As the title says, I'm trying to append items to a list called solution, below is the code:
(defun add-solution (n)
(let ((solution))
(do((current-node '() (next-state current-node n nil)))
((equal current-node '(0 0 0 0)) solution)
(if (goal-test current-node n)
(progn
(format t "Cur: ~S~%" current-node)
(setq solution (append solution (list current-node)))
(format t "Solution: ~S~%" solution)
)
)
)
)
)
Each time the new current-node is a something like:(1 7 8 14), (2 4 11 13), but when the loop returns it returns ((1) (2))..
What I need is (((1 7 8 14) (2 4 11 13)). Not sure what happened there??
edit:
I added the format functions just before and after the setq, and the output was like:
Cur: (1 7 8 14)
Solution: ((1 7 8 14))
Cur: (2 4 11 13)
Solution: ((1)(2 4 11 13))
and after the whole thing is done the returned value became ((1) (2)) again. I'm not really doing anything that modifies solution...
Looks like your error is somewhere else.
Style:
I would write/format the code like this:
(defun add-solution (n)
(do ((solution nil)
(current-node '() (next-state current-node n nil)))
((equal current-node '(0 0 0 0)) solution)
(when (goal-test current-node n)
(setq solution (append solution (list current-node))))))
Note that this is bad code, since you are APPENDING an item to the end of a list in a loop. This is a potentially very costly operation. Lisp lists are optimized for adding to the front, not to the end.
Related
I have a list who's length is divisible by two, and I'm looking for something similar to the answer to this question:
(loop for (a b) on lst while b
collect (+ a b))
However there is overlap between elements:
(1 2 3 4 5) -> (3 5 7 9)
adding 1 and 2 and then 2 and 3 etc.
Where as I have a list like (1 2 3 4) and am looking for something like
((1 2) (3 4))
as output. Is there a way to make loop step correctly over the list?
Another solution.
Something like this should work:
(let ((list '(1 2 3 4)))
(loop :for (a b) :on list :by #'cddr :while b
:collect (cons a b)))
Also a more verbose variant:
(let ((list '(1 2 3 4)))
(loop :for a :in list :by #'cddr
:for b :in (cdr list) :by #'cddr
:collect (cons a b)))
Another approach using the SERIES package.
See also the user manual from Richard C. Waters.
Setup
(ql:quickload :series)
(defpackage :stackoverflow (:use :series :cl))
(in-package :stackoverflow)
Code
(defun pairs (list)
(collect 'list
(mapping (((odd even) (chunk 2 2 (scan 'list list))))
(list odd even))))
scan the content of list as a "serie"
chunk it with M=2 and N=2:
This function has the effect of breaking up the input series items
into (possibly overlapping) chunks of length m. The starting positions
of successive chunks differ by n. The inputs m and n must both be
positive integers.
More precisely, (chunk 2 2 (scan '(1 2 3 4))) produces #Z(1 3) and #Z(2 4)
mapping in parallel over each odd and even element of those series, produce a series of couples, as done by (list odd even).
finally, collect the result, as a list.
Compilation
All the intermediate "series" are compiled away thanks to a stream-fusion mechanism. Here is the macro expansion when pointing at collect:
(LET* ((#:OUT-1120 LIST))
(LET (#:ELEMENTS-1117
(#:LISTPTR-1118 #:OUT-1120)
(#:COUNT-1113 0)
#:CHUNK-1114
#:CHUNK-1115
#:ITEMS-1123
(#:LASTCONS-1106 (LIST NIL))
#:LST-1107)
(DECLARE (TYPE LIST #:LISTPTR-1118)
(TYPE FIXNUM #:COUNT-1113)
(TYPE CONS #:LASTCONS-1106)
(TYPE LIST #:LST-1107))
(SETQ #:COUNT-1113 1)
(SETQ #:LST-1107 #:LASTCONS-1106)
(TAGBODY
#:LL-1124
(IF (ENDP #:LISTPTR-1118)
(GO SERIES::END))
(SETQ #:ELEMENTS-1117 (CAR #:LISTPTR-1118))
(SETQ #:LISTPTR-1118 (CDR #:LISTPTR-1118))
(SETQ #:CHUNK-1114 #:CHUNK-1115)
(SETQ #:CHUNK-1115 #:ELEMENTS-1117)
(COND ((PLUSP #:COUNT-1113) (DECF #:COUNT-1113) (GO #:LL-1124))
(T (SETQ #:COUNT-1113 1)))
(SETQ #:ITEMS-1123
((LAMBDA (ODD EVEN) (LIST ODD EVEN)) #:CHUNK-1114 #:CHUNK-1115))
(SETQ #:LASTCONS-1106
(SETF (CDR #:LASTCONS-1106) (CONS #:ITEMS-1123 NIL)))
(GO #:LL-1124)
SERIES::END)
(CDR #:LST-1107)))
CL-USER 156 > (loop with list = '(1 2 3 4)
while list
collect (loop repeat 2
while list
collect (pop list)))
((1 2) (3 4))
or
CL-USER 166 > (loop with list = '(1 2 3 4 5 6)
while (and list (cdr list))
collect (loop repeat 2 collect (pop list)))
((1 2) (3 4) (5 6))
CL-USER 167 > (loop with list = '(1 2 3 4 5 6 7)
while (and list (cdr list))
collect (loop repeat 2 collect (pop list)))
((1 2) (3 4) (5 6))
How to get the intersection of multiple lists using elisp? I'm a elisp newbie but I'm imagining there is some builtin function or a nicer solution using reduce. I cobbled this together, but it seems overly complicated.
