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Пишем свою Diffusion модель с нуля

Всем привет, думаю у вас на слуху разного рода Diffusion модели последние 2 года. На его основе генерируют реалистичные изображения и видео, поэтому мне захотелось копнуть поглубже и узнать какова кроличья нора...

Меня зовут Юра, я - разработчик, фаундер и временами ML энтузиаст. Я решил разобраться и понять, как устроена Diffusion модель внутри, понять ее математику и постараться объяснить и разложить ее на пальцах. Ну и конечно пописать код, который (спойлер) заработал. На гифке изображены примеры итоговых картинок на моей финальной модели.

Если вам тоже интересно, милости прошу под кат.

MNIST

Небольшое вступление

Для тех, кто не хочет ждать и сразу взглянуть на код - не буду тянуть, вот ссылка на github проект.

Какой примерно у нас план:

~~Вспоминаем матан и тервер~~. Разбираем математику прямого и обратного процесса
От математики приходим к простой функции потерь (быстрый переход)
Пишем код (быстрый переход)

Сразу напишу, что без этих людей: Jonathan Ho, Ajay Jain, Pieter Abbeel (авторы оригинального DDPM) не было бы ни бума генераций картинок, ни моей статьи. Поэтому респект и спасибо им и другим ресерчерам, на статьи и работы которых опираются в этом пейпере! Также большое спасибо Lil Log за ее блог с объяснением математики, без этого тоже было бы трудно. Все другие ссылки, на которые я опирался и использовал, укажу в конце статьи.

Также в статье возможно есть неточности или ошибки. Буду благодарен за любой ваш фидбек или найденные ошибки.

Ну что? Погнали!

Math

Прямой процесс

Когда впервые видишь или читаешь пейпер, испытываешь те же чувства, что у Бена снизу. Непонятный набор формул на каждой странице втапливает вас постепенно в землю, но не будем паниковать раньше времени. Попробуем разобраться степ бай степ.

Вообще, что такое Diffusion model? Цитата из оригинального пейпера:

What distinguishes diffusion models from other types of latent variable models is that the approximate posterior $q(x_{1:T} |x_0)$ , called the forward process or diffusion process, is fixed to a Markov chain that gradually adds Gaussian noise to the data according to a variance schedule β1 , . . . , βT

Для нас важно здесь, что есть некое апостериорное распределение $q(x_{1:T} |x_0)$ (или пространство вероятностей), которое представляет собой процесс диффузии. Выглядит оно так:

$q(x_{1:T} |x_0) := \prod_{t=1}^n q(x_t|x_t−1)$

$q(x_t|x_{t−1}) := \mathcal{N}(x_t, \sqrt{1 − β_t}x_{t−1}, β_tI)$

Вторая формула на пальцах: условное распределение $q(x_t|x_{t−1})$ является нормальным распределение x_t с мат ожиданием $\sqrt{1 − β_t}x_{t−1}$ и дисперсией β_t. Процесс перехода из одного распределения в другое мы делаем T раз. Запишем конечное распределение как $q(x_{1:T} |x_0) := \prod_{t=1}^n q(x_t|x_t−1)$ .

Читатель, который наверное в легком замешательстве, имеет только один вопрос: как генерация картинок, да и вообще картинки связаны с этими формулами?
Картинок бесконечно много в мире(их конечное число, но много), но для простоты их очень много. Поэтому мы можем говорить о неком распределении картинок. Начальное нами выбранное распределение картинок - это x_0 .

Давайте на примере попробуем применить эти формулы. Для простоты будем работать в одномерном пространстве (с картинками работать будет точно также, но их не удобно отображать на графике).

Какой наш аглоритм? Мы выбираем случайную точку из множества x_0 , затем x_1 - это случайная точка из нормального распределения $q(x_1|x_0) := \mathcal{N}(x_1, \sqrt{1 − β_1}x_{0}, β_1)$ . На всякий случай, нормальное распределение выглядит так:

Затем x_2 будет случайной точкой из нормального распределения c мат ожиданием $\sqrt{1 − β_2}x_{1}$ и $\beta_2$ и так далее. В конце мы получим x_T , которая тоже в свою очередь будет некой точкой, выбранная из последнего нормального распределения. В условиях, что у нас есть начальное распределение x_0 , то в конце мы получим распределение x_T , все точки которого пройдут T итераций.

В случае картинок каждая итерация предсталяет собой добавление шума к картинке. На человеческий взгляд это выглядит, как порча картинки. Через множество итераций вменяемое изображение становится черти пойми чем, но важное уточнение, что это для человека. А в рамках математики этот шум приводит нас к грандиозному результату!

Пример, как выглядит трансформация картинок с использованием формул

Напишем применение формул выше на питоне с разным набором $\beta_t$ , x_0 установим в 1(то есть одна единственная начальная точка) и смотреть будем графики на интервале -5 до 5 с 1000 значений. T (общее количество итераций) сделаем равным 800.

Небольшой код на питоне:

timestep = 800

xs = np.arange(-5, 5, 0.01)

y_results = norm.pdf(xs, 0, 1)

different_schedules = [
    # constant
    np.ones(timestep) * 0.1,
    np.ones(timestep) * 0.3,
    np.ones(timestep) * 0.5,
    np.ones(timestep) * 0.7,
    np.ones(timestep) * 0.9,
    # linear change (from paper)
    np.linspace(1e-4, 0.02, timestep)
]

x0 = 1

for bt in different_schedules:

    # Set xt to zero, which will be used as x(t-1) like in sqrt(1-b) * x(t-1)
    xt = x0

    q_results = None

    # A simple loop by timestep
    for t in range(timestep):

        # Calculate the mean as in the formula
        mean = math.sqrt(1 - bt[t]) * xt

        # Calculate the variance
        variance = bt[t]

        # Q is a normal distribution, so to get the next xt, we just get random element from our distribution
        xt = np.random.normal(mean, math.sqrt(variance))
    q_results = norm.pdf(xs, mean, math.sqrt(variance))
    plt.plot(xs, q_results, label=f"qt for bt {bt[0] if bt[0] == bt[-1] else 'linear'}")
plt.plot(xs, y_results, label="Y")
plt.legend()
plt.show()

Получаем это:

То есть в целом, не смотря на разные набор значений $\beta_t$ , мы приходим к нормальному распределению.

Но с точки зрения будущей нейросети есть проблема в одной строчке кода:

xt = np.random.normal(mean, math.sqrt(variance))

То есть мы берем случайное значение из распределения. Сделать производную x_t по $x_{t-1}$ будет нереально (потому что здесь выбираются случайные значения) и соответственно обратное распространение ошибки нам сделать нельзя.

Чтобы обойти это, используем небольшой трюк (в пейпере про VAE его назвали reparametrization trick)

$x_t \sim q(x_t | x_{t-1}) := \mathcal{N}(x_t, \mu_{t-1}, \delta_{t-1} ^2 )$

$x_t = \mu_{t-1} + \delta_{t-1} * \epsilon$ , где $\epsilon \sim \mathcal{N}(0, I)$

Для нашего случая:

$x_t = \sqrt{1 - \beta_t}x_{t-1} + \beta_t * \epsilon$ , где $\epsilon \sim \mathcal{N}(0, I)$

То есть случайную величину выносим в формулу и записываем x_t через сумму $x_{t-1}$ с коэффициентом некой epsilon, которая и будет представлять случайное значенеие из нормального распределения с ожиданием 0 и дисперсией 1.

