So i have this piece of code:
int* get_lotto_draw() //Returns an array of six random lottery numbers 1-49
{
int min = 1;
int max = 49;
int counter = 0;
srand(time(NULL));
int *arrayPointer = malloc(6 * sizeof(int));
for(counter = 0; counter <= 5; counter++)
{
arrayPointer[counter] = rand()%(max-min)+min;
}
return arrayPointer;
}
This gives me 6 int* but sometimes these int* values can be the same. How can i compare each of them to eachother, so that if they are the same number, it will re-calculate on of the values that are equal ? Thanks for your time.
make a search in the array for same number before storing the number as,
for(counter = 0; counter <= 5; counter++)
{
int x1 = 1;
while(x1)
{
int temp = rand()%(max-min)+min;
for(i = 0; i < counter; i++)
{
if(arrayPointer[i] == temp)
{
break;
}
}
if(i == counter)
{
x1 = 0;
arrayPointer[counter] = temp;
}
}
}
The problem is that random-number generators don't know or care about history (at least, not in the sense that you do--software random number generators use recent results to generate future results). So, your choices are generally to...
Live with the possibility of duplicates. Obviously, that doesn't help you.
Review previous results and discard duplicates by repeating the draw until the value is unique. This has the potential to never end, if your random numbers get stuck in a cycle.
Enumerate the possible values, shuffle them, and pick the values off the list as needed. Obviously, that's not sensible if you have an enormous set of possible values.
In your case, I'd pick the third.
Create an array (int[]) big enough to hold one of every possible value (1-49).
Use a for loop to initialize each array value.
Use another for loop to swap each entry with a random index. If you swap a position with itself, that's fine.
Use the first few (6) elements of the array.
You can combine the second and third steps, if you want, but splitting them seems more readable and less prone to error to me.
Related
I want to create a function that can return the number distinct values present in a given array. If for eg the array is
array[5] = { 1 3 4 1 3}, the return value should be 3(3 unique numbers in array).
I've so far only got this:
int NewFucntion(int values[], int numValues){
for (i=0; i<numValues; i++){
Im a new coder/New to C language and im stuck on how to proceed. Any guidance would be much appreciated. Thanks
Add elements from the array to the std::set<T> and since the set is not allowing duplicate elements, you can then only get the number of elements from the set which gives you the number of distinct elements.
For example:
#include<set>
int NewFucntion(int values[], int numValues){
std::set<int> set;
for(int i=0; i<numValues; i++){
set.insert(values[i]);
}
return set.size();
}
int distinct(int arr[], int arr_size){
int count = arr_size;
int current;
int i, j;
for (i = 0; i < arr_size; i++){
current = arr[i];
for (j = i+1; j < arr_size; j++) // checks values after [i]th element.
if (current == arr[j])
--count; // decrease count by 1;
}
if (count >= 0)
return count;
else return 0;
}
Here's the explanation.
The array with its size is passed as an argument.
current stores the element to compare others with.
count is the number that we need finally.
count is assigned the value of size of the array (i.e we assume that all elements are unique).
(It can also be the other way round)
A for loop starts, and the first (0th) element is compared with the elements after it.
If the element reoccurs, i.e. if (current==arr[j]), then the value of count is decremented by 1 (since we expected all elements to be unique, and because it is not unique, the number of unique values is now one less than what it was initially. Hence --count).
The loops go on, and the value is decremented to whatever the number of unique elements is.
In case our array is {1,1,1,1}, then the code will print 0 instead of a negative value.
Hope that helps.
Happy coding. :)
I like wdc's answer, but I am going to give an alternative using only arrays and ints as you seam to be coding in c and wdc's answer is a c++ answer:
To do this thing, what you need to do is to go through your array as you did, and store the new numbers you go over in a different array lets call it repArray where there wont be any repetition; So every time you add something to this array you should check if the number isn't already there.
You need to create it and give it a size so why not numValues as it cannot get any longer than that. And an integers specifying how many of it's indexes are valid, in other words how many you have written to let's say validIndexes. So every time you add a NEW element to repArray you need to increment validIndexes.
In the end validIndexes will be your result.
I want to generate random numbers without repeats till all gone, then again generating random numbers with the initial dataset.
I know keeping already generated numbers in an array and loopin through them to check whether it is alredy generated or the method deducting the numbers that are generated from the array and randomize numbers with the new array.
What I want is not those methods, if there is a way that is efficient using data structures will be quite nice, if it is any other method also ok
Thanks
Say you want to generate 1,000 unique random numbers and present them to some code one at a time. When you exhaust those numbers, you want to present the same numbers again, but in a different sequence.