;; get the intersection of these lists
;; result should be (3 4 5)
(setq test '((0 1 2 3 4 5) (2 3 4 5 6) (3 4 5 6 7)))
(require 'cl-lib)
(cl-remove-if-not
(lambda (x) (cl-every
(lambda (y) (> (length (memq x y) ) 0 ) )
(cdr test) ) )
(car test) )
;; ( 3 4 5)
There is a cl-intersection that takes only two operands:
(cl-intersection '(0 1 2 3 4 5) '(2 3 4 5 6))
You can use it do define your own intersection:
(defun my-intersection(l)
(cond ((null l) nil)
((null (cdr l)) (car l))
(t (cl-intersection (car l) (my-intersection (cdr l))))))
(my-intersection '((0 1 2 3 4 5) (2 3 4 5 6) (3 4 5 6 7)))
Updated
Thanks to the #Tobias comment below, you could have in the new function the same keyword parameters of cl-intersection, that is (:test :test-not :key) and propagate them to all the calls to it inside the recursion.
Here is the extended version:
(defun my-intersection(l &rest cl-keys)
(cond ((null l) nil)
((null (cdr l)) (car l))
(t (apply 'cl-intersection (car l) (apply 'my-intersection (cdr l) cl-keys) cl-keys))))
Install dash third-party list manipulation library (follow instructions to install it). Then you need:
(-reduce '-intersection '((1 2 3 4) (2 3 4 5) (3 4 5 6))) ; => (3 4)
If you need a function that accepts variable number of lists, instead of a single list of lists, wrap it in a function using &rest keyword, like that:
(defun -intersection* (&rest list-of-lists)
(-reduce '-intersection list-of-lists))
;; (-intersection* '(1 2 3 4) '(2 3 4 5) '(3 4 5 6)) ; => (3 4)
If it's the first time you use -reduce, it's a “fold” function: it takes a binary function, a list of elements, and reduces them to a final result one list element at a time. This answer explains the concept behind the fold.
I'm trying to get the largest sublist from a list using Common Lisp.
(defun maxlist (list)
(setq maxlen (loop for x in list maximize (list-length x)))
(loop for x in list (when (equalp maxlen (list-length x)) (return-from maxlist x)))
)
The idea is to iterate through the list twice: the first loop gets the size of the largest sublist and the second one retrieves the required list. But for some reason I keep getting an error in the return-from line. What am I missing?
Main problem with loop
There are a few problems here. First, you can write the loop as the following. There are return-from and while forms in Common Lisp, but loop defines its own little language that also recognizes while and return, so you can just use those:
(loop for x in list
when (equalp maxlen (list-length x))
return x)
A loop like this can actually be written more concisely with find though. It's just
(find maxlen list :key list-length :test 'equalp)
Note, however, that list-length should always return a number or nil, so equalp is overkill. You can just use eql, and that's the default for find, so you can even write
(find maxlen list :key list-length)
list-length and maximize
list-length is a lot like length, except that if a list has circular structure, it returns nil, whereas it's an error to call length with an improper list. But if you're using (loop ... maximize ...), you can't have nil values, so the only case that list-length handles that length wouldn't is one that will still give you an error. E.g.,
CL-USER> (loop for x in '(4 3 nil) maximize x)
; Evaluation aborted on #<TYPE-ERROR expected-type: REAL datum: NIL>.
(Actually, length works with other types of sequences too, so list-length would error if you passed a vector, but length wouldn't.) So, if you know that they're all proper lists, you can just
(loop for x in list
maximizing (length x))
If they're not all necessarily proper lists (so that you do need list-length), then you need to guard like:
(loop for x in list
for len = (list-length x)
unless (null len) maximize len)
A more efficient argmax
However, right now you're making two passes over the list, and you're computing the length of each sublist twice. Once is when you compute the maximum length, and the other is when you go to find one with the maximum value. If you do this in one pass, you'll save time. argmax doesn't have an obvious elegant solution, but here are implementations based on reduce, loop, and do*.
