I am creating an overloaded function that takes a list of numbers as a parameter (as well as a list of numbers and a single number in the overloaded signature). From this list the highest value must be found and returned. I am aware of the available "sort" method which automatically sorts the list for you but I am attempting to discover the highest value using recursion. Below is the code I am working with.
32 (defn my-max
33 ([list]
34 (if (empty? list)
35 -1
36 (my-max (pop list) (first list))) ; recursive call
37 )
38 ([list high]
39 (if (empty? list) ; base case
40 high)
41 (if (> (first list) high)
42 (my-max (pop list) (first list)))
43 (my-max (pop list) high))
44 )
45
46 (println (my-max '(1 2 3)))
As you can see, I have hard-coded a list containing 1, 2, and 3 for which 3 should be returned. when running the code I receive an error:
Exception in thread "main" java.lang.NullPointerException
at clojure.lang.Numbers.ops(Numbers.java:942)
at clojure.lang.Numbers.gt(Numbers.java:227)
at user$my_max.invoke(HelloWorld.clj:41)
at user$my_max.invoke(HelloWorld.clj:42)
at user$my_max.invoke(HelloWorld.clj:42)
at user$my_max.invoke(HelloWorld.clj:36)
at user$eval2.invoke(HelloWorld.clj:46)
at clojure.lang.Compiler.eval(Compiler.java:6619)
at clojure.lang.Compiler.load(Compiler.java:7064)
at clojure.lang.Compiler.loadFile(Compiler.java:7020)
at clojure.main$load_script.invoke(main.clj:294)
at clojure.main$script_opt.invoke(main.clj:356)
at clojure.main$main.doInvoke(main.clj:440)
at clojure.lang.RestFn.invoke(RestFn.java:408)
at clojure.lang.Var.invoke(Var.java:415)
at clojure.lang.AFn.applyToHelper(AFn.java:161)
at clojure.lang.Var.applyTo(Var.java:532)
at clojure.main.main(main.java:37)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:483)
at com.intellij.rt.execution.application.AppMain.main(AppMain.java:134)
This error occurs on the final recursive call where 3 has been "popped" from the list and is passed back to my-max as "high" the list is empty and:
(if (empty? list) ; line 39, base case
should evaluate true but is instead evaluating as false. Do you have any thoughts/solutions on this problem?
That statement (line 39-40) does nothing. No matter what it evaluates to (either high or nil), the value is not used. Then line 41 evaluates, and (first list) is nil if your list is empty, hence the null pointer exception.
the < function can only compare numbers so a very straightforward approach is to use an or to ensure they are numbers. Also the second and third if statements had the closing ) in the wrong spots.
user> (defn my-max
([list]
(if (empty? list)
-1
(my-max (pop list) (first list))))
([list high]
(if (empty? list) ; base case
high
(if (> (or (first list) 0) (or high 0))
(my-max (pop list) (first list))
(my-max (pop list) high)))))
#'user/my-max
user> (println (my-max '(1 2 3)))
3
nil
Related
Simple program for binary search.
elements contain no. of elements
then array contains those elements
then q contains no. of queries
search contains element to be searched.
Why this error is coming about high and low has no value after some iterations.
Kindly help :)
My Code :-
(setf elements (parse-integer (read-line)))
(setf array (make-array elements :fill-pointer 0))
(dotimes (i elements) (vector-push (parse-integer (read-line)) array))
(setf q (parse-integer (read-line)))
(defvar *mid*)
(dotimes (i q)
(setf search (parse-integer (read-line)))
(do ((low 0)
(high (- elements 1))
(mid (floor (+ low high) 2)
(floor (+ low high) 2)))
((>= low high) (setf *mid* nil))
(cond
((eql (elt array mid) search) (setf *mid* mid))
((< (elt array mid) search) (setf high (- mid 1)))
(t (setf low (+ mid 1)))))
(format t "~a" *mid*))
Your code is a fine example of an old adage:
the determined Real Programmer can write FORTRAN programs in any language.
