How to use function of multi parameters as function with one parameters?

Question

Suppose I have a function with three parameters:

f(x, y, z)
{
  return x*x + y*y + z*z;
}

and I have a minimum search function golden() only works for function with one parameter.

      //Given a function myFunc, and a bracketing triplet of abscissas ax bx cx(such that bx is between ax and cx, and myFunc(bx) is less than both myFunc(ax) and myFunc(cx)). This routine performs a golden section searhc for the minimum, isolating it to a fractional precision of about tol. The abscissa of the minumum is xmin.
      function golden(ax, bx, cx, myFunc, tol)
      {
        var r = 0.61803399;
        var c = 1.0 - r;
        var f1, f2, x0, x1, x2, x3, xmin;

        x0 = ax;            //At any given time we will keep track of four points, x0, x1, x2, x3.
        x3 = cx;
        if(Math.abs(cx - bx) > Math.abs(bx - ax))   //Make x0 to x1 the smaller segment
        {
            x1 = bx;
            x2 = bx + c * (cx - bx);                 //and fill in the new poit to be tried
        }else
        {
            x2 = bx;
            x1 = bx - c * (bx - ax);
        }
        f1 = myFunc(x1);        //the initial funciton evaluations. Note that we never neeed to evaluate the function at the original endpoints.
        f2 = myFunc(x2);
        while(Math.abs(x3 - x0) > tol * (Math.abs(x1) + Math.abs(x2)))
        {
          if(f2 < f1)           //One possible outcome,
          {
            x0 = x1; x1 = x2; x2 = r  * x1 + c * x3;    //its housekeeping,
            f1 = f2; f2 = myFunc(x2);                   //and a new funciton evaluation
          }else                 //The other outcome,
          {
            x3 = x2; x2 = x1; x1 = r * x2 + c * x0;
            f2 = f1; f1 = myFunc(x1);                   //and its new funciton evaluation.
          }
        }                       //Back to see if we are done.
        if(f1 < f2)             //We are done. Output the best of the two current values.
        {
            xmin = x1;
            //return f1;
        }else
        {
            xmin = x2;
            //return f2;
        }
        return xmin;
      }

How can I trans the f with three parameters to func with one parameter.

I have tryed to wrap the f like so:

wrapFunc(x)
{
  f(x, 0, 0);
}

But I use a constant y:0, z:0 here. I want to make the y, and z assiginable?

I need to search in x, y, z direction separately. And the search base is on the prior search.

For example. First base is (1,1,1) x direction search -> (0, 1, 1) then y directon search -> (0, 0, 1) then z direction search -> (0,0,0);

The programming language is javascript. Any help will be appreciated. Thanks

it depends on the calculation, you make. please add the rest as well and have a look here minimal reproducible example. — Nina Scholz
– Nina Scholz, Commented Dec 12, 2016 at 10:08

Alex Young · Accepted Answer · 2016-12-12 10:29:42Z

2

You can use currying. E.g.

function curry (y,z) {
    return function (x)
    {
       console.log(x +  y + z);
    }
}

var addToThis = curry(1,2);
addToThis(3); // 6
addToThis(5); //8

Edit: You added some more code, so more specifically...

function presetGoldenBxCx(bx, cx) {
    return function golden(ax, myFunc, tol)
          {
            var r = 0.61803399;
            var c = 1.0 - r;
            var f1, f2, x0, x1, x2, x3, xmin;

            x0 = ax;            //At any given time we will keep track of four points, x0, x1, x2, x3.
            x3 = cx;
            if(Math.abs(cx - bx) > Math.abs(bx - ax))   //Make x0 to x1 the smaller segment
            {
                x1 = bx;
                x2 = bx + c * (cx - bx);                 //and fill in the new poit to be tried
            }else
            {
                x2 = bx;
                x1 = bx - c * (bx - ax);
            }
            f1 = myFunc(x1);        //the initial funciton evaluations. Note that we never neeed to evaluate the function at the original endpoints.
            f2 = myFunc(x2);
            while(Math.abs(x3 - x0) > tol * (Math.abs(x1) + Math.abs(x2)))
            {
              if(f2 < f1)           //One possible outcome,
              {
                x0 = x1; x1 = x2; x2 = r  * x1 + c * x3;    //its housekeeping,
                f1 = f2; f2 = myFunc(x2);                   //and a new funciton evaluation
              }else                 //The other outcome,
              {
                x3 = x2; x2 = x1; x1 = r * x2 + c * x0;
                f2 = f1; f1 = myFunc(x1);                   //and its new funciton evaluation.
              }
            }                       //Back to see if we are done.
            if(f1 < f2)             //We are done. Output the best of the two current values.
            {
                xmin = x1;
                //return f1;
            }else
            {
                xmin = x2;
                //return f2;
            }
            return xmin;
          }
}

const golden11= presetGoldenBxCx(1, 1);
const answer = golden11(1);

edited Dec 12, 2016 at 10:29

answered Dec 12, 2016 at 10:23

Alex Young

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3 Comments

LF-DevJourney Over a year ago

is the const necessary?

LF-DevJourney Over a year ago

looks well, I'll check it. Thank you.

LF-DevJourney Over a year ago

thank you. Save me a lot of time, otherwise I've to reconstruct my cod.

Elad Douenias · Accepted Answer · 2016-12-12 10:09:23Z

1

you can just call f with one parameter. all the other params will have the value 'undefined':

f(5); // x=5, y=undefined, z=undefined

answered Dec 12, 2016 at 10:09

Elad Douenias

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2 Comments

Andre Pena Over a year ago

Came here to say this

LF-DevJourney Over a year ago

How I search in the y direction? And how can I search in the x directionn from the begin{x:4, y:5, z:4}?

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