Markov

AI 101

Setting the stage

• ImageNet, the big vision thing, came out in 2006.
  • AI without words
• The LLM was developed in 2017
  • AI with words
• ChatGPT became “prominent” in 2023
  • AI in the news

Looking at Google

• We’ll do Google relative - they competed in ImageNet, developed the LLM, and have a pretty major one (Gemini).
• How big is Google?
  • In 2006, 122 billion USD market cap
  • In 2017, 729 billion USD market cap
  • In 2026, 3680 billion USD market cap
• Why does this matter?

Scaling

• There are a lot of words.
  • And there are more words in a sentence than there are in general.
  • For example, that previous line contained “in” twice.
  • Basically, more than 9 inputs (for dice).
• The thing that held up LLMs was not having enough computing power to encode all words.

Our example

• We’ll use an extremely restricted example, inspired by a meme I saw once.

Punch into Colab

• I just typed these into Colab.
  • I used all lower case and no punctuation (and change “not” to “no” for length)
  • This is a similar “simplifying assumption” as regarded dots as only present or absent.

sar = "to do is to be"
soc = "to be is to do"
sha = "to be or no to be"
sin = "do be do be do"
all = [sar, soc, sha, sin]

Tokenizing

• There’s a core insight in computational linguistics called “tokenization”.
• Somehow, have to break up words into things that can be recognized by a “sensory neuron”.
• I treat individual words as tokens.
  • This could be an entire class.

words = [quote.split() for quote in all]
print(words)

[['to', 'do', 'is', 'to', 'be'], ['to', 'be', 'is', 'to', 'do'], ['to', 'be', 'or', 'no', 'to', 'be'], ['do', 'be', 'do', 'be', 'do']]

Sets

• I want to see how many unique words there are.
  • Obviously, “to” appears many times.
• I use a “set”, a mathematical object that is a (1) collection with (2) no duplicates.
  • They also aren’t ordered, which doesn’t really matter here.

sets = [set(quote) for quote in words]
print(sets)

[{'do', 'be', 'to', 'is'}, {'do', 'be', 'to', 'is'}, {'no', 'be', 'to', 'or'}, {'do', 'be'}]

Unique words

• How many words are there across all three phrases?
• I take the union of sets, which is all elements in at least one set.

unique = set.union(*sets)
print(unique)

{'is', 'or', 'do', 'be', 'to', 'no'}

• Only 6 words! Not too bad!
  • Only one more than our compressed dice.

Making a Network

• We can make something that looks an awful lot like a perceptron!

Edge Weights

• How do we determine edge weights?
• Or perhaps, what are we trying to do?
• Let’s imagine what our task is:
  • We want to generate text given some text, so…
  • Given a word, produce the next word.

Words-to-words

• So, what do we do?
  • Let’s take a look at our “quotes”.
  • Let’s see which words follow which other words.
    • Perhaps at which probability.
  • Let’s plug those in as edge weights.

First things first

• Let’s just look at one word - the first word we see.

words

[['to', 'do', 'is', 'to', 'be'],
 ['to', 'be', 'is', 'to', 'do'],
 ['to', 'be', 'or', 'no', 'to', 'be'],
 ['do', 'be', 'do', 'be', 'do']]

• Okay, that word is “to”.

Second things second

• What words can follow “to”.
  • Well, it looks to me like “do” and “be”
  • I can write some code to make sure.
  • The point of this class isn’t writing that code, but I will show it for the interested student!

Finding what’s next

• I loop over all quotes.
  • I loop over all words in the quotes.
    • If the current word is “to”, I save the next word.
• I look at all the next words.

next = []
for quote in words:
    for loc, word in enumerate(quote):
        if word == "to":
            next.append(quote[loc + 1])
print(next)

['do', 'be', 'be', 'do', 'be', 'be']

Do this for all words

• We aren’t restricted for doing this just for “to”
• Do it for all the words!
  • We have to add one special case.
  • We check quote length to make sure there is a next word.

nexts = []
for first in unique:
    next = []
    for quote in words:
        for loc, word in enumerate(quote):
            if word == first and loc + 1 < len(quote):
                next.append(quote[loc + 1])
    nexts.append(next)

Let’s see it

print(nexts)

[['to', 'to'], ['no'], ['is', 'be', 'be'], ['is', 'or', 'do', 'do'], ['do', 'be', 'be', 'do', 'be', 'be'], ['to']]

• Oh… that is tough to understand.
  • We’ll use another coding thing that I should mention.
  • I think it will be important to AI in the future, not seeing much usage now.

Key-Value Storage

A dictionary

• What we really need is something a lot like a dictionary.
  • Rather just have a list of lists of words, we want a list of words for each starting word.
  • Dictionaries also contain “lists of words” (definitions) for each word.
• In computing, we term this “key-value storage”.
  • Keys are words
  • Values are definitions.