Выглядит может и странно, но давайте проверим:

for bt in different_schedules:

    # Set xt to zero, which will be used as x(t-1) like in sqrt(1-b) * x(t-1)
    xt = x0

    q_results = None

    # A simple loop by timestep
    for t in range(timestep):

        # Calculate the mean as in the formula
        mean = math.sqrt(1 - bt[t]) * xt

        # Calculate the variance
        variance = math.sqrt(bt[t])

        # Get eps from normal distribution
        eps = np.random.normal(0, 1)
        
        # Calculate our xt
        xt = mean + variance * eps
    q_results = norm.pdf(xs, mean, math.sqrt(variance))
    plt.plot(xs, q_results, label=f"qt for bt {bt[0] if bt[0] == bt[-1] else 'linear'}")
plt.plot(xs, y_results, label="Y")
plt.legend()
plt.show()

Результаты очень похожи на те, что были ранее. Можно брать за правду!

Но есть еще одна проблемка: сейчас мы последовательно вычисляем x_t через $x_{t-1}$ . Держим в уме, что T может быть очень большим и мы будем работать не в одномерном пространстве(а n-мерном, как картинки например). Как итог - вычисления громоздкие, хочется облегчить себе жизнь. Давайте вспомним тервер и запишем пару формул.

$\alpha_t = 1 - \beta_t, \bar \alpha_t = \prod_{i=1}^t\alpha_i$

$x_t = \sqrt{\alpha_t}x_{t-1} + \sqrt{1 - \alpha_t} * \epsilon_t = \sqrt{\alpha_t} \sqrt{\alpha_{t-1}} x_{t-2} + \sqrt{\alpha_t * (1 - \alpha_{t-1})} * \epsilon_{t-1} + \sqrt{1 - \alpha_t} * \epsilon_t$

$\epsilon_t \sim \mathcal{N}(0, \delta_t^2 I)$ и $\epsilon_{t-1} \sim \mathcal{N}(0, \delta_{t-1}^2 I)$ . Их объединение будет давать: $\mathcal{N}(0, (\delta_t^2 + \delta_{t-1}^2)I)$ . Перепишем формулу выше:

$x_t=\sqrt{\alpha_t * \alpha_{t-1}} x_{t-2} + \sqrt{\alpha_t * (1 - \alpha_{t-1}) + 1 - \alpha_t} * \bar{\epsilon_{t-2}} = \sqrt{\alpha_t * \alpha_{t-1}} x_{t-2} + \sqrt{1 - \alpha_t * \alpha_{t-1}} * \bar{\epsilon_{t-2}}$

Продолжая, получим:

$x_t=\sqrt{\bar \alpha_t} x_0 + \sqrt{1 - \bar \alpha_t} * \bar{\epsilon}$

Для тех, кто тервер не помнит, вот хорошая ссылка на вспоминание. Можно пока поверить на слово и глянуть на результаты с новой формулой.

