Related
I'm looking for a function in ANSI C that would randomize an array just like PHP's shuffle() does. Is there such a function or do I have to write it on my own? And if I have to write it on my own, what's the best/most performant way to do it?
My ideas so far:
Iterate through the array for, say, 100 times and exchange a random index with another random index
Create a new array and fill it with random indices from the first one checking each time if the index is already taken (performance = 0 complexity = serious)
Pasted from Asmodiel's link to Ben Pfaff's Writings, for persistence:
#include <stdlib.h>
/* Arrange the N elements of ARRAY in random order.
Only effective if N is much smaller than RAND_MAX;
if this may not be the case, use a better random
number generator. */
void shuffle(int *array, size_t n)
{
if (n > 1)
{
size_t i;
for (i = 0; i < n - 1; i++)
{
size_t j = i + rand() / (RAND_MAX / (n - i) + 1);
int t = array[j];
array[j] = array[i];
array[i] = t;
}
}
}
EDIT: And here's a generic version that works for any type (int, struct, ...) through memcpy. With an example program to run, it requires VLAs, not every compiler supports this so you might want to change that to malloc (which will perform badly) or a static buffer large enough to accommodate any type you throw at it:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
/* compile and run with
* cc shuffle.c -o shuffle && ./shuffle */
#define NELEMS(x) (sizeof(x) / sizeof(x[0]))
/* arrange the N elements of ARRAY in random order.
* Only effective if N is much smaller than RAND_MAX;
* if this may not be the case, use a better random
* number generator. */
static void shuffle(void *array, size_t n, size_t size) {
char tmp[size];
char *arr = array;
size_t stride = size * sizeof(char);
if (n > 1) {
size_t i;
for (i = 0; i < n - 1; ++i) {
size_t rnd = (size_t) rand();
size_t j = i + rnd / (RAND_MAX / (n - i) + 1);
memcpy(tmp, arr + j * stride, size);
memcpy(arr + j * stride, arr + i * stride, size);
memcpy(arr + i * stride, tmp, size);
}
}
}
#define print_type(count, stmt) \
do { \
printf("["); \
for (size_t i = 0; i < (count); ++i) { \
stmt; \
} \
printf("]\n"); \
} while (0)
struct cmplex {
int foo;
double bar;
};
int main() {
srand(time(NULL));
int intarr[] = { 1, -5, 7, 3, 20, 2 };
print_type(NELEMS(intarr), printf("%d,", intarr[i]));
shuffle(intarr, NELEMS(intarr), sizeof(intarr[0]));
print_type(NELEMS(intarr), printf("%d,", intarr[i]));
struct cmplex cmparr[] = {
{ 1, 3.14 },
{ 5, 7.12 },
{ 9, 8.94 },
{ 20, 1.84 }
};
print_type(NELEMS(intarr), printf("{%d %f},", cmparr[i].foo, cmparr[i].bar));
shuffle(cmparr, NELEMS(cmparr), sizeof(cmparr[0]));
print_type(NELEMS(intarr), printf("{%d %f},", cmparr[i].foo, cmparr[i].bar));
return 0;
}
The following code ensures that the array will be shuffled based on a random seed taken from the usec time. Also this implements the Fisher–Yates shuffle properly. I've tested the output of this function and it looks good (even expectation of any array element being the first element after shuffle. Also even expectation for being the last).
void shuffle(int *array, size_t n) {
struct timeval tv;
gettimeofday(&tv, NULL);
int usec = tv.tv_usec;
srand48(usec);
if (n > 1) {
size_t i;
for (i = n - 1; i > 0; i--) {
size_t j = (unsigned int) (drand48()*(i+1));
int t = array[j];
array[j] = array[i];
array[i] = t;
}
}
}
I’ll just echo Neil Butterworth’s answer, and point out some trouble with your first idea:
You suggested,
Iterate through the array for, say, 100 times and exchange a random index with another random index
Make this rigorous. I'll assume the existence of randn(int n), a wrapper around some RNG, producing numbers evenly distributed in [0, n-1], and swap(int a[], size_t i, size_t j),
void swap(int a[], size_t i, size_t j) {
int temp = a[i]; a[i] = a[j]; a[j] = temp;
}
which swaps a[i] and a[j].
