I am working in Common Lisp, trying to make Windows game minesweeper.
I have a list (1 1 1 2 2 2 3 3 3) and want to print that like matrix
(1 1 1
2 2 2
3 3 3)
How to do that?
Edit
I am at the beginning of
(format t "Input width:")
(setf width (read))
(format t "Input height:")
(setf height (read))
(format t "How many mines:")
(setf brMina (read))
(defun matrica (i j)
(cond ((= 0 i) '())
(t (append (vrsta j) (matrica (1- i) j) ))))
(setf minefield (matrica width height))
(defun stampaj ()
(format t "~%~a" minefield ))
Another example, using the pretty-printer for fun:
(defun print-list-as-matrix
(list elements-per-row
&optional (cell-width (1+ (truncate (log (apply #'max list) 10)))))
(let ((*print-right-margin* (* elements-per-row (1+ cell-width)))
(*print-miser-width* nil)
(*print-pretty* t)
(format-string (format nil "~~<~~#{~~~ad~~^ ~~}~~#:>~%" cell-width)))
(format t format-string list)))
Works like this:
CL-USER> (print-list-as-matrix (loop for i from 1 to 9 collect i) 3)
1 2 3
4 5 6
7 8 9
NIL
CL-USER> (print-list-as-matrix (loop for i from 1 to 25 collect i) 5)
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
NIL
CL-USER> (print-list-as-matrix (loop for i from 1 to 16 collect i) 2)
1 2
3 4
5 6
7 8
9 10
11 12
13 14
15 16
Like this:
(defun print-list-as-grid (list rows cols)
(assert (= (length list) (* rows cols))
(loop for row from 0 below rows do
(loop for col from 0 below cols do
(princ (car list))
(princ #\space)
(setf list (cdr list)))
(princ #\newline)))
* (print-list-as-grid '(a b c d e f g h i) 3 3)
A B C
D E F
G H I
NIL
Related
I am using DrRacket.
How can I write a function for the difference between the maximum and minimum number in the list using accumulators and mutually recursive functions.
For instance, (list 10 2 3 -5 4 1 -6)) 9). The list has at least one element in the list.
Do I need two accumulators?
Version 1: An accumulator based solution that's mutually recursive. Since the input list is assumed to be non-empty, we start of with the first element being max and min. As we go through the list, we pick new max and mins by comparing the current element with the accumulators.
#lang racket
; [NEList-of Number] -> Number
(define (max-min-diff.v1 nelst)
(max-min-diff/t.v1 (rest nelst) (first nelst) (first nelst)))
; [List-of Number] -> Number
(define (max-min-diff/t.v1 l max min)
(cond [(empty? l) (- max min)]
[else (get-new-max-min (rest l) (first l) max min)]))
; [List-of Number] Number Number Number -> Number
(define (get-new-max-min rst fst max min)
(max-min-diff/t.v1 rst
(if (> fst max) fst max)
(if (< fst min) fst min)))
(max-min-diff.v1 '(0 33 2 32 4 8 3 3 5))
; => 33
(max-min-diff.v1'(1 3 9 4 7 2 2 5 11))
; => 10
Version 2: An accumulator based solution that's not mutually recursive. More abstract because we pass the comparators to a generic helper.
; [NEList-of Number] -> Number
(define (max-min-diff.v2 nelst)
(max-min-diff/t.v2 (rest nelst) (first nelst) (first nelst)))
; [List-of Number] -> Number
(define (max-min-diff/t.v2 l max min)
(cond [(empty? l) (- max min)]
[else (max-min-diff/t.v2 (rest l)
(f-if (first l) max >)
(f-if (first l) min <))]))
; X X [X X -> Boolean] -> X
(define (f-if n1 n2 func)
(if (func n1 n2) n1 n2))
(max-min-diff.v2 (list 0 33 2 32 4 8 3 3 5))
; => 33
(max-min-diff.v2 (list 1 3 9 4 7 2 2 5 11))
; => 10
Version 3: Small version. No explicit recursion.