Еще немного кода

```python for bt in different_schedules: alpha_t = 1 - bt cumprod_alpha_t = np.cumprod(alpha_t) mean = math.sqrt(cumprod_alpha_t[-1]) * x0 variance = math.sqrt(1 - cumprod_alpha_t[-1]) eps = np.random.normal(0, 1) variance = variance xt = mean + eps * variance q_results = norm.pdf(xs, mean, math.sqrt(variance)) plt.plot(xs, q_results, label=f"qt for bt {bt[0] if bt[0] == bt[-1] else 'linear'}") plt.plot(xs, y_results, label="Y") plt.legend() plt.show() ```

Видим, что такая формула ведет нас к нормальному распределению с ожиданием 0 и дисперсией 1 в итоге.

Кажется, мы разобрались с прямым процессом, переходим к обратному процессу.

Обратный процесс

Спойлер: очень много непростой математики. Прочитать и не понять ничего с первого раза - это нормально. Прочтите еще раз, что-то можно спросить LLM, что-то самому повыводить на бумаге.

Окей, у нас есть $q(x_t, x_{t-1})$ , а что насчет $q(x_{t-1}|x_t)$ ? На словах, добавляя шум к картинке, получаем шум (но мы уже знаем, что это нормальное распределение).

Но теперь самое интересное! Как получить из шума нормальную картинку?
Можно ли это вообще? Без спойлеров, все юзали Stable Diffusion, поэтому видимо можно. Давайте разберемся как.

$q(x_{t-1}|x_t)$ сходу кажется нельзя вывести. Давайте рассмотрим почему. Немного простого тервера.

$\begin{aligned}q(x_{t-1}|x_t, x_0) = \frac{q(x_{t-1}, x_t, x_0)}{q(x_t, x_0)}\end{aligned}$ (1)

$\begin{aligned}q(x_{t-1}|x_t)=\int_{x_0} q(x_{t-1}, x_0|x_t) \mathrm{d}x_0\end{aligned}$ (2)

$\begin{aligned}q(x_{t-1}, x_0| x_t) = \frac{q(x_{t-1}, x_0, x_t)}{q(x_t)} \end{aligned}$ (3)

Выводя одинаковый числитель из (1) и (3), получаем формулу:

$\begin{aligned} q(x_{t-1}|x_t, x_0) * q(x_t, x_0) = q(x_{t-1}, x_0| x_t) * q(x_t) => q(x_{t-1}, x_0| x_t) = q(x_{t-1}|x_t, x_0) * \frac{q(x_t, x_0)}{q(x_t)} \end{aligned}$ $\begin{aligned} q(x_{t-1}, x_0| x_t) = q(x_{t-1}|x_t, x_0) * q(x_0|x_t) \end{aligned}$

Теперь перепишем формулу (2):
$q(x_{t-1}|x_t)=\int q(x_{t-1}|x_t, x_0) * q(x_0|x_t) \mathrm{d}x_0$

С помощью формулы Баейса запишем q(x_0|x_t) , как $\begin{aligned} q(x_t|x_0) * \frac{q(x_0)}{q(x_t)} \end{aligned}$

То есть для вычисления $q(x_{t-1}|x_t)$ мы должны посчитать интеграл через все x_0 (все картинки в нашем датасете), вычисления которого будут долгими и дорогими.

Окей, $q(x_{t-1}|x_t)$ вывести нельзя, но можно вывести $q(x_{t-1}|x_t, x_0)$ . Что это означает? Мы пытаемся найти распределение $x_{t-1}$ , зная x_t и x_0 .

Считаем мат ожидание и дисперсию

Разбираемся с $q(x_{t-1}|x_t, x_0)$ .

Прежде всего возьмем за факт, что если $\beta$ - очень маленькое число и $q(x_t|x_{t-1})$ - Гауссово распределение, то $q(x_{t-1}|x_t)$ - тоже будет являться Гауссовым распределением. Цитата из пейпера:

"For both Gaussian and binomial diffusion, for continuous diffusion (limit of small step size β) the reversal of the diffusion process has the identical functional form as the forward process (Feller, 1949)."

Я попытался прочитать эту теорему и доказательство в оригинале и это было тяжело понимаемо, поэтому примем этот факт за правду.

Итак, если $q(x_{t-1}|x_t)$ - Гауссово распределение, то попытаемся найти $\mu$ и $\delta$ (то есть его мат ожидание и дисперсию).

Воспользовавшись правилом Байеса, мы имеем:

$\begin{aligned}q(x_{t-1}|x_t, x_0) = q(x_{t}|x_{t-1},x_0) * \frac{q(x_{t-1}|x_0)}{q(x_{t}|x_0)}\end{aligned}$

Далее, раскрываем каждый множитель в виде формулы Гауссова распределения (напомню, там будет перемножение 3х экспонент), затем подробно рассмотрим, что у нас будет происходит в экспонентах. Часть вычислений взял из Lil Log article

$\begin{aligned}\propto \exp (-\frac{1}{2} * (\frac{(x_t - \sqrt{\alpha_t}x_{t-1})^2}{\beta_t} + \frac{(x_{t-1} - \sqrt{\bar \alpha_{t-1}}x_0)^2}{1 - \bar \alpha_{t-1}} - \frac{(x_{t} - \sqrt{\bar \alpha_{t}}x_0)^2}{1 - \bar \alpha_t})) \end{aligned}$

$\begin{aligned} &= \exp (-\frac{1}{2} (\frac{\mathbf{x}_t^2 - 2\sqrt{\alpha_t} \mathbf{x}_t \mathbf{x}_{t-1} + \alpha_t \mathbf{x}_{t-1}^2 }{\beta_t} + \frac{ {\mathbf{x}_{t-1}^2} {- 2 \sqrt{\bar{\alpha}_{t-1}} \mathbf{x}_0} {\mathbf{x}_{t-1}} + \bar{\alpha}_{t-1} \mathbf{x}_0^2} {1-\bar{\alpha}_{t-1}} - \frac{(\mathbf{x}_t - \sqrt{\bar{\alpha}_t} \mathbf{x}_0)^2}{1-\bar{\alpha}_t} ) ) \\ &= \exp\Big( -\frac{1}{2} \big( {(\frac{\alpha_t}{\beta_t} + \frac{1}{1 - \bar{\alpha}_{t-1}})} \mathbf{x}_{t-1}^2 - (\frac{2\sqrt{\alpha_t}}{\beta_t} \mathbf{x}_t + \frac{2\sqrt{\bar{\alpha}_{t-1}}}{1 - \bar{\alpha}_{t-1}} \mathbf{x}_0) \mathbf{x}_{t-1} + C(\mathbf{x}_t, \mathbf{x}_0) \big) \Big) \end{aligned}$

Напомню, что:

$\begin{aligned} N(x, \mu, \delta^2) \sim x \propto \exp (-1/2 * \frac{(x - \mu) ^ 2}{\delta^2}) \end{aligned}$

Рассмотрим более детально:

$\begin{aligned}\frac{(x - \mu) ^ 2}{\delta^2} = \frac{(x^2 - 2x\mu + \mu^2)}{\delta^2} = \frac{1}{\delta^2} * x^2 - 2 * \frac{\mu}{\delta^2} * x + ... \end{aligned}$

Не сложно заметить, что если A * x ^ 2 , то $\delta^2 = 1 / A$ . Аналогично

$\mu: B * x => \mu = -\frac{1}{2} * \delta^2 * B$

Выведем мат ожидание и дисперсию для $q(x_{t-1}|x_t, x_0)$

$\begin{aligned} (\frac{\alpha_t}{\beta_t} + \frac{1}{1 - \bar{\alpha}_{t-1}}) \mathbf{x}_{t-1}^2 => \delta^2 = 1/A = 1 / (\frac{\alpha_t}{\beta_t} + \frac{1}{1 - \bar{\alpha}_{t-1}}) = \frac{1 - \bar \alpha_{t-1}}{1 - \bar \alpha_{t}} \beta_t \end{aligned}$ $\begin{aligned} -(\frac{2\sqrt{\alpha_t}}{\beta_t} \mathbf{x}_t + \frac{2\sqrt{\bar{\alpha}_{t-1}}}{1 - \bar{\alpha}_{t-1}} \mathbf{x}_0) \mathbf{x}_{t-1} => \mu = -\frac{1}{2} * \delta^2 * B = \frac{\sqrt{\alpha_t}(1- \bar \alpha_{t-1})}{1 - \bar \alpha_t}x_t + \frac{\sqrt{\bar \alpha_{t-1}} * \beta_t}{1 - \bar{\alpha_{t}}}x_0 \end{aligned}$

Сейчас $q(x_{t-1}|x_t, x_0) \sim N(x_{t-1}, \mu(x_t, x_0), \tilde \beta_tI)$

$\begin{aligned} \tilde \beta_t = \frac{1 - \bar \alpha_{t-1}}{1 - \bar \alpha_{t}} \beta_t \end{aligned}$ (4)

$\begin{aligned} \mu(x_t, x_0) = \frac{\sqrt{\alpha_t}(1- \bar \alpha_{t-1})}{1 - \bar \alpha_t}x_t + \frac{\sqrt{\bar \alpha_{t-1}} * \beta_t}{1 - \bar{\alpha_{t}}}x_0 \end{aligned}$ (5)

Вспомним, что $x_t=\sqrt{\bar \alpha_t} x_0 + \sqrt{1 - \bar \alpha_t} * \bar{\epsilon}$ , выразим x_0 через x_t

$\begin{aligned}x_0=\frac{1}{\sqrt{\bar \alpha_t}} (x_t - \sqrt{1 - \bar \alpha_t} * \bar{\epsilon}) \end{aligned}$

Затем выразим $\tilde \mu_t$ без использования x_0

$\begin{aligned} \tilde \mu_t = \frac{\sqrt{\alpha_t}(1- \bar \alpha_{t-1})}{1 - \bar \alpha_t}x_t + \frac{\sqrt{\bar \alpha_{t-1}} * \beta_t}{(1 - \bar{\alpha_{t}})\sqrt{\bar \alpha_t}}(x_t - \sqrt{1 - \bar \alpha_t} * \bar{\epsilon})=\frac{1}{\sqrt{\alpha_t}}x_t - \frac{1}{\sqrt{\alpha_t}} * \frac{\sqrt{1 - \bar \alpha_t}}{1 - \bar \alpha_t} * \beta_t * \bar {\epsilon} \end{aligned}$