Now let’s implement your suggestion:
void silly_shuffle(size_t n, int a[n]) {
for (size_t i = 0; i < n; i++)
swap(a, randn(n), randn(n)); // swap two random elements
}
Notice that this is not any better than this simpler (but still wrong) version:
void bad_shuffle(size_t n, int a[n]) {
for (size_t i = 0; i < n; i++)
swap(a, i, randn(n));
}
Well, what’s wrong? Consider how many permutations these functions give you: With n (or 2×_n_ for silly_shuffle) random selections in [0, n-1], the code will “fairly” select one of _n_² (or 2×_n_²) ways to shuffle the deck. The trouble is that there are n! = _n_×(n-1)×⋯×2×1 possible arrangements of the array, and neither _n_² nor 2×_n_² is a multiple of n!, proving that some permutations are more likely than others.
The Fisher-Yates shuffle is actually equivalent to your second suggestion, only with some optimizations that change (performance = 0, complexity = serious) to (performance = very good, complexity = pretty simple). (Actually, I’m not sure that a faster or simpler correct version exists.)
void fisher_yates_shuffle(size_t n, int a[n]) {
for (size_t i = 0; i < n; i++)
swap(a, i, i+randn(n-1-i)); // swap element with random later element
}
ETA: See also this post on Coding Horror.
There isn't a function in the C standard to randomize an array.
Look at Knuth - he has algorithms for the job.
Or look at Bentley - Programming Pearls or More Programming Pearls.
Or look in almost any algorithms book.
Ensuring a fair shuffle (where every permutation of the original order is equally likely) is simple, but not trivial.
Here a solution that uses memcpy instead of assignment, so you can use it for array over arbitrary data. You need twice the memory of original array and the cost is linear O(n):
void main ()
{
int elesize = sizeof (int);
int i;
int r;
int src [20];
int tgt [20];
for (i = 0; i < 20; src [i] = i++);
srand ( (unsigned int) time (0) );
for (i = 20; i > 0; i --)
{
r = rand () % i;
memcpy (&tgt [20 - i], &src [r], elesize);
memcpy (&src [r], &src [i - 1], elesize);
}
for (i = 0; i < 20; printf ("%d ", tgt [i++] ) );
}
The function you are looking for is already present in the standard C library. Its name is qsort. Random sorting can be implemented as:
int rand_comparison(const void *a, const void *b)
{
(void)a; (void)b;
return rand() % 2 ? +1 : -1;
}
void shuffle(void *base, size_t nmemb, size_t size)
{
qsort(base, nmemb, size, rand_comparison);
}
The example:
int arr[10] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
srand(0); /* each permutation has its number here */
shuffle(arr, 10, sizeof(int));
...and the output is:
3, 4, 1, 0, 2, 7, 6, 9, 8, 5
Assuming you may want to just access an array randomly instead of actually shuffling it, you can use the degenerative case of a linear congruential pseudo-random number generator
X_n+1 = (a Xn+c) mod N
where a is coprime to N
generates a random cycle over all values 0:N
Naturally you could store this sequence in an empty array.
uint32_t gcd ( uint32_t a, uint32_t b )
{
if ( a==0 ) return b;
return gcd ( b%a, a );
}
uint32_t get_coprime(uint32_t r){
uint32_t min_val = r>>1;
for(int i =0;i<r*40;i++){
uint64_t sel = min_val + ( rand()%(r-min_val ));
if(gcd(sel,r)==1)
return sel;
}
return 0;
}
uint32_t next_val(uint32_t coprime, uint32_t cur, uint32_t N)
{
return (cur+coprime)%N;
}
// Example output Array A in random order
void shuffle(float * A, uint32_t N){
uint32_t coprime = get_coprime(N);
cur = rand()%N;
for(uint32_t i = 0;i<N;i++){
printf("%f\n",A[cur]);
cur = next_val(coprime, cur, N);
}
Just run the following code first and modify it for your needs:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define arr_size 10
// shuffle array
void shuffle(int *array, size_t n) {
if (n > 1) {
for (size_t i = 0; i < n - 1; i++) {
size_t j = i + rand() / (RAND_MAX / (n - i) + 1);
int t = array[j];
array[j] = array[i];
array[i] = t;
}
}
}
// display array elements
void display_array(int *array, size_t n){
for (int i = 0; i < n; i++)
printf("%d ", array[i]);
}
int main() {
srand(time(NULL)); // this line is necessary
int numbers[arr_size] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
printf("Given array: ");
display_array(numbers, arr_size);
shuffle(numbers, arr_size);
printf("\nShuffled array: ");
display_array(numbers, arr_size);
return 0;
}
You would have something like:
You get different shuffled arrays every time you run the code:
The same answer like Nomadiq but the Random is kept simple.
