CODEWITHSUNDEEP
algorithmmath
user2894968
user2894968

Reputation: 23

A function that converts two unit interval numbers to another unit interval number

I need a function that takes two parameters as input: 0<=a<=1 and 0<=b<=1.

The output is another number c which falls in the interval [0,1] too. c must be greater than or equal to both a and b. The function must be monotonic as well, so if a2>a1 and b2>=b1, c2 must be greater than c1, unless c2=c1=1.

Thanks you guys, I've figured out a simple solution myself. simply make c=a+b-ab, all requirements are met.

Upvotes: 2

Views: 128

Answers (4)

Lutz Lehmann
Lutz Lehmann

Reputation: 25992

Another proposal:

1-(1-a)(1-b)=a+b-ab=a+b(1-a)=b+a(1-b)

is, by the first form, inside [0,1], by the third greater than a for b<1, and by the fourth greater than b for a<1.

Upvotes: 0

4pie0
4pie0

Reputation: 29724

The requirement that c is always greater than a and b cannot be met since a and b are allowed to be 1.0 and c cannot be greater than this.

Also the requirement that

The function must be monotonic as well, so if a2>a1 and b2>=b1, c2 must be greater than c1.

is bad formed. This is not a definition of monotonicity of a multivariate function. Nevertheless, I give an example how to fulfill

The function must be monotonic as well, so if a2>a1 and b2>=b1, c2 must be greater than c1 (or c = 1.0).

condition too, in function (3).

Here I assume that c is greater or equal to both parameters:

(1)

double f( double a, double b) {
  if ( (a == 1.0) || (b == 1.0)) return 1.0;
  return max(a,b) + (1.0 - max(a,b)) / 2;
}

You can eventually specify epsilon to make accuracy explicit.

(2)

double f( double a, double b, double epsilon) {
  if ( (a == 1.0) || (b == 1.0)) return 1.0;
  return min( 1.0, max(a,b) + epsilon);
}

Function that satisfies also (bad worded monotonicity)

The function must be monotonic as well, so if a2>a1 and b2>=b1, c2 must be greater than c1.

(3)

double f( double a, double b, double epsilon) {
  if ( (a == 1.0) || (b == 1.0)) return 1.0;
  double distance = 1.0 - max(a,b);
  return 1.0 - distance + max( epsilon, a * distance);
}

Upvotes: 1

Sneftel
Sneftel

Reputation: 41474

There is no such function possible. For a=0, b=1, the only valid output would be c=1. However, if a2=0.5, b2=1, then again the only valid output is c2=1, violating your second requirement.

Upvotes: 1

Aki Suihkonen
Aki Suihkonen

Reputation: 20027

One function to consider is bit interleaving or Morton numbers.

Example:

 a = 0.1 0 0 1 = 9/16 = 0.5625
 b = 0. 0 1 1 0 = 6/16 = 0.3750
 c = 0.10010110 = 150/256 = 0.5859375   c >= a, c >= b

 a = 1-epsilon = 0.1111111
 b = 1-epsilon = 0.1111111 --> c = 0.11111111 = 1-epsilon

The numbers a, b must be sorted however, i.e. a > b

Upvotes: 0

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