Optimization algorithm for calculating multiplier and divisor values

Question

I'm trying to optimize an algorithm and I can't think of a better way to do it.

There is one input (a clock frequency value) that will go through a combination of multipliers and divisors.

• The goal is to find the set of multiplier and divisor values that will produce the desired output value, given the input.

OutClk = (InClk * Mult1 * Mult2 * Mult3 * Mult4 / Div1 ) / Div2

My current (naive?) implementation is:

#define PRE_MIN 10000000
#define PRE_MAX 20000000

// Available values of the multipliers and divisors.
uint8_t mult1_vals[] = {1, 2};
uint8_t mult2_vals[] = {1, 2, 4, 8};
uint8_t mult3_vals[] = {3, 5, 7};
uint8_t div1_vals[] = {1, 2, 4};
uint8_t div2_vals[] = {1, 2, 4, 8};

bool exists_mults_divs(uint32_t in_val, uint32_t out_val)
{
    uint8_t i_m1, i_m2, i_m3, i_d1, i_d2;
    uint32_t calc_val;

    for (i_m1 = 0; i_m1 < sizeof(mult1_vals); i_m1++) {
    for (i_m2 = 0; i_m2 < sizeof(mult2_vals); i_m2++) {
    for (i_m3 = 0; i_m3 < sizeof(mult3_vals); i_m3++) {
    for (i_div1 = 0; i_div1 < sizeof(div1_vals); i_div1++) {

    calc_val = in_val * (mult1_vals[i_m1] * mult2_vals[i_m2] *
                         mult3_vals[i_m3] / div1_vals[i_div1]);
    if ((calc_val <= PRE_MIN) || (calc_val > PRE_MAX))
        continue; // Can this be refactored?

    for (i_div2 = 0; i_div2 < sizeof(div2_vals); i_div2++) {
        calc_val /= div2_vals[i_div2];
        if (calc_val == out_val)
            return true;
    }

    }
    }
    }
    }

    // No multiplier/divisor values found to produce the desired out_val.
    return false;
}

Is there any way to optimize this? Or use some algorithmic approach?

I'm using C, but any type of pseudo code is OK with me.

EDIT: Some examples for clarification. This will return true:

exists_mults_divs(2000000, 7000000); // in=2000000, out=7000000
// Iterating over the values internally:
// 1. in * 1 * 1 * 3 / 1 = 6000000
//    6000000 is not within PRE_MIN/MAX range of 10-20000000.
// 2. in * 1 * 1 * 5 / 1 = 10000000 is within range, try varying div2
//    2a. div2=1 => 10000000 / 1 = 10000000 != 7000000 not desired out
//    2b. div2=2 => 10000000 / 2 = 50000000 != 7000000
//    etc.
// 3. in * 1 * 1 * 7 / 1 = 7000000 not within range
// etc.
// 4. in * 1 * 2 * 7 / 1 = 14000000 is within range, try varying div2
//    4a. div2=1 => 14000000 / 1 != 7000000
//    4b. div2=2 => 14000000 / 2 == 7000000 IS desired out
//
// RETURN RESULT:
//    TRUE since a 2000000 in can generate a 7000000 out with
//    mult1=1, mult2=2, mult3=7, div1=1, div2=2

This will return false:

exists_mults_divs(2000000, 999999999);

Because there is no combination of divisor and multiplier with the available values that will result in getting the 999999999.

Answer 1

Reordering the formula, we have

OutClk/(Mult1*Mult2*Mult3) = InClk/(Div1*Div2);

• Take a look at the Mult1 = {1, 2} and Mult2 = {1, 2, 4, 8}, notice that, they are all power of two.

• Similarly, with Div1 and Div2, they also power of two.

• Mult3 = {3,5,7} , which are all prime numbers.

So, what we need to do is divide both InClk and OutClk by their greatest common divider (GCD)

int g = gcd(InClk, OutClk);

InClk /= g;

OutClk/= g;

In order for InClk == OutClk, we need to make both InClk/g and OutClk/g equal to 1.

And what is left in InClk after the division, we try to divide it by the largest element in each div_vals that InClk can be divided. (Because each element in div_vals all are power of two, so we need to pick the largest).

for(int i = sizeof(div1_vals) - 1; i>= 0; i--)
    if(InClk % div1_vals[i] == 0){
        InClk/= div1_vals[i];
        break;
    }

for(int i = sizeof(div2_vals) - 1; i>= 0; i--)
    if(InClk % div2_vals[i] == 0){
        InClk/= div2_vals[i];
        break;
    }

Similarly for OutClk

for(int i = sizeof(mul1_vals) - 1; i>= 0; i--)
    if(OutClk % mul1_vals[i] == 0){
        OutClk/= mul1_vals[i];
        break;
    }
....

In the end, if InClk == 1 and OutClk == 1, we return true, else false.

Time complexity is O(n) with n is maximum number of elements in all mul1_vals, ...