The Random will be the same if you call the function one after another:
#include <stdlib.h>
#include <time.h>
void shuffle(int aArray[], int cnt){
int temp, randomNumber;
time_t t;
srand((unsigned)time(&t));
for (int i=cnt-1; i>0; i--) {
temp = aArray[i];
randomNumber = (rand() % (i+1));
aArray[i] = aArray[randomNumber];
aArray[randomNumber] = temp;
}
}
I saw the answers and I've discovered an easy way to do it
#include <stdio.h>
#include <conio.h>
#include <time.h>
int main(void){
int base[8] = {1,2,3,4,5,6,7,8}, shuffled[8] = {0,0,0,0,0,0,0,0};
int index, sorted, discart=0;
srand(time(NULL));
for(index = 0; index<8; index++){
discart = 0;
while(discart==0){
sorted = rand() % 8;
if (shuffled[sorted] == 0){
//This here is just for control of what is happening
printf("-------------\n");
printf("index: %i\n sorted: %i \n", index,sorted);
printf("-------------\n");
shuffled[sorted] = base[index];
discart= 1;
}
}
}
//This "for" is just to exibe the sequence of items inside your array
for(index=0;index<8; index++){
printf("\n----\n");
printf("%i", shuffled[index]);
}
return 0;
}
Notice that this method doesn't allow duplicated items.
And at the end you can use either numbers and letters, just replacing them into the string.
This function will shuffle array based on random seed:
void shuffle(int *arr, int size)
{
srand(time(NULL));
for (int i = size - 1; i > 0; i--)
{
int j = rand() % (i + 1);
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
}
In the code example, I have a function that takes as parameters a pointer to an int ordered_array and a pointer to int shuffled_array and a number representing the length of both arrays. It picks in each loop a random number from the ordered_array and inserts it into the shuffled array.
void shuffle_array(int *ordered_array, int *shuffled_array, int len){
int index;
for(int i = 0; i < len; i++){
index = (rand() % (len - i));
shuffled_array[i] = ordered_array[index];
ordered_array[index] = ordered_array[len-i];
}
}
I didn't see it among answers so I propose this solution if it can help anybody:
static inline void shuffle(size_t n, int arr[])
{
size_t rng;
size_t i;
int tmp[n];
int tmp2[n];
memcpy(tmp, arr, sizeof(int) * n);
bzero(tmp2, sizeof(int) * n);
srand(time(NULL));
i = 0;
while (i < n)
{
rng = rand() % (n - i);
while (tmp2[rng] == 1)
++rng;
tmp2[rng] = 1;
arr[i] = tmp[rng];
++i;
}
}
I am creating a function which can compare two arrays. It returns 1 when they are the same and return 0 when they are not.
It required the program run as linear time, so i cannot use a for-for loop to compare it. Any suggestions for me?