To generate the numbers, use a hash table. In C#, it would look like this:
const int MaxNumbers = 1000;
HashSet<int> uniqueNumbers = new HashSet<int>();
Random rnd = new Random();
// generate random numbers until there are 1000 in the hash table
while (uniqueNumbers.Count < MaxNumbers)
{
uniqueNumbers.Add(rnd.Next());
}
// At this point you have 1,000 unique random numbers
// Put them in an array
int[] myNumbers = uniqueNumbers.ToArray();
This works because the HashSet.Add method rejects duplicates. And it's very fast because lookup in the hash table is O(1).
Now, you can serve them by setting a current index at 0 and increment it every time a number is requested. Something like:
int ixCurrent = 0;
int GetNextNumber()
{
if (ixCurrent < MaxNumbers)
{
++ixCurrent;
return myNumbers[ixCurrent-1];
}
But what to do when ixCurrent runs off the end of the array? Enter the Fisher-Yates Shuffle:
// out of numbers. Shuffle the array and return the first one.
for (int i = MaxNumbers-1; i > 0; --i)
{
int j = rnd.Next(i+1);
int temp = myNumbers[i];
myNumbers[i] = myNumbers[j];
myNumbers[j] = temp;
}
ixCurrent = 1;
return myNumbers[0];
}
If you know that the numbers you want to return are within a particular range (that is, you want to return the numbers 0-999 in random order), then you just fill an array with the values 0-999 and shuffle it.
I'm not sure what language you are using, but here's some C++ code that does what you're looking for. Instead of it searching an array, it just does a direct check of a specific section of memory for a set flag and if it isn't set then the number chosen is new and printed.
The section I marked as handler is the code that is first executed when a unique number is found. Change the 10's and the 11 to different numbers if you want a larger set of random numbers but you might have to wait forever for the output.
int main(int argc, char *argv[]){
char randn[10];
char randnset[10];
int n;
int ct=0;
memset(randnset,'1',10);
memset(randn,0,10);
while (ct < 10){
srand(time(NULL));
n=rand() % 11;
if (!randn[n]){
printf("%d\n",n); // handler
randn[n]='1';
ct++;
}
}
return 0;
}
Every random generator function takes a seed value as a parameter and uses it in its internal algorithm to generate random numbers. If you want to generate the same sequence of numbers, you have to use the same seed value. As an example you can achieve this in Java like this:
int seed = 10;
Random r = new Random(seed);
for(int i=0; i<10; i++){
System.out.println(r.nextInt());
}
The output is something like this (of course it will have different results in your system):
-1157793070
1913984760
1107254586
1773446580
254270492
-1408064384
1048475594
1581279777
-778209333
1532292428
and it gives me the same results each time I execute it.
I am writing a bucket sort program in C. After I merge small buckets into a large bucket, I need to remove the -1s I padded the small buckets with.
I'm pretty new to C, so there could be a very simple solution that I am overlooking.
This was my solution, which seems to return an array with 1 trailing junk value and trailing 0s that fill the array until it is the size of the untrimmed bucket (where the desired result is a bucket without -1s, junk values, and trailing 0s).
// A function to trim a bucket of a given size to a bucket containing no -1s
int* trimBucket(int* bucket, int size)
{
int n = 0, i = 0;
int* newBucket;
// This loop is to count the number of elements between 0 and 9999
for(n = 0; n < size; n++)
{
if(bucket[n] != -1 && bucket[n] < 10000)
i++;
}
// Create a new bucket equal to the number of elements counted
// Filled with -2 to differentiate from -1s contained in the bucket array
newBucket = allocateAndInitiateOneD(i, -2);
i = 0;
for(n = 0; n < size; n++)
{
// I only want values between 0-9999 to be put into the new array
if(bucket[n] != -1 && bucket[n] < 10000)
{
newBucket[i] = bucket[n];
i++;
}
}
free(bucket); // Am I doing this right?
return newBucket;
}
allocateAndInitiateOneD function:
// A function to allocate memory for a one dimensional array and fill it with the given value
int* allocateAndInitiateOneD(int x, int initialNum)
{
int runs = 0;
int* oneArray;
oneArray = malloc(sizeof(int) * x);
for(runs = 0; runs < x; runs++)
oneArray[runs] = initialNum;
return oneArray;
}
Could someone please help me understand what I'm doing wrong and how I could get the desired result?
Thanks for the help!
edit: I am compiling and running on a Unix system. Possibly not related, but this is a multi-processed program using the MPI library (this doesn't seem to be where the problem lies).
Looks like everything is working, but you need to return the new size from this function as well, without the idea of the length of the array, you can read right off the end of your newly chosen size and it will look like garbage (1 trailing junk value & all zeroes...).
