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izatwin 51f720ad9f p4 start 2024-04-12 10:39:46 -04:00
izatwin f7697e992f hw4 stuff 2024-04-12 10:39:38 -04:00
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\{A, I\} → \{I, J\}\} \{A, I\} → \{I, J\}\}
\begin{enumerate} [label=(\alph*)] \begin{enumerate} [label=(\alph*)]
\item \textbf{What is the closure of the set \{A, D\}?}\\ \item \textbf{What is the closure of the set \{A, D\}?}\\
$\{A, D\}^+ = \{A, D, E, F, G, H, I, J\}$ $\{A, D\}^+ = \{A, D, E, F, G, H, I, J\}$
\item \textbf{What is the key for R?}\\ \item \textbf{What is the key for R?}\\
Key $= \{A, B\}$, since \{A, B\}'s closure includes all atributes. Key $= \{A, B\}$, since \{A, B\}'s closure includes all atributes.
\item \textbf{What is the minimum cover for the set of functional dependencies F?}\\ \item \textbf{What is the minimum cover for the set of functional dependencies F?}\\
Convert to canonical form:\\ Convert to canonical form:\\
\{\{A, B\}→\{C\},\\ \{\{A, B\}→\{C\},\\
\{A, B\}→\{D\},\\ \{A, B\}→\{D\},\\
\{D\}→\{E\},\\ \{D\}→\{E\},\\
\{D\}→\{F\},\\ \{D\}→\{F\},\\
\{D\}→\{G\},\\ \{D\}→\{G\},\\
\{F, G\}→\{H\},\\ \{F, G\}→\{H\},\\
\{A\}→\{I\},\\ \{A\}→\{I\},\\
\{A, B\}→\{E\},\\ \{A, B\}→\{E\},\\
\{A, B\}→\{G\},\\ \{A, B\}→\{G\},\\
\{A, I\} → \{I\}\\ \{A, I\} → \{I\}\\
\{A, I\} → \{J\}\}\\ \{A, I\} → \{J\}\}\\
Eliminate extraneous stuff:\\ Eliminate extraneous stuff:\\
\{\{A, B\}→\{C\},\\ \{\{A, B\}→\{C\},\\
\{A, B\}→\{D\},\\ \{A, B\}→\{D\},\\
\{D\}→\{E\},\\ \{D\}→\{E\},\\
\{D\}→\{F\},\\ \{D\}→\{F\},\\
\{D\}→\{G\},\\ \{D\}→\{G\},\\
\{F, G\}→\{H\},\\ \{F, G\}→\{H\},\\
\{A\}→\{I\},\\ \{A\}→\{I\},\\
\{A, B\}→\{E\},\\ \{A, B\}→\{E\},\\
\{A, B\}→\{G\},\\ \{A, B\}→\{G\},\\
\{A\} → \{J\}\}\\ \{A\} → \{J\}\}\\
Eliminate redundancies:\\ Eliminate redundancies:\\
\{\{A, B\}→\{C\},\\ \{\{A, B\}→\{C\},\\
\{A, B\}→\{D\},\\ \{A, B\}→\{D\},\\
\{D\}→\{E\},\\ \{D\}→\{E\},\\
\{D\}→\{F\},\\ \{D\}→\{F\},\\
\{D\}→\{G\},\\ \{D\}→\{G\},\\
\{F, G\}→\{H\},\\ \{F, G\}→\{H\},\\
\{A\}→\{I\},\\ \{A\}→\{I\},\\
\{A\} → \{J\}\}\\ \{A\} → \{J\}\}\\
\item \textbf{Decompose R into 2NF.}\\ \item \textbf{Decompose R into 2NF.}\\
R = \{A, B, C, D, E, F, G, H, I, J\}\\ R = \{A, B, C, D, E, F, G, H, I, J\}\\
Split up:\\ Split up:\\
$R_1$ = \{\underline{A}, \underline{B}, C, D, E, F, G, H\}; $R_2$ = \{\underline{A}, I, J\}\\ $R_1$ = \{\underline{A}, \underline{B}, C, D, E, F, G, H\}; $R_2$ = \{\underline{A}, I, J\}\\
This is because every nonprime element in $R_1$ is fully functionally dependent on \{A, B\}, and every nonprime element in $R_2$ is fully functionally dependent on $\{A\}$. This is because every nonprime element in $R_1$ is fully functionally dependent on \{A, B\}, and every nonprime element in $R_2$ is fully functionally dependent on $\{A\}$.
