- #1
Ben Huke
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Dear number theorists and enthusiasts,
Following is a math-computing problem. If you are interested and want to be paid for solving it, please contact me (benhuke@gmail.com).
Suppose we have a set of n positive 32-bit integers: X = {x1, x2, … x_n}
Find a function f1 such that with T = f1(x1, x2, … x_n), it’s computationally infeasible to decompose T back to x_i's. That is, the time it takes to find the original component numbers will increase exponentially with T. However, also find a function f2 such that if r = f2(T, x) = 0 then we can conclude, with as high a probability (p) as possible, that x is in X.
For a rough example, we can have:
f1: T = x1 * x2 * … * x_n (In this case, with n > 20, is it computationally infeasible to decompose T back to x_i's ?)
f2: r = T mod x
Further requirements: The solution must have a higher p and have f2 more efficient (i.e. requiring less computation) than those in the example.
Please give my your opinion about this problem (e.g., how difficult is it? is it possibly solvable?).
Sincerely thanks,
Ben
Following is a math-computing problem. If you are interested and want to be paid for solving it, please contact me (benhuke@gmail.com).
Suppose we have a set of n positive 32-bit integers: X = {x1, x2, … x_n}
Find a function f1 such that with T = f1(x1, x2, … x_n), it’s computationally infeasible to decompose T back to x_i's. That is, the time it takes to find the original component numbers will increase exponentially with T. However, also find a function f2 such that if r = f2(T, x) = 0 then we can conclude, with as high a probability (p) as possible, that x is in X.
For a rough example, we can have:
f1: T = x1 * x2 * … * x_n (In this case, with n > 20, is it computationally infeasible to decompose T back to x_i's ?)
f2: r = T mod x
Further requirements: The solution must have a higher p and have f2 more efficient (i.e. requiring less computation) than those in the example.
Please give my your opinion about this problem (e.g., how difficult is it? is it possibly solvable?).
Sincerely thanks,
Ben