I have three nodes categories  (Person, Speech,TIME) and 6 different
interactions (3 intra layer) and (3 inter layer), my problem is how to
compute the diameter the hall graph (aggregation graph with the differents
relationships)





Best regards.



*----------------------------*
*MOURCHID Youssef*
---------------------------------------------------------------------------------------------------
*PhD Student - **Computer Sciences **and Telecommunications *
*                           Laboratory Research (LRIT)*
*Faculty of sciences  **University of Mohammed V Agdal-Rabat, Morocco *
-------------------------------------------------------------------------------

   - *    Phone : + 212 (0) 6 71 96 32 54*
   -
*    Mail : youssefm...@gmail.com <http://gmail.com>  *






On Mon, Sep 3, 2018 at 12:10 PM Analista de Sistemas Web/Mobile <
web2a...@gmail.com> wrote:

> So what are you cypher query?
>
> On Mon, Sep 3, 2018 at 8:06 AM Youssef Mourchid <youssefm...@gmail.com>
> wrote:
>
>> thank you, but where can i run this script, because i use cypher query in
>> neo4j !!
>>
>>
>>
>>
>>
>> Best regards.
>>
>>
>>
>> *----------------------------*
>> *MOURCHID Youssef*
>>
>> ---------------------------------------------------------------------------------------------------
>> *PhD Student - **Computer Sciences **and Telecommunications *
>> *                           Laboratory Research (LRIT)*
>> *Faculty of sciences  **University of Mohammed V Agdal-Rabat, Morocco *
>>
>> -------------------------------------------------------------------------------
>>
>>    - *    Phone : + 212 (0) 6 71 96 32 54*
>>    -
>> *    Mail : youssefm...@gmail.com <http://gmail.com>  *
>>
>>
>>
>>
>>
>>
>> On Mon, Sep 3, 2018 at 12:03 PM Analista de Sistemas Web/Mobile <
>> web2a...@gmail.com> wrote:
>>
>>> public double getDistancia(double latitude, double longitude, double
>>> latitudePto, double longitudePto){ double dlon, dlat, a, distancia;
>>> dlon = longitudePto - longitude; dlat = latitudePto - latitude; a = Math.
>>> pow(Math.sin(dlat/2),2) + Math.cos(latitude) * Math.cos(latitudePto) *
>>> Math.pow(Math.sin(dlon/2),2); distancia = 2 * Math.atan2(Math.sqrt(a),
>>> Math.sqrt(1-a)); return 6378140 * distancia; /* 6378140 is the radius
>>> of the Earth in meters*/ }
>>>
>>> https://en.wikipedia.org/wiki/Distance_(graph_theory)
>>>
>>>
>>> On Mon, Sep 3, 2018 at 7:02 AM Youssef Mourchid <youssefm...@gmail.com>
>>> wrote:
>>>
>>>> Hello,
>>>>
>>>> if can anyone help me to find a way to compute the diameter of a graph
>>>> with different categories of edges i will be very thankful.
>>>>
>>>> In my case my graph contains three categories of nodes (Person, speech,
>>>> Time) and 6 relationships.
>>>>
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>>> --
>>> JOSÉ RIBAMAR FERREIRA JUNIOR
>>>           ANALISTA DE SISTEMAS
>>>                     Joinville - SC
>>>                    (47) 9844-23634
>>>                    (47) 3436-4774
>>>
>>> web2a...@gmail.com
>>> web2a...@hotmail.com
>>>
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>>>
>>>
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> --
> JOSÉ RIBAMAR FERREIRA JUNIOR
>           ANALISTA DE SISTEMAS
>                     Joinville - SC
>                    (47) 9844-23634
>                    (47) 3436-4774
>
> web2a...@gmail.com
> web2a...@hotmail.com
>
> https://www.linkedin.com/in/josé-r-f-junior-b72b6b1a/
>
>
> --
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