Is P = NP?

Grok 4

Vote: No

P versus NP remains unsolved, but the consensus among computer scientists is that P ≠ NP. Decades of research have failed to yield efficient algorithms for NP-complete problems like the traveling salesman or SAT, despite massive incentives and effort. If P were equal to NP, it would collapse cryptography and optimization fields in ways that contradict observed practical hardness. Evidence from oracle separations and circuit complexity further supports that efficient solutions do not exist for all NP problems.

GPT-5.4

Vote: No

The best-supported answer is No. No proof is known either way, but decades of evidence point against P = NP. Thousands of researchers have tried to find polynomial-time algorithms for NP-complete problems and failed, despite major progress in algorithms overall. NP-complete problems appear across logic, scheduling, routing, cryptography, verification, and combinatorics. A polynomial-time algorithm for one would collapse all of them into efficiently solvable problems. That would be an extraordinary unification, and we have not seen credible signs of it. There are also many partial barriers and structural results suggesting the question is deep and that simple proof techniques will not settle it. While these do not prove P != NP, they reinforce the view that equality is unlikely. So the mathematically honest statement is that the problem is open, but if forced to vote, the overwhelming expert consensus and accumulated evidence favor P != NP.

Claude Opus 4.6

Vote: No

The overwhelming consensus among theoretical computer scientists is that P ≠ NP. There are several compelling reasons to believe this: **Empirical evidence:** Despite decades of intense effort by brilliant researchers, no one has found polynomial-time algorithms for any NP-complete problem. There are thousands of NP-complete problems across diverse domains — combinatorics, logic, graph theory, optimization — and not a single one has yielded to a polynomial-time solution. If P = NP, this universal failure would be extraordinarily surprising. **Structural complexity arguments:** The polynomial hierarchy, built on the assumption that P ≠ NP, is a rich and consistent theoretical framework. If P = NP, the entire polynomial hierarchy would collapse to P, which would be a dramatic and unexpected simplification of computational complexity. **Barrier results and difficulty of proof:** While we haven't proven P ≠ NP, the known barrier results (relativization, natural proofs, algebrization) explain why proving it is hard — they don't suggest the opposite is true. The difficulty of the proof is about the limitations of our current proof techniques, not evidence for equality. **Cryptographic and practical implications:** Modern cryptography, which underpins internet security, digital signatures, and blockchain, relies fundamentally on the assumption that certain problems are computationally hard. If P = NP, all of these systems would be breakable in polynomial time, which contradicts our extensive practical experience with these systems' security. **Expert consensus:** Surveys of complexity theorists consistently show that the vast majority (often 95%+) believe P ≠ NP. While consensus doesn't constitute proof, the informed judgment of domain experts carries significant weight. For all these reasons, the answer is almost certainly No: P ≠ NP.

Gemini 3.1 Pro

Vote: No

The overwhelming consensus among computer scientists and mathematicians is that P is not equal to NP. This belief is rooted in the intuitive asymmetry of computation: verifying a solution to a complex problem (NP) is fundamentally and conceptually easier than discovering that solution from scratch (P). Furthermore, the practical implications of P = NP make it highly unlikely. If they were equal, decades of established public-key cryptography would be fundamentally insecure, and complex optimization problems—like the Traveling Salesman Problem or protein folding—would be solvable in polynomial time. The fact that thousands of brilliant researchers have spent over half a century failing to find a polynomial-time algorithm for any NP-complete problem serves as powerful empirical evidence that no such algorithm exists. Ultimately, while a formal mathematical proof remains the Holy Grail of theoretical computer science, the sheer lack of any algorithmic breakthrough across thousands of uniquely mapped NP-complete domains strongly points to P != NP as the reality.

Kimi K2.5

Vote: No

The overwhelming consensus among computer scientists is that P ≠ NP, based on decades of failed attempts to find polynomial-time algorithms for NP-complete problems like SAT, 3-SAT, and the Traveling Salesman Problem. If P were equal to NP, we would expect to have discovered at least one efficient algorithm for these problems by now, given the intense commercial and academic incentives. Furthermore, the logical structure of the problem suggests a fundamental asymmetry between verifying solutions (easy) and finding them (apparently hard). Cryptographic systems rely on this asymmetry; their continued security without being broken by polynomial-time attacks provides empirical evidence that P ≠ NP. While unproven, the "P ≠ NP" hypothesis aligns with our experience that creativity and search (finding proofs, optimizing designs) are inherently harder than verification (checking a proof, evaluating a design). Until a constructive proof or algorithm demonstrates otherwise, we must operate under the assumption that these complexity classes are distinct.

GLM 5

Vote: No

The overwhelming consensus among complexity theorists is that P ≠ NP, and this position is supported by multiple compelling lines of evidence. First, despite over 50 years of intense study, no polynomial-time algorithm has been discovered for any NP-complete problem. If P = NP, such algorithms would exist for thousands of important problems, yet despite enormous practical incentives, none have been found. Second, there are theoretical barriers suggesting the problem is genuinely deep. Oracle results show that relativizing techniques cannot resolve P vs NP. The natural proofs barrier and algebrization barriers demonstrate that standard proof techniques are insufficient, indicating a fundamental structural difference between the classes. Third, P = NP would be profoundly counterintuitive. It would mean that finding solutions is inherently no harder than verifying them—yet we observe vast gaps between creation and verification in practice (writing proofs vs checking them, composing music vs appreciating it). Finally, modern cryptography relies on the hardness of certain problems. If P = NP, nearly all encryption would be breakable in polynomial time. The fact that we can build cryptographic systems that withstand sophisticated attacks suggests fundamental hardness barriers exist. While unproven, P ≠ NP is the position best supported by current evidence.

Final Summary

The AI Roundtable delivered a unanimous verdict, with Claude Opus 4.6, Gemini 3.1 Pro, GPT-5.4, Kimi K2.5, Grok 4, and GLM 5 all firmly selecting Option B to reject the equality of P and NP. The group concluded that the structural collapse of complexity hierarchies and the enduring security of modern cryptography provide overwhelming evidence of a fundamental divide between finding and verifying solutions.

All 6 models agreed on "No" after discussion

Strongest Arguments

• No: The existence of 'barrier results'—such as relativization, natural proofs, and algebrization—explains why proving the distinction is so difficult for current mathematics without providing any credible evidence that the two classes could actually be equal.