Related
The "finding the digits problem" is this:
Find unique decimal digits A, B, C such that
CCC
+ BBB
+ AAA
= CAAB
To solve it using recursion in Common Lisp, I've written this code:
(defun find! ()
(found? 0 ;; initially point to the number 1
'(1 2 3) ;; initial list
'() ;; initially no numbers found
3 ;; numbers list width is 3
) )
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ( (j 1 (1+ j) ) )
( (> j 9) lst)
(unless (some (lambda (x) (= x j)) occupied)
(setf (nth index lst) j)
(push j occupied)
(if (found? (1+ index) lst occupied width) ;; recursion happens here
lst
(setf occupied (remove j occupied)))))
(do ( (j 1 (1+ j) ) )
( (> j 9) lst)
(unless (some (lambda (x) (= x j)) occupied)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar
(lambda (x y) (* x y))
'(1000 100 10 1)
(list (third lst) (first lst)
(first lst) (second lst))))))
(if (= lefthnd rghthnd)
lst
'nil))))))
The delivered result (lst) is (9 9 9)
The expected result (lst) is (9 8 1) meaning A=9, B=8, C=1 so that the equation CCC + BBB + AAA = CAAB holds i.e.
111 ; CCC
+ 888 ; BBB
+ 999 ; AAA
= 1998 ; CAAB
Which parts of the code should I change so that it gives the expected result? Can someone fix the code? Note that using recursion is a must. Only one line of recursion is enough i.e. like the line where the ;; recursion happens here comment is.
What is the minimal edit to fix this code?
The minimal edit needed to make your code work is the following three small changes (marked with ;;;; NB in the comments):
You are not allowed to surgically modify the structure of a quoted list, as you do. It must be freshly allocated, for that.
(defun find! ()
(found? 0 ;; initially point to the number 1
(list 1 2 3) ;; initial list ;;;; NB freshly allocated!
'() ;; initially no numbers found
3 ;; numbers list width is 3
) )
You must change the structure of the code (moving one closing paren one line up) to always undo the push of j into occupied:
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ( (j 1 (1+ j) ) )
( (> j 9) lst)
(unless (some (lambda (x) (= x j)) occupied)
(setf (nth index lst) j)
(push j occupied)
(if (found? (1+ index) lst occupied width) ;; recursion happens here
lst) ;;;; NB
(setf occupied (remove j occupied)))) ;;;; NB _always_ undo the push
(do ( (j 1 (1+ j) ) )
( (> j 9) lst)
(unless (some (lambda (x) (= x j)) occupied)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar
(lambda (x y) (* x y))
'(1000 100 10 1)
(list (third lst) (first lst)
(first lst) (second lst))))))
(if (= lefthnd rghthnd)
(return-from found? lst) ;;;; NB actually return here
'nil))))))
You also must actually return the result, once it is found (seen in the above snippet as well).
If you change the return-from line to print the result instead of returning it, you will get all of them printed.
If you want to get them all in a list instead of being printed, you can surgically append each of the results to some list defined in some outer scope (or cons onto the front and reverse it when it's all done, if you prefer).
Or in general, you can change this code to accept a callback and call it with each result, when it is found, and let this callback to do whatever it does with it -- print it, append it to an external list, whatever.
Remarks: your code follows a general recursive-backtracking approach, creating three nested loops structure through recursion. The actual result is calculated -- and put into lst by surgical manipulation -- at the deepest level of recursion, corresponding to the innermost loop of j from 1 to 9 (while avoiding the duplicates).
There's lots of inconsequential code here. For instance, the if in (if (found? ...) lst) isn't needed at all and can be just replaced with (found? ...). I would also prefer different names -- occupied should really be named used, lst should be res (for "result"), index is canonically named just i, width is just n, etc. etc. (naming is important)... But you did request the smallest change.
This code calculates the result lst gradually, as a side effect on the way in to the innermost level of the nested loops, where it is finally fully set up.
Thus this code follows e.g. an example of Peter Norvig's PAIP Prolog interpreter, which follows the same paradigm. In pseudocode:
let used = []
for a from 1 to 9:
if a not in used:
used += [a]
for b from 1 to 9:
if b not in used:
used += [b]
for c from 1 to 9:
if c not in used and valid(a,b,c):
return [a,b,c] # or:
# print [a,b,c] # or:
# call(callback,[a,b,c]) # etc.
remove b from used
remove a from used
Here's your code re-structured, renamed, and streamlined:
(defun find2 ( &aux (res (list 0 0 0))
(used '()) (n (length res)))
(labels
((f (i)
(do ((d 1 (1+ d))) ; for d from 1 to 9...
((> d 9) 'FAIL) ; FAIL: no solution!
(unless (member d used) ; "d" for "digit"
(setf (nth i res) d) ; res = [A... B... C...]
(cond
((< i (- n 1)) ; outer levels
(setf used (cons d used))
(f (1+ i)) ; recursion! going in...
(setf used (cdr used))) ; and we're out.
(T ; the innermost level!
(let ((left (* 111 (reduce #'+ res)))
(rght (reduce #'+
(mapcar #'* '(1000 100 10 1)
(list (third res) ; C A A B
(first res)
(first res)
(second res))))))
(if (= left rght)
(return-from find2 res))))))))) ; success!
