I've completed the Graham's exercise Chapter 5.8,and my code is:
(defun max-min (vec &key (start 0) (end (length vec)))
(cond
((eql start (1- end)) (values (elt vec start) (elt vec (1- end))))
((zerop end) (values nil nil))
(t
(multiple-value-bind (x y) (max-min vec :start (1+ start) :end end)
(let* ((maxx x)(minn y))
(values (max maxx (elt vec start)) (min minn (elt vec start))))))))
You don't need to worry about the details, basically it just returns the max and min of a given vector in a "value" form.
I use the above recursion to solve the problem, but my teachers marked my function as "almost done" with such critique:
"If a function takes start and end, then length is neither needed nor correct. Length could be > 0 but it's whether start < end or not that matters. Testing end all by itself is not relevant at all."
I am not very clear at this point, I tried getting rid of the (length vec) default value for "end", but then the default value for end becomes nil.
We have clear instrution that "length" should at most be called once.
Could you please give me some hint on this? Thanks.
Your lambda list is OK. The problem is the base case: (zerop end) should be modified so that you also get a sensible result if called like (min-max myvec :start 5 :end 3).
The next critique is about these two lines:
(multiple-value-bind (x y) (max-min vec :start (1+ start) :end end)
(let* ((maxx x) (minn y))
;; ...
If you want the results of the recursive call to be named maxx and minn, why don't you name them like that directly?
(multiple-value-bind (maxx minn) (max-min vec :start (1+ start) :end end)
;; ...
By the way, you can call them max and min (there are separate namespaces for variables and functions), or max-of-rest and min-of-rest (to be more descriptive).
I don't know some of the lisp features you're using, but it looks to me very much like your teacher is incorrect.
"If a function takes start and end, then length is neither needed nor correct. Length could be 0 but it's whether start < end or not that matters. Testing end all by itself is not relevant at all."
It looks to me like length is the default value for end, and that you are only testing end, not testing length at all... His complaint already appears to be addressed.
That said, there are a other things to complain about in that code. For example, what happens if you call max-min with end < start? It looks to me like you'll recurse by incrementing start, until you run out of vector, then you'll get some kind of exception.
Related
CL-USER> (a-sum 0 3)
->> 6
I wrote this program :
(defun a-sum (x y)
(if (and (> x -1) (> y -1))
(do ((i 0 (1+ i))
(sum 0)
(num x))
((equal i (+ (- y x) 1)))
(setq sum (+ sum num))
(setq num (+ num 1))
sum)
(print " NOPE")))
put if I run it in the terminal it returns nil and not the answer stated above;
can someone help with the problem so it returns the value then Boolean.
DO,DO* Syntax
The entry for DO,DO* says that the syntax is as follows:
do ({var | (var [init-form [step-form]])}*)
(end-test-form result-form*)
declaration*
{tag | statement}*
The body is used as a list of statements and no intermediate value in this body is used as the result form of the do form. Instead, the do form evaluates as the last expression in result-form*, which defaults to nil.
(do ((i 0 (1+ i))
(sum 0)
(num x))
((equal i (+ (- y x) 1))
;;; RESULT FORMS HERE
)
(setq sum (+ sum num)) ;; (*)
(setq num (+ num 1)) ;; (*)
sum ;; (*)
)
All the expressions marked commented (*) above are used for side-effects only: the result of their evaluation is unused and discarded.
Problem statement
It is not clear to me what Σpi=ni means, and your code does not seem to compute something that could be expressed as that mathematical expression.
One red flag for example is that if (+ (- y x) 1) is negative (i.e. if y < x-1, for example y=1,x=3), then your loop never terminates because i, which is positive or null, will never be equal to the other term which is negative.
I would try to rewrite the problem statement more clearly, and maybe try first a recursive version of your algorithm (whichever is easier to express).
Remarks
Please indent/format your code.
