从 N*N*N 到 N 的双射函数

Question

Can anyone help me in finding a bijective mathematical function from N * N * N → N that takes three parameters x, y, and z and returns a number n?

I would like to know the function f and its inverse f' in a way that if I have n I will be able to determine x, y, z by applying f'(n).

Answer 1

Defining f as a composition of a simpler function g

Suppose g is a bijection from N × N to N and let g ^-1 be its inverse. Then we can define f in terms of g as follows.

f(x, y, z) = g(g(x, y), z) = n

f ^-1 (n) = (x, y, z) where g ^-1 (n) = (w, z) and g ^-1 (w) = (x, y)

Defining g as a bijection from N × N to N

We now have the much simpler problem of defining g.

g(x, y) = (x + y)(x + y + 1) / 2 + y = n

g ^-1 (n) = (x, y) where m = ⌊(2n) ^1/2 ⌋ and exactly one of the following two conditions hold.

• x + y = m and y = n - m(m + 1) / 2

• x + y = m - 1 and y = n - m(m - 1) / 2

Python implementation

def f(x, y, z):
    return g(g(x, y), z)

def f_inv(n):
    w, z = g_inv(n)
    x, y = g_inv(w)
    return (x, y, z)

def g(x, y):
    return (x + y) * (x + y + 1) / 2 + y

def g_inv(n):
    m = math.floor(math.sqrt(2 * n))
    while True:
        y = n - m * (m + 1) / 2
        if y >= 0:
            break
        m -= 1
    x = m - y
    return x, y

Answer 2

F(x,y,z) = 2^x*3^y*5^z

In fact you can choose any distinct set of prime numbers. And inverse is simply by factorizing to corresponding prime factors.

Answer 3

你的函数不是满射的，让 p 是一个素数，我们在 N 中找不到任何 x,y,z 使得 p=2^x 3^y 5^z ...