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Welcome to CityTech Sites. This is your first post. Edit or delete it, then start blogging!
Our goal is to make the OpenLab accessible for all users.
Our goal is to make the OpenLab accessible for all users.
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1)The 1st run:
After the 1st run of the program, the smaller partition have 2 positions.
2)2nd run:
a) Therefore, I made changes in the program to delete 2 rows in the second run.
b) After 2nd run of the program, the smaller partition have 3 positions.
3)3rd run:
a) Therefore, I made changes in the program to delete 2+3 =5 rows in the 3rd run because the input matrix in this run is still the full original matrix consisting of 17 rows, no matter I have deleted 2 rows in previous step.
Is it correct procedure?
Response to the following email:
You just need to delete it once.
After you run my code your matrix should have actually 13 rows and columns. All 4 zero rows and columns would be deleted automatically. You just need to keep track of the species with the rows and columns.
Next you save the code in a different name and identify the smallest cluster and delete those species from the matrix inside the code. Since it is essentially the same code the original competition matrix still will have 17 species but in your final answer should have 11 eigenvalues and the corresponding eigenvectors.
No need to delete any species after this. Analyze and interpret.
Thanks,
Urmi
>>> “Mariusz.Nuckowski@mail.citytech.cuny.edu” 12/12/12 8:41 AM >>>
Dear Professor
I want to be sure to procedure:
1)The 1st run:
After the 1st run of the program, the smaller partition have 2 positions.
2)2nd run:
a) Therefore, I made changes in the program to delete 2 rows in the second run.
b) After 2nd run of the program, the smaller partition have 3 positions.
3)3rd run:
a) Therefore, I made changes in the program to delete 2+3 =5 rows in the 3rd run because the input matrix in this run is still the full original matrix consisting of 17 rows, no matter I have deleted 2 rows in previous step.
Is it correct procedure?