yeah...first search in 2^0 and 2^1....if elemtn range is between this thern
search in this length with Binary search else
searchin 2^1 and 1^2 and apply same procedure until element range is found
and if element is not found in range then it means element not founddddd

On Tue, Oct 25, 2011 at 3:12 PM, Bittu Sarkar <[email protected]> wrote:

> @Saurabh I think you did not get the question clearly. It's
> a theoretical question because in practice you cannot have an infinite
> length array of non-decreasing numbers anyways. Yes the sequence 1,1,1,...
> is infinite and the algorithm cannot just know all by itself what the
> largest number is, just because it has processed millions of 1s. In this
> case no algorithm can find the required number (>1) in which case the
> algorithm does not terminate.
>
>
> On 25 October 2011 13:46, saurabh singh <[email protected]> wrote:
>
>> Really?????????????
>> the array is infinite length......
>> what if the sequence is
>> 1,1,1,1,1,1,1,1...........................infinity?
>> We need to know the max in order the algorithm is terminating..........
>>
>> On Tue, Oct 25, 2011 at 11:32 AM, Bittu Sarkar <[email protected]>wrote:
>>
>>> @Saurabh There is no biggest number in an infinite array [?]
>>>
>>>
>>> On 25 October 2011 09:07, saurabh singh <[email protected]> wrote:
>>>
>>>> suppose the element doesn't lies in the array and is bigger than the
>>>> biggest number........:D
>>>> everything  will fail.......
>>>>
>>>>
>>>> On Mon, Oct 24, 2011 at 9:43 PM, ravindra patel <
>>>> [email protected]> wrote:
>>>>
>>>>> using power of 2 approach doubles the scope of search each time.
>>>>> How about using approximation. Say I have lower bound lb and upper
>>>>> bound ub. Now -
>>>>>
>>>>> initially lb = 0, ub = 1;
>>>>>
>>>>> while (a[ub] < k)
>>>>> {
>>>>>     lb = ub;
>>>>>     ub = (ub*k) / a[ub];
>>>>> }
>>>>>
>>>>> after end of this loop we'll have a lower bound and upper which should
>>>>> provide a narrow scope.
>>>>>
>>>>> Feedback welcome :-),
>>>>> - Ravindra
>>>>>
>>>>> On Mon, Oct 24, 2011 at 7:09 PM, Bittu Sarkar <[email protected]>wrote:
>>>>>
>>>>>> @Ankur Don't think there's any major reason for using powers of 2
>>>>>> except the fact that computing the powers of 2 can be done very 
>>>>>> efficiently
>>>>>> than computing the powers of any other number. Complexity in any case
>>>>>> remains the same.
>>>>>>
>>>>>>
>>>>>> On 24 October 2011 10:29, rahul sharma <[email protected]>wrote:
>>>>>>
>>>>>>> +1 ankur
>>>>>>>
>>>>>>>
>>>>>>> On Mon, Oct 24, 2011 at 1:26 AM, Ankur Garg <[email protected]>wrote:
>>>>>>>
>>>>>>>> Use Binary Search
>>>>>>>>
>>>>>>>> start = 2^n-1 high =2^n where n=0,1....
>>>>>>>>
>>>>>>>> On Mon, Oct 24, 2011 at 12:28 AM, sunny agrawal <
>>>>>>>> [email protected]> wrote:
>>>>>>>>
>>>>>>>>> hint 1: try to find 2 indexes i, j such that a[i] <= K <= a[j]
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> On Sun, Oct 23, 2011 at 11:23 PM, Ankuj Gupta <[email protected]
>>>>>>>>> > wrote:
>>>>>>>>>
>>>>>>>>>> Given a sorted array of Infinite size, find an element ‘K’ in the
>>>>>>>>>> array without using extra memory in O (lgn) time
>>>>>>>>>>
>>>>>>>>>> --
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>>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> --
>>>>>>>>> Sunny Aggrawal
>>>>>>>>> B.Tech. V year,CSI
>>>>>>>>> Indian Institute Of Technology,Roorkee
>>>>>>>>>
>>>>>>>>>
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>>>>>> --
>>>>>> Bittu Sarkar
>>>>>> 5th Year Dual Degree Student
>>>>>> Department of Computer Science & Engineering
>>>>>> Indian Institute of Technology Kharagpur
>>>>>>
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>>>>
>>>> --
>>>> Saurabh Singh
>>>> B.Tech (Computer Science)
>>>> MNNIT ALLAHABAD
>>>>
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>>>
>>>
>>> --
>>> Bittu Sarkar
>>> 5th Year Dual Degree Student
>>> Department of Computer Science & Engineering
>>> Indian Institute of Technology Kharagpur
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>>
>>
>> --
>> Saurabh Singh
>> B.Tech (Computer Science)
>> MNNIT ALLAHABAD
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>
>
>
> --
> Bittu Sarkar
> 5th Year Dual Degree Student
> Department of Computer Science & Engineering
> Indian Institute of Technology Kharagpur
>
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