Examples of arrays for which scrambled should return 1:
a = {10,15,20}, b = {10,15,20}
a = {1,2,3,4,5}, b = {5,3,4,2,1}
a = {}, b = {} (i.e. len = 0)
a = {2,1,3,4,5}, b = {1,2,4,3,5}
Examples of arrays for which scrambled should return 0:
a = {1,1}, b = {1,2}
a = {10,15,20}, b = {10,15,21}
a = {1,2,3,4,5}, b = {5,3,4,2,2}
If you can specify a maximum value for the array elements, you can compare them in linear time pretty easily, by just looping through each one and counting the values which are present, like so:
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#define MAX_ARRAY_VALUE 100
bool compare_arrays(int * arr1, size_t arr1_size,
int * arr2, size_t arr2_size)
{
/* Set up array to count elements */
int * table = calloc(MAX_ARRAY_VALUE + 1, sizeof * table);
if ( !table ) {
perror("couldn't allocate memory");
exit(EXIT_FAILURE);
}
/* Increment index if element is present in first array... */
for ( size_t i = 0; i < arr1_size; ++i ) {
table[arr1[i]]++;
}
/* ...and decrement index if element is present in second array. */
for ( size_t i = 0; i < arr2_size; ++i ) {
table[arr2[i]]--;
}
/* If any elements in array are not zero, then arrays are not equal */
for ( size_t i = 0; i < MAX_ARRAY_VALUE + 1; ++i ) {
if ( table[i] ) {
free(table);
return false;
}
}
free(table);
return true;
}
int main(void) {
int a1[] = {10, 20, 30, 10};
int a2[] = {20, 10, 10, 30};
int a3[] = {1, 4, 5};
int a4[] = {1, 3, 5};
if ( compare_arrays(a1, 4, a2, 4) ) {
puts("a1 and a2 are equal"); /* Should print */
}
else {
puts("a1 and a2 are not equal"); /* Should not print */
}
if ( compare_arrays(a3, 3, a4, 3) ) {
puts("a3 and a4 are equal"); /* Should not print */
}
else {
puts("a3 and a4 are not equal"); /* Should print */
}
if ( compare_arrays(a1, 4, a4, 3) ) {
puts("a1 and a4 are equal"); /* Should not print */
}
else {
puts("a1 and a4 are not equal"); /* Should print */
}
return 0;
}
which outputs:
paul#horus:~/src/sandbox$ ./lincmp
a1 and a2 are equal
a3 and a4 are not equal
a1 and a4 are not equal
paul#horus:~/src/sandbox$
Without specifying the maximum value, you can loop through each array and find the maximum. It'll still be linear time, but without an upper bound you might end up with a huge index table.
Because the comparison of the two arrays is independent of the order of the elements, both must be sorted before they can be compared. Because of this, you can't do this in linear time. The best you can do is O(n log n), as that is the best order of most sorting algorithms.
My C is pretty rusty so the may be some memory problem with the script below. However, the basic task is to sort the 2 arrays, compare them element by element. If they all match, scramble should return 1, other it's 0.
The script below works with GCC, haven't tested it with any other compiler. It compares arrays of equal length. The case of unequal length is left for the OP as an minor exercise.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int compare_ints(const void * a , const void * b)
{
const int * ia = (const int *)a;
const int * ib = (const int *)b;
return *ia > *ib;
}
int scramble(const int * a, const int * b, unsigned int len)
{
int * sorted_a = malloc(len * sizeof(int));
int * sorted_b = malloc(len * sizeof(int));
memcpy(sorted_a, a, len * sizeof(int));
memcpy(sorted_b, b, len * sizeof(int));
qsort(sorted_a, len, sizeof(int), compare_ints);
qsort(sorted_b, len, sizeof(int), compare_ints);
for (int i = 0; i < len; i++)
{
if (sorted_a[i] != sorted_b[i])
{
free(sorted_a);
free(sorted_b);
return 0;
}
}
free(sorted_a);
free(sorted_b);
return 1;
}
int main (int argc, char const *argv[])
{
int a[3] = {20, 10, 15};
int b[3] = {10, 15, 20};
int is_equal = scramble(a, b, 3);
printf("is_equal = %i\n", is_equal);
return 0;
}
FYI: you can't do it in linear time. qsort has O(n log n).
At a certain extent, you can obtain the sum of all the hashed values of the first array, and compare it to the sum of the hashed values of the second array. It'll work, but I don't know precise it is.
Here is my attempt, all my tests have given positive results so far :
#include <stdio.h>
unsigned int hash(unsigned int x) {
x = ((x >> 16) ^ x) * 0x45d9f3b;
x = ((x >> 16) ^ x) * 0x45d9f3b;
x = ((x >> 16) ^ x);
return x;
}
int compareArrays(int *arr1, int *arr2, int size) {
if (size == 0) return 1;
unsigned long sum1 = 0;
unsigned long sum2 = 0;
for (int i = 0; i < size; ++i) {
sum1 += hash(arr1[i]);
}
for (int i = 0; i < size; ++i)
sum2 += hash(arr2[i]) ;
return sum1 == sum2;
}
int main(void) {
int a[] = {1,2,3,4,5,6,7,8,255};
int b[] = {1,2,3,4,5,6,7,8,9};
int size = 9;
printf("Are they the same? %s.\n", compareArrays(a, b, size) ? "yes" : "no");
return 0;
}
You can use memcmp function with this format:
int memcmp ( const void * ptr1, const void * ptr2, size_t num );
For example;
/* memcmp example */
#include <stdio.h>
#include <string.h>
int main ()
{
char buffer1[] = "abcde";
char buffer2[] = "abcde";
int n;
n=memcmp ( buffer1, buffer2, sizeof(buffer1) );
if (n>0) printf ("'%s' is greater than '%s'.\n",buffer1,buffer2);
else if (n<0) printf ("'%s' is less than '%s'.\n",buffer1,buffer2);
else printf ("'%s' is the same as '%s'.\n",buffer1,buffer2);
return 0;
}
output is:
'abcde' is same as 'abcde'.
