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An easy problem and a hard problem

As we mentioned at the end of the last handout, cryptography exploits the dichotomy between easy computational problems and hard computational problems. We saw in that handout that the following computational problem is easy: even for input numbers with hundreds of digits, a computer can calculate the output in a reasonable amount of time.

modular exponentiation problem
input: modulus m, standard name b, positive integer k
output: standard name c such that

displaymath683

We next give an example of a computational problem that is believed to be hard.

modular logarithm problem
input: modulus m, standard name b, standard name c
output: nonnegative integer k such that tex2html_wrap_inline693 if such an integer exists; otherwise, the answer ``none exists''.

If there exists an integer k such that tex2html_wrap_inline693 , we say that k is a mod-m-logarithm (base b) of c. In this class, we sometimes write this mathematically as

displaymath705

This notation is deceptive in that it suggests that tex2html_wrap_inline707 is a function. In fact, as our description of the computational problem indicates, there is sometimes no modular logarithm for a given m,b, and c. Moreover, Euler's Theorem (described in Handout 7) can be used to show that if there is a modular logarithm, there are an infinite number. Suppose k is the mod m logarithm base b of c. This means that

displaymath723

By Euler's Theorem, as long as b is relatively to prime to m,

displaymath729

Multiplying these two equations, we get

displaymath731

Finally, using the first law of exponentiation, we get

displaymath733

That is, tex2html_wrap_inline735 is also a mod m logarithm base b of c.

Thus tex2html_wrap_inline707 fails to be a function in two ways: for some inputs, it has no answers and for some inputs it has many answers. Nevertheless it is often useful to view it as a function. We can get away with this behavior because in many cases where it arises, (1) we know from the start that there is some answer, and (2) it doesn't matter which answer we get.




next up previous
Next: How modular logarithms differ Up: Contents Previous: Contents

Lisa Eckstein
Thu Nov 21 01:20:27 EST 1996