NVidia Number Test

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Re: NVidia Number Test [Re: Doug] #162303 10/22/07 17:57 10/22/07 17:57
Joined: May 2002 Posts: 7,441 ventilator Senior Expert
ventilator Senior Expert Joined: May 2002 Posts: 7,441	you still will only get 5 different values. it's like TWO's try.

Re: NVidia Number Test [Re: ventilator] #162304 10/22/07 18:25 10/22/07 18:25
Joined: Jan 2003 Posts: 4,615 Cambridge Joey Expert
Joey Expert Joined: Jan 2003 Posts: 4,615 Cambridge	Quote: no, the latest two both don't give equal probabilities and i think the range of returned numbers is wrong too. why that? Code: unsigned short int rand7() { static unsigned short int temp = 0; return (temp += rand5(), temp %= 7); } Last edited by Joey; 10/22/07 18:27.

Re: NVidia Number Test [Re: Joey] #162305 10/22/07 18:33 10/22/07 18:33
Joined: May 2002 Posts: 7,441 ventilator Senior Expert
ventilator Senior Expert Joined: May 2002 Posts: 7,441	julzmighty has explained the problem with the more likely combinations. ... i had a new idea: Code: s = 1 def rand7(): global s s = 1 - s result = 0 if rand5() > (2+s): result += 1<<0 if rand5() > (2+s): result += 1<<1 if rand5() > (2+s): result += 1<<2 return result but it uses a global variable and 0 and 7 are about 15% more probable than the rest of the numbers. ... i wonder how much time the candidates have for this question at nvidia and how many come up with a correct solution. or is this some very well known problem you learn about if you study computer science at an university?

Re: NVidia Number Test [Re: ventilator] #162306 10/22/07 18:41 10/22/07 18:41
Joined: May 2002 Posts: 7,441 ventilator Senior Expert
ventilator Senior Expert Joined: May 2002 Posts: 7,441	Code: temp = 0 def rand7(): global temp temp += rand5() return temp % 7 at a first glance joey's last one looks like the most elegant solution so far.

Re: NVidia Number Test [Re: ventilator] #162307 10/22/07 18:46 10/22/07 18:46
Joined: Jan 2003 Posts: 4,615 Cambridge Joey Expert
Joey Expert Joined: Jan 2003 Posts: 4,615 Cambridge	but some numbers seem predictable, such if you get a 1 p.ex. you can be sure not to retreive a 7 or one in the next round. but if you're caching any value inside a random function you can only reduce such an effect, never extinguish it, by making the function more complex.

Re: NVidia Number Test [Re: Joey] #162308 10/22/07 19:07 10/22/07 19:07
Joined: May 2002 Posts: 7,441 ventilator Senior Expert
ventilator Senior Expert Joined: May 2002 Posts: 7,441	such predictability is bad... darn, i already thought it's THE solution after i let it run through my python test.

Re: NVidia Number Test [Re: ventilator] #162309 10/22/07 20:04 10/22/07 20:04
Joined: Jul 2000 Posts: 8,973 Bay Area Doug Senior Expert
Doug Senior Expert Joined: Jul 2000 Posts: 8,973 Bay Area	Quote: you still will only get 5 different values. it's like TWO's try. I missed the "integer" part. My updated answer (just off the top of my head): Code: ans = (rand5()+rand5()+rand5()+rand5()+rand5()+rand5()+rand5()) / 7; 6 adds, 1 div, and 7 function calls. Not a great answer, just a first draft. Conitec's Free Resources: User Magazine \|\| Docs and Tutorials \|\| WIKI

Re: NVidia Number Test [Re: ventilator] #162310 10/22/07 20:13 10/22/07 20:13
Joined: Jun 2005 Posts: 4,875 broozar Expert
broozar Expert Joined: Jun 2005 Posts: 4,875	ok, at first i solved the probabilities: each arrow from ran5() is 1/5, each from rand2() is 0.5 rand5() is known rand2() is (rand()%2)+6 so the function is (not elegant, but correct, i assume): rand7(){ switch ((rand()%7)){ //first level: determin rand5() or rand2() case 0: case 1: rand2(); break; // 2/7 prob case 2: case3: case 4: case 5: case 6: rand5(); break; // 5/7 prob }} homepage \| shameless advertising \| Linux supporter

Re: NVidia Number Test [Re: broozar] #162311 10/22/07 20:19 10/22/07 20:19
Joined: May 2002 Posts: 7,441 ventilator Senior Expert
ventilator Senior Expert Joined: May 2002 Posts: 7,441	doug, this way the probabilites aren't equal. see julzmighty's explanation. and you made a small mistake. after -7 the range is 0..28. Code: (rand5()+rand5()+rand5()+rand5()+rand5() - 2) / 3 gives this distribution after 1 million runs: {1: 6864, 2: 72170, 3: 243728, 4: 355145, 5: 243124, 6: 72255, 7: 6714} did anyone have a thought about the one i posted earlier? Code: (rand5() + 5rand5() + 25rand5() + 125rand5() + 625rand5() + 3125rand5() + 15625rand5() - 19531) % 7) + 1 is there some hidden flaw? the result seems to be correct: {1: 143263, 2: 142785, 3: 142579, 4: 142533, 5: 143609, 6: 142520, 7: 142711}

Re: NVidia Number Test [Re: ventilator] #162312 10/22/07 20:30 10/22/07 20:30
Joined: Jun 2005 Posts: 4,875 broozar Expert
broozar Expert Joined: Jun 2005 Posts: 4,875	Quote: rand5() + 5rand5() + 25rand5() + 125rand5() + 625rand5() + 3125rand5() + 15625rand5() - 19531) % 7) + 1 is nothing but Code: (very_large_nuber)%7+1 which is Code: rand()%7+1 of course that's correct, but why would you need 7 function calls, 6 multiplicatios... for such a simple task? you do not really use rand5(), it's only a factor to make your factor in the brackets bigger, you could have used rand6(), rand9() etc. without any change in the result. homepage \| shameless advertising \| Linux supporter

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