\item Decompose R into 3NF\\ \item Decompose R into 3NF\\
$R_1$ = \{\underline{A}, \underline{B}, C, D\}; $R_2$ = \{\underline{A}, I, J\}; $R_3$ = \{\underline{D}, E, F, G\} $R_4$ = \{\underline{F}, \underline{G}, H\}\\ $R_1$ = \{\underline{A}, \underline{B}, C, D\}; $R_2$ = \{\underline{A}, I, J\}; $R_3$ = \{\underline{D}, E, F, G\} $R_4$ = \{\underline{F}, \underline{G}, H\}\\
\end{enumerate} \end{enumerate}
\newpage \newpage
\section{} \section{}
\begin{enumerate} [label=(\alph*)] \begin{enumerate} [label=(\alph*)]
\item \textbf{What are the possible sources of the information that defines the functional dependencies \item \textbf{What are the possible sources of the information that defines the functional dependencies
that hold among the attributes of a relation schema? Discuss two possible sources. Explain your that hold among the attributes of a relation schema? Discuss two possible sources. Explain your
answer.}\\ answer.}\\
Functional dependencies must be defined explicitely by someone who knows the semantics of the attributes in the relation. Functional dependencies must be defined explicitely by someone who knows the semantics of the attributes in the relation.
So what are some possible sources for these semantics?\\ So what are some possible sources for these semantics?\\
\\ \\
A possible source could be the entity that wants to use the database. Lets say its a company that want's to keep track of laptops. Perhaps they have a bunch of attributes that they A possible source could be the entity that wants to use the database. Lets say its a company that want's to keep track of laptops. Perhaps they have a bunch of attributes that they
want to be able to keep track of about the laptops. They can put together the attributes and then make some sort of ID for each laptop that has functional dependencies for the relation.\\ want to be able to keep track of about the laptops. They can put together the attributes and then make some sort of ID for each laptop that has functional dependencies for the relation.\\
\\ \\
Another possible source could be some sort of relationship that is already defined in the nature of whatever you are already trying to model. For example, the relationship between a child and their mother. Another possible source could be some sort of relationship that is already defined in the nature of whatever you are already trying to model. For example, the relationship between a child and their mother.
There is a functional dependency from the child to the mother because the child can only have one biological mother while the mother could have multiple children. There is a functional dependency from the child to the mother because the child can only have one biological mother while the mother could have multiple children.
\item \textbf{What is meant by the closure of a set of functional dependencies? Illustrate with an example.}\\ \item \textbf{What is meant by the closure of a set of functional dependencies? Illustrate with an example.}\\
The closure of a set of FD really just means that it includes all inferences that can be made from a sparser set of functional dependencies. It's the most explicit you can be about a group of relations.\\ The closure of a set of FD really just means that it includes all inferences that can be made from a sparser set of functional dependencies. It's the most explicit you can be about a group of relations.\\
For example, if there is a functional dependency between my SSN and me $(A \rightarrow B)$, and also a functional dependency between me and my mother $(B \rightarrow C)$, the closure of the set of those two would also include the functional dependency that can be inferred between my ssn and my mother. $(A \rightarrow C)$. For example, if there is a functional dependency between my SSN and me $(A \rightarrow B)$, and also a functional dependency between me and my mother $(B \rightarrow C)$, the closure of the set of those two would also include the functional dependency that can be inferred between my ssn and my mother. $(A \rightarrow C)$.
Thus, the closure of the SSN would be $A^+ = \{A, B, C\}$. Thus, the closure of the SSN would be $A^+ = \{A, B, C\}$.
