@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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>>>
>>>
>>> --
>>> Umer
>>>
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>
>

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