 While L-Gamal provides a way of signing documents digitally, there is a problem. The L-Gamal signature scheme requires evaluating high powers mod n. This takes computational time and power, and so can we reduce the computational burden. So one of the problems is that exponent. And so a useful thing to remember is the Euler-Vermont theorem. If the greatest common divisor of a and n is 1, then the least positive value x that solves a to power x congruent to 1 mod n is a divisor of phi of n. Now one important consequence of this is that exponent's mod n can be reduced mod phi of n. But we might even be able to go further. This suggests that the right choice of a allows us to work with exponent's mod q, where q is actually a divisor of phi of n. Let's see how that might work. So given that x equals 210 is the least positive solution to 2 to the x congruent to 1 mod 211, let's find 8 to power 153 mod 211. And so here we note that 2 to power 210, well that's really the same as 2 to power 3 to power 70. Now you might wonder why we split up the exponent that way, and that's because 2 to power 3 is just 8. So that means 8 to power 70 is congruent to 1. And this leads to the following useful reduction. Remember that in general exponents mod n can be reduced mod phi of n, but because 8 to power 70 is going to be congruent to 1, this means we can reduce the exponents of 8 mod 70. And so we have our exponent that we want, 153, well that's congruent to 13 mod 70. And so 8 to power 153, well that's congruent to 8 to power 13 mod 211. And here it's important to remember that we wanted to find 8 to power 153 mod 211. We can reduce the exponent mod 70, but we're still working mod 211. And so instead of having to evaluate this very high exponent, we can evaluate a much more modest exponent and get 198. And so that suggests the following simplification. The Elgamol signature algorithm relies on public base A and public modulus n. We can shorten the computation time by using base A to power r, where r is a divisor of phi of n. And this allows our exponents to be reduced mod phi of n over r. So let's see what we can do. We assume Alice has an Elgamol system with public modulus n and public base A. So let phi of n equal q times r and pick g equal to a to power r mod n. Now since g to power q is then going to be a to power q r, which is going to be congruent to 1 mod n, then the exponents of g can be reduced mod q. And what this suggests is that for digital signatures, she used g as a new base. She might make all of her communication use base g, or she can restrict this to just for digital signatures. That's up to her. And so she can sign a message m by announcing g as a public base, announcing p congruent to g to power x mod n as her public key. This is her half key for the encryption process. She'll then choose k and evaluate s1 congruent to g to power k mod n. This is her half signature. And then as before, she'll find s2 congruent to k inverse m minus xs1. And because these are exponents of g, we can reduce them mod q. And as before, p to power s1 times s1 to power s2 is going to be g to power m mod n. And so Bob can verify Alice's signature by checking to see that these two are in fact the same mod n. Now this gives us a small gain. Unfortunately, while this reduces the exponents, it doesn't reduce the modulus. So what can we do? Well, let's think about this. Since s1 appears as part of the expression for s2, which is an exponent, could we reduce s1 mod q? And the answer is we could, but we'd have to change a few other things. So let's take a look at that next.