Flow Matching in 100 LOC

flow_matching.py

#!/usr/bin/env python

import math
import matplotlib.pyplot as plt
import torch
import torch.nn as nn

from sklearn.datasets import make_moons
from torch import Tensor
from tqdm import tqdm
from typing import *
from zuko.utils import odeint


def log_normal(x: Tensor) -> Tensor:
    return -(x.square() + math.log(2 * math.pi)).sum(dim=-1) / 2


class MLP(nn.Sequential):
    def __init__(
        self,
        in_features: int,
        out_features: int,
        hidden_features: List[int] = [64, 64],
    ):
        layers = []

        for a, b in zip(
            (in_features, *hidden_features),
            (*hidden_features, out_features),
        ):
            layers.extend([nn.Linear(a, b), nn.ELU()])

        super().__init__(*layers[:-1])


class CNF(nn.Module):
    def __init__(self, features: int, freqs: int = 3, **kwargs):
        super().__init__()

        self.net = MLP(2 * freqs + features, features, **kwargs)

        self.register_buffer('freqs', torch.arange(1, freqs + 1) * torch.pi)

    def forward(self, t: Tensor, x: Tensor) -> Tensor:
        t = self.freqs * t[..., None]
        t = torch.cat((t.cos(), t.sin()), dim=-1)
        t = t.expand(*x.shape[:-1], -1)

        return self.net(torch.cat((t, x), dim=-1))

    def encode(self, x: Tensor) -> Tensor:
        return odeint(self, x, 0.0, 1.0, phi=self.parameters())

    def decode(self, z: Tensor) -> Tensor:
        return odeint(self, z, 1.0, 0.0, phi=self.parameters())

    def log_prob(self, x: Tensor) -> Tensor:
        I = torch.eye(x.shape[-1], dtype=x.dtype, device=x.device)
        I = I.expand(*x.shape, x.shape[-1]).movedim(-1, 0)

        def augmented(t: Tensor, x: Tensor, ladj: Tensor) -> Tensor:
            with torch.enable_grad():
                x = x.requires_grad_()
                dx = self(t, x)

            jacobian = torch.autograd.grad(dx, x, I, create_graph=True, is_grads_batched=True)[0]
            trace = torch.einsum('i...i', jacobian)

            return dx, trace * 1e-2

        ladj = torch.zeros_like(x[..., 0])
        z, ladj = odeint(augmented, (x, ladj), 0.0, 1.0, phi=self.parameters())

        return log_normal(z) + ladj * 1e2


class FlowMatchingLoss(nn.Module):
    def __init__(self, v: nn.Module):
        super().__init__()

        self.v = v

    def forward(self, x: Tensor) -> Tensor:
        t = torch.rand_like(x[..., 0, None])
        z = torch.randn_like(x)
        y = (1 - t) * x + (1e-4 + (1 - 1e-4) * t) * z
        u = (1 - 1e-4) * z - x

        return (self.v(t.squeeze(-1), y) - u).square().mean()


if __name__ == '__main__':
    flow = CNF(2, hidden_features=[64] * 3)

    # Training
    loss = FlowMatchingLoss(flow)
    optimizer = torch.optim.Adam(flow.parameters(), lr=1e-3)

    data, _ = make_moons(16384, noise=0.05)
    data = torch.from_numpy(data).float()

    for epoch in tqdm(range(16384), ncols=88):
        subset = torch.randint(0, len(data), (256,))
        x = data[subset]

        loss(x).backward()

        optimizer.step()
        optimizer.zero_grad()

    # Sampling
    with torch.no_grad():
        z = torch.randn(16384, 2)
        x = flow.decode(z)

    plt.figure(figsize=(4.8, 4.8), dpi=150)
    plt.hist2d(*x.T, bins=64)
    plt.savefig('moons_fm.pdf')

    # Log-likelihood
    with torch.no_grad():
        log_p = flow.log_prob(data[:4])

    print(log_p)

Adaptive ODE solvers modify their integration step size according to an estimation of the integration error. If the error is too large, the step is rejected and the step size is reduced. The "if" can only be evaluated on CPU, and hence requires CPU-GPU synchronization.

I thought it was reverse that flow matching is the umbrella framework and score-based is one of the special case.

You can view this either way. The main difference is that flow-matching approximates an ODE while score-matching approximates an SDE.

Thanks for the nice implementation! If I understand correctly, in line 88-89, you implement conditional flow matching loss (CMF) based on Equation 23 in the flow matching paper (https://arxiv.org/pdf/2210.02747.pdf). However, shouldn't it be as following?

y = (1 - (1 - 1e-4) * t) * z + t * x
u = x - (1 - 1e-4) * z

However, if I change the code, the CFM model won't work. Could you please help me with that. Thanks a lot!

Hello @yuyangw 👋 As mentioned earlier in the thread, the 0 and 1 extremities of the time are reversed in this implementation, which is why the loss is slightly different. If you want to change the loss, you also have to switch the initial and final times ($0 \leftrightarrow 1$) of the odeint calls (in encode, decode and log_prob).

Hello @yuyangw 👋 As mentioned earlier in the thread, the 0 and 1 extremities of the time are reversed in this implementation, which is why the loss is slightly different. If you want to change the loss, you also have to switch the initial and final times (0↔1) of the odeint calls (in encode, decode and log_prob).

Hi @francois-rozet, thank you so much for the reply!

Hi @francois-rozet 👋, could you give a bit more details about the probability calculation? In particular where does the added 1e2 (1e-2) come from?

Hello @radiradev, as explained in the first comment,

Adaptive ODE solvers choose their step size according to an estimation of the integration error. For the trace-augmented ODE, odeint over estimates the integration error because the trace has large(r) absolute values, which leads to small step sizes. To mitigate this without significant loss of accuracy, I multiply the trace by a factor $10^{-2}$.

To paraphrase, I don't want the computation of log-absolute-determinant of the Jacobian (ladj) to influence the step size of the solver. But because the trace has high magnitude compared to the derivative (dx), it does influence it (and makes it much slower) in practice. To mitigate this, I multiply trace by a factor $10^{-2}$, and at the end multiply the ladj by the inverse factor $10^2$.

To paraphrase, I don't want the computation of log-absolute-determinant of the Jacobian (ladj) to influence the step size of the solver. But because the trace has high magnitude compared to the derivative (dx), it does influence it (and makes it much slower) in practice. To mitigate this, I multiply trace by a factor 10−2, and at the end multiply the ladj by the inverse factor 102.

Hi @francois-rozet, should I change this factor when dealing with other data (e.g. image embedding), or keep it the same ?

Hello @thangld201, the best would be to try different values for the factor (basically its a tradeoff between log-prob accuracy and efficiency) and pick what suits your needs. Note that this code expects x to be a vector or a batch of vectors. If x has the shape of an image it will likely not work.

@francois-rozet Thanks for your answer. So if the factor is lower (e.g. 1e-6), it gets less accurate but faster ?

Exactly, but potentially much less accurate, while being marginally faster. That's why you should try a few values (with the same input, to compare the results).

For decoding - I don't see anything that necessitates z being from a normal distribution. Does this mean z can be sampled from any probability distribution?

@jenkspt I would think so, I am aware of at least one study (in the context of data unfolding in High Energy Physics) that does data to data with this formulation. https://arxiv.org/abs/2311.17175
I have to think a bit deeply if that makes sense, though. (Results look good nonetheless)

@jenkspt As long as the distribution of $z$ is the same during training and sampling, I think it should work.