(f 0)))
This is now closely resembling the C++ code you once had in your question, where the working function (here, f) also received just one argument, indicating the depth level of the nested loop -- corresponding to the index of the digit being tried, -- and the rest of the variables were in an outer scope (there, global; here, the auxiliary variables in the containing function find2).
By the way, you aren't trying any 0s for some reason.
You seem to be able to solve the problem using another language, so I won't spend too long talking about the problem/algorithm used (you already know how to do it). However, as it seems that you are learning Common Lisp, I am going to provide a typical StackOverflow answer, and give a lot of advice that you haven't asked for !
Fix your parentheses/indentation, this will make the code clearer for you.
Split your code in more, smaller functions. You are solving a problem using a recursive function, with several parameters, and the function is more than twenty lines long. This makes it really hard to read and to debug.
Use built-in functions: (some (lambda (x) (= x j)) occupied) == (member j occupied :test #'=), and in that case, it still works without specifying the test (this is technically wrong, the two functions do not return the same thing, but you only ever use the result as a boolean so this is effectively the same thing here).
(mapcar (lambda (x y) (* x y)) ...) is just a longer way to write (mapcar #'* ...)
'nil == nil, you don't need to quote it. It is also (arguably) good style to use () instead of nil to represent the empty list (as opposed to a boolean value), but this really is a minor point.
As far as the algorithm is concerned, I will gladly help if you rewrite it using smaller functions. At the moment, it really is unnecessarily hard to read and understand.
EDIT:
I still tried to take the time to rewrite the code and come up with a cleaner solution.
TL;DR: this is the final result, with "minimal" modifications to your code:
(defun find! ()
(found? 0 (list 1 2 3) () 3))
(defun compute-lefthand (list)
(* 111 (reduce #'+ list)))
(defun compute-righthand (list)
(reduce #'+ (mapcar #'*
'(1000 100 10 1)
(list (third list)
(first list)
(first list)
(second list)))))
(defun check-solution (list)
(when (= (compute-lefthand list)
(compute-righthand list))
list))
(defun try-solution (j index list occupied width)
(unless (member j occupied)
(setf (nth index list) j)
(found? (1+ index)
list
(cons j occupied)
width)))
(defun found? (index lst occupied width)
(if (= index width)
(check-solution lst)
(dotimes (j 10)
(when (try-solution j index lst occupied width)
(return lst)))))
Your initial code, on top of style issues already mentioned in my initial answer, had shaky control flow. It was somewhat hard to determine what was really returned by each recursive call, because you do not have smaller functions and so it was not clear what the goal of each part was, how the information was transmitted from the recursive call to the parent, which objects where modified and so on.
Now, my code is not the cleanest, and I would probably not use this strategy to solve the problem, but I tried to stay as close as possible to your initial code. Main differences:
I split things into smaller functions. This makes everything clearer, and above all, easier to test. Each function returns something clear. For example, check-solution returns the list if it represents a proper solution, and nil otherwise; this is made clear by the fact that I use a when instead of an if control structure.
I replace do by dotimes which is also clearer; the variable that is changing, and how it is changing at each step, is now immediately visible.
I do not use the &optional return argument to the do/dotimes macro, and instead use an explicit return. It is then clear to determine what is being returned, and when.
I do not use push/pop to modify my lists. You are using a recursive strategy, and so your "modifications" should take the form of different arguments passed to functions. Once again, it makes reasoning about the program easier, by knowing exactly what each function does to each argument. An even better solution would also be to remove the call to setf and instead use (cons <smtg> lst) as the argument of the recursive call, but it's fine.
The error in your initial program is probably coming from the fact that your function does not return what you think, because you have several consecutive expressions, each invoked under different circumstances, whose return value is itself wrong because they are not in the right order and modify objects and return them at the wrong time using do's optional return value.
TL;DR: split things up; make each function do a single thing.
Your code
(defun find! ()
(found? 0 ;; initially show the number 1
'(1 2 3) ;; initial list
'() ;; initially no numbers found
3 ;; numbers list width is 3
) )
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ( (j 1 (1+ j) ) )
( (> j 9) lst)
(unless (some (lambda (x) (= x j)) occupied)
(setf (nth index lst) j)
(push j occupied)
(if (found? (1+ index) lst occupied width) ;; recursion
lst
(setf occupied (remove j occupied)))))
(do ( (j 1 (1+ j) ) )
( (> j 9) lst)
(unless (some (lambda (x) (= x j)) occupied)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+ (mapcar (lambda (x y) (* x y))
'(1000 100 10 1)
(list (third lst) (first lst) (first lst) (second lst))
))))
(if (= lefthnd rghthnd)
lst
'nil))))))
Indentation and comment style: end-of-line comments use a single semicolon,
align non-body arguments, indent bodies by two spaces
(defun find! ()
(found? 0 ; initially show the number 1
'(1 2 3) ; initial list
'() ; initially no numbers found
3)) ; numbers list width is 3
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (some (lambda (x) (= x j)) occupied)
(setf (nth index lst) j)
(push j occupied)
(if (found? (1+ index) lst occupied width) ; recursion
lst
(setf occupied (remove j occupied)))))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (some (lambda (x) (= x j)) occupied)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar (lambda (x y) (* x y))
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(if (= lefthnd rghthnd)
lst
'nil))))))
Use more telling predicates: find or member. Don't wrap * in a lambda doing
nothing else. (I'll leave aside find! hereafter.)
