Your question seems to come from the fact that you're not immediately convinced that the current problem can't be harder. The current problem can't be harder than any NP-complete problem because it is in NP.

If you want to be convinced that the notion of NP-completeness even exists (i.e. that you can reduce anything in NP to an NP-complete problem) you should study the Cook-Levin theorem which states that SAT is NP-complete.

Since NP-completeness states something about being the target of a reduction from any problem in NP, the proof works by encoding the Turing machine which was the verifier of the original problem as a formula. This formula is satisfiable if and only if there exists an input which lets the Turing machine accept. You could in principle use this proof to reduce your current problem to SAT, and after that to every other NP-complete problem via transitivity of polynomial-time reductions.

(I actually wanted to just add a link to the Cook-Levin theorem in a comment, but this is my first stackexchange post and I think it doesn't let me comment yet.)

Your question seems to come from the fact that you're not immediately convinced that the current problem can't be harder. The current problem can't be harder than any NP-complete problem because it is in NP.

If you want to be convinced that the notion of NP-completeness even exists (i.e. that you can reduce anything in NP to an NP-complete problem) you should study the Cook-Levin theorem which states that SAT is NP-complete.

Since NP-completeness states something about being the target of a reduction from any problem in NP, the proof works by encoding the Turing machine which was the verifier of the original problem as a formula. This formula is satisfiable if and only if there exists an input which lets the Turing machine accept. You could in principle use this proof to reduce your current problem to SAT, and after that to every other NP-complete problem via transitivity of polynomial-time reductions.

Your question seems to come from the fact that you're not immediately convinced that the current problem can't be harder. The current problem can't be harder than any NP-complete problem because it is in NP.

If you want to be convinced that the notion of NP-completeness even exists (i.e. that you can reduce anything in NP to an NP-complete problem) you should study the Cook-Levin theorem which states that SAT is NP-complete.

Since NP-completeness states something about being the target of a reduction from any problem in NP, the proof works by encoding the Turing machine which was the verifier of the original problem as a formula. This formula is satisfiable if and only if there exists an input which lets the Turing machine accept. You could in principle use this proof to reduce your current problem to SAT, and after that to every other NP-complete problem via transitivity of polynomial-time reductions.

Your question seems to come from the fact that you're not immediately convinced that the current problem can't be harder. The current problem can't be harder than any NP-complete problem because it is in NP.

If you want to be convinced that the notion of NP-completeness even exists (i.e. that you can reduce anything in NP to an NP-complete problem) you should study the Cook-Levin theorem which states that SAT is NP-complete.

Since NP-completeness states something about being the target of a reduction from any problem in NP, the proof works by encoding the Turing machine which was the verifier of the original problem as a formula. This formula is satisfiable if and only if there exists an input which lets the Turing machine accept. You could in principle use this proof to reduce your current problem to SAT, and after that to every other NP-complete problem via transitivity of polynomial-time reductions.

(I actually wanted to just add a link to the Cook-Levin theorem in a comment, but this is my first stackexchange post and I think it doesn't let me comment yet.)