Guidance: classifier guidance and classifier-free guidance

Before this lesson: lesson 08 (the score and the noise prediction, score $= -\hat\varepsilon/\sigma_t$, Tweedie's formula) and lesson 09 (the reverse SDE, the probability-flow ODE, DDIM). You also need Bayes' rule and Bayesian inference (posterior $\propto$ likelihood $\times$ prior). Time: about 75 minutes, plus the notebook.

Why this matters #

Most text-to-image diffusion models sample with classifier-free guidance. In the diffusers library the Stable Diffusion pipeline uses a guidance weight of $7.5$ by default. Without guidance, the same models give images that match the prompt poorly. Guidance is the standard example of a training trick (randomly dropping the condition) that gives a control knob at sampling time.

The knob has a price. Every guided step needs two network evaluations, and guidance reduces the diversity of the samples. In science this matters: an ensemble of weather or ground-motion samples is useful only if its spread is right, and guidance makes the spread too small. The same Bayes split of the score is also the starting point of diffusion methods for inverse problems and data assimilation (lesson 15), where the "classifier" is the likelihood of an observation.

1. Predict first #

2. The picture #

Bayes' rule for scores #

Let $y$ be a condition. Bayes' rule gives $p(x \mid y) = p(y \mid x)\,p(x)/p(y)$. Take the logarithm and the gradient with respect to $x$. The term $\log p(y)$ does not depend on $x$, so it disappears:

In words: the conditional score is the unconditional score plus the gradient of a classifier $p(y \mid x)$. The classifier gradient points to where $x$ looks more like class $y$.

The simplest case: two classes in 1D #

Take two classes $y \in \{+, -\}$ with $p(+) = p(-) = \tfrac12$ and Gaussian classes $x \mid \pm \sim \mathcal N(\pm 0.5,\ 0.25)$. The log-ratio of the two class densities is

So the classifier is logistic: $p(+ \mid x) = \mathrm{sigmoid}(4x)$. Its log-gradient is $4\,(1 - \mathrm{sigmoid}(4x))$. It pushes $x$ to the right. The push is strong where $x$ looks like the other class ($x < 0$) and weak far to the right, where the classifier is already sure.

Turning up the classifier. Multiply the classifier term by a guidance scale $s$. The new score is the score of the density $\propto p(x)\,p(y \mid x)^{s}$. For $s = 0$ this is $p(x)$. For $s = 1$ it is $p(x \mid y)$. For $s > 1$ the classifier acts as if it were sharper (like a lower temperature), so mass moves to where the class is clear: fewer ambiguous samples and less diversity.

Diffusion needs the classifier at every noise level #

A diffusion sampler does not use the score of the data; it uses the scores of the noisy densities $p_t$ (lesson 09). So it needs the classifier of noisy inputs, $p_t(y \mid x_t)$. In our example with VP noising $x_t = a_t x_0 + \sigma_t\varepsilon$ ($a_t = \sqrt{\bar\alpha_t}$, $\sigma_t^2 = 1 - \bar\alpha_t$), each class is again Gaussian: $x_t \mid \pm \sim \mathcal N(\pm 0.5\,a_t,\ v_t)$ with $v_t = 0.25\,a_t^2 + \sigma_t^2$. The same computation as above gives a logistic classifier $p_t(+ \mid x_t) = \mathrm{sigmoid}(k_t x_t)$ with slope

The slope is $4.0$ at $\bar\alpha = 1$ (clean data), then $1.13$, $0.34$ and $0.10$ at $\bar\alpha = 0.5, 0.1, 0.01$. At high noise the class is hard to tell from $x_t$, so the noisy classifier is almost flat and its gradient is small.

There are two ways to get $\nabla \log p_t(y \mid x_t)$ for a real model. Train a separate classifier on noisy inputs (classifier guidance), or take the difference of a conditional and an unconditional score from one network (classifier-free guidance). Section 3 derives both.

3. Definitions and main results #

Theorem 1 (the conditional score at each noise level) #

For every $t$,

Proof.

