Suppose we want to modify the differentiation program so that it works with ordinary mathematical notation, in which
In the previous post we dealt with part (a) of this exercise which allowed us to ignore the associativity of the operations and just handle infix expressions with two operands. In this post we not only need to deal with variable numbers of operands (which we dealt with for prefix notation in exercise 2.57)... The example above contains a mix of operators at the same level within the expression. This means we'll also need to deal with operator precedence.+ and * are infix rather than prefix operators. Since the differentiation program is defined in terms of abstract data, we can modify it to work with different representations of expressions solely by changing the predicates, selectors, and constructors that define the representation of the algebraic expressions on which the differentiator is to operate.
- Show how to do this in order to differentiate algebraic expressions presented in infix form, such as
(x + (3 * (x + (y + 2)))). To simplify the task, assume that+and*always take two arguments and that expressions are fully parenthesized. - The problem becomes substantially harder if we allow standard algebraic notation, such as
(x + 3 * (x + y + 2)), which drops unnecessary parentheses and assumes that multiplication is done before addition. Can you design appropriate predicates, selectors, and constructors for this notation such that our derivative program still works?
Dealing With Precedence
At the moment our program only deals with three different kinds of operations: exponentiation, multiplication and addition. The normal order of precedence for these operations is the order in which they've just been listed. I.e. exponentiation binds more tightly than multiplication, which binds more tightly than addition. Of course we've also got parenthesized sub-expressions to consider - but, as we're representing our expressions as Scheme lists this doesn't actually cause us any additional precedence headaches. Provided we simply treat the sub-expression (which is a list) as an operand of the operator to which it is bound, and then process the sub-expression itself as if it were its own expression, all will be fine.
So how can we deal with standard algebraic notation and still have
deriv work then? Well, one way is to treat the expression as if it were appropriately parenthesized according to the precedence rules, binding the operands for the highest precedence operators first. For example, we can treat the expression (a + b * c ** d * e + f ** g) as if it were parenthesized as (a + (b * (c ** d) * e) + (f ** g)). I.e. we should treat this expression as a sum with three operands - the symbol a, a sub-expression that's the product b * (c ** d) * e, and the exponentiation f ** g. We should then also be able to apply the reduction process we built in exercise 2.57 (or a variant of it) to this to simplify expressions as much as is possible (or at least as much as is possible using reduce!).In order to do this we first need a way to identify what the lowest precedence operator in an expression is (ignoring sub-expressions). This effectively provides us with our groupings. For example, the lowest precedence operator in
(a + b * c ** d * e + f ** g) is +. If we then divide the expression up into sub-expressions around the occurrences of + we get the expression (a + (b * c ** d * e) + (f ** g)). We can then repeat this for the sub-expressions we've created: the lowest precedence operation in (b * c ** d * e) is *, so that allows us to divide that sub-expression up into (b * (c ** d) * e), while the lowest precedence operation in (f ** g) is **, so there's no further division to do there.Let's start by defining our operators in a list, ordered by precedence:
(define operators '(+ * **))We can then get the precedence of an operator by getting its index in the list. First, let's define a general procedure for getting the index of an element in a list. This can simply iterate through the list, accumulating a count until it finds a matching element. I'll use zero-based indexes, and have my procedure return -1 if the element isn't present:
(define (get-index element alist)
(define (find-element-index index remaining)
(cond ((null? remaining) -1)
((eq? element (car remaining)) index)
(else (find-element-index (+ index 1) (cdr remaining)))))
(find-element-index 0 alist))
Here it is in action:
> (get-index '+ operators) 0 > (get-index '** operators) 2 > (get-index 'wibble operators) -1We can then wrap
get-index in a procedure that gets the precedence of a given operator, or reports an error if it's not a valid operator:
(define (precedence op)
(let ((index (get-index op operators)))
(if (>= index 0)
index
(error "Unknown operation -- PRECEDENCE" op))))
So let's test that with the same symbols:
> (precedence '+) 0 > (precedence '**) 2 > (precedence 'wibble) . . Unknown operation -- PRECEDENCE wibbleOf course it's worth noting that, were we dealing with a larger or different set of operations then this simplistic approach to operator precedence would not necessarily work. This is due to some operations, such as subtraction or division, sharing the same precedence as other operations: subtraction has the same precedence as addition while division has the same precedence as multiplication. We could solve this by redefining
operators as a list of lists such as '((+ -) (* /) (**)), and then find the index of the list that the operator is a member of, rather than just finding the index of the matching element. However, to keep everything simple, both here and in the rest of the exercise, I'm going to stick to our limited set of three operations!Determining an Expression's Operation Type
As noted above, we want to treat expressions as if they were appropriately parenthesized according to the precedence rules, and to do this we want to identify what the lowest precedence operator in an expression is. We've produced
precedence, which will tell us what the precedence of an operator is. All we need to do now is iterate through the elements of an expression, find all the operators in it and figure out which has the lowest precedence.First, let's build a helper procedure to figure out whether or not a value is an operator. We can do this simply by asking for its index in the list of
operators and checking that it's non-negative:
(define (operator? op) (>= (get-index op operators) 0))We can then use
accumulate (introduced in the Sequence Operations sub-section in section 2.2.3) to process the expression. For each element of the expression this needs to check whether or not it's an operator and, if it is, check its precedence against that of the lowest precedence operator found so far. If it's got a lower precedence then we change the accumulated value to this new operator, otherwise we keep the current operator.