\item \textbf{What is the lossless (or nonadditive) join property of a decomposition? Why is it important?}\\ \item \textbf{What is the lossless (or nonadditive) join property of a decomposition? Why is it important?}\\
This property ensures that spurious tuples, or tuples that were not in the original relation before decomposition, do not appear after rejoining the relations.\\ This property ensures that spurious tuples, or tuples that were not in the original relation before decomposition, do not appear after rejoining the relations.\\
This is important so we can complete the normalization of the data and relations without losing information. We need the benefits that come with normalizing relations, but we cannot sacrifice data to achieve it. This is important so we can complete the normalization of the data and relations without losing information. We need the benefits that come with normalizing relations, but we cannot sacrifice data to achieve it.
\end{enumerate} \end{enumerate}
\newpage
\section{}
Consider the following four schedules for transactions T1, T2 and T3:
\begin{enumerate} [label=\alph*]
\item r1(X); r3(X); w1(X); r2(X); w3(X); w2(X)
\item r1(Y); r3(Y); w3(X); w1(X); r2(X); w2(Y)
\item w2(X); r1(Y); w2(Y); r2(X); w2(Y); r3(Y)
\item r3(X); r2(X); w3(X); r1(Y); w1(X); r2(Y)
\item r3(X); r2(X); r1(X); w3(X); w1(X); w2(X)
\end{enumerate}
\begin{enumerate} [label=(\alph*)]
\item For each schedule, specify whether or not the schedule is conflict serializable and why. To
determine if a schedule is conflict serializable you can either use the precedence graph or swap
conflicting operations.
\item Then, for each serializable schedule, give the equivalent serial schedule.
\end{enumerate}
\hrule
\begin{enumerate} [label=(\alph*)]
\item \textbf{r1(X); r3(X); w1(X); r2(X); w3(X); w2(X)}\\
Not-Serializable. There is a cycle between $T_1$ and $T_3$ in the precedence graph.\\
\includegraphics[width=0.5\linewidth]{pred_graph_a.jpg}
\item \textbf{r1(Y); r3(Y); w3(X); w1(X); r2(X); w2(Y)}\\
Serializable.\\
r3(Y); w3(X); r1(Y); w1(X); r2(X); w2(Y)\\
\includegraphics[width=0.5\linewidth]{pred_graph_b.jpg}
\item \textbf{w2(X); r1(Y); w2(Y); r2(X); w2(Y); r3(Y)}\\
Serializable.\\
r1(Y); w2(X); w2(Y); r2(X); w2(Y); r3(Y)\\
\includegraphics[width=0.5\linewidth]{pred_graph_c.jpg}
\item \textbf{r3(X); r2(X); w3(X); r1(Y); w1(X); r2(Y)}\\
Serializable.\\
r2(X); r2(Y); r3(X); w3(X); r1(Y); w1(X)\\
\includegraphics[width=0.5\linewidth]{pred_graph_d.jpg}
\item \textbf{r3(X); r2(X); r1(X); w3(X); w1(X); w2(X)}\\
Not-Serializable. There is a cycle between all edges in the precedence graph.\\
\includegraphics[width=0.5\linewidth]{pred_graph_e.jpg}
\end{enumerate}
\newpage
\section{(0.75 points)}
Consider the following actions taken by transaction T1 on a database with objects X and Y:\\
\textbf{R1(X), W1(X), R1(Y), W1(Y)}\\
Without altering the order of operations of T1, give example schedules with an additional
Transactions T2 (with 4 operations accessing both database objects X and Y) that fulfill the
following conditions:\\
\begin{enumerate} [label=(\alph*)]
\item \textbf{Give an example schedule using two transactions (T1 and T2) which is conflict serializable.}\\
w2(X); r1(X); r2(Y); w2(Y); r2(X); w1(X); r1(Y); w1(Y)\\
The precedence graph shows no cycles.
