Assignment 3 COMP 250 solution

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2 Building Management
2.1 Introduction
You are having coffee with one of your friends named Paul. Paul recently
graduated and is now working for a Town Housing Office. He seems to be
enjoying his work : the buildings that they are managing are quite old now,
so they need to be renovated. He tells you more about this and then asks
what you are up to. So you tell him about that amazing class (COMP250)
that you are following. You tell him about binary search trees.
Paul looks puzzled and interrupts you to ask : “That’s cute, but what’s
the point of this ?”. You answer : “you are using software and database
in your everyday work life, I am learning how to write them”. But Paul is
looking as if he does not believe you, so you add “let me show you. I will
show you how the database you use at the office works”.
“Ok, so you talked a great deal about how old your buildings are, this
is how I am going to organize the database. What else is important on
your buildings ?”. Paul gets overly enthusiastic, so you restrict the list of
important features to just a few : “Paul, I am not going to include the
guy who designed it or what techniques were used to build the building. I
only want to give you an idea of how the software you are using everyday is
working !”.
2.2 Organization of the code you are provided with :
The database of all buildings is going to be represented by a tertiary tree. We
have separated the data associated to one building and the tree structure.
The class OneBuilding represents the data associated to a specific building. A OneBuilding has several fields : first its name (it can represent the
address or the name for instance). A OneBuilding also has a year of construction (yearOfConstruction), a height (height), a year when it is likely
to need to be repaired (yearForRepair) and the projected cost of those
repairs (costForRepair).
A Building bears the tree structure. A Building has a field data bearing the information. A Building has also three fields of type Building
: older, same and younger. These are going to allow you to construct a
tertiary tree representing all the buildings. As the names suggest, given a
Building b, the Building b.older and all its children were built before
Building b was built, the Building b.younger and all its children were
built after Building b was built and the Building b.same and all its children were built the same year as Building b was built. We also ask that
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the buildings of same year of construction are sorted by decreasing height.
You are also provided with a tester. The tester is VERY partial. Even
if your code seems to work, it does not mean that your code is correct : it
only means that your code is working ON AN EXAMPLE. You NEED
to expand the tester.
You do not need to import any other code (and if you do, you will get 0).
All the material needed for this assignment was covered during class.
2.3 Building your database
First thing to do is to add buildings in your tree structure.
Question 1. (14 points) First, code addBuilding. The argument is the
OneBuilding that you want to add to the tree structure.
Given a Building b and a OneBuilding obToAdd, b.addBuilding(obToAdd)
adds the OneBuilding obToAdd in the tree with root b accordingly to the
rules explained before : if obToAdd is older (resp. younger) than b.data,
it has to be added (in the appropriate place) to the subtree b.older (resp.
b.younger). if obToAdd is the same age as b.data, then it depends on the
height : if obToAdd is strictly higher than b.data, it becomes the new root
and you have to adapt the other fields accordingly.
To clarify, let us give some examples. At the nodes of the tree, you will see
(for instance) b1(1987, 50). This means Building b1 has data.yearOfConstruction
1987 and data.height 50. When the fields older, same or younger are not
null, we represent them with an arrow to the corresponding Building. Let
us start with the following tree :
b1(1987, 50)
b1.older
v
b1.younger
(
b6(1985, 60)
b6.same

b2(1988, 60)
b2.younger
(
b4(1985, 50)
b4.same

b8(1995, 55)
b5(1985, 45)
Now, if we have a One Building ob3 with yearOfConstruction 1987, height
55, the command b1.addBuilding (ob3) should return a Building where
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the root b3 has data ob3 and the fields should have been updated to :
b3(1987, 55)
b3.older
v
b3.same

b3.younger
(
b6(1985, 60)
b6.same

b1(1987, 50) b2(1988, 60)
b2.younger
(
b4(1985, 50)
b4.same

b8(1995, 55)
b5(1985, 45)
If OneBuilding ob7 has yearOfConstruction 1986 and height 55, then
the command b3.addBuilding(ob7) should return b3 and the tree structure
should be updated to :
b3(1987, 55)
b3.older
s
b3.same

b3.younger
(
b6(1985, 60)
b6.same

b6.younger
(
b1(1987, 50) b2(1988, 60)
b2.younger
(
b4(1985, 50)
b4.same

b7(1986, 55) b8(1995, 55)
b5(1985, 45)
with b7.data = ob7.
Question 2. (12 points) You can now code addBuildings. The argument
is a Building. Given a Building b and a Building bToAdd, the command
b.addBuildings(bToAdd) first adds the OneBuilding bToAdd.data to b,
then goes on by adding the Building bToAdd.older, then bToAdd.Same and
finally bToAdd.younger and returns the root of the tree structure as in the
case for addBuilding.
For instance, if we have the two following Buildings :
b3(1987, 55)
b3.older
w
b3.same

b3.younger

b4(1985, 50)
b4.same

b1(1987, 50) b2(1988, 60)
b5(1985, 45)
b6(1985, 60)
b6.younger

b7(1986, 55)
b7.younger

b8(1995, 55)
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The command b1.addBuildings(b6) will return b1 and update the tree
structure to :
b3(1987, 55)
b3.older
s
b3.same