(define (get-operation expression)
(accumulate
(lambda (x y)
(if (and (operator? x)
(or (null? y) (< (precedence x) (precedence y))))
x
y))
'()
expression))
You'll note that I've used '() as the initial value. I'm not overly happy with this, and it probably betrays my C/C++/Java background somewhat... I could have made the assumption that every expression has a valid operator in it and used the highest precedence operator (i.e. **) as the initial value. If I'd done this then I could omit the null? check here. However, it feels more correct to me to start from a position of not making such an assumption and not knowing what the operator is. To do this we need a value which is easily testable and is guaranteed not to be an operator. I really want to use a null value here (like I say, C/C++/Java background), but as Scheme doesn't, as far as I'm aware, support these, I'm using the empty list instead...Anyway, here it is in action:
> (get-operation '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) '+ > (get-operation '(1 * 3)) '* > (get-operation '(1 ** 3 * 2)) '* > (get-operation '(1 + x * 3)) '+ > (get-operation '(1 ** y)) '**We can then redefine
sum?, product? and exponentiation in terms of get-operation, as all they need to do is check whether or not the operator returned by get-operation is the correct one for the type in question. As all three procedures perform the same test, differing only in the operation symbol, I extracted a helper procedure, is-operation? and had all three procedures delegate to this:
(define (is-operation? op expression) (and (pair? expression) (eq? (get-operation expression) op))) (define (sum? x) (is-operation? '+ x)) (define (product? x) (is-operation? '* x)) (define (exponentiation? x) (is-operation? '** x))Using the same expressions as before we can put this to the test:
> (sum? '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) #t > (sum? '(1 * 3)) #f > (sum? '(1 ** 3 * 2)) #f > (sum? '(1 + x * 3)) #t > (sum? '(1 ** y)) #f > (product? '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) #f > (product? '(1 * 3)) #t > (product? '(1 ** 3 * 2)) #t > (product? '(1 + x * 3)) #f > (product? '(1 ** y)) #f > (exponentiation? '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) #f > (exponentiation? '(1 * 3)) #f > (exponentiation? '(1 ** 3 * 2)) #f > (exponentiation? '(1 + x * 3)) #f > (exponentiation? '(1 ** y)) #tSelectors
In exercise 2.57 we are told that we can cope with arbitrary numbers of terms by having the selector for the left-hand side of an operator return the first term and the selector for the right hand side of an operator returning the rest of the terms: "For example, the addend of a sum would be the first term, and the augend would be the sum of the rest of the terms."
To do this with infix expressions we need to identify the first occurrence of the operator in a given expression and return the sub-list either preceding or following this element as appropriate. We can then use these to iterate through an expression with a known lowest precedence operator and divide it up into sub-expressions. Of course we'll need to cope with the situation that will then occur when we get to the last sub-expression - there will be no occurrence of the lowest precedence operator left. In this case the sub-list preceding the "first occurrence" of the operator should be the last sub-expression, while the sub-list following the "first occurrence" of the operator should be the empty list. This is consistent with exercise 2.57 and allows us to easily test when we've processed the last sub-expression (i.e. we test whether or not the sub-list following the first occurrence of the operator is null).
