random data in structure - checking for no double values

[Erich Pul]

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("\nGezog en wurden folgende Zahlen: %d %d %d %d %d %d",
ziehungT.z1,zie hungT.z2,ziehun gT.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,
--
EP

[Richard Bos]

"Erich Pul" <er*******@blac kbox.net> wrote:

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
No, ziehung() is the function. Tipp * is the type of its return value.
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;
So that won't work, then.
is there a possibility to circumvent packing an array with the values
and running through several loops?

No. Is there any problem with filling an array of 45 elements with 1-45,
shuffling it (takes only a single loop of at most 44 iterations - in
this case you need only 6) and then taking the first 6 elements of the
shuffled array?

Richard

[Erich Pul]

> > Tipp* ziehung() // Tipp* is the function

No, ziehung() is the function. Tipp * is the type of its return value.
yep, i meant that ; )

[...]

ziehungT.z4 = (rand() % 45) +1;
ziehungT.z5 = (rand() % 45) +1;
ziehungT.z6 = (rand() % 45) +1;

So that won't work, then.

nah, that works perfectly - the problem is that there *MAY* be values
that occur twice or more (sometimes they do, sometimes not)

No. Is there any problem with filling an array of 45 elements with 1-45,
shuffling it (takes only a single loop of at most 44 iterations - in
this case you need only 6) and then taking the first 6 elements of the
shuffled array?

basically not, but that is just a variation of the same problem,
because there is no proof that there won't be values that occur twice.

greetings,
--
Erich Pul

[Vladimir Oka]

Erich Pul wrote:

Don't snip attributions (who said what). This was by Richard Bos:

No. Is there any problem with filling an array of 45 elements with 1-45,
shuffling it (takes only a single loop of at most 44 iterations - in
this case you need only 6) and then taking the first 6 elements of the
shuffled array?

basically not, but that is just a variation of the same problem,
because there is no proof that there won't be values that occur twice.

I don't think you read Richard's suggestion carefully. It is guaranteed
to give all different numbers. E.g.:

0 1 2 3 4 5 6 7 8 9

<shuffle>

1 7 3 5 6 2 4 9 8 0

<take first N>

Probably the easiest way to shuffle is to swap N pairs of randomly
chosen elements.

[Erich Pul]

Vladimir Oka schrieb:

I don't think you read Richard's suggestion carefully. It is guaranteed
to give all different numbers. E.g.:

0 1 2 3 4 5 6 7 8 9

<shuffle>

1 7 3 5 6 2 4 9 8 0

<take first N>

Probably the easiest way to shuffle is to swap N pairs of randomly
chosen elements.

well, you are right - sorry - must use two eyes from now on. thanks,
i'll try that.

E

[Duncan Muirhead]

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("\nGezog en wurden folgende Zahlen: %d %d %d %d %d %d",
ziehungT.z1,zie hungT.z2,ziehun gT.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(in t 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)ran d()/(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

[Erich Pul]

Duncan Muirhead schrieb:

I think this is off topic for this news group. However I'd suggest
uhm.. why?
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(in t 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)ran d()/(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

thank you, i will focus on that for my program

E

[Vladimir Oka]

Erich Pul wrote:

Duncan Muirhead schrieb:

I think this is off topic for this news group. However I'd suggest

uhm.. why?

Because c.l.c deals in Standard C language issues only, not general
algorithm questions (that's what comp.programmin g is for). If the
question tingles the imagination sufficiently, or is easily dealt with
you can expect to get answers, though (but don't try to abuse this).

[Erich Pul]

Vladimir Oka schrieb:

Because c.l.c deals in Standard C language issues only, not general
algorithm questions (that's what comp.programmin g is for). If the
question tingles the imagination sufficiently, or is easily dealt with
you can expect to get answers, though (but don't try to abuse this).

ok, maybe i should have mentioned that i'm currently learning C, so if
i make a post here, expect the volume of my question to be quite...
unpredictable.. . : )

but thx anyway for the replies

E

[Vladimir Oka]

Richard Heathfield wrote:

Vladimir Oka said:

Probably the easiest way to shuffle is to swap N pairs of randomly
chosen elements.

I chose N = 6, and ran 1,000,000 trials. Using your method, I got a standard
deviation of 9.63, whereas by swapping N pairs of elements chosen randomly
between i and N, where i was the loop control, I got a standard deviation
of 9.15.

I said "easiest" not "best".

(I also did not mean "N" to be the number of random numbers the OP
wanted.)