In this page there is a comment after the post that gives a very short implementation of amb as a procedure:
(define (amb-backtrack)
(error "no solution found"))
(define (amb . args)
(call/cc (lambda (return)
(let ((backtrack amb-backtrack))
(map (lambda (x)
(call/cc (lambda (k)
(set! amb-backtrack k)
(return x))))
args)
(backtrack 'fail)))))
But I usually see amb implemented as a macro -- in the schemers.org FAQ, and also in Dorai Sitaram's book:
(define amb-fail '*)
(define initialize-amb-fail
(lambda ()
(set! amb-fail
(lambda ()
(error "amb tree exhausted")))))
(initialize-amb-fail)
(define-macro amb
(lambda alts...
`(let ((+prev-amb-fail amb-fail))
(call/cc
(lambda (+sk)
,#(map (lambda (alt)
`(call/cc
(lambda (+fk)
(set! amb-fail
(lambda ()
(set! amb-fail +prev-amb-fail)
(+fk 'fail)))
(+sk ,alt))))
alts...)
(+prev-amb-fail))))))
So -- the macro version is longer, and a little harder to understand. I could not see any advantages of it over the procedure version, and of course I would rather use a procedure than a macro. Is there anything I missed?
The difference is that a procedure call always evaluates all the arguments.
(amb 1 (very-expensive-computation))
will, with the procedure version of amb, perform the very-expensive-computation, then yield 1. If yielding 1 suffices for all further computation, then you've wasted a lot of time on a computation whose result value is never used. Even worse things happen, as #Eli Barzilay mentions in a comment, when amb is used to model an infinite non-determinism such as generating all natural numbers.
The macro version avoids this and its behavior is therefore closer to that of non-deterministic programming languages such as Prolog.
Related
Currently, i have been reading the book the scheme programming language written by Kent Dybvig.
In section 5.7, it implements memorized lazy eval in scheme using the scheme macro system.
The source code as
(define-syntax delay
(syntax-rules ()
[(_ expr) (make-promise (lambda () expr))]))
(define make-promise
(lambda (p)
(let ([val #f] [set? #f])
(lambda ()
(unless set?
(let ([x (p)])
(unless set?
(set! val x)
(set! set? #t))))
val))))
(define force
(lambda (promise)
(promise)))
But i can not understand why the variable set? need to be tested twice in the procedure make-promise.
This is the reason from the book
The second test of the variable set? in make-promise is necessary in the event that, as a result of
applying p, the promise is recursively forced. Since a promise must always return the same value, the result of
the first application of p to complete is returned.
which i can not understand
Could anyone explain it to me? Thanks!
The key is that force may reenter itself. Maybe an example can help you understand this:
(define x 5)
(letrec ([f (delay
(if (zero? x)
0
(begin
(set! x (- x 1))
(+ (force f) 1))))])
(force f))
The result will be 0, because the inner force call returns 0.
If without the second test, the result will be 5. In this situation, every (force f) returns different values.
Looking at Practical Common Lisp, we're looking at a simple automated unit test framework. We're trying to write a macro to be used as such:
(check (= (+ 1 2) 3) (= (- 1 4) 9))
This should expand to something using a previously defined function report-result. The suggested implementation is:
(defmacro check (&body forms)
`(progn
,#(loop for f in forms collect `(report-result ,f ',f))))
However, that expansion seems rather procedural to me. I wanted to replace the loop with a mapcar to expand to something like this:
(mapcar #'(lambda (form) (report-result form 'form)) (list form-1 ... form-n))
However, I'm clearly lacking the macro-writing skills to do so. Can someone come up with one such macro?
In case it's relevant, this is the definition of report-result:
(defun report-result (result form)
(format t "~:[FAIL~;pass~] ... ~a~%" result form))
It's indeed fairly simple: you just place the collect expression into the body of your mapcar:
(defmacro check (&body forms)
`(progn
,#(mapcar #'(lambda (form)
`(report-result ,form ',form))
forms)))
You don't really need to know anything about the "macro-y" stuff that's going on, in order to do the replacement you want, which is simply replacing a loop with some other equivalent expression: it will work just as well in a macro context as it would outside.
If you want to expand to a mapcar you can, but there's no real reason to do so, since the list's size is known at compile time. Here's what that would look like:
(defmacro check (&body forms)
`(let ((results (list ,#(mapcar #'(lambda (form)
`(list ,form ',form))
forms))))
(mapcar #'(lambda (result)
(report-result (car result) (cadr result)))
results)))
Which expands like so
> (macroexpand-1 '(check (+ 1 2) (* 2 3)))
(let ((results (list (list (+ 1 2) '(+ 1 2))
(list (* 2 3) '(* 2 3)))))
(mapcar #'(lambda (result) (report-result (car result) (cadr result)))
results))
Which as you can see is rather awkward: the macro already has the forms like (+ 1 2) available to it, but in order to preserve them to runtime for the mapcar lambda to see, you have to emit the input form twice. And you have to produce the whole list to map over, rather than just producing a list that's "finished" to begin with. Additionally, this produces a list as output, and requires having all the inputs and outputs in memory at once: the original macro with progn produced the inputs and outputs one at a time, and discarded them when finished.