Given an array of numbers, print the each and every range available.
For example
Array : 9, 3, 5, 7, 4, 8, 1
Output: 1, 3-5, 7-9
Note: Please execute this problem without using an additional array.
How do i proceed?
*
#include<stdio.h>
int main()
{
int a[]={9,8,8,7,6,5,14};
int n= sizeof(a) / sizeof(a[0]);
int i,j;
int temp;
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if(a[i]>a[j])
{
temp=a[i];
a[i]=a[j];
a[j]=temp;
}
}
}
}
*
1st i will sort in ascending order, i don't know what to do next?
P.S : I am coding this in C.
The next step is to identify sequences. Try the following loop (not fully debugged):
first= next= a[0];
for (i=1; i<n; i++) {
if (a[i] > next+1) {
if (next>first)
printf("%d-%d,", first, next);
else printf("%d,", first);
first= next= a[i];
}
else next++;
}
If you may to change the original array that is if you may to sort it then the program can look like
#include <stdlib.h>
#include <stdio.h>
int cmp( const void *lhs, const void *rhs )
{
int a = *( const int * )lhs;
int b = *( const int * )rhs;
return ( b < a ) - ( a < b );
}
int main()
{
int a[] = { 9, 8, 8, 7, 6, 5, 14 };
const size_t N = sizeof( a ) / sizeof( *a );
qsort( a, N, sizeof( int ), cmp );
/*
for ( size_t i = 0; i < N; i++ ) printf( "%d ", a[i] );
printf( "\n" );
*/
int *p = a;
int *start = a, *end = a;
do
{
if ( ++p == a + N || *p != *end + 1 )
{
printf( "{ %d", *start );
start == end ? printf( " }\n" ) : printf( ", %d }\n", *end );
start = end = p;
}
else
{
end = p;
}
} while ( p != a + N );
}
The program output is
{ 5, 8 }
{ 8, 9 }
{ 14 }
I wrote a simple, readable function for you, take a look:
void printRange(int sortedArray[], int len) {
int i, current, next, printStart, printEnd, startIndex = 0;
bool print = false;
for (i = 0; i < len; i++) {
printStart = sortedArray[startIndex];
printEnd = sortedArray[i];
current = sortedArray[i];
if(i < len -1) {
next = sortedArray[i + 1];
} else
next = current;
if (next - current != 1) {
startIndex = i + 1;
print = true;
}
if (print) {
if (printStart - printEnd == 0) {
printf("%d,", printStart);
} else {
printf("%d-%d,", printStart, printEnd);
}
print = false;
}
}
}
Run live.
Note, for good understanding variable current is declared whereas current and printEnd is same. You may replace current by printEnd.
There are already some pretty good answers for this task in here but maybe the sorting part in the beginning is worth to talk a little bit more about. Especially if you need something like that in school, university or in a job interview.
The easiest sorting technique/algorithm would be something like BubbleSort which can easily be implemented with 2 for loops.
void BubbleSort (int a[], int length)
{
int i, j, temp;
for (i = 0; i < length; i++)
{
for (j = 0; j < length - i - 1; j++)
{
if (a[j + 1] < a[j])
{
temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
}
Source and more information
The best way of sorting such arrays with integers (or any kind of number) is QuickSort. The algorithm is pretty advanced but if you watch a good video on Youtube or read this article you definitely know how it works.
void quick(int array[], int start, int end){
if(start < end){
int l=start+1, r=end, p = array[start];
while(l<r){
if(array[l] <= p)
l++;
else if(array[r] >= p)
r--;
else
swap(array[l],array[r]);
}
if(array[l] < p){
swap(array[l],array[start]);
l--;
}
else{
l--;
swap(array[l],array[start]);
}
quick(array, start, l);
quick(array, r, end);
}
}
Source and more information
Note: QuickSort uses a technique called recursion. If you are not familiar which that you can have a look here:
In computer science, recursion is a method of solving a problem where the solution depends on solutions to smaller instances of the same problem. Such problems can generally be solved by iteration, but this needs to identify and index the smaller instances at programming time. Recursion solves such recursive problems by using functions that call themselves from within their own code. The approach can be applied to many types of problems, and recursion is one of the central ideas of computer science.