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(push j occupied)
(if (found? (1+ index) lst occupied width) ; recursion
lst
(setf occupied (remove j occupied)))))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(if (= lefthnd rghthnd)
lst
'nil))))))
The body of a do doesn't return anything. There is a lot of dead code,
which we remove now:
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(push j occupied)
(unless (found? (1+ index) lst occupied width) ; recursion
(setf occupied (remove j occupied)))))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)))))
Instead of pushing and then conditionally removing, we can conditionally push:
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(when (found? (1+ index) lst occupied width) ; recursion
(push j occupied))))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)))))
While it makes a difference in performance, putting the outer conditional
into the inner body makes it more readable here:
(defun found? (index lst occupied width)
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(when (and (< index (1- width))
(found? (1+ index) lst occupied width)) ; recursion
(push j occupied)))))
This does nothing except count to 9 a few times, which seems to be congruent
to your findings.
I guess that you wanted to return something from the dead code. You might
want to use return-from for that.
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(push j occupied)
(if (found? (1+ index) lst occupied width) ; recursion
(return-from found? lst)
(setf occupied (remove j occupied)))))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(when (= lefthnd rghthnd)
(return-from found? lst)))))))
This returns (1 2 9), which is wrong. The problem seems to be that you
return the list even when you run over 9, but you want to return nil then,
because you didn't find anything.
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) nil) ; <- nothing found
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(push j occupied)
(if (found? (1+ index) lst occupied width) ; recursion
(return-from found? lst)
(setf occupied (remove j occupied)))))
(do ((j 1 (1+ j)))
((> j 9) nil) ; <- nothing found
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(when (= lefthnd rghthnd)
(return-from found? lst)))))))
This returns (9 8 1), which is correct. Now that I seem to understand what
you're trying to do, let's refactor a bit more. Instead of pushing and
removing from the occupied list, just create a new list with the new element
in front transiently:
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) nil)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(when (found? (1+ index) ; recursion
lst
(cons j occupied)
width)
(return-from found? lst))))
(do ((j 1 (1+ j)))
((> j 9) nil)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(when (= lefthnd rghthnd)
(return-from found? lst)))))))
I think that using loop instead of do makes this much more readable:
(defun found? (index lst occupied width)
(if (< index (1- width))
(loop :for j :from 1 :to 9
:unless (find j occupied :test #'=)
:do (setf (nth index lst) j)
(when (found? (1+ index) ; recursion
lst
(cons j occupied)
width)
(return-from found? lst)))
(loop :for j :from 1 :to 9
:unless (find j occupied :test #'=)
:do (setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(when (= lefthnd rghthnd)
(return-from found? lst))))))
Since the loop is rather elaborate, I'd want to write and read it only once,
so move the outer condition inside:
(defun found? (index lst occupied width)
(loop :for j :from 1 :to 9
:unless (find j occupied :test #'=)
:do (setf (nth index lst) j)
(if (< index (1- width))
(when (found? (1+ index) ; recursion
lst
(cons j occupied)
width)
(return-from found? lst))
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(when (= lefthnd rghthnd)
(return-from found? lst))))))
Did you see that occupied is just the first one or two elements of lst,
reversed? Instead of setting list elements, we can build up lst through the
recursion. We actually need to return the recursive results for that, so
this is better referential transparency.
(defun find! ()
(found? 0 ; initially show the number 1
'() ; initially no numbers found
3)) ; numbers list width is 3
(defun found? (index part width)
(loop :for j :from 1 :to 9
:unless (find j part :test #'=)
:do (if (< index (1- width))
(let ((solution (found? (1+ index) ; recursion
(cons j part)
width)))
(when solution
(return-from found? solution)))
(let* ((full (cons j part))
(lefthnd (* 111 (reduce #'+ full)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third full)
(first full)
(first full)
(second full))))))
(when (= lefthnd rghthnd)
(return-from found? full))))))
Index and width are now only used for counting, so we only need one number,
which we can count towards zero. This also makes apparent that we should
probably move the base case out of the looping:
(defun find! ()
(found? '() ; initially no numbers found
3)) ; numbers list width is 3
(defun found? (part count)
(if (zerop count)
(let* ((full part) ; just rename to show that the number is complete
(lefthnd (* 111 (reduce #'+ full)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third full)
(first full)
(first full)
(second full))))))
(when (= lefthnd rghthnd)
(return-from found? full)))
(loop :for j :from 1 :to 9
:unless (find j part :test #'=)
:do (let ((solution (found? (cons j part)
(1- count))))
(when solution
(return-from found? solution))))))
I think this more or less is what you can do if you keep it to a single
function. Now you'd probably want to separate the generation of
permutations from the actual code. There are for example some functions to
deal with such things in the widely used library alexandria.
I am working on a complicated macro and have run into a roadblock.