Assume that the macro would take the boolean types a and b . If a is nil, then the macro should return nil (without ever evaluating b), otherwise it returns b. How do you do this?
This really depends on what you can use.
E.g., is or available? if? cond?
Here is one example:
(defmacro and (a b)
`(if ,a ,b nil)
EDIT. In response to a comment, or is more complicated because we have to avoid double evaluation:
(defmacro or (a b)
(let ((v (gensym "OR")))
`(let ((,v ,a))
(if ,v ,v ,b))))
sds's answer is nice and concise, but it has two limitations:
It only works with two arguments, whereas the built in and and or take any number of arguments. It's not too hard to update the solution to take any number of arguments, but it would be a bit more complicated.
More importantly, it's based very directly in terms of delayed operations that are already present in the language. I.e., it takes advantage of the fact that if doesn't evaluate the then or else parts until it has first evaluated the condition.
It might be a good exercise, then, to note that when a macro needs to delay evaluation of some forms, it's often the simplest strategy (in terms of implementation, but not necessarily the most efficient) to use a macro that expands to a function call that takes a function. For instance, a naive implementation of with-open-file might be:
(defun %call-with-open-file (pathname function)
(funcall function (open pathname)))
(defmacro my-with-open-file ((var pathname) &body body)
`(%call-with-open-file
,pathname
(lambda (,var)
,#body)))
Using a technique like this, you can easily get a binary and (and or):
(defun %and (a b)
(if (funcall a)
(funcall b)
nil))
(defmacro my-and (a b)
`(%and (lambda () ,a)
(lambda () ,b)))
CL-USER> (my-and t (print "hello"))
"hello" ; printed output
"hello" ; return value
CL-USER> (my-and nil (print "hello"))
NIL
or is similar:
(defun %or (a b)
(let ((aa (funcall a)))
(if aa
aa
(funcall b))))
(defmacro my-or (a b)
`(%or (lambda () ,a)
(lambda () ,b)))
To handle the n-ary case (since and and or actually take any number of arguments), you could write a function that takes a list of lambda functions and calls each of them until you get to one that would short circuit (or else reaches the end). Common Lisp actually already has functions like that: every and some. With this approach, you could implement and in terms of every by wrapping all the arguments in lambda functions:
(defmacro my-and (&rest args)
`(every #'funcall
(list ,#(mapcar #'(lambda (form)
`(lambda () ,form))
args))))
For instance, with this implementation,
(my-and (listp '()) (evenp 3) (null 'x))
expands to:
(EVERY #'FUNCALL
(LIST (LAMBDA () (LISTP 'NIL))
(LAMBDA () (EVENP 3))
(LAMBDA () (NULL 'X))))
Since all the forms are now wrapped in lambda functions, they won't get called until every gets that far.
The only difference is that and is specially defined to return the value of the last argument if all the preceding ones are true (e.g., (and t t 3) returns 3, not t, whereas the specific return value of every is not specified (except that it would be a true value).
With this approach, implementing or (using some) is no more complicated than implementing and:
(defmacro my-or (&rest args)
`(some #'funcall ,#(mapcar #'(lambda (form)
`(lambda () ,form))
args)))
I'm reading Peter Norvig's Paradigms of AI. In chapter 6.2, the author uses code like below (not the original code, I picked out the troubling part):
Code Snippet:
(progv '(op arg) '(1+ 1)
(eval '(op arg)))
As the author's original intent, this code should return 2, but in sbcl 1.1.1, the interpreter is apparently not looking up op in the environment, throwing out op: undefined function.
Is this implementation specific? Since the code must have been tested on some other lisp.
p.s Original code
You probably mean
(progv '(op arg) '(1+ 1)
(eval '(funcall op arg)))
Edit(2013-08-21):
PAIP was written in pre-ANSI-Common-Lisp era, so it's possible the code
there contains a few noncompliances wrt the standard. We can make
the examples work with the following revision:
(defun match-if (pattern input bindings)
"Test an arbitrary expression involving variables.
The pattern looks like ((?if code) . rest)."
