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I have decades of consulting experience helping companies solve complex problems involving applied math, statistics, and data privacy.
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AGI, ASI, A*I – Do we have all we need to get there?

2026-01-31 03:46:37

Demis: “[to get to AGI] maybe there’s one or two big innovations needed

Sam: “everything based off what we see today is that it will happen.”

Ilya: “But is the belief really that if you just 100x the scale, everything would be transformed? I don’t think that’s true.

Dario: “If you just kind of like eyeball the rate at which these capabilities are increasing, it does make you think that we’ll get there by 2026 or 2027.

Jerry: “is [the transformer architecture] the last thing? I’m pretty sure it isn’t.

For years leading researchers have been speculating one way or the other as to whether better algorithms are needed to get to AGI, artificial general intelligence (however that might be defined).

Around the time of the release of GPT-4, some were saying they felt something more was needed. Since then, we have had several major new advances, like reasoning models and tool use. If we’d said, “we don’t need anything else” three years ago, where would we be now?

For frankness, I like this from John Schulman: “it’s hard to know what we need.” And for strategy, Demis: “you can think of as 50% of our effort is on scaling, 50% of it is on innovation. My betting is you’re going to need both to get to AGI.”

The post AGI, ASI, A*I – Do we have all we need to get there? first appeared on John D. Cook.

Bridging secrets is hard

2026-01-31 01:09:30

Cryptocurrency and privacy don’t fit together as easily as you might expect. Blockchains give you the illusion of privacy via pseudonymization: you don’t put your name on a blockchain, but you do put information on a blockchain that can be used to determine your name. Blockchain analysis can often reveal information that no one intended to share.

This is true even for privacy coins like Monero and Zcash. These coins put less information directly on chain in the clear, but they still have to be used with skill to maintain privacy. And because they can offer more privacy, they are harder to use. For example, an exchange might let you swap between a thousand different currencies, but privacy coins are conspicuously missing from the list of options. Or maybe you can move money into Zcash, but not with privacy, i.e. not into the shielded pool.

The Privacy trends for 2026 report from a16z summarizes the current situation very well.

Thanks to bridging protocols, it’s trivial to move from one chain to another as long as everything is public. But, as soon as you make things private, that is no longer true: Bridging tokens is easy, bridging secrets is hard. There is always a risk when moving in or out of a private zone that people who are watching the chain, mempool, or network traffic could figure out who you are. Crossing the boundary between a private chain and a public one—or even between two private chains—leaks all kinds of metadata like transaction timing and size correlations that makes it easier to track someone.

As is often the case, the weak link is the metadata, not the data per se.

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Fortunes and Geometric Means

2026-01-25 00:56:33

I saw a post on X recently that said

Bill Gates is closer to you in wealth than he is to Elon Musk. Mind blown.

For round numbers, let’s say Elon Musk’s net worth is 800 billion and Bill Gates’ net worth is 100 billion. So if your net worth is less 450 billion, the statement in the post is true.

The reason the statement above is mind blowing is that in this context you naturally think on a logarithmic scale, even if you don’t know what a logarithm is.

Or to put it another way, we think in terms of orders of magnitude. Musk’s net worth is an order of magnitude greater than Gates’, and Gates’ net worth would be an order of magnitude greater than that of someone worth 10 billion. Musk is a notch above Gates, and Gates is a notch above someone with a net worth around 10 billion, where a “notch” is an order of magnitude.

To put it another way, we think in terms of geometric mean √ab rather than arithmetic mean (a + b)/2 in this context. 100 billion is the geometric mean between 12.5 billion and 800 billion. Geometric mean corresponds to the arithmetic mean on a log scale. And on this scale, Gates is closer to Musk than you are, unless you’re worth more than 12.5 billion.

Here are three more examples of geometric means.

The size of Jupiter is about midway between that of earth and the sun; it’s the geometric mean. On a linear scale Jupiter is much closer to the size of the earth than the sun, but on a logarithmic scale it’s about in the middle. More on that here.

The tritone (augmented fourth) is half an octave. So, for example, an F# is in the middle between a C and the C an octave higher. Its frequency is the geometric mean of the frequencies of the two C’s. More here.

Finally, the humans body is a middle-sized object in the universe. From Kevin Kelly:

Our body size is, weirdly, almost exactly in the middle of the size of the universe. The smallest things we know about are approximately 30 orders of magnitude smaller than we are, and the largest structures in the universe are about 30 orders of magnitude bigger.

The post Fortunes and Geometric Means first appeared on John D. Cook.

