I'm doing an algorithm that randomizes a TSP (array of citys) based on 1 TSP.
(do ((i 0 (+ i 1)))
((= i n-population))
(setf (aref population i) (shuffle TSP 100))
)
And as far as I know im filling up i positions of the array population with (shuffle TSP 100) that is beeing called each iteration, but the algorithm is setting all array positions and not just i position.
[Note. An earlier version of this answer contained a mistake which would badly alter the statistics of the shuffling: please check below for the corrected version and a note as to what the problem was.]
Given your code, slightly elaborated to turn it into a function:
(defun fill-array-with-something (population n-population TSP)
(do ((i 0 (+ i 1)))
((= i n-population))
(setf (aref population i) (shuffle TSP 100))))
Then each element of population from 0 to (1- n-population) will be set to the result of (shuffle TSP 100). There are then two possibilities:
(shuffle TSP 100) returns a fresh object from each call;
(shuffle TSP 100) returns the same object – probably TSP – from each call.
In the first case, each element of the array will have a distinct value. In the second case, all elements below n-population will have the same value.
Without knowing what your shuffle function does, here is an example of one which will give the latter behaviour:
(defun shuffle (vec n)
;; shuffle pairs of elts of VEC, N times.
(loop with max = (length vec)
repeat n
do (rotatef (aref vec (random max))
(aref vec (random max)))
finally (return vec)))
And we can test this:
> (let ((pop (make-array 10))
(tsp (vector 0 1 2 3 4 5 6 7 8 9 )))
(fill-array-with-something pop (length pop) tsp)
pop)
#(#(2 8 7 1 3 9 5 4 0 6) #(2 8 7 1 3 9 5 4 0 6) #(2 8 7 1 3 9 5 4 0 6)
#(2 8 7 1 3 9 5 4 0 6) #(2 8 7 1 3 9 5 4 0 6) #(2 8 7 1 3 9 5 4 0 6)
#(2 8 7 1 3 9 5 4 0 6) #(2 8 7 1 3 9 5 4 0 6) #(2 8 7 1 3 9 5 4 0 6)
#(2 8 7 1 3 9 5 4 0 6))
As you can see all the elements are mysteriously the same thing, which is because my shuffle simply returned its first argument, having modified it in place.
You can check this by either explicitly checking the result of shuffle, or by, for instance, using *print-circle* to see the sharing. The latter approach is pretty neat:
> (let ((*print-circle* t)
(pop (make-array 10))
(tsp (vector 0 1 2 3 4 5 6 7 8 9 )))
(fill-array-with-something pop (length pop) tsp)
(print pop)
(values))
#(#1=#(4 6 7 0 1 2 5 9 3 8) #1# #1# #1# #1# #1# #1# #1# #1# #1#)
And now it's immediately apparent what the problem is.
The solution is to make sure either that shuffle returns a fresh object, or to copy its result. With my shuffle this can be done like this:
(defun fill-array-with-something (population n-population tsp)
(do ((i 0 (+ i 1)))
((= i n-population))
(setf (aref population i) (shuffle (copy-seq TSP) 100))))
Note that a previous version of this answer had (copy-seq (shuffle TSP 100)): with my version of shuffle this is a serious mistake, as it means that the elements in population are related to each other but get increasingly shuffled as you go along. With (shuffle (copy-seq TSP) 100) each element gets the same amount of shuffling, independently.
And now
> (let ((*print-circle* t)
(pop (make-array 10))
(tsp (vector 0 1 2 3 4 5 6 7 8 9 )))
(fill-array-with-something pop (length pop) tsp)
(print pop)
(values))
#(#(8 3 4 1 6 9 2 5 0 7) #(8 6 5 1 3 0 4 2 9 7) #(5 0 4 7 1 6 9 3 2 8)
#(3 0 7 6 2 9 4 5 1 8) #(8 2 5 1 7 3 9 0 4 6) #(0 5 6 3 8 7 2 1 4 9)
#(4 1 3 7 8 0 5 2 9 6) #(6 9 1 5 0 7 4 2 3 8) #(2 7 5 8 0 9 6 3 4 1)
#(5 4 8 9 6 7 2 0 1 3))
I suspect that the problem is in OP function SHUFFLE which has not yet been shared; my suspicion is that SHUFFLE is shuffling the *TSP* array itself in place instead of creating a shuffled copy of that array. The POPULATION values are then all referencing the same shuffled *TSP* array.