b3.younger
(
b6(1985, 60)
b6.same

b6.younger
(
b1(1987, 50) b2(1988, 60)
b2.younger
(
b4(1985, 50)
b4.same

b7(1986, 55) b8(1995, 55)
b5(1985, 45)
Question 3. (24 points) Now you can code removeBuilding. Its argument is a OneBuilding. The command b.removeBuilding (ob) should
return either b if no children of b has data corresponding to ob, or the root
of the tree corresponding to b where the OneBuilding ob has been removed
otherwise.
You should not remove any of the children of the Building with data
corresponding to ob. For the rest of the explanation, let us assume that
b.data == ob (you can easily adapt what is going to be said to the case
where the Building to remove is not the root). In this case, there are three
cases to consider. First, if b.same != null, then the new root becomes
b.same and the subtrees of b should be correctly attached to the new root.
Second, if b.same == null and b.older != null, then the new root becomes b.older and you need to add the OneBuildings in b.younger to this
new root. Finally, if b.same == null and b.older == null, then the new
root is b.younger (note that this also covers the case when b is a leaf).
hint : you just coded addBuildings
On an example, consider
b3(1987, 55)
b3.older
s
b3.same

b3.younger
(
b6(1985, 60)
b6.same

b6.younger
(
b1(1987, 50) b2(1988, 60)
b2.younger
(
b4(1985, 50)
b4.same

b7(1986, 55) b8(1995, 55)
b5(1985, 45)
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The command b3.removeBuilding(b3.data) should return the new root
b1 with fields updated to :
b1(1987, 50)
b1.older
v
b1.younger
(
b6(1985, 60)
b6.same

b6.younger
(
b2(1988, 60)
b2.younger
(
b4(1985, 50)
b4.same

b7(1986, 55) b8(1995, 55)
b5(1985, 45)
The command b1.removeBuilding(b1.data) should now return the new
root b6 with fields updated to :
b6(1985, 60)
b6.same

b6.younger
(
b4(1985, 50)
b4.same

b7(1986, 55)
b7.younger
(
b5(1985, 45) b2(1988, 60)
b2.younger
(
b8(1995, 55)
2.4 Showing Paul how powerful this is
Congratulations, you can now construct a database, you are now going to
exploit this to find out useful information. In all the remaining questions,
you should not modify the tree structure anymore.
Paul : “Those numbers are ridiculous, that’s not how high a building
is or how much it costs to repair a building !”
you : “I do not work on buildings, if you are unhappy about those numbers, you can either change them or imagine they are in a different unit.”
Paul : “Oh, OK. So what now ?”
you : “well, I can compute things. For instance, I can easily tell you when
the oldest building was constructed.”
Question 4. (8 points) Code oldest. When called on a Building b, the
integer b.oldest() should be the year when the oldest Building in the tree
with root b was built. Do not modify the tree structure.
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Let us give you an example. At the nodes of the tree, you will see for instance b1(1950, 10). This means Building b1 has data.yearOfConstruction
1950 and data.height 10.
b3(1987, 55)
b3.older
s
b3.same

b3.younger
(
b6(1985, 60)
b6.same

b6.younger
(
b1(1987, 50) b2(1988, 60)
b2.younger
(
b4(1985, 50)
b4.same

b7(1986, 55) b8(1995, 55)
b5(1985, 45)
b3.oldest() should return 1985 and b1.oldest() should return 1987.
Paul : “Can you tell me how high is the highest building then ?”
you : “Yes. But it is missing the point of having my database organized
by year of construction though.”
Question 5. (9 points) Code highest. When called on a Building b,
the integer b.highest() should be the height of the highest Building in the
tree with root b. Do not modify the tree structure.
On the example given to you in question ??, b3.highest() should return
60 and b4.highest() should return 50.
Paul : “You said that I was missing a point on the organization of the
database.”
you : “Well, I decided to organize buildings by their years of construction, so it is very efficient to answer questions based on this parameter. For
instance, instead of how high the highest building is, I can tell you which one
is the highest building constructed on a certain year.”
Question 6. (8 points) Code highestFromYear. When called on a Building b,
the OneBuilding b.highestFromYear(year) should be the OneBuilding
that was the highest built in year year and that is in the tree with root b
(and null if there are none). Do not modify the tree structure.
On the example from question ??, b3.highestFromYear(1985) should
return b6.data, and so should b6.highestFromYear(1985), but b4.highestFromYear(1985)
should return b4.data.
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Paul : “Can you answer questions based on multiple years ?”
you : “Yes, for instance, I can tell you how many buildings were built
during the time from one year to another one.”
Question 7. (10 points) Code numberFromYears. When called on a
Building b, the integer b.numberFromYears(yearMin, yearMax) should
be the number of Buildings built from year yearMin to yearMax (including
those two years) that is in the tree with root b. If yearMin > yearMax, this
should return 0. Do not modify the tree structure.
On the example from question ??, b3.numberFromYears(1986, 1988)
should return 4 (corresponding to buildings b7, b3, b1 and b2).
Paul : “That’s pretty cool ! Can this also help us plan the renovation of
the buildings ?”
you : “yes”
Question 8. (15 points) Code costPlanning. When called on a Building b,
the array b.costPlanning(n) should have in its i-th cell, how much you
should spend on year (2018 + i) to repair all the scheduled buildings in the
tree with root b. The argument n will always be a positive integer. Do not
modify the tree structure.
Let us modify the example given in question ?? to only display the year
planned for repairs and the costs for such repairs.
b3(2024, 23)
b3.older
s
b3.same

b3.younger
(
b6(2024, 26)
b6.same

b6.younger
(
b1(2025, 35) b2(2023, 50)
b2.younger
(
b4(2022, 5)
b4.same

b7(2024, 11) b8(2019, 46)
b5(2021, 42)
On this example, b3.costPlanning(7) should return the array {0, 46, 0, 42, 5, 50, 60} :
in year 2018, there is no cost to plan for repairing the buildings, but in year
2024, buildings b3, b6 and b7 will need to be repaired, for a cost of 23
+ 26 + 11 = 60. Similarly, b2.costPlanning(7) should return the array
{0, 46, 0, 0, 0, 50, 0}.
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