i am not sure about the solutions so far since i am lazy and it takes too long to fully think them through. they all seem wrong to me at a first glance.

i came up with something pretty simple now:
Code:

---

int rand7()
{
  return (rand5()+rand5()+rand5()+rand5()+rand5() - 2) / 3;
}

---

the five rand5() give the range 5..25, after -2 the range will be 3..23, after the integer division by 3 it will be 1..7! or did i miss something?

when i woke up this morning i realised that mine's still wrong -- it still suffers from dice syndrome.

ventilator, as elegant as your solution is, it suffers from the same problem.
there are, of course, 7 different possible results, but there are 5*5*5*5*5 = 3125 different ways to get those results. 7 is not a factor of 3125. of course, that's only a little sign that something does NOT have 7 results with equal probability. if it were divisible by 7, it wouldn't necessarily have avoided the problem i'm about to explain as follows:

a "1" will be returned whenever rand5()+...rand5()-2 < 6 (if i'm right in assuming that type conversion to integer doesn't round, but instead truncates the number). this will happen when rand5()+...rand5() < 8. now, it would take way too long to count out the number of combinations. but, it does require a result between 5-7, and that's the bottom range of this whole process. 5 can only be obtained the following way:
1+1+1+1+1
6 is much more likely to be obtained:
1+1+1+1+2
1+1+1+2+1
1+1+2+1+1
1+2+1+1+1
2+1+1+1+1
and as we get closer to the middle ((25-5+1)/2+5=15.5, which should be exactly halfway between 5 and 25) we get more possible combinations, and so the number with the highest probability of being obtained is (15-2)/3 = ~4, (16-2)/3 = ~4, because its rand5()+...+rand5() results need to be between 14-16, which have the highest number of combinations attributed to them.

which makes sense, of course, because 4 is exactly halfway between 1 and 7.

this basically explains the "dice syndrome" i mentioned earlier, which we're all yet to overcome.

julz

i've got it:

rand5() + rand5()*5 has a result between 6 and 30, with only one possible combination for each answer <-- that's really nice, but what can we do with the 25 different results?

we can take just 21 of them. if we get higher than the 21st result, just do it again.
Code:

---

int rand7()
{
  int answer = 27;
  while(answer>26)
    answer = rand5() + rand5()*5;
  return (answer-5)%7+1;
}

---

i think that's it.

julz

edited to return the answer

you are right! mine doesn't give equal probabilites. here is a simple python test script:
Code:

---

import random

def rand5():
    return int((random.random() * 5) + 1)

##def rand7():
##    return (rand5()+rand5()+rand5()+rand5()+rand5() - 2) / 3

def rand7():
    answer = 27
    while answer > 26:
        answer = rand5() + rand5() * 5
    return (answer - 5) % 7 + 1

count = {}
for i in range(1000000):
    r = rand7()
    if count.has_key(r):
        count[r] += 1
    else:
        count[r] = 1

print count

---

your version works! but is a version without while or if possible?



<edit>hm... i am not sure but this one seems to work:
Code:

---

def rand7():
    return ((rand5() +
             5*rand5() +
             25*rand5() +
             125*rand5() +
             625*rand5() +
             3125*rand5() +
             15625*rand5() - 19531) % 7) + 1

---

i only found this with some trial and error though! </edit>

good question -- branches and loops are ugly!

if we can find a power of 5 that is divisible by 7, then my version could be changed a little to do without the while-loop. the trouble is that there aren't any

i'm glad it works though i have loads of study to do for tomorrow and i couldn't justify taking the time to test it. would you mind explaining how your test works? i don't know anything about python, so that last bit is gibberish to me. is it counting how many of each number is returned?

julz

i tend to think it's not possible at all since both 5 and 7 are prime numbers and rand5 can't return 0.

edit: ok, i found a way... at least all numbers are equally probable here. not very elegant, though.
Code:

---

short int rand7()
{
	short int temp = 0;

	for (int i = 0; i < 8; i ++) {
		temp += rand5()-1;
	}

        if (temp == 32) {
                return rand7();
        }

        return temp/4+1;
}

---

no, the latest two both don't give equal probabilities and i think the range of returned numbers is wrong too.

I'm must be missing something because this sounds too easy:

Code:

---

ans = (((rand5()-1) / 4f)*6f) + 1;

---

You could reduce this down, but I left it long so you can follow the logic.

Code:

---

(rand5()-1) -> random number 0-4
 / 4        -> random number 0-1
 * 6        -> random number 0-6
 + 1        -> random number 1-7

---