robust-linear-solve/linsolve.js

Exactly solve a low dimensional linear system

"use strict"

var determinant = require("robust-determinant")

var NUM_EXPAND = 6

function generateSolver(n) {
  var funcName = "robustLinearSolve" + n + "d"
  var code = ["function ", funcName, "(A,b){return ["]
  for(var i=0; i<n; ++i) {
    code.push("det([")
    for(var j=0; j<n; ++j) {
      if(j > 0) {
        code.push(",")
      }
      code.push("[")
      for(var k=0; k<n; ++k) {
        if(k > 0) {
          code.push(",")
        }
        if(k === i) {
          code.push("+b[", j, "]")
        } else {
          code.push("+A[", j, "][", k, "]")
        }
      }
      code.push("]")
    }
    code.push("]),")
  }
  code.push("det(A)]}return ", funcName)
  var proc = new Function("det", code.join(""))
  if(n < 6) {
    return proc(determinant[n])
  }
  return proc(determinant)
}

function robustLinearSolve0d() {
  return [ 0 ]
}

function robustLinearSolve1d(A, b) {
  return [ [ b[0] ], [ A[0][0] ] ]
}

var CACHE = [
  robustLinearSolve0d,
  robustLinearSolve1d
]

function generateDispatch() {
  while(CACHE.length < NUM_EXPAND) {
    CACHE.push(generateSolver(CACHE.length))
  }
  var procArgs = []
  var code = ["function dispatchLinearSolve(A,b){switch(A.length){"]
  for(var i=0; i<NUM_EXPAND; ++i) {
    procArgs.push("s" + i)
    code.push("case ", i, ":return s", i, "(A,b);")
  }
  code.push("}var s=CACHE[A.length];if(!s)s=CACHE[A.length]=g(A.length);return s(A,b)}return dispatchLinearSolve")
  procArgs.push("CACHE", "g", code.join(""))
  var proc = Function.apply(undefined, procArgs)
  module.exports = proc.apply(undefined, CACHE.concat([CACHE, generateSolver]))
  for(var i=0; i<NUM_EXPAND; ++i) {
    module.exports[i] = CACHE[i]
  }
}

generateDispatch()