(define (max-min-diff.v3 nelst)
(- (apply max nelst) (apply min nelst)))
(max-min-diff.v3 (list 0 33 2 32 4 8 3 3 5))
; => 33
(max-min-diff.v3 (list 1 3 9 4 7 2 2 5 11))
; => 10
I have this board with atom T and I wanna get is position in list and sub-list
(defun board ()
"position of T: i=0 e j=9"
'(
;; 0 1 2 3 4 5 6 7 8 9
(96 25 54 89 21 8 36 14 41 T) ;; 0
(78 47 56 23 5 NIL 13 12 26 60) ;; 1
(0 27 17 83 34 93 74 52 45 80) ;; 2
(69 9 77 95 55 39 91 73 57 30) ;; 3
(24 15 22 86 1 11 68 79 76 72) ;; 4
(81 48 32 2 64 16 50 37 29 71) ;; 5
(99 51 6 18 53 28 7 63 10 88) ;; 6
(59 42 46 85 90 75 87 43 20 31) ;; 7
(3 61 58 44 65 82 19 4 35 62) ;; 8
(33 70 84 40 66 38 92 67 98 97);; 9
)
)
Function to get line and cell from board
(defun line (x board)
(nth x board))
(defun cell-board (x y board)
(nth y (line x board)))
(defun column (index board)
(cond ((not (numberp index)) nil)
((< index 0) nil)
(t (mapcar #'(lambda (line &aux (n-column (nth index line))) n-column) board))))
Function that receives the board and returns the position (i j) where the "T" is. If "T" is not on the board, NIL should be returned.
(defun find-T-position (board)
)
you can teste and see the result here https://ideone.com/GQIePI
(print "position : " (find-T-position (board)))
the result correct should be
(0 9)
The board function tries to call a literal list as-if it was a function. The quote is misplaced.
The find-t-position function has no body.
If you add more code and an actual question you will have better feedback.
Hint: either T is in current row (car board), or you need to search the board (cdr board); test often to spot errors.
(defun find-t (rows)
(let* ((col nil)
(row (position-if (lambda (r) (setf col (position t r))) rows)))
(values row col)))
Some tests:
[1]> (find-t nil)
NIL ;
NIL
[2]> (find-t '(()))
NIL ;
NIL
[3]> (find-t '((0)))
NIL ;
NIL
[4]> (find-t '((t)))
0 ;
0
[5]> (find-t '((0 t)))
0 ;
1
[6]> (find-t '((0 t 0)))
0 ;
1
[7]> (find-t '((0 0 t)))
0 ;
2
[8]> (find-t '((0 0 0)))
NIL ;
NIL
[9]> (find-t '((0 0 0)
(t 0 0)))
1 ;
0
[10]> (find-t '((0 0 0)
(t 0 t)))
1 ;
0
[11]> (find-t '((0 0 0)
(0 0 t)))
1 ;
2
[12]> (find-t '((0 0 t)
(0 0 t)))
0 ;
2
I find the answer in this question Lisp position of nested list element with children and it´s work perfectly
(defun my-position (elm tree &optional (start 0))
"find the generalized position of elm inside tree.
Parameters: elm - element to be found
tree - a list of atoms and lists in which to search the element
start - the tentative position"
(cond ((null tree) nil) ; element not present => nil
((atom (first tree)) ; if the first element is an atom, then
(if (eql elm (first tree)) ; if equal to element, found
(list start) ; return position start
;; otherwise, recur on rest of list incrementing the tentative position
(my-position elm (rest tree) (1+ start))))
;; otherwise, the first element is a list,
;; try to find it inside, with a recursive call
(t (let ((pos (my-position elm (first tree) 0)))
(if pos ; if not nil the element has been found
(cons start pos) ; return the current position followed by the position inside the list
; otherwise recur on rest of list incrementing the tentative position
(my-position elm (rest tree) (1+ start)))))))
and my function find-t-position just call the function my-position
with element 'T and the board and return the position of element 'T in
list
(defun find-T-position (board)
(my-position ('T board))
you can see the correct result https://ideone.com/DOIOoB
I have two functions in my program. The one that is commented out mods every element in a list by five. And the second function counts how many times an element appears in a list. How do I combine both of these to get my desired result of determining how many elements of a list are divisible by five?
