Homework 10 Solutions

• hw10.zip

Solution Files

You can find the solutions in the hw10.sql file.

Usage

First, check that a file named sqlite_shell.py exists alongside the assignment files. If you don't see it, or if you encounter problems with it, scroll down to the Troubleshooting section to see how to download an official precompiled SQLite binary before proceeding.

You can start an interactive SQLite session in your Terminal or Git Bash with the following command:

python3 sqlite_shell.py

While the interpreter is running, you can type .help to see some of the commands you can run.

To exit out of the SQLite interpreter, type .exit or .quit or press Ctrl-C. Remember that if you see ...> after pressing enter, you probably forgot a ;.

You can also run all the statements in a .sql file by doing the following: (Here we're using the lab13.sql file as an example.)

1. Runs your code and then exits SQLite immediately afterwards.

   python3 sqlite_shell.py < lab13.sql

2. Runs your code and then opens an interactive SQLite session, which is similar to running Python code with the interactive -i flag.

   python3 sqlite_shell.py --init lab13.sql

To check your progress, you can run sqlite3 directly by running:

python3 sqlite_shell.py --init hw10.sql

You should also check your work using ok:

python3 ok

Questions

Getting Started Videos

Dog Data

In each question below, you will define a new table based on the following tables.

CREATE TABLE parents AS
  SELECT "abraham" AS parent, "barack" AS child UNION
  SELECT "abraham"          , "clinton"         UNION
  SELECT "delano"           , "herbert"         UNION
  SELECT "fillmore"         , "abraham"         UNION
  SELECT "fillmore"         , "delano"          UNION
  SELECT "fillmore"         , "grover"          UNION
  SELECT "eisenhower"       , "fillmore";

CREATE TABLE dogs AS
  SELECT "abraham" AS name, "long" AS fur, 26 AS height UNION
  SELECT "barack"         , "short"      , 52           UNION
  SELECT "clinton"        , "long"       , 47           UNION
  SELECT "delano"         , "long"       , 46           UNION
  SELECT "eisenhower"     , "short"      , 35           UNION
  SELECT "fillmore"       , "curly"      , 32           UNION
  SELECT "grover"         , "short"      , 28           UNION
  SELECT "herbert"        , "curly"      , 31;

CREATE TABLE sizes AS
  SELECT "toy" AS size, 24 AS min, 28 AS max UNION
  SELECT "mini"       , 28       , 35        UNION
  SELECT "medium"     , 35       , 45        UNION
  SELECT "standard"   , 45       , 60;

Your tables should still perform correctly even if the values in these tables change. For example, if you are asked to list all dogs with a name that starts with h, you should write:

SELECT name FROM dogs WHERE "h" <= name AND name < "i";

Instead of assuming that the dogs table has only the data above and writing

SELECT "herbert";

The former query would still be correct if the name grover were changed to hoover or a row was added with the name harry.

Q1: By Parent Height

Create a table by_parent_height that has a column of the names of all dogs that have a parent, ordered by the height of the parent dog from tallest parent to shortest parent.

-- All dogs with parents ordered by decreasing height of their parent
CREATE TABLE by_parent_height AS
  SELECT child FROM parents, dogs WHERE name = parent ORDER BY height desc;

For example, fillmore has a parent eisenhower with height 35, and so should appear before grover who has a parent fillmore with height 32. The names of dogs with parents of the same height should appear together in any order. For example, barack and clinton should both appear at the end, but either one can come before the other.

sqlite> select * from by_parent_height;
herbert
fillmore
abraham
delano
grover
barack
clinton

Use Ok to test your code:

python3 ok -q by_parent_height
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Hint Video

Q2: Size of Dogs

The Fédération Cynologique Internationale classifies a standard poodle as over 45 cm and up to 60 cm. The sizes table describes this and other such classifications, where a dog must be over the min and less than or equal to the max in height to qualify as a size.

Create a size_of_dogs table with two columns, one for each dog's name and another for its size.

-- The size of each dog
CREATE TABLE size_of_dogs AS
  SELECT name, size FROM dogs, sizes
    WHERE height > min AND height <= max;

The output should look like the following:

sqlite> select * from size_of_dogs;
abraham|toy
barack|standard
clinton|standard
delano|standard
eisenhower|mini
fillmore|mini
grover|toy
herbert|mini

Use Ok to test your code:

python3 ok -q size_of_dogs
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Hint Video

Q3: Sentences

There are two pairs of siblings that have the same size. Create a table that contains a row with a string for each of these pairs. Each string should be a sentence describing the siblings by their size.

-- Filling out this helper table is optional
CREATE TABLE siblings AS
  SELECT a.child AS first, b.child AS second FROM parents AS a, parents AS b
    WHERE a.parent = b.parent AND a.child < b.child;
-- Sentences about siblings that are the same size
CREATE TABLE sentences AS
  SELECT "The two siblings, " || first || " plus " || second || " have the same size: " || a.size
    FROM siblings, size_of_dogs AS a, size_of_dogs AS b
    WHERE a.size = b.size AND a.name = first AND b.name = second;

Each sibling pair should appear only once in the output, and siblings should be listed in alphabetical order (e.g. "barack plus clinton..." instead of "clinton plus barack..."), as follows:

sqlite> select * from sentences;
The two siblings, barack plus clinton have the same size: standard
The two siblings, abraham plus grover have the same size: toy

Hint: First, create a helper table containing each pair of siblings. This will make comparing the sizes of siblings when constructing the main table easier.

Hint: If you join a table with itself, use AS within the FROM clause to give each table an alias.

Hint: In order to concatenate two strings into one, use the || operator.

Hint Video

Use Ok to test your code:

python3 ok -q sentences
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Q4: Low Variance

We want to create a table which contains the height range (difference between maximum and minimum height) of all dogs that share a fur type. However, we'll only consider fur types where each dog with that fur type is within 30% of the average height of all dogs with that fur type.

For example, if the average height for short-haired dogs is 10, then in order to be included in our output, all dogs with short hair must have a height of at most 13 and at least 7.

To achieve this, we can use MIN, MAX, and AVG. For this problem, we'll want to find the average height and make sure that:

• There are no heights smaller than 0.7 of the average.
• There are no heights greater than 1.3 of the average.

Your output should first include the fur type and then the height range for the fur types that meet this criteria.

-- Height range for each fur type where all of the heights differ by no more than 30% from the average height
CREATE TABLE low_variance AS
  SELECT fur, MAX(height) - MIN(height) FROM dogs GROUP BY fur
      HAVING MIN(height) > .7 * AVG(height) AND MAX(height) < 1.3 * AVG(height);
-- Example:
SELECT * FROM low_variance;
-- Expected output:
-- curly|1

Explanation: The average height of long-haired dogs is 39.7, so the low variance criterion requires the height of each long-haired dog to be between 27.8 and 51.6. However, abraham is a long-haired dog with height 26, which is outside this range. For short-haired dogs, barack falls outside the valid range (check!). Thus, neither short nor long haired dogs are included in the output. There are two curly haired dogs: fillmore with height 32 and herbert with height 31. This gives a height range of 1.

Use Ok to test your code:

python3 ok -q low_variance
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Submit

Make sure to submit this assignment by uploading any files you've edited to the appropriate Gradescope assignment. For a refresher on how to do this, refer to Lab 00.

Exam Practice

The following are some SQL exam problems from previous semesters that you may find useful as additional exam practice.

1. Fall 2019 Final, Question 10: Big Game
2. Summer 2019 Final, Question 8: The Big SQL
3. Fall 2018 Final, Question 7: SQL of Course