(defmacro for-each-hashtable-band (body vars on &optional counter name)
`(block o
(with-hash-table-iterator (next-entry ,on)
(destructuring-bind
,(apply #'append vars)
(let ((current-band (list ,#(mapcar #'not (apply #'append vars)))))
(for (i 1 ,(length (apply #'append vars)) 2)
(multiple-value-bind
(succ k v) (next-entry)
(if succ
(progn
(setf (nth i current-band) k)
(setf (nth (+ 1 i) current-band) v))
(return-from o nil))))
current-band)
,#body))))
im getting "Evaluation aborted on #<UNDEFINED-FUNCTION NEXT-ENTRY {100229C693}>"
i dont understand why next-entry appears to be invisible to the macro i have created.
I've tried stripping down this to a small replicable example but i couldnt find a minimal scenario without the macro i created where next-entry would be invisible besides this scenario no matter what I tried, i've always managed to find a way to call next-entry in my other examples so im stumped as to why i cannot get it working here
I've tested the for macro ive created and it seems to generally work in most cases but for some reason it cannot see this next-entry variable. How do i make it visible?
In your code there are multiple places where the macro generates bindings in a way that is subject to variable capture (pdf).
(defmacro for-each-hashtable-band (body vars on &optional counter name)
`(block o ;; VARIABLE CAPTURE
(with-hash-table-iterator (next-entry ,on) ;; VARIABLE CAPTURE
(destructuring-bind ,(apply #'append vars)
(let ((current-band ;;; VARIABLE CAPTURE
(list ,#(mapcar #'not (apply #'append vars)))))
(for
(i ;;; VARIABLE CAPTURE
1 ,(length (apply #'append vars)) 2)
(multiple-value-bind (succ k v) ;;; VARIABLE CAPTURE
,(next-entry) ;;; WRONG EVALUATION TIME
(if succ
(progn
(setf (nth i current-band) k)
(setf (nth (+ 1 i) current-band) v))
(return-from o nil))))
current-band)
,#body))))
A simplified example of such a capture is:
`(let ((x 0)) ,#body)
Here above, the x variable is introduced, but if the code is expanded in a context where xis already bound, then body will not be able to reference that former x binding and will always see x bound to zero (you generally don't want this behavior).
Write a function instead
Instead of writing a big macro for this, let's first try understanding what you want to achieve and write instead a higher-order function, ie. a function that calls user-provided functions.
If I understand correctly, your function iterates over a hash-table by bands of entries. I assume vars holds a list of (key value) pairs of symbols, for example ((k1 v1) (k2 v2)). Then, body works on all the key/value pairs in the band.
In the following code, the function map-each-hashtable-band accepts a function, a hash-table, and instead of vars it accepts a size, the width of the band (the number of pairs).
Notice how in your code, you only have one loop, which builds a band using the hash-table iterator. But then, since the macro is named for-each-hashtable-band, I assume you also want to loop over all the bands. The macro with-hash-table-iterator provides an iterator but does not loop itself. That's why here I have two loops.
(defun map-each-hashtable-band (function hash-table band-size)
(with-hash-table-iterator (next-entry hash-table)
(loop :named outer-loop :do
(loop
:with key and value and next-p
:repeat band-size
:do (multiple-value-setq (next-p key value) (next-entry))
:while next-p
:collect key into current-band
:collect value into current-band
:finally (progn
(when current-band
(apply function current-band))
(unless next-p
(return-from outer-loop)))))))
For example:
(map-each-hashtable-band (lambda (&rest band) (print `(:band ,band)))
(alexandria:plist-hash-table
'(:a 0 :b 1 :c 2 :d 3 :e 4 :f 5 :g 6))
2)
NB. Iterating over a hash-table happens in an arbitrary order, there is no guarantee that you'll see the entries in any particular kind of order, this is implementation-dependant.
With my current version of SBCL this prints the following:
(:BAND (:A 0 :B 1))
(:BAND (:C 2 :D 3))
(:BAND (:E 4 :F 5))
(:BAND (:G 6))
Wrap the function in a macro
The previous function might not be exactly the behavior you want, so you need to adapt to your needs, but once it does what you want, you can wrap a macro around it.
