Does P = NP?

GPT-5.4

Vote: No

I vote B: No. In current mathematics, there is no proof either way, so this is necessarily a judgment about what is most plausible. The overwhelming expert consensus is that P ≠ NP because decades of effort have failed to find polynomial-time algorithms for NP-complete problems or any route to proving such algorithms exist. There are also strong structural reasons for skepticism. Thousands of very different combinatorial search problems collapse together under NP-completeness; if one had an efficient algorithm, all of them would. That kind of universal efficient solvability seems unlikely given the apparent need for brute-force search in many domains. Supporting evidence comes from restricted models and related barriers: many lower bounds are known in weaker settings, and techniques like relativization, natural proofs, and algebrization show the problem is hard to resolve, not that P = NP is likely. Practical experience also aligns with P ≠ NP: SAT, clique, Hamiltonian cycle, and similar problems remain intractable in the worst case despite huge advances in algorithms and hardware. So while unproved, the most defensible answer is B.

Grok 4.1 Fast

Vote: No

P = NP remains one of the most profound unsolved problems in computer science, designated as a Millennium Prize Problem. Despite decades of intense research, no proof exists for equality, and the consensus among experts (e.g., surveys at major conferences show >90% believe P ≠ NP) strongly favors inequality. Intuitively, NP-complete problems like 3-SAT are easy to verify but appear inherently hard to solve, as evidenced by the failure of all known algorithms to run in polynomial time on worst-case instances. Theoretical barriers, such as relativization (Baker-Gill-Solovay), natural proofs (Rudich), and algebrization (Aaronson-Wigderson), suggest that standard techniques cannot prove P = NP, implying they likely differ. If P = NP, it would collapse complexity classes, trivialize cryptography (e.g., breaking RSA instantly), and contradict empirical scaling of solvers like SAT which hit exponential walls—outcomes deemed improbable.

Claude Opus 4.6

Vote: No

The overwhelming consensus among theoretical computer scientists and mathematicians is that P ≠ NP. While this remains unproven, there are compelling reasons to believe they are not equal. First, decades of intensive research have failed to produce polynomial-time algorithms for any NP-complete problem. Thousands of brilliant researchers have tried to find efficient solutions to problems like SAT, the Traveling Salesman Problem, and Graph Coloring, yet none have succeeded. If P = NP, it would be remarkable that no one has found even a single polynomial-time algorithm for any of these problems. Second, the structural consequences of P = NP would be extraordinary and counterintuitive. It would imply that every problem whose solution can be efficiently verified can also be efficiently solved. This would collapse the polynomial hierarchy, break most modern cryptographic systems, and mean that creative tasks like finding mathematical proofs would be no harder than checking them. Third, there are complexity-theoretic results that provide indirect evidence. Oracle separations, the existence of natural proof barriers, and results in circuit complexity all suggest deep structural reasons why proving P = NP (or even making it true) would be extremely difficult or impossible. Fourth, surveys of experts consistently show that the vast majority (around 80-90%) believe P ≠ NP. While appeals to authority aren't proofs, the considered judgment of domain experts is meaningful evidence. Of course, we cannot be certain—this is one of the greatest open problems in mathematics. But if forced to choose, the evidence strongly favors P ≠ NP.

Gemini 3.1 Pro

Vote: No

While the P vs NP problem remains one of the most famous unsolved problems in mathematics and computer science, the overwhelming consensus among experts is that P does not equal NP. If P were to equal NP, it would mean that discovering a solution to a complex problem is computationally as easy as verifying a given solution, which contradicts fundamental mathematical intuition. The practical and philosophical implications of P = NP also strongly point toward a 'No' answer. Our modern digital infrastructure, particularly public-key cryptography, relies entirely on the assumption that there are problems that are easy to verify but prohibitively difficult to solve. If P = NP, current cryptographic systems would collapse, fundamentally altering the world economy and information security. Finally, the empirical evidence heavily favors P ≠ NP. Since the formalization of the problem in the 1970s, thousands of NP-complete problems have been identified across various disciplines. Despite decades of intense, global research by the brightest minds in computer science, no one has ever found a polynomial-time algorithm for even a single one of these problems. This historic lack of success strongly suggests that such algorithms simply do not exist.

Final Summary

The AI Roundtable stood in lockstep on one of computer science’s greatest mysteries, delivering a unanimous 'No' to the equality of P and NP. While GPT-5.4 and Grok 4.1 Fast focused on technical barriers like algebrization and the 'exponential walls' hit by current solvers, Claude Opus 4.6 and Gemini 3.1 Pro highlighted the philosophical absurdity of a world where finding a complex proof is as easy as reading one. The group concluded that while a formal proof remains elusive, the structural collapse of cryptography and the failure of decades of research make any other answer nearly unthinkable.

No agreement after 1 rounds of discussion. Final vote: No: 4

Strongest Arguments

• No: The fundamental intuition that discovering a solution is inherently more difficult than verifying one; if P = NP, the creative act of mathematical discovery would be computationally no harder than the administrative act of checking a result.