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These are replies submitted by Teep


This seems to have done the job - thank you so much!

I like your routine ... with this, I can move on.


Thanks again.


Yes - I can relabel the vertices as you proposed. The Help link was very useful ... thanks.

However, I might be on the wrong track here. I wish to assign labels (letters) to the matrix rows and numbers to the matrix columns as mentioned in the problem statement. The sources lie in the row number; the sinks lie in the columns that have entries=1. 



Thank you for the suggestion - much appreciated.

However, I tried to apply this to the matrix B graph ... I can't seem to find a way to replicate your method.


These are very useful insights you provided - I appreciate you taking the time and effort to share these. 

And of course, you are completely right -  the worksheet source file should have been attached rather than the pdf. I'll remember that in future also!


Thanks again...

@Carl Love 

It is great to get this workaround - thank you! It makes life so much easier ... now I can visualise the relationships of the solutions.

Thanks (again!) for your generous support.

Thank you, Carl. 

I tried that - but doesn't the adjacency matrix has 2 non-zero diagonal entries? If so - then the routine won't run.

Perhaps I am misinterpreting your suggestion (I am new to Graph Theory) - in this instance, the solution to my LP problem is given by the square matrix,Y, and will change as certain conditions / values are altered in the model. In other words, the entries will vary on a case-by-case basis and I simply wish to show these on a graph.







Thanks again to all for addressing this question - as I don't use these methods and techniques, I find that the support you all gave is generous and well-constructed. I have learned a lot today.

Last question .... is there a convenient way to allocate a name to each i and j node?

Thanks for this - this is a great help and is very much appreciated. Nice solution!

However, can we separate each node i in the graph? For example, can we show node 2 (i=2) connected to the j-nodes, 3, 4 6 and 9 alone?

In the same way, I'd like to show node 3 (i=3) simply connected to the j-nodes and in the case where there are no connections from i to j, can we simply omit these? 

I'm only interested in the plots that show the active connections, i.e. in the case where the entries are 1's and not 0's.


@Carl Love 

Thanks for the insights here. I'll give this a go also!



Yes ... that does the trick!

Thanks a lot - I'm grateful.

@Carl Love 

Yes indeed .... I frequently lost my connection also for small p.

I figured it was pc-related rather than numerical computation issues, so I removed firewall, re-booted, etc. but it still made no difference.  


Nice model. I varied the parameter values across my range of interest and it produces outputs in quick time.

Thanks very much .... and thanks again to all who worked on this.

@Carl Love 

This looks very interesting for small p values. I'll apply this to my problem. 

Thanks for the creative input (as usual!).



Interesting and useful routine .. thanks for this insight.



At least some good arises!

I'm glad you picked this up, Erik - thanks for informing us.

Is it possible to obtain a pre-release version of the fix before the next release?


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