Unfortunately Lisp programmers are generally quiche-eating hippies: so here is one quiche-eater's solution to this problem, using notions not present when FORTRAN IV was handed down to us from above on punched stones. These notions are therefore clearly heretical, but nonetheless useful.
Assuming this is homework, you probably will not be able to submit this answer.
Reading the data
First of all we'll write some functions which read the specification of the problem from a stream or file. I have inferred what it is from your code.
(defun stream->search-spec (stream)
;; Read a search vector from a stream: return a vector to be searched
;; and a vector of elements to search for.
;;
;; This function defines what is in files: each line contains an
;; integer, and the file contains a count followed by that many
;; lines, which specifies first the vector to be searched, and then
;; the things to search for.
;;
;; This relies on PARSE-INTEGER & READ-LINE to puke appropriately.
(flet ((read-vector ()
(let* ((elts (parse-integer (read-line stream)))
(vec (make-array elts :element-type 'integer))) ;won't help
(dotimes (i elts vec)
(setf (aref vec i) (parse-integer (read-line stream)))))))
(values (read-vector) (read-vector))))
(defun file->search-spec (file)
;; Read a search vector from a file. This is unused below but is
;; useful to have.
(with-open-file (in file)
(stream->search-spec in)))
(defun validate-sorted-vector (v)
;; check that V is a sorted vector
(dotimes (i (- (length v) 1) v)
(unless (<= (aref v i) (aref v (1+ i)))
(return-from validate-sorted-vector nil))))
The last function is used below to sanity check the data, since the search algorithm assumes the vector is sorted.
The search function
This implements binary search in the same way yours tries to do. Rather than doing it with loops and explicit assignemnt it does it using a local recursive function, which is far easier to understand. There are also various sanity checks and optionally debugging output. In any implementation which optimises tail calls this will be optimised to a loop; in implementations which don't then there will be a few extra function calls but stack overflow problems are very unlikely (think about why: how big would the vector need to be?).
(defun search-sorted-vector-for (vector for &key (debug nil))
;; search a sorted vector for some value. If DEBUG is true then
;; print what we're doing. Return the index, or NIL if FOR is not
;; present.
(when debug
(format *debug-io* "~&* ~D:~%" for))
(labels ((search (low mid high)
(when debug
(format *debug-io* "~& ~10D ~10D ~10D~%" low mid high))
(if (<= low mid high)
;; more to do
(let ((candidate (aref vector mid)))
(cond ((= candidate for)
;; found it
mid)
((< candidate for)
;; look higher
(search (1+ mid) (floor (+ high mid 1) 2) high))
((> candidate for)
;; look lower
(search low (floor (+ low mid) 2) (1- mid)))
(t
;; can't happen
(error "mutant death"))))
;; low = high: failed
nil)))
(let ((high (1- (length vector))))
(search 0 (floor high 2) high))))
Putting the previous two things together.
search-sorted-vector-with-search-vector will repeatedly search using the two vectors that the *->search-spec functions return. stream->search-results uses stream->search-spec and then calls this on its values. file->search-results does it all from a file.
(defun search-sorted-vector-with-search-vector (vector searches &key (debug nil))
;; do a bunch of searches, returning a vector of results.
(let ((results (make-array (length searches))))
(dotimes (i (length searches) results)
(setf (aref results i) (search-sorted-vector vector (aref searches i)
:debug debug)))))
(defun stream->search-results (stream &key (debug nil))
;; Read search specs from a stream, and search according to them.
;; Return the vector of results, the vector being searched and the
;; vector of search specifications.
