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Two philosophers A and B, two forks numbered "1", "2"; A needs both "1" and "2" for eating, so does B.
Is this theoretical a dining philosopher problem? I'm questioning about the forks philosopher A needs all come from B. However, the DP problem usually gives an example of 5 philosophers, in which case, for example, philosophers A requests 2 forks from two different neighbors.
Short answer: yes, though it could be considered a "degenerate case" (that is, a case that ends up being somewhat different from most others).
Long answer: the Dining Philosophers problem is really about having $n$ philosophers, and coming up with an algorithm that works for any $n$. A good algorithm should work for even "degenerate cases" like zero philosophers or one philosopher—by hard-coding them if necessary ("if there's only one philosopher, eat whenever you feel like"). So $n=2$ is a perfectly valid Dining Philosophers problem.
