Finding the maximum of a floating point counter - c++

Question

My apologies if this has been asked before, but I cannot find it.
I was wondering if there is a way to calculate the point at which a single precision floating point number that is used as a counter will reach a 'maximum' (the point at which it is no longer able to add another value due to loss of precision).
For example, if I continuously add 0.1f to a float I will eventually reach a point where the value does not change:
const float INCREMENT = 0.1f;
float value = INCREMENT;
float prevVal = 0.0f;
do {
prevVal = value;
value += INCREMENT;
} while (value != prevVal);
cout << value << endl;
On GCC this outputs 2.09715e+06
Is there a way to compute this mathematically for different values of INCREMENT? I believe it should in theory be when the exponent portion of the float requires a shift beyond 23 bits, resulting in losing the mantissa and simply adding 0.

Answer 1

Given some positive y used as an increment, the smallest X for which adding y does not produce a result greater than X is the least power of 2 not less than y divided by half the “epsilon” of the floating-point format. It can be calculated by:
Float Y = y*2/std::numeric_limits<Float>::epsilon();
int e;
std::frexp(Y, &e);
Float X = std::ldexp(.5, e);
if (X < Y) X *= 2;
A proof follows. I assume IEEE-754 binary floating-point arithmetic using round-to-nearest-ties-to-even.
When two numbers are added in IEEE-754 floating-point arithmetic, the result is the exact mathematical result rounded to the nearest representable value in a selected direction.
A note about notation: Text in source code format represents floating-point values and operations. Other text is mathematical. So x+y is the exact mathematical sum of x and y, x is x in floating-point format, and x+y is the result of adding x and y in a floating-point operation. Also, I will use Float for the floating-point type in C++.
Given a floating-point number x, consider adding a positive value y using floating-point arithmetic, x+y. Under what conditions will the result exceed x?
Let x1 be the next value greater than x representable in the floating-point format, and let xm be the midpoint between x and x1. If the mathematical value of x+y is less than xm, then the floating-point calculation x+y rounds down, so it produces x. If x+y is greater than xm, either it rounds up and produces x1, or it produces some greater number because y is large enough to move the sum beyond x1. If x+y equals xm, the result is whichever of x or x1 has an even low digit. For reasons we will see, this is always x in the situations relevant to this question, so the calculation rounds down.
Therefore, x+y produces a result greater than x if and only if x+y exceeds xm, meaning that y exceeds half the distance from x to x1. Note that the distance from x to x1 is the value of 1 in the low digit of the significand of x.
In a binary floating-point format with p digits in its significand, the position value of the low digit is 21−p times the position value of the high digit. For example, if x is 2e, the highest bit in its significand represents 2e, and the lowest bit represents 2e+1−p.
The question asks, given a y, what is the least x for which x+y does not produce a result greater than x? It is the least x for which y does not exceed half the value of the low digit of the significand of x.
Let 2e be the position value of the high bit of the significand of x. Then y ≤ ½•2e+1−p = 2e−p, so y•2p ≤ 2e.
Therefore, given some positive y, the least x for which x+y does not produce a result greater than x has its leading bit, 2e, equal to or exceeding y•2p. And in fact it must be exactly 2e because all other floating-point numbers whose leading bit has position value 2e have other bits set in their significands, so they are greater. 2e is the least number for which the leading bit represents 2e.
Therefore, x is the least power of two that equals or exceeds y•2p.
In C++, std::numeric_limits<Float>::epsilon() (from the <limits> header) is the step from 1 to the next representable value, meaning it is 21−p. So y•2p equals y*2/std::numeric_limits<Float>::epsilon(). (This operation is exact unless it overflows to ∞.)
Let’s assign this to a variable:
Float Y = y*2/std::numeric_limits<Float>::epsilon();
We can find the position value represented by the highest bit of Y’s significand by using frexp (from the <cmath> header) to extract the exponent from the floating-point representation of Y and ldexp (also <cmath>) to apply that exponent to a new significand (.5 because of the scale that frexp and ldexp use):
int e;
std::frexp(Y, &e);
Float X = std::ldexp(.5, e);
Then X is a power of two, and it is less than or equal to Y. It is in fact the greatest power of two not greater than Y, since the next greater power of 2, 2X, is greater than Y. However, we want the least power of two not less than Y. We can find this with:
if (X < Y) X *= 2;
The resulting X is the number sought by the question.

Answer 2

Marek's Answer is pretty close, and a decent way to find it using a program (that is more efficient than the one I originally posted). However, I don't necessarily need the answer in a program form, just a mathematical one.
From what I can tell, the answer comes down to the exponent of the delta used, and the number of mantissa bits. We need to round to the nearest power of 2, which is kind of complicated. Basically if the mantissa is 0, we do nothing, otherwise we add 1 to the exponent. So, assuming we now have the delta as a power of 2, represented as 1.0 x 2exp, and a mantissa of N bits, the maximum value is 1.0 x 2(N + exp). Note that FLT_EPSILON in C is equal to 1.0 x 2-N. So we can also find this by dividing our nearest power of 2 by FLT_EPSILON.
For a delta of 0.1, the nearest power of 2 is 0.125, or 1.0 x 2-3. Therefore we want 1.0 x 2(23 + (-3)) or 1.0 x 221 which is equal to 2097152.

Answer 3

Yes it is possible.
there is std::numeric_limits::epsilon() which defines smallest value which can increase value 1.0.
Using this you can calculate this limit for any number.
In C there is DBL_EPSILON
So in your case this goes like this:
template<class T>
auto maximumWhenAdding(T delta) -> T
{
static_assert(std::is_floating_point_v<T>, "Works only for floating points.");
int power2= std::ilogb(delta);
float roudedDelta = ldexp(T { 1.0 }, power2);
if (roudedDelta != delta) {
roudedDelta *= 2;
}
return 2 * roudedDelta / std::numeric_limits<T>::epsilon();
}
live example C++
Note in live test examples delta fails to increase maxForDelta, but subtraction is successful, so this is exactly what you need.