(multiple-value-bind (to-search search-specs) (stream->search-spec stream)
(when debug
(format *debug-io* "~&searching ~S~% for ~S~&" to-search search-specs))
(assert (validate-sorted-vector to-search) (to-search) "not sorted")
(values (search-sorted-vector-with-search-vector to-search search-specs
:debug debug)
to-search search-specs)))
(defun file->search-results (file &key (debug nil))
;; sort from a file
(with-open-file (in file)
(stream->search-results in :debug debug)))
Using it
Given a file /tmp/x.dat with:
9
1
10
100
101
102
103
200
201
400
6
10
102
200
1
400
99
then:
> (file->search-results "/tmp/x.dat" :debug t)
searching #(1 10 100 101 102 103 200 201 400)
for #(10 102 200 1 400 99)
* 10:
0 4 8
0 2 3
0 1 1
* 102:
0 4 8
* 200:
0 4 8
5 6 8
* 1:
0 4 8
0 2 3
0 1 1
0 0 0
* 400:
0 4 8
5 6 8
7 7 8
8 8 8
* 99:
0 4 8
0 2 3
0 1 1
2 1 1
#(1 4 6 0 8 nil)
#(1 10 100 101 102 103 200 201 400)
#(10 102 200 1 400 99)
You can see that the last search failed (99 is not in the vector).
The floor Hyperspec article on dotimes has this example:
(defun palindromep (string &optional
(start 0)
(end (length string)))
(dotimes (k (floor (- end start) 2) t)
(unless (char-equal (char string (+ start k))
(char string (- end k 1)))
(return nil))))
If floor returns two values, e.g. (floor 5 2) -> 2 and 1, how does dotimes know to just use the first value and disregard the second for its count-form?
It's a general mechanism and not specific to dotimes.
If one calls a function or sets a variable, then only the first value will be passed:
CL-USER 52 > (defun foo (x) x)
FOO
CL-USER 53 > (foo (floor 5 2))
2
CL-USER 54 > (let ((foo (floor 5 2)))
foo)
2
To do the equivalent (calling functions, binding variables) with multiple values, one needs to use special constructs:
CL-USER 55 > (multiple-value-call #'list
(floor 5 2) (floor 7 3))
(2 1 2 1)
CL-USER 56 > (multiple-value-bind (foo0 foo1)
(floor 5 2)
(list foo0 foo1))
(2 1)
From 7.10.1,
Normally multiple values are not used. Special forms are required both to produce multiple values and to receive them. If the caller of a function does not request multiple values, but the called function produces multiple values, then the first value is given to the caller and all others are discarded; if the called function produces zero values, then the caller gets nil as a value.
Unless you specifically do something to deal with the multiple values (such as by multiple-value-call or one of the various macros equipped to handle them), all except the first value will be ignored.
I started programming with lisp yesterday so please excuse if I am making some really newbie mistake. I am trying to create a function which calculates the bell numbers using the bell triangle and my recursive triangle function is not working properly. I am also sure if I got my recursive triangle function working that my recursive bell function is somehow also broken.
When I test my triangle function I get the output:
(defun bell(l n)
(if(< n 1)(list 1))
(if (= n 1)(last l))
(bell (triangle (reverse l) (last l) (list-length l)) (- n 1))
)
(defun triangle(pL nL i)
(if(<= i 0)
(write "equals zero!")
(reverse nL)
)
(triangle pL (append (list (+ (nth i pL) (nth i nL))) nL) (- i 1))
)
(write (triangle '(1) '(1) 0))
=>
"equals zero!""equals zero!"
*** - NTH: -1 is not a non-negative integer
For some reason, it is printing my debug code twice even though the function should be meeting my base case on the first call.
For some reason, it is printing my debug code twice even though the function should be meeting my base case on the first call.
It is printed twice because if is not doing what you think it does. The first if test is true, therefore equals zero! is printed. After that, a recursive call to triangle function is invoked. The test is again true (-1 <= 0), so equals zero! is again printed. Finally, you get an error because nthcdr function is called with -1. I strongly recommend you a good lisp debugger. The one from Lispworks is pretty good.
I honestly don't get the logic of what you were trying to achieve with your code. so I wrote mine:
(defun generate-level (l &optional (result))
"given a list l that represents a triangle level, it generates the next level"
(if (null l) result
(if (null result)
(generate-level l (list (car (last l))))
(generate-level (cdr l) (append result
(list (+ (car l)
(car (last result)))))))))
(defun bell (levels &optional (l))
"generate a bell triangle with the number of labels given by the first parameter"
(unless (zerop levels)
(let ((to-print (if (null l) (list 1) (generate-level l))))
(print to-print)
(bell (1- levels) to-print))))
Things to understand the implementation:
&optional (parameter): this parameter is optional and nil by default.
append concatenates two lists. I'm using it to insert in the back of the list.
let ((to-print x)) creates a new variable binding (local variable) called to-print and initialized to x.
