Homework 10 Solutions
You can find the solutions in the hw10.sql file.
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:
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
.quit or press
Ctrl-C. Remember that if you see
...> after pressing enter, you probably
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.)
Runs your code and then exits SQLite immediately afterwards.
python3 sqlite_shell.py < lab13.sql
Runs your code and then opens an interactive SQLite session, which is similar to running Python code with the interactive
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
Getting Started Videos
These videos may provide some helpful direction for tackling the coding problems on this assignment.
To see these videos, you should be logged into your berkeley.edu email.
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
The former query would still be correct if the name
grover were changed to
hoover or a row was added with the name
Q1: By Parent Height
Create a table
by_parent_height that has a column of the names of all dogs that have
parent, ordered by the height of the parent dog from tallest parent to shortest
-- All dogs with parents ordered by decreasing height of their parent CREATE TABLE by_parent_height ASSELECT child FROM parents, dogs WHERE name = parent ORDER BY height desc;
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,
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
We need information from both the
parents and the
dogs table. This time, the
only rows that make sense are the ones where a child is matched up with their
parent. Finally, we order the result by descending height.
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
height to qualify as a
size_of_dogs table with two columns, one for each dog's
another for its
-- The size of each dog CREATE TABLE size_of_dogs ASSELECT 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
We know that at a minimum, we need information from both the
table. Finally, we filter and keep only the rows that make sense: a size that
corresponds to the size of the dog we're currently considering.
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 ASSELECT 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 ASSELECT "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
FROMclause to give each table an alias.
Hint: In order to concatenate two strings into one, use the
Use Ok to test your code:
python3 ok -q sentences
Roughly speaking, there are two tasks we need to solve here:
Figure out which dogs are siblings
A sibling is someone you share a parent with. This will probably involve the
It might be tempting to join this with
dogs, but there isn't any extra
information provided by a dogs table that we need at this time. Furthermore, we
still need information on sibling for a given dog, since the
just associates each dog to a parent.
The next step, therefore, is to match all children to all other children by joining the parents table to itself. The only rows here that make sense are the rows that represent sibling relationships since they share the same parent.
Remember that we want to avoid duplicates! If dog A and B are siblings, we don't want both A/B and B/A to appear in the final result. We also definitely don't want A/A to be a sibling pair. Enforcing ordering on the sibling names ensures that we don't have either issue.
Construct sentences based on sibling information
After determining the siblings, constructing the sentences just requires us to
get the size of each sibling. We could join on the
sizes tables as
we did in an earlier problem, but there's no need to redo that work. Instead,
we'll reuse our
size_of_dogs table to figure out the size of each sibling in
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
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 ASSELECT 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
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.
The following are some SQL exam problems from previous semesters that you may find useful as additional exam practice.