Вспоминая, что $\beta_t = 1 - \alpha_t$ , запишем

$\begin{aligned} \tilde \mu_t = \frac{1}{\sqrt{\alpha_t}}(x_t - \frac{1 - \alpha_t}{\sqrt{1 - \bar \alpha_t}} \bar {\epsilon}) \end{aligned}$ (6)

Вводим модель

Как мы увидели ранее, мы не можем посчитать $q(x_{t-1}|x_t)$ , поэтому введем модель $p_\theta(x_{t-1}|x_t)$ , которая апроксимирует наше неизвестное распределение q. Также ранее мы доказали, что $q(x_{t-1}|x_t)$ - Гауссово распределение, значит $p_\theta(x_{t-1}|x_t)$ - тоже Гауссово. Запишем

$p_\theta(x_{t-1}|x_t) = N(x_{t-1}, \mu_\theta(x_t, t), \Sigma_\theta(x_t, t)) \quad p(x_{0...T}) = p(x_{0:T}) = p(x_T) * \prod_{t=1}^T p_\theta(x_{t-1}|x_t)$

У нас есть модель $p_\theta(x_{t-1}|x_t)$ , которое описывает некое распределение и у нас есть $q(x_{t-1}|x_t)$ . Наша задача найти оптимальное $\theta$ , при котором $p_\theta$ будет наиболее приближено к . Звучит так, что нам нужно найти функцию ошибки.

Чтобы сделать это, запишем функцию правдоподобия для $p_\theta$

$p_\theta(x) = \int p_\theta(x_{0:T})\mathrm{d}x_1,...x_T$ `

Наша задача максимизировать его, давайте запишем логарифм от функции правдоподобия(который точно также нужно максимизировать):

$\begin{aligned} \log p_\theta(x) &= \log \int p_\theta(x_{0:T})\mathrm{d}x_1,...x_T \\ &= \log \int p_\theta(x_{0:T}) * \frac{q(x_{1:T}|x_0)}{q(x_{1:T}|x_0)}\mathrm{d}x_1,...x_T \\ &= \log \int q(x_{1:T}|x_0) * \frac{p_\theta(x_{0:T})}{q(x_{1:T}|x_0)}\mathrm{d}x_1,...x_T \\ &= \log E_{q(x_{1:T}|x_0)} \frac{p_\theta(x_{0:T})}{q(x_{1:T}|x_0)} \\ \end{aligned}$

где E - мат ожидание от функции $\frac{p_\theta(x_{0:T})}{q(x_{1:T}|x_0)}$

Запишем еще формулу неравенства Jensen:
f(E(x)) <= E(f(x)) , если f(x) - выпуклая функция и

f(E(x)) >= E(f(x)) , если f(x) - вогнутая.

log(x) - вогнутая, поэтому получаем:

log(E(x)) >= E(log(x))

Перепишем форумулу выше:

$\begin{aligned} \log p_\theta(x) &= \log E_{q(x_{1:T}|x_0)} \frac{p_\theta(x_{0:T})}{q(x_{1:T}|x_0)} \\ &= E_{q(x_{1:T}|x_0)}[ \log \frac {p_\theta(x_T) * p_\theta(x_0|x_1) * \prod_{t=2}^{T} p_\theta(x_{t-1}|x_t) } { q(x_1|x_0) * \prod_{t=2}^{T} q(x_t|x_{t-1}) } ] \\ &= E_{q(x_{1:T}|x_0)}[ \log \frac {p_\theta(x_T) * p_\theta(x_0|x_1) } { q(x_1|x_0) } + \log \prod_{t=2}^{T} \frac { p_\theta(x_{t-1}|x_t) } { q(x_t|x_{t-1}) } ] \\ &= E_{q(x_{1:T}|x_0)}[ \log \frac {p_\theta(x_T) * p_\theta(x_0|x_1) } { q(x_1|x_0) } + \log \prod_{t=2}^{T} \frac { p_\theta(x_{t-1}|x_t) } { \frac { q(x_{t-1}|x_t, x_0) * \cancel { q(x_t|x_0) } } { \cancel { q(x_{t-1} | x_0) } } } ] \\ &= E_{q(x_{1:T}|x_0)}[ \log \frac {p_\theta(x_T) * p_\theta(x_0|x_1) } { q(x_1|x_0) } + \log \frac {q_(x_1|x_0) } { q(x_T|x_0) } + \log \prod_{t=2}^{T} \frac { p_\theta(x_{t-1}|x_t) } { q(x_{t-1}|x_t, x_0) } ] \\ &= E_{q(x_{1:T}|x_0)}[ \log \frac{p_\theta(x_T) * p_\theta(x_0|x_1)}{q(x_T|x_0) }] + E_{q(x_{1:T}|x_0)}[ \log \prod_{t=2}^{T} \frac {p_\theta(x_{t-1}|x_{t})}{q(x_{t-1}|x_{t}, x_0)} ] \\ &= E_{q(x_{1:T}|x_0)} \log p_\theta(x_0|x_1) + E_{q(x_{1:T}|x_0)} \log \frac{p_\theta(x_T)}{q(x_T|x_{0}) } + E_{q(x_{1:T}|x_0)}[ \sum_{t=2}^{T} \log \frac {p_\theta(x_{t - 1}|x_{t})}{q(x_{t-1}|x_{t},x_0)} ] \\ &= E_{q(x_1|x_0)} \log p_\theta(x_0|x_1) + E_{q(x_T|x_0)} \log \frac{p_\theta(x_T)}{q(x_T|x_0) } + \sum_{t=2}^{T} [E_{q(x_{t-1},x_t|x_0)} \log \frac {p_\theta(x_{t-1}|x_{t})}{q(x_{t-1}|x_{t},x_0)} ] \\ &= E_{q(x_1|x_0)} \log p_\theta(x_0|x_1) - D_{KL} (q(x_T|x_0) || p_\theta(x_T)) - \sum_{t=2}^{T} [E_{q(x_{t-1}|x_0)} D_{KL} (q(x_{t-1}|x_t, x_0) || p_\theta(x_{t-1}|x_t)) ] \end{aligned}$

Для тех, кого смущает, что такое $D_{KL}$ . $D_{KL}(P||Q)$ - KL Divergence или по русски "Расстояние Кульбака — Лейблера", метрика, которая показывает разницу между вероятностным распределением P и Q. В случае если они одинаковые, то $D_{KL}=0$ . Ссылка на вики

$\log p_\theta(x) >= E_{q(x_1|x_0)} \log p_\theta(x_0|x_1) - D_{KL} (q(x_T|x_0) || p_\theta(x_T)) - \sum_{t=2}^{T} [E_{q(x_{t-1}|x_0)} D_{KL} (q(x_{t-1}|x_t, x_0) || p_\theta(x_{t-1}|x_t)) ]$ (7)

То что мы вывели выше, называется Evidence Lower Bound (ELBO) или по русски Нижняя граница правдоподобия. Отдельное спасибо Calvin Luo с его замечательной статьей с объяснениями выведения этой формулы.

В формуле выше 1й и 2й член не зависят от $p_\theta(x_{t-1}|x_t)$ , поэтому мы можем убрать их и работать только с последней суммой. Наша цель - максимизировать правдоподобие, то есть смотря на формулу 7, мы должны минимизировать

$\sum_{t=2}^{T} [E_{q(x_{t-1}|x_0)} D_{KL} (q(x_{t-1}|x_t, x_0) || p_\theta(x_{t-1}|x_t)) ]$ (8)

$D_{KL}$ всегда положительное, поэтому наша цель - минимизировать его.

Кажется мы нашли нашу функцию потерь!

Функция потерь

Если подытожить очень кратко, то все вычисления, которые мы делали ранее, нужны для того, чтобы понять, как нам сделать распределение $p_\theta(x_{t-1}|x_t)$ максимально приближенным к $q(x_{t-1}|x_t)$ , которое как мы доказали ранее невозможно посчитать. Наши вычисления привели к тому, что нужно минимизировать некое $D_{KL}$ расстояние, которое сейчас мы и выведем.

Конечно же, к $D_{KL}$ расстоянию мы пришли не просто так, а потому, что есть формула, как посчитать расстояние между 2мя гауссовыми распределениями. Для любознательных вот линк.

$\begin{aligned} N_0(\mu_0, \Sigma_0); \: N_1(\mu_1, \Sigma_1); \; D_{KL}(N_0 || N_1) = \frac{1}{2}(tr(\Sigma_1^{-1}\Sigma_0) - k + (\mu_1 - \mu_0)^T\Sigma_1^{-1}(\mu_1 - \mu_0) + ln \frac{det\Sigma_1}{det\Sigma_0}) \end{aligned}$

Напомним, что $p_{theta} \sim N(\mu_{\theta}(x_t, t), \sigma_{\theta}(x_t, t)); \ q(x_{t-1}|x_t, x_0) \sim N(x_t, \mu(x_t, x_0), \sigma(t))$

Вычислим $D_{KL} (q(x_{t-1}|x_t, x_0) || p_\theta(x_{t-1}|x_t))$

$\begin{aligned} D_{KL} (q(x_{t-1}|x_t, x_0) || p_\theta(x_{t-1}|x_t)) &= \frac{1}{2}(k-k + (\mu(x_t, x_0) - \mu_{\theta}(x_t, t)) \Sigma_1^{-1}(\mu(x_t, x_0) - \mu_{\theta}(x_t, t)) + ln1) \\ &= \frac{1}{2}(\mu(x_t, x_0) - \mu_{\theta}(x_t, t))\frac{1}{\sigma_q^2(t)}(\mu(x_t, x_0) - \mu_{\theta}(x_t, t)) \\ &= \frac{1}{2\sigma_q^2(t)} ||\mu(x_t, x_0) - \mu_{\theta}(x_t, t)||_2^2 \end{aligned}$

Теперь перепишем нашу формулу 5:

$\begin{aligned} \mu(x_t, x_0)=\frac{\sqrt{\alpha_t}(1- \bar \alpha_{t-1})}{1 - \bar \alpha_t}x_t + \frac{\sqrt{\bar \alpha_{t-1}} * \beta_t}{1 - \bar{\alpha_{t}}}x_0 \\ \mu_\theta(x_t, t)=\frac{\sqrt{\alpha_t}(1- \bar \alpha_{t-1})}{1 - \bar \alpha_t}x_t + \frac{\sqrt{\bar \alpha_{t-1}} * \beta_t}{1 - \bar{\alpha_{t}}}x_{\theta} \\ D_{KL} = \frac{1}{2\sigma_q^2(t)} * \frac{\bar \alpha_{t-1} * (1 - \alpha_t)^2}{(1 - \bar{\alpha_{t}})^2} ||x_0 - x_{\theta}||_2^2 \end{aligned}$