I almost forgot to mention how if works in common lisp:
(if (= x 1) y z) means if x is equal to 1 then return y, otherwise z.
Now if you call the function to create a Bell triangle of 7 levels:
CL-USER 9 > (bell 7)
(1)
(1 2)
(2 3 5)
(5 7 10 15)
(15 20 27 37 52)
(52 67 87 114 151 203)
(203 255 322 409 523 674 877)
NIL
It would be nicer to print it with the appropiate padding, like this:
1
1 2
2 3 5
5 7 10 15
15 20 27 37 52
52 67 87 114 151 203
203 255 322 409 523 674 877
but I left that as an exercise to the reader.
Your ifs don't have any effect. They're evaluated, and produce results, but then you discard them. Just like
(defun abc ()
'a
'b
'c)
would evaluate 'a and 'b to produce the symbols a and b, and then would evaluate 'c to produce the symbol c, which would then be returned. In the case of
(if(<= i 0)
(write "equals zero!") ; then
(reverse nL) ; else
)
you're comparing whether i is less than or equal to zero, and if it is, you print equals zero, and if it's not, you (non-destructively) reverse nL and discard the result. Then you finish the function by making a call to triangle. It seems like you probably want to return the reversed nL when i is less than or equal to zero. Use cond instead, since you can have multiple body forms, as in:
(cond
((<= i 0) (write ...) (reverse nL))
(t (triangle ...)))
You could also use if with progn to group the forms:
(if (<= i 0)
(progn
(write ...)
(reverse nL))
(triangle ...))
Your other function has the same problem. If you want to return values in those first cases, you need to use a form that actually returns them. For instance:
(if (< n 1)
(list 1)
(if (= n 1)
(last l)
(bell #| ... |#)))
More idiomatic would be cond, and using list rather than l, which looks a lot like 1:
(cond
((< n 1) (list 1))
((= n 1) (last list))
(t (bell #| ... |#)))
Thank you all for the explanations. I eventually arrived at the code below. I realized that the if block worked something like..
(if (condition) (execute statement) (else execute this statement))
(defun bell(l n)
(if (< n 2)(last l)
(bell (triangle l (last l) 0) (- n 1))
)
)
(defun triangle(pL nL i)
(if(= i (list-length pL)) nL
(triangle pL (append nL (list (+ (nth i pL) (nth i nL)))) (+ i 1))
)
)
(write (bell (list 1) 10))
I am working on program related to the different of dealing with even numbers in C and lisp , finished my c program but still having troubles with lisp
isprime function is defined and I need help in:
define function primesinlist that returns unique prime numbers in a lis
here what i got so far ,
any help with that please?
(defun comprimento (lista)
(if (null lista)
0
(1+ (comprimento (rest lista)))))
(defun primesinlist (number-list)
(let ((result ()))
(dolist (number number-list)
(when (isprime number)
( number result)))
(nreverse result)))
You need to either flatten the argument before processing:
(defun primesinlist (number-list)
(let ((result ()))
(dolist (number (flatten number-list))
(when (isprime number)
(push number result)))
(delete-duplicates (nreverse result))))
or, if you want to avoid consing up a fresh list, flatten it as you go:
(defun primesinlist (number-list)
(let ((result ()))
(labels ((f (l)
(dolist (x l)
(etypecase x
(integer (when (isprime x)
(push x result)))
(list (f x))))))
(f number-list))
(delete-duplicates (nreverse result))))
To count distinct primes, take the length of the list returned by primesinlist.
Alternatively, you can use count-if:
(count-if #'isprime (delete-duplicates (flatten number-list)))
It sounds like you've already got a primality test implemented, but for sake of completeness, lets add a very simple one that just tries to divide a number by the numbers less than it up to its square root:
(defun primep (x)
"Very simple implementation of a primality test. Checks
for each n above 1 and below (sqrt x) whether n divides x.
