{"id":1280,"date":"2024-01-05T03:24:36","date_gmt":"2024-01-05T03:24:36","guid":{"rendered":"https:\/\/www.writemyessays.app\/blog\/questions\/use-information-from-the-modular-background-readings-as-well-as-the-given-resources-also-you-could-use-any-good-quality-resource-you-can-find-please-cite-all-sources-and-provide-a-reference-list-at\/"},"modified":"2024-01-05T03:24:36","modified_gmt":"2024-01-05T03:24:36","slug":"use-information-from-the-modular-background-readings-as-well-as-the-given-resources-also-you-could-use-any-good-quality-resource-you-can-find-please-cite-all-sources-and-provide-a-reference-list-at","status":"publish","type":"questions","link":"https:\/\/www.writemyessays.app\/blog\/questions\/use-information-from-the-modular-background-readings-as-well-as-the-given-resources-also-you-could-use-any-good-quality-resource-you-can-find-please-cite-all-sources-and-provide-a-reference-list-at\/","title":{"rendered":"Use information from the modular background readings as well as the given resources. Also, you could use any good quality resource you can find. Please cite all sources and provide a reference list at the end of your paper."},"content":{"rendered":"

Use information from the modular background readings as well as the given resources. Also, you could use any good quality resource you can find. Please cite all sources and provide a reference list at the end of your paper.<\/span><\/p>\n


<\/span><\/div>\n
<\/p>\n

The following items will be assessed in particular:<\/p>\n

    \n
  1. Your ability to consolidate ideas from reading materials and your understanding of the materials.<\/li>\n
  2. Your ability to write a report with strong argument.<\/li>\n
  3. Some in-text references to modular background readings.<\/li>\n<\/ol>\n
    <\/div>\n
    \n

    Data Encryption Standards: DES and Triple DES (Part I)<\/strong><\/p>\n

    DES Encryption<\/strong><\/p>\n

    The overall scheme for DES encryption is illustrated in Figure 1. As with any encryption scheme, there are two inputs to the encryption function: the plaintext to be encrypted and the key. In this case, the plaintext must be 64 bits in length and the key is 56 bits in length.<\/p>\n

    <\/p>\n

    Looking at the left-hand side of the figure, we can see that the processing of the plaintext proceeds in three phases. First, the 64-bit plaintext passes through an initial permutation (IP) that rearranges the bits to produce the permuted input. This is followed by a phase consisting of sixteen rounds of the same function, which involves both permutation and substitution functions. The output of the last (sixteenth) round consists of 64 bits that are a function of the input plaintext and the key. The left and right halves of the output are swapped to produce the preoutput. Finally, the preoutput is passed through a permutation [IP -1] that is the inverse of the initial permutation function, to produce the 64-bit ciphertext.<\/p>\n

    The right-hand portion of Figure 1 shows the way in which the 56-bit key is used. Initially, the key is passed through a permutation function. Then, for each of the sixteen rounds, a subkey (Ki) is produced by the combination of a left circular shift and a permutation. The permutation function is the same for each round, but a different subkey is produced because of the repeated shifts of the key bits.<\/p>\n

    DES works on bits, or binary numbers\u2014the 0s and 1s common to digital computers. Each group of four bits makes up a hexadecimal, or base 16, number. Binary “0001” is equal to the hexadecimal number “1”, binary “1000” is equal to the hexadecimal number “8”, “1001” is equal to the hexadecimal number “9”, “1010” is equal to the hexadecimal number “A”, and “1111” is equal to the hexadecimal number “F”.<\/p>\n

    DES works by encrypting groups of 64 message bits, which is the same as 16 hexadecimal numbers. To do the encryption, DES uses “keys” which are also apparently<\/i> 16 hexadecimal numbers long, or apparently<\/i> 64 bits long. However, every 8th key bit is ignored in the DES algorithm, so that the effective key size is 56 bits. But, in any case, 64 bits (16 hexadecimal digits) is the round number upon which DES is organized.<\/p>\n

    For example, if we take the plaintext message “8787878787878787”, and encrypt it with the DES key “0E329232EA6D0D73”, we end up with the ciphertext “0000000000000000”. If the ciphertext is decrypted with the same secret DES key “0E329232EA6D0D73”, the result is the original plaintext “8787878787878787”.<\/p>\n

    A DES Example<\/b><\/p>\n

    We now work through an example and consider some of its implications. Although you are not expected to duplicate the example by hand, you will find it informative to study the hex patterns that occur from one step to the next.<\/p>\n

