The complexity class NP (nondeterministic polynomial time) is the set decision problems for which a given "yes" instance can be verified in polynomial time by a deterministic Turing machine, given the appropriate certificate (also known as a witness or solution).

To prove that the language COLOR is in NP, we need to demonstrate that for any given graph G, if G a member of COLOR (i.e., G can be colored 4 colors in such a way that no two adjacent vertices have the same color), then there exists some certificate that a deterministic Turing machine can use to verify G’s membership in COLOR in polynomial time.

Here's how we can prove COLOR is in NP:

. Certificate or Witness: The certificate in this case would be the actual color assignment for each vertex in the graph G.

2. Verification in Polynomial Time: Given a certificate (a color assignment), we need to check the following two conditions to verify that it is correct:
   • Condition 1: Verify that two adjacent vertices are assigned the same color. This can be done by looking at each edge in the graph once and checking the colors of the vertices it connects. Since the number of edges in a graph is at quadratic in the number of vertices, this check can be done in polynomial time.
   • Condition 2: Verify that at most 4 distinct colors are used. As look at the graph, we can keep track of used colors. Since the certificate claims the use of only 4 colors, we're only checking a constant number of colors, which doesn't depend on the size of the graph and thus can be done polynomial time.

Since both the certificate (color assignment) and the checks (conditions 1 and 2) can be conducted in polynomial time, COLOR a member of NP. We don't need to provide an algorithm to find the coloring just the ability to verify a correct coloring quickly, which is the requirement for a problem to be in NP.

The solution describes a Turing machine algorithm that calculates the square of a number x, provided in unary representation (i.e., the number 3 is represented as 111), using three tapes. The function f(x) = x^2 is computed through a series of steps that essentially add x to itself x times. Here's a breakdown of the steps described:

1. Initial Input on Tapes 1 and 2: The input w (representing the number x in unary) is printed on both Tape 1 and Tape 2. If w is 111 (which represents 3), both Tape 1 and Tape 2 will look like this:

   Tape 1: 111
   Tape 2: 111
   Tape 3: (empty)

2. Process Tape 1: The machine begins by crossing out the left-most '1' on Tape 1. This signifies that we're starting the process of adding w to itself and will do this x times. After crossing one '1' on Tape 1, the contents of Tape 2 are appended to Tape 3.

   Progression example after one crossing out and appending:

   Tape 1: X11 (X denotes the crossed out 1)
   Tape 2: 111
   Tape 3: 111 (We've just added the contents of Tape 2 to Tape 3)

3. Repeat Process for All 1's on Tape 1: This step is repeated, crossing out one '1' on Tape 1 and appending the contents of Tape 2 to Tape 3 every time, until there are no '1's left to cross out on Tape 1.

   Continuing the example, the machine will do two more cycles:

   Second cycle:
   Tape 1: XX1
   Tape 2: 111
   Tape 3: 111111
   
   Third cycle:
   Tape 1: XXX
   Tape 2: 111
   Tape 3: 111111111

4. Result on Tape 3: At the end of the computation, Tape 3 contains w^2 in unary. In our example with w = 111, the final state of the tapes is:

   Tape 1: XXX   (all 1's crossed out, meaning we've added w to itself w times)
   Tape 2: 111   (remains unchanged)
   Tape 3: 111111111 (9 1's, which is 3^2)

   The machine effectively squares the number by repeated addition, and since this is being done in unary, each addition is equivalent to string concatenation—a process that is easily done with a three-tape Turing machine.

It's important to remember that this methodology benefits from the simplicity of unary representation. If the number were in binary, the process would be more complex due to carrying over during addition and the more complex representation of numbers. However, even in unary, the algorithm runs in quadratic time relative to the number being squared because it involves adding w a total of w times.