To solve this problem, SHUFFLE should return a shuffled array instead of shuffling the array in place. Here is a function that performs a Fisher-Yates shuffle on a vector:
(defun shuffle-vector (vect)
"Takes a vector argument VECT and returns a shuffled vector."
(let ((result (make-array (length vect) :fill-pointer 0)))
(labels ((shuffle (v)
(if (zerop (length v))
result
(let* ((i (random (length v)))
(x (elt v i)))
(vector-push x result)
(shuffle (concatenate 'vector
(subseq v 0 i)
(subseq v (1+ i))))))))
(shuffle vect))))
Testing in the REPL:
CL-USER> (defvar *TSP* #("Village" "Town" "City" "Metropolis" "Megalopolis"))
*TSP*
CL-USER> (defvar *n-population* 5)
*N-POPULATION*
CL-USER> (defvar *population* (make-array *n-population*))
*POPULATION*
CL-USER> (dotimes (i *n-population*)
(setf (aref *population* i) (shuffle-vector *TSP*)))
NIL
CL-USER> *population*
#(#("Megalopolis" "City" "Metropolis" "Town" "Village")
#("Megalopolis" "Metropolis" "Town" "City" "Village")
#("City" "Megalopolis" "Town" "Village" "Metropolis")
#("City" "Megalopolis" "Village" "Metropolis" "Town")
#("Megalopolis" "Town" "Metropolis" "City" "Village"))
I want to make an experiment where I create a list of many lists of randomly generated sequences that all contain every digit 0 to 9 inclusive, that is, the generation function is to generate random numbers and place them in a list of integers while there is at least 1 digit not found in the list.
The intention for the experiment is to try to make some generalizations about things like expected number # of digits in such a function, how long can a sequence get(can my program loop indefinitely and never find that last digit?), and other interesting things(for me).
I am using PERL for the experiment.
The idea seemed simple at first, I sat down, created a list, and figured I can just make a loop that runs an arbitrary amount of times (I decided to choose 100 times), which calls a function generate_sequence(input: none, output: list of numbers that contains at least 1 of every digit) and adds it to the list.
I quickly realized that I struggle cleanly specifying what it means, pragmatically, to generate a list of numbers that contains one of every digit.
My original attempt was to make a list of digits(0..9), and as I generate numbers, I would search the list for that digit if it is in the list, and remove it. This way, it would generate numbers until the list of digits "still needed" is empty. This approach seems unappealing and can involve a lot of redundant tasks such as checking whether the digit generated is in the list of digits needed every single time a number is generated...
Is there a more elegant solution to such a problem? I am really unhappy with the way I am approaching the function.
In general, I need a function F that accepts nothing, and returns a list of randomly generated numbers that contains every digit 1..9, that is, it stops as soon as every digit from 1 to 9 inclusive is generated.
Thanks ahead of time.
Well, the problem is if you 'roll randomly' you don't actually know how many iterations you're going to need - in theory it could be infinite.
If you're doing it in perl you're probably much better off using the List::Util module and shuffle - feed it a list of elements you want to shuffle.
E.g.
#!/usr/bin/env perl
use strict;
use warnings;
use List::Util qw( shuffle );
my #shuffled = shuffle ( 0..9 );
print #shuffled;
You could reproduce this quite easily, but why bother when List::Util is core as of 5.7.3
However it does sound like you're trying to generate a list, that might contain repeats, until you hit a terminate condition.
I'm not entirely sure why, but that would be best done using a hash, and counting occurences. (And terminate when your 'keys' is complete).
E.g.:
#!/usr/bin/env perl
use strict;
use warnings;
my %seen;
my #list_of_numbers;
while ( keys %seen < 10 ) {
my $gen = int rand ( 10 );
$seen{$gen}++;
push ( #list_of_numbers, $gen );
}
print #list_of_numbers;
Note - there's actually an extremely small chance of this rolling extremely long sequences, because of the nature of 'random' - it means in theory you might have a very long 'streak' of not rolling a 6.
For bonus points in %seen you have a frequency spread of your generated numbers.
