I am printing a triangle of numbers in Lisp and I want to make sure that the input upon calling the function is an integer. If it's a string or a decimal, it should return a message not accepting the input. This is my code for the numbers.
(defun nested-loop (n)
(loop for i from 1 to n doing
(loop for j from 1 to i collecting
(progn
(prin1 j)))
(format t "~%")))
(nested-loop 5)
Use the macro CHECK-TYPE:
CL-USER 9 > (let ((n "10"))
(check-type n integer))
Error: The value "10" of N is not of type INTEGER.
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 need a function that will check if all digits in some number on even positions are even. The least significant digit is on position 1, starting from right to left. The function need to be written in lisp.
Examples:
245 -> true, since 4 is even
238456 -> false, since 5 is odd and 8 and 2 are even
and so on...
Here`s what I got:
(defun check(number fac)
(cond
((= (/ number fac) 0) t)
((= (mod (/ number fac) 2 ) 0) (check number (* 100 fac) ) )
(nil)))
The initial value for fac is 10, we divide the number with 10, extract the second digit, check if it is even, if so proceed and divide number with 1000 to extract the 4-th digit and so on until we get over all digits, than the function returns true, meanwhile if some digit is odd the function should return nil.
But something is wrong and the function return nil all the time , when I call it like (check 22 10) for example.
Any thoughts?
Here is a non recursive solution that checks for the correctness of the parameter:
(defun check(num)
(assert (integerp num))
(loop for i = (truncate num 10) then (truncate i 100) until (zerop i)
always (evenp i)))
Just another variant. Basicly I'm converting number to list (through string though, maybe not the best way), then reverse it, select every second element and check it all for being even.
;; Helper for getting every
(defun get-all-nth (list period)
"Get all NTH element in the list"
(remove-if-not
(let ((iterator 0))
(lambda (x)
(declare (ignore x))
(= 0 (mod (incf iterator) period)))) list))
(defun check-evens (num)
"Checks if all digits in some number on even positions are even.
Goes Rigth-to-left."
(assert (integerp num))
(every #'evenp
(get-all-nth
(reverse
(map 'list #'digit-char-p
(prin1-to-string num))) 2)))
Some test cases:
CL-USER> (check-evens 123)
T
CL-USER> (check-evens 238456)
NIL
CL-USER> (check-evens 238446)
T
CL-USER> (check-evens 23844681)
T
I'm trying to write a function in Common Lisp to convert a base 10 number into a base 8 number, represented as a list, recursively.
Here's what I have so far:
(defun base8(n)
(cond
((zerop (truncate n 8)) (cons n nil))
((t) (cons (mod n 8) (base8 (truncate n 8))))))
This function works fine when I input numbers < 8 and > -8, but the recursive case is giving me a lot of trouble. When I try 8 as an argument (which should return (1 0)), I get an error Undefined operator T in form (T).
Thanks in advance.
Just for fun, here's a solution without recursion, using built-in functionality:
(defun base8 (n)
(reverse (coerce (format nil "~8R" n) 'list)))
It seems you have forgotten to (defun t ...) or perhaps it's not the function t you meant to have in the cond? Perhaps it's t the truth value?
The dual namespace nature of Common Lisp makes it possible for t to both be a function and the truth value. the difference is which context you use it and you clearly are trying to apply t as a function/macro.
Here is the code edited for the truth value instead of the t function:
(defun base8(n)
(cond
((zerop (truncate n 8)) (cons n nil))
(t (cons (mod n 8) (base8 (truncate n 8))))))
(base8 8) ; ==> (0 1)
I'm in the process of reading a flat file - to use the characters read I want to convert them into numbers. I wrote a little function that converts a string to a vector:
(defun string-to-vec (strng)
(setf strng (remove #\Space strng))
(let ((vec (make-array (length strng))))
(dotimes (i (length strng) vec)
(setf (svref vec i) (char strng i)))))
However this returns a vector with character entries. Short of using char-code to convert unit number chars to numbers in a function, is there a simple way to read numbers as numbers from a file?
In addition to Rainer's answer, let me mention read-from-string (note that Rainer's code is more efficient than repeated application of read-from-string because it only creates a stream once) and parse-integer (alas, there is no parse-float).
Note that if you are reading a CSV file, you should probably use an off-the-shelf library instead of writing your own.
Above is shorter:
? (map 'vector #'identity (remove #\Space "123"))
#(#\1 #\2 #\3)
You can convert a string:
(defun string-to-vector-of-numbers (string)
(coerce
(with-input-from-string (s string)
(loop with end = '#:end
for n = (read s nil end)
until (eql n end)
unless (numberp n) do (error "Input ~a is not a number." n)
collect n))
'vector))
But it would be easier to read the numbers directly form the file. Use READ, which can read numbers.
Note that read-like functions are affected by reader macros.
Pick an example:
* (defvar *foo* 'bar)
*FOO*
* (read-from-string "#.(setq *foo* 'baz)")
BAZ
19
* *foo*
BAZ
As you can see read-from-string can implicitly set a variable. You can disable the #. reader macro by setting *read-eval* to nil but anyway if you have only integers on the input then consider using parse-integer instead.
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.