CPSC457 Assignment 4-Deadlock-Detector-main Solved 

Q1 – Deadlock Detector

For this question you will write a C++ function that detects a deadlock in a system state with a single instance per resource type. The system state will be provided to this function as an ordered sequence of request- and assignment-edges. Your function will start by initializing an empty system state. Then, your function will process the edges one at a time. For each edge, your function will update the system state, and run a deadlock detection algorithm. If a deadlock is detected after processing an edge, your function will stop processing any more edges and return results immediately.

Below is the signature of the detect_deadlock function you need to implement:

The parameter edges[] is an ordered list of strings, each representing an edge. The function returns an instance of Result containing two fields as described below.

Your function will start with an empty system state – by initializing an empty graph data structure. For each string in edges[] it will parse it to determine if it represents an assignment edge or request edge, and update the graph accordingly. After inserting each edge, the function will run an algorithm that will look for a deadlock (by detecting if a cycle is present in the graph). If deadlock is detected, your function will stop processing any more edges and immediately return Result:

• with edge_index set to the index of the edge in edges[] that caused the deadlock; and
• with dl_procs[] containing process names that are involved in a deadlock. Order is arbitrary.If no deadlock is detected after processing all edges, your function will indicate this by returning Result with edge_index=-1 and an empty dl_procs[].

  Edge description

  Your function will be given the edges as a vector of strings, where each edge will represent and edge. A request edge will have the format “(P) -> (R)”, and assignment edge will be of the form
  “(P) <- (R)”, where (P) and (R) are the names of the process and resource, respectively. Here is a sample input, and its graphical representation:

edges =
[ “2

]

The input above represents a system with three processes: “plato”, “socrates” and “2”, and two resources: “fork1” and “2”. The first line “2 <- fork1” represents an assignment edge, and it denotes process “2” currently holding resource “fork1”. The second line “plato -> fork1” is a request edge, meaning that process “plato” is waiting for resource “fork1”. The resource allocation graph on the right is a graphical representation of the input. Process and resource names are independent from each other, and it is therefore possible for processes and resources to have the same names.

Notice that each individual string may contain arbitrary number of spaces. Feel free to use the provided simplify() and/or split() functions from common.cpp to help you parse these strings.

Starter code

Start by downloading the following starter code:

You need to implement detect_deadlock() function my modifying deadlock_detector.cpp. The rest of the starter code contains a driver code (main.cpp) and some convenience functions you may use in your implementation (common.cpp). Only modify file deadlock_detector.cpp, and do not modify any other files.

The included driver will read edge descriptions from standard input. It parses them into an array of strings, and then calls your detect_deadlock(), and afterwards prints out the results. The driver will ensure that the input passed to your function is syntactically valid, i.e. every string in edges[] will contain a valid edge. Here is how you run it on file test1.txt:

Please note that the above output on the left is incorrect because the starter code has an incomplete implementation of the deadlock detector that returns hard-coded values. Once implemented correctly, the output of the program will look like the one on the right. The correct output indicates no deadlock because there is no loop in the graph.

” plato
“plato
“socrates -> fork1”, “socrates <- 2 ”

<- fork1″, -> fork1″, -> 2″,

$ ./deadlock < test1.txt Reading in lines from stdin… Running detect_deadlock()…

edge_index : 6
dl_procs   : [12,7,7]
real time  : 0.0000s

Few more examples:

$ cat test2.txt 5 -> 1
5 <- 3
5 -> 2

7 <- 1
12 <- 2
7 -> 3
$ ./deadlock < test2.txt edge_index : 5

  dl_procs   : [5,7]

$ cat test4.txt 12 -> 1
12 <- 1000
7 -> 1000

  7 <- 1
  2 -> 3
  2 <- 432
  77 -> 432
  77 <- 3

$ ./deadlock < test4.txt edge_index : 3
dl_procs : [12,7]

Limits

You may assume the following limits on input:

$ cat test3.txt p7<-r7 p7->r1 p3->r7 p3<-r1

p12 <- r1
$ ./deadlock < test3.txt edge_index : 3
dl_procs : [p7,p3]

$ cat test5.txt 12->1
12 <- 1000
7 -> 1000 2->3 2<-432

77 -> 432
77<-3
7<-1
$ ./deadlock < test5.txt edge_index : 6

dl_procs   : [2,77]

• Both process and resource names will contain at most 40 alphanumeric characters (digits, upper- and lower-case letters).
• Number of edges will be in the range [0 … 30,000].Your solution should be efficient enough to run on any input within the above limits in less than 10s, which means you should implement an efficient cycle-detection algorithm (see appendix for hints). Remember, you are responsible for designing your own test cases.

