Assuming the original instance is not easy in NP-hardness reduction Partition problem: given a non-empty finite set $P = \{p_i : i ∈ I = \{1,\dots,m\}\}$ of $m$ positive integers such that $\sum_{i=1}^m p_i = 2T$, can $I$ be partition into two disjoint subsets $I_1, I_2$ such that $\sum_{ i \in I _ 1} p_i = \sum_{ i \in I _ 2} p_i = T$?
My question is: when we deriving reduction from Partition to another problem, can we assume that $\max_{ i\in I} p_i \leq T$?
A polynomial time reduction from PARTITION to a language $L$ is a polynomial time function $f$ such that if $x$ is a Yes instance of PARTITION then $f(x) \in L$, and if $x$ is a No instance of PARTITION then $f(x) \notin L$.
If $x$ is an &quot;easy&quot; instance — belongs to some polynomially solvable class of instances — then $f$ can just solve it, outputting a fixed Yes or No instance of $L$. Therefore you can assume that $x$ is a &quot;hard&quot; instance.