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Re: on a self-complementary graph problem

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Ben Bacarisse <ben.usenet@bsb.me.uk> writes:

> Daniel Bastos <dbastos@toledo.com> writes:
>
>> Ben Bacarisse <ben.usenet@bsb.me.uk> writes:
>>
>>> Daniel Bastos <dbastos@toledo.com> writes:
>>>
>>>> I'm looking for a strategy on how to solve this problem.  I appreciate
>>>> any direction you may give me.
>>>>
>>>> Problem.  A simple graph G is self-complementary if G =~ G^c.  Show if G
>>>> is self-complementary, then v = 0 (mod 4) or v = 1 (mod 4).
>>>
>>> Let's say G has n vertices.  G^c is K_n with the edges of G removed.
>>> What do you know about e(G) and e(G^c)?  What does that tell you about
>>> e(K_n)?  Now consider exercise 1.2.7.
>>
>> By definition, e(G^c) = e(K_n) - e(G).  I know e(G) and e(G^c) have the
>> same number of vertices and edges because G =~ G^c.  Also, from exercise
>> 1.1.3, I know
>>
>>    e(G^c) <= (v 2)
>>    e(G)   <= (v 2)
>>
>> By exercise 1.2.7, I know 
>>
>>    e(K_n) = (v 2)
>>
>> Combining the definition of G^c with these facts, I get    
>>
>>   e(G^c) + e(G) = (v 2) = 1/2 v (v - 1).
>>
>> (*) Attempt 1
>>
>> Let G be self-complementary.  I must show v = 0 or 1 mod 4. 
>>
>> Let e := e(G) = e(G^c).  Then from
>>
>>   e(G^c) + e(G) = (v 2) = 1/2 v (v - 1) 
>>
>> we get 
>>
>>   2e = 1/2 v (v - 1)
>>
>> implying
>>
>>   4e = v (v - 1)
>>
>> Which numbers v satisfy this equation?  I can't honestly deduce the
>> result here.  At least not at the moment.
>
> That's great.  It's just arithmetic now.  Take both sides mod 4:
>
>   0 = v(v - 1) mod 4
>
> Can you see what v mod 4 must be for v(v - 1) to be 0 mod 4?  

I can.  The integers modulo 4 are 0, 1, 2, 3.  The only ones that can
satisfy the equation are 0 and 1 --- as you showed below.

> If not, here's how I'd show it (using rot13 and some spoiler alert
> lines)

How nice to see someone using rot13! :-)

> I'd probably just list the cases:
>
>   v = 0 mod 4 <=> v-1 = 3 mod 4 <=> v(v-1) = 0 mod 4
>   v = 1 mod 4 <=> v-1 = 0 mod 4 <=> v(v-1) = 0 mod 4
>   v = 2 mod 4 <=> v-1 = 1 mod 4 <=> v(v-1) = 2 mod 4
>   v = 3 mod 4 <=> v-1 = 2 mod 4 <=> v(v-1) = 6 mod 4 <=> 2 mod 4
>
> so v(v - 1) = 0 mod 4 if and only if v = 0 mod 4 or v = 1 mod 4.
>
> (There's probably some nice simple rule for solving things like
>
>   a = x(x - b) mod n

[...]

I wonder that myself.  I think what we want here is to claim we have an
integral domain, an algebraic structure to the effect that a product 
xy = 0 if both x, y are zero.  Had we this structure, we'd be looking
for solutions of

      v = 0 mod 4
  v - 1 = 0 mod 4

But I think the structure here is Z/4Z, a quotient ring.  Since 4 is not
prime, there are factors x, y such that 4 = xy.  Since 4 = 0 mod 4, it
turns out that x = 2, y = 2 satisfy 4 = xy = 0, but both x,y are
nonzero.  Therefore Z/4Z is not an integral domain and so the argument
laid out doesn't apply here.

Nevertheless, since you listed out all cases, I think that proves it.

>> Thanks!
>
> You're welcome.  How nice to have a real question from a serious student.

How nice to have someone to investigate these matters with!  Thank you!

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Thread

on a self-complementary graph problem Daniel Bastos <dbastos@toledo.com> - 2017-09-09 15:48 -0300
  Re: on a self-complementary graph problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2017-09-09 21:05 +0100
    Re: on a self-complementary graph problem Daniel Bastos <dbastos@toledo.com> - 2017-09-16 11:13 -0300
      Re: on a self-complementary graph problem Ben Bacarisse <ben.usenet@bsb.me.uk> - 2017-09-16 16:10 +0100
        Re: on a self-complementary graph problem Daniel Bastos <dbastos@toledo.com> - 2017-09-17 07:32 -0300

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