Source and more information
We have an array, int array={1,2,3};
Display all the possible permutations which will be
{1,3,2} ,
{2,1,3} ,
{2,3,1} ,
{3,1,2} ,
{3,2,1} etc.
all n! possibilities.
I know both ways direct recursive and backtracking too.
Is there any better way to do the same ?
Thank You.
You can use the methods described here.
You can modify his algo slightly for your need:
#include <stdio.h>
void print(const int *v, const int size)
{
if (v != 0) {
for (int i = 0; i < size; i++) {
printf("%c", i ? ',':'{');
printf("%d", v[i] );
}
printf("}\n");
}
} // print
void permute(int *v, const int start, const int n)
{
if (start == n-1) {
print(v, n);
}
else {
for (int i = start; i < n; i++) {
int tmp = v[i];
v[i] = v[start];
v[start] = tmp;
permute(v, start+1, n);
v[start] = v[i];
v[i] = tmp;
}
}
}
int main(void)
{
int v[] = {1, 2, 3};
permute(v, 0, sizeof(v)/sizeof(int));
}
Live example here
As described in other answer, C++ stl library provides an implmentation of next_permutation. You can peep inside stl code and make necessary changes to port it to C code.
I'm looking for a function in ANSI C that would randomize an array just like PHP's shuffle() does. Is there such a function or do I have to write it on my own? And if I have to write it on my own, what's the best/most performant way to do it?
My ideas so far:
Iterate through the array for, say, 100 times and exchange a random index with another random index
Create a new array and fill it with random indices from the first one checking each time if the index is already taken (performance = 0 complexity = serious)
Pasted from Asmodiel's link to Ben Pfaff's Writings, for persistence:
#include <stdlib.h>
/* Arrange the N elements of ARRAY in random order.
Only effective if N is much smaller than RAND_MAX;
if this may not be the case, use a better random
number generator. */
void shuffle(int *array, size_t n)
{
if (n > 1)
{
size_t i;
for (i = 0; i < n - 1; i++)
{
size_t j = i + rand() / (RAND_MAX / (n - i) + 1);
int t = array[j];
array[j] = array[i];
array[i] = t;
}
}
}
EDIT: And here's a generic version that works for any type (int, struct, ...) through memcpy. With an example program to run, it requires VLAs, not every compiler supports this so you might want to change that to malloc (which will perform badly) or a static buffer large enough to accommodate any type you throw at it:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
/* compile and run with
* cc shuffle.c -o shuffle && ./shuffle */
#define NELEMS(x) (sizeof(x) / sizeof(x[0]))
/* arrange the N elements of ARRAY in random order.
* Only effective if N is much smaller than RAND_MAX;
* if this may not be the case, use a better random
* number generator. */
static void shuffle(void *array, size_t n, size_t size) {
char tmp[size];
char *arr = array;
size_t stride = size * sizeof(char);
if (n > 1) {
size_t i;
for (i = 0; i < n - 1; ++i) {
size_t rnd = (size_t) rand();
size_t j = i + rnd / (RAND_MAX / (n - i) + 1);
memcpy(tmp, arr + j * stride, size);
memcpy(arr + j * stride, arr + i * stride, size);
memcpy(arr + i * stride, tmp, size);
}
}
}
#define print_type(count, stmt) \
do { \
printf("["); \
for (size_t i = 0; i < (count); ++i) { \
stmt; \
} \
printf("]\n"); \
} while (0)
struct cmplex {
int foo;
double bar;
};
int main() {
srand(time(NULL));
int intarr[] = { 1, -5, 7, 3, 20, 2 };
print_type(NELEMS(intarr), printf("%d,", intarr[i]));
shuffle(intarr, NELEMS(intarr), sizeof(intarr[0]));
print_type(NELEMS(intarr), printf("%d,", intarr[i]));
struct cmplex cmparr[] = {
{ 1, 3.14 },
{ 5, 7.12 },
{ 9, 8.94 },
{ 20, 1.84 }
};
print_type(NELEMS(intarr), printf("{%d %f},", cmparr[i].foo, cmparr[i].bar));
shuffle(cmparr, NELEMS(cmparr), sizeof(cmparr[0]));
print_type(NELEMS(intarr), printf("{%d %f},", cmparr[i].foo, cmparr[i].bar));
return 0;
}
The following code ensures that the array will be shuffled based on a random seed taken from the usec time. Also this implements the Fisher–Yates shuffle properly. I've tested the output of this function and it looks good (even expectation of any array element being the first element after shuffle. Also even expectation for being the last).
