CareerCup

Hello,
does that mean for each execution it should give a different answer?

#include<stdio.h>
main()
{
int myrand5();
printf("Rand %d\n",myrand5());
}
int myrand5()
{
int a;
return (-1*((int)(&a)%5)+1);
}

using this code you can find random no. from 1-5

----------------------------------------------------------------------------------
#include<stdio.h>
main()
{
int myrand5();
printf("Rand %d\n",(myrand5()+2)%7+1);
}
int myrand5()
{
int a;
return (-1*((int)(&a)%5)+1);
}

this will give you random no.s 1-7

<pre lang="java" line="1" title="CodeMonkey10628" class="run-this">int rand15()
{
return (rand() % 5) + 1;
}

int rand17()
{
return ((rand15() + 7 + rand15()) % 7) + 1;
}

</pre><pre title="CodeMonkey10628" input="yes">
</pre>

why not below one

rand7(){
return rand5()*7/5;
}

reasoning
for rand5()=0 rand7() will be 0 too.
for rand5()=5 rand7() will be 7
rest of rand5() ; rand7() should be distributed equally.

x1=rand5();
x2=rand5();
x3 = (x1*10+x2)
finalRand = x3%7 + 1.

frequencies for each digit between 1 and 7 :: [4,3,3,4,4,3,4]

simple...make a random func that generates a rand no b/w 0 n 2 and it with rand5...

int rand17()
{
   return rand15() + abs(rand15()-3);
}

It is impossible to write a random generator to give you 1-7 number with 1/7 probability using 1/5 generator.

- write a function Rand12 which return 1 or 2 with equal prob (50/50 chance) using the 1 to 5 generator you have. Just ignore any 3, 4, or 5 until you get 1 or 2.

- draw a number 1 - 5, call it X, then if Rand12 is 2, add 5 to X. Now X is uniform 1 - 10. Keep doing this step until your 1 - 10 number is not 8, 9, or 10.

Not too efficient but it works.

rand5()+ rand5()%4 -1;

min value=1+1-1=1
max value=5+3-1=7

Lets represent 7 in base 5. It will require 2 bits. So, call Rand5() for selecting the zeroth bit and rand5() for calling 1st bit. If the result is greater than 7, discard it. Note that the first bit doesn't depend on the second bit. So, probably of each bit is random, hence the final number is also random. I think this approach works for all these sort of examples.

Please correct me if I am wrong.

http: // stackoverflow .com/ questions/137783/expand-a-random-range-from-1-5-to-1-7

in the main program write the below code.. outside the main(global variable)

static int a[5]={1,2,3,4,5};

then call this function rand1_5()..each time it will give a number between 1 to 5.

int rand1_5()
{
static int i=0;
i++;
if(i==4)
i--;

return arr[i];

}

editing ..

int rand1_5()
{
static int i=0;
i++;
if(i==4)
i--;

if(i==0)
i++;

return arr[i];

}

It is impossible for me not to understand why won't this be a good solution:

rand15()-1 -random btw. 0 and 4
(rand15()-1)/2.0 -random btw. 0 and 2
((rand15()-1)/2.0)*3 -random btw. 0 and 6
1+((rand15()-1)/2.0)*3 -random btw. 1 and 7

1. rand5() will give result from 0 to 5.
2. Probablity of getting an even number or an odd number (assuming rand5() is perfect) will be .5 each
3. 7 when converted into binary has 3 set bits.
4. call rand5() 3 times if result is even then set the corresponding bit otherwise make that bit 0.

Basically here I am trying to generate all the bits with equal probablity

psuedo code:

int multiplicationFactor = 1;
int rand7Number = 0;
for(int i = 0 ; i < 3 ; i++)
{
      int j = rand5();
      rand7Number += (j%2) *  multiplicationFactor ;
      multiplicationFactor *= 2;
}
return rand7Number ;

it could be genOne2Seven() = ( genOne2Five() -1) * (7-1) / (5-1) +1;

( ( rand15() + rand15() ) % 7 ) +1

i think this would work ...

for rand7, each number between 1 and 7 has roughly 15% probability to be selected.
for rand5, each number between 1 and 5 has 20% probability to be selected.
What about we use rand5() to control the probabilities for 1-7 to be selected?
Like for rand5, the probability of a number is not being selected is 80%.
code:
int a = rand5(), b = rand5();
if (a!=1 && b != 1) return 1;
else if (a!=1 && b != 2) return 2;
else if (a!=1 && b != 3) return 3;
else if (a!=1 && b != 4) return 4;
else if (a!=1 && b != 5) return 5;
else if (a!=2 && b != 1) return 6;
else if (a!=3 && b != 1) return 7;