(defmacro for-each-hashtable-band (vars hash-table &body body)
`(map-each-hashtable-band (lambda ,(apply #'append vars) ,#body)
,hash-table
,(length vars)))
For example:
(let ((test (alexandria:plist-hash-table '(:a 0 :b 1 :c 2 :d 3 :e 4 :f 5))))
(for-each-hashtable-band ((k1 v1) (k2 v2)) test
(format t "~a -> ~a && ~a -> ~a ~%" k1 v1 k2 v2)))
This prints:
A -> 0 && B -> 1
C -> 2 && D -> 3
E -> 4 && F -> 5
Macro-only solution, for completeness
If you want to have only one, single macro, you can start by inlining the body of the above function in the macro, you don't need to use apply anymore, but instead you need to establish bindings around the body, using destructuring-bind as you did. A first draft would be to simply as follows, but notice that this is not a proper solution:
(defmacro for-each-hashtable-band (vars hash-table &body body)
(let ((band-size (length vars)))
`(with-hash-table-iterator (next-entry ,hash-table)
(loop :named outer-loop :do
(loop
:with key and value and next-p
:repeat ,band-size
:do (multiple-value-setq (next-p key value) (next-entry))
:while next-p
:collect key into current-band
:collect value into current-band
:finally (progn
(when current-band
(destructuring-bind ,(apply #'append vars) current-band
,#body))
(unless next-p
(return-from outer-loop))))))))
In order to be free of variable capture problems with macros, each temporary variable you introduce must be named after a symbol that cannot exist in any context you expand your code. So instead we first unquote all the variables, making the macro definition fail to compile:
(defmacro for-each-hashtable-band (vars hash-table &body body)
(let ((band-size (length vars)))
`(with-hash-table-iterator (,next-entry ,hash-table)
(loop :named ,outer-loop :do
(loop
:with ,key and ,value and ,next-p
:repeat ,band-size
:do (multiple-value-setq (,next-p ,key ,value) (,next-entry))
:while ,next-p
:collect ,key into ,current-band
:collect ,value into ,current-band
:finally (progn
(when ,current-band
(destructuring-bind ,(apply #'append vars) ,current-band
,#body))
(unless ,next-p
(return-from ,outer-loop))))))))
When compiling the macro, the macro is supposed to inject symbols into the code, but here we have a compilation error that says undefined variables:
;; undefined variables: CURRENT-BAND KEY NEXT-ENTRY NEXT-P OUTER-LOOP VALUE
So now, those variables should be fresh symbols:
(defmacro for-each-hashtable-band (vars hash-table &body body)
(let ((band-size (length vars)))
(let ((current-band (gensym))
(key (gensym))
(next-entry (gensym))
(next-p (gensym))
(outer-loop (gensym))
(value (gensym)))
`(with-hash-table-iterator (,next-entry ,hash-table)
(loop :named ,outer-loop :do
(loop
:with ,key and ,value and ,next-p
:repeat ,band-size
:do (multiple-value-setq (,next-p ,key ,value) (,next-entry))
:while ,next-p
:collect ,key into ,current-band
:collect ,value into ,current-band
:finally (progn
(when ,current-band
(destructuring-bind ,(apply #'append vars) ,current-band
,#body))
(unless ,next-p
(return-from ,outer-loop)))))))))
This above is a bit verbose, but you could simplify that.
Here is what the previous for-each-hashtable-band example expands into with this new macro:
(with-hash-table-iterator (#:g1576 test)
(loop :named #:g1578
:do (loop :with #:g1575
and #:g1579
and #:g1577
:repeat 2
:do (multiple-value-setq (#:g1577 #:g1575 #:g1579) (#:g1576))
:while #:g1577
:collect #:g1575 into #:g1574
:collect #:g1579 into #:g1574
:finally (progn
(when #:g1574
(destructuring-bind
(k1 v1 k2 v2)
#:g1574
(format t "~a -> ~a && ~a -> ~a ~%" k1 v1 k2
v2)))
(unless #:g1577 (return-from #:g1578))))))
Each time you expand it, the #:gXXXX variables are different, and cannot possibly shadow existing bindings, so for example, the body can use variables named like current-band or value without breaking the expanded code.
I was trying to make a recursive function to split a list into two lists according the number of elements one wants.
Ex:
(split 3 '(1 3 5 7 9)) ((1 3 5) (7 9))
(split 7 '(1 3 5 7 9)) ((1 3 5 7 9) NIL)
(split 0 '(1 3 5 7 9)) (NIL (1 3 5 7 9))
My code is like this:
(defun split (e L)
(cond ((eql e 0) '(() L))
((> e 0) (cons (car L) (car (split (- e 1) (cdr L))))))))
I don't find a way to join the first list elements and return the second list.
Tail recursive solution
(defun split (n l &optional (acc-l '()))
(cond ((null l) (list (reverse acc-l) ()))
((>= 0 n) (list (reverse acc-l) l))
(t (split (1- n) (cdr l) (cons (car l) acc-l)))))
Improved version
(in this version, it is ensured that acc-l is at the beginning '()):
(defun split (n l)
(labels ((inner-split (n l &optional (acc-l '()))
(cond ((null l) (list (reverse acc-l) ()))
((= 0 n) (list (reverse acc-l) l))
(t (inner-split (1- n) (cdr l) (cons (car l) acc-l))))))
(inner-split n l)))
Test it:
(split 3 '(1 2 3 4 5 6 7))
;; returns: ((1 2 3) (4 5 6 7))
(split 0 '(1 2 3 4 5 6 7))
;; returns: (NIL (1 2 3 4 5 6 7))
(split 7 '(1 2 3 4 5 6 7))
;; returns ((1 2 3 4 5 6 7) NIL)
(split 9 '(1 2 3 4 5 6 7))
;; returns ((1 2 3 4 5 6 7) NIL)
(split -3 '(1 2 3 4 5 6 7))
;; returns (NIL (1 2 3 4 5 6 7))
In the improved version, the recursive function is placed one level deeper (kind of encapsulation) by using labels (kind of let which allows definition of local functions but in a way that they are allowed to call themselves - so it allows recursive local functions).
How I came to the solution:
Somehow it is clear, that the first list in the result must result from consing one element after another from the beginning of l in successive order. However, consing adds an element to an existing list at its beginning and not its end.
