Did most of HW3
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Uzair Mohammed
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hw3/bayesian_network_1_hw3.png
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hw3/bayesian_network_1_hw3.svg
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hw3/bayesian_network_1_hw3.svg
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hw3/data/family0.csv
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hw3/data/family0.csv
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name,mother,father,trait
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Harry,Lily,James,
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James,,,1
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Lily,,,0
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hw3/data/family1.csv
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hw3/data/family1.csv
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name,mother,father,trait
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Arthur,,,0
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Charlie,Molly,Arthur,0
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Fred,Molly,Arthur,1
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Ginny,Molly,Arthur,
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Molly,,,0
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Ron,Molly,Arthur,
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hw3/data/family2.csv
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hw3/data/family2.csv
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name,mother,father,trait
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Arthur,,,0
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Hermione,,,0
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Molly,,,
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Ron,Molly,Arthur,0
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Rose,Ron,Hermione,1
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hw3/heredity.py
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hw3/heredity.py
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import csv
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import itertools
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import sys
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PROBS = {
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# Unconditional probabilities for having gene
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"gene": {
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2: 0.01,
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1: 0.03,
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0: 0.96
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},
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"trait": {
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# Probability of trait given two copies of gene
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2: {
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True: 0.65,
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False: 0.35
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},
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# Probability of trait given one copy of gene
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1: {
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True: 0.56,
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False: 0.44
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},
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# Probability of trait given no gene
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0: {
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True: 0.01,
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False: 0.99
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}
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},
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# Mutation probability
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"mutation": 0.01
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}
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def main():
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# Check for proper usage
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if len(sys.argv) != 2:
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sys.exit("Usage: python heredity.py data.csv")
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people = load_data(sys.argv[1])
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# Keep track of gene and trait probabilities for each person
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probabilities = {
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person: {
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"gene": {
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2: 0,
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1: 0,
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0: 0
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},
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"trait": {
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True: 0,
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False: 0
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}
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}
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for person in people
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}
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# Loop over all sets of people who might have the trait
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names = set(people)
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for have_trait in powerset(names):
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# Check if current set of people violates known information
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fails_evidence = any(
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(people[person]["trait"] is not None and
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people[person]["trait"] != (person in have_trait))
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for person in names
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)
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if fails_evidence:
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continue
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# Loop over all sets of people who might have the gene
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for one_gene in powerset(names):
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for two_genes in powerset(names - one_gene):
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# Update probabilities with new joint probability
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p = joint_probability(people, one_gene, two_genes, have_trait)
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update(probabilities, one_gene, two_genes, have_trait, p)
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# Ensure probabilities sum to 1
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normalize(probabilities)
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# Print results
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for person in people:
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print(f"{person}:")
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for field in probabilities[person]:
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print(f" {field.capitalize()}:")
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for value in probabilities[person][field]:
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p = probabilities[person][field][value]
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print(f" {value}: {p:.4f}")
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def load_data(filename):
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"""
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Load gene and trait data from a file into a dictionary.
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File assumed to be a CSV containing fields name, mother, father, trait.
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mother, father must both be blank, or both be valid names in the CSV.
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trait should be 0 or 1 if trait is known, blank otherwise.
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"""
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data = dict()
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with open(filename) as f:
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reader = csv.DictReader(f)
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for row in reader:
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name = row["name"]
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data[name] = {
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"name": name,
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"mother": row["mother"] or None,
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"father": row["father"] or None,
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"trait": (True if row["trait"] == "1" else
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False if row["trait"] == "0" else None)
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}
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return data
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def powerset(s):
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"""
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Return a list of all possible subsets of set s.
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"""
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s = list(s)
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return [
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set(s) for s in itertools.chain.from_iterable(
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itertools.combinations(s, r) for r in range(len(s) + 1)
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)
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]
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def joint_probability(people, one_gene, two_genes, have_trait):
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"""
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Compute and return a joint probability.
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Args:
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- people is a dictionary of people. The keys represent names, and
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the values are dictionaries that contain mother and father keys.
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You may assume that either mother and father are both blank (no
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parental information in the data set), or mother and father will
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both refer to other people in the people dictionary.
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- one_gene is a set of all people for whom we want to compute the
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probability that they have one copy of the gene.
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- two_genes is a set of all people for whom we want to compute the
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probability that they have two copies of the gene.
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- have_trait is a set of all people for whom we want to compute the
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probability that they have the trait.
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The probability returned should be the probability that
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- everyone in set `one_gene` has one copy of the gene, and
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- everyone in set `two_genes` has two copies of the gene, and
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- everyone not in `one_gene` or `two_gene` does not have the gene, and
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- everyone in set `have_trait` has the trait, and
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- everyone not in set` have_trait` does not have the trait.
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and trait = {"Harry", "James"} should calculate the probability that
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Lily has zero copies of the gene, Harry has one copy of the gene,
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James has two copies of the gene, Harry exhibits the trait, James
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exhibits the trait, and Lily does not exhibit the trait.
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For anyone with no parents listed in the data set, use the probability
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distribution PROBS["gene"] to determine the probability that they have
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a particular number of the gene.
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For anyone with parents in the data set, each parent will pass one of
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their two genes on to their child randomly, and there is a
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PROBS["mutation"] chance that it mutates (goes from being the gene to not
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being the gene, or vice versa).
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Use the probability distribution PROBS["trait"] to compute the probability
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that a person does or does not have a particular trait.