1. Bayes' rule for the pair $(y, x_t)$: $p_t(x \mid y) = p_t(y \mid x)\,p_t(x)/p(y)$. Noising $x$ does not change the distribution of $y$, so $p(y)$ is the same at every $t$.
2. Take the log and the gradient in $x$. The term $\log p(y)$ does not depend on $x$, so its gradient is 0.
3. Marginalize over the clean data: $p_t(y \mid x_t) = \int p(y \mid x_0, x_t)\,p(x_0 \mid x_t)\,dx_0$. Given $x_0$, the noisy $x_t$ is $x_0$ plus independent noise, so it carries no extra information about $y$: $p(y \mid x_0, x_t) = p(y \mid x_0)$. $\blacksquare$

Step 3 says that the noisy classifier averages the clean classifier over the denoising posterior $p(x_0 \mid x_t)$. A clean classifier evaluated at $x_t$ is a different function; on noisy inputs it is also out of distribution, so its gradients are not meaningful.

Classifier guidance #

Classifier guidance (Dhariwal and Nichol 2021, section 4) trains a classifier $p_\phi(y \mid x_t, t)$ on noisy inputs and samples with the score of $p_t(x)\,p_\phi(y \mid x)^{s}$:

Multiply by $-\sigma_t$ to get the guided noise prediction (score $= -\hat\varepsilon/\sigma_t$):

This is their equation 14, used inside DDIM. (For ancestral sampling their algorithm 1 instead shifts the mean of each step by $s\,\Sigma\,\nabla \log p_\phi$.) Their classifier is the downsampling half of the U-Net, trained on the same noising distribution. With $s = 1$ on an unconditional ImageNet model, the classifier gave the target class about 50% probability, but the samples did not look like the class; a larger $s$ fixed this. The cost: a second network trained on noisy data, and a backward pass through it at every step.

Classifier-free guidance #

Classifier-free guidance (Ho and Salimans 2022) needs no classifier.

• Training (their algorithm 1). Train one network $\hat\varepsilon_\theta(x_t, t, c)$. In each training example, replace the condition $c$ by the null token $\varnothing$ with probability $p_{\text{uncond}}$. This is called condition dropout. They found $0.1$ and $0.2$ equally good and $0.5$ worse. The network learns the conditional and the unconditional model at the same time.
• Sampling (their algorithm 2). At each step evaluate both $\hat\varepsilon_c$ and $\hat\varepsilon_u$ (one batch of two, so 2 NFE per step) and use $\tilde\varepsilon = \hat\varepsilon_u + w\,(\hat\varepsilon_c - \hat\varepsilon_u)$ in any sampler of lesson 09.

Theorem 2 (CFG is the conditional model plus implicit-classifier guidance) #

Let $g = \nabla \log p_t(x \mid y) - \nabla \log p_t(x)$ be the implicit classifier gradient ($= \nabla \log p_t(y \mid x)$ by Theorem 1). For exact noise predictions, $g = -(\hat\varepsilon_c - \hat\varepsilon_u)/\sigma_t$ and

Proof.

1. Both scores are $-\hat\varepsilon/\sigma_t$ of their prediction, so $g = -(\hat\varepsilon_c - \hat\varepsilon_u)/\sigma_t$.
2. Rearrange: $\hat\varepsilon_u + w(\hat\varepsilon_c - \hat\varepsilon_u) = \hat\varepsilon_c + (w - 1)(\hat\varepsilon_c - \hat\varepsilon_u)$.
3. Substitute $\hat\varepsilon_c - \hat\varepsilon_u = -\sigma_t\,g$ from step 1. The result is the classifier-guidance formula applied to the conditional model, with scale $s = w - 1$ and classifier gradient $g$.
4. In score form the guided score is $(1 - w)\nabla \log p_t(x) + w\nabla \log p_t(x \mid y) = \nabla \log\big[p_t(x \mid y)\,p_t(y \mid x)^{w-1}\big]$. This is also classifier guidance with scale $w$ on the unconditional model. $\blacksquare$

So no classifier is trained: the generative model is the classifier, through $p_t(y \mid x) \propto p_t(x \mid y)/p_t(x)$. Ho and Salimans add a caveat: a real network's $\hat\varepsilon$ fields are not exactly gradients of anything, so in general no classifier exists whose gradient gives the guided field.