(and (eval (reduce (lambda (code binding)
(destructuring-bind (var . val) binding
(subst val var code)))
bindings :initial-value (second (first pattern))))
(pat-match (rest pattern) input bindings)))
;; CL-USER> (pat-match '(?x ?op ?y is ?z (?if (eql (?op ?x ?y) ?z))) '(3 + 4 is 7))
;; ((?Z . 7) (?Y . 4) (?OP . +) (?X . 3) (T . T))
;; CL-USER> (pat-match '(?x ?op ?y (?if (?op ?x ?y))) '(3 > 4))
;; NIL
Elements in first positions are not looked up as values, but as functions and there is no concept of dynamic binding in the function namespace.
I'd say after a quick look that the original code was designed to evaluate in a context like
(progv '(x y) '(12 34)
(eval '(> (+ x y) 99)))
i.e. evaluating a formula providing substitution for variables, not for function names.
The other answers so far are right, in that the actual form being evaluated is not the variables being bound by progv (simply (op arg)), but none have mentioned what is being evaluated. In fact, the comments in the code you linked to provide a (very) short explanation (this is the only code in that file that uses progv):
(defun match-if (pattern input bindings)
"Test an arbitrary expression involving variables.
The pattern looks like ((?if code) . rest)."
;; *** fix, rjf 10/1/92 (used to eval binding values)
(and (progv (mapcar #'car bindings)
(mapcar #'cdr bindings)
(eval (second (first pattern))))
(pat-match (rest pattern) input bindings)))
The idea is that a call to match-if gets called like
(match-if '((?if code) . rest) input ((v1 val1) (v2 val2) ...))
and eval is called with (second (first pattern)), which the value of code. However, eval is called within the progv that binds v1, v2, &c., to the corresponding val1, val2, &c., so that if any of those variables appear free in code, then they are bound when code is evaluated.
Problem
The problem that I see here is that, by the code we can't tell if the value is to be saved as the variable's symbol-value or symbol-function. Thus when you put a + as a value to some corresponding variable, say v, then it'll always be saved as the symbol-value of var, not it's symbol-function.
Therefore when you'll try to use it as, say (v 1 2) , it won't work. Because there is no function named v in the functions' namespace(see this).
So, what to do?
A probable solution can be explicit checking for the value that is to be bound to a variable. If the value is a function, then it should be bound to the variable's function value. This checking can be done via fboundp.
So, we can make a macro functioner and a modified version of match-if. functioner checks if the value is a function, and sets it aptly. match-if does the dynamic local bindings, and allows other code in the scope of the bound variables.
(defmacro functioner (var val)
`(if (and (symbolp ',val)
(fboundp ',val))
(setf (symbol-function ',var) #',val)
(setf ,var ,val)))
(defun match-if (pattern input bindings)
(eval `(and (let ,(mapcar #'(lambda (x) (list (car x))) bindings)
(declare (special ,# (mapcar #'car bindings)))
(loop for i in ',bindings
do (eval `(functioner ,(first i) ,(rest i))))
(eval (second (first ',pattern))))
(pat-match (rest ',pattern) ',input ',bindings))))
I'm a Lisp beginner. I'm trying to memoize a recursive function for calculating the number of terms in a Collatz sequence (for problem 14 in Project Euler). My code as of yet is:
(defun collatz-steps (n)
(if (= 1 n) 0
(if (evenp n)
(1+ (collatz-steps (/ n 2)))
(1+ (collatz-steps (1+ (* 3 n)))))))
(defun p14 ()
(defvar m-collatz-steps (memoize #'collatz-steps))
(let
((maxsteps (funcall m-collatz-steps 2))
(n 2)
(steps))
(loop for i from 1 to 1000000
do
(setq steps (funcall m-collatz-steps i))
(cond
((> steps maxsteps)
(setq maxsteps steps)
(setq n i))
(t ())))
n))
(defun memoize (fn)
(let ((cache (make-hash-table :test #'equal)))
#'(lambda (&rest args)
(multiple-value-bind
(result exists)
(gethash args cache)
(if exists
result
(setf (gethash args cache)
(apply fn args)))))))
The memoize function is the same as the one given in the On Lisp book.
This code doesn't actually give any speedup compared to the non-memoized version. I believe it's due to the recursive calls calling the non-memoized version of the function, which sort of defeats the purpose. In that case, what is the correct way to do the memoization here? Is there any way to have all calls to the original function call the memoized version itself, removing the need for the special m-collatz-steps symbol?
EDIT: Corrected the code to have
(defvar m-collatz-steps (memoize #'collatz-steps))
which is what I had in my code.