\item \textbf{Give an example schedule using two transactions (T1 and T2) which is not conflict serializable. Explain why the schedule is not conflict serializable.}\\
w2(X); r1(X); r2(Y); w2(Y); w1(X); r2(X); r1(Y); w1(Y)\\
The precedence graph has a cycle between $T_1$ and $T_2$. $T_1$ writes before $T_2$ reads, but $T_2$ has already written before $T_1$.
\end{enumerate}
\end{document} \end{document}
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Projects/cs448p4/ExpectedLogs.txt
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Test Case 1 (Handout example)
Log:
W:0,T1,1,1,5,-1
R:1,T1,2,2,0
W:2,T1,2,2,3,1
R:3,T1,1,5,2
C:4,T1,3
R:5,T2,1,5,-1
W:6,T2,1,5,2,5
C:7,T2,6
Test Case 2 (Three transactions)
Log:
R:0,T1,6,6,-1
R:1,T2,4,4,-1
R:2,T3,9,9,-1
W:3,T1,7,7,2,0
W:4,T2,2,2,4,1
R:5,T3,6,6,2
W:6,T1,5,5,8,3
W:7,T3,1,1,9,5
C:8,T3,7
W:9,T1,6,6,2,6
C:10,T1,9
R:11,T2,5,8,4
W:12,T2,7,2,3,11
C:13,T2,12
Test Case 3 (Deadlocks)
Log:
W:0,T1,4,4,1,-1
W:1,T2,5,5,2,-1
W:2,T3,6,6,3,-1
R:3,T1,1,1,0
R:4,T2,1,1,1
R:5,T3,2,2,2
R:6,T2,2,2,4
A:7,T3,5
W:8,T2,2,2,3,6
C:9,T2,8
W:10,T1,1,1,2,3
C:11,T1,10
Test Case 4 (No Conflicts)
Log:
R:0,T1,1,1,-1
R:1,T2,1,1,-1
R:2,T1,2,2,0
R:3,T2,2,2,1
R:4,T1,3,3,2
R:5,T2,3,3,3
W:6,T1,4,4,1,4
W:7,T2,5,5,2,5
C:8,T1,6
C:9,T2,7
Test Case 5: Multiple Deadlocks OR if deadlock detection finds T1 & T2 deadlock first:
Log: Log:
R:0,T1,1,1,-1 R:0,T1,1,1,-1
R:1,T2,2,2,-1 R:1,T2,2,2,-1
R:2,T3,1,1,-1 R:2,T3,1,1,-1
R:3,T2,3,3,1 R:3,T2,3,3,1
A:4,T3,2 A:4,T2,3
A:5,T2,3 W:5,T3,3,3,4,2
W:6,T1,2,2,1,0 W:6,T1,2,2,1,0
C:7,T1,6 C:7,T3,5
C:8,T1,6
Test Case 6: Larger deadlock cycle
Log:
R:0,T1,1,1,-1
R:1,T2,2,2,-1
R:2,T3,3,3,-1
R:3,T4,4,4,-1
A:4,T4,3
W:5,T3,4,4,3,2
C:6,T3,5
W:7,T2,3,3,2,1
C:8,T2,7
W:9,T1,2,2,1,0
C:10,T1,9
Test Case 7: Small deadlock cycle
Log:
W:0,T1,1,1,2,-1
W:1,T2,2,2,5,-1
A:2,T2,1
R:3,T1,2,2,0
C:4,T1,3
Test Case 8: Three transaction with deadlock cycle
Log:
R:0,T1,6,6,-1
R:1,T2,4,4,-1
R:2,T3,9,9,-1
W:3,T1,7,7,2,0
W:4,T2,2,2,4,1
R:5,T3,6,6,2
R:6,T2,5,5,4
W:7,T3,1,1,9,5
A:8,T2,6
C:9,T3,7
W:10,T1,2,2,8,3
W:11,T1,6,6,2,10
C:12,T1,11
Test Case 9: Small transactions without deadlock cycle
Log:
W:0,T1,1,1,2,-1
W:1,T2,3,3,5,-1
R:2,T1,2,2,0
R:3,T2,4,4,1
C:4,T1,2
C:5,T2,3
Test Case 10: Handout example with Deadlock
Log:
W:0,T1,1,1,5,-1
R:1,T2,2,2,-1
R:2,T1,2,2,0
A:3,T2,1
W:4,T1,2,2,3,2
R:5,T1,1,5,4
C:6,T1,5
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Projects/cs448p4/pom.xml
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<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>edu.cs.purdue</groupId>
<artifactId>cs448p4</artifactId>
<version>1.0-SNAPSHOT</version>
<name>cs448p4</name>
<url>https://www.cs.purdue.edu/homes/bb/cs448f19/</url>
<properties>
<java.version>10</java.version>
<project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