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"""
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probability = 1
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for person in people:
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# Check if the person has one copy of the gene
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if person in one_gene:
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# Calculate the probability of having one copy of the gene
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gene_prob = PROBS["gene"][1]
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# Check if the person has two copies of the gene
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elif person in two_genes:
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# Calculate the probability of having two copies of the gene
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gene_prob = PROBS["gene"][2]
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else:
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# Calculate the probability of not having the gene
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gene_prob = PROBS["gene"][0]
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# Check if the person has the trait
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if person in have_trait:
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# Calculate the probability of having the trait
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trait_prob = PROBS["trait"][gene_prob][True]
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else:
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# Calculate the probability of not having the trait
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trait_prob = PROBS["trait"][gene_prob][False]
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# Multiply the gene and trait probabilities
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probability *= gene_prob * trait_prob
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return probability
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def update(probabilities, one_gene, two_genes, have_trait, p):
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"""
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Add to `probabilities` a new joint probability `p`.
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Each person should have their "gene" and "trait" distributions updated.
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Which value for each distribution is updated depends on whether
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the person is in `have_gene` and `have_trait`, respectively.
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Args:
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- probabilities is a dictionary of people. Each person is mapped to a
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"gene" distribution and a "trait" distribution.
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- one_gene is a set of people with one copy of the gene in the current
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joint distribution.
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- two_genes is a set of people with two copies of the gene in the
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current joint distribution.
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- have_trait is a set of people with the trait in the current joint
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distribution.
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- p is the probability of the joint distribution.
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For each person person in `probabilities`, the function should update
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the probabilities[person]["gene"] distribution and probabilities[person]["trait"]
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distribution by adding p to the appropriate value in each distribution.
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All other values should be left unchanged.
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For example, if "Harry" were in both two_genes and in have_trait, then p would
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be added to probabilities["Harry"]["gene"][2] and to
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probabilities["Harry"]["trait"][True].
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The function should not return any value: it just needs to update the
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probabilities dictionary.
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"""
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for person in probabilities:
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if person in one_gene:
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probabilities[person]["gene"][1] += p
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elif person in two_genes:
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probabilities[person]["gene"][2] += p
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else:
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probabilities[person]["gene"][0] += p
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if person in have_trait:
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probabilities[person]["trait"][True] += p
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else:
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probabilities[person]["trait"][False] += p
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def normalize(probabilities):
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"""
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Update `probabilities` such that each probability distribution
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is normalized (i.e., sums to 1, with relative proportions the same).
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Args:
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- probabilities is a dictionary of people. Each person is mapped to a
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"gene" distribution and a "trait" distribution.
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For both of the distributions for each person in probabilities, this
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function should normalize that distribution so that the values in the
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distribution sum to 1, and the relative values in the distribution are the same.
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For example, if probabilities["Harry"]["trait"][True] were equal to 0.1 and
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probabilities["Harry"]["trait"][False] were equal to 0.3, then your function
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should update the former value to be 0.25 and the latter value to be 0.75: the
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numbers now sum to 1, and the latter value is still three times larger than
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the former value.
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The function should not return any value: it just needs to update the
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probabilities dictionary.
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"""
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for person in probabilities:
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# Calculate the sum of the gene probabilities
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gene_sum = sum(probabilities[person]["gene"].values())
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# Calculate the sum of the trait probabilities
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trait_sum = sum(probabilities[person]["trait"].values())
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# Normalize the gene probabilities
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for gene in probabilities[person]["gene"]:
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probabilities[person]["gene"][gene] /= gene_sum
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# Normalize the trait probabilities
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for trait in probabilities[person]["trait"]:
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probabilities[person]["trait"][trait] /= trait_sum
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if __name__ == "__main__":
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main()
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hw3/hw3.pdf
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hw3/motorcade.png
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(generated)
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"main.log"
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main.fls
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PWD c:\Users\uzair\OneDrive\Documents\Programming\classes\spring-2024\CSC_665_Homework
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PWD d:\uzair\projects\sfsu\csc-665-homework
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INPUT d:\uzair\projects\sfsu\csc-665-homework\main.tex
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OUTPUT main.log
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538
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@ -18,9 +18,9 @@ My current workflow is as follows:
|
||||
|
||||
\begin{enumerate}
|
||||
\item Read the homework instructions and try to understand the assignment.
|
||||
|
||||
|
||||
\item Work by hand, on my tablet, in the Samsung Notes app.
|
||||
|
||||
|
||||
\item Beautify my work by transcribing it here using \LaTeX.
|
||||
\end{enumerate}
|
||||
When it's time to submit an assignment, I will export this document as a PDF file and turn in just the relevant pages. Please let me know what you think of this format!
|
||||
@ -34,157 +34,157 @@ When it's time to submit an assignment, I will export this document as a PDF fil
|
||||
\begin{enumerate}
|
||||
\item[a.] My pronouns are he/him.
|
||||
\item[b.] I've taken a lot of math and computer science courses. I'm not sure if I can list them, as I took most of them at the College of San Mateo and their course numbers are different. However, off the top of my head, I've taken:
|
||||
\begin{itemize}
|
||||
\item Calculus 1
|
||||
\item Calculus 2
|
||||
\item Calculus 3
|
||||
\item Discrete Mathematics
|
||||
\item Linear Algebra
|
||||
\item Analysis of Algorithms
|
||||
\item Data Structures
|
||||
\end{itemize}
|
||||
This list consists of the courses I think are relevant to this class; I've taken other CS courses, of course.
|
||||
\begin{itemize}
|
||||
\item Calculus 1
|
||||
\item Calculus 2
|
||||
\item Calculus 3
|
||||
\item Discrete Mathematics
|
||||
\item Linear Algebra
|
||||
\item Analysis of Algorithms
|
||||
\item Data Structures
|
||||
\end{itemize}
|
||||
This list consists of the courses I think are relevant to this class; I've taken other CS courses, of course.
|
||||
\item [c.] Yes, I am okay with being called on. I won't always know the answer, and I might embarrass myself sometimes, but that's alright because I believe it will force me to pay more attention and learn better, and as a result be more prepared for assignments and exams.
