|Short Graph Coloring Question email@example.com (Stephen Molloy) (2002-03-31)|
|Re: Short Graph Coloring Question firstname.lastname@example.org (Daniel Berlin) (2002-04-06)|
|Re: Short Graph Coloring Question email@example.com (Mark Lacey \[MSFT\]) (2002-04-06)|
|Re: Short Graph Coloring Question firstname.lastname@example.org (Amal Banerjee) (2002-04-06)|
|Re: Short Graph Coloring Question Sid-Ahmed-Ali.TOUATI@inria.fr (Sid Ahmed Ali TOUATI) (2002-04-13)|
|From:||Amal Banerjee <email@example.com>|
|Date:||6 Apr 2002 23:41:32 -0500|
|Organization:||The University of Texas at Austin; Austin, Texas|
|Posted-Date:||06 Apr 2002 23:41:32 EST|
It looks like you have all the nodes on the stack. So, by
Chaitin's algorithm, pop them off the stack and color them.
On 31 Mar 2002, Stephen Molloy wrote:
> Graph Coloring Short Question
> Interference Information
> Variable | Interferes With
> a | b,c,d,e
> b | a,c,e
> c | a,b,d
> d | a,c
> e | a,b
> Show with graph coloring how we can put them in three registers?
> Remove any node with less than K edges (K=3)
> 1. remove e
> 2. remove d
> 3. remove b
> 4. remove a
> 5. remove c
> How do we get from here to register allocation?
> I'm looking for the method not the answer, just a part of the method
> I'm looking for - the bit where you go from reducing the graph to
> allocating into registers.
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