Где $x_{\theta}$ - изображение, которое мы получаем от нейросети.

Возьмем формулу 6 и перепишем наше $D_{KL}$ расстояние:

$\begin{aligned} \tilde \mu_t(x_t) = \frac{1}{\sqrt{\alpha_t}}(x_t - \frac{1 - \alpha_t}{\sqrt{1 - \bar \alpha_t}} \bar {\epsilon}) \\ \tilde \mu_{\theta}(x_t) = \frac{1}{\sqrt{\alpha_t}}(x_t - \frac{1 - \alpha_t}{\sqrt{1 - \bar \alpha_t}} \epsilon_{\theta}) \\ D_{KL} = \frac{1}{2\sigma_q^2(t)} * \frac{(1 - \alpha_t)^2}{\alpha_t (1 - \bar{\alpha_{t}})} ||\epsilon - \epsilon_{\theta}||_2^2 \end{aligned}$

В нашем случае с картинками $\epsilon$ - шум, который мы добавляем к картинке, $\epsilon_{\theta}$ - шум, который нам предсказывает нейросеть.

Теперь наша функция ошибки выглядит так:

$\begin{aligned} Loss(\theta) = \arg \min_{\theta} \sum_{t=2}^{T}E_{q(x_t|x_0)} [\ \frac{1}{2\sigma_q^2(t)} * \frac{(1 - \alpha_t)^2}{\alpha_t (1 - \bar{\alpha_{t}})} ||\epsilon - \epsilon_{\theta}||_2^2 \ ] \end{aligned}$

Можно использовать более простую формулу:

$\begin{aligned} Loss_{simple}(\theta) = \arg \min_{\theta} \sum_{t=2}^{T}E_{q(x_t|x_0)} ||\epsilon - \epsilon_{\theta}||_2^2 \end{aligned}$

То есть простыми словами, при обучении на каждую картинку мы будем накладывать шум и будем учить нейросеть определять шум, который мы наложили. Шум будем накладывать по формуле, которую мы записали в самом начале:

$x_t=\sqrt{\bar \alpha_t} x_0 + \sqrt{1 - \bar \alpha_t} * \bar{\epsilon}$

При этом t будет разный для разных картинок и будет выбираться случайно на этапе обучения.

Как сгенерировать картинку после обучения

Напомним, что $p_\theta(x_{t-1}|x_t) = N(x_t, \mu_\theta(x_t), \Sigma_\theta(x_t, t))$ и $\Sigma_\theta = \beta_t$ в нашем случае, поэтому:

$\begin{aligned} x_{t-1} &= \mu_\theta(x_t) + \epsilon * \Sigma_\theta, \epsilon \sim N(0, I) \\ &= \frac{1}{\sqrt{\alpha_t}}(x_t - \frac{1 - \alpha_t}{\sqrt{1 - \bar \alpha_t}} \bar {\epsilon}) + \epsilon * \beta_t \end{aligned}$

То есть в цикле по t нам на каждом шагу надо вычислить x_t по формуле выше и в итоге мы придем к x_0 . При этом x_t - случайный шум (если более точно, случайное значение, взято из нормального распределения с ожиданием 1 и дисперсией 0), $\bar {\epsilon}$ - посчитанный шум нейросетью на шаге t и ${\epsilon}$ - еще один случайный шум.

Итог математики

Поздравляю, тех, кто полностью прочитал и ~~проклинает математику~~ выжил. Для тех, кто начал страдать, что он ничего не понял, просьба не сильно унывать. Без должной подготовки, это точно не простое чтиво. Что-то нужно перечитать 10-50 раз и потом придет осознание, где-то пописать формулы, вообщем рецепт у каждого свой!

Ну самое тяжелое на бумаге позади, предлагаю переходить к коду.

Пишем код

Писать код на бумаге - точно не лучшая затея. Поэтому прикрепляю ссылку на репозиторий. А я подсвечу некоторые интересные моменты.

Пойдем от большого к маленькому.

Diffusion model

DDPM(Denoising Diffusion Probalistic Model) или проще наша Diffusion модель. Код.

У модели есть метод forward, который используется для обучения и sample для генерации. Интересный момент в конструкторе этого класса:

self.T = T
self.eps_model = eps_model.to(device)
self.device = device
beta_schedule = torch.linspace(1e-4, 0.02, T + 1, device=device)
alpha_t_schedule = 1 - beta_schedule
bar_alpha_t_schedule = torch.cumprod(alpha_t_schedule.detach().cpu(), 0).to(device)
sqrt_bar_alpha_t_schedule = torch.sqrt(bar_alpha_t_schedule)
sqrt_minus_bar_alpha_t_schedule = torch.sqrt(1 - bar_alpha_t_schedule)
self.register_buffer("beta_schedule", beta_schedule)
self.register_buffer("alpha_t_schedule", alpha_t_schedule)
self.register_buffer("bar_alpha_t_schedule", bar_alpha_t_schedule)
self.register_buffer("sqrt_bar_alpha_t_schedule", sqrt_bar_alpha_t_schedule)
self.register_buffer("sqrt_minus_bar_alpha_t_schedule", sqrt_minus_bar_alpha_t_schedule)
self.criterion = nn.MSELoss()

Можно сказать, что эта вся основная математика, которая вообще есть в Diffusion модели. То есть мы считаем alpha и beta для вычисления картинки с шумом. Напомню формулу:

$x_t=\sqrt{\bar \alpha_t} x_0 + \sqrt{1 - \bar \alpha_t} * \bar{\epsilon}$

И напомню, что из alpha можно вывести beta и наоборот.

Сразу же здесь создаем нашу функцию ошибки или потерь, которая по сути является разницей(или L2 расстоянием) между шумом, который мы будем накладывать на картинку во время обучения и шум, который должна предсказать наша нейросеть. Также передаем на вход конструктора модель, которая и должна будет вычислять этот шум. Ее опишем ниже.

Выводили функцию потерь мы здесь.

Обучение

Напишем еще раз нашу формулу ошибки:

$\begin{aligned} Loss_{simple}(\theta) = \arg \min_{\theta} \sum_{t=2}^{T}E_{q(x_t|x_0)} ||\epsilon - \epsilon_{\theta}||_2^2 \end{aligned}$

Теперь давайте реализуем это в методе forward у ddpm:

t = torch.randint(low=1, high=self.T+1, size=(imgs.shape[0],), device=self.device)
noise = torch.randn_like(imgs, device=self.device)

# get noise image as: sqrt(alpha_t_bar) * x0 + noise * sqrt(1 - alpha_t_bar)
batch_size, channels, width, height = imgs.shape
noise_imgs = self.sqrt_bar_alpha_t_schedule[t].view((batch_size, 1, 1 ,1)) * imgs \
    + self.sqrt_minus_bar_alpha_t_schedule[t].view((batch_size, 1, 1, 1)) * noise

# get predicted noise from our model
pred_noise = self.eps_model(noise_imgs, t.unsqueeze(1))

# calculate of Loss simple ||noise - pred_noise||^2, which is MSELoss
return self.criterion(pred_noise, noise)

Генерация изображений

Вспомним формулу для генерации:

$\begin{aligned} x_{t-1} = \frac{1}{\sqrt{\alpha_t}}(x_t - \frac{1 - \alpha_t}{\sqrt{1 - \bar \alpha_t}} \bar {\epsilon}) + \epsilon * \beta_t \end{aligned}$

И код для него

# создаем случайный шум
x_t = torch.randn(n_samples, *size, device=self.device)
# вычисляем x_(t-1) на каждой итерации
for t in range(self.T, 0, -1):
    t_tensor = torch.tensor([t], device=self.device).repeat(x_t.shape[0], 1)
    pred_noise = self.eps_model(x_t, t_tensor)

    z = torch.randn_like(x_t, device=self.device) if t > 0 else 0

    x_t = 1 / torch.sqrt(self.alpha_t_schedule[t]) * \
        (x_t - pred_noise * (1 - self.alpha_t_schedule[t]) / self.sqrt_minus_bar_alpha_t_schedule[t]) + \
        torch.sqrt(self.beta_schedule[t]) * z
return x_t

По сути наш скелет готов, переходим на уровень ниже.

UNet модель

Как мы видели выше, у нас была загадочная eps_model, которая вычисляла шум, принимая на вход шум текущей итерации и время t.

