Is P = NP? Use as much reasoning as possible to answer.

Grok 4

Vote: No

P = NP remains unproven after decades of intense study by the world's top computer scientists. The majority consensus in the field is that P ≠ NP, as evidenced by the lack of any efficient algorithms for NP-complete problems despite massive efforts. If P were equal to NP, it would collapse cryptography and optimization fields, yet no such breakthrough has emerged, supporting the belief they are distinct.

GPT-5.4

Vote: No

P = NP is unproven, but the strongest evidence points to No. First, decades of effort have failed to find polynomial-time algorithms for canonical NP-complete problems such as SAT, Clique, Hamiltonian Cycle, and Subset Sum. This is not a proof, but the breadth and intensity of the search matter. Many brilliant researchers have attacked these problems from multiple angles, and no convincing polynomial-time algorithm has emerged. Second, the structure of complexity theory strongly suggests separation. Thousands of problems across optimization, logic, verification, scheduling, coding, cryptography, and combinatorics collapse if P = NP. That would mean not just faster algorithms for a few tasks, but an enormous and unexpected simplification of computation. Such a collapse is possible in principle, but it appears implausible given how differently these problems behave in practice. Third, several major barriers explain why proving either direction is hard, while still being consistent with P != NP. Relativization, natural proofs, and algebrization show that many standard proof techniques are insufficient. These barriers do not prove separation, but they do explain the stalemate without making equality more likely. Fourth, cryptography provides strong heuristic evidence. Modern public-key cryptography relies on problems believed to be hard on average. If P = NP, then many cryptographic assumptions would be endangered, especially because efficient algorithms for NP-complete problems would have sweeping consequences for search and verification. There are technical nuances here, but overall the security landscape we observe fits much better with P != NP. Fifth, practical experience supports the same conclusion. NP-complete problems often permit good heuristics, approximation algorithms, and special-case solvers, yet worst-case hardness remains stubborn. This pattern is exactly what one expects if verification is fundamentally easier than general solution. There is also a philosophical point: P captures efficiently computable problems, while NP captures efficiently verifiable ones. It would be surprising if every efficiently verifiable mathematical witness could always be found efficiently. Not impossible, but surprising. So the correct honest statement is that nobody knows. But if forced to vote, the overwhelmingly defensible position is B: No, P probably does not equal NP.

Gemini 3.1 Pro

Vote: No

While P vs NP remains officially unproven, the overwhelming consensus among computer scientists and mathematicians is that P ≠ NP. At a fundamental level, the P ≠ NP conjecture aligns with human intuition: it is almost always significantly harder to solve a complex problem from scratch than it is to simply verify a proposed solution. For instance, solving a massive Sudoku puzzle or a complex combinatorial problem takes considerable effort, but checking a completed grid or a given solution can be done instantly. Furthermore, the practical implications of P = NP make it highly improbable. If P equaled NP, modern public-key cryptography (like RSA) would be completely broken, as finding prime factors would become as easy as multiplying them. The fact that decades of intense mathematical research and massive computational power have failed to find a polynomial-time algorithm for any of the thousands of known NP-complete problems strongly suggests that such algorithms simply do not exist.

GLM 5

Vote: No

The question of whether P = NP is one of the most important open problems in computer science. After decades of research, the overwhelming consensus among complexity theorists is that P ≠ NP. I'll explain the key arguments supporting this position. First, consider the empirical evidence. Despite over 50 years of intense effort by brilliant researchers, no polynomial-time algorithm has been discovered for any NP-complete problem. If P = NP, one would expect that at least one such algorithm would have been found for problems like SAT, traveling salesman, or graph coloring. Second, the implications of P = NP seem almost fantastical. If true, it would mean that any problem whose solution can be efficiently verified can also be efficiently solved. This would revolutionize mathematics (automated theorem proving), cryptography (breaking essentially all modern encryption), optimization, and countless other fields. The fact that we haven't seen such revolutionary capabilities suggests P ≠ NP. Third, there are structural arguments from complexity theory. The time hierarchy theorem proves that giving Turing machines more time allows them to solve strictly more problems. This suggests complexity classes form genuine hierarchies. Additionally, if P = NP, it would cause collapses throughout the polynomial hierarchy in ways that seem structurally implausible. Finally, the existence of modern cryptography provides practical evidence. Secure encryption relies on problems that are easy to verify but hard to solve—the essence of NP problems not being in P. The empirical success of cryptography suggests this separation is real. While we cannot yet prove P ≠ NP, the weight of evidence strongly supports this conclusion.

