mpz_urandomb problem

carmen cojocaru

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Nov 26, 2011, 8:13:48 AM11/26/11

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Hi,
I am trying to generate a sequence of random numbers and I need them
to have the same length(100 bits, not 99, not 98).
But my code doesn't do what I was expecting to do. Can someone tell me
where's the problem?
Thanks in advance.
Regards.

Code:
#include <stdio.h>
#include <time.h>
#include "mpir.h"
int main()
{
int i;
mpz_t number;
gmp_randstate_t rs;
time_t seed;

FILE *f;

time(&(seed));
mpz_init(number);

gmp_randinit_default(rs);

gmp_randseed_ui(rs, (long)seed);

for(i = 0; i < 5; n++) {

mpz_urandomb(number, rs, 100);
mpz_out_str(f, 2, number);

printf("\n");
}
return 0;
}
Output:

110110101010101010100001111001100101101110111100011001000110010001100110110011110100000011010011100
1111000111100110111000111100100011110100000110110010111111001000001101111000101110000011110000100101
1010010101110010010111111000100111100111111110110110011101011011000000110111111101011110111000100110
1111010111111110111011011001100110101011110110100110001011100100000010111110001110010001001101011
11110110101101111111011100100111000011011111110100101011011101010111110001001111001100110001110001

Andrew Haley

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Nov 26, 2011, 9:39:09 AM11/26/11

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carmen cojocaru <carmenn....@yahoo.com> wrote:
> Hi,
> I am trying to generate a sequence of random numbers and I need them
> to have the same length(100 bits, not 99, not 98).
> But my code doesn't do what I was expecting to do. Can someone tell me
> where's the problem?

I don't think there is one. All of the numbers are 100 bits long, but
the output conversion routine doesn't print the leftmost zeroes.

Andrew.

carmen cojocaru

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Nov 26, 2011, 11:30:03 AM11/26/11

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On Nov 26, 4:39 pm, Andrew Haley <andre...@littlepinkcloud.invalid>
wrote:

> carmen cojocaru <carmenn.cojoc...@yahoo.com> wrote:
> > Hi,
> > I am trying to generate a sequence of random numbers and I need them
> > to have the same length(100 bits, not 99, not 98).
> > But my code doesn't do what I was expecting to do. Can someone tell me
> > where's the problem?
>
> I don't think there is one. All of the numbers are 100 bits long, but
> the output conversion routine doesn't print the leftmost zeroes.
>
> Andrew.

My apologies. I forgot about this case. Let me give you a more
appropriate output.

101010011000011001001110010111001010000101111111100011111010001101011111110110000100100111110010100
1110010100111111011101011000111101110000011001101111111011110110110101001000110000111001111011010110
1110000100011011010000001010111011101100000011110000001110111101110010101011111010110001000001100110
1001111001000110010101011010000011110000100100010000011100000111100111011100011001100001011010101
1111010101111100010100001110010010101101010100100110111101111110111100111100001101110000110010100

Taking a good look at the last string, it's obvious that the three
leftmost bits are zeroes, so why printing two zero-bits and not all
five bits since they are all zeroes?
Is this also an output conversion?
I'll be waiting for your reply. Thanks.
Carmen.

Andrew Haley

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Nov 27, 2011, 5:25:00 AM11/27/11

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carmen cojocaru <carmenn....@yahoo.com> wrote:
> On Nov 26, 4:39?pm, Andrew Haley <andre...@littlepinkcloud.invalid>

> wrote:
>> carmen cojocaru <carmenn.cojoc...@yahoo.com> wrote:
>> > Hi,
>> > I am trying to generate a sequence of random numbers and I need them
>> > to have the same length(100 bits, not 99, not 98).
>> > But my code doesn't do what I was expecting to do. Can someone tell me
>> > where's the problem?
>>

>> I don't think there is one. ?All of the numbers are 100 bits long, but

>> the output conversion routine doesn't print the leftmost zeroes.
>

> My apologies. I forgot about this case. Let me give you a more
> appropriate output.
>
> 101010011000011001001110010111001010000101111111100011111010001101011111110110000100100111110010100
> 1110010100111111011101011000111101110000011001101111111011110110110101001000110000111001111011010110
> 1110000100011011010000001010111011101100000011110000001110111101110010101011111010110001000001100110
> 1001111001000110010101011010000011110000100100010000011100000111100111011100011001100001011010101
> 1111010101111100010100001110010010101101010100100110111101111110111100111100001101110000110010100
>
> Taking a good look at the last string, it's obvious that the three
> leftmost bits are zeroes, so why printing two zero-bits and not all
> five bits since they are all zeroes?

I don't understand what you're asking.

Andrew.