|
||||
\end{enumerate}
|
||||
|
||||
\subsection{Optimization}
|
||||
\begin{itemize}
|
||||
\item [a.] Let a, b, and c be positive real numbers. Consider the quadratic function
|
||||
$$f(\theta) = a{\theta}^2 + b\theta + c$$
|
||||
Note that $\theta$ here is a real number. What value of $\theta$ minimizes $f(\theta)$?\\[0.1in]
|
||||
\underline{Solution}:
|
||||
Since this is a positive parabola, we can use the formula for the vertex of a parabola, $\frac{-b}{2a}$, to find the minimum:\\
|
||||
$$
|
||||
\boxed{\theta = \frac{-b}{2a}}
|
||||
$$
|
||||
$$f(\theta) = a{\theta}^2 + b\theta + c$$
|
||||
Note that $\theta$ here is a real number. What value of $\theta$ minimizes $f(\theta)$?\\[0.1in]
|
||||
\underline{Solution}:
|
||||
Since this is a positive parabola, we can use the formula for the vertex of a parabola, $\frac{-b}{2a}$, to find the minimum:\\
|
||||
$$
|
||||
\boxed{\theta = \frac{-b}{2a}}
|
||||
$$
|
||||
|
||||
\fbox{\parbox{\textwidth}{For the following problems, I'm going to use the general approach of:
|
||||
\begin{enumerate}
|
||||
\item Take derivative of equation
|
||||
\item Set derivative equal to 0
|
||||
\item Solve for $\theta$
|
||||
\end{enumerate}}}
|
||||
|
||||
\fbox{\parbox{\textwidth}{For the following problems, I'm going to use the general approach of:
|
||||
\begin{enumerate}
|
||||
\item Take derivative of equation
|
||||
\item Set derivative equal to 0
|
||||
\item Solve for $\theta$
|
||||
\end{enumerate}}}
|
||||
|
||||
\item [b.] Let $x_1,\dots,x_n$ be real numbers. Consider the quadratic function
|
||||
$$g(\theta) = \sum^n_{i=1}(\theta - x_i)^2.$$
|
||||
What value of $\theta$ minimizes $g(\theta)$?\\
|
||||
\underline{Solution}:
|
||||
$$g(\theta) = \sum^n_{i=1}(\theta - x_i)^2.$$
|
||||
What value of $\theta$ minimizes $g(\theta)$?\\
|
||||
\underline{Solution}:
|
||||
|
||||
\begin{align*}
|
||||
g'(\theta) & = \sum_{i=1}^{n} 2(\theta -x_i) \\
|
||||
0 & = \sum_{i=1}^{n} 2(\theta -x_i) \\
|
||||
\theta \sum_{i=1}^{n} 2 & = \sum_{i=1}^{n} 2x_i \\
|
||||
\Aboxed{\theta & = \frac{\sum_{i=1}^{n}x_i}{n}}
|
||||
\end{align*}
|
||||
|
||||
\begin{align*}
|
||||
g'(\theta) &= \sum_{i=1}^{n} 2(\theta -x_i)\\
|
||||
0 &= \sum_{i=1}^{n} 2(\theta -x_i)\\
|
||||
\theta \sum_{i=1}^{n} 2 &= \sum_{i=1}^{n} 2x_i\\
|
||||
\Aboxed{\theta &= \frac{\sum_{i=1}^{n}x_i}{n}}
|
||||
\end{align*}
|
||||
|
||||
|
||||
\item [c.] Let $x_1,\dots,x_n$ again be real numbers, and let $w_1,\dots,w_n$ be positive real numbers that we can interpret as representing the importance of each of the $x_i$'s. Consider the weighted quadratic function
|
||||
$$h(\theta) = \sum_{i=1}^{n}w_i(\theta - x_i)^2.$$
|
||||
What value of $\theta$ minimizes $h(\theta)$?\\
|
||||
\underline{Solution}:
|
||||
$$h(\theta) = \sum_{i=1}^{n}w_i(\theta - x_i)^2.$$
|
||||
What value of $\theta$ minimizes $h(\theta)$?\\
|
||||
\underline{Solution}:
|
||||
|
||||
\begin{align*}
|
||||
h'(\theta) &= 2 * \sum_{i=1}^{n} w_i (\theta - x_i)\\
|
||||
0 &= 2 * \sum_{i=1}^{n} w_i (\theta - x_i)\\
|
||||
\frac{0}{2} &= \frac{\cancel{2}*\sum_{i=1}^{n} w_i (\theta - x_i)}{\cancel{2}}\\
|
||||
0 &= \sum_{i=1}^{n} w_i (\theta - x_i)\\
|
||||
0 &= \sum_{i=1}^{n} (w_i \theta - w_i x_i)\\
|
||||
0 &= \sum_{i=1}^{n} w_i \theta - \sum_{i=1}^{n} w_i x_i\\
|
||||
\sum_{i=1}^{n} w_i x_i &= \sum_{i=1}^{n} w_i \theta\\
|
||||
\Aboxed{\frac{\sum_{i=1}^{n} w_i x_i}{\sum_{i=1}^{n} w_i } &= \theta}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
h'(\theta) & = 2 * \sum_{i=1}^{n} w_i (\theta - x_i) \\
|
||||
0 & = 2 * \sum_{i=1}^{n} w_i (\theta - x_i) \\
|
||||
\frac{0}{2} & = \frac{\cancel{2}*\sum_{i=1}^{n} w_i (\theta - x_i)}{\cancel{2}} \\
|
||||
0 & = \sum_{i=1}^{n} w_i (\theta - x_i) \\
|
||||
0 & = \sum_{i=1}^{n} (w_i \theta - w_i x_i) \\
|
||||
0 & = \sum_{i=1}^{n} w_i \theta - \sum_{i=1}^{n} w_i x_i \\
|
||||
\sum_{i=1}^{n} w_i x_i & = \sum_{i=1}^{n} w_i \theta \\
|
||||
\Aboxed{\frac{\sum_{i=1}^{n} w_i x_i}{\sum_{i=1}^{n} w_i } & = \theta}
|
||||
\end{align*}
|
||||
|
||||
\item [d.] What issue could arise in the minimization of $h$ if some of the $w_i$'s are negative?\\
|
||||
\underline{Solution}: If some of the $w_i$'s are negative, it can lead to convoluted results and a possible divergence towards $\infty$
|
||||
\underline{Solution}: If some of the $w_i$'s are negative, it can lead to convoluted results and a possible divergence towards $\infty$
|
||||
\end{itemize}
|
||||
|
||||
\subsection{Probability}
|
||||
|
||||
\begin{itemize}
|
||||
\item [a.] Consider a standard 52-card deck of cards with 13 card values (Ace, King, Queen, Jack, and
|
||||
2-10) in each of the four suits (clubs, diamonds, hearts, spades). If a card is drawn at random, what
|
||||
is the probability that it is a spade or a two?