    DES is a block cipher <\/span><\/i><\/b>— meaning it operates on plaintext blocks of a given size (64-bits) and returns ciphertext blocks of the same size. Thus DES results in a permutation<\/i><\/b> among the 2^64 (read this as: “2 to the 64th power”) possible arrangements of 64 bits, each of which may be either 0 or 1. Each block of 64 bits is divided into two blocks of 32 bits each, a left half block L <\/span><\/b>and a right half R<\/b>. (This division is only used in certain operations.)<\/p>\n

    Example:<\/b> Let M<\/b> be the plain text message M<\/b> = 0123456789ABCDEF, where M<\/b> is in hexadecimal (base 16) format. Rewriting M<\/b> in binary format, we get the 64-bit block of text:<\/p>\n

    M<\/b> = 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111
    L<\/b> = 0000 0001 0010 0011 0100 0101 0110 0111
    R<\/b> = 1000 1001 1010 1011 1100 1101 1110 1111<\/p>\n

    The first bit of M<\/b> is “0”. The last bit is “1”. We read from left to right.<\/p>\n

    DES operates on the 64-bit blocks using key<\/i> sizes of 56-bits. The keys are actually stored as being 64-bits long, but every 8th bit in the key is not used (i.e., bits numbered 8, 16, 24, 32, 40, 48, 56, and 64). However, we will nevertheless number the bits from 1 to 64, going left to right, in the following calculations. But, as you will see, the eight bits just mentioned get eliminated when we create subkeys.<\/p>\n

    Example:<\/b> Let K<\/b> be the hexadecimal key K<\/b> = 133457799BBCDFF1. This gives us as the binary key (setting 1 = 0001, 3 = 0011, etc., and grouping together every eight bits, of which the last one in each group will be unused):<\/p>\n

    K<\/b> = 00010011 00110100 01010111 01111001 10011011 10111100 11011111 11110001<\/p>\n

    The DES algorithm uses the following steps:<\/p>\n

    Step 1: Create 16 subkeys, each of which is 48-bits long.<\/strong><\/p>\n

    The 64-bit key is permuted according to the following table, PC-1<\/b>. Since the first entry in the table is “57”, this means that the 57th bit of the original key K<\/b> becomes the first bit of the permuted key K<\/b>+. The 49th bit of the original key becomes the second bit of the permuted key. The 4th bit of the original key is the last bit of the permuted key. Note only 56 bits of the original key appear in the permuted key.<\/p>\n

    <\/p>\n

    Example:<\/b> From the original 64-bit key<\/p>\n

    K<\/b> = 00010011 00110100 01010111 01111001 10011011 10111100 11011111 11110001<\/p>\n

    we get the 56-bit permutation<\/p>\n

    K<\/b>+ = 1111000 0110011 0010101 0101111 0101010 1011001 1001111 0001111<\/p>\n

    Next, split this key into left and right halves, C0<\/sub><\/i><\/b> and D0<\/sub><\/i><\/b>, where each half has 28 bits.<\/p>\n

    Example:<\/b> From the permuted key K<\/b>+, we get<\/p>\n

    C0<\/sub><\/i><\/b> = 1111000 0110011 0010101 0101111 
    D0<\/sub><\/i><\/b> = 0101010 1011001 1001111 0001111<\/p>\n

    With C0<\/sub><\/i><\/b> and D0<\/sub><\/i><\/b> defined, we now create sixteen blocks Cn<\/sub><\/i><\/b> and Dn<\/sub><\/i><\/b>, 1<=n<\/i><\/b><=16. Each pair of blocks Cn<\/sub><\/i><\/b> and Dn<\/sub><\/i><\/b> is formed from the previous pair Cn-1<\/sub><\/i><\/b> and Dn-1<\/sub><\/i><\/b>, respectively, for n<\/i><\/b> = 1, 2, …, 16, using the following schedule of “left shifts” of the previous block. To do a left shift, move each bit one place to the left, except for the first bit, which is cycled to the end of the block<\/p>\n

    This means, for example, C3<\/sub><\/i><\/b> and D3<\/sub><\/i><\/b> are obtained from C2<\/sub><\/i><\/b> and D2<\/sub><\/i><\/b>, respectively, by two left shifts, and C16<\/sub><\/i><\/b> and D16<\/sub><\/i><\/b> are obtained from C15<\/sub><\/i><\/b> and D15<\/sub><\/i><\/b>, respectively, by one left shift. In all cases, by a single left shift is meant a rotation of the bits one place to the left, so that after one left shift the bits in the 28 positions are the bits that were previously in positions 2, 3,…, 28, 1.<\/p>\n

    Example:<\/b> From original pair <\/span>C0<\/sub><\/i><\/b> and D0<\/sub><\/i><\/b> we obtain:<\/p>\n

    C0<\/sub><\/i><\/b> = 1111000011001100101010101111
    D0<\/sub><\/i><\/b> = 0101010101100110011110001111<\/p>\n

