Recall the \(\textsf{PARTITION}\) decision problem (and variations thereof) which asks: given some set of integers, is there two disjoint subsets of these integers such that their sums are equal? As it turns out, this decision problem is \(\textsf{NP}\)-hard and in \(\textsf{NP}\), making it an \(\textsf{NP}\)-complete language. In other words, we’re unsure if there is a polynomial time decider computing this language (and we think that \(\textsf{P} \neq \textsf{NP}\), so such a decider probably doesn’t exist). However, consider the following new language: \(\textsf{BOUNDED-PARTITION}\), which has the same specifications as before except now the magnitude of our integers are polynomially bounded. In this case, we can show a non-deterministic log-space decider for this new language (left as an exercise), which thus puts this language in \(\textsf{P}\), since \(\textsf{NL} \subseteq \textsf{P}\).

This is quite peculiar: seemingly, both nondeterministic deciders have mostly the same protocol. The only difference is that in \(\textsf{BOUNDED-PARTITION}\), their respective inputs are length/size-bounded, and immediately this puts one solvable in polynomial time but the other in (conjectured) exponential time!

A few observations: \(\textsf{PARTITION}\) is not the only language where this phenomena arises. For instance, \(\textsf{3SAT}\) is \(\textsf{NP}\)-complete, but \(\textsf{2SAT}\) is not (in fact, \(\textsf{2SAT}\) is \(\textsf{NL}\)-complete). \(\textsf{3COL}\) of graphs is also hard, but \(\textsf{2COL}\) reduces to bipartite matching.