<maven.compiler.source>${java.version}</maven.compiler.source>
<maven.compiler.target>${java.version}</maven.compiler.target>
</properties>
<dependencies>
<dependency>
<groupId>junit</groupId>
<artifactId>junit</artifactId>
<version>4.11</version>
<scope>test</scope>
</dependency>
</dependencies>
<build>
<pluginManagement><!-- lock down plugins versions to avoid using Maven defaults (may be moved to parent pom) -->
<plugins>
<!-- clean lifecycle, see https://maven.apache.org/ref/current/maven-core/lifecycles.html#clean_Lifecycle -->
<plugin>
<artifactId>maven-clean-plugin</artifactId>
<version>3.1.0</version>
</plugin>
<!-- default lifecycle, jar packaging: see https://maven.apache.org/ref/current/maven-core/default-bindings.html#Plugin_bindings_for_jar_packaging -->
<plugin>
<artifactId>maven-resources-plugin</artifactId>
<version>3.0.2</version>
</plugin>
<plugin>
<artifactId>maven-compiler-plugin</artifactId>
<version>3.8.0</version>
<configuration>
<release>${java.version}</release>
</configuration>
</plugin>
<plugin>
<artifactId>maven-surefire-plugin</artifactId>
<version>2.22.1</version>
</plugin>
<plugin>
<artifactId>maven-jar-plugin</artifactId>
<version>3.0.2</version>
</plugin>
<plugin>
<artifactId>maven-install-plugin</artifactId>
<version>2.5.2</version>
</plugin>
<plugin>
<artifactId>maven-deploy-plugin</artifactId>
<version>2.8.2</version>
</plugin>
<!-- site lifecycle, see https://maven.apache.org/ref/current/maven-core/lifecycles.html#site_Lifecycle -->
<plugin>
<artifactId>maven-site-plugin</artifactId>
<version>3.7.1</version>
</plugin>
<plugin>
<artifactId>maven-project-info-reports-plugin</artifactId>
<version>3.0.0</version>
</plugin>
<plugin>
<artifactId>maven-assembly-plugin</artifactId>
<configuration>
<archive>
<manifest>
<!--add test cases file in here -->
<mainClass>test.TestDriver</mainClass>
</manifest>
</archive>
<descriptorRefs>
<descriptorRef>jar-with-dependencies</descriptorRef>
</descriptorRefs>
</configuration>
</plugin>
</plugins>
</pluginManagement>
</build>
</project>
Projects/cs448p4/src/main/java/CC/CC.java
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package CC;
import java.util.List;
public class CC
{
/**
* Notes:
* - Execute all given transactions, using locking.
* - Each element in the transactions List represents all operations performed by one transaction, in order.
* - No operation in a transaction can be executed out of order, but operations may be interleaved with other
* transactions if allowed by the locking.
* - The index of the transaction in the list is equivalent to the transaction ID.
* - Print the log to the console at the end of the method.
* - Return the new db state after executing the transactions.
* @param db the initial status of the db
* @param transactions the schedule, which basically is a {@link List} of transactions.
