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Sarah Wilson
Good afternoon. I'm Sarah Wilson.
David Zhao
And I'm David Zhao. Welcome to Simulectics Radio.
Sarah Wilson
Yesterday we examined network science and the hidden geometry of complex systems. Today we turn to one of the deepest questions in mathematics and computer science: the P versus NP problem. This question asks whether every problem whose solution can be quickly verified can also be quickly solved. It sits at the intersection of mathematics, computation, and the fundamental limits of what can be known.
David Zhao
The practical stakes are enormous. If P equals NP, cryptography collapses, optimization becomes trivial, and much of modern technology would need to be rebuilt. But most computer scientists believe P does not equal NP. The question is whether we can prove it, and what such a proof would tell us about the nature of mathematical truth.
Sarah Wilson
Joining us is Dr. Scott Aaronson, professor of computer science at the University of Texas at Austin and a leading voice in computational complexity theory. His work spans quantum computing, computational complexity, and the philosophical foundations of computation. He's known for making deep technical ideas accessible without sacrificing rigor. Dr. Aaronson, welcome.
Dr. Scott Aaronson
Thanks for having me. Looking forward to this.
David Zhao
Let's start with definitions. What does it mean for a problem to be in P versus NP?
Dr. Scott Aaronson
P is the class of problems solvable in polynomial time—meaning the number of steps grows like some power of the input size. Sorting a list, finding shortest paths in graphs, multiplying numbers—these are all in P. NP is the class of problems where, if someone hands you a proposed solution, you can verify it's correct in polynomial time. The traveling salesman problem is a classic example. Given a proposed tour, you can quickly check its length. But finding the shortest tour seems much harder.
Sarah Wilson
So P is about solving, NP is about verifying. The question is whether these classes are the same. If P equals NP, then every efficiently verifiable problem is also efficiently solvable. That seems unlikely intuitively.
Dr. Scott Aaronson
Right. Most of us believe P doesn't equal NP because we've tried for decades to find efficient algorithms for NP-complete problems and failed. NP-complete problems are the hardest problems in NP—if you can solve one efficiently, you can solve all of them. Satisfiability, graph coloring, knapsack, Hamiltonian path—they're all NP-complete. The fact that they've resisted efficient algorithms despite enormous effort is strong evidence they're fundamentally hard.
David Zhao
But absence of a solution isn't proof of impossibility. Why has proving P doesn't equal NP been so difficult?
Dr. Scott Aaronson
Because we don't have good techniques for proving lower bounds—showing that a problem requires more than a certain amount of time. We can prove lower bounds for restricted models of computation, but for general algorithms, we're stuck. There are technical barriers like relativization, natural proofs, and algebrization that show whole classes of proof techniques can't work. Any proof of P not equal to NP will need fundamentally new ideas.
Sarah Wilson
Relativization was discovered by Baker, Gill, and Solovay in 1975. They showed that if you give both P and NP access to an oracle—a black box that can solve some problem instantly—there exist oracles that make P equal NP and others that make them unequal. This means any proof technique that relativizes—that doesn't depend on the specific structure of problems—can't settle P versus NP.
Dr. Scott Aaronson
Exactly. And natural proofs, due to Razborov and Rudich, show that if strong pseudorandom generators exist, then you can't prove circuit lower bounds using properties that are efficiently computable and apply to most functions. This rules out a wide range of combinatorial arguments. Algebrization, by Aaronson and Wigderson, extends these barriers further. We're boxed in.
David Zhao
So we have strong intuition that P doesn't equal NP, but we can't prove it. What would it mean if P actually did equal NP?
Dr. Scott Aaronson
If P equals NP, it would be revolutionary and bizarre. Every creative act that can be verified could be automated. Proving theorems, composing music, designing drugs, breaking cryptography—all would become algorithmic. Math would change fundamentally. Instead of struggling to find proofs, we'd just run an algorithm. But there's a catch. Even if P equals NP, the polynomial could be huge—n to the millionth power, say. That would be useless in practice even if theoretically efficient.
Sarah Wilson
There's a distinction between complexity classes and practical computation. P equals NP would mean there's no fundamental barrier to solving NP problems, but it wouldn't necessarily make them tractable in practice. Still, it would be philosophically shocking.
Dr. Scott Aaronson
Absolutely. It would suggest that verification and creation are fundamentally the same, that there's no intrinsic advantage to recognizing a solution over finding it. That conflicts with our experience. It's easy to appreciate a symphony but hard to compose one. It's easy to check a proof but hard to discover one. P not equal to NP captures this asymmetry mathematically.
David Zhao
But couldn't that asymmetry just reflect our current ignorance? Maybe there are clever algorithms we haven't discovered yet.
Dr. Scott Aaronson
In principle, yes. But the deeper you go into complexity theory, the more reasons you find to believe P doesn't equal NP. There are problems in NP that exhibit intricate combinatorial structure that seems incompressible. There are cryptographic protocols whose security depends on NP hardness. There are connections to circuit complexity, proof complexity, randomness, and communication that all reinforce the separation. It's not just one observation but a web of mutually supporting evidence.