Colab has dictionaries

• There so happens to be something called a dictionary (well, a dict) I can use.

d = dict()
print(d)

{}

Adding keys

• We add things to dictionary as key-value pairs
  • We take the name of the dictionary, like my_values
  • We add box brackets []
  • Within those brackets, we give the key (e.g. the word for which we are storing the definition
  • We use single-equals assignment to set the value of the key within the dictionary.
• Like this:

d["to"] = ['do', 'be', 'be', 'do', 'be', 'be']

Seeing values

• It is easy enough to see a value from here.
• Same as setting a value, just without single-equals assignment.
  • We take the name of the dictionary, like my_values
  • We add box brackets []
  • Within those brackets, we give the key (e.g. the word for which we are storing the definition

print(d["to"])

['do', 'be', 'be', 'do', 'be', 'be']

Back To Work

Use A dict

• Same as before, but we use a dictionary.
  • The key is “before” or “first” word.
  • The value is the “next” or “second” word.

nexts = dict()
for first in unique:
    next = []
    for quote in words:
        for loc, word in enumerate(quote):
            if word == first and loc + 1 < len(quote):
                next.append(quote[loc + 1])
    nexts[first] = next

Easier to see

• Recall - we are doing this to find edge weights

print(nexts)

{'is': ['to', 'to'], 'or': ['no'], 'do': ['is', 'be', 'be'], 'be': ['is', 'or', 'do', 'do'], 'to': ['do', 'be', 'be', 'do', 'be', 'be'], 'no': ['to']}

• A bit easier visually with some formatting.

for first in nexts:
    print(first, nexts[first])

is ['to', 'to']
or ['no']
do ['is', 'be', 'be']
be ['is', 'or', 'do', 'do']
to ['do', 'be', 'be', 'do', 'be', 'be']
no ['to']

Takeaways

for first in nexts:
    print(first, nexts[first])

is ['to', 'to']
or ['no']
do ['is', 'be', 'be']
be ['is', 'or', 'do', 'do']
to ['do', 'be', 'be', 'do', 'be', 'be']
no ['to']

• “no” only precedes “to”.
• “or” only precedes “no”
• “is” only precedes “to”
• Others are more complex.

Early sketch

• “no” only precedes “to”.
• “or” only precedes “no”
• “is” only precedes “to”

Meaning of weights

• We have a problem now.
• Historically, we have used determinism
  • Every single time we see a six-sided die, we say it represents a six.
• Now we must use non-determinism
  • Sometimes when we see a “to”, it is followed by “be”
  • Sometimes it is followed by “do”.
  • How do we handle this?

Calculate Weights

• Let’s take an edge weight to be the probability.
• “to” will have an edge of weight \(\frac{2}{3}\) going to “be”.

Let’s look at “be”

• The way I think of this is:
  • After a “be”, at 25% probability see an “is”
  • After a “be”, at 25% probability see an “or”
  • After a “be”, at 50% probability see an “do”

Generating Text

• So, if we want to generate text.
  • We look at the current word.
  • We flip a coin or roll a die.
• For “be” we use a four-sided die.
  • And two numbers correspond to “do”.

Rolling Die in Colab

• We have already seen random numbers, when setting up our perceptron.
• It is easy enough to.
  • Generate a random number between zero and one.
  • Multiply it by the number of “next words”
  • Round or truncate to a whole number
  • Look up the number in that position, which is now the next number.

In Colab

• Here’s the code - don’t worry about understanding it unless you want.
  • nexts is the dictionary we made earlier.

def next_word(prev_word):
    die = np.random.rand()
    possible_words = nexts[prev_word]
    position = die * len(possible_words)
    position = int(position)
    return possible_words[position]

Try it out

• We’ll ask for the next word several times and hope to get different answers.

next_word("be")

'do'

next_word("be")

'do'

next_word("be")

'do'

next_word("be")

'do'

next_word("be")

'do'

Generating text

• We can now use a neural network to generate text!
• We just provide a starting word.
  • We will also provide a length, though there’s ways around that!

word = "to"
print(word)
for _ in range(4):
    word = next_word(word)
    print(word)

to
do
be
do
is

Clean it up

• Rather than printing each word, we’ll add words together into one big thing to print.
• We’ll also put all the code in a function.
• We’ll allow the function to accept a starting word and a length.

def make_text(first_word, num_words):
    text = first_word
    word = first_word
    for _ in range(num_words - 1):
        word = next_word(word)
        text = text + " " + word
    print(text)

Some examples

make_text("to", 5)

to be do is to

make_text("to", 5)

to be do is to

make_text("to", 5)

to be do be or

make_text("to", 5)

to be is to do

make_text("to", 5)

to do be do is

make_text("to", 5)

to do be do be

make_text("to", 5)

to be is to be

More examples

• Longer.

make_text("to", 10)

to be is to be or no to be or

make_text("to", 10)

to do be or no to be do is to

• Start with other words.

make_text("be", 5)

be is to be or

make_text("no", 5)

no to be is to

make_text("is", 5)

is to be do is

Bonuses

If Time

• We can look at start and stop tokens.
• We can discuss attention.

Summary

What we learned

• You can generate text with neural networks.
• We used a single layer, but of course…
  • It seems an awful lot like you can do anything by stacking them.

Fin