Now Scheme provides a procedure,
list-tail, that returns the tail of a list from a particular index onwards... and we've already built a procedure, get-index, that returns the index of an element in a list. We can plug these together to produce a procedure that will give us what we need:
(define (get-list-expression-after-op op expression)
(let ((index (get-index op expression)))
(if (< index 0)
'()
(list-tail expression (+ index 1)))))
Note that we need to put in an explicit check of the index as, under DrRacket at least, calling list-tail with a negative number returns the entire list, whereas we want it to return the empty list in such a case.Also note that this procedure always returns a list. However, to keep our selectors for the right-hand sides of operators consistent with the preceding exercises, if there's only a single element in the list we need it to return the element itself, not the list. While we could modify the procedure appropriately, I'm actually going to produce a wrapper procedure that gives us what we need (as we'll need the original procedure when we come on to modifying
reduce):
(define (get-expression-after-op op expression)
(let ((tail (get-list-expression-after-op op expression)))
(if (and (pair? tail) (null? (cdr tail)))
(car tail)
tail)))
Unfortunately, scheme doesn't appear to provide a complementary procedure to list-tail for getting the head of a list. However, it's not that difficult to build up a procedure that gets the first n items of a list. As before we'll have two procedures here: one to get the list preceding the first occurrence of an operation (or the whole list if it doesn't appear), and one to get that list and, if it only contains one element, return the element instead of the list:
(define (get-list-expression-before-op op expression)
(define (build-expression l)
(if (or (null? l) (eq? (car l) op))
'()
(cons (car l) (build-expression (cdr l)))))
(build-expression expression))
(define (get-expression-before-op op expression)
(let ((head (get-list-expression-before-op op expression)))
(if (null? (cdr head))
(car head)
head)))
Let's put these to the test:
> (get-expression-before-op '+ '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) '(1 * 3 * 5 ** 7) > (get-expression-after-op '+ '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) '(3 ** 4 + 4) > (get-expression-before-op '* '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) 1 > (get-expression-after-op '* '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) '(3 * 5 ** 7 + 3 ** 4 + 4) > (get-expression-before-op '** '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) '(1 * 3 * 5) > (get-expression-after-op '** '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) '(7 + 3 ** 4 + 4) > (get-expression-before-op '+ '(1 * 3 * 5 ** 7 + 3)) '(1 * 3 * 5 ** 7) > (get-expression-after-op '+ '(1 * 3 * 5 ** 7 + 3)) 3 > (get-expression-before-op '- '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) '(1 * 3 * 5 ** 7 + 3 ** 4 + 4) > (get-expression-after-op '- '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) '()It's worth noting the last case here - if the operator is not present in the expression then the whole expression is "before" it, and nothing is "after" it.
Okay, now we can easily build our selectors using these:
(define (addend s) (get-expression-before-op '+ s)) (define (augend s) (get-expression-after-op '+ s)) (define (multiplier s) (get-expression-before-op '* s)) (define (multiplicand s) (get-expression-after-op '* s)) (define (base s) (get-expression-before-op '** s)) (define (exponent s) (get-expression-after-op '** s))And here they are in action:
> (sum? '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) #t > (addend '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) '(1 * 3 * 5 ** 7) > (augend '(1 * 3 * 5 ** 7 + 3 ** 4 + 4)) '(3 ** 4 + 4) > (addend '(3 ** 4 + 4)) '(3 ** 4) > (augend '(3 ** 4 + 4)) 4 > (product? '(1 ** 3 * 2)) #t > (multiplier '(1 ** 3 * 2)) '(1 ** 3) > (multiplicand '(1 ** 3 * 2)) 2 > (exponentiation? '(1 ** y)) #t > (base '(1 ** y)) 1 > (exponent '(1 ** y)) 'yReduce, Reuse, Recycle
In exercise 2.57 we produced the procedure
reduce that would reduce an expression to a simpler form. We should be able to reuse it here, with some adjustments, to perform the same reductions for infix expressions. We can reuse the main body of the procedure and count-accumulate directly, as these deal with expressions as atomic units - they don't deal with any of the internals of the representation. However, both generate and count-accumulate need a few changes. Let's start with generate.We still want
generate to check whether or not we have any expression counts and just return the accumulation of the numeric values if we don't. However, we're trying to generate a different expression now. I had a think about it and decided that a better way to approach the production of expressions was to:
- No longer delegate the insertion of operators to passed in procedures - they're all using infix notation, so rather than repeat the same pattern of insertion in each constructor we can simply have the constructor pass in the appropriate symbols for the two operations we need to know about: the normal operation to insert between each sub-expression and the operation to use when promoting a sub-expression that occurs multiple times. I've assumed these are in scope as
operationandpromoted-oprespectively. - Include the accumulation into the sub-expression counts. If we append it to the end of the list of counts as having occurred once then we can process all of the sub-expressions and the accumulation at once.
- Map the counts across into a list of sub-expressions (where they occurred once) and promoted sub-expressions (where they occurred multiple times). I.e. we can use
mapto examine each {sub-expression, count} tuple and, if the count is 1, map it to the sub-expression, otherwise promote it. - Iterate through the mapped list and insert the operation's symbol between each mapped sub-expression. To do this we can use
fold-leftfrom exercise 2.38, start with an initial value of a list containing the first mapped sub-expression only and then, for each subsequent sub-expression, append the operator and the sub-expression to this list.
