There are 706 possible assignments of gray levels to two adjacent pixels that satisfy the given rule.
Let's consider the gray levels of two adjacent pixels as A and B. The rule states that the absolute difference between A and B cannot be more than one. There are three possible cases:
∣A−B∣=0 (the gray levels are the same)
∣A−B∣=1 (the gray levels differ by one)
Now, let's count the number of possibilities for each case:
Case
∣A−B∣=0: There are 16 choices for both A and B (since they can be any of the 16 gray levels). So, there are
16×16=256 possibilities for this case.
Case
∣A−B∣=1: There are two sub-cases for this:
a. A<B: In this case,
A can be any of the 15 gray levels from 0 to 14, and
B can be any of the 15 gray levels from 1 to 15. So, there are
15×15=225 possibilities for this sub-case.
b. A>B: Similarly,
A can be any of the 15 gray levels from 1 to 15, and
B can be any of the 15 gray levels from 0 to 14. Again, there are
15×15=225 possibilities for this sub-case.
Now, add up the possibilities for both sub-cases of
∣A−B∣=1:
225+225=450.
Finally, sum up the possibilities for both cases:
256+450=706
So, there are 706 possible assignments of gray levels to two adjacent pixels that satisfy the given rule.