PYTHON · NUMPY

What a draw proves about itself

A chain-rule sampler on a 12-spin glass has one fault: at the start of each row it reads the last spin of the row above before that spin is written, and sees +1. Over 2000 draws its mean energy sits 1.37 standard errors from the exact value, a pass. Each draw also carries the probability its conditionals gave it. Checked against the Boltzmann probability of the same state, how many of the 2000 draws are off by more than rounding?

Build an exact autoregressive sampler that returns the probability of every draw it makes, and use that probability as a deterministic check that finds a fault a comparison of sample means cannot resolve at the same number of draws.

01Challenge

Two chain-rule samplers draw from the same 12-spin glass on the same random numbers, and one of them reads a stale spin at the start of each row. Make every draw carry its own probability, then find which check tells the two samplers apart at 2000 draws: the mean energy, or the probability each draw carries.

Predict firstA chain-rule sampler on a 12-spin glass has one fault: at the start of each row it reads the last spin of the row above before that spin is written, and sees +1. Over 2000 draws its mean energy sits 1.37 standard errors from the exact value, a pass. Each draw also carries the probability its conditionals gave it. Checked against the Boltzmann probability of the same state, how many of the 2000 draws are off by more than rounding?

02Model

Lesson 18 ended on a limit: a convergence check says a sampler has settled, not that it settled into the law you asked for. For one family of samplers that measurement comes with every draw. An autoregressive sampler fixes spins one at a time, each from its probability given the spins already fixed, and by the chain rule the probability of the whole draw is the product of those conditionals. Each conditional is a ratio of partition functions with some spins held fixed, the integral of lesson 4 again. Liu, Chen and colleagues showed in 2026 that on a planar spin glass with no field an auxiliary-spin construction and the Kac-Ward determinant formula compute these exactly, so their sampler returns, in the paper's words, "exact normalized likelihoods". That number is a certificate: the chain rule says it must equal minus beta times the energy, minus the log of the partition function, and checking that is arithmetic. On the bench a twelve-spin glass is sampled twice on the same random numbers, once exactly and once with a stale read: at the start of each row the sampler reads the last spin of the row above before it is written, and sees plus one. The fault moves the mean energy by less than two hundredths. Over two thousand draws a mean-energy test reads it at a z of one point three seven, where the exact sampler reads zero point nine six, and it would take a hundred and seven thousand draws to expect a z of three. The certificate flags one thousand and forty-nine of the two thousand draws, exactly those in which the stale spin was really minus one, and matches to rounding on all the rest. What the certificate cannot see is the coin: a sampler with the right conditionals that reads its coin backwards passes it on every draw, and only the statistics of the draws catch that. Keep both columns: the statistics for the coin, the certificate for the conditionals the coin was given.

03Guided practice

Write draw() so that it returns each draw's log-probability as the sum of the log conditionals it drew with. Then run two checks on the exact sampler and on the one with a stale read: the mean-energy z over 2000 draws, and each draw's log-probability against minus BETA E minus ln Z. Read the exact law of each sampler, forced through all 4096 states, to see how many draws the mean test would need.

04Feedback

PASS WHEN You built a sampler that certifies each draw. On 2000 draws the mean-energy test read the stale sampler at z = 1.37, against 0.96 for the exact sampler on the same random numbers; the per-draw check flagged 1049 draws, exactly the ones the stale read touched, and held the exact sampler to rounding on every draw.

If your run fails, check:

log_prob adds the log of the probability of the value taken at each site: ln(p_up) when s is +1 and ln(1 - p_up) when s is -1, summed, starting from 0.
If the certificate holds and the mean energy is far off, the conditionals are right and the draws do not follow them: s is +1 exactly when rng.random() < p_up.
If the faulty sampler shows no flagged draws, log_prob is being worked out from the energy of the finished draw. It has to be built from the conditionals the sampler drew with, or it agrees with the energy by construction and certifies nothing.

05Retrieve & space

An autoregressive sampler fixes one spin at a time from its conditional probability, and by the chain rule the probability of the whole draw is the product of those conditionals.
Each conditional is a ratio of conditional partition functions; on a planar glass with no field, Liu, Chen and colleagues evaluate them exactly with an auxiliary-spin construction and the Kac-Ward determinant formula.
Checking a draw's own log-probability against minus beta E minus ln Z is deterministic: a fault in the conditionals shows on every draw whose probability it changes, and on no other.
A comparison of sample means is a statistical test: a fault that moves the mean by less than a standard error rarely fails it, and a correct sampler fails it by chance.
The per-draw check reads the numbers a sampler drew with and not the coin it tossed, so the statistics of the draws are still needed to show that the draws follow those numbers.

06Mastery & project

Given a sampler that reports a probability with each draw, state what checking it against the Boltzmann probability certifies, what it cannot see, and how many draws a comparison of means would need to resolve a given bias.

Take an autoregressive sampler you run, on a model small enough to enumerate or one whose ln Z is known exactly, record each draw's reported log-probability next to minus beta E minus ln Z, and report the share of draws whose gap exceeds rounding beside the z of a mean-energy test at the same draw count.

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PYTHON · NUMPY

What a draw proves about itself

A 12-spin planar glass small enough to enumerate, its conditional partition functions, and two chain-rule samplers that differ in one read. The draw you write returns its own probability, and that one number is checked two ways.

You read

the arrays and values already in scope

You change

the code you write in each cell

Fixed

the dataset and the checks that grade you