Transcript Group_B3_Final_Prese..

University of Macau
Faculty of Science and Technology
Programming Languages Architecture
SFTW 241 spring 2004
Class B Group 3
SFTW241
Programming
Language
Architecture
Grouping Method
Analysis by Program
Presentation
Contents
• For this presentation, we have the
following three parts:
• 1)A4’s initial idea
• 2)The final version of B3
• 3)The conclusion
Part 1
A4’s initial idea
A4’s initial idea
• There are 21 students in class A and each
student file structure include the following
information:
1.Student ID
2.Student name
3.Firest, second and third choice
4.Choose = 0
A4’s initial idea
• There is one array and four stacks in this
program
• All the information of each student is stored in
this array
• And each stacks store all the members’
information of the corresponding group
• In each student’s data structure, if the choose is
equal to 0, it meant that this student has not
been selected; if it is 1, it meant that this student
has been selected
A4’s initial idea
• First load all the students’ information into
the array
• Then we choose four leaders random and
load their information into the
corresponding stack for each group
• And in this process there is a very
important detail is that if the information
is push into the stack, the choose in the
array become 1 at once.
A4’s initial idea
captain
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captain
captain
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0 1 2 3 4 5 6 7 ……
A4’s initial idea
• Then load the captain’s first choice’s ID number
and compare with the student ID number in the
array
• If the ID number is not equal, compare with the
next one
• If the ID number is equal and the choose is equal
to 1, compare with the next one
• If the ID number is equal and the choose is equal
to 0, push the student’s information into the
corresponding stack
A4’s initial idea
Eric
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Yedda WFZH
ZDF
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LJQ
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Choose = 0
Walter
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ZDF
Walter
Eric
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A4’s initial idea
• If someone’s all three choices are
selected, then the corresponding
team will stop adding new member.
• And we use the randomly method to
add new members
A4’s initial idea
• Their idea is use stack to store each group’s
information
• As we know, its advantage is that it is very easy
to access data
• But its disadvantage is that we do not know who
has been selected and who has not, and if you
need to random select one person, you must do
the selection from all the people, then the time
you spend is very expensive
Improvement
In the presentation between us and
group A4, we got lots of good
suggestions, and we also got some
ideas from their algorithm.
Idea & Suggestion
• The most important suggestions:
(a)One array is enough to carry out
our algorithm
(b)Swapping structures which contain
huge data will cost lots of time
(c)There is no need to swap or delete
by using link-list
Part 2
Final Version
of B3
Suggestion Analysis
• I think our program should contain
the following two advantages:
(a)Efficient
(b)Reusable
(c)Informational
Reusable
• The data in the main memory can be used
again even after the grouping.
• Compare with the memory operation, I/O
operation (file processing) is very low.
• We have to do the grouping from the
beginning if some special case happen.
Efficient
• It is necessary to divide the data into
two parts: unused and used
• Just use one array to store the
information
• The choice should be the integer
instead of string for checking easy
and less space
Informational
• Easy to get the information of a team:
the number of the member; identity
of the member
Data Structure
• Data is stored as the following form:
class Record{
private:
int id;
string name;
int first, second, third;
}
Unit
Student Unit
Student ID
Student Name
First Choice
Second Choice
Third Choice
Improvements
• Id number is added, it is easy to get
the corresponding student’s
information
• Actually, id number is not the real ID
of the student
• The favorites were stored as integer
Index array
• Index array, we also use the index array
• Advantages:
(a) just swap the index when process a
piece of information
(b) protect the original data for the reuse
Length of the Index
Array
• Let N = number of the students
• Let M = the number of the teams
• Let IA = index array
• Then IA.Length = int (N/M) *M
• If there are 18 students, and we will divide them
into 4 groups, then the length of the index array
should be 20.
Differences
• The length of the index array may be
not the same as the number of the
students
• IndexArray.Length >= S_R.Length
Processing of IA
• Our grouping just operate on the IA,
S_R (student records) is just used to
store the information
Processing of IA
• A count (integer) is used to divide
the IA into two parts: used and
unused
• When a new member was selected,
it’s index will be moved to last
position of the unused part, then the
count will decrease one unit
used
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Index i
Yedda
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Counter
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used
THL
SUN WFZH
Index i
UN
Yedda
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Counter
LJQ
ZDF
Walter ERIC
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Searching
Searching
How to get the information
of a group
ID12 ID11 ID10 ID9 ID1
ID2
ID3
ID6
ID8
ID7
ID4
ID5
Special case 1
• If N/M == integer, it is the best case
• Else, the length of the IA will be
longer than the length of the S_R
• The processing will be a little
different at the end
Swapping
Solution of the special
case 1
• 10 students
ID11 ID12
ID1
ID2
Index ( i )
ID3
ID9
ID10
ID6
ID8
Member Counter
ID7
ID4
ID5
Special case 2
• At the beginning of the grouping, we
will choose the leaders by chance.
• If one leader’s three choices are the
leaders either, then it will choose the
leaders again
Implementation
Implementation
• Show our program:
Part 3
Conclusion
Conclusion
•
Both of two methods are base on the
index array
•
Advantages:
a) Reusable
b) Efficient
c) Informational
Comparison
• Two versions
1. Pointer version (A4’s algorithm)
2. Array version (Our algorithm)
Comparison
• Comparison
a)It’s faster and easier to get the information of
a group by using pointer
b)Easy to add a new member or delete a wrong
record by using pointer
c)Pointer version have to use two pointer
operations when a member is added into a
group
Comparison
d) Array version have to do some
comparison when it want to get the
information of a group
e) Array version use less space
Space Comparison
• Pointer version: each element has to
contain pointer, it meant that the
pointer version must spend more
space than the array version
• Array Version: It has the best and
the worst case
Best case: N / M == Integer
Worst case: (N – 1)/ M == Integer
Best & Worst Case
• Array Version:
 Best Case: Each group has the same number
of member
 Worst Case: There are one student more
• Pointer Version:
 There are not best and worst case in the
pointer version
Conclusion
• Actually, both the two versions are
efficient
• Running time are both O(N)
• Their have their own advantages and
disadvantages
Conclusion
• What we done ?