@sourabh....There are O(n^2) elements in the matrix of size nXn. Yes
we can find patterns in a substring with n elements in O(n) but can
you do that in O(sqrt(n)) (To complete your analogy),
Beside's can you allocate a 10^6X10^6 matrix in an array?


On 3/15/12, saurabh singh <saurab...@gmail.com> wrote:
> On 3/15/12, Sourabh Singh <singhsourab...@gmail.com> wrote:
>> @rahul
>>
>> may be this will help you..
>>
>> /* Given a binary matrix, find out the maximum size square sub-matrix
>> with
>> all 1s.
>>    1. O(n^3) sol is very obvious
>>    2. O(n^2) sol [ this file]
>>    3. O(n)   sol [ possible but we need to know tucker matrix, etc
>> advanced
>> set theory's]
>> */
>>
>>
>> #include<iostream>
>> #include<conio.h>
>>
>> int b[6][6];
>> using namespace std;
>> main()
>> {     int i,j,t;
>>       int a[6][6]=
>>               { 0,0,0,1,0,1,
>>                 0,1,1,1,0,0,
>>                 1,1,1,1,0,1,
>>                 0,1,1,1,1,0,
>>                 1,0,0,1,0,0,
>>                 0,1,0,1,1,0,};
>>       for(i=0;i<6;i++)
>>       for(j=0;j<6;j++)
>>       {               if(a[i][j]==1)
>>                       {
>>
>> b[i][j]=min(min(b[i-1][j],b[i][j-1]),b[j-1][i-1])+1;
>>                       }
>>                       else
>>                                     b[i][j]=0;
>>       }
>>       for(i=0;i<6;i++)
>>       {   for(j=0;j<6;j++)
>>           {    printf(" %d",a[i][j]);
>>           }
>>           printf("\n");
>>       }
>>       printf("\n\n\n\n\n");
>>       for(i=0;i<6;i++)
>>       {   for(j=0;j<6;j++)
>>           {    printf(" %d",b[i][j]);
>>           }
>>           printf("\n");
>>       }
>>       getch();
>>       return 0;
>> }
>>
>>
>> On Wed, Mar 14, 2012 at 11:56 AM, Sourabh Singh
>> <singhsourab...@gmail.com>wrote:
>>
>>> @atul
>>>
>>> 1) it won't work for large dimention's coz their is a limit to size of
>>> array we can declare on stack.
>>>    ( which is typically 10^6 as far as i know :-)  ).
>>>
>>> 2) the algo i m trying to find would work in linear time. while this one
>>> is more then O(n^2)
>>>     fo rvery very large input this algo would be very very slow.. making
>>> it impractial..
>>>
>>> ( it's like if u can find substring's in linear time then why use an
>>> O(n^2) algo ;-) )
>>>
>>> NOTE: sry if m getting any fact's wrong m in mid of exam's (so a bit
>>> short
>>> on time to check implemention of  your algo right now )
>>>
>>>
>>>
>>> On Wed, Mar 14, 2012 at 9:07 AM, atul anand
>>> <atul.87fri...@gmail.com>wrote:
>>>
>>>> @rahul: i have alreday explained it in the provided link.
>>>> @sourbh : why it would not work for large dimension
>>>> On 14 Mar 2012 19:39, "rahul sharma" <rahul23111...@gmail.com> wrote:
>>>>
>>>>> @atul..plz tell me an example for square matrix...actually i faced it
>>>>> first tym...it executes...but explain plz..
>>>>>
>>>>> On Wed, Mar 14, 2012 at 6:56 PM, Sourabh Singh
>>>>> <singhsourab...@gmail.com
>>>>> > wrote:
>>>>>
>>>>>> @atul
>>>>>>
>>>>>> Also the histogram algo and algo given by you can't work on very very
>>>>>> big dimentions. say 10^5 x 10^5 matrix.
>>>>>>  but if we can find a DP then  we just need to keep 2 row's at a
>>>>>> time.
>>>>>> :-)
>>>>>>
>>>>>>
>>>>>>
>>>>>> On Tue, Mar 13, 2012 at 1:03 PM, Sourabh Singh <
>>>>>> singhsourab...@gmail.com> wrote:
>>>>>>
>>>>>>> @atul
>>>>>>>
>>>>>>> read it ..
>>>>>>>
>>>>>>> it's good but more or less like the histogram algo.. i wanted a DP.
>>>>>>> approach..
>>>>>>>