Instead of adding setq statements in the body, try to see if you can define them in the iteration clauses of the loop (since I'm not sure what you are trying to achieve, the following example is only a rewrite of your code):
(do ((i 0 (1+ i))
(sum 0 (+ sum num)
(num x (1+ num))
(... sum))
Consider what value(s) a function returns. It's the value of the last form evaluated. In your case, that appears to be a do or maybe a setq or print (It's difficult to read as it's formatted now, and I don't have question edit privileges).
In short, the form that's returning the value for the function looks to be one evaluated for side-effects instead of returning a value.
I am bad at Lisp. Help me please to find a syntax error. I need to write a function which swaps two elements in list. This function must consist loop-cycle. Here is what if have so far.
(defun swap-two-element(z x y)
(let ((newlist nil) (newlist2 nil) (copyz z) (copyz2 z) (newx nil))
(loop
(when (= (- (length z) (length copyz2)) y)
(return (set newx car z)))
(setq newlist2 (append newlist2(car copyz2))
copyz2 (cdr copyz2)))))
Call example: (swap-two-element '(a b c d) 2 3)
Replace the word set with the word values and you are good to go.
PS. You need to address the warnings though, and explain what the function is supposed to do so that we could help you with the algorithm.
You really need to tidy up your question. The title says nothing, the code is badly formatted and you really need to play around with loop to get started. I won't give you your solution since you need to learn this by trying. Here is an example you can make use of to do your assignment.
;; this orders a list by their odd index
;; NB: index starts at zero so first element is even
(defun order-odd-index (list)
(loop :for element :in list ; iterates your list
:for index :from 0 ; starts from 0 and goes on
:if (oddp index) ; if in a loop
:collect element :into odd-list ; variable is automatically created
:else ; else in a loop
:collect element :into even-list
:finally (return (append odd-list even-list)))) ; what to do in the end
(order-odd-index '(4 67 3 2 7 9)) ; ==> (67 2 9 4 3 7)
I use keywords (like :for instead of for) to indicate what symbols are loop keywords and what are not. It's optional but I think it looks a lot cleaner.
Now your problem can be solved with collecting element into 5 variables. Two of them is when index is equal to one of the places (given as arguments) to be switched the other 3 are before, in between and greater. In the finally you can just append those variables in the correct order and you're done.
For Project Euler Problem 8, I am told to parse through a 1000 digit number.
This is a brute-force Lisp solution, which basically goes through every 5 consecutive digits and multiplies them from start to finish, and returns the largest one at the end of the loop.
The code:
(defun pep8 ()
(labels ((product-of-5n (n)
(eval (append '(*)
(loop for x from n to (+ n 5)
collect (parse-integer
1000digits-str :start x :end (+ x 1)))))))
(let ((largestproduct 0))
(do ((currentdigit 0 (1+ currentdigit)))
((> currentdigit (- (length 1000digits-str) 6)) (return largestproduct))
(when (> (product-of-5n currentdigit) largestproduct)
(setf largestproduct (product-of-5n currentdigit)))))))
It compiles without any warnings, but upon running it I get:
no non-whitespace characters in string "73167176531330624919225119674426574742355349194934...".
[Condition of type SB-INT:SIMPLE-PARSE-ERROR]
I checked to see if the local function product-of-5n was working by writing it again as a global function:
(defun product-of-5n (n)
(eval (append '(*)
(loop for x from n to (+ n 5)
collect (parse-integer
1000digits-str :start x :end (+ x 1))))))
This compiled without warnings and upon running it, appears to operate perfectly. For example,
CL_USER> (product-of-5n 1) => 882
Which appears to be correct since the first five digits are 7, 3, 1, 6 and 7.
As for 1000digits-str, it was simply compiled with defvar, and with Emacs' longlines-show-hard-newlines, I don't think there are any white-space characters in the string, because that's what SBCL is complaining about, right?
I don't think there are any white-space characters in the string, because that's what SBCL is complaining about, right?
The error-message isn't complaining about the presence of white-space, but about the absence of non-white-space. But it's actually a bit misleading: what the message should say is that there's no non-white-space in the specific substring to be parsed. This is because you ran off the end of the string, so were parsing a zero-length substring.