По сути это может быть какая угодно модель, но в оригинальном пейпере (и даже в самом Stable Diffusion) используется UNet модель. Многим известная картинка из оригинального пейпера UNet, выглядит она схематично так:

На практике часто использовалась для детектирования изображения. В нашем же случае она немного поменялась, чтобы научиться работать с t.

Парочка интересных моментов:

def __init__(
    self,
    in_channels,
    out_channels,
    T,
    steps=(1, 2, 4),
    hid_size = 128,
    attn_step_indexes = [1],
    has_residuals=True,
    num_resolution_blocks=2,
    is_debug = False
):

На входе мы указываем steps: это кол-во слоев downscale и upscale(смотрим на картинку выше), где каждый элемент - общее число ResnetBlock на этом уровне. Также указываем слои с Attention(который всем нужен), чтобы ускорить обучение модели.

Для самого t, мы создаем embedding:

self.time_embedding = nn.Sequential(
    PositionalEmbedding(T=T, output_dim=hid_size),
    nn.Linear(hid_size, time_emb_dim),
    nn.ReLU(),
    nn.Linear(time_emb_dim, time_emb_dim)
)

И этот embedding мы будем прокидывать во все ResNet блоки при прохождении данных:

time_emb = self.time_embedding(t)

x = self.first_conv(x)
hx = []
for down_block in self.down_blocks:
    x = down_block(x, time_emb)
    if not isinstance(down_block, DownBlock):
        hx.append(x)
x = self.backbone(x, time_emb)

А вот как выглядит этот код внутри Resnet блока:

h = self.conv_1(x)
if t is None:
    return self.conv_3(x) + self.conv_2(h)
t = self.time_emb(t)
batch_size, emb_dim = t.shape 
t = t.view(batch_size, emb_dim, 1, 1)
if self.is_residual:
    return self.conv_3(x) + self.conv_2(h + t)
else:
    return self.conv_2(h + t)

Как видно эмбеддинг времени просто суммируется с внутренним представлением x, пропущенного через Convolutional слой. Но можно найти примеры, где обработка t происходит по-другому.

Результаты

ЭХЕЙ, можем поздравить друг друга! Есть кто еще живой?

Во-первых, спасибо, что прочитали. Я очень рад, что смог сам пройти этот путь осознания с нуля, но главное попытаться донести это в более простой форме для других.

Давайте оглянемся назад и вспомним, что мы узнали:

Мы рассмотрели Diffusion модель с точки зрения как и почему это работает
Прошлись через самые дебри математики, которая в итоге нас вывела к довольно простой функции потерь
Разобрали основные моменты кода и переноса математики в этот код

Если сравнивать результаты модели из пейпера и моей обученной на датасете CIFAR 10, то видно, что до бенчмарков еще далековато, но визуально "на глаз" очень похоже. Я обучал примерно 5-7 часов на RTX 3090 на runpod.io. Для улучшения бенчмарков стоит увеличить обучение модели, еще раз прочекать глубину слоев UNet и возможно пройтись по оптимизациям блоков UNet.

Model	FID (CIFAR 10)
Original Diffusion	3.17
My Diffusion	30

References

Если вам понравилась моя статья и вам интересно читать другие материалы от меня, то приглашаю подписаться на мой ТГ канал Jurassimo Park, там я пишу заметки о разработке стартапов, немного о нейронках и других технологиях.

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21 нояб. 2024 г.

07.09.23 16:24 CherryTeam

Cherry Team atlyginimų skaičiavimo programa yra labai naudingas įrankis įmonėms, kai reikia efektyviai valdyti ir skaičiuoti darbuotojų atlyginimus. Ši programinė įranga, turinti išsamias funkcijas ir patogią naudotojo sąsają, suteikia daug privalumų, kurie padeda supaprastinti darbo užmokesčio skaičiavimo procesus ir pagerinti finansų valdymą. Štai keletas pagrindinių priežasčių, kodėl Cherry Team atlyginimų skaičiavimo programa yra naudinga įmonėms: Automatizuoti ir tikslūs skaičiavimai: Atlyginimų skaičiavimai rankiniu būdu gali būti klaidingi ir reikalauti daug laiko. Programinė įranga Cherry Team automatizuoja visą atlyginimų skaičiavimo procesą, todėl nebereikia atlikti skaičiavimų rankiniu būdu ir sumažėja klaidų rizika. Tiksliai apskaičiuodama atlyginimus, įskaitant tokius veiksnius, kaip pagrindinis atlyginimas, viršvalandžiai, premijos, išskaitos ir mokesčiai, programa užtikrina tikslius ir be klaidų darbo užmokesčio skaičiavimo rezultatus. Sutaupoma laiko ir išlaidų: Darbo užmokesčio valdymas gali būti daug darbo jėgos reikalaujanti užduotis, reikalaujanti daug laiko ir išteklių. Programa Cherry Team supaprastina ir pagreitina darbo užmokesčio skaičiavimo procesą, nes automatizuoja skaičiavimus, generuoja darbo užmokesčio žiniaraščius ir tvarko išskaičiuojamus mokesčius. Šis automatizavimas padeda įmonėms sutaupyti daug laiko ir pastangų, todėl žmogiškųjų išteklių ir finansų komandos gali sutelkti dėmesį į strategiškai svarbesnę veiklą. Be to, racionalizuodamos darbo užmokesčio operacijas, įmonės gali sumažinti administracines išlaidas, susijusias su rankiniu darbo užmokesčio tvarkymu. Mokesčių ir darbo teisės aktų laikymasis: Įmonėms labai svarbu laikytis mokesčių ir darbo teisės aktų, kad išvengtų baudų ir teisinių problemų. Programinė įranga Cherry Team seka besikeičiančius mokesčių įstatymus ir darbo reglamentus, užtikrindama tikslius skaičiavimus ir teisinių reikalavimų laikymąsi. Programa gali dirbti su sudėtingais mokesčių scenarijais, pavyzdžiui, keliomis mokesčių grupėmis ir įvairių rūšių atskaitymais, todėl užtikrina atitiktį reikalavimams ir kartu sumažina klaidų riziką. Ataskaitų rengimas ir analizė: Programa Cherry Team siūlo patikimas ataskaitų teikimo ir analizės galimybes, suteikiančias įmonėms vertingų įžvalgų apie darbo užmokesčio duomenis. Ji gali generuoti ataskaitas apie įvairius aspektus, pavyzdžiui, darbo užmokesčio paskirstymą, išskaičiuojamus mokesčius ir darbo sąnaudas. Šios ataskaitos leidžia įmonėms analizuoti darbo užmokesčio tendencijas, nustatyti tobulintinas sritis ir priimti pagrįstus finansinius sprendimus. Pasinaudodamos duomenimis pagrįstomis įžvalgomis, įmonės gali optimizuoti savo darbo užmokesčio strategijas ir veiksmingai kontroliuoti išlaidas. Integracija su kitomis sistemomis: Cherry Team programinė įranga dažnai sklandžiai integruojama su kitomis personalo ir apskaitos sistemomis. Tokia integracija leidžia automatiškai perkelti atitinkamus duomenis, pavyzdžiui, informaciją apie darbuotojus ir finansinius įrašus, todėl nebereikia dubliuoti duomenų. Supaprastintas duomenų srautas tarp sistemų padidina bendrą efektyvumą ir sumažina duomenų klaidų ar neatitikimų riziką. Cherry Team atlyginimų apskaičiavimo programa įmonėms teikia didelę naudą - automatiniai ir tikslūs skaičiavimai, laiko ir sąnaudų taupymas, atitiktis mokesčių ir darbo teisės aktų reikalavimams, ataskaitų teikimo ir analizės galimybės bei integracija su kitomis sistemomis. Naudodamos šią programinę įrangą įmonės gali supaprastinti darbo užmokesčio skaičiavimo procesus, užtikrinti tikslumą ir atitiktį reikalavimams, padidinti darbuotojų pasitenkinimą ir gauti vertingų įžvalgų apie savo finansinius duomenis. Programa Cherry Team pasirodo esanti nepakeičiamas įrankis įmonėms, siekiančioms efektyviai ir veiksmingai valdyti darbo užmokestį. https://cherryteam.lt/lt/
