Implement the **forward pass of Dropout**, a simple regularization technique that randomly zeroes some activations during training to reduce overfitting. You’ll apply **inverted dropout**, which scales kept activations so the expected output stays the same at train time.

$$
Y = \frac{X \odot M}{1 - p}, \quad M_{ij} \sim \text{Bernoulli}(1-p)
$$

## Requirements

Implement the function
```python
import numpy as np

def dropout_forward(X, p, training=True, seed=123):
    \"""
    Computes the forward pass of inverted dropout.

    Args:
        X (np.ndarray): Input activations of shape (n, d).
        p (float): Dropout probability (fraction to drop), e.g., 0.2.
        training (bool): If True, apply dropout; if False, return X unchanged.
        seed (int): Random seed for reproducibility.

    Returns:
        np.ndarray: Output activations after dropout, same shape as X.
    \"""
    # Your solution below
```

**Rules:**
- If `training=True`, sample a dropout mask `M` with keep probability `1 - p` and apply **inverted dropout** scaling.
- If `training=False`, return `X` unchanged (no mask, no scaling).
- Use `np.random.seed(seed)` to make the mask deterministic.
- Do not use any prebuilt dropout layers (e.g., from PyTorch/TensorFlow).
- Return a **NumPy array**.

## Example

```python
X = np.array([
    [1.0, 2.0, 3.0],
    [4.0, 5.0, 6.0]
])
p = 0.5

dropout_forward(X, p, training=True, seed=123)
```

Output:
```python
array([
    [0.0, 4.0, 0.0],
    [8.0, 0.0, 12.0]
])
```

Subarray Sum Equals K

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Dropout forward pass

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828. Dropout forward pass

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Constraints

Similar Questions

Argument	Type
X	np.ndarray
p	float
seed	int
training	bool