Also, product-of-5n is not defined quite right. It's just happenstance that (product-of-5n 1) returns the product of the first five digits. Strings are indexed from 0, so (product-of-5n 1) starts with the second character; and the function iterates from n + 0 to n + 5, which is a total of six characters; so (product-of-5n 1) returns 3 × 1 × 6 × 7 × 1 × 7, which happens to be the same as 7 × 3 × 1 × 6 × 7 × 1.
EVAL is not a good idea.
Your loop upper bound is wrong.
Otherwise I tried it with the number string and it works.
It's also Euler 8, not 9.
This is my version:
(defun euler8 (string)
(loop for (a b c d e) on (map 'list #'digit-char-p string)
while e maximize (* a b c d e)))
since I don't know common lisp, I slightly modified your code to fit with elisp. As far as finding bugs go and besides what have been said ((product-of-5n 1) should return 126), the only comment I have is that in (pep8), do length-4 instead of -6 (otherwise you loose last 2 characters). Sorry that I don't know how to fix your parse-error (I used string-to-number instead), but here is the code in case you find it useful:
(defun product-of-5n (n) ;take 5 characters from a string "1000digits-str" starting with nth one and output their product
(let (ox) ;define ox as a local variable
(eval ;evaluate
(append '(*) ;concatenate the multiplication sign to the list of 5 numbers (that are added next)
(dotimes (x 5 ox) ;x goes from 0 to 4 (n is added later to make it go n to n+4), the output is stored in ox
(setq ox (cons ;create a list of 5 numbers and store it in ox
(string-to-number
(substring 1000digits-str (+ x n) (+ (+ x n) 1) ) ;get the (n+x)th character
) ;end convert char to number
ox ) ;end cons
) ;end setq
) ;end dotimes, returns ox outside of do, ox has the list of 5 numbers in it
) ;end append
) ;end eval
) ;end let
)
(defun pep8 () ;print the highest
(let ((currentdigit 0) (largestproduct 0)) ;initialize local variables
(while (< currentdigit (- (length 1000digits-str) 4) ) ;while currentdigit (cd from now on) is less than l(str)-4
;(print (cons "current digit" currentdigit)) ;uncomment to print cd
(when (> (product-of-5n currentdigit) largestproduct) ;when current product is greater than previous largestproduct (lp)
(setq largestproduct (product-of-5n currentdigit)) ;save lp
(print (cons "next good cd" currentdigit)) ;print cd
(print (cons "with corresponding lp" largestproduct)) ;print lp
) ;end when
(setq currentdigit (1+ currentdigit)) ;increment cd
) ;end while
(print (cons "best ever lp" largestproduct) ) ;print best ever lp
) ;end let
)
(setq 1000digits-str "73167176531330624919")
(product-of-5n 1)
(pep9)
which returns (when ran on the first 20 characters)
"73167176531330624919"
126
("next good cd" . 0)
("with corresponding lp" . 882)
("next good cd" . 3)
("with corresponding lp" . 1764)
("best ever lp" . 1764)
I've done this problem some time ago, and there's one thing you are missing in the description of the problem. You need to read consequent as starting at any offset into a sting, not only the offsets divisible by 5. Therefore the solution to the problem will be more like the following:
(defun pe-8 ()
(do ((input (remove #\Newline
"73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450"))
(tries 0 (1+ tries))
(result 0))
((= tries 5) result)
(setq result
(max result
(do ((max 0)
(i 0 (+ 5 i)))
((= i (length input)) max)
(setq max
(do ((j i (1+ j))
(current 1)
int-char)
((= j (+ 5 i)) (max current max))
(setq int-char (- (char-code (aref input j)) 48))
(case int-char
(0 (return max))
(1)
(t (setq current (* current int-char))))))))
input (concatenate 'string (subseq input 1) (subseq input 0 1)))))
It's a tad ugly, but it illustrates the idea.
EDIT sorry, I've confused two of your functions. So that like was incorrect.