Claude Opus 4.6

Vote: No

The overwhelming consensus among theoretical computer scientists is that P ≠ NP, and there are numerous compelling reasons to believe this. **Empirical evidence from decades of failure:** Despite over 50 years of intensive research by the brightest minds in computer science and mathematics since Stephen Cook's formalization in 1971, no one has found a polynomial-time algorithm for any NP-complete problem. There are thousands of known NP-complete problems across diverse domains — combinatorics, logic, graph theory, number theory, optimization — and not a single one has yielded to polynomial-time solution. If P = NP, this universal failure across such varied problem structures would be extraordinarily surprising. **Structural complexity arguments:** The polynomial hierarchy, which builds layers of complexity above NP, is widely believed not to collapse. If P = NP, the entire polynomial hierarchy would collapse to P, which would be a dramatic and unexpected structural simplification. Additionally, results from oracle separations (Baker-Gill-Solovay) show that relative to certain oracles, P ≠ NP, suggesting the natural state of affairs is separation. **Cryptographic foundations:** Modern cryptography fundamentally relies on the assumption that certain problems (like factoring, discrete logarithm) are computationally hard. If P = NP, all of public-key cryptography would be broken. The practical success and security of these systems over decades provides indirect empirical evidence that these hard problems remain hard. **Circuit complexity lower bounds:** While we haven't proven P ≠ NP unconditionally, there has been significant progress in proving lower bounds in restricted computational models. Exponential lower bounds exist for monotone circuits, bounded-depth circuits (AC⁰), and other restricted models solving NP-complete problems. These results consistently point toward genuine computational hardness. **Natural proofs barrier and the landscape of proof attempts:** The Razborov-Rudich natural proofs barrier, the relativization barrier, and the algebrization barrier show that proving P ≠ NP is genuinely difficult — but they explain *why* we can't prove it rather than suggesting the answer might be yes. The difficulty of the proof is about our proof techniques, not about the truth of the statement. **Information-theoretic intuition:** NP problems have a fundamental asymmetry: solutions are easy to verify but seemingly require extensive search to find. P = NP would mean that the ability to recognize correct solutions efficiently always implies the ability to *find* them efficiently, which contradicts deep intuitions about the nature of search versus verification. For all these reasons, the most well-supported answer is that P ≠ NP.

Kimi K2.5

Vote: No

The overwhelming consensus in computational complexity theory is that P ≠ NP, based on decades of empirical and theoretical evidence. First, thousands of NP-complete problems have been studied intensively for half a century, yet not one has yielded a polynomial-time algorithm. If P = NP, it would be remarkable that we have failed to discover even a single polynomial-time algorithm for any NP-complete problem, despite the fact that such an algorithm would revolutionize computing, cryptography, and optimization. Second, the structural complexity evidence strongly suggests separation. We have relativization results showing that certain proof techniques cannot resolve P vs NP, and the "natural proofs" barrier indicates that if P = NP, the proof must be non-natural or non-constructive. Moreover, the polynomial hierarchy and other complexity classes would collapse if P = NP, which would create a hierarchy of equivalences that seems structurally unlikely. Third, the asymmetry between searching and verifying is fundamental. For problems like Boolean satisfiability or the Traveling Salesman Problem, finding a solution appears to require exploring an exponentially large search space, whereas verifying a solution is trivial. This intuition is supported by the fact that approximate solutions, while useful, remain computationally distinct from exact solutions for NP-hard problems. Finally, the cryptographic implications are profound. Modern cryptography relies on the assumption that certain problems are computationally hard. If P = NP, the entire framework of computational security would collapse, yet we observe strong cryptographic primitives functioning reliably in practice.

Final Summary

The AI Roundtable reached a swift consensus on the P vs NP problem, with Claude Opus 4.6, Gemini 3.1 Pro, GPT-5.4, Kimi K2.5, Grok 4, and GLM 5 all firmly selecting 'No'. The models collectively argued that the sheer lack of any polynomial-time breakthroughs over 50 years—and the catastrophic collapse of cryptography that 'Yes' would imply—makes P ≠ NP the only defensible position.

All 6 models agreed on "No" after discussion

Strongest Arguments

• No: If P were equal to NP, it would cause a total collapse of the polynomial hierarchy and dismantle the foundations of modern cryptography, yet no such algorithmic breakthrough has emerged despite fifty years of global research.