How to find the highest 5 adjacent pixels in a matrix?

Hi, I am trying to find the 5 highest value pixels in a matrix. The drawback is that the pixels need to be connected. So, if I have:

X=
       9.1113       64.762       23.623       77.029       25.644
       57.621       67.902        11.94       35.022       61.346
       68.336       63.579        60.73       66.201       58.225
       54.659       94.517       45.014       41.616       54.074
       42.573       20.893       45.873       84.193       86.994
       64.444       70.928       66.194       83.292       26.478

I would want something like:

B= 
       0       0        0       0       0
       0       0        0       0       0
       0       0        0       0       0
       0       0        0       0       0
       0       0       45.873   84.193  86.994
       0       0       66.194   83.292  0

Any ideas on how to achieve this?

Thank you in advance!

[Guillaume]

You need to explain what these particular 5 values are chosen (and not 5 that are connected to the max 94.517 for example).

You also need to state what degree of connectivity you want. 4-connected or 8-connected?

[turningpoint]

I want the 5 connected pixels whose sum is the highest in the matrix. That's why the pixel with the value 94 is not include in the answer. I want 8-connected pixs.

Answer 1

[Guillaume]

The only way I can think of solving this is to calculate the convolution of your matrix with all possible 5 pixel patterns. You only need to generate the patterns in one quadrant, but with 8-connected components, it's still a lot of patterns:

conv2(yourimage, [1 1 1 1 1], 'valid')
conv2(yourimage, [1 1 1 1 0; 0 0 0 0 1], 'valid')
...
conv2(yourimage, eye(5), 'valid');
...
conv2(yourimage, [1 0; 1 0; 1 0; 1 0; 0 1], 'valid')
conv2(yourimage, [1; 1; 1; 1; 1], 'valid')

The location of the maximum of each convolution is the top-left corner of the corresponding 5 pixel pattern with the maximum sum. The maximum of these maximum is what you're looking for.

[turningpoint]

I guess that's my best option. Eventhough I was trying to avoid it. Thank you!