So, successively consing the car of the list will lead to a reversed order.
Thus, it is clear that in the last step, when the first list is returned, it hast to be reversed. The second list is simply (cdr l) of the last step so can be added to the result in the last step, when the result is returned.
So I thought, it is good to accumulate the first list into (acc-l) - the accumulator is mostly the last element in the argument list of tail-recursive functions, the components of the first list. I called it acc-l - accumulator-list.
When writing a recursive function, one begins the cond part with the trivial cases. If the inputs are a number and a list, the most trivial cases - and the last steps of the recursion, are the cases, when
the list is empty (equal l '()) ---> (null l)
and the number is zero ----> (= n 0) - actually (zerop n). But later I changed it to (>= n 0) to catch also the cases that a negative number is given as input.
(Thus very often recursive cond parts have null or zerop in their conditions.)
When the list l is empty, then the two lists have to be returned - while the second list is an empty list and the first list is - unintuitively - the reversed acc-l.
You have to build them with (list ) since the list arguments get evaluated shortly before return (in contrast to quote = '(...) where the result cannot be evaluated to sth in the last step.)
When n is zero (and later: when n is negative) then nothing is to do than to return l as the second list and what have been accumulated for the first list until now - but in reverse order.
In all other cases (t ...), the car of the list l is consed to the list which was accumulated until now (for the first list): (cons (car l) acc-l) and this I give as the accumulator list (acc-l) to split and the rest of the list as the new list in this call (cdr l) and (1- n). This decrementation in the recursive call is very typical for recursive function definitions.
By that, we have covered all possibilities for one step in the recursion.
And that makes recursion so powerful: conquer all possibilities in ONE step - and then you have defined how to handle nearly infinitely many cases.
Non-tail-recursive solution
(inspired by Dan Robertson's solution - Thank you Dan! Especially his solution with destructuring-bind I liked.)
(defun split (n l)
(cond ((null l) (list '() '()))
((>= 0 n) (list '() l))
(t (destructuring-bind (left right) (split (1- n) (cdr l))
(list (cons (car l) left) right)))))
And a solution with only very elementary functions (only null, list, >=, let, t, cons, car, cdr, cadr)
(defun split (n l)
(cond ((null l) (list '() '()))
((>= 0 n) (list '() l))
(t (let ((res (split (1- n) (cdr l))))
(let ((left-list (car res))
(right-list (cadr res)))
(list (cons (car l) left-list) right-list))))))
Remember: split returns a list of two lists.
(defun split (e L)
(cond ((eql e 0)
'(() L)) ; you want to call the function LIST
; so that the value of L is in the list,
; and not the symbol L itself
((> e 0)
; now you want to return a list of two lists.
; thus it probably is a good idea to call the function LIST
; the first sublist is made of the first element of L
; and the first sublist of the result of SPLIT
; the second sublist is made of the second sublist
; of the result of SPLIT
(cons (car L)
(car (split (- e 1)
(cdr L)))))))
Well let’s try to derive the recursion we should be doing.
(split 0 l) = (list () l)
So that’s our base case. Now we know
(split 1 (cons a b)) = (list (list a) b)
But we think a bit and we’re building up the first argument on the left and the way to build up lists that way is with CONS so we write down
(split 1 (cons a b)) = (list (cons a ()) b)
And then we think a bit and we think about what (split 0 l) is and we can write down for n>=1:
(split n+1 (cons a b)) = (list (cons a l1) l2) where (split n b) = (list l1 l2)
So let’s write that down in Lisp:
(defun split (n list)
(ecase (signum n)
(0 (list nil list))
(1 (if (cdr list)
(destructuring-bind (left right) (split (1- n) (cdr list))
(list (cons (car list) left) right))
(list nil nil)))))
The most idiomatic solution would be something like:
(defun split (n list)
(etypecase n
((eql 0) (list nil list))
(unsigned-integer
(loop repeat n for (x . r) on list
collect x into left
finally (return (list left r))))))
my task is to count all element within a list, which have duplicates, eg
( 2 2 (3 3) 4 (3)) will result in 2 (because only 2 and 3 have duplicates)
Searchdeep - just returns a nill if WHAT isn't find in list WHERE
Count2 - go through the single elements and sub-lists
If it finds atom he will use SEARCHDEEP to figure out does it have duplicates, then list OUT will be checked (to make sure if this atom was not already counted (e.g. like ( 3 3 3), which should return 1, not 2)
, increase counter and add atom to the OUT list.
However, i don't understand why, but it constantly returns only 1. I think it is some kind of logical mistake or wrong use of function.