An array in C has two pieces, always. Start pointer, and size. Sometimes the size is implicit, but it needs to be there somehow, or you'll just keep reading forever.
If you need to return multiple things from a function, either:
return (one or both) via a pointer parameter
return them both through a struct
I have a big 2-D array, array[length][2]. the length= 500000.
In array[i][0]= hex number, array[i][1]= 0 or 1, which represents some information related to each hex number. Like this:
array[i][0] array[i][1]
e05f56f8 1
e045ac44 1
e05f57fc 1
e05f57b4 1
e05ff8dc 0
e05ff8ec 0
e05ff900 1
I want to get a new array which stores: the hex number,# of occurance, the sum of array[i][1] of the same hex number.
I write the code like this:
//First Sort the array according to array[][0]
int x,y,temp1,temp2;
for (x=lines_num1-2;x>=0;x--)
{
for (y=0;y<=x;y++)
{
if(array[y][0]>array[y+1][0])
{
temp1=array[y][0];
array[y][0]=array[y+1][0];
array[y+1][0]=temp1;
temp2=array[y][1];
array[y][1]=array[y+1][1];
array[y+1][1]=temp2;
}
}
}
// generate the new_array[][]
int new_array[length][3];
int n=0;
for (n=0; n<length; n++){
new_array[n][0]=0;
new_array[n][1]=0;
new_array[n][2]=0;
}
int prev = array[0][0];
new_array[0][0]=array[0][0];
new_array[0][1]=1;
new_array[0][2]=array[0][2];
for (k=1;k<length;k++)
{
if (array[k][0] == prev)
{
new_array[n][1]=new_array[n][1]+1;
new_array[n][2]=new_array[n][2]+array[k][0];
}else{
prev = array[k][0];
new_array[n+1][0]=array[k][0];
new_array[n+1][1]=new_array[n+1][1]+1;
new_array[n+1][2]=new_array[n+1][2]+array[k][0];
n++;
}
}
But the code seems not work as I expected. First the sorting is so slow. And It seems cannot generate the correct new_array. Any suggestion on how to deal with this.
Personally, I would write a hash function to index the result array with the hexadecimal value directly. Then it is simple:
struct {
unsigned int nocc;
unsigned int nsum;
} result[/* ... */];
/* calculate the results */
for (i = 0; i < LENGTH; ++i) {
int *curr = &array[i];
unsigned int index = hash(curr[0]);
result[index].nocc++;
result[index].nsum += curr[1];
}
If you want to sort your array, don't reinventing the wheel: use qsort from the standard C library.
Sorting is slow because you're using bubble sort to sort the data. Bubble sort has quadratic average complexity, which means it has to perform more then 100 billion comparisons and swaps to sort your array. For this reason, never use bubble sort. Instead, learn to use the qsort library function and apply it to your problem.
Also, your sorting code has at least one bug: when exchanging values for the second column of the array, you are getting the value with the wrong column index, [3] instead of [1].
For your scenario insertion sort is the right solution, while doing the insertion itself you could make the #count and the sum. When the sort is finished, you will have your result array as well.
The code might look something like this
int hex = 0, count = 0, sum = 0, iHole;
for (i=1; i < lines_num1 -1; i++)
{
hex = array[i][0];
count = array[i][1];
sum = array[i][2];
iHole = i
// keep moving the hole to next smaller index until A[iHole - 1] is <= item
while (iHole > 0 and array[iHole - 1][0] > hex)
{
// move hole to next smaller index
A[iHole][0] = A[iHole - 1][0];
A[iHole][1] = A[iHole - 1][1];
A[iHole][2] = A[iHole - 1][2];
iHole = iHole - 1
}
// put item in the hole
if (array[iHole][0] == hex)
{
array[iHole][1]++;
array[iHole][2] += array[iHole][0];
}
else
{
array[iHole][0] = hex;
array[iHole][1] = 1;
array[iHole][2] = hex;
}
}
So the cost of making the second array is cost of the sorting itself. O(n) best case, O(n^2) worst case, and you don't have to travel again to make the sum and count.
Remember this sort is a inplace sort. If you don't want to affect your original array that could be done as well with iHole pointing to the new array. The iHole should point to the tail of new array instead of "i"
I am trying to write C code to randomly select 10 random sites from a grid of 10x10. The way I am considering going about this is to assign every cell a random number between zero and RAND_MAX and then picking out the 10 smallest/largest values. But I have very little idea about how to actually code something like that :/
I have used pseudo-random number generators before so I can do that part.
Just generate 2 random numbers between 0 and 9 and the select the random element from the array like:
arr[rand1][rand2];
Do that 10 times in a loop. No need to make it more complicated than that.