Before the edit I had erroneously put:
(defvar collatz-steps (memoize #'collatz-steps))
Seeing that error gave me another idea, and I tried using this last defvar itself and changing the recursive calls to
(1+ (funcall collatz-steps (/ n 2)))
(1+ (funcall collatz-steps (1+ (* 3 n))))
This does seem to perform the memoization (speedup from about 60 seconds to 1.5 seconds), but requires changing the original function. Is there a cleaner solution which doesn't involve changing the original function?
I assume you're using Common-Lisp, which has separate namespaces for variable and function names. In order to memoize the function named by a symbol, you need to change its function binding, through the accessor `fdefinition':
(setf (fdefinition 'collatz-steps) (memoize #'collatz-steps))
(defun p14 ()
(let ((mx 0) (my 0))
(loop for x from 1 to 1000000
for y = (collatz-steps x)
when (< my y) do (setf my y mx x))
mx))
Here is a memoize function that rebinds the symbol function:
(defun memoize-function (function-name)
(setf (symbol-function function-name)
(let ((cache (make-hash-table :test #'equal)))
#'(lambda (&rest args)
(multiple-value-bind
(result exists)
(gethash args cache)
(if exists
result
(setf (gethash args cache)
(apply fn args)))))))
You would then do something like this:
(defun collatz-steps (n)
(if (= 1 n) 0
(if (evenp n)
(1+ (collatz-steps (/ n 2)))
(1+ (collatz-steps (1+ (* 3 n)))))))
(memoize-function 'collatz-steps)
I'll leave it up to you to make an unmemoize-function.
something like this:
(setf collatz-steps (memoize lambda (n)
(if (= 1 n) 0
(if (evenp n)
(1+ (collatz-steps (/ n 2)))
(1+ (collatz-steps (1+ (* 3 n))))))))
IOW: your original (non-memoized) function is anonymous, and you only give a name to the result of memoizing it.
Note a few things:
(defun foo (bar)
... (foo 3) ...)
Above is a function that has a call to itself.
In Common Lisp the file compiler can assume that FOO does not change. It will NOT call an updated FOO later. If you change the function binding of FOO, then the call of the original function will still go to the old function.
So memoizing a self recursive function will NOT work in the general case. Especially not if you are using a good compiler.
You can work around it to go always through the symbol for example: (funcall 'foo 3)
(DEFVAR ...) is a top-level form. Don't use it inside functions. If you have declared a variable, set it with SETQ or SETF later.
For your problem, I'd just use a hash table to store the intermediate results.
Changing the "original" function is necessary, because, as you say, there's no other way for the recursive call(s) to be updated to call the memoized version.
Fortunately, the way lisp works is to find the function by name each time it needs to be called. This means that it is sufficient to replace the function binding with the memoized version of the function, so that recursive calls will automatically look up and reenter through the memoization.
huaiyuan's code shows the key step:
(setf (fdefinition 'collatz-steps) (memoize #'collatz-steps))
This trick also works in Perl. In a language like C, however, a memoized version of a function must be coded separately.
Some lisp implementations provide a system called "advice", which provides a standardized structure for replacing functions with enhanced versions of themselves. In addition to functional upgrades like memoization, this can be extremely useful in debugging by inserting debug prints (or completely stopping and giving a continuable prompt) without modifying the original code.
This function is exactly the one Peter Norvig gives as an example of a function that seems like a good candidate for memoization, but which is not.
See figure 3 (the function 'Hailstone') of his original paper on memoization ("Using Automatic Memoization as a Software Engineering Tool in Real-World AI Systems").
So I'm guessing, even if you get the mechanics of memoization working, it won't really speed it up in this case.
A while ago I wrote a little memoization routine for Scheme that used a chain of closures to keep track of the memoized state:
(define (memoize op)
(letrec ((get (lambda (key) (list #f)))
(set (lambda (key item)
(let ((old-get get))
(set! get (lambda (new-key)
(if (equal? key new-key) (cons #t item)
(old-get new-key))))))))
(lambda args
(let ((ans (get args)))
(if (car ans) (cdr ans)
(let ((new-ans (apply op args)))
(set args new-ans)
new-ans))))))
This needs to be used like so:
(define fib (memoize (lambda (x)
(if (< x 2) x
(+ (fib (- x 1)) (fib (- x 2)))))))
I'm sure that this can be ported to your favorite lexically scoped Lisp flavor with ease.
I'd probably do something like:
(let ((memo (make-hash-table :test #'equal)))
(defun collatz-steps (n)
(or (gethash n memo)
(setf (gethash n memo)
(cond ((= n 1) 0)
((oddp n) (1+ (collatz-steps (+ 1 n n n))))
(t (1+ (collatz-steps (/ n 2)))))))))
It's not Nice and Functional, but, then, it's not much hassle and it does work. Downside is that you don't get a handy unmemoized version to test with and clearing the cache is bordering on "very difficult".