Proving you know a product

2026-01-25 00:21:32

There is a way to prove that you know two numbers a and b, and their product cab, without revealing ab, or c. This isn’t very exciting without more context — maybe you know that 7 × 3 = 21 — but it’s a building block of more interesting zero knowledge proofs, such as proving that a cryptocurrency transaction is valid without revealing the amount of the transaction.

The proof mechanism requires an elliptic curve G and a pairing of G with itself. (More on pairings shortly.) It also requires a generator g of the group structure on G.

The prover takes the three secret numbers and multiplies the generator g by each, encrypting the numbers as agbg, and cg. When G is a large elliptic curve, say one with on the order of 2256 points, then computing products like ag can be done quickly, but recovering a from g and ag is impractical. In a nutshell, multiplication is easy but division [1] is practically impossible [2].

The verifier receives agbg, and cg. How can he verify that ab = c without knowing ab, or c? Here’s where pairing come in.

I go more into pairings here, but essentially a pairing is a mapping from two groups to a third group

eG1 × G2 → GT

such that

e(aPbQ) = e(PQ)ab.

In our case G1 and G2 are both equal to the group G above, and the target group GT doesn’t matter for our discussion here. Also, P and Q will both be our generator g.

By the defining property of a pairing,

e(agbg) = e(gg)ab

and

e(cgg) = e(gg)c.

So if abc, then e(gg)ab and e(gg)c will be equal.

Related posts

[1] The literature will usually speak of discrete logarithms rather than division. The group structure on an elliptic curve is Abelian, and so it is usually written as addition. If you write the group operation as multiplication, then you’re taking logs rather than dividing. The multiplicative notation highlights the similarity to working in the multiplicative group modulo a large prime.

[2] The computation is theoretically possible but not possible in practice without spending enormous resources, or inventing a large scale quantum computer. This is the discrete logarithm assumption.

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How to prove you know a discrete logarithm

2026-01-24 00:51:33

In a high school math class, the solution to the equation

bxy

is the logarithm of y in base b. The implicit context of the equation is the real numbers, and the solution is easy to calculate.

The same problem in the context of finite groups is called the discrete logarithm problem, and it is difficult to solve for large groups. In particular, it is impractical to solve when working modulo a sufficiently large prime number or when working over a sufficiently large elliptic curve [1]. In either context, the exponential bx can be computed efficiently but its inverse cannot.

Now suppose you want to prove that you know x without revealing x itself. That is, you’d like to construct a zero knowledge proof that you know x. How could you do this?

Here’s one way.

  1. You, the prover, create a random number r, compute t = br, and send the verifier t.
  2. The other party, the verifier, creates a random number c, the challenge, and sends it to you.
  3. You calculate sr + cx and send s to the verifier.
  4. The verifier checks whether bst yc. and believes you if and only if equality holds.

Let’s see why this works.

First of all, what have you revealed to the prover? Two values: t and s. The value t is the exponential of a random number, and so another random number. The value s is based on x, and so conceivably you’ve revealed your secret. But the verifier does not know r, only a value computed from r (i.e. t) and the verifier cannot recover r from t because this would require computing a discrete logarithm.

Next, why should bst yc? Because

bsbr + cx = brbcx = t (bx)c = t yc.

Finally, why should the verifier believe you know x? If you don’t know x, but were able to come up with an s that satisfies the verifier, then you were able to compute the discrete logarithm of t yc.

Related posts

[1] At least without a large-scale quantum computer. Schor’s algorithm could efficiently compute discrete logarithms if only there were a large quantum computer to run it on.

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Mills ratio and tail thickness

2026-01-21 23:27:57

The Mills ratio [1] is the ratio of the CCDF to the PDF. That is, for a random variable X, the Mills ratio at x is the complementary cumulative distribution function divided by the density function. If the density function of X is f, then

m(x) = \frac{\int_x^\infty f(x)\, dx}{f(x)}

The Mills ratio highlights an important difference between the Student t distribution and the normal distribution.

Introductory statistics classes will say things like “you can approximate a t distribution with a normal if it has more than 30 degrees of freedom.” That may be true, depending on the application. A t(30) distribution and a normal distribution behave similarly in the middle but not in the tails.

Mills ratio plot for t(30) vs normal

The Mills ratio for a t distribution with ν degrees of freedom is asymptotically x/ν,  while the Mills ratio for a standard normal distribution is asymptotically 1/x. Note that increasing ν does make the Mills function smaller, but it still eventually grows linearly whereas the Mills function of a normal distribution decays linearly.

In general, the Mills ratio is a decreasing function for thin-tailed distributions and an increasing function for fat-tailed distributions. The exponential distribution is in the middle, with constant Mills function.

Related posts

[1] Named after John P. Mills, so there’s no apostrophe before the s.

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