    C1<\/sub><\/i><\/b> = 1110000110011001010101011111
    D1<\/sub><\/i><\/b> = 1010101011001100111100011110<\/p>\n

    C2<\/sub><\/i><\/b> = 1100001100110010101010111111
    D2<\/sub><\/i><\/b> = 0101010110011001111000111101<\/p>\n

    C3<\/sub><\/i><\/b> = 0000110011001010101011111111
    D3<\/sub><\/i><\/b> = 0101011001100111100011110101<\/p>\n

    C4<\/sub><\/i><\/b> = 0011001100101010101111111100
    D4<\/sub><\/i><\/b> = 0101100110011110001111010101<\/p>\n

    C5<\/sub><\/i><\/b> = 1100110010101010111111110000
    D5<\/sub><\/i><\/b> = 0110011001111000111101010101<\/p>\n

    C6<\/sub><\/i><\/b> = 0011001010101011111111000011
    D6<\/sub><\/i><\/b> = 1001100111100011110101010101<\/p>\n

    C7<\/sub><\/i><\/b> = 1100101010101111111100001100
    D7<\/sub><\/i><\/b> = 0110011110001111010101010110<\/p>\n

    C8<\/sub><\/i><\/b> = 0010101010111111110000110011
    D8<\/sub><\/i><\/b> = 1001111000111101010101011001<\/p>\n

    C9<\/sub><\/i><\/b> = 0101010101111111100001100110
    D9<\/sub><\/i><\/b> = 0011110001111010101010110011<\/p>\n

    C10<\/sub><\/i><\/b> = 0101010111111110000110011001
    D10<\/sub><\/i><\/b> = 1111000111101010101011001100<\/p>\n

    C11<\/sub><\/i><\/b> = 0101011111111000011001100101
    D11<\/sub><\/i><\/b> = 1100011110101010101100110011<\/p>\n

    C12<\/sub><\/i><\/b> = 0101111111100001100110010101
    D12<\/sub><\/i><\/b> = 0001111010101010110011001111<\/p>\n

    C13<\/sub><\/i><\/b> = 0111111110000110011001010101
    D13<\/sub><\/i><\/b> = 0111101010101011001100111100<\/p>\n

    C14<\/sub><\/i><\/b> = 1111111000011001100101010101
    D14<\/sub><\/i><\/b> = 1110101010101100110011110001<\/p>\n

    C15<\/sub><\/i><\/b> = 1111100001100110010101010111
    D15<\/sub><\/i><\/b> = 1010101010110011001111000111<\/p>\n

    C16<\/sub><\/i><\/b> = 1111000011001100101010101111
    D16<\/sub><\/i><\/b> = 0101010101100110011110001111<\/p>\n

    We now form the keys Kn<\/sub><\/i><\/b>, for 1<=n<\/i><\/b><=16, by applying the following permutation table to each of the concatenated pairs <\/span>Cn<\/sub>Dn<\/sub><\/i><\/b>. Each pair has 56 bits, but PC-2<\/b> only uses 48 of these.<\/p>\n<\/div>\n

    <\/span><\/div>\n","protected":false},"excerpt":{"rendered":"

    Use information from the modular background readings as well as the given resources. Also, you could use any good quality resource you can find. Please cite all sources and provide a reference list at the end of your paper. The following items will be assessed in particular: Your ability to consolidate ideas from reading materials […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"closed","template":"","meta":[],"disciplines":[63],"paper_types":[],"tagged":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.writemyessays.app\/blog\/wp-json\/wp\/v2\/questions\/1280"}],"collection":[{"href":"https:\/\/www.writemyessays.app\/blog\/wp-json\/wp\/v2\/questions"}],"about":[{"href":"https:\/\/www.writemyessays.app\/blog\/wp-json\/wp\/v2\/types\/questions"}],"author":[{"embeddable":true,"href":"https:\/\/www.writemyessays.app\/blog\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/www.writemyessays.app\/blog\/wp-json\/wp\/v2\/comments?post=1280"}],"version-history":[{"count":0,"href":"https:\/\/www.writemyessays.app\/blog\/wp-json\/wp\/v2\/questions\/1280\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.writemyessays.app\/blog\/wp-json\/wp\/v2\/media?parent=1280"}],"wp:term":[{"taxonomy":"disciplines","embeddable":true,"href":"https:\/\/www.writemyessays.app\/blog\/wp-json\/wp\/v2\/disciplines?post=1280"},{"taxonomy":"paper_types","embeddable":true,"href":"https:\/\/www.writemyessays.app\/blog\/wp-json\/wp\/v2\/paper_types?post=1280"},{"taxonomy":"tagged","embeddable":true,"href":"https:\/\/www.writemyessays.app\/blog\/wp-json\/wp\/v2\/tagged?post=1280"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}