isnt it similar to youngs tabulaeau as gvn in CLRS

On Wed, Nov 25, 2009 at 5:18 PM, Arun <[email protected]> wrote:

> how about splitting the matrix into 4 squares and considering the rightmost
> bottom and comparing it to topmost left corner of each square. in every
> iteration you will eliminate atleast one square. so worst case, you have to
> repeat this divide and conquer on the three other squares.
> not good as as binary search though(in the sense of eliminating 2 squares).
>
>
> On Wed, Nov 25, 2009 at 3:12 AM, Aditya Shankar <
> [email protected]> wrote:
>
>>
>>
>> On Wed, Nov 25, 2009 at 2:16 PM, Bharath <[email protected]> wrote:
>>
>>> You can actually do it in O(logn) complexity. Binary Search on diagonal
>>> and then on a row.
>>
>> Will that always work?
>> 1 2 3
>> 3 5 6
>> 4 7 8
>>
>> Find(6) fails with the usual binary search algorithm.
>>
>>
>>>
>>>
>>> On Tue, Nov 24, 2009 at 10:33 PM, chitta koushik <
>>> [email protected]> wrote:
>>>
>>>> Start from top right or bottom left corner and move according if the
>>>> element to be searched is lesser or greater than current.
>>>>
>>>>
>>>> --Koushik C
>>>> Pablo 
>>>> Picasso<http://www.brainyquote.com/quotes/authors/p/pablo_picasso.html> - 
>>>> "Computers are useless. They can only give you answers."
>>>>
>>>>
>>>> On Tue, Nov 24, 2009 at 7:27 PM, Rohit Saraf <
>>>> [email protected]> wrote:
>>>>
>>>>> A nice problem that i encountered :
>>>>> In O(n) search for a value x in a sorted NxN matrix.
>>>>> Definition of sorted matrix:  All rows and all columns are sorted in
>>>>> ascending order.
>>>>>
>>>>>  So thought of sharing ..
>>>>>
>>>>>
>>>>> Rohit Saraf
>>>>> Sophomore
>>>>> Computer Science and Engineering
>>>>> IIT Bombay
>>>>>
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>>>
>>>
>>>
>>> --
>>> <<Bharath>>
>>>
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