no equal distribution here, neither (@ lostclimate).

another problem is that when you have a look at the output of rand5 bit-wise you'll notice that the least significant bit is set 60% of time, bit 2 and 3 both 40% of time. for our rand7 we need (beginning at the least significant bit) p0=4/7, p1=4/7, p2=4/7. now try combining the first probabilities to retreive the second ones... you'll notice that it's impossible since x/7 is a prime fraction which can't be the result of a multiplication without another seven-fraction (i don't know the word in english, i hope you get my point). (in fact, not with an and, an or or an xor, since all outcome-probabilities are just additions or multiplications).
so forget about combining the results of rand5 bitwise.

i tend to think that if you want no predictability (wrong in my example) and a perfectly equal distribution you'll need at least one if-branch.

joey.

I'd like to post a suggestion, even though I'm new to programming (and I've been doing Java which is slightly different)...

(Sorry if the syntax is wrong)

Just off the top of my head:
Code:

---

int rand7(){
int a = rand5()*rand5(); // Min = 1, Max = 25
     while (7<a<1) {
          int b = a - rand5()*rand5();
                    // If a = 1 => b = -24 to 0. If a = 25 =>  b = 0 to 24.. 
               }
          a = b;
}

---

I had to include a second variable, since I'm not sure how C++ would cope with;
a = a - rand5()*rand5();
And I'm not sure how to get that function to return, so anyway, the final answer is stored in b or a. Anyone that knows more about C++ is welcome to tidy up/comment on the code.

yes, a = a - rand5()*rand5(); is allowed in C++
and values are returned simply by:
Code:

---

return name_of_variable;

---

now, in your while loop 'a' doesn't actually get changed, so making "b = a" at the end is just going to make "b = whatever 'a' was at the beginning". and since 'a' doesn't change, if the condition was true (which it won't be, as i explain later), it would be an infinite loop.

also, you don't need to have "int" in front of each reference; in fact you aren't allowed, because then it tries to declare another integer of the same name. basically what i mean is that last line should be "b = a;", not "int b = a;".

another thing is the "7<a<1" -- that means "seven is less than 'a' which is less than one", but since 7>1, that will never be true.

additionally, i don't actually know if C++ would handle a two-part comparison like that. i think i would normally do "((1<a)&&(a<7))", but "(1<a<7)" might work. i've just never tried.

and lastly, rand5()*rand5() won't deal with every number between 1 and 25, because some are primes (7, 11, 13, 17, 19, 23), and some just won't be reached by multiplying two numbers between 1-5, even if they aren't primes (such as 18 and 24). the numbers that you DO get have unequal probability; for example, you can only obtain a 25 by getting 5*5, but you can obtain 4 by getting 1*4, 4*1, and 2*2.

and before i go, don't take this the wrong way, like i'm picking apart your code. i know you said:

Quote:

---

I'm new to programming (and I've been doing Java which is slightly different)

---

good onya for making an attempt anyway

julz

(1<a<7) gets compiled as ((1<a)<7) which is always true, since both true and false are smaller than 7.
does anyone know the "solution" nvidia wanted to see?

Ahh, thanks for the explainations guys. I knew what I was trying to get at.
I didn't realise C++ didn't like two part comparisions.

(And JulzMighty, I don't take it the wrong way. Anything helps me learn. )

I might give it another go when I've thought about it more. Maths is normally my strong point so I'm disappointed I forgot about the equal probabilities.

Edit: oh and the last line was supposed to be "a = b;" not "int b = a;.. I'll change the first post to suit.

OT:
@ JulzyMighty:

Just out of curiousity: What do you mean by "dice-syndrome"?

when you throw two dice, there are 12 possible answers, but not with equal probability, because there are actual 36 different combinations.

only 1 combination (1+1) will give "2", and only 1 combination (6+6) will give us "12", but 6 combinations (1+6, 2+5, 3+4, 4+3, 5+2, 6+1) can give us "7", so "7" is the most probable result.

i use the term "dice-syndrome" to describe adding different "equal-probability" events together to come up with an unequal-probability event, such as using rand5()+rand5().

julz

Thanks, so learned a bit more! Wasn't aware of such effect before!