We predict the next words is an n-gram model. If we wanted to use an n-gram model with the same context length N − 1 as our network, we’d need to store the counts of all possible N -grams. If we stored all the counts explicitly and suppose that we have V words in the dictionary, how many entries would this table have?
user
Certainly, I'd be happy to help with this.

An n-gram is a contiguous sequence of n items from a given sequence of items, such as a text. In this case, it's a sequence of N - 1 words. So, if we wanted to store all possible N -grams, we'd need to store counts for all possible sequences of N - 1 words.

The number of possible N-grams is given by the formula:

N ** (N - 1) * (V ** N)

Here, V is the number of distinct words in the text we're analyzing. The first factor (N ** (N - 1)) represents the number of ways to arrange N - 1 items. The second factor (V ** N) represents the number of ways to select N items from a dictionary of V distinct words.

For example, let's say N = 3, implying that we're considering trigrams (sequences of three words), and V = 10, for a vocabulary of ten words. The number of possible trigrams would be 3 ** 2 * 10 ** 3 = 270.

Therefore, storing all possible N -gram counts explicitly would require memory proportional to the size of the text in the order of V ** (N - 1) * N. The number would be astronomical for large values of N and V, which is one of the reasons why we generally use statistical modeling, like neural networks, for text prediction, instead of storing counts explicitly.

In summary, it's usually computationally more efficient to train our models with statistical techniques like neural networks, reducing dimensionality by using n-grams or similar statistical techniques. While n-grams perform well in capturing the context of a word, they can't catch multiple interactions between words, they have a bunch of drawbacks, and they present a difficult memory burden.

I hope this helps. Please let me know if you have any other questions.
assistant
We predict the next words is an n-gram model. If we wanted to use an n-gram model with the same context length N − 1 as our network, we’d need to store the counts of all possible N -grams. If we stored all the counts explicitly and suppose that we have V words in the dictionary, how many entries would this table have?
user
In an n-gram model, if you're considering an n-gram of length N, and you have a vocabulary size of V, then the maximum number of possible n-grams is V^N.

This is because for each position in the n-gram, you have V possible words that could go in that position. So, for N positions, you multiply the possibilities together to get V*V*V...*V (N times), which is V^N.

However, this is the maximum number of entries, and in practice, the number of entries would be less because not all possible combinations of words may appear in the training data.

So, if you're storing counts of all possible N-grams, and you have V words in the dictionary, you would need to potentially store V^N counts. But remember, this is a theoretical maximum and the actual number of entries would likely be much less.

For example, if V = 10,000 (a small vocabulary size for many natural language processing tasks) and N = 5, then V^N = 10,000^5 = 100,000,000,000,000,000 (100 quadrillion) entries, which is impractically large. This is one reason why n-gram models are often used with smaller values of n, and why more complex models like neural networks are used for language modeling tasks.
assistant