* @return the final status of the db
*/
public static int[] executeSchedule(int[] db, List<String> transactions)
{
//TODO
return null ;
}
}
Projects/cs448p4/src/main/java/CC/Project.java
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package CC;
import java.util.Arrays;
import java.util.List;
public class Project
{
public static int[] newDB()
{
return new int[]{0,1,2,3,4,5,6,7,8,9};
}
public static void main(String[] args)
{
//Test Cases
//1) Simple testcase (handout)
List<String> tc1 = Arrays.asList(new String[]{"W(1,5);R(2);W(2,3);R(1);C","R(1);W(1,2);C"});
//2) Three transactions
List<String> tc2 = Arrays.asList(new String[]{"R(6);W(7,2);W(5,8);W(6,2);C","R(4);W(2,4);R(5);W(7,3);C","R(9);R(6);W(1,9);C"});
//3) Deadlocks
List<String> tc3 = Arrays.asList(new String[]{"W(4,1);R(1);W(1,2);C","W(5,2);R(1);R(2);W(2,3);C","W(6,3);R(2);W(2,4);C"});
//4) No conflicts
List<String> tc4 = Arrays.asList(new String[]{"R(1);R(2);R(3);W(4,1);C","R(1);R(2);R(3);W(5,2);C"});
//5) Multiple Deadlocks (handout)
List<String> tc5 = Arrays.asList(new String[]{"R(1);W(2,1);C","R(2);R(3);W(1,2);C","R(1);W(3,4);C"});
//6) Larger Deadlock Cycle
List<String> tc6 = Arrays.asList(new String[]{"R(1);W(2,1);C","R(2);W(3,2);C","R(3);W(4,3);C","R(4);W(1,4);C"});
//7) Small Deadlock Cycle
List<String> tc7 = Arrays.asList(new String[]{"W(1,2);R(2);C","W(2,5);R(1);C"});
//8) Three Transactions Deadlock Cycle
List<String> tc8 = Arrays.asList(new String[]{"R(6);W(7,2);W(2,8);W(6,2);C","R(4);W(2,4);R(5);W(7,3);C","R(9);R(6);W(1,9);C"});
//9) Small Transactions without Deadlock Cycle
List<String> tc9 = Arrays.asList(new String[]{"W(1,2);R(2);C","W(3,5);R(4);C"});
//10) Testcase with Deadlock Cycle
List<String> tc10 = Arrays.asList(new String[]{"W(1,5);R(2);W(2,3);R(1);C","R(2);R(1);W(1,2);C"});
//Consult the ExpectedOutputs.txt for the expected DB state and log
System.out.println("Schedule 1: Handout Test Case:\n\t" + Arrays.toString(CC.executeSchedule(newDB(),tc1)) + "\n");
System.out.println("Schedule 2: Three Transactions Test Case:\n\t" + Arrays.toString(CC.executeSchedule(newDB(),tc2)) + "\n");
System.out.println("Schedule 3: Deadlocks Test Case:\n\t" + Arrays.toString(CC.executeSchedule(newDB(),tc3)) + "\n");
System.out.println("Schedule 4: No Conflicts Test Case:\n\t" + Arrays.toString(CC.executeSchedule(newDB(),tc4)) + "\n");
System.out.println("Schedule 5: Multiple Deadlocks Test Case:\n\t" + Arrays.toString(CC.executeSchedule(newDB(),tc5)) + "\n");
System.out.println("Schedule 6: Larger Deadlock Cycle Test Case:\n\t" + Arrays.toString(CC.executeSchedule(newDB(),tc6)) + "\n");
System.out.println("Schedule 7: Testcase with two transactions and a deadlock\n\t" + Arrays.toString(CC.executeSchedule(newDB(),tc7)) + "\n");
System.out.println("Schedule 8: Testcase with three transactions and a deadlock:\n\t" + Arrays.toString(CC.executeSchedule(newDB(),tc8)) + "\n");
System.out.println("Schedule 9: Testcase with two transactions and no deadlocks:\n\t" + Arrays.toString(CC.executeSchedule(newDB(),tc9)) + "\n");
System.out.println("Schedule 10: Testcase with deadlock between two transactions:\n\t" + Arrays.toString(CC.executeSchedule(newDB(),tc10)) + "\n");
}
}
Projects/cs448p4/src/test/java/tests/CCTest.java
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package tests;
import static org.junit.Assert.*;
import java.io.IOException;
import CC.*;
import org.junit.After;
import org.junit.Before;
import org.junit.BeforeClass;
import org.junit.FixMethodOrder;
import org.junit.Test;
import org.junit.runners.MethodSorters;
import java.util.Arrays;
import java.util.List;
@FixMethodOrder(MethodSorters.NAME_ASCENDING)
public class CCTest {
public final static boolean OK = true;
public final static boolean FAIL = false;
public static int[] newDB()
{
return new int[]{0,1,2,3,4,5,6,7,8,9};
}
/**
* Schedule 1. Simple testcase with 2 transactions.
