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Lattice Algebra: Problems with Flat and Orderless

  • To: mathgroup at
  • Subject: [mg14896] Lattice Algebra: Problems with Flat and Orderless
  • From: immtwg at (Torben Winther Graversen)
  • Date: Wed, 25 Nov 1998 17:48:18 -0500
  • Organization: UNI-C
  • Sender: owner-wri-mathgroup at


I'm trying to implement an operator, called say "p", from a lattice 
algebra. It is thus commutative, associative, distributive and 
idempotent. I use the Flat and Orderless attributes to implement 
associativity and commutativity, but I also need to have a "top" 
element, called say "a", with this property:

p[a, __] = a;

so that e.g. "p[a,b,c]" gives just "a". This is obviously similar to 
multiplication with zero.

If I implement this before I set the Flat and Orderless attributes it 

In[1]:= p[a, __] = a;

In[2]:= p[a,b,c]
Out[2]= a

If I implement it after I set the attributes the pattern matcher will 
never let me reduce the expression to just "a" but rather "p[a]":

In[3]:= ClearAll[p];

In[4]:= Attributes[p] = {Orderless, Flat};

In[5]:= p[a, __] = a;

In[6]:= p[a,b,c]
Out[6]= p[a]

Question 1:
How do I get around that? Adding a definition like p[x_]=x has the same 
problem, and an infinite recursion. The only way I have found to do it 
was to implement the rule before the attributes, and then set the 

In[7]:= ClearAll[p];

In[8]:= p[a, __] = a;

In[9]:= Attributes[p] = {Orderless, Flat};

In[10]:= p[a,b,c]
Out[10]= a

But if I now repeat the definition of the rule, I get two seamingly 
identical rules for p:

In[11]:= p[a, __] = a;

In[12]:- ?p


Attributes[p] = {Flat, Orderless}
p[a, __] = a
p[a, __] = a

But it's still the original rule that is used:

In[13]:= p[a,b,c]
Out[13]= a

Question 2:
What exactly is going on here?

Best regards,

  Torben Winther Graversen

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