  Marking

  Your submission will be marked both on correctness and speed for several test files. To get full marks, your program will need to finish under 10s on all cases.

  About 80% of the marks will be based on tests with less than 10,000 edges, which should be easy to achieve. For example, your program should be able to finish test6.txt under 10s on linuxlab machines.

  The remaining 20% will be awarded only to submissions that can finish under 10s for ~30,000 edges, which is more difficult. You can test your code on the large test7.txt file to see if it is fast enough to get full marks.

Q2 – Scheduler simulation

For this question you will implement a round-robin CPU scheduling simulator for a set of processes and a time slice. Each process will be described by an id, arrival time and CPU burst. Your simulator will simulate RR scheduling on these processes and for each process it will calculate its start time and finish time. Your simulator will also compute a condensed execution sequence of all processes. You will implement your simulator as a function simulate_rr() with the following signature:

The parameter quantum will contain a positive integer describing the length of the time slice and max_seq_len will contain the maximum length of execution order to be reported. The array processes will contain the description of processes, where struct Process is defined in scheduler.h as:

The fields id, arrival_time and burst are the inputs to your simulator. Do not modify these.

You must populate the start_time and finish_time for each process with computed values. You must also report the condensed execution sequence of the processes via the output parameter seq[]. You need to make sure the reported order contains at most the first max_seq_len entries. The entries in seq[] will contain either process ids, or -1 to denote idle CPU.

A condensed execution sequence is similar to a regular execution sequence, except consecutive repeated numbers are condensed to a single value. For example, if a regular sequence was [-1,-1,-1,1,1,2,1,2,2,2], then the condensed equivalent would be [-1,1,2,1,2].

Starter code

Download the starter code and compile it:

You need to implement the simulate_rr() function in scheduler.cpp. The rest of the starter code contains a driver code (main.cpp) and some convenience functions you may use in your implementation (common.cpp). Do not modify any files except scheduler.cpp.

Using the driver on external files

The starter code includes a driver that parses command lines arguments to obtain the time slice and the maximum execution sequence length. It then parses standard input for the description of processes, where each process is specified on a separate line. Each input line contains 2 integers: the first one

CPSC 457: Assignment 4 4

denotes the arrival time of the process, and the second one denotes the CPU burst length. For example, the file test1.txt contains information about 3 processes: P0, P1 and P2:

The 2nd line “3 5” means that process P1 arrives at time 3 and it has a CPU burst of 5 seconds.

Once the driver parses the command line and standard input, it calls your simulator, and then prints out the results. For example, to run your simulator with quantum=3 and max_seq_len=20 on a file test1.txt, you would invoke the driver like this:

Please note that the output above is incorrect, as the starter code contains an incomplete implementation of the scheduling algorithm. The correct results should look like this:

Important – If your simulation detects that an existing process exceeds its time slice at the same time as a new process arrives, you need to insert the existing process into the ready queue before inserting the newly arriving process.

Limits

You may make the following assumptions about the inputs:

• The processes are sorted by their arrival time, in ascending order. Processes arriving at the same time must be inserted into the ready queue in the order they are listed.
• Process IDs will be consecutively numbered starting with 0.
• All processes are 100% CPU-bound, i.e., a process will never be in the waiting state.
• There will be between 0 and 30 processes.

CPSC 457: Assignment 4 5

• Time slice and CPU bursts will be integers in the range [1 … 262]
• Process arrival times will be integers in the range [0 … 262]
• The finish time of every process will fit into int64_t.The git repository includes some test files and the README.md contains several sample results. Please remember to also design your own test data to make sure your program works correctly and efficiently for all of the above limits.