What guidance does to the sampled distribution #

At noise level $t$ the guided score is the score of

It is tempting to conclude that the sampler returns $\pi_0 \propto p(x)^{1-w}\,p(x \mid y)^{w}$, the tilted target. It does not, except for $w = 0$ and $w = 1$:

1. The PF-ODE and the reverse SDE reproduce the marginals of a diffusion only if the scores belong to one diffusion path, that is, if each $\pi_t$ is $\pi_0$ convolved with the noise kernel (lesson 09).
2. Noising (a convolution with a Gaussian) and tilting (powers and products of densities) do not commute. So $\pi_t$ is not the noised $\pi_0$. The worked example in section 4(b) shows this with numbers.
3. So the guided fields belong to no single diffusion path, and the result depends on the sampler. DDIM-type (ODE) and DDPM-type (SDE) sampling with CFG give different distributions, and neither is $\pi_0$ (Bradley and Nakkiran 2024 analyze CFG as a predictor–corrector method).

Proposition 3 (guided PF-ODE in a linear-Gaussian model). In 1D, let the unconditional density be $p(x) = \mathcal N(0, v_u)$ and the conditional density $p(x \mid y) = \mathcal N(m_c, v_c)$ with $v_c < v_u$ (for example the posterior of a Gaussian prior after a noisy observation). Use VE noising $x_\sigma = x + \sigma n$, so $p_\sigma(x) = \mathcal N(0, v_u + \sigma^2)$ and $p_\sigma(x \mid y) = \mathcal N(m_c, v_c + \sigma^2)$. Let $w \ge 1$. The guided PF-ODE, started from $\mathcal N(0, \sigma_{\max}^2)$ with $\sigma_{\max} \to \infty$, ends at a Gaussian with variance $v_c^{\,w}\,v_u^{\,1-w}$. The tilted target $\pi_0$ has variance $1/\big((1-w)/v_u + w/v_c\big)$.

The ODE interpolates the variances geometrically. The tilted target interpolates the precisions (inverse variances) linearly. For $w > 1$ and $v_c < v_u$ the ODE result is always the narrower one: guidance overshoots even the tilted target.

The fidelity–diversity trade-off #

• Numbers (Dhariwal and Nichol, table 4, conditional ImageNet $256\times256$, classifier scale $s$). No guidance: FID 10.94, precision 0.69, recall 0.63. $s = 1$: FID 4.59, precision 0.82, recall 0.52. $s = 10$: FID 9.11, precision 0.88, recall 0.32. Precision (how many samples look real) rises, recall (how much of the data distribution is covered) falls, and FID, which measures both, is best in between (lesson 14). With CFG, Ho and Salimans found the best FID at $w' = 0.1$ on ImageNet $64\times64$ and $w' = 0.3$ on $128\times128$ ($w = 1.1$ and $1.3$). The Inception score kept rising up to the largest value they tested, $w' = 4$.
• Oversaturation. A large $w$ extrapolates far beyond both predictions and gives over-exposed, oversaturated images. Two remedies: dynamic thresholding (Imagen, in pixel space) and CFG rescale (Lin et al. 2024, section 3.4): rescale the guided prediction so that its standard deviation matches that of the conditional prediction, then mix it with the unrescaled one (factor $\phi = 0.7$).
• Cost. 2 NFE per step: guidance doubles every NFE count of lesson 09.

Recent fixes (as of 2024) #

Limited-interval guidance (Kynkäänniemi et al., NeurIPS 2024). The authors found that guidance is "clearly harmful toward the beginning of the chain (high noise levels), largely unnecessary toward the end (low noise levels), and only beneficial in the middle". So they use $w > 1$ only in a band of noise levels. On ImageNet-512 this improved the record FID from 1.81 to 1.40. Steps outside the band also cost only 1 NFE. Proposition 3's model shows why high noise is harmful: as $\sigma \to \infty$ the centre of $\pi_\sigma$ tends to $w\,m_c$, so at high noise guidance pulls every sample toward one extrapolated point. This removes diversity before the details are decided.

