Suppose Alice and Bob are going to communicate using AES in CBC mode. Unfortunately Alice's message length (in bytes) is not a multiple of 16. Suppose the last block of her message is just a single zero byte. How can she pad out the last block so that she can use CBC mode? Since this needs to be a reversible operation, how does Bob recognize the padding and remove it?

Suppose Alice and Bob are going to communicate using AES in CBC mode. Unfortunately Alice's message length (in bytes) is not a multiple of 16. Suppose the last block of her message is just a single zero byte. How can she pad out the last block so that she can use CBC mode? Since this needs to be a reversible operation, how does Bob recognize the padding and remove it?

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter8: Data And Network Communication Technology

Section: Chapter Questions

Problem 41VE

See similar textbooks

Similar questions

When Bob wants to send Alice a message M on the Internet, he breaks M into n data packets, numbers the packets consecutively, and injects them into the network. When the packets arrive at Alice’s computer, they may be out of order, so Alice must assemble the sequence of n packets in order before she can be sure she has the entire message. Describe an efficient scheme for Alice to do this. What is the running time of this algorithm? From Data Structures and Algorithms C-7.56
We use DES in cipher feedback mode (CFB) to encrypt a plaintext m = m1m2 ...m100 into a ciphertext c1c2 ...c100, where each mi is 8-bit long. The ciphertext is sent to Bob. If c15 and c25 are missing and c8 and c88 are received as c8' and c88' wrongly, what mi’s can B compute correctly from the received ciphertext?
Alice and Bob decided to use HMAC to authenticate the communication between them. They already share a key k for the MAC scheme. To sign a message M (which is always of an even bit length), the sender parses the message as M=M_L||M_R, i.e. the message is parsed as a left half and a right half (both halves have equal length), then the sender computes X-M₁MR (where denotes XOR). The sender then sends M and T=HMACK(X) to the other party. Does this construction provide existential unforgeability? Yes No
The token bucket algorithm is a popular method for traffic shaping (recapthe algorithm in textbook Ch. 5.3.2). A computer uses a token bucket with a capacity of 400megabytes (MB), and a rate of 10 MB per second. The machine starts generating 20 MB persecond continuously when the bucket contains 200 MB. How long will it take to send 1000 MB?
A messaging system reads message as 4-digit positive integer d1d2d3d4 and stores encoding of the message e1e2e3e4 for security reason. The encoding process is as follows: ei = (di+i) mod 3 For example, if message is 4590, the encoded message is 2101. Read any arbitrary message and display the encoded message in the following format: Original message = Encoded message = language : C
Create a MIPS assembly program using MARS to do RSA encryption/decryption. Get a numeric message from the user and encrypt it. Also, get a numeric message from the user and decrypt it. An RSA public key consists of a pair of numbers (n, e), and the message m is encrypted to a ciphertext c by calculating c = me (mod n). The RSA private key consists of a pair of numbers (n, d), and a ciphertext c is decrypted by calculating m = cd (mod n). For this, use n=21733, e=257, d=1403. Note that since me and cd can get rather large (hundreds or thousands of digits for our examples), use the following pseudocode to handle encryption and decryption: doRSA( n, exp, msg ) cmsg = msg Initialize ciphertext, requires odd exponent mpow = msg message to the power 1, 2, 4, 8, 16, … tempexp = exp temporary exponent while ( tempexp > 1 ) mpow = ( mpow * mpow ) % n square the message tempexp = tempexp / 2 update the power if ( tempexp % 2 == 1 ) 1 bit in the binary representation of e? cmsg = ( cmsg…
Please solve it correctly and please provide explanation of your answers. Please answer parts g, h and i. A PC and a Web Server are communicating over a TCP connection. The PC had started the three way handshake with the initial sequence number of 3069 . The Web Server's initial sequence number is 4830 . The window size of the PC is 815 bytes and the window size of the Web Server is 463 bytes. Using the third TCP handshake ack segment the PC sends the http request of the size 396 bytes to the Web Server. Then the Web Server answers with 3 segments containing the requested data. The first segment size is 42 bytes and the second segment size is 276 bytes and the third segment size is 146 bytes respectively. The PC receives all three segments within the timer. But unfortunately the second segment was corrupted. So the PC immediately sends an acknowledgement segment. Assume that the PC uses Selective Repeat/Reject ARQ. g) The second segment from webserver was corrupted. So the PC…
Used C PLUS PLUS LANGUAGE. all instructions are given in question. Kindly follow it. When a message is transmitted in secret code over a transmission channel, it is usually sent as a sequence of bits, that is, 0s and 1s. Due to noise in the transmission channel, the transmitted message may become corrupted. That is, the message received at the destination is not the same as the message transmitted; some of the bits may have been changed. There are several techniques to check the validity of the transmitted message at the destination. One technique is to transmit the same message twice. At the destination, both copies of the message are compared bit by bit. If the corresponding bits are the same, the message received is error-free. Response Required: Write a program to check whether the message received at the destination is error-free. For simplicity, assume that the secret code representing the message is a sequence of digits (0 to 9) and the maximum length of the message is 250…
The function f(R, K) = P(S(E(R XOR K)) of the DES cryptographic protocol manipulates different sizes of bit-strings. The bit-string sizes of P-SEK in this order are: 32-48-48-32 32-32-48 48 48-48-32-32 64-32-32 48 64 48 32-64 SAVE
Computer A uses the Go-back-N ARQ protocol to send a 110 Mbytes file to computer B with a window size of 15. Given each frame carries 100K bytes data. How long does it take to send the whole file (the total time taken from A sending the first bit of the file until A receiving the last acknowledgment)? Given that the transmission rate of the link is 500 Mbps and the propagation time between A and B is 15ms. Assume no data or control frame is lost or damaged and ignore the overhead due to header and trailer.
Next time you snap a picture, you realize you are capturing millions of pixels into a buffer. The buffer data is read and converted into JPEG in real time. Each pixel in that buffer is an unsigned int (four bytes ): Alpha, Blue, Green, and Red. Let us ignore alpha for now. As you know, a unsigned byte can have a value 0 to 255. In remote sensing jargons, it is called blue channel, green channel, and red channel. Each channel provides valuable information such as, say farm lands, forest fire, drought, landscape, diseases , If a pixel has a value (say in hex) = 0x00a1b1c1 , then 0xc1 is the red pixel, 0xb1 is the green pixel, 0xa1 is the blue and 00 is the alpha. #define RED 1 #define GREEN 2 #define BLUE 3 then, develop a function void calculateSum ( unsigned int *ptr , int count , unsigned char channel , unsigned int *sum, float *average ) { *sum = 0; if (channel == RED ) calculate sum and average for red channel else if (channel == GREEN )…
Please solve it correctly and please provide explanation of your answers. A PC and a Web Server are communicating over a TCP connection. The PC had started the three way handshake with the initial sequence number of 3069 . The Web Server's initial sequence number is 4830 . The window size of the PC is 815 bytes and the window size of the Web Server is 463 bytes. Using the third TCP handshake ack segment the PC sends the http request of the size 396 bytes to the Web Server. Then the Web Server answers with 3 segments containing the requested data. The first segment size is 42 bytes and the second segment size is 276 bytes and the third segment size is 146 bytes respectively. The PC receives all three segments within the timer. But unfortunately the second segment was corrupted. So the PC immediately sends an acknowledgement segment. Assume that the PC uses Selective Repeat/Reject ARQ. (d) What will be the window size of PC A after receiving only the first segment from webserver? (e)…