  Marking

  Your submission will be marked both on correctness and speed for a number of different test files. To get full marks, your program will need to finish under 10s on all test cases. About half of the test cases will include inputs with small values for arrival times and CPU bursts. But there will be some test cases with very large arrival times, and/or CPU bursts, and very small time-slices.

  Start with a simulation loop that increments current time by 1. This should make your simulator work fast enough for small arrival times and bursts. Once you are convinced that it works correctly, you can try to improve it to run fast for large burst and arrival numbers. To do this, you will need to determine the optimal value of the amount by which you increment the current time for each simulation loop iteration. There are more detailed hints in the appendix on how this can be achieved.

The field adj_list is a hash table of dynamic arrays, representing an adjacency list. Insert nodes into it so that adj_list[“node”] will contain a list of all nodes with edges pointing towards “node”. The out_counts field is a hash table of integers, representing the number of outgoing edges for every node (outdegrees). Populate it so that out_counts[“node”] contains the number of edges pointing out from “node”.

With these data structures you can implement a simple yet efficient topological sort (pseudo-code):

To get run times under 10s on large files, such as test7.txt, you can use the same algorithm as above, but you will need to switch to more efficient data structures than hash tables. The problem with hash tables is they spend considerable time on calculating hashes on strings, and also on collisions. To avoid this overhead, you can pre-convert all strings (process and resource names) into consecutive integers, and then use fast dynamic arrays instead of hash tables to store the adjacency list and outdegree counts:

You can use the provided Word2Int class to help you with converting strings to unique consecutive integers. The topological sort algorithm would remain the same as above. Do not forget to convert the integers back to strings at the end, to correctly populate dl_procs[].

Appendix 2 – hints for Q2

Here are some suggestions for keeping track of the current state of the simulation:

• The current time
o A 64-bit integer should do the trick, e.g. int64_t curr_time =0;

• The remaining time slice
CPSC 457: Assignment 4 7

o e.g. int64_t remaining_slice; • Currently executing process

o an index into processes[] should suffice, where ‘-1’ could represent idle CPU

o e.g. int cpu = -1;

• Contents of the Ready Queue (RQ) and Job Queue (JQ)o I suggest std::vector of indices into processes[]

  o e.g.std::vector<int>rq,jq;

• You also probably need to keep track of the remaining bursts for all processes

o e.g. std::vector<int64_t> remaining_bursts;
I suggest you start with an implementation where you increment current time by 1. Once you are done

with this basic implementation, you can implement additional optimizations. Here are some ideas for simple optimizations:

• see if you can adjust your simulation loop so that at the top of the loop, your CPU is always idle;
• if you start executing a process, you can execute full time slice, just make sure that before you put the process back into RQ, you check to see if any processes arrived during that time and ifthey did, put them into RQ before
• any time you start executing a process, and the process will finish before it exceeds the quantum,you can finish the process and advance current time to the end time of the process. If any

  processes arrive during this time, insert those into RQ.

• if you are executing last process (i.e., RQ is empty, and JQ is empty), then you can finishexecuting the current process, and end simulation.
• if RQ is empty and CPU is idle, then you can skip current time to the arrival of the next process.
• if RQ is empty but CPU has a process, then you can execute multiple time slices as long as youdon’t increment current time past the next arrival time.

  The hardest part is to optimize the case when CPU is idle, but RQ is not empty, and there are processes still arriving in the future. This is especially needed when the time slice is tiny but arrival times and/or burst times are very large. Here are some hints for an optimization that you can do after you have already populated the execution sequence, and after every process in RQ has a recorded start time:

• ask yourself this question: could you execute one quantum for every process in RQ without any processes finishing, and without going past the arrival time of the next process? If you could, that would not change the order of processes in the RQ, and you would not miss the end time for any processes, and you would not miss arrival time for any process. In other words, you could increment current time by rq.size()*quantum without changing the state of the system, as long as you update the remaining time for each process in the RQ by subtracting quantum.
• next question is: how many times could you do the above step? i.e. what is the maximum safe N such that you can increment current time by N*rq.size()*quantum? The N needs to be as big as possible, but small enough that no processes will finish during this time, and no processes will arrive during this time.