To simplify slightly, treat the 10x10 array as an equivalent linear array of 100 elements. Now the problem becomes that of picking 10 distinct numbers from a set of 100. To get the first index, just pick a random number in the range 0 to 99.
int hits[10]; /* stow randomly selected indexes here */
hits[0] = random1(100); /* random1(n) returns a random int in range 0..n-1 */
The second number is almost as easy. Choose another number from the 99 remaining possibilities. Random1 returns a number in the continuous range 0..99; you must then map that into the broken range 0..hits[0]-1, hits[0]+1..99.
hits[1] = random1(99);
if (hits[1] == hits[0]) hits[1]++;
Now for the second number the mapping starts to get interesting because it takes a little extra work to ensure the new number is distinct from both existing choices.
hits[2] = random1(98);
if (hits[2] == hits[0]) hits[2]++;
if (hits[2] == hits[1]) hits[2]++;
if (hits[2] == hits[0]) hits[2]++; /* re-check, in case hits[1] == hits[0]+1 */
If you sort the array of hits as you go, you can avoid the need to re-check elements for uniqueness. Putting everything together:
int hits[10];
int i, n;
for (n = 0; n < 10; n++) {
int choice = random1( 100 - n ); /* pick a remaining index at random */
for (i = 0; i < n; i++) {
if (choice < hits[i]) /* find where it belongs in sorted hits */
break;
choice++; /* and make sure it is distinct *
/* need ++ to preserve uniform random distribution! */
}
insert1( hits, n, choice, i );
/* insert1(...) inserts choice at i in growing array hits */
}
You can use hits to fetch elements from your 10x10 array like this:
array[hits[0]/10][hits[0]%10]
for (int i = 0; i < 10; i++) {
// ith random entry in the matrix
arr[rand() % 10][rand() % 10];
}
Modified this from Peter Raynham's answer - I think the idea in it is right, but his execution is too complex and isn't mapping the ranges correctly:
To simplify slightly, treat the 10x10 array as an equivalent linear array of 100 elements. Now the problem becomes that of picking 10 distinct numbers from a set of 100.
To get the first index, just pick a random number in the range 0 to 99.
int hits[10]; /* stow randomly selected indexes here */
hits[0] = random1(100); /* random1(n) returns a random int in range 0..n-1 */
The second number is almost as easy. Choose another number from the 99 remaining possibilities. Random1 returns a number in the continuous range 0..99; you must then map that into the broken range 0..hits[0]-1, hits[0]+1..99.
hits[1] = random1(99);
if (hits[1] >= hits[0]) hits[1]++;
Note that you must map the complete range of hits[0]..98 to hits[0]+1..99
For another number you must compare to all previous numbers, so for the third number you must do
hits[2] = random1(98);
if (hits[2] >= hits[0]) hits[2]++;
if (hits[2] >= hits[1]) hits[2]++;
You don't need to sort the numbers! Putting everything together:
int hits[10];
int i, n;
for (n = 0; n < 10; n++) {
int choice = random1( 100 - n ); /* pick a remaining index at random */
for (i = 0; i < n; i++)
if (choice >= hits[i])
choice++;
hits[i] = choice;
}
You can use hits to fetch elements from your 10x10 array like this:
array[hits[0]/10][hits[0]%10]
If you want your chosen random cells from grid to be unique - it seems that you really want to construct random permutations. In that case:
Put cell number 0..99 into 1D array
Take some shuffle algorithm and toss that array with it
Read first 10 elements out of shuffled array.
Drawback: Running time of this algorithm increases linearly with increasing number of cells. So it may be better for practical reasons to do as #PaulP.R.O. says ...
There is a subtle bug in hjhill's solution. If you don't sort the elements in your list, then when you scan the list (inner for loop), you need to re-scan whenever you bump the choice index (choice++). This is because you may bump it into a previous entry in the list - for example with random numbers: 90, 89, 89.
The complete code:
int hits[10];
int i, j, n;
for (n = 0; n < 10; n++) {
int choice = random1( 100 - n ); /* pick a remaining index at random */
for (i = 0; i < n; i++) {
if (choice >= hits[i]) { /* find its place in partitioned range */
choice++;
for (j = 0; j < i; j++) { /* adjusted the index, must ... */
if (choice == hits[j]) { /* ... ensure no collateral damage */
choice++;
j = 0;
}
}
}
}
hits[n] = choice;
}
I know it's getting a little ugly with five levels of nesting. When selecting just a few elements (e.g., 10 of 100) it will have better performance than the sorting solution; when selecting a lot of elements (e.g., 90 of 100), performance will likely be worse than the sorting solution.