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Problem 13

# Let $$M O D E X P=\\{\langle a, b, c, p\rangle \mid a, b, c$$, and $$p$$ are positive binary integers such that $$\left.a^{b} \equiv c(\bmod p)\right\\}$$. Show that $$M O D E X P \in P$$. (Note that the most obvious algorithm doesn't run in polynomial time. Hint: Try it first where $$b$$ is a power of 2 .)

Expert verified
The problem $$MODEXP$$, defined as $$MODEXP =\\{\langle a, b, c, p\rangle \mid a, b, c, and p$$ are binary integers such that $$a^{b} \equiv c(\bmod p)\right\\}$$, can be solved using an algorithm running in polynomial time. In the case where $$b$$ is a power of 2, the result can be efficiently computed by successively squaring the base mod $$p$$. This concept can be extended to arbitrary exponents $$b$$ with a divide-and-conquer approach, breaking down the exponentiation into smaller components and utilizing the squaring method. This approach, known as "exponentiation by squaring", evaluates even and odd exponents differently: for even exponents, we compute $$(a^{(b/2)} \cdot a^{(b/2)}) \pmod{p}$$, and for odd exponents, we compute $$(a^{(b-1)} \cdot a) \pmod{p}$$. This algorithm shows that $$MODEXP$$ is indeed a class $$P$$ problem, thus, the problem $$MODEXP \in P$$, as required.
See the step by step solution

## Step 1: Understanding the Modular Exponentiation Problem

Begin by observing that we're trying to calculate $$a^b \equiv c \pmod{p}$$, which means that we need to find the remainder when $$a^b$$ is divided by $$p$$. Our task is to prove that an algorithm for this problem exists that can operate in polynomial time with respect to the input length. Step 2:

## Step 2: Employing the Squaring Technique

For the case where $$b$$ is a power of 2, we can efficiently find the result using the squaring technique. Essentially, this method computes the modular exponentiation by successively squaring the base modulo $$p$$. Here's how the algorithm would look: 1. Initialize the result as $$a$$ and the exponent counter as $$k=1$$. 2. While $$k < b$$: a. Set the result as the square of the current result, modulo $$p$$. b. Double the exponent counter: $$k = 2k$$. At the end of this process, the result will be equal to $$a^b \pmod{p}$$. Step 3:

## Step 3: Extending to Arbitrary Exponents

For general exponents $$b$$, we need to adjust the algorithm to accommodate any non-power-of-two values. We can do this by employing a divide-and-conquer strategy, breaking down the exponentiation into smaller components and reusing the squaring method. The adapted algorithm, known as "exponentiation by squaring," can be outlined as follows: 1. If $$b = 0$$, return 1 (because $$a^0 = 1$$). 2. If $$b = 1$$, return $$a$$ (since $$a^1 = a$$). 3. If $$b$$ is even, compute $$a^{b/2}$$, then square the result and take the modulo $$p$$: $$(a^{(b/2)} \cdot a^{(b/2)}) \pmod{p}$$. 4. If $$b$$ is odd, compute $$a^{(b-1)}$$, then multiply it by $$a$$ modulo $$p$$: $$(a^{(b-1)} \cdot a) \pmod{p}$$\$. Applying the algorithm recursively, we can evaluate the modular exponentiation in polynomial time with respect to the input length. Step 4:

## Step 4: Showing that $$MODEXP$$ is a Class P Problem

Since we've successfully designed a polynomial-time algorithm for evaluating the modular exponentiation operation using the "exponentiation by squaring" technique, it demonstrates that $$MODEXP$$ is indeed a class P problem. Now we've shown that $$MODEXP \in P$$, as required by the problem.

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