|
||||
2-10) in each of the four suits (clubs, diamonds, hearts, spades). If a card is drawn at random, what
|
||||
is the probability that it is a spade or a two?
|
||||
|
||||
\underline{Solution}:\\
|
||||
$\rightarrow$ \underline{Let} $P(S)$ be the probability of drawing a spade = $\frac{13}{52} = \frac{1}{4}$\\
|
||||
$\rightarrow$ \underline{Let} $P(T)$ be the probability of drawing a card that is a two = $\frac{4}{52} = \frac{1}{13}$\\
|
||||
|
||||
$$
|
||||
P(S \cap T) = \frac{1}{52}
|
||||
$$
|
||||
\underline{Solution}:\\
|
||||
$\rightarrow$ \underline{Let} $P(S)$ be the probability of drawing a spade = $\frac{13}{52} = \frac{1}{4}$\\
|
||||
$\rightarrow$ \underline{Let} $P(T)$ be the probability of drawing a card that is a two = $\frac{4}{52} = \frac{1}{13}$\\
|
||||
|
||||
$$
|
||||
P(S \cap T) = \frac{1}{52}
|
||||
$$
|
||||
|
||||
The overall probability of drawing a spade or a two is:
|
||||
\begin{align*}
|
||||
P(S \cup T) & = (P(S) + P(T)) - P(S \cap T) \\
|
||||
& = (\frac{1}{4} + \frac{1}{13}) - \frac{1}{52} \\
|
||||
\Aboxed{ & = \frac{4}{13} \approx 30.7\%}
|
||||
\end{align*}
|
||||
|
||||
The overall probability of drawing a spade or a two is:
|
||||
\begin{align*}
|
||||
P(S \cup T) &= (P(S) + P(T)) - P(S \cap T)\\
|
||||
&= (\frac{1}{4} + \frac{1}{13}) - \frac{1}{52}\\
|
||||
\Aboxed{&= \frac{4}{13} \approx 30.7\%}
|
||||
\end{align*}
|
||||
|
||||
|
||||
\item [b.] Two factories — Factory A and Factory B — design batteries to be used in mobile phones.
|
||||
Factory A produces 60\% of all batteries, and Factory B produces the other 40\%. 2\% of Factory A's
|
||||
batteries have defects, and 4\% of Factory B's batteries have defects. What is the probability that a
|
||||
battery is both made by Factory A and defective?
|
||||
Factory A produces 60\% of all batteries, and Factory B produces the other 40\%. 2\% of Factory A's
|
||||
batteries have defects, and 4\% of Factory B's batteries have defects. What is the probability that a
|
||||
battery is both made by Factory A and defective?
|
||||
|
||||
\underline{Solution}:\\
|
||||
$\rightarrow$ \underline{Let} $P(A) = 0.6$ (probability battery is from Factory A)\\
|
||||
$\rightarrow$ \underline{Let} $P(D|A) = 0.02$ (probability battery is defective if it's from Factory A)\\
|
||||
$\rightarrow$ \underline{Let} $P(A)' = 0.4$ (probability battery is not from Factory A)\\
|
||||
$\rightarrow$ \underline{Let} $P(D|A)' = 0.04$ (probability battery is defective if it's not from Factory A)
|
||||
|
||||
$\rightarrow$ \underline{Find} $P(A \cap D)$.
|
||||
\underline{Solution}:\\
|
||||
$\rightarrow$ \underline{Let} $P(A) = 0.6$ (probability battery is from Factory A)\\
|
||||
$\rightarrow$ \underline{Let} $P(D|A) = 0.02$ (probability battery is defective if it's from Factory A)\\
|
||||
$\rightarrow$ \underline{Let} $P(A)' = 0.4$ (probability battery is not from Factory A)\\
|
||||
$\rightarrow$ \underline{Let} $P(D|A)' = 0.04$ (probability battery is defective if it's not from Factory A)
|
||||
|
||||
$$\begin{gathered}$$
|
||||
...\\...\\...\\P(A \cap D) = 0.012 \approx 1.2\%
|
||||
$$\end{gathered}$$
|
||||
$\rightarrow$ \underline{Find} $P(A \cap D)$.
|
||||
|
||||
$$\begin{gathered}$$
|
||||
...\\...\\...\\P(A \cap D) = 0.012 \approx 1.2\%
|
||||
$$\end{gathered}$$
|
||||
|
||||
|
||||
\item [c.] Consider the following (made up) facts about COVID incidence and testing:
|
||||
\begin{itemize}
|
||||
\item In the absence of any special information, the probability that a person has COVID is 1\%.
|
||||
\item If a person has COVID, the probability that a test will correctly read positive is 80\%.
|
||||
\item If a person does not have COVID, the probability that a test will incorrectly produce a false positive is 10\%.
|
||||
\end{itemize}
|
||||
\begin{itemize}
|
||||
\item In the absence of any special information, the probability that a person has COVID is 1\%.
|
||||
\item If a person has COVID, the probability that a test will correctly read positive is 80\%.
