Consider 7-bit messages that we encode into 11-bit codewords using 4 parity bits. Thus our codewords look like: ⟨a, b, c, d, e, f, g, p1, p2, p3, p4⟩ where a, b, c, d, e, f, g are the original 7-bits of the message, and p1, p2, p3, p4 are 4 parity bits. Recall the two properties from Tutorial 2: (0) each bit of the original message appears in at least two parity bits, and (1) no two bits of the original message appear in exactly the same set of parity bits. For the following proposed parity bits, select whether they satisfy property (0), property (1), both, or neither. p1 = a ⊕ d ⊕ e p2 = a ⊕ b ⊕ e ⊕ f p3 = b ⊕ c ⊕ f ⊕ g p4 = c ⊕ d ⊕ g   Question 1 options: a neither property b both properties c only property (0) d only property (1)

Suppose we are working with an error-correcting code that will allow all single-bit errors to be corrected for memory words of length 7. We have already calculated that we need 4 check bits, and the length of all code words will be 11. Code words are created according to the Hamming algorithm presented in the text. We now receive the following code word: 1 0 1 0 1 0 1 0 1 1 0  Assuming odd parity, is this a legal code word? If not, according to our error-correcting code, where is the error?

In this problem we will work through a round of DES. For notational simplicity, assume it is the first round. Please enter all answers as strings of 0's and 1's. The input is the 64 bit block 0000000000000010000000000000001000000100000001000100000000000000 Suppose that the subkey for the current round is this 48 bit number: 000000000000000001000000000000000000000000000000 What does the 64 bit state look like after the IP transformation is applied to the input? 00000000000110000100000000000000 Now find Lo and Ro, the left and right halves of the state. Lo 01000000010000000000000000010000 = Ro=00000000000000100000000000000000 What is the result of applying the expansion box to R₁? E(Ro) = 00000000000000100000000000000000 What is the result of XORing the subkey with E(Ro)? k₁ E(Ro) = 00000000000000100000000000000000 We now apply the S-box transformation. S(k₁ E(Ro)) = 11011000110110001101111110111100 Finally we apply the permutation box to complete the function f. f(Ro) = P(S(k₁ + E(Ro)))…

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 )…