Partition Problem:
Input: $A:=$ {$a_{1}, ..., a_{n} $}. $a_{i} \in \mathbb{N}$ $\forall i \in$ $\{1, \ldots, n\}$.
Question: Exists a subset $A_{1} \subset A$ with: $\sum_{a_{i} \in A_{1}} a_{i} = \sum_{a_{i} \in A\setminus A_{1}} a_{i}$ ?
Output: Yes or No.
Makespan Problem:
Input: Jobs $J$ and $b \in \mathbb{N}$.
Question: Exists a Schedule $S$ with $\text{FinishTime}(S) \leq b$ ?
Output: Yes or No.
So I have to change the Makespan Problem in a way that the Partition Problem gives me the correct answers for the Makespan Problem. $Partition \leq_{p} Makespan$
First Problem: When the Makespan Problem has $m$ machines and not two, I don't know how I can use Partition for that scenario. I could recursive iterate the jobs (so that I split Makespan in several Partition Problems) but then I had to use the Partition Problem several times and with my understanding, I am only allowed to change the Input for a Polynomial-Time Reduction.
But let's say we have $m= 2$ machines. If Partition has the Output "Yes", I know that there exists an optimal solution for the Makespan, so that I have an optimal Schedule with value= $\dfrac{1}{m} \cdot \sum_{a_{i} \in A}$. If $b \leq \dfrac{1}{m} \cdot \sum_{a_{i} \in A}$, the result for Makespan is also Yes.
Second problem: But what happens if there is no optimal solution? I only receive "No" from Partition but I don't know how bad the Partition result is.
I would be really happy if someone has an idea for this, I don't know how to I could find a solution for this.