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790a5941c3
This function returns the number of leaves in the parse tree of an expression. The motivation is to provide an initial complexity measure that could be used to decide whether or not to apply some simplification rules. Docs, embedded docs, and test cases are provided. Resolves #2389.
60 lines
1.6 KiB
JavaScript
60 lines
1.6 KiB
JavaScript
import { factory } from '../../utils/factory.js'
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const name = 'leafCount'
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const dependencies = [
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'parse',
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'typed'
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]
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export const createLeafCount = /* #__PURE__ */ factory(name, dependencies, ({
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parse,
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typed
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}) => {
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// This does the real work, but we don't have to recurse through
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// a typed call if we separate it out
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function countLeaves (node) {
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let count = 0
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node.forEach(n => { count += countLeaves(n) })
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return count || 1
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}
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/**
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* Gives the number of "leaf nodes" in the parse tree of the given expression
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* A leaf node is one that has no subexpressions, essentially either a
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* symbol or a constant. Note that `5!` has just one leaf, the '5'; the
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* unary factorial operator does not add a leaf. On the other hand,
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* function symbols do add leaves, so `sin(x)/cos(x)` has four leaves.
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*
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* The `simplify()` function should generally not increase the `leafCount()`
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* of an expression, although currently there is no guarantee that it never
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* does so. In many cases, `simplify()` reduces the leaf count.
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*
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* Syntax:
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*
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* leafCount(expr)
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*
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* Examples:
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*
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* math.leafCount('x') // 1
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* math.leafCount(math.parse('a*d-b*c')) // 4
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* math.leafCount('[a,b;c,d][0,1]') // 6
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*
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* See also:
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*
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* simplify
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*
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* @param {Node|string} expr The expression to count the leaves of
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*
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* @return {number} The number of leaves of `expr`
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*
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*/
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return typed(name, {
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string: function (expr) {
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return this(parse(expr))
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},
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Node: function (expr) {
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return countLeaves(expr)
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}
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})
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})
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