Here is my code:
(defun divide-bye-five (lst)
(loop for x in lst collect (mod x 5)))
(defun counter (a lst)
(cond ((null lst) 0)
((equal a (car lst)) (+ 1 (counter a (cdr lst))))
(t (counter a (cdr lst)))))
(counter '0 '(0 0 0 20 0 0 0 0 0 5 31))
If you need just select all elemets in list, which are dividable by five, you can use remove-if-not.
(defun dividable-by-5 (num)
(zerop (mod num 5))
CL-USER> (remove-if-not #'dividable-by-5 '(1 2 3 10 15 30 31 40))
(10 15 30 40)
But I'm not sure, do you want select this elements, or just count them? Of course you can count them by calling length on resulting list, or of you don't need all elements, but just a number, you can use count-if.
CL-USER> (count-if #'dividable-by-5 '(1 2 3 10 15 30 31 40))
4
If you have two functions where the result of one is what you would like as the input for the second you can combine them like this:
(second-fun (first-fun first-fun-arg ...))
So specifically using your provided functions it should work doing:
(counter 0 (divide-bye-five '(1 2 3 4 5 6 7 8 9 10))) ; ==> 2
If you'd like to abstract it you make it a function:
(defun count-dividable-with-five (lst)
(counter 0 (divide-bye-five lst)))
ANSI Common Lisp.
Why I get an other answer in the last case?
(list 1 2 3 nil) ; (1 2 3 nil)
(funcall (function list) 1 2 3 nil) ; (1 2 3 nil)
(apply (function list) '(1 2 3 nil)) ; (1 2 3 nil)
(apply (function list) 1 2 3 nil) ; (1 2 3)
APPLY expects as arguments:
a function
zero ... n arguments
and then a list of arguments at the end
The function will basically be called with the result of (list* 0-arg ... n-arg argument-list)
Note that (list* '(1 2 3)) evaluates to just (1 2 3).
The arguments are called spreadable argument list in Common Lisp.
CL-USER 60 > (apply (function list) 1 2 3 nil)
(1 2 3)
CL-USER 61 > (apply (function list) (list* 1 2 3 nil))
(1 2 3)
CL-USER 62 > (apply (function list) (list* '(1 2 3)))
(1 2 3)
APPLY uses such a spreadable argument list by design. For example (... 1 2 3 '(4 5)). With FUNCALL you have to write the arguments as usual: (... 1 2 3 4 5).
APPLY has a single purpose in Common Lisp: it allows functions to be called with computed argument lists. To make that a bit more convenient, this idea of the spreadable argument list has been used. It works the same for example in Emacs Lisp.
Imagine that you have a list of arguments and you want to add two arguments in front.
CL-USER 64 > (apply '+ args)
60
CL-USER 65 > (apply '+ 1 2 args)
63
CL-USER 66 > (apply '+ (append (list 1 2) args))
63
Is it possible to tell dolist to start at (or even better after) a certain element in the given list? As I may not want to evaluate all the elements before.
If there is no way to do so, is there any other macro which might do the job?
Considering this example:
(defvar *liste* #(1 2 3 4 5 6))
(dolist (x *liste* :start-after: '4)
(FORMAT t "~a~%" x))
resulting in:
5
6
Which Lisp dialect are we talking about?
Assuming Common Lisp.
#(1 2 3 4 5 6) is not a list. It is a vector.
CL-USER > (let ((v #(1 2 3 4 5 6)))
(loop for i from 4 below (length v)
do (print (aref v i))))
5
6
NIL
With a list:
CL-USER 1 > (mapc #'print (nthcdr 4 '(1 2 3 4 5 6)))
5
6
(5 6)
What's wrong with NTHCDR?
http://www.lispworks.com/documentation/HyperSpec/Body/f_nthcdr.htm#nthcdr