I want to prove the NP hardness of a planning problem. The problem seems to be NP-hard in the ordinary sense of the term, so I try with the problem of partition, precisely the partition with equal cardinality (ECP). So we have:

**(ECP):** Let $ X = {x_1, x_2, points, x_ {2n} } $ is a set of positive integers, is there a partition of $ X $ in two subassemblies $ X_1 $ and $ X_2 $, such as $ sum_ {x_i in X_1} x_i = sum_ {x_i in X_2} x_i = B $, or $ B $ is a positive integer, and such that $ | X_1 | = | X_2 | $?

The entries for my planning problem are a set of $ n $ works, where each job has a processing time $ p_i $ and a date of expiry $ d_i $. Thus, my instance has jobs for which the processing times are related to integers of the partition problem, ie. $ p_i = x_i $.

The question I have is this:

If I assign a common expiry date for all jobs, that is to say $ d_i = d $then the problem is not NP-difficult. So, how can I generate the instance with unequal deadlines? For example, can I use the same integers $ x_i $ for the due dates (eg. $ d_i = B-x_i $)? Is this correct if I use the work *login* ($ i $) in the due dates (for example $ d_i = B-2i ^ 2 $)?

p.s. I realized that I can not use the $ x_i $ from one job to the date of expiry of another job because these jobs will be related to each other, which is not correct. In fact, this makes even the problems very simple to solve.