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Suppose f is a function of two variables: $f = f(x,y)\quad$ and $L$ is a list. Could someone help me in defining a list of the form: $D = \{f(x,y)\ |\ x \in L,\ y \in L\}$

I tried this:

D = List[ f[x,y], {x,y} ∈ KroneckerProduct[L,L]]

But I got syntax errors.

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1  
What syntax errors did you get? It could just be that you were trying to assign to D which is a defined symbol already in Mathematica. – Verbeia♦ Aug 10 '12 at 20:26
Also I formatted your code. You might want to read this page which shows how to do it yourself. – Verbeia♦ Aug 10 '12 at 20:27
Are you looking for functioning code or are you trying to typeset something? – Sjoerd C. de Vries Aug 10 '12 at 20:29
Sorry about ambiguities. The syntax error was: D = Table[f[x, y], {x, y} \[Element] Tuples[L, 2]] Table::itform: Argument {x,y}\[Element]Tuples[L,2] at position 2 does not have the correct form for an iterator. >> where L = {{0, 0, 0, 1}, {0, 1, 1, 0}}. In my question, i didn't need x and y to be the same elements, so Tuples[L,2] is more appropriate. I am looking for a functioning code. Let me state it simply, similar to: Flatten[Table[f[x,y], {x, 0, 10}, {y, 0, 10}]] can't we have a similar definition where x and y belong to different lists, say L1 and L2 respectively ? – Pavithran Iyer Aug 10 '12 at 21:21
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In future questions could you please post the code that you actually used? Apparently you used Table instead of List, Tuples instead of KroneckerProduct and x and y need not equal vs. need to be equal. – Sjoerd C. de Vries Aug 10 '12 at 22:13

3 Answers

up vote 5 down vote accepted

Your original definition of D can be given with these lines:

In[1]:= L = {a, b, c, d};
In[2]:= Outer[f, L, L] // Flatten
Out[2]= {f[a, a], f[a, b], f[a, c], f[a, d], f[b, a], f[b, b], 
 f[b, c], f[b, d], f[c, a], f[c, b], f[c, c], f[c, d], f[d, a], 
 f[d, b], f[d, c], f[d, d]}]

If f is symmetric, then you can:

In[3]:= SetAttributes[f, Orderless]
In[4]:= Outer[f, L, L] // Flatten // Union
Out[4]= {f[a, a], f[a, b], f[a, c], f[a, d], f[b, b], f[b, c], 
 f[b, d], f[c, c], f[c, d], f[d, d]}
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up vote 8 down vote

If f is a function of two-variables, a direct way to define your list (we call it Dlt since D is a built-in Mathematica function) would be this :

L = {a, b, c, d, e};

Dlt = f @@@ Tuples[ L, 2]

{f[a, a], f[a, b], f[a, c], f[a, d], f[a, e],
 f[b, a], f[b, b], f[b, c], f[b, d], f[b, e],
 f[c, a], f[c, b], f[c, c], f[c, d], f[c, e],
 f[d, a], f[d, b], f[d, c], f[d, d], f[d, e],
 f[e, a], f[e, b], f[e, c], f[e, d], f[e, e]}

Symbol f @@@ expr is a shorthand for Apply[ f, expr, {1}]. Tuples[L, 2 ] generates all pairs of elements of list L ( e.g. {a,b} is different from {b,a}).

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In case you want to define a list of values f over a cartesian product of two sets K and L I recommend this f @@@ Tuples[{K, L}]. You should also look at Scan. – Artes Aug 10 '12 at 22:14
up vote 2 down vote

I think you intend to do something like this:

L = {a, b, c, d};
f /@ Transpose[{L, L}]

{f[{a, a}], f[{b, b}], f[{c, c}], f[{d, d}]}

With /@ a prefix shorthand for the Map function.

You might want to look up KroneckerProduct in the manual. It doesn't work as you seem to think. 

KroneckerProduct[L, L]

{{a^2, a b, a c, a d}, {a b, b^2, b c, b d}, {a c, b c, c^2, c d}, {a d, b d, c d, d^2}}

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