Homework 10: SQL
Due by 11:59pm on Thursday, December 5
Instructions
Download hw10.zip.
Submission: When you are done, submit the assignment by uploading all code files you've edited to Gradescope. You may submit more than once before the deadline; only the final submission will be scored. Check that you have successfully submitted your code on Gradescope. See Lab 0 for more instructions on submitting assignments.
Using Ok: If you have any questions about using Ok, please refer to this guide.
Grading: Homework is graded based on
correctness. Each incorrect problem will decrease the total score by one point.
This homework is out of 2 points.
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
Visualizing SQL
If you would like some support with visualizing SQL, please navigate to code.cs61a.org and select Start SQL Interpreter.
From there, you can either practice on tables you make yourself, existing tables from the CS 61A database (which can be found by typing .tables
), or on the tables from the assignment you are working on by copying and pasting the entire CREATE TABLE ...
command.
Required Questions
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.
SQL
Dog Data
In each question below, you will define a new table based on the following tables.
CREATE TABLE parents AS
SELECT "ace" AS parent, "bella" AS child UNION
SELECT "ace" , "charlie" UNION
SELECT "daisy" , "hank" UNION
SELECT "finn" , "ace" UNION
SELECT "finn" , "daisy" UNION
SELECT "finn" , "ginger" UNION
SELECT "ellie" , "finn";
CREATE TABLE dogs AS
SELECT "ace" AS name, "long" AS fur, 26 AS height UNION
SELECT "bella" , "short" , 52 UNION
SELECT "charlie" , "long" , 47 UNION
SELECT "daisy" , "long" , 46 UNION
SELECT "ellie" , "short" , 35 UNION
SELECT "finn" , "curly" , 32 UNION
SELECT "ginger" , "short" , 28 UNION
SELECT "hank" , "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";
In other words, you should not assume that the dogs
table has only the data in the table above by writing:
SELECT "hank";
The former query would still be correct if the name ginger
were changed to
gigi
or a row was added with the name harry
. Contrastingly, writing SELECT "hank";
would not.
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 "REPLACE THIS LINE WITH YOUR SOLUTION";
For example, finn
has a parent ellie
with height 35, and so
should appear before ginger
who has a parent finn
with height 32.
The names of dogs with parents of the same height should appear together in any
order. For example, bella
and charlie
should both appear at the end, but
either one can come before the other.
The by_parent_height
table should look like this:
+----------+
| chil |
+----------+
| hank |
| finn |
| ace |
| daisy |
| ginger |
| bella |
| charlie |
+----------+
Use Ok to test your code:
python3 ok -q by_parent_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
the max
in height
to qualify as 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 "REPLACE THIS LINE WITH YOUR SOLUTION";
The size_of_dogs
table should look like this:
+----------+----------+
| name | size |
+----------+----------+
| ace | toy |
| bella | standard |
| charlie | standard |
| daisy | standard |
| ellie | mini |
| finn | mini |
| ginger | toy |
| hank | mini |
+----------+----------+
Use Ok to test your code:
python3 ok -q size_of_dogs
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.
-- [Optional] Filling out this helper table is recommended
CREATE TABLE siblings AS
SELECT "REPLACE THIS LINE WITH YOUR SOLUTION";
-- Sentences about siblings that are the same size
CREATE TABLE sentences AS
SELECT "REPLACE THIS LINE WITH YOUR SOLUTION";
Each sibling pair should appear only once in the output, and siblings should be
listed in alphabetical order (e.g. "bella and charlie..."
instead of
"charlie and bella..."
), as follows:
sqlite> SELECT * FROM sentences;
The two siblings, bella and charlie, have the same size: standard
The two siblings, ace and ginger, 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 theFROM
clause to give each table an alias.Hint: In order to concatenate two strings into one, use the
||
operator, e.g.SELECT "hello" || "world";
will returnhelloworld
.
Use Ok to test your code:
python3 ok -q sentences
Q4: Low Variance
We want to create a table that contains the height range (defined as the 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; we call this the low variance criterion.
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 (inclusive).
Hint:
MIN
,MAX
, andAVG
will be useful here. Hint: You may want to first find the average height and make sure that:* There are no heights smaller than 0.7 (i.e. 70%) of the average. * There are no heights greater than 1.3 (i.e. 130%) of the average.
-- 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 "REPLACE THIS LINE WITH YOUR SOLUTION";
Your output should have two columns, in this order: the fur
type and the height_range
for the fur
types that meet this criteria. It should look like this:
+------------+--------------+
| fur | height_range |
+------------+--------------+
| Curly | 1 |
+------------+--------------+
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, ace
is a long-haired dog with height 26, which is outside this range. For short-haired dogs, bella
falls outside the valid range (check!). Thus, neither short nor long haired dogs are included in the output. There are two curly haired dogs: finn
with height 32 and hank
with height 31. This gives a height range of 1.
Use Ok to test your code:
python3 ok -q low_variance
Submit Assignment
Submit this assignment by uploading any files you've edited to the appropriate Gradescope assignment. Lab 00 has detailed instructions.
Make sure to submit hw10.sql
to the autograder!
Exam Practice
The following are some SQL exam problems from previous semesters that you may find useful as additional exam practice.