Example:
(mapcar 'primep '(2 3 4 5 6 7 8 9 10 11 12 13))
;=> (T T NIL T NIL T NIL NIL NIL T NIL T)
"
(do ((sqrt-x (sqrt x))
(i 2 (1+ i)))
((> i sqrt-x) t)
(when (zerop (mod x i))
(return nil))))
Now, you need a way to flatten a potentially nested list of lists into a single list. When approaching this problem, I usually find it a bit easier to think in terms of trees built of cons-cells. Here's an efficient flattening function that returns a completely new list. That is, it doesn't share any structure with the original tree. That can be useful, especially if we want to modify the resulting structure later, without modifying the original input.
(defun flatten-tree (x &optional (tail '()))
"Efficiently flatten a tree of cons cells into
a list of all the non-NIL leafs of the tree. A completely
fresh list is returned.
Examples:
(flatten-tree nil) ;=> ()
(flatten-tree 1) ;=> (1)
(flatten-tree '(1 (2 (3)) (4) 5)) ;=> (1 2 3 4 5)
(flatten-tree '(1 () () 5)) ;=> (1 5)
"
(cond
((null x) tail)
((atom x) (list* x tail))
((consp x) (flatten-tree (car x)
(flatten-tree (cdr x) tail)))))
Now it's just a matter of flatting a list, removing the number that are not prime, and removing duplicates from that list. Common Lisp includes functions for doing these things, namely remove-if-not and remove-duplicates. Those are the "safe" versions that don't modify their input arguments. Since we know that the flattened list is freshly generated, we can use their (potentially) destructive counterparts, delete-if-not and delete-duplicates.
There's a caveat when you're removing duplicate elements, though. If you have a list like (1 3 5 3), there are two possible results that could be returned (assuming you keep all the other elements in order): (1 3 5) and (1 5 3). That is, you can either remove the the later duplicate or the earlier duplicate. In general, you have the question of "which one should be left behind?" Common Lisp, by default, removes the earlier duplicate and leaves the last occurrence. That behavior can be customized by the :from-end keyword argument. It can be nice to duplicate that behavior in your own API.
So, here's a function that puts all those considerations together.
(defun primes-in-tree (tree &key from-end)
"Flatten the tree, remove elements which are not prime numbers,
using FROM-END to determine whether earlier or later occurrences
are kept in the list.
Examples:
(primes-in-list '(2 (7 4) ((3 3) 5) 6 7))
;;=> (2 3 5 7)
(primes-in-list '(2 (7 4) ((3 3) 5) 6 7) :from-end t)
;;=> (2 7 3 5)"
;; Because FLATTEN-TREE returns a fresh list, it's OK
;; to use the destructive functions DELETE-IF-NOT and
;; DELETE-DUPLICATES.
(delete-duplicates
(delete-if-not 'primep (flatten-tree list))
:from-end from-end))
I want my function to print each item in the list and sublist without quotes and return the number of items. The output of the list also needs to be in order, but my function is printing in reverse. I'm not sure why, is there any reasons why? Any suggestions to how I can recursively count the number of items and return that number? In addition why is the last item printed is supposed to be 9.99 instead of 100.999?
Edit: Thanks for the help so far. Just last question: Is there a way to make any output like DAY to be in lower case (day), or is that something that can't be done?
My function:
(defun all-print (inlist)
(cond
((not (listp inlist))
(format t "Error, arg must be a list, returning nil")
())
((null inlist) 0)
((listp (car inlist))
(ffn (append (car inlist)(cdr inlist))))
(t
(format t "~a " (car inlist) (ffn (cdr inlist))))))
My output example:
CL-USER 1 > (all-print (list 5 "night" 3 (list 9 -10) (quote day) -5.9 (* 100.999)))
100.999 -5.9 DAY -10 9 3 night 5
NIL
What it's suppose to output example:
5 night 3 9 -10 day -5.9 9.99 ;print
8 ;returns
It looks like all-print is supposed to be called ffn, since it looks like those are supposed to be recursive calls. In the rest of this answer, I'm just going to use ffn since it's shorter.