My code is:
(SETQ OUT NIL)
(SETQ X (LIST 2 -3 (LIST 4 3 0 2) (LIST 4 -4) (LIST 2 (LIST 2 0 2))-5))
(SETQ count 0)
(DEFUN SEARCHDEEP (WHAT WHERE) (COND
((NULL WHERE) NIL)
(T (OR
(COND
((ATOM (CAR WHERE)) (EQUAL WHAT (CAR WHERE)))
(T (SEARCHDEEP WHAT (CAR WHERE)))
)
(SEARCHDEEP WHAT (CDR WHERE))
)
)
)
)
(DEFUN Count2 ( input)
(print input)
(COND
((NULL input) NIL)
(T
(or
(COND
((ATOM (CAR input))
(COND
(
(and ;if
(SEARCHDEEP (CAR INPUT) (CDR INPUT))
(NOT (SEARCHDEEP (CAR INPUT) OUT))
)
(and ;do
(Setq Count (+ count 1))
(SETQ OUT (append OUT (LIST (CAR INPUT))))
(Count2 (CDR input))
)
)
(t (Count2 (CDR input)))
)
)
(T (Count2 (CAR input)))
)
(Count2 (CDR input))
)
)
)
)
(Count2 x)
(print count)
First, your code has some big style issues. Don't write in uppercase (some, like myself, like to write symbols in uppercase in comments and in text outside of code, but the code itself should be written in lowercase), and don't put parentheses on their own lines. So the SEARCHDEEP function should look more like
(defun search-deep (what where)
(cond ((null where) nil)
(t (or (cond ((atom (car where)) (equal what (car where)))
(t (searchdeep what (car where))))
(searchdeep what (cdr where))))))
You also should not use SETQ to define variables. Use DEFPARAMETER or DEFVAR instead, although in this case you should not use global variables in the first place. You should name global variables with asterisks around the name (*X* instead of x, but use a more descriptive name).
For the problem itself, I would start by writing a function to traverse a tree.
(defun traverse-tree (function tree)
"Traverse TREE, calling FUNCTION on every atom."
(typecase tree
(atom (funcall function tree))
(list (dolist (item tree)
(traverse-tree function item))))
(values))
Notice that TYPECASE is more readable than COND in this case. You should also use the mapping or looping constructs provided by the language instead of writing recursive loops yourself. The (values) at the end says that the function will not return anything.
(let ((tree '(2 -3 (4 3 0 2) (4 -4) (2 (2 0 2)) -5)))
(traverse-tree (lambda (item)
(format t "~a " item))
tree))
; 2 -3 4 3 0 2 4 -4 2 2 0 2 -5
; No values
If you were traversing trees a lot, you could hide that function behind a DO-TREE macro
(defmacro do-tree ((var tree &optional result) &body body)
`(progn (traverse-tree (lambda (,var)
,#body)
,tree)
,result))
(let ((tree '(2 -3 (4 3 0 2) (4 -4) (2 (2 0 2)) -5)))
(do-tree (item tree)
(format t "~a " item)))
; 2 -3 4 3 0 2 4 -4 2 2 0 2 -5
;=> NIL
Using this, we can write a function that counts every element in the tree, returning an alist. I'll use a hash table to keep track of the counts. If you're only interested in counting numbers that will stay in a small range, you might want to use a vector instead.
(defun tree-count-elements (tree &key (test 'eql))
"Count each item in TREE. Returns an alist in
form ((item1 . count1) ... (itemn . countn))"
(let ((table (make-hash-table :test test)))
(do-tree (item tree)
(incf (gethash item table 0)))
(loop for value being the hash-value in table using (hash-key key)
collect (cons key value))))
(let ((tree '(2 -3 (4 3 0 2) (4 -4) (2 (2 0 2)) -5)))
(tree-count-elements tree))
;=> ((2 . 5) (-3 . 1) (4 . 2) (3 . 1) (0 . 2) (-4 . 1) (-5 . 1))
The function takes a keyword argument for the TEST to use with the hash table. For numbers or characters, EQL works.
Now you can use the standard COUNT-IF-function to count the elements that occur more than once.
(let ((tree '(2 -3 (4 3 0 2) (4 -4) (2 (2 0 2)) -5)))
(count-if (lambda (item)
(> item 1))
(tree-count-elements tree)
:key #'cdr))
;=> 3
Let's say I have the floating point number 1234.9
I want to format it as 1.234,90
Is there a format directive combination for that? ~D ,which can handle the grouping and the group char, handles only integers. ~F doesn't handle grouping at all. And none as far as I know can change the decimal point from . to ,
The only solution I see is to use ~D for the integer part digit grouping and concatenate it with , and the decimal part. Any better ideas?
You can define a function to be called with tilde-slash, which most of the other answers have already done, but in order to get output similar to ~F, but with comma chars injected, and with the decimal point replaced, I think it's best to call get the output produced by ~F, and then modify it and write it to the string. Here's a way to do that, using a utility inject-comma that adds a comma character at specified intervals to a string. Here's the directive function:
(defun print-float (stream arg colonp atp
&optional
(point-char #\.)
(comma-char #\,)
(comma-interval 3))
"A function for printing floating point numbers, with an interface
suitable for use with the tilde-slash FORMAT directive. The full form
is
~point-char,comma-char,comma-interval/print-float/
The point-char is used in place of the decimal point, and defaults to
#\\. If : is specified, then the whole part of the number will be
grouped in the same manner as ~D, using COMMA-CHAR and COMMA-INTERVAL.
If # is specified, then the sign is always printed."