void shuffle(int *array, size_t n) {
struct timeval tv;
gettimeofday(&tv, NULL);
int usec = tv.tv_usec;
srand48(usec);
if (n > 1) {
size_t i;
for (i = n - 1; i > 0; i--) {
size_t j = (unsigned int) (drand48()*(i+1));
int t = array[j];
array[j] = array[i];
array[i] = t;
}
}
}
I’ll just echo Neil Butterworth’s answer, and point out some trouble with your first idea:
You suggested,
Iterate through the array for, say, 100 times and exchange a random index with another random index
Make this rigorous. I'll assume the existence of randn(int n), a wrapper around some RNG, producing numbers evenly distributed in [0, n-1], and swap(int a[], size_t i, size_t j),
void swap(int a[], size_t i, size_t j) {
int temp = a[i]; a[i] = a[j]; a[j] = temp;
}
which swaps a[i] and a[j].
Now let’s implement your suggestion:
void silly_shuffle(size_t n, int a[n]) {
for (size_t i = 0; i < n; i++)
swap(a, randn(n), randn(n)); // swap two random elements
}
Notice that this is not any better than this simpler (but still wrong) version:
void bad_shuffle(size_t n, int a[n]) {
for (size_t i = 0; i < n; i++)
swap(a, i, randn(n));
}
Well, what’s wrong? Consider how many permutations these functions give you: With n (or 2×_n_ for silly_shuffle) random selections in [0, n-1], the code will “fairly” select one of _n_² (or 2×_n_²) ways to shuffle the deck. The trouble is that there are n! = _n_×(n-1)×⋯×2×1 possible arrangements of the array, and neither _n_² nor 2×_n_² is a multiple of n!, proving that some permutations are more likely than others.
The Fisher-Yates shuffle is actually equivalent to your second suggestion, only with some optimizations that change (performance = 0, complexity = serious) to (performance = very good, complexity = pretty simple). (Actually, I’m not sure that a faster or simpler correct version exists.)
void fisher_yates_shuffle(size_t n, int a[n]) {
for (size_t i = 0; i < n; i++)
swap(a, i, i+randn(n-1-i)); // swap element with random later element
}
ETA: See also this post on Coding Horror.
There isn't a function in the C standard to randomize an array.
Look at Knuth - he has algorithms for the job.
Or look at Bentley - Programming Pearls or More Programming Pearls.
Or look in almost any algorithms book.
Ensuring a fair shuffle (where every permutation of the original order is equally likely) is simple, but not trivial.
Here a solution that uses memcpy instead of assignment, so you can use it for array over arbitrary data. You need twice the memory of original array and the cost is linear O(n):
void main ()
{
int elesize = sizeof (int);
int i;
int r;
int src [20];
int tgt [20];
for (i = 0; i < 20; src [i] = i++);
srand ( (unsigned int) time (0) );
for (i = 20; i > 0; i --)
{
r = rand () % i;
memcpy (&tgt [20 - i], &src [r], elesize);
memcpy (&src [r], &src [i - 1], elesize);
}
for (i = 0; i < 20; printf ("%d ", tgt [i++] ) );
}
The function you are looking for is already present in the standard C library. Its name is qsort. Random sorting can be implemented as:
int rand_comparison(const void *a, const void *b)
{
(void)a; (void)b;
return rand() % 2 ? +1 : -1;
}
void shuffle(void *base, size_t nmemb, size_t size)
{
qsort(base, nmemb, size, rand_comparison);
}
The example:
int arr[10] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
srand(0); /* each permutation has its number here */
shuffle(arr, 10, sizeof(int));
...and the output is:
3, 4, 1, 0, 2, 7, 6, 9, 8, 5
Assuming you may want to just access an array randomly instead of actually shuffling it, you can use the degenerative case of a linear congruential pseudo-random number generator
X_n+1 = (a Xn+c) mod N
where a is coprime to N
generates a random cycle over all values 0:N
Naturally you could store this sequence in an empty array.