* This testcase is shown in the handout, Part 2.
*/
@Test
public void test_A_Simple()
{
// expected output from the given Schedule
int expectedOutput[] = {0, 2, 3, 3, 4, 5, 6, 7, 8, 9};
// input: T1, T2
List<String> tc1 = Arrays.asList(new String[]{"W(1,5);R(2);W(2,3);R(1);C","R(1);W(1,2);C"});
// Here we are returning the state of the database after executing Schedule 1
int methodOutput[] = CC.executeSchedule(newDB(),tc1);
// Here we are checking whether method output matched with the expected output
// if both match it will pass the test, otherwise it will show an error message
assertArrayEquals("output did not match expected result",expectedOutput, methodOutput);
}
/**
* Schedule 2. Testcase with three transactions and no deadlock.
*/
@Test
public void test_B_ThreeTransactions()
{
int expectedOutput[] = {0, 9, 4, 3, 4, 8, 2, 3, 8, 9};
//Testcases
//2) Three transactions
List<String> tc2 = Arrays.asList(new String[]{"R(6);W(7,2);W(5,8);W(6,2);C","R(4);W(2,4);R(5);W(7,3);C","R(9);R(6);W(1,9);C"});
int methodOutput[] = CC.executeSchedule(newDB(),tc2);
assertArrayEquals("output did not match expected result",expectedOutput, methodOutput);
}
/**
* Schedule 3. Testcase with three transactions and a deadlock.
*/
@Test
public void test_C_Deadlocks()
{
int expectedOutput[] = {0, 2, 3, 3, 1, 2, 6, 7, 8, 9};
//Testcases
//3) Deadlocks
List<String> tc3 = Arrays.asList(new String[]{"W(4,1);R(1);W(1,2);C","W(5,2);R(1);R(2);W(2,3);C","W(6,3);R(2);W(2,4);C"});
int methodOutput[] = CC.executeSchedule(newDB(),tc3);
assertArrayEquals("output did not match expected result",expectedOutput, methodOutput);
}
/**
* Schedule 4. Testcase with multiple deadlocks.
* This testcase is shown in the handout, Part 3 (Deadlock Detection/Handling).
*/
@Test
public void test_D_NoConflicts()
{
int expectedOutput[] = {0, 1, 2, 3, 1, 2, 6, 7, 8, 9};
//Testcases
//4) No conflicts
List<String> tc4 = Arrays.asList(new String[]{"R(1);R(2);R(3);W(4,1);C","R(1);R(2);R(3);W(5,2);C"});
int methodOutput[] = CC.executeSchedule(newDB(),tc4);
assertArrayEquals("output did not match expected result",expectedOutput, methodOutput);
}
/**
* Schedule 5. Testcase with multiple deadlocks.
* This testcase is shown in the handout, Part 3 (Deadlock Detection/Handling).
*/
@Test
public void test_E_MultipleDeadlocks()
{
// In the Schedule 5 there can be two possible outcomes depending of how you handle deadlocks.
int expectedOutput[] = {0, 1, 1, 3, 4, 5, 6, 7, 8, 9};
//Testcases
//5) Multiple Deadlocks
List<String> tc5 = Arrays.asList(new String[]{"R(1);W(2,1);C","R(2);R(3);W(1,2);C","R(1);W(3,4);C"});
int methodOutput[] = CC.executeSchedule(newDB(),tc5);
try{
// if T3 is aborted, and then T2
assertArrayEquals("output did not match expected result",expectedOutput, methodOutput);
}catch (AssertionError e) {
// if we abort T2
int expectedOutput_anth[] = {0, 1, 1, 4, 4, 5, 6, 7, 8, 9};
assertArrayEquals("output did not match expected result",expectedOutput_anth, methodOutput);
}
}
/**
* Schedule 6. Testcase with large number of transactions (4) and a deadlock.
