Re: random data in structure - checking for no double values

On Thu, 22 Jun 2006 01:00:45 -0700, Erich Pul wrote:

hi!

i got a structure, which should be filled with random integer values
(it is in fact a generator for numbers like in a lotto), but these
values must not be recurring, so no double occurrences are desired.

my code:

<code>
Tipp* ziehung() // Tipp* is the function
{
Node *bewK; // a node that is my index
Tipp ziehungT; // a new struct for storing the random values
int x;

x = 0;
srand(time(NULL)); // init for the rand()
while(x<=20) // for testing, gimme 20 values
{
ziehungT.z1 = (rand() % 45) +1; // data input in the
struct's members
ziehungT.z2 = (rand() % 45) +1;
ziehungT.z3 = (rand() % 45) +1;
ziehungT.z4 = (rand() % 45) +1;
ziehungT.z5 = (rand() % 45) +1;
ziehungT.z6 = (rand() % 45) +1;

printf("\nGezogen wurden folgende Zahlen: %d %d %d %d %d %d",
ziehungT.z1,ziehungT.z2,ziehungT.z3,ziehungT.z4,ziehungT.z5,ziehungT.z6);
x++;
}
}

</code>

is there a possibility to circumvent packing an array with the values
and running through several loops?

TIA,
I think this is off topic for this news group. However I'd suggest
creating an array holding 1..45, shuffling it (search for Knuth
Shuffling algorithm) and then taking the last 6 entries from the array.
For example:
void KnuthShuffle(int n, int* deck)
{ int r;
while( --n>0)
{ /* r = "random" number between 0 and n inclusive */
/* swap deck[n] and deck[r] */
}
}
If speed is very important, then you could just go round the loop 6 times,
because the last 6 entries won't change after that.

By the way, generating a random integer between 1 and 45 is better (if
somewhat more slowly) done by
1 + 45*((double)rand()/(RAND_MAX+1.0)) than by rand() % 45) +1.
The latter will be more sensitive to the low bits being highly
correlated between calls to rand().
Duncan



.

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