(define (generate counts accumulation)
(if (null? counts)
accumulation
(let ((full-reduction (if (= accumulation identity-element)
counts
(append counts (list (list accumulation 1))))))
(let ((promoted-reduction (map (lambda (x) (if (= (cadr x) 1)
(car x)
(list (car x)
promoted-op
(cadr x))))
full-reduction)))
(if (null? (cdr promoted-reduction))
(car promoted-reduction)
(fold-left (lambda (x y) (append x (list operation y)))
(list (car promoted-reduction))
(cdr promoted-reduction)))))))
Now let's move on to process... First note that we've managed to simplify the null case. The updated generate now copes with the case when the accumulation equals the identity element, so we can simplify process's handling so that it just calls generate. Numbers can just be dealt with as before, as can the case when the head of the expression is not a sub-expression of the operation type we're reducing for. The only remaining case is when the head of the expression is a sub-expression of the operation type we're reducing for. In this case we can use is-operation? as the test - after all we've made the assumption that the operation we're processing for is in scope as operation, so there's no need to have an explicit procedure to test this passed in! The append that takes place is slightly different though. We're no longer dealing with prefix operators, so we can't just lop the top off the sub-expression and append that in front of the tail of the expression. Instead we need to split the sub-expression around the operator, and append both of these sub-expressions, in order, in front of the tail of the expression.As usual, this can be expressed more cleaning in code than in English:
(define (process expression counts accumulation)
(cond ((null? expression) (generate counts accumulation))
((number? (car expression))
(process (cdr expression)
counts
(accumulator (car expression) accumulation)))
((is-operation? operation (car expression))
(process (append (get-list-expression-before-op operation (car expression))
(get-list-expression-after-op operation (car expression))
(cdr expression))
counts
accumulation))
(else (process (cdr expression)
(count-accumulate (car expression) counts)
accumulation))))
Putting all this together gives us a new version of reduce as follows:
(define (reduce operation promoted-op accumulator identity-element expression)
(define (generate counts accumulation)
(if (null? counts)
accumulation
(let ((full-reduction (if (= accumulation identity-element)
counts
(append counts (list (list accumulation 1))))))
(let ((promoted-reduction (map (lambda (x) (if (= (cadr x) 1)
(car x)
(list (car x)
promoted-op
(cadr x))))
full-reduction)))
(if (null? (cdr promoted-reduction))
(car promoted-reduction)
(fold-left (lambda (x y) (append x (list operation y)))
(list (car promoted-reduction))
(cdr promoted-reduction)))))))
(define (count-accumulate expression counts)
(cond ((null? counts) (list (list expression 1)))
((equal? expression (caar counts))
(cons (list expression (+ (cadar counts) 1)) (cdr counts)))
(else (cons (car counts) (count-accumulate expression (cdr counts))))))
(define (process expression counts accumulation)
(cond ((null? expression) (generate counts accumulation))
((number? (car expression))
(process (cdr expression)
counts
(accumulator (car expression) accumulation)))
((is-operation? operation (car expression))
(process (append (get-list-expression-before-op operation (car expression))
(get-list-expression-after-op operation (car expression))
(cdr expression))
counts
accumulation))
(else (process (cdr expression)
(count-accumulate (car expression) counts)
accumulation))))
(if (and (pair? expression) (not (null? (cdr expression))))
(process expression '() identity-element)
(error "invalid expression -- REDUCE" expression)))
It's then a simple matter to modify make-sum and make-product to use this:
(define (make-sum . a)
(reduce '+ '* + 0 a))
(define (make-product . p)
(let ((reduction (reduce '* '** * 1 p)))
(if (and (pair? reduction)
(number? (last reduction))
(= (last reduction) 0))
0
reduction)))
And to complete the set we can modify make-exponentiation so that it too uses infix notation:
(define (make-exponentiation base exponent)
(cond ((=number? base 1) 1)
((=number? exponent 0) 1)
((=number? exponent 1) base)
((and (number? base) (number? exponent)) (expt base exponent))
(else (list base '** exponent))))
Let's make some expressions!
> (make-sum 'a 'b 'c 1 2 3 'c 'd 'e 4 5 6 'a 'd 'f) '((a * 2) + b + (c * 2) + (d * 2) + e + f + 21) > (make-product 'a 'b 'c 1 2 3 'c 'd 'e 4 5 6 'a 'd 'f) '((a ** 2) * b * (c ** 2) * (d ** 2) * e * f * 720) > (make-sum 1 2 3) 6 > (make-product 4 5 6) 120 > (make-sum 'a 'a 'a) '(a * 3) > (make-product (make-sum 'a 'b) (make-sum 'a 'b)) '((a + b) ** 2)Derive Me Crazy!
And, without further ado (or any changes to
deriv), here it is in action:
> (deriv (make-product 'x 'y (make-sum 'x 3)) 'x) '((x * y) + (y * (x + 3))) > (deriv (make-product (make-product 'x 'y) (make-sum 'x 3)) 'x) '((x * y) + (y * (x + 3))) > (deriv (make-exponentiation 'a (make-sum 'a 'b)) 'a) '((a + b) * (a ** (a + b + -1)))
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