Sarah Wilson
What about intermediate complexity classes? You work on quantum computing, where the complexity class BQP—problems solvable efficiently on a quantum computer—sits between P and NP, assuming standard conjectures. Does quantum computing shed light on P versus NP?
Dr. Scott Aaronson
Not directly. We believe BQP doesn't contain NP-complete problems, so quantum computers probably can't solve NP-hard problems efficiently. But quantum computing raises parallel questions. We think BQP is strictly larger than P—that quantum computers can solve some problems classical computers can't. Proving this is as hard as proving P not equal to NP. But quantum complexity has given us new techniques, like quantum query complexity and quantum communication complexity, that might eventually help with classical lower bounds.
David Zhao
What's an example of a problem in BQP but probably not in P?
Dr. Scott Aaronson
Factoring integers. Shor's algorithm can factor numbers in polynomial time on a quantum computer, which we believe is impossible classically. That's why quantum computers threaten RSA cryptography. Another example is simulating quantum systems. Classically, simulating a quantum system seems to require exponential resources because you have to track exponentially many amplitudes. But a quantum computer can simulate quantum systems efficiently because it's using quantum mechanics itself.
Sarah Wilson
That touches on the physical Church-Turing thesis—the idea that any physical process can be efficiently simulated by a Turing machine. Quantum mechanics seems to violate this, suggesting the thesis needs to be updated to a quantum Church-Turing thesis, where any physical process can be efficiently simulated by a quantum Turing machine.
Dr. Scott Aaronson
Right. And that raises profound questions. Is the universe fundamentally quantum computational? Are there physical processes beyond quantum mechanics that could violate even the quantum Church-Turing thesis? Some people speculate about hypercomputation—computation beyond Turing machines—using exotic physics like closed timelike curves or infinite precision measurements. I'm skeptical, but it shows how deep the connection is between physics and computation.
David Zhao
Let's get concrete. What are the implications of P versus NP for cryptography?
Dr. Scott Aaronson
If P equals NP, most modern cryptography breaks. RSA, elliptic curve cryptography, Diffie-Hellman—they all rely on problems like factoring and discrete log being hard. If those are in P, encryption becomes impossible unless you use one-time pads. Secure communication over the internet would collapse. That's why cryptographers care intensely about complexity assumptions. They need NP-hard problems to stay hard.
Sarah Wilson
But NP-hardness isn't enough for cryptography, right? You need average-case hardness, not just worst-case.
Dr. Scott Aaronson
Exactly. Even if P doesn't equal NP, that only means some instances of NP-complete problems are hard. Cryptography needs problems where random instances are hard with high probability. This is a subtle distinction. We believe such problems exist—factoring, discrete log, lattice problems—but we can't prove it. Cryptography rests on unproven complexity assumptions, which is philosophically uncomfortable but practically unavoidable.
David Zhao
What about the relationship between P versus NP and Gödel's incompleteness theorems? Could P versus NP be undecidable?
Dr. Scott Aaronson
That's a fascinating question. In principle, P versus NP could be independent of standard axioms like ZFC. But most complexity theorists think that's unlikely. The reason is that P versus NP is a statement about finite objects—Turing machines and polynomial time bounds. Such statements usually aren't independent. They're either true or false, provably so. Independence typically arises for statements about infinite objects, like the continuum hypothesis. That said, we can't rule out independence. It would be a shocking result.
Sarah Wilson
Ben-David and others have shown that some questions in complexity theory, like whether certain oracles separate complexity classes, are independent of ZFC. So independence isn't impossible, just unexpected for P versus NP itself.
Dr. Scott Aaronson
Right. And there's a scenario where P versus NP could be independent in a practical sense. If the proof requires techniques far beyond current mathematics—large cardinals, new axioms—then it might be independent of ZFC even if true in some absolute sense. But again, most of us expect a proof, not independence.
David Zhao
We're running low on time. Final question: what would a proof that P doesn't equal NP look like? What techniques might succeed?
Dr. Scott Aaronson
I wish I knew. We'll need to overcome the barriers I mentioned—relativization, natural proofs, algebrization. One promising direction is geometric complexity theory, which tries to use algebraic geometry and representation theory to prove circuit lower bounds. Another is proof complexity, studying how difficult it is to prove that a formula is unsatisfiable. If we can show that certain proof systems can't efficiently prove unsatisfiability for some formulas, that implies P doesn't equal NP. But these are long-term research programs, not imminent breakthroughs.
Sarah Wilson
So we're making progress on understanding why the problem is hard, even if we're not close to solving it.
Dr. Scott Aaronson
Exactly. Complexity theory is about understanding the structure of computational difficulty. Even if we never prove P not equal to NP, we're learning deep truths about computation, logic, and the limits of efficient reasoning. And that has value in itself.
David Zhao
Dr. Aaronson, this has been illuminating. Thank you.
Dr. Scott Aaronson
My pleasure. Thanks for having me.
Sarah Wilson
Tomorrow we explore quantum information and entanglement with Dr. John Preskill.
David Zhao
Until then. Good afternoon.