A python implementation:
from random import randint
s = set(range(10))
def f():
result = []
t = set()
while 1:
n = randint(0, 9)
result.append(n)
t.add(n)
if t == s:
return result
For example:
for i in range(10):
print(len(f()))
....:
20
34
69
22
23
25
20
29
30
32
This should work (python):
import random
nums = []
while any([ i not in set(nums) for i in range(1,11)]):
nums.append(random.randrange(1, 11, 1))
or more specific to what you are trying to do:
import random
lengths = []
for i in range(1000):
nums = []
while any([ i not in set(nums) for i in set(range(1,11))]):
nums.append(random.randrange(1, 11, 1))
lengths.append(len(nums))
This approach counts the iterations needed to fill a dictionary of digits:
import random
c = 0
d = dict()
while len(d.keys()) <10:
d[random.randint(0,9)] = 1
c += 1
print c
Wrote this before you switched to just Perl...
from random import randrange
def F():
todo = set(range(10))
nums = []
while todo:
r = randrange(10)
nums.append(r)
todo.discard(r)
return nums
>>> F()
[8, 2, 2, 3, 1, 0, 3, 9, 3, 4, 7, 4, 7, 5, 0, 9, 5, 5, 6]
Another:
def F():
done = 0
nums = []
while done < 1023:
r = randrange(10)
nums.append(r)
done |= 1 << r
return nums
In Clojure, I am keeping track of both the random list and the existing values, thus avoiding a search on the growing list.
(defn random-list [ up-to ]
(loop [ n [] tries [] ]
(if (> (count n) (dec up-to))
tries
(let [i (rand-int up-to) n-tries (conj tries i)]
(if (some #{i} n )
(recur n n-tries)
(recur (conj n i) n-tries))))))
We can define similar functions:
(defn random-list-to-10 []
(random-list 10))
(random-list-to-10)
; [3 6 9 0 8 0 5 7 3 8 1 8 4 3 4 2]
We can also take only a few random elements:
(take 5 (random-list 10))
; (6 1 0 9 5)
Here is a possible Perl implementation that counts the iterations needed to fill a hash with the 10 digts:
#!/usr/bin/perl
my $count = 0;
my %dict = ();
while (scalar keys %dict < 10) {
$dict{int(rand(10))} = 1;
$count ++;
}
print $count;
(see online demo)
In clojure (though probably not the most elegant):
(loop [n []
s (set (range 0 10))]
(if (= s (set n))
n
(recur (conj n (rand-int 10)) s)))
Sample output:
user=> (loop [n [] s (set (range 0 10))] (if (= s (set n)) n (recur (conj n (rand-int 10)) s)))
[0 6 2 8 5 2 0 0 9 3 0 3 0 1 7 5 0 4]
user=> (loop [n [] s (set (range 0 10))] (if (= s (set n)) n (recur (conj n (rand-int 10)) s)))
[2 1 7 7 3 2 8 8 4 7 5 0 1 3 0 3 0 4 0 0 3 7 3 4 5 8 1 3 8 5 3 5 5 9 4 0 2 1 2 7 8 3 9 7 8 6]
user=> (loop [n [] s (set (range 0 10))] (if (= s (set n)) n (recur (conj n (rand-int 10)) s)))
[7 1 8 3 1 1 0 6 8 4 9 7 0 0 2 7 4 0 1 1 8 8 4 3 9 8 4 2 8 3 2 8 4 6 0 9 9 7 2 3 0 3 0 4 2 4 0 5]
user=> (loop [n [] s (set (range 0 10))] (if (= s (set n)) n (recur (conj n (rand-int 10)) s)))
[9 1 9 0 9 5 3 0 3 8 4 0 1 6 3 0 1 8 0 3 8 3 5 4 3 9 8 8 8 8 2 2 8 9 9 3 9 2 5 1 1 3 4 6 3 1 4 0 2 6 7]
user=> (loop [n [] s (set (range 0 10))] (if (= s (set n)) n (recur (conj n (rand-int 10)) s)))
[4 1 5 5 5 5 2 2 5 5 3 1 5 3 5 1 4 2 4 2 3 1 4 7 1 9 3 8 0 8 4 0 9 3 4 9 9 1 8 8 0 6]
user=> (loop [n [] s (set (range 0 10))] (if (= s (set n)) n (recur (conj n (rand-int 10)) s)))
[0 4 0 9 1 8 4 8 6 6 6 9 8 4 9 0 9 3 3 7 6 1 4 3 8 1 1 4 9 5 1 4 1 2]
user=>
I am trying to make a (somehow) delicate sorting function in Lisp. I know there is lambda operator that should make my work a lot easier, but I couldn't find anything helpful so I hope you can help me.