>>>>>>> here is some of wat i heard from a senior in colg..
>>>>>>>
>>>>>>> 1. at every index we can keep 4 variable
>>>>>>>
>>>>>>> ht: height of max rectangle possible at index above current
>>>>>>>  wt width   "   "          "             "             "
>>>>>>> "           "
>>>>>>>  hl:height of max rectangle possible at index left of  current
>>>>>>> wl:   "            "        "               "             "
>>>>>>> "            "
>>>>>>>
>>>>>>>
>>>>>>> now problem is which one to take for current... index
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> On Tue, Mar 13, 2012 at 10:52 AM, atul anand
>>>>>>> <atul.87fri...@gmail.com>wrote:
>>>>>>>
>>>>>>>> @ Sourabh: check solution i have posted in below link
>>>>>>>>
>>>>>>>>
>>>>>>>> http://groups.google.com/group/algogeeks/browse_thread/thread/91a17f7c78c2319e/991d1c2625a62ff0?hl=en&lnk=gst&q=rectangle+of+max+sum+MS+Q#991d1c2625a62ff0
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> On Tue, Mar 13, 2012 at 10:26 PM, Sourabh Singh <
>>>>>>>> singhsourab...@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> @ ALL
>>>>>>>>>
>>>>>>>>>  finding square matrix is quite a standard question and nw an easy
>>>>>>>>> one as everyone knows the reccussence atul has given.
>>>>>>>>>  but  i wanted to find max rectangle..
>>>>>>>>>
>>>>>>>>> i know there is a DP for it. in O(n^2). for nxn matrix..don't know
>>>>>>>>> the whole approach .but  here is what i remember..
>>>>>>>>>
>>>>>>>>> 1. aproach is simple to keep track of max rectangle which can be
>>>>>>>>> formed from any point taking that point as top  left corner of max
>>>>>>>>> rectangle and
>>>>>>>>>     proceed further down .
>>>>>>>>>
>>>>>>>>> can someone suggest how can be proceed further..
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> [ NOTE: problem occurs mainly when their are more than one
>>>>>>>>> rectangles which can be formed from same point ]
>>>>>>>>>
>>>>>>>>> plz.. don't suggest the histogram method it's just a dirty way of
>>>>>>>>> avoiding to work on getting this DP right. :-)
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> On Mon, Mar 12, 2012 at 11:29 PM, atul anand <
>>>>>>>>> atul.87fri...@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>>> here is the recurrence for solving this
>>>>>>>>>>
>>>>>>>>>> R[i, j] = (M[i,j] == 0 ? 0 : 1 + min( R[i-1, j], R[i-1, j-1],
>>>>>>>>>> R[i,,j-1] ) );
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> On Tue, Mar 13, 2012 at 11:48 AM, rahul sharma <
>>>>>>>>>> rahul23111...@gmail.com> wrote:
>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>  April 4, 2010
>>>>>>>>>>>
>>>>>>>>>>> Given a binary matrix, find out the maximum size square
>>>>>>>>>>> sub-matrix
>>>>>>>>>>> with all 1s.
>>>>>>>>>>>
>>>>>>>>>>> For example, consider the below binary matrix.
>>>>>>>>>>>
>>>>>>>>>>>    0  1  1  0  1
>>>>>>>>>>>    1  1  0  1  0
>>>>>>>>>>>    0  1  1  1  0
>>>>>>>>>>>    1  1  1  1  0
>>>>>>>>>>>    1  1  1  1  1
>>>>>>>>>>>    0  0  0  0  0
>>>>>>>>>>>
>>>>>>>>>>> The maximum square sub-matrix with all set bits is
>>>>>>>>>>>
>>>>>>>>>>>     1  1  1
>>>>>>>>>>>     1  1  1
>>>>>>>>>>>     1  1  1
>>>>>>>>>>>
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>
>
> --
> Saurabh Singh
> B.Tech (Computer Science)
> MNNIT
> blog:geekinessthecoolway.blogspot.com
>


-- 
Saurabh Singh
B.Tech (Computer Science)
MNNIT
blog:geekinessthecoolway.blogspot.com

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