yeah, you are right. It comes from 2 to 6. But is there any way to solve it
on paper?
-Regards
Amit Agarwal
Contact: 09765348182
www.amitagrwal.com



On Mon, May 3, 2010 at 3:02 PM, Sundeep Singh <singh.sund...@gmail.com>wrote:

> oops ....
>
> On Sat, May 1, 2010 at 5:50 PM, Sundeep Singh <singh.sund...@gmail.com>wrote:
>
>> Hi Amit,
>>
>> here's the answer: (I am assuming in your equation "lg" implies log to the
>> base 10)
>> n < 8 log(n)
>> => n/8 < log(n)
>> => 10 ^(n/8) < n
>>
>
> The final deduction was incorrect!!
> for log base 10, the answer is:
> 2 <= n <= 6
>
> --Sudneep.
>
>
>
>> => n > 8
>>
>> --Sundeep.
>>
>>
>>
>> On Sat, May 1, 2010 at 10:43 AM, Amit Agarwal <lifea...@gmail.com> wrote:
>>
>>> I could not get you properly. This is an equation comes from the problem
>>> statement where I need to find out cut-off value of n between insertion and
>>> merge sort. I think equation is part of basic mathematics but I don't
>>> remember how do I solve it.
>>>
>>>
>>> -Regards
>>> Amit Agarwal
>>> Contact: 09765348182
>>> www.amitagrwal.com
>>>
>>>
>>>
>>>
>>> On Sat, May 1, 2010 at 9:13 AM, abhijith reddy <abhijith200...@gmail.com
>>> > wrote:
>>>
>>>> binary search on n
>>>>
>>>> On Fri, Apr 30, 2010 at 10:15 PM, Amit Agarwal <lifea...@gmail.com>wrote:
>>>>
>>>>> how do I compute n from this equation.
>>>>> n < 8lg(n)
>>>>>
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