|
||||
\item If a person does not have COVID, the probability that a test will incorrectly produce a false positive is 10\%.
|
||||
\end{itemize}
|
||||
|
||||
Suppose you take a COVID test and it reads positive. Given the facts above, what is the probability
|
||||
that you have COVID?
|
||||
Suppose you take a COVID test and it reads positive. Given the facts above, what is the probability
|
||||
that you have COVID?
|
||||
|
||||
\underline{Solution}:\\
|
||||
$\rightarrow$ \underline{Let} $C$ = COVID\\
|
||||
$\rightarrow$ \underline{Let} $T$ = Positive Test\\[0.5cm]
|
||||
$\rightarrow$ \underline{Let} $P(C) = 0.01$ (probability of having COVID)\\
|
||||
$\rightarrow$ \underline{Let} $P(T|C)$ = $0.8$ (probability of a positive test if a person has COVID)\\
|
||||
$\rightarrow$ \underline{Let} $P(T|C)' = 0.1$ (probability of a false positive)\\
|
||||
$\rightarrow$ \underline{Find} $P(C|T)$ (probability of having COVID with a positive test result)
|
||||
\underline{Solution}:\\
|
||||
$\rightarrow$ \underline{Let} $C$ = COVID\\
|
||||
$\rightarrow$ \underline{Let} $T$ = Positive Test\\[0.5cm]
|
||||
$\rightarrow$ \underline{Let} $P(C) = 0.01$ (probability of having COVID)\\
|
||||
$\rightarrow$ \underline{Let} $P(T|C)$ = $0.8$ (probability of a positive test if a person has COVID)\\
|
||||
$\rightarrow$ \underline{Let} $P(T|C)' = 0.1$ (probability of a false positive)\\
|
||||
$\rightarrow$ \underline{Find} $P(C|T)$ (probability of having COVID with a positive test result)
|
||||
|
||||
\begin{align*}
|
||||
P(T) &= P(T|C) * (P(C) + (P(T|C)' * P(C)'))\\
|
||||
P(T) &= 0.8 * (0.01 + (0.1 * 0.99))\\
|
||||
P(T) &= 0.0872
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
P(T) & = P(T|C) * (P(C) + (P(T|C)' * P(C)')) \\
|
||||
P(T) & = 0.8 * (0.01 + (0.1 * 0.99)) \\
|
||||
P(T) & = 0.0872
|
||||
\end{align*}
|
||||
|
||||
\rule{4.4in}{0.5pt}
|
||||
\rule{4.4in}{0.5pt}
|
||||
|
||||
\begin{align*}
|
||||
P(C|T) &= \frac{(0.8*0.01)}{0.0872}\\
|
||||
&= \frac{0.008}{0.0872}\\
|
||||
\Aboxed{&= \frac{10}{109}\approx 9.17\%}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
P(C|T) & = \frac{(0.8*0.01)}{0.0872} \\
|
||||
& = \frac{0.008}{0.0872} \\
|
||||
\Aboxed{ & = \frac{10}{109}\approx 9.17\%}
|
||||
\end{align*}
|
||||
|
||||
\item [d. ] Suppose you repeatedly roll a fair six-sided die until you roll a 1 (and then you stop). Every
|
||||
time you roll a 3, you win a points, and every time you roll a 6, you lose b points. You do not win or
|
||||
lose any points if you roll a 2, 4, or 5. What is the expected number of points (as a function of a and
|
||||
b) you will have when you stop?
|
||||
time you roll a 3, you win a points, and every time you roll a 6, you lose b points. You do not win or
|
||||
lose any points if you roll a 2, 4, or 5. What is the expected number of points (as a function of a and
|
||||
b) you will have when you stop?
|
||||
\end{itemize}
|
||||
|
||||
\subsection{Counting}
|
||||
|
||||
\begin{itemize}
|
||||
\item [a.] $n \times n$ grid, rectangle, Big O?\\
|
||||
\underline{Solution}: $O(n^2)$
|
||||
\underline{Solution}: $O(n^2)$
|
||||
|
||||
\item [b.] Three rectangles, possible ways?\\
|
||||
\underline{Solution}: $n^2 * n^2 * n^2 = n^6$
|
||||
\underline{Solution}: $n^2 * n^2 * n^2 = n^6$
|
||||
\end{itemize}
|
||||
|
||||
\subsection{Programming in Python}
|
||||
@ -198,127 +198,275 @@ See attached .py file
|
||||
|
||||
\subsection{Grid City}
|
||||
|
||||
Modeled as a search problem:
|
||||
\begin{itemize}
|
||||
\item $s_0 = (0,0)$
|
||||
\item Actions$(s) = \lbrace{(+1,0), (-1,0), (0,+1), (0,-1)}\rbrace$
|
||||
\item Succ$(s,a) = s+a$
|
||||
\item Cost$((x,y), a) = 1+$ max$(x,0)$ (it is more expensive as you go further to the right)
|
||||
\item IsEnd$(s) = \begin{cases}
|
||||
\text{True} &\text{if\ } s = (m,n)\\
|
||||
\text{False} &\text{otherwise}
|
||||
\end{cases}$
|
||||
\end{itemize}
|
||||
Modeled as a search problem:
|
||||
\begin{itemize}
|
||||
\item $s_0 = (0,0)$
|
||||
\item Actions$(s) = \lbrace{(+1,0), (-1,0), (0,+1), (0,-1)}\rbrace$
|
||||
\item Succ$(s,a) = s+a$
|
||||
\item Cost$((x,y), a) = 1+$ max$(x,0)$ (it is more expensive as you go further to the right)
|
||||
\item IsEnd$(s) = \begin{cases}
|
||||
\text{True} & \text{if\ } s = (m,n) \\
|
||||
\text{False} & \text{otherwise}
|
||||
\end{cases}$
|
||||
\end{itemize}
|
||||
|
||||
\begin{itemize}
|
||||
\item[a.] Minimum cost of reaching location $(m,n)$ starting from $(0,0)$? Describe possible path achieving the minimum cost. Is it unique?