(defun argmax (fn list &key (predicate '>) (key 'identity))
(destructuring-bind (first &rest rest) list
(car (reduce (lambda (maxxv x)
(destructuring-bind (maxx . maxv) maxxv
(declare (ignore maxx))
(let ((v (funcall fn (funcall key x))))
(if (funcall predicate v maxv)
(cons x v)
maxxv))))
rest
:initial-value (cons first (funcall fn (funcall key first)))))))
(defun argmax (function list &key (predicate '>) (key 'identity))
(loop
for x in list
for v = (funcall function (funcall key x))
for maxx = x then maxx
for maxv = v then maxv
when (funcall predicate v maxv)
do (setq maxx x
maxv v)
finally (return maxx)))
(defun argmax (function list &key (predicate '>) (key 'identity))
(do* ((x (pop list)
(pop list))
(v (funcall function (funcall key x))
(funcall function (funcall key x)))
(maxx x)
(maxv v))
((endp list) maxx)
(when (funcall predicate v maxv)
(setq maxx x
maxv v))))
They produce the same results:
CL-USER> (argmax 'length '((1 2 3) (4 5) (6 7 8 9)))
(6 7 8 9)
CL-USER> (argmax 'length '((1 2 3) (6 7 8 9) (4 5)))
(6 7 8 9)
CL-USER> (argmax 'length '((6 7 8 9) (1 2 3) (4 5)))
(6 7 8 9)
Short variant
CL-USER> (defparameter *test* '((1 2 3) (4 5) (6 7 8 9)))
*TEST*
CL-USER> (car (sort *test* '> :key #'length))
(6 7 8 9)
Paul Graham's most
Please, consider also Paul Graham's most function:
(defun most (fn lst)
(if (null lst)
(values nil nil)
(let* ((wins (car lst))
(max (funcall fn wins)))
(dolist (obj (cdr lst))
(let ((score (funcall fn obj)))
(when (> score max)
(setq wins obj
max score))))
(values wins max))))
This is the result of test (it also returns value that's returned by supplied function for the 'best' element):
CL-USER> (most #'length *test*)
(6 7 8 9)
4
extreme utility
After a while I came up with idea of extreme utility, partly based on Paul Graham's functions. It's efficient and pretty universal:
(declaim (inline use-key))
(defun use-key (key arg)
(if key (funcall key arg) arg))
(defun extreme (fn lst &key key)
(let* ((win (car lst))
(rec (use-key key win)))
(dolist (obj (cdr lst))
(let ((test (use-key key obj)))
(when (funcall fn test rec)
(setq win obj rec test))))
(values win rec)))
It takes comparison predicate fn, list of elements and (optionally) key parameter. Object with the extreme value of specified quality can be easily found:
CL-USER> (extreme #'> '(4 9 2 1 5 6))
9
9
CL-USER> (extreme #'< '(4 9 2 1 5 6))
1
1
CL-USER> (extreme #'> '((1 2 3) (4 5) (6 7 8 9)) :key #'length)
(6 7 8 9)
4
CL-USER> (extreme #'> '((1 2 3) (4 5) (6 7 8 9)) :key #'cadr)
(6 7 8 9)
7
Note that this thing is called extremum in alexandria. It can work with sequences too.
Using recursion:
(defun maxim-list (l)
(flet ((max-list (a b) (if (> (length a) (length b)) a b)))
(if (null l)
nil
(max-list (car l) (maxim-list (cdr l))))))
The max-list internal function gets the longest of two list. maxim-list is getting the longest of the first list and the maxim-list of the rest.
What I have to do is removing some elements from the list,the 1st,2nd,4th,8th,elements on positions power of 2.I figured out that the easyest way for me to solve this is to construct how the result list should look like without destroying the original list.Here's my code but it doesn't work yet,I'm getting a type error.I'm using contor to know with which element of the list I'm working with an counter to specify only the position from which the elements should be removed.My question is what am I doing wrong and how can it be fixed?
(defun remo(l)
(defparameter e ())
(setq contor 0)
(setq counter 0)
(dolist (elem l) (
(cond
(
((or (< (expt 2 contor) counter) (> (expt 2 contor) counter))
((push elem e) (setq contor (+ 1 contor))))
))
(setq counter (+1 counter))
)
)
(print e)
)
(defun remo (l)
(do ((power-of-2 1)
(counter 1 (1+ counter))
(result ())
(sublist l (cdr sublist)))
((null sublist) (nreverse result))
(if (= counter power-of-2)
(setq power-of-2 (* 2 power-of-2))
(push (car sublist) result))))
(remo '(1 2 3 4 5 6 7 8 9 10))
=> (3 5 6 7 9 10)
I already improved another of your attempts at https://stackoverflow.com/a/20711170/31615, but since you stated the real problem here, I propose the following solution:
(defun remove-if-index-power-of-2 (list)
(loop :for element :in list
:for index :upfrom 1 ; correct for language: "1st" is index 0
:unless (power-of-2-p index)
:collect element))
(defun power-of-2-p (number)
"Determines whether number, which is assumed to be a nonnegative
integer, is a power of 2 by counting the bits."
(declare (type (integer 0 *) number))
(= 1 (logcount number)))
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.