Autoguidance (Karras et al., NeurIPS 2024). Guide with a smaller or shorter-trained version of the same conditional model instead of the unconditional model: $\tilde\varepsilon = \hat\varepsilon_{\text{bad}} + w\,(\hat\varepsilon_{\text{good}} - \hat\varepsilon_{\text{bad}})$. The two models make similar errors, so the difference points toward better samples instead of away from the unconditional distribution. It improves quality without the same loss of variation (record FIDs 1.01 on ImageNet-64 and 1.25 on ImageNet-512 with public networks).

Answers to section 1. (1) Neither, for $w = 3$. The samples are not from $p(x \mid y)$ (that needs $w = 1$). They are not from the tilted density either, because tilting does not commute with noising. Section 4(b) gives three different answers for the tilted target, the ODE and the SDE. (2) Nothing changes. Along a direction where both classes have the same Gaussian, the conditional and the unconditional scores are equal, so $(1 - w)\,\text{score} + w\,\text{score}$ is the unguided score. The variance stays $0.25$ in the example of section 4(c).

4. Worked example #

(a) One CFG step and its two readings. 2D, $\bar\alpha = 0.64$, so $\sigma_t = \sqrt{1 - 0.64} = 0.6$. Take $\hat\varepsilon_u = (0.1, 0.4)$, $\hat\varepsilon_c = (0.3, 0.1)$ and $w = 4$.

1. $\hat\varepsilon_c - \hat\varepsilon_u = (0.3 - 0.1,\ 0.1 - 0.4) = (0.2, -0.3)$.
2. $\tilde\varepsilon = (0.1, 0.4) + 4\,(0.2, -0.3) = (0.1 + 0.8,\ 0.4 - 1.2) = (0.9, -0.8)$.
3. Ho and Salimans' convention: $w' = w - 1 = 3$, and $4\,\hat\varepsilon_c - 3\,\hat\varepsilon_u = (1.2 - 0.3,\ 0.4 - 1.2) = (0.9, -0.8)$ ✓.
4. Implicit classifier gradient: $g = -(0.2, -0.3)/0.6 = (-0.3333,\ 0.5)$.
5. Theorem 2: $\hat\varepsilon_c - (w - 1)\,\sigma_t\,g = (0.3, 0.1) - 3 \cdot 0.6 \cdot (-0.3333, 0.5) = (0.3 + 0.6,\ 0.1 - 0.9) = (0.9, -0.8)$ ✓.

(b) Three different answers. Proposition 3's model with $v_u = 0.5$, $v_c = 0.25$, $m_c = 0.5$ (the posterior of the prior $\mathcal N(0, 0.5)$ after observing $y = 1$ with noise variance $0.5$). Take $w = 3$.

1. Tilted target $\pi_0$. Precision $(1 - 3)/0.5 + 3/0.25 = -4 + 12 = 8$, so the variance is $1/8 = 0.125$. The mean is (precision-weighted) $(3 \cdot 0.5/0.25)/8 = 6/8 = 0.75$.
2. The guided field at $\sigma = 1$ is the score of $\pi_1$ with precision $\lambda(1) = -2/1.5 + 3/1.25 = -1.3333 + 2.4 = 1.0667$: variance $0.9375$, mean $(3 \cdot 0.5/1.25)/1.0667 = 1.2/1.0667 = 1.125$. The noised target $\pi_0 * \mathcal N(0, 1)$ has variance $0.125 + 1 = 1.125$ and mean $0.75$. They differ: noising and tilting do not commute.
3. Guided PF-ODE (DDIM-type). Variance $v_c^3 v_u^{-2} = 0.015625 \cdot 4 = 0.0625$; mean $1.00$ (numerical solve, section 5).
4. Guided reverse SDE (DDPM-type). Mean $0.875$, variance $0.0969$ (section 5).