|
||||
|
||||
\underline{Solution}: Since $(m,n) \geq 0$, we can safely say that $(m,n)$ can't be at $(0,0)$ in the minimum cost path. The next closest location to $(0,0)$ that $(m,n)$
|
||||
could be is $(1,1)$, as it satisfies the condition listed earlier. With this in mind, the minimum cost would be:
|
||||
\begin{align*}
|
||||
\text{Cost}((x,y),a) &= 1 + \text{max}(x,0) \\
|
||||
&= (1 + \text{max}(m,0)) + (1 + \text{max}(n,0)) \\
|
||||
&= 1 + m + 1 + n \\
|
||||
\Aboxed{&= 2 + m + n}
|
||||
\end{align*}
|
||||
|
||||
\item[a.] Minimum cost of reaching location $(m,n)$ starting from $(0,0)$? Describe possible path achieving the minimum cost. Is it unique?
|
||||
|
||||
\underline{Solution}: Since $(m,n) \geq 0$, we can safely say that $(m,n)$ can't be at $(0,0)$ in the minimum cost path. The next closest location to $(0,0)$ that $(m,n)$
|
||||
could be is $(1,1)$, as it satisfies the condition listed earlier. With this in mind, the minimum cost would be:
|
||||
\begin{align*}
|
||||
\text{Cost}((x,y),a) & = 1 + \text{max}(x,0) \\
|
||||
& = (1 + \text{max}(m,0)) + (1 + \text{max}(n,0)) \\
|
||||
& = 1 + m + 1 + n \\
|
||||
\Aboxed{ & = 2 + m + n}
|
||||
\end{align*}
|
||||
|
||||
\item[b.] True or false (and explain): UCS will never terminate on this problem because the number of states is infinite.
|
||||
|
||||
\underline{Solution}: False, because UCS explores nodes with the lowest cost first. If $(m,n)$ is available at the minimum cost from $(0,0)$, then UCS will terminate, and that too fairly quickly.
|
||||
|
||||
\underline{Solution}: False, because UCS explores nodes with the lowest cost first. If $(m,n)$ is available at the minimum cost from $(0,0)$, then UCS will terminate, and that too fairly quickly.
|
||||
|
||||
\item[c.] True or false (and explain): UCS will return the minimum cost path and explore only locations between $(0, 0)$ and $(m, n)$; that is, locations $(x, y)$ such that $0 \leq x \leq m$ and $0 \leq y \leq n$.
|
||||
|
||||
\underline{Solution}: True, if $(m,n)$ is available at the minimum cost from $(0,0)$, because UCS will terminate very quickly and therefore won't explore other paths. This is because UCS prioritizes the lowest cost first.
|
||||
|
||||
\underline{Solution}: True, if $(m,n)$ is available at the minimum cost from $(0,0)$, because UCS will terminate very quickly and therefore won't explore other paths. This is because UCS prioritizes the lowest cost first.
|
||||
|
||||
|
||||
\item[d.] True or false (and explain): UCS will return the minimum cost path and explore only
|
||||
locations whose path costs are strictly less than the minimum cost from $(0, 0)$ to $(m, n)$.
|
||||
locations whose path costs are strictly less than the minimum cost from $(0, 0)$ to $(m, n)$.
|
||||
|
||||
\underline{Solution}: True, because UCS focuses on cumulative costs that are less than the optimal cost to reach $(m,n)$.
|
||||
|
||||
\underline{Solution}: True, because UCS focuses on cumulative costs that are less than the optimal cost to reach $(m,n)$.
|
||||
|
||||
\end{itemize}
|
||||
|
||||
Now consider UCS running on an arbitrary graph.
|
||||
|
||||
\begin{itemize}
|
||||
\item[e.] True or false (and explain): If you add an edge between two nodes, the cost of the min-cost path cannot go up.
|
||||
|
||||
\underline{Solution}: True, because adding an edge can only maintain or reduce the cost of the min-cost path.
|
||||
|
||||
\underline{Solution}: True, because adding an edge can only maintain or reduce the cost of the min-cost path.
|
||||
|
||||
\item[f.] True or false (and explain): If you make the cost of an action from some state small enough (possibly negative), you can guarantee that that action will show up in the minimum cost path.
|
||||
|
||||
\underline{Solution}: False, because that action isn't guaranteed to lead us to the goal, no matter how cheap we make it.
|
||||
|
||||
\underline{Solution}: False, because that action isn't guaranteed to lead us to the goal, no matter how cheap we make it.
|
||||
|
||||
\item[g.] True or false (and explain): If you increase the cost of every action by 1, the minimum cost path does not change (even though its cost does).
|
||||
|
||||
\underline{Solution}: True. Since we are raising the cost of \underline{all} costs equally, the minimum cost path will remain the the same even despite the costs going up.
|
||||
|
||||
\subsection{Six Degrees}
|
||||
See degrees.py
|
||||
\underline{Solution}: True. Since we are raising the cost of \underline{all} costs equally, the minimum cost path will remain the the same even despite the costs going up.
|
||||
|
||||
\subsection{Tic Tac Toe}
|
||||
See tictactoe.py
|
||||
\subsection{Six Degrees}
|
||||
See degrees.py
|
||||
|
||||
\subsection{Tic Tac Toe}
|
||||
See tictactoe.py
|
||||
|
||||
\end{itemize}
|
||||
|
||||
\section{Homework 2}
|
||||
|
||||
\subsection{Cake}
|
||||
\subsection{Cake}
|
||||
|
||||
Alice and Bob are sharing a cake....
|
||||
Alice and Bob are sharing a cake....