I've done the Graham Common Lisp Chapter 5 Exercise 5, which requires a function that takes an object X and a vector V, and returns a list of all the objects that immediately precede X in V.
It works like:
> (preceders #\a "abracadabra")
(#\c #\d #r)
I have done the recursive version:
(defun preceders (obj vec &optional (result nil) &key (startt 0))
(let ((l (length vec)))
(cond ((null (position obj vec :start startt :end l)) result)
((= (position obj vec :start startt :end l) 0)
(preceders obj vec result
:startt (1+ (position obj vec :start startt :end l))))
((> (position obj vec :start startt :end l) 0)
(cons (elt vec (1- (position obj vec :start startt :end l)))
(preceders obj vec result
:startt (1+ (position obj vec
:start startt
:end l))))))))
It works correctly, but my teachers gives me the following critique:
"This calls length repeatedly. Not so bad with vectors, but still unnecessary. More efficient and more flexible (for the user) code is to define this like other sequence processing functions. Use :start and :end keyword parameters, the way the other sequence functions do, with the same default initial values. length should need to be called at most once."
I am consulting the Common Lisp textbook and google, but there seem to be of little help on this bit: I don't know what he means by "using :start and :end keyword parameters", and I have no clue of how to "call length just once". I would be grateful if you guys could give me some idea how on to refine my code to meet the requirement that my teacher posted.
UPDATE:
Now I have come up with the following code:
(defun preceders (obj vec
&optional (result nil)
&key (start 0) (end (length vec)) (test #'eql))
(let ((pos (position obj vec :start start :end end :test test)))
(cond ((null pos) result)
((zerop pos) (preceders obj vec result
:start (1+ pos) :end end :test test))
(t (preceders obj vec (cons (elt vec (1- pos)) result)
:start (1+ pos) :end end :test test)))))
I get this critique:
"When you have a complex recursive call that is repeated identically in more than one branch, it's often simpler to do the call first, save it in a local variable, and then use the variable in a much simpler IF or COND."
Also,for my iterative version of the function:
(defun preceders (obj vec)
(do ((i 0 (1+ i))
(r nil (if (and (eql (aref vec i) obj)
(> i 0))
(cons (aref vec (1- i)) r)
r)))
((eql i (length vec)) (reverse r))))
I get the critique
"Start the DO at a better point and remove the repeated > 0 test"
a typical parameter list for such a function would be:
(defun preceders (item vector
&key (start 0) (end (length vector))
(test #'eql))
...
)
As you can see it has START and END parameters.
TEST is the default comparision function. Use (funcall test item (aref vector i)).
Often there is also a KEY parameter...
LENGTH is called repeatedly for every recursive call of PRECEDERS.
I would do the non-recursive version and move two indexes over the vector: one for the first item and one for the next item. Whenever the next item is EQL to the item you are looking for, then push the first item on to a result list (if it is not member there).
For the recursive version, I would write a second function that gets called by PRECEDERS, which takes two index variables starting with 0 and 1, and use that. I would not call POSITION. Usually this function is a local function via LABELS inside PRECEDERS, but to make it a bit easier to write, the helper function can be outside, too.
(defun preceders (item vector
&key (start 0) (end (length vector))
(test #'eql))
(preceders-aux item vector start end test start (1+ start) nil))
(defun preceders-aux (item vector start end test pos0 pos1 result)
(if (>= pos1 end)
result
...
))
Does that help?
Here is the iterative version using LOOP:
(defun preceders (item vector
&key (start 0) (end (length vector))
(test #'eql))
(let ((result nil))
(loop for i from (1+ start) below end
when (funcall test item (aref vector i))
do (pushnew (aref vector (1- i)) result))
(nreverse result)))
Since you already have a solution that's working, I'll amplifiy Rainer Joswig's solution, mainly to make related stylistic comments.