(let* ((sign (if (minusp arg) "-" (if (and atp (plusp arg)) "+" "")))
(output (format nil "~F" arg))
(point (position #\. output :test 'char=))
(whole (subseq output (if (minusp arg) 1 0) point))
(fractional (subseq output (1+ point))))
(when colonp
(setf whole (inject-comma whole comma-char comma-interval)))
(format stream "~A~A~C~A"
sign whole point-char fractional)))
Here are some examples:
(progn
;; with # (for sign) and : (for grouping)
(format t "~','.2#:/print-float/ ~%" 12345.6789) ;=> +1.23.45,679
;; with no # (no sign) and : (for grouping)
(format t "~'.'_3:/print-float/ ~%" 12345.678) ;=> 12_345.678
;; no # (but sign, since negative) and : (for grouping)
(format t "~'.'_3:/print-float/ ~%" -12345.678) ;=> -12_345.678
;; no # (no sign) and no : (no grouping)
(format t "~'.'_3#/print-float/ ~%" 12345.678)) ;=> +12345.678 (no :)
Here are the examples from coredump-'s answer, which actually helped me catch a bug with negative numbers:
CL-USER> (loop for i in '(1034.34 -223.12 -10.0 10.0 14 324 1020231)
do (format t "~','.:/print-float/~%" i))
1.034,34
-223,12
-10,0
10,0
14,0
324,0
1.020.231,0
NIL
Here's inject-comma, with some examples:
(defun inject-comma (string comma-char comma-interval)
(let* ((len (length string))
(offset (mod len comma-interval)))
(with-output-to-string (out)
(write-string string out :start 0 :end offset)
(do ((i offset (+ i comma-interval)))
((>= i len))
(unless (zerop i)
(write-char comma-char out))
(write-string string out :start i :end (+ i comma-interval))))))
(inject-comma "1234567" #\, 3)
;;=> "1,234,567"
(inject-comma "1234567" #\. 2)
;;=> "1.23.45.67"
As the comment of jkiiski suggests, you could use the ~/func/ directive.
This is just an example, you can elaborate more with the function:
CL-USER> (defun q(stream arg &rest args)
(declare (ignore args))
(format stream
"~,,'.,:D,~a"
(truncate arg)
(let ((float-string (format nil "~f" arg)))
(subseq float-string (1+ (position #\. float-string))))))
Q
CL-USER> (format t "~/q/~%" 1024.36)
1.024,36
NIL
CL-USER> (format t "~/q/~%" -1024.36)
-1.024,36
NIL
Edited
The first version had round, which is wrong, truncate is the right operator to use.
If you don't mind splitting integer and fractional part, you can do the following:
(multiple-value-bind (int rest) (floor 1234.56)
(let ((rest (round (* rest 1000))))
(format t "~,,'.,:D,~D~%" int rest)))
1.234,560
The multiplication before rounding tells how many digits after comma you would like to print. Not sure if this approach lands itself nicely into automatic control of precision printing, i.e. 1.5 printed as "1,5" and not as "1,500".
Other answers currently use round, which is probably not the intended behavior when rounding up (positive numbers) or down (negative numbers). Here is another approach for a ~/custom/ directive, derived mostly from Renzo's answer.
(defun custom (stream number &rest args)
(declare (ignore args))
(multiple-value-bind (integer decimal) (truncate number)
(format stream "~,,'.,:D~#[,~a~]"
integer
(unless (zerop decimal)
(let ((decimal-string (princ-to-string (abs decimal))))
(subseq decimal-string (1+ (position #\. decimal-string))))))))
TESTS
(loop for i in '(1034.34 -223.12 -10.0 10.0 14 324 1020231)
collect (custom nil i))
=> ("1.034,33996582" "-223,11999512" "-10" "10" "14" "324" "1.020.231")
I've come to this little solution for positive numbers.
(defun comma-point (stream arg &rest args)
(declare (ignore args))
(multiple-value-bind (i r) (truncate arg)
(format stream "~,,',,:D.~2,'0D" i (truncate (* 100 r)))))
;; ^ ^
;; | `Decimal point
;; `Thousands separator
(defun point-comma (stream arg &rest args)
(declare (ignore args))
(multiple-value-bind (i r) (truncate arg)
(format stream "~,,'.,:D,~2,'0D" i (truncate (* 100 r)))))
(defun space-comma (stream arg &rest args)
(declare (ignore args))
(multiple-value-bind (i r) (truncate arg)
(format stream "~,,' ,:D,~2,'0D" i (truncate (* 100 r)))))
The testing numbers:
(dolist (value '(1034.34 -223.12 -10.0 10.0 14 324 1020231.099))
(format t "~16#A" (format nil "~/comma-point/" value))
(format t "~16#A" (format nil "~/point-comma/" value))
(format t "~16#A~%" (format nil "~/space-comma/" value)))
;; 1,034.33 1.034,33 1 034,33
;; -223.-11 -223,-11 -223,-11
;; -10.00 -10,00 -10,00
;; 10.00 10,00 10,00
;; 14.00 14,00 14,00
;; 324.00 324,00 324,00
;; 1,020,231.12 1.020.231,12 1 020 231,12
The second test number shows that does not work for negative numbers (-223.11 => -223,-11). Also, using truncate (or other similar functions) implies that a loss of accuracy appears, as can be seen in the last test number (1020231.099 => 1.020.231,12).