|
||||
|
||||
\begin{center}
|
||||
\includegraphics[scale=0.3]{hw2/gameTree.png}
|
||||
\end{center}
|
||||
|
||||
|
||||
Write down the utility Alice should expect to receive in each of the game states given the following knowlege of player strategies:
|
||||
|
||||
\begin{itemize}
|
||||
\item [a.] Alice and Bob playing adversarially, each trying to maximize the cake they receive.
|
||||
\begin{enumerate}
|
||||
\item 1/2
|
||||
\item 1/2
|
||||
\item 1/2
|
||||
\end{enumerate}
|
||||
\item [b. ] Alice still trying to maximize, Bob now playing collaboratively and helping Alice.
|
||||
\begin{enumerate}
|
||||
\item 2/3
|
||||
\item 1/2
|
||||
\item 2/3
|
||||
\end{enumerate}
|
||||
\item [c. ] Random play.
|
||||
\begin{enumerate}
|
||||
\item Node 2 or 3
|
||||
\item Doesn't matter because of constant utility
|
||||
\item Node 3
|
||||
\end{enumerate}
|
||||
\item [d. ] Alpha Beta pruning
|
||||
\begin{enumerate}
|
||||
\item Leaf notes with utility 1/3, 1/3, 1/3, 1/3 on left, and right leaf node with utility 1/6
|
||||
\end{enumerate}
|
||||
\item [e. ] Alpha Beta pruning with tree elements swapped.
|
||||
\begin{enumerate}
|
||||
\item No nodes are pruned
|
||||
\end{enumerate}
|
||||
\end{itemize}
|
||||
|
||||
\begin{itemize}
|
||||
\item [2. ] Logical Formulas
|
||||
\begin{itemize}
|
||||
\item [a. ] $(P \land \neg Q) \lor (\neg P \land Q)$
|
||||
\item [b. ] $(\exists x \text{Course}(x) \land \text{Enrolled}(A, x) \land \text{Enrolled}(B,x))$
|
||||
\item [c. ] $(\forall x \text{Course}(x) \land \text{Enrolled}(A, x) \rightarrow \text{Enrolled}(B,x))$
|
||||
\item [d. ] $\forall x \forall y (\text{Enrolled}(x,y) \land \text{Course}(y) \rightarrow \text{Student}(x))$
|
||||
\item [e. ] $\forall x (\text{Course}(x) \rightarrow \exists y (\text{Student}(y) \land \text{Enrolled}(y,x)))$
|
||||
\end{itemize}
|
||||
\end{itemize}
|
||||
|
||||
\subsection{Knights and Knaves}
|
||||
See attached puzzle.py file
|
||||
|
||||
\subsection{Odds and Evens}
|
||||
See attached submission.py file
|
||||
|
||||
\section{Homework 3}
|
||||
\subsection{Probability}
|
||||
|
||||
\begin{itemize}
|
||||
\item[a.] An urn contains w white balls and b black balls...
|
||||
\begin{enumerate}
|
||||
\item What is the probability that the first ball is white?
|
||||
$$
|
||||
P(\text{first ball is white}) = \boxed{\frac{w}{w + b}}
|
||||
$$
|
||||
|
||||
\item What is the probability that the second ball is white?
|
||||
$$
|
||||
\begin{aligned}
|
||||
\text{If first ball was white}: \\
|
||||
P(\text{second ball white | first ball white}) = \Aboxed{\frac{w-1}{w + b - 1}} \\
|
||||
\text{If first ball was black}: \\
|
||||
P(\text{second ball white | first ball black}) = \Aboxed{\frac{w}{w + b + 1}} \\
|
||||
\end{aligned}
|
||||
$$
|
||||
\end{enumerate}
|
||||
\item[b.] Alice has two brothers, Bob and Charlie...
|
||||
\begin{enumerate}
|
||||
\item What is the probability that Alice is older than Charlie? \\[0.1in]
|
||||
There are $3! = 6$ possible scenarios, and Alice is older in 3 of them. So, $\boxed{\frac{3}{6} \ \text{or} \ 0.5}$.
|
||||
\item Alice tells you that she is older than Bob \\[0.1in]
|
||||
In this case, the probability of Alice being older than Charlie is $\boxed{\frac{2}{3} \ \text{or} \ 0.66}$.
|
||||
\end{enumerate}
|
||||
\item[c.] The inhabitants of a small island... \\[0.1in]
|
||||
Let:
|
||||
\begin{itemize}
|
||||
\item T = first person tells truth
|
||||
\item L = first person lies
|
||||
\item Y = second person says yes
|
||||
\end{itemize}
|
||||
|
||||
We know that:
|
||||
\begin{itemize}
|
||||
\item P(T) = 1/3
|
||||
\item P(L) = 2/3
|
||||
\item P(Y | T) = 1/3
|
||||
\item P(Y | L) = 2/3
|
||||
\end{itemize}
|
||||
|
||||
We find: $P(T | Y) = \frac{P(Y | T) \cdot P(T)}{P(Y)}$
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
P(Y) & = P(Y | T) \cdot P(T) + P(Y | L) \cdot P(L) \\
|
||||
& = \frac{1}{3} \cdot \frac{1}{3} + \frac{2}{3} \cdot \frac{2}{3} \\
|
||||
& = \frac{1}{9} + \frac{4}{5} \\
|
||||
& = \frac{5}{9}
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
P(T | Y) & = \frac{\frac{1}{3} \cdot \frac{1}{3}}{\frac{5}{9}} \\
|
||||
& = \Aboxed{\frac{1}{5} \ \text{or} \ 0.2}
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
\item[d.] Bag contains one ball... \\[0.1in]
|
||||
Let:
|
||||
\begin{itemize}
|
||||
\item W = original ball was white
|
||||
\item B = original ball was black
|
||||
\item D = white ball is drawn
|
||||
\end{itemize}
|
||||
|
||||
We know:
|
||||
\begin{itemize}
|
||||
\item P(W) = 1/2
|
||||
\item P(B) = 1/2
|
||||
\item P(D | W) = 1
|
||||
\item P(D | B) = 1/2
|
||||
\end{itemize}
|
||||
|
||||
We find: $P(W | D) = \frac{P(D | W) \cdot P(W)}{P(D)}$
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
P(D) & = P(D | W) \cdot P(W) + P(D | B) \cdot P(B) \\
|
||||
& = 1 \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} \\
|
||||
& = \frac{1}{2} + \frac{1}{4} \\
|
||||
& = \frac{3}{4}
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
P(W | D) & = \frac{1 \cdot \frac{1}{2}}{\frac{3}{4}} \\
|
||||
\Aboxed{ & = \frac{2}{3}}
|
||||
\end{aligned}
|
||||
$$
|
||||
\end{itemize}
|
||||
|
||||
\subsection{Bayesian Networks}
|
||||
\begin{itemize}
|
||||
\item[a.] Given the following Bayesian network, write down a factorization of the joint probability distribution P(A,B, C,D, E) as a product of local conditional distributions, one for each random variable.
