⚙️ Utilisation d'un meilleur algorithme d'inversion

Pas pour autant lourd : une fonction de cent lignes
pull/138/head
mama 2017-11-06 20:12:08 +01:00
parent d3fb1e80be
commit a0f62a9058
3 changed files with 133 additions and 32 deletions

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@ -36,6 +36,8 @@ import {
applyOrEmpty
} from './traverse-common-functions'
import uniroot from './uniroot'
let nearley = () => new Parser(Grammar.ParserRules, Grammar.ParserStart)
/*
@ -403,34 +405,13 @@ export let computeRuleValue = (formuleValue, isApplicable) =>
? formuleValue
: isApplicable === false ? 0 : formuleValue == 0 ? 0 : null
let computeInversion = (objective, computeGivenInput, currentValue) => {
let v = currentValue || objective, // notre première approximation est l'objectif lui-même (on suppose donc qu'ils sont du même ordre de grandeur, ce qui est vrai pour les salaires mais pas forcément pour d'autres variables évidemment)
here = computeGivenInput(v)
console.log('coucou', v, here)
if (Math.abs(here - objective) < 20 ) {
return v
}
let
ascend = computeGivenInput(v + 10),
descend = computeGivenInput(v - 10)
if (Math.abs(ascend - objective) < Math.abs(descend - objective))
return computeInversion(objective, computeGivenInput, v + 10)
else
return computeInversion(objective, computeGivenInput, v - 10)
}
export let treatRuleRoot = (rules, rule) => {
let evaluate = (situationGate, parsedRules, r) => {
let inversions = r['inversions possibles']
if (inversions) {
/*
Quel inversion possible est renseignée dans la situation courante ?
Quelle inversion possible est renseignée dans la situation courante ?
Ex. s'il nous est demandé de calculer le salaire de base, est-ce qu'un candidat à l'inversion, comme
le salaire net, a été renseigné ?
*/
@ -439,14 +420,18 @@ export let treatRuleRoot = (rules, rule) => {
if (fixedObjective != null) {
let
fixedObjectiveRule = findRuleByName(parsedRules, fixedObjective),
nodeValue = computeInversion(
situationGate(fixedObjective),
i =>
evaluateNode(
n => (r.name === n || n === 'sys.filter') ? i : situationGate(n),
parsedRules,
fixedObjectiveRule
).nodeValue
fx = x => evaluateNode(
n => (r.name === n || n === 'sys.filter') ? x : situationGate(n), //TODO pourquoi doit-on nous préoccuper de sys.filter ?
parsedRules,
fixedObjectiveRule
).nodeValue,
tolerancePercentage = 0.00001,
nodeValue = uniroot(
x => fx(x) - situationGate(fixedObjective),
0,
1000000000,
tolerancePercentage * situationGate(fixedObjective),
100
)
return {nodeValue}

116
source/engine/uniroot.js Normal file
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@ -0,0 +1,116 @@
/**
* Searches the interval from <tt>lowerLimit</tt> to <tt>upperLimit</tt>
* for a root (i.e., zero) of the function <tt>func</tt> with respect to
* its first argument using Brent's method root-finding algorithm.
*
* Translated from zeroin.c in http://www.netlib.org/c/brent.shar.
*
* Copyright (c) 2012 Borgar Thorsteinsson <borgar@borgar.net>
* MIT License, http://www.opensource.org/licenses/mit-license.php
*
* @param {function} function for which the root is sought.
* @param {number} the lower point of the interval to be searched.
* @param {number} the upper point of the interval to be searched.
* @param {number} the desired accuracy (convergence tolerance).
* @param {number} the maximum number of iterations.
* @returns an estimate for the root within accuracy.
*
*/
export default function uniroot(func, lowerLimit, upperLimit, errorTol, maxIter) {
var a = lowerLimit,
b = upperLimit,
c = a,
fa = func(a),
fb = func(b),
fc = fa,
tol_act, // Actual tolerance
new_step, // Step at this iteration
prev_step, // Distance from the last but one to the last approximation
p, // Interpolation step is calculated in the form p/q; division is delayed until the last moment
q
errorTol = errorTol || 0
maxIter = maxIter || 1000
while (maxIter-- > 0) {
prev_step = b - a
if (Math.abs(fc) < Math.abs(fb)) {
// Swap data for b to be the best approximation
(a = b), (b = c), (c = a)
;(fa = fb), (fb = fc), (fc = fa)
}
tol_act = 1e-15 * Math.abs(b) + errorTol / 2
new_step = (c - b) / 2
if (Math.abs(new_step) <= tol_act || fb === 0) {
return b // Acceptable approx. is found
}
// Decide if the interpolation can be tried
if (Math.abs(prev_step) >= tol_act && Math.abs(fa) > Math.abs(fb)) {
// If prev_step was large enough and was in true direction, Interpolatiom may be tried
var t1, cb, t2
cb = c - b
if (a === c) {
// If we have only two distinct points linear interpolation can only be applied
t1 = fb / fa
p = cb * t1
q = 1.0 - t1
} else {
// Quadric inverse interpolation
(q = fa / fc), (t1 = fb / fc), (t2 = fb / fa)
p = t2 * (cb * q * (q - t1) - (b - a) * (t1 - 1))
q = (q - 1) * (t1 - 1) * (t2 - 1)
}
if (p > 0) {
q = -q // p was calculated with the opposite sign; make p positive
} else {
p = -p // and assign possible minus to q
}
if (
p < 0.75 * cb * q - Math.abs(tol_act * q) / 2 &&
p < Math.abs(prev_step * q / 2)
) {
// If (b + p / q) falls in [b,c] and isn't too large it is accepted
new_step = p / q
}
// If p/q is too large then the bissection procedure can reduce [b,c] range to more extent
}
if (Math.abs(new_step) < tol_act) {
// Adjust the step to be not less than tolerance
new_step = new_step > 0 ? tol_act : -tol_act
}
(a = b), (fa = fb) // Save the previous approx.
;(b += new_step), (fb = func(b)) // Do step to a new approxim.
if ((fb > 0 && fc > 0) || (fb < 0 && fc < 0)) {
(c = a), (fc = fa) // Adjust c for it to have a sign opposite to that of b
}
}
}
/*
var test_counter;
function f1 (x) { test_counter++; return (Math.pow(x,2)-1)*x - 5; }
function f2 (x) { test_counter++; return Math.cos(x)-x; }
function f3 (x) { test_counter++; return Math.sin(x)-x; }
function f4 (x) { test_counter++; return (x + 3) * Math.pow(x - 1, 2); }
[
[f1, 2, 3],
[f2, 2, 3],
[f2, -1, 3],
[f3, -1, 3],
[f4, -4, 4/3]
].forEach(function (args) {
test_counter = 0;
var root = uniroot.apply( pv, args );
;;;console.log( 'uniroot:', args.slice(1), root, test_counter );
})
*/

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@ -42,12 +42,12 @@ describe("inversions", () => {
- nom: brut
format: euro
inversions possibles:
inversions possibles:
- net
`,
rules = yaml.safeLoad(rawRules).map(enrichRule),
analysis = analyseSituation(rules, "brut")(stateSelector)
expect(analysis.nodeValue).to.be.closeTo(2570, 0.001)
expect(analysis.nodeValue).to.be.closeTo(2000/(77/100), 0.0001*2000)
})
})