(defun preceders (obj seq &key (start 0) (end (length seq)) (test #'eql))
(%preceders obj seq nil start end test))
The main reason to have separate helper function (which I call %PRECEDERS, a common convention for indicating that a function is "private") is to eliminate the optional argument for the result. Using optional arguments that way in general is fine, but optional and keyword arguments play horribly together, and having both in a single function is a extremely efficient way to create all sorts of hard to debug errors.
It's a matter of taste whether to make the helper function global (using DEFUN) or local (using LABELS). I prefer making it global since it means less indentation and easier interactive debugging. YMMV.
A possible implementation of the helper function is:
(defun %preceders (obj seq result start end test)
(let ((pos (position obj seq :start start :end end :test test)))
;; Use a local binding for POS, to make it clear that you want the
;; same thing every time, and to cache the result of a potentially
;; expensive operation.
(cond ((null pos) (delete-duplicates (nreverse result) :test test))
((zerop pos) (%preceders obj seq result (1+ pos) end test))
;; I like ZEROP better than (= 0 ...). YMMV.
(t (%preceders obj seq
(cons (elt seq (1- pos)) result)
;; The other little bit of work to make things
;; tail-recursive.
(1+ pos) end test)))))
Also, after all that, I think I should point out that I also agree with Rainer's advice to do this with an explicit loop instead of recursion, provided that doing it recursively isn't part of the exercise.
EDIT: I switched to the more common "%" convention for the helper function. Usually whatever convention you use just augments the fact that you only explicitly export the functions that make up your public interface, but some standard functions and macros use a trailing "*" to indicate variant functionality.
I changed things to delete duplicated preceders using the standard DELETE-DUPLICATES function. This has the potential to be much (i.e., exponentially) faster than repeated uses of ADJOIN or PUSHNEW, since it can use a hashed set representation internally, at least for common test functions like EQ, EQL and EQUAL.
A slightly modofied variant of Rainer's loop version:
(defun preceders (item vector
&key (start 0) (end (length vector))
(test #'eql))
(delete-duplicates
(loop
for index from (1+ start) below end
for element = (aref vector index)
and previous-element = (aref vector (1- index)) then element
when (funcall test item element)
collect previous-element)))
This makes more use of the loop directives, and among other things only accesses each element in the vector once (we keep the previous element in the previous-element variable).
Answer for your first UPDATE.
first question:
see this
(if (foo)
(bar (+ 1 baz))
(bar baz))
That's the same as:
(bar (if (foo)
(+ 1 baz)
baz))
or:
(let ((newbaz (if (foo)
(+ 1 baz)
baz)))
(bar newbaz))
Second:
Why not start with I = 1 ?
See also the iterative version in my other answer...
The iterative version proposed by Rainer is very nice, it's compact and more efficient since you traverse the sequence only one time; in contrast to the recursive version which calls position at every iteration and thus traverse the sub-sequence every time. (Edit: I'm sorry, I was completely wrong about this last sentence, see Rainer's comment)
If a recursive version is needed, another approach is to advance the start until it meets the end, collecting the result along its way.
(defun precede (obj vec &key (start 0) (end (length vec)) (test #'eql))
(if (or (null vec) (< end 2)) nil
(%precede-recur obj vec start end test '())))
(defun %precede-recur (obj vec start end test result)
(let ((next (1+ start)))
(if (= next end) (nreverse result)
(let ((newresult (if (funcall test obj (aref vec next))
(adjoin (aref vec start) result)
result)))
(%precede-recur obj vec next end test newresult)))))
Of course this is just another way of expressing the loop version.
test:
[49]> (precede #\a "abracadabra")
(#\r #\c #\d)
[50]> (precede #\a "this is a long sentence that contains more characters")
(#\Space #\h #\t #\r)
[51]> (precede #\s "this is a long sentence that contains more characters")
(#\i #\Space #\n #\r)
Also, I'm interested Robert, did your teacher say why he doesn't like using adjoin or pushnew in a recursive algorithm?
I have some code which collects points (consed integers) from a loop which looks something like this:
(loop
for x from 1 to 100
for y from 100 downto 1
collect `(,x . ,y))
My question is, is it correct to use `(,x . ,y) in this situation?
