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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