Finding the inverse of an expression

[sfopeano]

Hi there...

I'm trying to create an equation that can be used in cryptography, so I'm trying to figure out how to get the inverse of the function. Here is what I have for the encryption function:

cipher text = k^l/(1/2k-1)^1/3

Thanks for any help...

 

[stapel]

I'm trying to figure out how to get the inverse of the function.

cipher text = k^l/(1/2k-1)^1/3

Is the power on "k" the variable "l" (being lower-case "L")? If so, on what basis have you concluded that one can find the "inverse" of this two-variable function?

Also, is the functional expression any one of the following?

. . . . .\(\displaystyle f(k,\,L)\, =\, \dfrac{k^L}{\left(\dfrac{1}{2k\, -\, 1}\right)^{\frac{1}{3}}}\)

. . . . .\(\displaystyle f(k,\,L)\, =\, \dfrac{k^L}{\left(\dfrac{1}{2}k\, -\, 1\right)^{\frac{1}{3}}}\)

. . . . .\(\displaystyle f(k,\,L)\, =\, \dfrac{k^L}{\left(\dfrac{1}{2k}\, -\, 1\right)^{\frac{1}{3}}}\)

Or something else?

When you reply, please include a clear listing of your thoughts and efforts so far. Thank you!

 

[Ishuda]

Hi there...

I'm trying to create an equation that can be used in cryptography, so I'm trying to figure out how to get the inverse of the function. Here is what I have for the encryption function:

cipher text = k^l/(1/2k-1)^1/3

Thanks for any help...

Although there is a way to compensate, the cube root function does not always admit integer (or even less than 4 'decimal' place) solutions so that your cypher text is not necessarily an integer [or even necessarily an integer if multiplied by a large number] as is usually expected.

If k is the clear text and l is a positive number, you shouldn't have trouble with uniqueness if the clear text is at least several characters long and you can always ensure that. However, the lth root may again require some 'compensation' because, again, the deciphered clear text will not generally be an integer for most choices of l.

 

[sfopeano]

Is the power on "k" the variable "l" (being lower-case "L")? If so, on what basis have you concluded that one can find the "inverse" of this two-variable function?

Also, is the functional expression any one of the following?

. . . . .\(\displaystyle f(k,\,L)\, =\, \dfrac{k^L}{\left(\dfrac{1}{2k\, -\, 1}\right)^{\frac{1}{3}}}\)

. . . . .\(\displaystyle f(k,\,L)\, =\, \dfrac{k^L}{\left(\dfrac{1}{2}k\, -\, 1\right)^{\frac{1}{3}}}\)

. . . . .\(\displaystyle f(k,\,L)\, =\, \dfrac{k^L}{\left(\dfrac{1}{2k}\, -\, 1\right)^{\frac{1}{3}}}\)

Or something else?

When you reply, please include a clear listing of your thoughts and efforts so far. Thank you!

stapel,

Thanks for your response!

My intention was for the equation to be

. . . . .\(\displaystyle f(k,\,L)\, =\, \dfrac{k^L}{\left(\dfrac{1}{2}k\, -\, 1\right)^{\frac{1}{3}}}\)

I should have specified that K is the key, and l (yes, lower case L) is the letter you are encrypting in a numeric format, a=1, b=2 etc.

And I have not concluded that the function is invertable. I'm not familiar with a test to determine the invertability of a function.