We were discussing identity management without login screens.
We elaborated on the following three components as necessary for the alternatives:
1) user knowledge based component
2) server issued dynamic component
3) convenience factor
The last component above is the determining factor in why dynamic challenge questions cannot be a single and complete replacement to logins and passwords. The questions can be generated dynamically by the server such as from a list of ten questions that only the user can answer and the questions do not even need to have static answers, yet they cannot be as simple as the username and password. Moreover, people like to use their email or phone numbers for their username, something that does not seem likely to change although it adds to vulnerability by tying a centralized option for resetting all accounts.
We now consider blockchain technologies for identity management as per IBM. We discussed earlier that we can persist a token/passcode/password/private-keys/HTTP-Links for the user to login in a blockchain like database, we will have no need to maintain or manage these ourselves. Alternatively, we could consider using encryption-decryption-based and server issued-and-verified claims but without requiring a centralized server.
The benefits of identity with blockchain for consumers include:
 - convenience for accessing services and resources
 - better protection of privacy
 - greater control of personal data
The benefits for businesses include:
 - reduced risk and cost of data breach
 - efficient compliance management and governance
 - easier onboarding of customers
The benefits for regulators include:
- standardized processes
- prompt auditing
- compliance control
Find the largest rectangular sub-matrix having sum divisible by k.
The naive approach is to exhaust all the rectangular sub matrices to find the sums. check that the sums are divisible by k and then update the max of the rectangular size found so far.
This appears something like
int GetLargest(int[,] M, int R, int C, int k)
{
int max = 0;
// Every element forms the top left corner of a rectangular area of different sizes
for (int i = 0; i < R; i++)
  for (int j = 0; j < C; j++)
{
   int kx = i;
   int ky = j;


    for (int m = kx; m < R; m++)
         for (int n = ky; n < C; n++)
    {
           int sum = GetSum(M, i,j,m,n);
           if (sum % k == 0)
           {
                    int size = GetSize(i,j,m,n);
                    if (size > max)
                        max = size;
           }
    }
return max;
}