08.10.23 01:30 davec8080

The "Shibarium for this confirmed rug pull is a BEP-20 project not related at all to Shibarium, SHIB, BONE or LEASH. The Plot Thickens. Someone posted the actual transactions!!!! https://bscscan.com/tx/0xa846ea0367c89c3f0bbfcc221cceea4c90d8f56ead2eb479d4cee41c75e02c97 It seems the article is true!!!! And it's also FUD. Let me explain. Check this link: https://bscscan.com/token/0x5a752c9fe3520522ea88f37a41c3ddd97c022c2f So there really is a "Shibarium" token. And somebody did a rug pull with it. CONFIRMED. But the "Shibarium" token for this confirmed rug pull is a BEP-20 project not related at all to Shibarium, SHIB, BONE or LEASH.
24.06.24 04:31 tashandiarisha

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26.06.24 18:46 Jacobethannn098

LEGAL RECOUP FOR CRYPTO THEFT BY ADRIAN LAMO HACKER
26.06.24 18:46 Jacobethannn098

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04.07.24 04:49 ZionNaomi

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13.07.24 21:13 michaelharrell825

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17.07.24 02:26 thompsonrickey

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18.07.24 20:13 austinagastya

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27.08.24 12:50 James889900

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27.08.24 13:06 James889900

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02.09.24 20:24 [email protected]

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06.09.24 01:35 Celinagarcia

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06.09.24 01:44 Celinagarcia

HOW TO RECOVER MONEY LOST IN BITCOIN/USDT TRADING OR TO CRYPTO INVESTMENT !! Hi all, friends and families. I am writing From Alberton Canada. Last year I tried to invest in cryptocurrency trading in 2023, but lost a significant amount of money to scammers. I was cheated of my money, but thank God, I was referred to Hack Recovery Wizard they are among the best bitcoin recovery specialists on the planet. they helped me get every penny I lost to the scammers back to me with their forensic techniques. and I would like to take this opportunity to advise everyone to avoid making cryptocurrency investments online. If you have already lost money on forex, cryptocurrency or Ponzi schemes, please contact [email protected] or WhatsApp: +1 (757) 237–1724 at once they can help you get back the crypto you lost to scammers. BEST WISHES. Celina Garcia.
16.09.24 00:10 marcusaustin

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16.09.24 00:11 marcusaustin

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23.09.24 18:56 matthewshimself

At first, I was admittedly skeptical about Worldcoin (ref: https://worldcoin.org/blog/worldcoin/this-is-worldcoin-video-explainer-series), particularly around the use of biometric data and the WLD token as a reward mechanism for it. However, after following the project closer, I’ve come to appreciate the broader vision and see the value in the underlying tech behind it. The concept of Proof of Personhood (ref: https://worldcoin.org/blog/worldcoin/proof-of-personhood-what-it-is-why-its-needed) has definitely caught my attention, and does seem like a crucial step towards tackling growing issues like bots, deepfakes, and identity fraud. Sam Altman’s vision is nothing short of ambitious, but I do think he & Alex Blania have the chops to realize it as mainstay in the global economy.
01.10.24 14:54 Sinewclaudia

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02.10.24 22:27 Emily Hunter

Can those who have fallen victim to fraud get their money back? Yes, you might be able to get back what was taken from you if you fell prey to a fraud from an unregulated investing platform or any other scam, but only if you report it to the relevant authorities. With the right plan and supporting documentation, you can get back what you've lost. Most likely, the individuals in control of these unregulated platforms would attempt to convince you that what happened to your money was a sad accident when, in fact, it was a highly skilled heist. You should be aware that there are resources out there to help you if you or someone you know has experienced one of these circumstances. Do a search using (deftrecoup (.) c o m). Do not let the perpetrators of this hoaxes get away with ruining you mentally and financially.
18.10.24 09:34 freidatollerud

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31.10.24 00:13 ytre89

Can those who have fallen victim to fraud get their money back? Yes, you might be able to get back what was taken from you if you fell prey to a fraud from an unregulated investing platform or any other scam, but only if you report it to the relevant authorities. With the right plan and supporting documentation, you can get back what you've lost. Most likely, the individuals in control of these unregulated platforms would attempt to convince you that what happened to your money was a sad accident when, in fact, it was a highly skilled heist. You should be aware that there are resources out there to help you if you or someone you know has experienced one of these circumstances. Do a search using (deftrecoup (.) c o m). Do not let the perpetrators of this hoaxes get away with ruining you mentally and financially.
02.11.24 14:44 diannamendoza732

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12.11.24 00:50 TERESA

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17.11.24 09:31 Vivianlocke223

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19.11.24 03:06 [email protected]

