How might one measure the difficulty of formulae?
Vladimir Batagelj
vladimir.batagelj at FMF.UNI-LJ.SI
Wed Jun 16 02:53:39 EDT 2010
PS. In the proposed measure
complexity =
average depth of tree leaves X average degree of internal nodes
the degree is out-degree (number of sons of a node).
For example
(-b+sqrt(b^2-4ac))/(2a)
out-degrees of
depth internal nodes
0 / 2
/ \
1 + * 2 2
/ \ / \
2 - sqrt 2 a 1 1
/ \
3 b - 2
/ \
4 ^2 * 1 3
/ / | \
5 b 4 a c
Therefore
average depth = (3+5+5+5+5+2+2)/7 = 27/7
average degree = (2+2+2+1+1+2+1+3)/8 = 14/8 = 7/4
complexity = 27/7 X 7/4 = 27/4 = 6.75
For an expression consisting of single variable or constant we set
complexity = 1.
f(g(h(x)))
f
|
g
|
h
|
x complexity = 3 X 1 = 3
f(x,y,z)
f
/ | \
x y z complexity = 1 X 3 = 3
There are two possibilities how to treat associative cases such as
a + b + c + d
The first option is
+
/ / \ \
a b c d complexity = 1 X 4 = 4
and the second
+
/ \
a +
/ \
b +
/ \
c d complexity = 9/4 X 2 = 4.5
If the names of variables and functions are of different length
the measure can be extended by the additional length term. For
example:
complexity =
average depth of tree leaves X
average degree of internal nodes X
sqrt(average length of the names)
Note also that
leaves = variables and constants
internal nodes = operations and functions
Vlado
--
Vladimir Batagelj, University of Ljubljana, FMF, Department of Mathematics
Jadranska 19, 1000 Ljubljana, Slovenia
http://vlado.fmf.uni-lj.si
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