Re: number of Trangles in a graph-network

• To: mathgroup at smc.vnet.net
• Subject: [mg99446] Re: number of Trangles in a graph-network
• From: Peter Pein <petsie at dordos.net>
• Date: Wed, 6 May 2009 05:21:08 -0400 (EDT)
• References: <gtefaf\$197\$1@smc.vnet.net> <gtnk4g\$3m4\$1@smc.vnet.net>

```Peter Pein schrieb:
> Luca Cinacchio schrieb:
>> Greetings,
>>
>> having a graph (network, i.e. one created with RandomGraph) wich can have
>> not connected nodes, I would like to count the total number of triangles
>> inside the graph.
>> I gave a look to Combinatorica and its related book by Pemmaraju Skiena,
>> but I did'nt find any solution (maybe I am wrong). Do you know if there is
>> a easy way to answer this problem with Mathematica and/or Combinatorica?
>>
>
> It seems I'm writing too fast today :(
>
> I've got a slightly faster method:
>
> In[2]:= SeedRandom[123];
> rg=RandomGraph[32,1/2];
> In[4]:= (* this is from my 1st posting *)
> Timing@Total[ReplaceList[rg[[1]],{___,{{a_,b_}},___,{{a_,c_}},___,{{b_,c_}},___}:>1]]
> Out[4]= {5.98437,573}
>
> and faster but a little bit more brain-twisting:
>
> In[5]:=
> countTriangles[g_Graph]:=Total@ReplaceList[g[[1]],{___,{{a_,b_}},r___/;Length[{r}]>1}:>Count[Split[Sort[Reverse/@Flatten[{r},1]],#1[[1]]===#2[[1]]&][[All,All,2]],{___,a,___,b,___}]]
> In[6]:= Timing@countTriangles[rg]
> Out[6]= {0.100006,573}
>
> don't try this with my first method:
>
> In[7]:= Timing[countTriangles[RandomGraph[100,2/3]]]
> Out[7]= {16.705,45453}
>
>
> Cheers,
> Peter
>

Hi Luca,

sometimes it seems I'm sitting on my eyes ...

Here is a natural and really speedy way to get the desired result:

SeedRandom[123];
rg=RandomGraph[100,2/3];

this is my last vesion:

countTriangles[rg]//AbsoluteTiming
{24.137618,49063}

And the trace of the third power of the adjacency counts every triangle six
times (abc,bca,cab,cba,acb,bac):