Quiz 4 Solutions
Solutions: You can find the file with solutions for all questions here.
Quiz submissions were graded automatically for correctness. Implementations did not need to be efficient, as long as they were correct.
In addition to the doctests provided to students, we also used extra doctests to check for corner cases. These extra test cases are highlighted below.
To complete this homework assignment, you will need to use SQLite version 3.8.3 or greater. See Lab 12 for setup and usage instructions.
To check your progress, you can run
sqlite3 directly by running:
sqlite3 -init quiz04.sql
You should also check your work using
In each question below, you will define a new table based on the following
tables. The first defines the names, opening, and closing hours of great pizza
places in Berkeley. The second defines typical meal times (for college
students). A pizza place is open for a meal if the meal time is at or within
create table pizzas as select "Pizzahhh" as name, 12 as open, 15 as close union select "La Val's" , 11 , 22 union select "Sliver" , 11 , 20 union select "Cheeseboard" , 16 , 23 union select "Emilia's" , 13 , 18; create table meals as select "breakfast" as meal, 11 as time union select "lunch" , 13 union select "dinner" , 19 union select "snack" , 22;
Your tables should still perform correctly even if the values in these tables were to change. Don't just hard-code the output to each query.
Question 1If two meals are more than 6 hours apart, then there's nothing wrong with going to the same pizza place for both, right? Create a
doubletable with three columns. The first columns is the earlier meal, the second is the later meal, and the third is the name of a pizza place. Only include rows that describe two meals that are more than 6 hours apart and a pizza place that is open for both of the meals. The rows may appear in any order.
-- Two meals at the same place create table double as select a.meal, b.meal, name from meals as a, meals as b, pizzas where open <= a.time and a.time <= close and open <= b.time and b.time <= close and b.time > a.time + 6; -- Example: select * from double where name="Sliver"; -- Expected output: -- breakfast|dinner|Sliver
Test your solution with OK:
python3 ok -q double
Question 2For each meal, list all the pizza options. Create a table
optionsthat has one row for every meal and three columns. The first column is the meal, the second is the total number of pizza places open for that meal, and the last column is a comma-separated list of open pizza places in alphabetical order. Assume that there is at least one pizza place open for every meal. Order the resulting rows by meal time.
-- Pizza options for every meal create table options as with lists(meal, time, names, last, n) as ( select meal, time, name, name, 1 from pizzas, meals where open <= time and time <= close union select meal, time, names || ", " || name, name, n+1 from lists, pizzas where open <= time and time <= close and name > last ) select meal, max(n), names from lists group by meal order by time; -- Example: select * from options where meal="dinner"; -- Expected output: -- dinner|3|Cheeseboard, La Val's, Sliver
Hint: Define a recursive table in a
with statement that includes all
partial lists of options, then use the
max aggregate function to pick the
full list for each meal.
Test your solution with OK:
python3 ok -q options