|
||||
\begin{center}
|
||||
\includegraphics[scale=0.3]{hw2/gameTree.png}
|
||||
\includegraphics[scale=0.3]{hw3/bayesian_network_1_hw3.png}
|
||||
\end{center}
|
||||
|
||||
|
||||
Write down the utility Alice should expect to receive in each of the game states given the following knowlege of player strategies:
|
||||
$$
|
||||
P(A, B, C, D, E) = \ \boxed{P(A) \cdot P(B) \cdot P(C | A, B) \cdot P(D | B) \cdot P(E | C)}
|
||||
$$
|
||||
|
||||
\begin{itemize}
|
||||
\item [a.] Alice and Bob playing adversarially, each trying to maximize the cake they receive.
|
||||
\begin{enumerate}
|
||||
\item 1/2
|
||||
\item 1/2
|
||||
\item 1/2
|
||||
\end{enumerate}
|
||||
\item [b. ] Alice still trying to maximize, Bob now playing collaboratively and helping Alice.
|
||||
\begin{enumerate}
|
||||
\item 2/3
|
||||
\item 1/2
|
||||
\item 2/3
|
||||
\end{enumerate}
|
||||
\item [c. ] Random play.
|
||||
\begin{enumerate}
|
||||
\item Node 2 or 3
|
||||
\item Doesn't matter because of constant utility
|
||||
\item Node 3
|
||||
\end{enumerate}
|
||||
\item [d. ] Alpha Beta pruning
|
||||
\begin{enumerate}
|
||||
\item Leaf notes with utility 1/3, 1/3, 1/3, 1/3 on left, and right leaf node with utility 1/6
|
||||
\end{enumerate}
|
||||
\item [e. ] Alpha Beta pruning with tree elements swapped.
|
||||
\begin{enumerate}
|
||||
\item No nodes are pruned
|
||||
\end{enumerate}
|
||||
\end{itemize}
|
||||
|
||||
\begin{itemize}
|
||||
\item [2. ] Logical Formulas
|
||||
\item[b.] For the same Bayesian network above, list all the random variables that are independent of A, assuming that none of the variables have been observed.
|
||||
\begin{itemize}
|
||||
\item [a. ] $(P \land \neg Q) \lor (\neg P \land Q)$
|
||||
\item [b. ] $(\exists x \text{Course}(x) \land \text{Enrolled}(A, x) \land \text{Enrolled}(B,x))$
|
||||
\item [c. ] $(\forall x \text{Course}(x) \land \text{Enrolled}(A, x) \rightarrow \text{Enrolled}(B,x))$
|
||||
\item [d. ] $\forall x \forall y (\text{Enrolled}(x,y) \land \text{Course}(y) \rightarrow \text{Student}(x))$
|
||||
\item [e. ] $\forall x (\text{Course}(x) \rightarrow \exists y (\text{Student}(y) \land \text{Enrolled}(y,x)))$
|
||||
\item B
|
||||
\end{itemize}
|
||||
\end{itemize}
|
||||
|
||||
\subsection{Knights and Knaves}
|
||||
See attached puzzle.py file
|
||||
\item[c.] For the same Bayesian network above, assume that E is now observed. List all pairs of random variables (not including E) that are conditionally independent given E.
|
||||
\begin{itemize}
|
||||
\item (A, B)
|
||||
\item (A, D)
|
||||
\item (B, D)
|
||||
\end{itemize}
|
||||
|
||||
\subsection{Odds and Evens}
|
||||
See attached submission.py file
|
||||
|
||||
\item[d.] Consider the following model for traffic jams in a small town, which can be caused by either a car accident or by a visit from the president (and the accompanying security motorcade).
|
||||
\begin{center}
|
||||
\includegraphics[scale=0.3]{hw3/motorcade.png}
|
||||
\end{center}
|
||||
|
||||
Compute P(Accident = 1 | Traffic = 1) and P(Accident = 1 | Traffic = 1, President = 1). \\
|
||||
|
||||
Let:
|
||||
\begin{itemize}
|
||||
\item A = event of accident
|
||||
\item P = event of president's visit
|
||||
\item T = event of traffic jam
|
||||
\end{itemize}
|
||||
|
||||
We're given:
|
||||
\begin{itemize}
|
||||
\item P(A) = prior probability of an accident
|
||||
\item P(P) = prior probability of the president's visit
|
||||
\item P(T | A) = probability of a traffic jam given an accident
|
||||
\item P(T | P) = probability of a traffic jam given the president's visit
|
||||
\end{itemize}
|
||||
|
||||
$$
|
||||
P(\text{Accident} = 1 | \text{Traffic} = 1) \rightarrow P(A | T) = \frac{P(T | A) \cdot P(A)}{P(T)}
|
||||
$$
|
||||
|
||||
|
||||
\end{itemize}
|
||||
|
||||
\end{document}
|
||||
Loading…
Reference in New Issue
Block a user