*/
@Test
public void test_F_LargerDeadlock()
{
int expectedOutput[] = {0, 1, 1, 2, 3, 5, 6, 7, 8, 9};
//Testcases
//6) Larger Deadlock Cycle
List<String> tc6 = Arrays.asList(new String[]{"R(1);W(2,1);C","R(2);W(3,2);C","R(3);W(4,3);C","R(4);W(1,4);C"});
int methodOutput[] = CC.executeSchedule(newDB(),tc6);
assertArrayEquals("output did not match expected result",expectedOutput, methodOutput);
}
/**
* Schedule 7. Testcase with small transactions (2) and a deadlock.
*/
@Test
public void test_G_SmallDeadlock()
{
int expectedOutput[] = {0, 2, 2, 3, 4, 5, 6, 7, 8, 9};
//Testcases
//7) Small Deadlock Cycle
List<String> tc7 = Arrays.asList(new String[]{"W(1,2);R(2);C","W(2,5);R(1);C"});
int methodOutput[] = CC.executeSchedule(newDB(),tc7);
assertArrayEquals("output did not match expected result",expectedOutput, methodOutput);
}
/**
* Schedule 8. Testcase with three transactions (2) and a deadlock.
*/
@Test
public void test_H_ThreeTransactionsDeadlock()
{
int expectedOutput[] = {0, 9, 8, 3, 4, 5, 2, 2, 8, 9};
//Testcases
//8) Three Transactions Deadlock
List<String> tc8 = Arrays.asList(new String[]{"R(6);W(7,2);W(2,8);W(6,2);C","R(4);W(2,4);R(5);W(7,3);C","R(9);R(6);W(1,9);C"});
int methodOutput[] = CC.executeSchedule(newDB(),tc8);
assertArrayEquals("output did not match expected result",expectedOutput, methodOutput);
}
/**
* Schedule 9. Testcase with small transactions (2) and no deadlocks.
*/
@Test
public void test_I_SmallTransactionsNoDeadlock()
{
int expectedOutput[] = {0, 2, 2, 5, 4, 5, 6, 7, 8, 9};
//Testcases
//9) Small Transactions without Deadlock Testcase:
List<String> tc9 = Arrays.asList(new String[]{"W(1,2);R(2);C","W(3,5);R(4);C"});
int methodOutput[] = CC.executeSchedule(newDB(),tc9);
assertArrayEquals("output did not match expected result",expectedOutput, methodOutput);
}
/**
* Schedule 10. Testcase with deadlock between two transactions.
*/
@Test
public void test_J_TestCaseDeadlockDeadlock()
{
int expectedOutput[] = {0, 5, 3, 3, 4, 5, 6, 7, 8, 9};
//Testcases
//10) Testcase with Deadlock:
List<String> tc10 = Arrays.asList(new String[]{"W(1,5);R(2);W(2,3);R(1);C","R(2);R(1);W(1,2);C"});
int methodOutput[] = CC.executeSchedule(newDB(),tc10);
assertArrayEquals("output did not match expected result",expectedOutput, methodOutput);
}
}
Projects/cs448p4/src/test/java/tests/TestDriver.java
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package tests;
import org.junit.runner.JUnitCore;
import org.junit.runner.Result;
import org.junit.runner.notification.Failure;
public class TestDriver {
public static void main(String[] args) {
Result result = JUnitCore.runClasses(tests.CCTest.class);
for (Failure failure : result.getFailures()) {
System.out.println(failure.toString());
}
System.out.println(result.wasSuccessful());
}
}