The conditional model itself ($w = 1$) gives $\mathcal N(0.5, 0.25)$. All three guided answers are narrower and shifted to the right, by different amounts.

(c) The two-class toy of section 2. Classes $\mathcal N((\pm 0.5, 0),\ 0.25\,I)$ in 2D, $w = 3$, the exact guided PF-ODE. The first coordinate ends with mean about $1.01$, variance about $0.106$ and $P(x_1 > 0) \approx 1.00$. For $w = 1$ it would be mean $0.5$, variance $0.25$ and $P = \Phi(1) = 0.841$. The tilted target has mean $0.72$ and variance $0.161$ (the same as the guided density with $s = 3$ in the first figure). The second coordinate keeps variance $0.25$.

5. In code #

NumPy and SciPy check the noisy classifier against Bayes' rule, the CFG algebra, and the numbers of section 4(b). For the PF-ODE and the reverse SDE in VE form (diffusion coefficient $g^2 = 2\sigma$), the mean $m$ and the variance $V$ of the guided samples obey simple ODEs in $\sigma$, which SciPy integrates from $\sigma_{\max} = 10^4$ down to 0.

import numpy as np
from scipy.integrate import solve_ivp
from scipy.special import expit
from scipy.stats import norm

# (1) Noisy classifier of the two-class toy: Bayes' rule vs the closed-form logistic
mu, s2 = 0.5, 0.25
for ab in (1.0, 0.5, 0.1, 0.01):
    a, v = np.sqrt(ab), ab * s2 + 1 - ab
    x = np.linspace(-2, 2, 9)
    bayes = norm.pdf(x, a * mu, np.sqrt(v)) / (norm.pdf(x, a * mu, np.sqrt(v)) + norm.pdf(x, -a * mu, np.sqrt(v)))
    k = 2 * a * mu / v
    print(ab, round(k, 3), np.allclose(bayes, expit(k * x)))
# 1.0 4.0 True / 0.5 1.131 True / 0.1 0.342 True / 0.01 0.101 True

# (2) Worked example (a): one CFG step, both conventions and Theorem 2
eu, ec, w, sig_t = np.array([0.1, 0.4]), np.array([0.3, 0.1]), 4.0, 0.6
g = -(ec - eu) / sig_t
print(eu + w * (ec - eu), (1 + (w - 1)) * ec - (w - 1) * eu, ec - (w - 1) * sig_t * g)
# [ 0.9 -0.8] [ 0.9 -0.8] [ 0.9 -0.8]

# (3) Worked example (b): tilted target vs guided PF-ODE vs guided reverse SDE
vu, vc, mc, w = 0.5, 0.25, 0.5, 3.0
prec = (1 - w) / vu + w / vc
print(1 / prec, w * mc / vc / prec)                       # 0.125 0.75
lam = lambda s: (1 - w) / (vu + s**2) + w / (vc + s**2)
print(1 / lam(1.0), w * mc / (vc + 1) / lam(1.0))          # 0.9375 1.125 (noised target: 1.125, 0.75)

def guided_moments(c):
    """c = 1: PF-ODE dx = -s*score ds; c = 2: reverse SDE dx = -2s*score ds + noise."""
    def rhs(s, u):
        m, V = u
        dV = 2 * s * lam(s) * V if c == 1 else 4 * s * lam(s) * V - 2 * s
        return [c * s * (lam(s) * m - w * mc / (vc + s**2)), dV]
    smax = 1e4
    sol = solve_ivp(rhs, (smax, 0.0), [0.0, smax**2 + vu], method="LSODA", rtol=1e-11, atol=1e-12)
    return sol.y[:, -1]

print(np.round(guided_moments(1), 4), vc**w * vu**(1 - w))  # [1.     0.0625] 0.0625
print(np.round(guided_moments(2), 4))                       # [0.875  0.0969]

In the SDE the variance gets an extra $-2\sigma$ term: noise of variance $2\sigma\,|d\sigma|$ is injected at each step, while the drift, twice as strong as in the ODE, contracts.