uint32_t gcd ( uint32_t a, uint32_t b )
{
if ( a==0 ) return b;
return gcd ( b%a, a );
}
uint32_t get_coprime(uint32_t r){
uint32_t min_val = r>>1;
for(int i =0;i<r*40;i++){
uint64_t sel = min_val + ( rand()%(r-min_val ));
if(gcd(sel,r)==1)
return sel;
}
return 0;
}
uint32_t next_val(uint32_t coprime, uint32_t cur, uint32_t N)
{
return (cur+coprime)%N;
}
// Example output Array A in random order
void shuffle(float * A, uint32_t N){
uint32_t coprime = get_coprime(N);
cur = rand()%N;
for(uint32_t i = 0;i<N;i++){
printf("%f\n",A[cur]);
cur = next_val(coprime, cur, N);
}
Just run the following code first and modify it for your needs:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define arr_size 10
// shuffle array
void shuffle(int *array, size_t n) {
if (n > 1) {
for (size_t i = 0; i < n - 1; i++) {
size_t j = i + rand() / (RAND_MAX / (n - i) + 1);
int t = array[j];
array[j] = array[i];
array[i] = t;
}
}
}
// display array elements
void display_array(int *array, size_t n){
for (int i = 0; i < n; i++)
printf("%d ", array[i]);
}
int main() {
srand(time(NULL)); // this line is necessary
int numbers[arr_size] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
printf("Given array: ");
display_array(numbers, arr_size);
shuffle(numbers, arr_size);
printf("\nShuffled array: ");
display_array(numbers, arr_size);
return 0;
}
You would have something like:
You get different shuffled arrays every time you run the code:
The same answer like Nomadiq but the Random is kept simple.
The Random will be the same if you call the function one after another:
#include <stdlib.h>
#include <time.h>
void shuffle(int aArray[], int cnt){
int temp, randomNumber;
time_t t;
srand((unsigned)time(&t));
for (int i=cnt-1; i>0; i--) {
temp = aArray[i];
randomNumber = (rand() % (i+1));
aArray[i] = aArray[randomNumber];
aArray[randomNumber] = temp;
}
}
I saw the answers and I've discovered an easy way to do it
#include <stdio.h>
#include <conio.h>
#include <time.h>
int main(void){
int base[8] = {1,2,3,4,5,6,7,8}, shuffled[8] = {0,0,0,0,0,0,0,0};
int index, sorted, discart=0;
srand(time(NULL));
for(index = 0; index<8; index++){
discart = 0;
while(discart==0){
sorted = rand() % 8;
if (shuffled[sorted] == 0){
//This here is just for control of what is happening
printf("-------------\n");
printf("index: %i\n sorted: %i \n", index,sorted);
printf("-------------\n");
shuffled[sorted] = base[index];
discart= 1;
}
}
}
//This "for" is just to exibe the sequence of items inside your array
for(index=0;index<8; index++){
printf("\n----\n");
printf("%i", shuffled[index]);
}
return 0;
}
Notice that this method doesn't allow duplicated items.
And at the end you can use either numbers and letters, just replacing them into the string.
This function will shuffle array based on random seed:
void shuffle(int *arr, int size)
{
srand(time(NULL));
for (int i = size - 1; i > 0; i--)
{
int j = rand() % (i + 1);
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
}
In the code example, I have a function that takes as parameters a pointer to an int ordered_array and a pointer to int shuffled_array and a number representing the length of both arrays. It picks in each loop a random number from the ordered_array and inserts it into the shuffled array.
void shuffle_array(int *ordered_array, int *shuffled_array, int len){
int index;
for(int i = 0; i < len; i++){
index = (rand() % (len - i));
shuffled_array[i] = ordered_array[index];
ordered_array[index] = ordered_array[len-i];
}
}
I didn't see it among answers so I propose this solution if it can help anybody:
static inline void shuffle(size_t n, int arr[])
{
size_t rng;
size_t i;
int tmp[n];
int tmp2[n];
memcpy(tmp, arr, sizeof(int) * n);
bzero(tmp2, sizeof(int) * n);
srand(time(NULL));
i = 0;
while (i < n)
{
rng = rand() % (n - i);
while (tmp2[rng] == 1)
++rng;
tmp2[rng] = 1;
arr[i] = tmp[rng];
++i;
}
}