My entire existence fell apart when a malevolent hacker recently gained access to my online accounts. I felt violated and extremely uneasy after discovering that the digital platforms I depended on for communication, employment, and finances had been compromised. Regaining control and restoring my digital security was an overwhelming task in the immediate aftermath. To help me navigate the difficult process of recovering my accounts and getting my peace of mind back, TRUST GEEKS HACK EXPERT came into my life as a ray of hope. They immediately put their highly skilled professionals to work, thoroughly examining the vulnerability and methodically preventing unwanted access. They guided me through each stage soothingly, explaining what was occurring and why, so I never felt lost or alone. They communicated with service providers to restore my legitimate access while skillfully navigating the complex labyrinth of account recovery procedures. My digital footprint was cleaned and strengthened against future attacks thanks to their equally amazing ability to remove any remaining evidence of the hacker's presence. However, TRUST GEEKS HACK EXPERT actual worth went beyond its technical aspects. They offered constant emotional support during the ordeal, understanding my fragility and sense of violation. My tense nerves were calmed by their comforting presence and kind comments, which served as a reminder that I wasn't alone in this struggle. With their help, I was able to reestablish my sense of security and control, which enabled me to return my attention to the significant areas of my life that had been upended. Ultimately, TRUST GEEKS HACK EXPERT all-encompassing strategy not only recovered my online accounts but also my general peace of mind, which is a priceless result for which I am incredibly appreciative of their knowledge and kindness. Make the approach and send a message to TRUST GEEKS HACK EXPERT Via Web site <> www://trustgeekshackexpert.com/-- E>mail: Trustgeekshackexpert(At)fastservice..com -- TeleGram,<> Trustgeekshackexpert
19.11.24 03:07 [email protected]

My entire existence fell apart when a malevolent hacker recently gained access to my online accounts. I felt violated and extremely uneasy after discovering that the digital platforms I depended on for communication, employment, and finances had been compromised. Regaining control and restoring my digital security was an overwhelming task in the immediate aftermath. To help me navigate the difficult process of recovering my accounts and getting my peace of mind back, TRUST GEEKS HACK EXPERT came into my life as a ray of hope. They immediately put their highly skilled professionals to work, thoroughly examining the vulnerability and methodically preventing unwanted access. They guided me through each stage soothingly, explaining what was occurring and why, so I never felt lost or alone. They communicated with service providers to restore my legitimate access while skillfully navigating the complex labyrinth of account recovery procedures. My digital footprint was cleaned and strengthened against future attacks thanks to their equally amazing ability to remove any remaining evidence of the hacker's presence. However, TRUST GEEKS HACK EXPERT actual worth went beyond its technical aspects. They offered constant emotional support during the ordeal, understanding my fragility and sense of violation. My tense nerves were calmed by their comforting presence and kind comments, which served as a reminder that I wasn't alone in this struggle. With their help, I was able to reestablish my sense of security and control, which enabled me to return my attention to the significant areas of my life that had been upended. Ultimately, TRUST GEEKS HACK EXPERT all-encompassing strategy not only recovered my online accounts but also my general peace of mind, which is a priceless result for which I am incredibly appreciative of their knowledge and kindness. Make the approach and send a message to TRUST GEEKS HACK EXPERT Via Web site <> www://trustgeekshackexpert.com/-- E>mail: Trustgeekshackexpert(At)fastservice..com -- TeleGram,<> Trustgeekshackexpert
21.11.24 04:14 ronaldandre617

Being a parent is great until your toddler figures out how to use your devices. One afternoon, I left my phone unattended for just a few minutes rookie mistake of the century. I thought I’d take a quick break, but little did I know that my curious little genius was about to embark on a digital adventure. By the time I came back, I was greeted by two shocking revelations: my toddler had somehow managed to buy a $5 dinosaur toy online and, even more alarmingly, had locked me out of my cryptocurrency wallet holding a hefty $75,000. Yes, you heard that right a dinosaur toy was the least of my worries! At first, I laughed it off. I mean, what toddler doesn’t have a penchant for expensive toys? But then reality set in. I stared at my phone in disbelief, desperately trying to guess whatever random string of gibberish my toddler had typed as a new password. Was it “dinosaur”? Or perhaps “sippy cup”? I felt like I was in a bizarre game of Password Gone Wrong. Every attempt led to failure, and soon the laughter faded, replaced by sheer panic. I was in way over my head, and my heart raced as the countdown of time ticked away. That’s when I decided to take action and turned to Digital Tech Guard Recovery, hoping they could solve the mystery that was my toddler’s handiwork. I explained my predicament, half-expecting them to chuckle at my misfortune, but they were incredibly professional and empathetic. Their confidence put me at ease, and I knew I was in good hands. Contact With WhatsApp: +1 (443) 859 - 2886 Email digital tech guard . com Telegram: digital tech guard recovery . com website link :: https : // digital tech guard . com Their team took on the challenge like pros, employing their advanced techniques to unlock my wallet with a level of skill I can only describe as magical. As I paced around, anxiously waiting for updates, I imagined my toddler inadvertently locking away my life savings forever. But lo and behold, it didn’t take long for Digital Tech Guard Recovery to work their magic. Not only did they recover the $75,000, but they also gave me invaluable tips on securing my wallet better like not leaving it accessible to tiny fingers! Who knew parenting could lead to such dramatic situations? Crisis averted, and I learned my lesson: always keep my devices out of reach of little explorers. If you ever find yourself in a similar predicament whether it’s tech-savvy toddlers or other digital disasters don’t hesitate to reach out to Digital Tech Guard Recovery. They saved my funds and my sanity, proving that no challenge is too great, even when it involves a toddler’s mischievous fingers!
21.11.24 08:02 Emily Hunter

If I hadn't found a review online and filed a complaint via email to support@deftrecoup. com , the people behind this unregulated scheme would have gotten away with leaving me in financial ruins. It was truly the most difficult period of my life.