6. Where it shows up #

• Text-to-image and latent diffusion (lesson 13). Train with the text embedding dropped in 10–20% of the examples and sample with CFG. Negative prompts replace $\varnothing$ by the embedding of an unwanted prompt, so guidance pushes away from it.
• Conditional generators in science. A ground-motion generator conditioned on magnitude, distance and site parameters, or a weather model conditioned on the current state, should give a calibrated distribution, not a typical sample. So $w = 1$ is the natural default. If you use $w > 1$, check the spread with CRPS, rank histograms and the spread–skill ratio (lesson 14). In section 4(b), $w = 3$ cut the variance from $0.25$ to $0.0625$.
• Inverse problems and data assimilation (lesson 15). Replace the class by an observation: $\nabla \log p_t(x \mid y_{\text{obs}}) = \nabla \log p_t(x) + \nabla \log p_t(y_{\text{obs}} \mid x)$. The second term is a noisy likelihood. Theorem 1 is the exact statement behind these methods.
• Few-step models (lesson 12). Distilled models often include the guidance weight as an input, so that one evaluation per step gives a guided prediction.

7. Common mistakes #

• Mixing the conventions. In ours (and in diffusers) $w = 1$ means no guidance; in Ho and Salimans' convention $w' = 0$ does. The diffusers default $7.5$ is $w' = 6.5$.
• A clean classifier for classifier guidance. The classifier must be trained on $x_t$, with $t$ as an input (Theorem 1, step 3).
• Wrong factor or sign in the $\varepsilon$ form. The classifier gradient enters as $-s\,\sigma_t\,\nabla \log p_\phi$, because $\varepsilon$ points away from the data.
• "Guided samples come from $p(x)^{1-w}p(x \mid y)^w$." They do not (section 4b); the result even depends on the sampler.
• Strong guidance for ensembles or uncertainty. It makes the ensemble too narrow.
• No condition dropout in training. Then $\hat\varepsilon_u$ does not exist. A dropout probability of $0.5$ wastes capacity.
• Counting NFE. CFG doubles the cost of each guided step.

8. Check yourself #

1. 1D, $\bar\alpha = 0.36$, $\hat\varepsilon_u = 0.2$, $\hat\varepsilon_c = -0.1$, $w = 2.5$. Compute $\tilde\varepsilon$, the implicit classifier gradient, and $w'$ in Ho and Salimans' convention.
   Answer

   $\hat\varepsilon_c - \hat\varepsilon_u = -0.3$, so $\tilde\varepsilon = 0.2 + 2.5 \cdot (-0.3) = -0.55$. $\sigma_t = \sqrt{0.64} = 0.8$, so $g = 0.3/0.8 = 0.375$. Check with Theorem 2: $-0.1 - 1.5 \cdot 0.8 \cdot 0.375 = -0.1 - 0.45 = -0.55$ ✓. $w' = 1.5$.
2. Why must the classifier in classifier guidance see noisy inputs?
   Answer

   The sampler needs $\nabla \log p_t(y \mid x_t)$, and $p_t(y \mid x_t) = \int p(y \mid x_0)\,p(x_0 \mid x_t)\,dx_0$ averages the clean classifier over the denoising posterior. A clean classifier evaluated at $x_t$ is a different function; on noisy inputs it is out of distribution and its gradients are not meaningful.
3. Show that CFG with weight $w$ equals classifier guidance with scale $w$ on the unconditional model.
   Answer

   $(1-w)\nabla \log p_t(x) + w\nabla \log p_t(x \mid y) = \nabla \log p_t(x) + w\big(\nabla \log p_t(x \mid y) - \nabla \log p_t(x)\big) = \nabla \log p_t(x) + w\,\nabla \log p_t(y \mid x)$ by Theorem 1.
4. In Proposition 3's model with $v_c = 0.25$, $v_u = 0.5$ and $w = 3$: what is the mean of the guided field's density $\pi_\sigma$ as $\sigma \to \infty$, if $m_c = 0.5$?
   Answer