Edit: This sample is not about generating a table of 100x100 items, the code here just illustrate the use of two loop variables and the consing of their values. I have edited the loop to make this clear. The actual loop I use depends on several other functions (and is part of one itself) so it made more sense to replace the calls with literal integers and to pull the loop out of the function.
It would be much 'better' to just do (cons x y).
But to answer the question, there is nothing wrong with doing that :) (except making it a tad slower).
I think the answer here is resource utilization (following from This post)
for example in clisp:
[1]> (time
(progn
(loop
for x from 1 to 100000
for y from 1 to 100000 do
collect (cons x y))
()))
WARNING: LOOP: missing forms after DO: permitted by CLtL2, forbidden by ANSI
CL.
Real time: 0.469 sec.
Run time: 0.468 sec.
Space: 1609084 Bytes
GC: 1, GC time: 0.015 sec.
NIL
[2]> (time
(progn
(loop
for x from 1 to 100000
for y from 1 to 100000 do
collect `(,x . ,y)) ;`
()))
WARNING: LOOP: missing forms after DO: permitted by CLtL2, forbidden by ANSI
CL.
Real time: 0.969 sec.
Run time: 0.969 sec.
Space: 10409084 Bytes
GC: 15, GC time: 0.172 sec.
NIL
[3]>
dsm: there are a couple of odd things about your code here. Note that
(loop for x from 1 to 100000
for y from 1 to 100000 do
collect `(,x . ,y))
is equivalent to:
(loop for x from 1 to 100
collecting (cons x x))
which probably isn't quite what you intended. Note three things: First, the way you've written it, x and y have the same role. You probably meant to nest loops. Second, your do after the y is incorrect, as there is not lisp form following it. Thirdly, you're right that you could use the backtick approach here but it makes your code harder to read and not idiomatic for no gain, so best avoided.
Guessing at what you actually intended, you might do something like this (using loop):
(loop for x from 1 to 100 appending
(loop for y from 1 to 100 collecting (cons x y)))
If you don't like the loop macro (like Kyle), you can use another iteration construct like
(let ((list nil))
(dotimes (n 100) ;; 0 based count, you will have to add 1 to get 1 .. 100
(dotimes (m 100)
(push (cons n m) list)))
(nreverse list))
If you find yourself doing this sort of thing a lot, you should probably write a more general function for crossing lists, then pass it these lists of integers
If you really have a problem with iteration, not just loop, you can do this sort of thing recursively (but note, this isn't scheme, your implementation may not guaranteed TCO). The function "genint" shown by Kyle here is a variant of a common (but not standard) function iota. However, appending to the list is a bad idea. An equivalent implementation like this:
(defun iota (n &optional (start 0))
(let ((end (+ n start)))
(labels ((next (n)
(when (< n end)
(cons n (next (1+ n))))))
(next start))))
should be much more efficient, but still is not a tail call. Note I've set this up for the more usual 0-based, but given you an optional parameter to start at 1 or any other integer. Of course the above can be written something like:
(defun iota (n &optional (start 0))
(loop repeat n
for i from start collecting i))
Which has the advantage of not blowing the stack for large arguments. If your implementation supports tail call elimination, you can also avoid the recursion running out of place by doing something like this:
(defun iota (n &optional (start 0))
(labels ((next (i list)
(if (>= i (+ n start))
nil
(next (1+ i) (cons i list)))))
(next start nil)))
Hope that helps!
Why not just
(cons x y)
By the way, I tried to run your code in CLISP and it didn't work as expected. Since I'm not a big fan of the loop macro here's how you might accomplish the same thing recursively:
(defun genint (stop)
(if (= stop 1) '(1)
(append (genint (- stop 1)) (list stop))))
(defun genpairs (x y)
(let ((row (mapcar #'(lambda (y)
(cons x y))
(genint y))))
(if (= x 0) row
(append (genpairs (- x 1) y)
row))))
(genpairs 100 100)