Why the output is in reverse
At present, your final cond clause makes the recursive call before doing any printing, because your recursive call is an argument to format:
(format t "~a " (car inlist) (ffn (cdr inlist)))
; ------------ -----------------
; 3rd 4th
All the arguments to format, including the 4th in this case, are evaluated before format is called. The 4th argument here will print the rest of the list, and then format will finally print the first element of the list. Your last cond clause should do the printing, and then make the recursive call:
(cond
…
(t
(format t "~a " (car inlist))
(ffn (cdr inlist))))
Why you get 100.999 rather than 9.99
You're getting 100.999 in your output rather than 9.99 (or something close to it) because the value of (* 100.999) is simply the value of 100.999. I'm guessing that you wanted (* 10 0.999) (note the space between 10 and 0.99). That still won't be quite 9.99 because of floating point arithmetic, though, but it will be close.
How to get the number of elements printed
uselpa's answer provides a good solution here. If you're supposed to return the number of elements printed, then every return value from this function should be a number. You have four cases,
not a list — returning nil is not a great idea. If this can't return a number (e.g., 0), then signal a real error (e.g., with (error "~A is not a list" inlist).
inlist is empty — return 0 (you already do)
(car inlist) is a list — here you make a recursive call to ffn. Since the contract says that it will return a count, you're fine. This is one of the reasons that it's so important in the first case (not a list) that you don't return a non-number; the contract depends on every call that returns returning an number.
In the final case, you print one item, and then make a recursive call to ffn. That recursive call returns the number of remaining elements that are printed, and since you just printed one, you need to add one to it. Thus the final cond clause should actually be something like the following. (Adding one to something is so common that Common Lisp has a 1+ function.)
(cond
…
(t
(format t "~a " (car inlist))
(1+ (ffn (cdr inlist))))) ; equivalent to (+ 1 (ffn (cdr inlist)))
A more efficient solution
We've addressed the issues with your original code, but we can also ask whether there are better approaches to the problem.
Don't append
Notice that when you have input like ((a b c) d e f), you create the list (a b c d e f) and recurse on it. However, you could equivalently recurse on (a b c) and on (d e f), and add the results together. This would avoid creating a new list with append.
Don't check argument types
You're checking that the input is a list, but there's really not much need to do that. If the input isn't a list, then using list processing functions on it will signal a similar error.
A new version
This is somewhat similar to uselpa's answer, but I've made some different choices about how to handle certain things. I use a local function process-element to handle elements from each input list. If the element is a list, then we pass it to print-all recursively, and return the result of the recursive call. Otherwise we return one and print the value. (I used (prog1 1 …) to emphasize that we're returning one, and printing is just a side effect. The main part of print-all is a typical recursion now.
(defun print-all (list)
(flet ((process-element (x)
(if (listp x)
(print-all x)
(prog1 1
(format t "~A " x)))))
(if (endp list)
0
(+ (process-element (first list))
(print-all (rest list))))))
Of course, now that we've pulled out the auxiliary function, the iteration is a bit clearer, and we see that it's actually a case for reduce. You might even choose to do away with the local function, and just use a lambda function:
(defun print-all (list)
(reduce '+ list
:key (lambda (x)
(if (listp x)
(print-all x)
(prog1 1
(format t "~A " x))))))
Here's my suggestion on how to write this function:
(defun all-print (lst)
(if (null lst)
0 ; empty list => length is 0
(let ((c (car lst))) ; bind first element to c
(if (listp c) ; if it's a list
(+ (all-print c) (all-print (cdr lst))) ; recurse down + process the rest of the list
(progn ; else
(format t "~a " c) ; not a list -> print item, then
(1+ (all-print (cdr lst)))))))) ; add 1 and process the rest of the list
then
? (all-print (list 5 "night" 3 (list 9 -10) (quote day) -5.9 (* 100.999)))
5 night 3 9 -10 DAY -5.9 100.999
8