Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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Now unmount the filesystem.
Create a sparse image file, called book.img, of the 2GiB hard drive.
Use a hexdump program to look at book.img and work out the offsets that store the contents of the
book youdownloaded. Record the offsets as a sequence of one or more ranges of hexadecimal
numbers. For example, in the following hexdump the text The rain in Spain falls mainly on the
plain. is stored at 0x10-0x1f, 0x30-0x4b (inclusive).
00000000: 0000 0000 0000 0000 0000 0000 0000 0000 .................00000010: 5468 6520 7261
696e 2069 6e20 5370 6169 The rain in Spai00000020: 0000 0000 0000 0000 0000 0000 0000
0000.................00000030: 6e20 6661 6c6c 7320 6d61 696e 6c79 206f n falls mainly 000000040:
6e20 7468 6520 706c 6169 6e2e 0000 0000 n the plain........
Compress the filesystem using gzip by running:
#gzip book.img
This should leave you with a file book.img.gz. You will use scp to transfer this file in a later task.
Enter your ranges in answers2.json as the answer for question1. To…
Task 4: Examine floppy.img
Mount floppy.img at a suitable mountpoint inside your VM.
Look inside the mountpoint directory. You should see a number of files in the top-level directory.
Look inside your answers2.json at the strings labelled contentA and contentB. Exactly one of those two strings
appears as the content of a file in the floppy disk's filesystem. I.e., there is a file that contains contentA or a file
that contains contentB but not both.
Additionally, the other string has been written into unused space of the floppy disk, so the data is in floppy.img
but cannot be seen inside any of the files at the mountpoint. However it can be seen in a hexdump of floppy.img.
I.e., if there is a file that contains contentA then contentB has been written into unused space.
You must search through your mountpoint directory to find out which of contentA and contentB is stored in a file
of the filesystem and which is stored in unused space. Suppose that contentA is stored in a file, and…
Download your personalized assignment files, answers2.json and floppy.img, into your Kali Linux Vm. Use
wget as before. For example, you'd use "wget
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- It is possible to sort an array of n values using pipeline of n filter processes.The first process inputs all the values one at a time, keep the minimum, and passes the others on to the next process. Each filter does the same thing; it receives a stream of values from the previous process, keep the smallest, and passes the others to the next process. Assume each process has local storage for only two values--- the next input value and the minimum it has seen so far. (a) Developcode for filter processes. Declare the channels and use asynchronous message passing. Hint:Define an array of channels value[n] (int), and a set of filter processes Filter[i = 0 ton-1]. Each process Filter[i] (where 0 <= i <= n-2) receives a stream of integers through channelvalue[i], keeps the smallest, and sends all other integers to channel value[i+1]. The last processFilter[n-1] receives only one integer through channel value[n-1] and does not need to send anyinteger further.arrow_forwardI need help: Challenge: Assume that the assigned network addresses are correct. Can you deduce (guess) what the network subnet masks are? Explain while providing subnet mask bits for each subnet mask. [Hint: Look at the addresses in binary and consider the host ids]arrow_forwardI would like to know if my answer statment is correct? My answer: The main difference is how routes are created and maintained across different networks. Static routing establishes router connections to different networks from the far left and far right. The Dynamic routing focus emphasizes immediate connection within the router while ignoring the other connections from different networks. Furthermore, the static routing uses the subnet mask to define networks such as 25.0.0.0/8, 129.60.0.0/16, and 200.100.10.0/30, which correspond to 255.0.0.0, 255.255.0.0, and 255.255.255.252. On the other hand, dynamic routing uses the wildcard mask to inverse the subnet mask, where network bits become 0 and host bits become 1, giving us 0.0.0.255, 0.0.255.255, and 0.0.0.3. Most importantly, the CLI commands used for Static and Dynamic routing are also different. For static routing, the “IP route” corresponds with the network, subnet mask, and next-hop IP address. In contrast, dynamic routing uses…arrow_forward
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