthized-expression? expr) (deparenthize expr) expr)))
    (cond ((if (= (length expr) 1) (variable? (car expr)) false) (car expr))
          ((if (= (length expr) 1) (number? (car expr)) false) (car expr))
          ((sum? expr) (make-sum (addend expr)
                                 (augend expr)))
          ((product? expr) (make-product (multiplier expr)
                                         (multiplicand expr))))))

(define (deparenthize expr) (car expr))

;;'(+ x y)
(define (sum? x)
  (and (pair? x) (in? '+ x)))

(define (addend s)
  (process-expr (car (split-on '+ s))))

(define (augend s)
  (process-expr (cdr (split-on '+ s))))

(define (make-sum a1 a2)
  (cond ((=number? a1 0) a2)
        ((=number? a2 0) a1)
        ((and (number? a1) (number? a2)) (+ a1 a2))
        (else (list a1 '+ a2))))

;;'(* x y)
(define (product? x)
  (and (pair? x) (not (in? '+ x)) (in? '* x)))

(define (multiplier p)
  (process-expr (car (split-on '* p))))

(define (multiplicand p)
  (process-expr (cdr (split-on '* p))))

(define (make-product m1 m2)
  (cond ((or (=number? m1 0) (=number? m2 0)) 0)
        ((=number? m1 1) m2)
        ((=number? m2 1) m1)
        ((and (number? m1) (number? m2)) (* m1 m2))
        (else (list m1 '* m2))))

(define (deriv expr var)
  (cond ((number? expr) 0)
        ((variable? expr)
         (if (same-variable? expr var) 1 0))
        ((sum? expr)
         (make-sum (deriv (addend expr) var)
                   (deriv (augend expr) var)))
        ((product? expr)
         (make-sum
           (make-product (multiplier expr)
                         (deriv (multiplicand expr) var))
           (make-product (deriv (multiplier expr) var)
                         (multiplicand expr))))
        (else
         (error "unknown expression type -- DERIV" expr))))

>(deriv '(x + 3 * (x + y + 2)) 'x)
4

Posted 09/05/2010 by Emmanuel Jacyna in SICP

Tagged with , , , , , ,

SICP – Section 2.1.4   Leave a comment

I’ve been silent for the past couple days, but not without good reason :)

I decided to finish all my solutions to the interval arithmetic exercise
before posting, because I can share as much insight as I have gained, in one
hit. ;)

In this extended exercise, we are helping Alyssa P. Hacker
(note the pun :)  to develop an interval arithmetic system for
her fellow electrical engineering compatriots.

The answers to exercises 2.7 to 2.10 are fairly clear, so here they are:

Exercise 2.7

(define (make-interval a b) (cons a b))

(define (lower-bound i) (car i))
(define (upper-bound i) (cdr i))

Exercise 2.8

(define (sub-interval x y)
(make-interval (- (lower-bound x) (lower-bound y))
(- (upper-bound x) (upper-bound y))))

Exercise 2.9

Let’s add some intervals and note the relation between the width of the
operands and the width of the result.


(define a (make-center-width 5 1))
(define b (make-center-width 10 2))
(define x (mul-interval a b))
(width x) = 3

From this it is fairly obvious that the width of the result of addition can
be described like so:
w(x*y) = w(x)+w(y)

The inverse holds true of division:
w(x/y) = w(x)-w(y)

Hmm… anyone else see the relation to logarithms?

Exercise 2.10
This one is also fairly simple.


(define (zero-span? i)
(if (and (<= (lower-bound i) 0) (>= (upper-bound i) 0))
true
false))

(define (div-interval x y)
(if (zero-span? y)
(error "Attempt to divide by an interval that spans 0")
(mul-interval x
(make-interval (/ 1.0 (upper-bound y))
(/ 1.0 (lower-bound y))))))


Exercise 2.11

To be honest, I didn’t get Ben’s cryptic comment. After reading
Eli Bendersky’s post on the topic I understood, but previously I had
gone off on a wrong tangent.

Regarding this, let me state my policy regarding my completion of these exercises. I firmly believe that cheating cheats the cheater, and while it is not necessarily cheating to check an answer to get hints, this is usually not necessary. There have been a few exercises where I have encountered a mindblock, but my advice to others is to go do something else, completely unrelated to computing or using your brain. Hop on your bike, go for a run or eat a pie, whatever works for you :) Usually when you get a mindblock it shows that you’ve been working your brain too long, too hard, and you just need to let it relax. This also happens a bit in programming, at least when you’re starting out, and in “How To Think Like A Computer Scientist” the author also suggests taking a break when you get frustrated, as this can be a source of unpleasant code and bugs.

I tend to read Eli’s posts after completing a section, because they are quite enlightening and informative. There is nothing quite like a veteran hackers view of the topic :)

This leads me to a shout out to Eli. Thanks for writing such detailed, interesting, and informative posts! I believe that, along with watching the lectures, doing the exercises and reading the book, your posts are a very complementary aid in understanding the material.


Exercise 2.12


(define (make-center-percent c p)
(make-interval (- c (* c p))
(+ c (* c p))))

(define (percent i)
(/ (- (upper-bound i)
(center i))
(center i)))

Again, this is fairly simple. The percent selector may have caused
some trouble, but it’s not too hard to figure out.

Exercise 2.13

;;Let's try:
(define a (make-center-percent 1 .1))
(define b (make-center-percent 1 .2))
(define x (* a b))
(percent x)
.294

So we can indeed see that (percent x) ~= (percent a) + (percent b)
This follows from exercise 2.9, except here we see that the percentage
is only approximate.

Exercise 2.14

To show why, we will use:


(define A (make-center-percent 8 .01))
(define B (make-center-percent 9 .05))

We will process A/A by hand, in order to see what’s going on.

(mul-interval A (make-interval (/ 1 8.08) (/ 1 7.92)))
;;p1 = .980
;;p2 = 1
;;p3 = 1
;;p4 = 1.02

The result is: (make-interval .980 1.02)
Instead of (make-interval 1 1) which you might have expected.

Let’s also do (/ A B) as suggested

;;(mul-interval A (make-interval (/ 1 9.45) (/ 1 (8.55))))
;;p1 = .838
;;p2 = .926
;;p3 = .855
;;p4 = .945

So the result is: (make-interval .838 .945) instead of the more accurate
(make-interval .855 .926)

Here’s my explanation.

Interval A/A is not an interval spanning 1 <-> 1 as we would expect in normal arithmetic, but an interval spanning .980 <-> 1.02. The width of this interval is a factor of the initial uncertainty .01, where the new uncertainty is .04. Interval A/B is also not the .855 <-> .926 that we might expect, but instead .838 <-> .945. We can see from these examples that arithmetic operations increase the innacuracy/uncertainty of the answer. The error in the answer is therefore directly related to the uncertainty in the intervals given.

Exercise 2.15

She is correct. As stated above, the amount of error in the answer is directly related to the uncertainty in the intervals given. Because there are more intervals in par1, there is a higher error in the answer. That is why par2 is a better method.

Reading Eli’s solutions to this extended exercise also gives a different view on the matter, he talks about it in terms of the fact that interval arithmetic is not a one-to-one mapping of simple numerical arithmetic.

Posted 27/03/2010 by Emmanuel Jacyna in Code, Scheme, SICP

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