As an input I have a nested list like this one:
((o1 10 15 20) (o2 5 14 20) (o3 7 8 8))
The output should be a nested list like this one:
((o1 1 1 1) (o2 3 2 1) (o3 2 3 3))
To be more specific, the first element from o1 is compared to the first element from o2 and o3 and the return should be it's position (in the example above, 10 is greater than 5 and 7 so it will be on the first position in the resulting list) and so on.
The highest number will get the first position.
(it's like a score function. Some students make an application and their features numbers are compared. The one with highest number of features will get the first place, but when comparing the number of different technologies used, he may get the 2nd or 3rd place).
Thanks and I hope you can help me
A little exploration is in order.
[3]> (setq a '((o1 10 15 20) (o2 5 14 20) (o3 7 8 8)))
((O1 10 15 20) (O2 5 14 20) (O3 7 8 8))
[4]> (setq b (apply #'mapcar #'list a))
((O1 O2 O3) (10 5 7) (15 14 8) (20 20 8))
[5]> (setq c (mapcar #'(lambda(x)(sort x #'>)) (cdr b)))
((10 7 5) (15 14 8) (20 20 8))
[6]> (mapcar #'(lambda(g)(cons (car g) (mapcar #'1+
(mapcar #'position (cdr g) c)))) a)
((O1 1 1 1) (O2 3 2 1) (O3 2 3 3))
Now stir, form, and bake at 190 C until ready.
When I do (/ 7 2), what should I do to get the result 3? If I do (/ 7 2.0), I get 3.5, which is as expected.
(floor 7 2)
Ref: http://rosettacode.org/wiki/Basic_integer_arithmetic#Common_Lisp
See FLOOR, CEILING and TRUNCATE in ANSI Common Lisp.
Examples (see the positive and negative numbers):
CL-USER 218 > (floor -5 2)
-3
1
CL-USER 219 > (ceiling -5 2)
-2
-1
CL-USER 220 > (truncate -5 2)
-2
-1
CL-USER 221 > (floor 5 2)
2
1
CL-USER 222 > (ceiling 5 2)
3
-1
CL-USER 223 > (truncate 5 2)
2
1
Usually for division to integer TRUNCATE is used.
You can use the floor function:
(floor 7 2)
3
1
Note that it returns multiple values, and you only need the first one. Since floor returns multiple values, that can be done with multiple-value-bind as follows:
(multiple-value-bind (q r) (floor 7 2) q)
=> 3
Edit: As Rainer notes in his comment, you can just pass the result of floor as an argument if all you need is the quotient.
[1]> (floor 7 2)
3 ;
1
[2]> (+ (floor 7 2) 5)
8
[3]>
I'm leaving the reference to multiple-value-bind in the answer, since it's an important function to be familiar with.
Use the floor function. In SBCL:
* (floor (/ 7 2))
3
1/2
Two values are returned, the integer part and the fractional part.
If I had a N lists each of length M, how could I write a nice clean function to return a single list of length M, where each element is the sum of the corresponding elements in the N lists?
(starting to learn lisp - go easy!)
This is a job for the map and apply functions. Here is a way to do it, with an EDIT suggested by Nathan Sanders:
(define (add-lists . more)
(apply map + more))
For a more matlab like syntax:
(define (piecewise func)
(lambda more
(apply map func more)))
(define pw piecewise)
((pw +) '(1 2 3 4 5) '(6 7 8 9 0))
((pw -) '(1 2 3 4 5) '(6 7 8 9 0))
((pw *) '(1 2 3 4 5) '(6 7 8 9 0))
((pw /) '(1 2 3 4 5) '(6 7 8 9 0.1))
outputs:
(7 9 11 13 5)
(-5 -5 -5 -5 5)
(6 14 24 36 0)
(1/6 2/7 3/8 4/9 50.0)
Just this works in MIT scheme.
(map + '(1 2 3) '(4 5 6) '(7 8 9))
;Value 28: (12 15 18)