@ sohinee basak
it won't matter much . +1 or -1 will get u right range.. : ...
On Thu, Jun 21, 2012 at 9:54 PM, sohinee basak <sohine...@gmail.com> wrote:
> i have a doubt ... is the range of rand7 from 0 to 6 or from 1 to 7
>
> On Fri, Jun 22, 2012 at 10:13 AM, Anant Sharma <black.b...@gmail.com> wrote:
>>
>>
>> The reason why finding a solution to this question is difficult is because
>> the combinations of rand5() function which we use ( eg. rand5() + rand5()%2
>> ) leads to some numbers being generated more frequently than others. So the
>> problem would be solved if we could find a function which leads to each
>> output being generated an equal number of times ( once, or maybe more. )
>>
>> 5 * rand5() + rand5()
>>
>> Looking at this function, it is easy to see that each value from 0 to 24
>> will be generated only once for every respective value that first and second
>> rand5() returns.
>>
>> Suppose the first rand5() returns 0, then the second rand5() can return
>> any value from 0 - 4. Now see that no value from 0-4 could be generated by
>> any other combination. When the value returned by the second rand5() is 1,
>> corresponding to the value returned by second rand5(), we could get 5 - 9.
>>
>>
>> Now we have each number from 0-24 appearing once. But simply returning the
>> mod of this value with 7 wouldn't lead to equal probability distribution as
>> the values 0 - 3 would be returned more times than 4-6. So to fix this
>> inequality, we ignore when the value of the function is more than 20. Now
>> doing mod with 7 will result in every number 0 - 6 being returned with equal
>> probability.
>>
>> We could even have ignored the values above 6, or the values above 13, the
>> only difference being that it would take more number of calls to return a
>> rand7() result.
>>
>> This way is non-deterministic i.e. we cannot say how many times the
>> function will be called before we get the rand7() result, but we can be
>> sure that whenever the function returns a value, it would have been as
>> probable as any other value from 0 - 6. Taking the mod of the result of the
>> function, there would be 3 ways in which each digit 0 through 6 can be
>> generated.
>>
>>
>> Implementation:
>>
>> int rand7 ( ) {
>> int num = 5 * rand5 ( ) + rand ( ) ;
>> if ( num > 20 )
>> return rand7 ( ) ;
>> else
>> retutrn num % 7 ;
>> }
>>
>> On Thu, Jun 21, 2012 at 12:44 PM, Abhishek Sharma <abhi120...@gmail.com>
>> wrote:
>>>
>>> @umer
>>> it's not random,but biased
>>>
>>>
>>> On Wed, Jun 20, 2012 at 11:16 AM, Umer Farooq <the.um...@gmail.com>
>>> wrote:
>>>>
>>>> Hmmm, true. That's what I expected in this solution. Similarly, we can
>>>> get 3 by (3,3) (1,2)
>>>>
>>>> However, did you take a look at the other solution I proposed? I guess
>>>> that solves the problem to some extent.
>>>>
>>>>
>>>> On Tue, Jun 19, 2012 at 7:18 PM, Sourabh Singh
>>>> <singhsourab...@gmail.com> wrote:
>>>>>
>>>>> @Umer and Navin :
>>>>> 1 is generated by (1,3) only.
>>>>> 2 is generated by (1,1) and (2,3).
>>>>>
>>>>> so given solution is wrong
>>>>>
>>>>>
>>>>> On Tue, Jun 19, 2012 at 5:17 AM, Sourabh Singh
>>>>> <singhsourab...@gmail.com> wrote:
>>>>>>
>>>>>> @ ALL
>>>>>>
>>>>>> please. post along with your method .
>>>>>> proof than it make's equal distribution over the given range.
>>>>>>
>>>>>> On Tue, Jun 19, 2012 at 4:47 AM, Navin Kumar
>>>>>> <algorithm.i...@gmail.com> wrote:
>>>>>>>
>>>>>>> @Umer:
>>>>>>>
>>>>>>> rand(5) + (rand(5)%2): => it will never give 6 because for rand(7)
>>>>>>> range will be 0-6.
>>>>>>> So better try this: rand(5) + (rand(5)%3).
>>>>>>>
>>>>>>>
>>>>>>> On Tue, Jun 19, 2012 at 2:45 PM, Umer Farooq <the.um...@gmail.com>
>>>>>>> wrote:
>>>>>>>>
>>>>>>>> rand(5) + (rand(5)%2);
>>>>>>>>
>>>>>>>>
>>>>>>>> On Tue, Jun 19, 2012 at 12:30 PM, Sourabh Singh
>>>>>>>> <singhsourab...@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>> @ sry
>>>>>>>>> condition should be:
>>>>>>>>>
>>>>>>>>> if(20*prob <= 500/7) :-)
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> On Tue, Jun 19, 2012 at 12:26 AM, Sourabh Singh
>>>>>>>>> <singhsourab...@gmail.com> wrote:
>>>>>>>>>>
>>>>>>>>>> @ALL
>>>>>>>>>>
>>>>>>>>>> Given a random number generator say r(5) generates number between
>>>>>>>>>> 1-5 uniformly at random , use it to in r(7) which should generate a
>>>>>>>>>> random
>>>>>>>>>> number between 1-7 uniformly at random
>>>>>>>>>>
>>>>>>>>>> i have seen this on many site's but not a single correct solution.
>>>>>>>>>> all solution's posted got rejected by someone else.:
>>>>>>>>>> plz.. suggest some algo :
>>>>>>>>>>
>>>>>>>>>> my aprroach:
>>>>>>>>>>
>>>>>>>>>> let's assume a rectangle :
>>>>>>>>>>
>>>>>>>>>> 100 |___________________
>>>>>>>>>> |___________________|______
>>>>>>>>>> 500/7 | | |
>>>>>>>>>> | | |
>>>>>>>>>> |___________________|______|
>>>>>>>>>> 0 1 2 3 4 5 6 7
>>>>>>>>>> now :
>>>>>>>>>>
>>>>>>>>>> let : num = rand(5);
>>>>>>>>>> prob = rand(5);
>>>>>>>>>>
>>>>>>>>>> if(prob <= rand(5))
>>>>>>>>>> print num
>>>>>>>>>> else
>>>>>>>>>> print 5 + num*(2/5)
>>>>>>>>>>
>>>>>>>>>>
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>>>>>>>>>
>>>>>>>>>
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>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> --
>>>>>>>> Umer
>>>>>>>>
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>>>>>>
>>>>>>
>>>>>
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>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> Umer
>>>>
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>>>
>>>
>>>
>>>
>>> --
>>> Abhishek Sharma
>>> Under-Graduate Student,
>>> PEC University of Technology
>>>
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>>
>>
>>
>>
>> --
>> Anant Sharma
>> IIIrd year
>> Computer Enginering
>> Delhi Technological University
>> (formerly DCE)
>>
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>
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