# What does binary NP-Hard exactly mean?

Can someone tell me what binary NP-Hard exactly means. My understanding was that if the problem is encoded in binary and if the problem is binary NP-Hard, its complexity of solving is NP-Hard. Is this the correct interpretation?

I also wanted to know, if binary NP-Hard also says something about Strong or Weak NP-Completeness of the problem.

One of the go-to examples here is the Knapsack problem. Given $$n$$ items with (integer) weights $$w_1,...,w_n$$, and two integers $$W,V$$, decide whether there exists a subset $$S\subseteq [n]$$ such that $$V\le \sum\limits_{i\in S}w_i\le W$$. We know that the problem can be solved in time $$O(nW)$$ in the word-ram model. This means that assuming $$P\neq NP$$, knapsack is not NP complete if the input is of size $$\Omega(nW)$$, hence it is not complete when $$W$$ is encoded in unary. Thus, when talking about the NP completeness of knapsack, it is important to specify that we're talking about the version where $$W$$ is encoded in binary, or "binary knapsack" for short.