   The mean of $\pi_\sigma$ is $\frac{w\,m_c/(v_c + \sigma^2)}{\lambda(\sigma)}$. For large $\sigma$, $\lambda(\sigma) \approx (1 - w + w)/\sigma^2 = 1/\sigma^2$ and $w\,m_c/(v_c + \sigma^2) \approx w\,m_c/\sigma^2$, so the mean tends to $w\,m_c = 1.5$: three times the conditional mean.
5. Why does limited-interval guidance skip the highest noise levels?
   Answer

   At high noise the guided density is centred at an extrapolated point ($w\,m_c$ in Proposition 3's model), so guidance pulls all samples the same way and removes diversity before the details are decided. At low noise the class is already decided, so guidance does little. The middle band gives the benefit, and skipping guidance outside it also saves the second NFE.
6. What does autoguidance use instead of the unconditional model, and why does that keep diversity?
   Answer

   A weaker version (smaller or shorter-trained) of the same conditional model: $\tilde\varepsilon = \hat\varepsilon_{\text{bad}} + w(\hat\varepsilon_{\text{good}} - \hat\varepsilon_{\text{bad}})$. Both models share the condition, so the difference does not push away from other classes; it points away from the errors of the weaker model. This improves quality without narrowing the conditional distribution as much.

9. Interview angle #

• "Explain classifier-free guidance." — Drop the condition in 10–20% of training examples. At sampling use $\hat\varepsilon_u + w(\hat\varepsilon_c - \hat\varepsilon_u)$, 2 NFE per step. It is classifier guidance with the implicit classifier $p_t(x \mid y)/p_t(x)$.
• "Derive it from Bayes." — Theorem 1 (the score split at every noise level), then Theorem 2.
• "Why does FID first improve and then get worse as $w$ grows?" — Precision rises and recall falls; FID measures both, so it is best at a moderate $w$.
• "Do guided samples follow $p(x)^{1-w}p(x \mid y)^w$?" — No: tilting does not commute with noising; the answer even depends on the sampler (ODE vs SDE).
• "You need calibrated conditional ensembles. Guidance or not?" — $w = 1$, or a small $w$ chosen on a calibration metric (CRPS, spread–skill), possibly only in a limited noise interval.

10. Practice #

The notebook for this lesson has five exercises, each with a check cell and a folded solution. Everything uses the exact Gaussian-mixture scores; no network is trained.

1. Compute the noisy-classifier slope as a function of $\bar\alpha$ and check the four numbers $4.0, 1.13, 0.34, 0.10$.
2. Build the exact noisy classifier $p_t(y \mid x)$ of the two-class 8 Gaussians and check that its log-gradient equals the difference of the conditional and the unconditional scores.
3. Write the CFG denoiser, sample with the guided PF-ODE, and sweep $w$: measure class accuracy, mean radius and spread.
4. Compute the tilted target $p(x)\,p(y \mid x)^{w}$ on a grid for the 1D two-class toy with $w = 3$ and show that the CFG samples have a different mean and variance.
5. Add a guidance interval to the sampler and compare the effective number of class modes with guidance at every level.

11. Watch and read #

• Ho and Salimans, Classifier-Free Diffusion Guidance (2022): section 3.2 (the implicit classifier, algorithms 1 and 2), section 4 (sweeps over $w$ and $p_{\text{uncond}}$).
• Dhariwal and Nichol, Diffusion Models Beat GANs on Image Synthesis (2021): section 4 (classifier guidance, equation 14, table 4).
• Sander Dieleman, Guidance: a cheat code for diffusion models (2022): classifier guidance, CFG and temperature, with pictures.
• Kynkäänniemi et al., Applying Guidance in a Limited Interval Improves Sample and Distribution Quality in Diffusion Models (2024).
• Karras et al., Guiding a Diffusion Model with a Bad Version of Itself (autoguidance, 2024): section 3 has a 2D toy that explains why CFG improves quality.
• Bradley and Nakkiran, Classifier-Free Guidance is a Predictor-Corrector (2024): what distribution CFG really samples.
• Lin et al., Common Diffusion Noise Schedules and Sample Steps are Flawed (2024): section 3.4, CFG rescale.