Preamble
Suppose we have some symbols x, y, ... and wildcards (ξ), (ζ), ... which have any meaning. Wildcards are said to ‘match’ any symbol: wildcard (ξ) matches symbol x under the mapping {(ξ): x}. Now suppose we can construct directed trees from these symbols and wildcards. The notion of matching should extend to these trees, like this:
x x
/ \ / \
a y matches (ξ) y under the mapping {(ξ): a, (ζ): b}.
/ \ / \
z b z (ζ)
It is easy to implement an algorithm which takes two trees like those above and gives the necessary mapping. (Does it have a name?)
I want to introduce certain ‘equivalence relations’ <n> between trees. For instance, if I wanted to capture the notion of the symbol f being commutative and associative, I could write:
f f
<1>: / \ = / \ # commutativity f(ξ, ζ) = f(ζ, ξ)
(ξ) (ζ) (ζ) (ξ)
f f
/ \ / \
<2>: f (χ) = (ξ) f # associativity f(f(ξ, ζ), χ) = f(ξ, f(ζ, ξ))
/ \ / \
(ξ) (ζ) (ζ) (χ)
This is where things get interesting, because I now want to implement a more intelligent matching algorithm which, given a set of equivalence relations like those above, is able to perform ‘indirect’ matches like this:
f f
/ \ (indirectly) matches / \ under the mapping {(ξ): x}...
y x (ξ) y
f <1> f f
...because / \ = / \ which directly matches / \ .
y x x y (ξ) y
Question
How would one go about finding an algorithm which is able to transform a given tree (via given rules) so that it ‘matches’ another tree?
Is there a name for this kind of problem?
More examples
Matches aren’t always unique. Ideally, the algorithm should be able to find all of them.
f f
/ \ / \
x f matches f (ξ) under {(ξ): y, (ζ): z} or {(ξ): z, (ζ): y}...
/ \ / \
y z (ζ) x
...using rules <1> and <2>.
A quintessential demonstrative example:
For brevity, define the symbols 1, 2, 3, ... in terms of 0 and a unary symbol s:
1 = s(0), 2 = s(s(0)), 3 = s(s(s(0)))...
Now, define two transformation rules s(n) = +(n, 1) and s(+(n, m)) = +(n, s(m)):
+ s
<1>: / \ = |
(n) 1 (n)
s +
<2>: | = / \
+ (n) s
/ \ |
(n) (m) (m)
Check this out:
+
5 matches / \ under {(ξ): 2}, because...
3 (ξ)
<1> <2>
...5 = s(4) = s(s(3)) = s(+(3, 1)) = +(3, s(1)) = +(3, 2) which matches 5 with {(ξ): 2}.
(as trees:)
s s <1> s <2> + + +
...5 = | = | = | = / \ = / \ which matches / \ with {(ξ): 2}.
4 s + 3 s 3 2 3 (ξ)
| / \ |
3 3 1 1
(I apologise for the size of this question. I don’t have a background, so feel free to improve this post.)
Perhaps you can direct me to some reading, or set me off in the right direction? Thanks heaps!
