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    zareenZ
    @Muhammad-Arif https://cyberian.pk/topic/1116/eng101-assignment-2-solution-and-discussion/9
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    zareenZ
    Q.1[image: AmOsgwE.png] Q.2 [image: tS465OP.png]
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    [image: dlR77xh.png] [image: DIUIHJx.png]
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    zareenZ
    https://youtu.be/VSHrNQftyHQ
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    zareenZ
    Answer to Q#1 Modern transponders can carry many different types of communications traffic. They can also receive signals from multiple ground stations, combining (multiplexing) or splitting (de-multiplexing) them for onwards transmission to other multiple ground stations. This method, by which many users share a common satellite resource, is called Multiple Access. There are several schemes for accomplishing this, each with its benefits and drawbacks. TDMA (Time Division Multiple Access) allocates a time slot to the user in a repetitive time frame. The signal is digitized and the data bits are stored in a buffer in a compressed time frame until their allocated time slot comes around when they are transmitted during their allocated time. At the receiver end of the link the bits are rearranged, spreading them out to reassemble the original digital signal and converted back to analogue form. The signal occupies the entire transponder bandwidth, but only during its allocated time slot. The rest of the time the bandwidth is available to other users. Digital signals typically have better noise immunity than analogue signals. FDMA (Frequency Division Multiple Access) shares the bandwidth between the users, with each user allocated a unique, narrower section of the available bandwidth. It works with analogue signals and all users have uninterrupted use of their own narrow frequency band or channel with all users occupying the available bandwidth simultaneously, each within their own narrow channel. The sender’s signal, called the baseband signal, is frequency shifted into the allocated frequency band for transmission and the receiver restores it back to the baseband. CDMA (Code Division Multiple Access) also known a Spread Spectrum, modulates the user’s signal with a pseudorandom code so that it occupies the full available spectrum, appearing as noise. The receiver uses the same pseudorandom code in an auto correlator device, which only recognizes a signal modulated with the same auto code and thus separates it from the noise. CDMA is more complex but has better noise immunity and provides greater security than the other two systems. Answer to Q#2 • Mobile Service Satellites (MSS) • Example: Satellite Phones
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    A
    @zaasmi thank you so much sir
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    zareenZ
    Q. 1 Solution Idea Devices Visibility Consistency Constraints Affordance Keyboard Keyboard design is a good design all the Alphabet and numeric are in different section also the special function key are arranged in top of the keyboard, specially delete button is very good deployed at right top, which cannot be accidentally pressed during any typing. Keyboard has strict constraint i.e if we want to used special character which are on numeric buttons these cannot be activated without pressing the shift button. Or some button on keyboard especially on laptop is only work with pressing “fn” key. Atm machine ATM has good visibility because it have a limited amount of button on his left and right side and only some limited amount of processing screen. ATM has low consistence because there is no similar function to perform. ATM machine have strongly constraint because without the limited amount of action the individual can nothing to do. ATM Machine have no good affordance because naive users have facing a lot of difficulty while using the ATM. Digital watch Digital watch have no good visibility because its screen is not visible in full sun light and in darken area. It is not consistence device because every button has its function, we cannot perform similar task with different button or method. Digital watch has less affordance because its button have no label, so the naïve user can not configure the time or date. Digital camera It has consistence function because every time we press the button the same work will be done against mentioned button. Good constraint because a person cannot capture video except listed format. Also we cannot change function or setting of camera. Less affordance because the new user can used Digital camera with facing some difficulties. Q.2 Solution Idea: [image: pK2ZpIj.png]
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    zareenZ
    Q1. Write a subroutine that will find the first even number from an array of your VU ID and calculate its factorial. (10 Marks) Note: Skip 0’s in your VU ID as shown below, VU ID: BC190206435 After skipping 0’s, array would be: Array: 1, 9, 2, 6, 4, 3, 5 The first even number is 2 in the array so its factorial will be calculated and saved in AX register. Solution: [org 0x100] jmp start data: dw 1, 9,2,6,4,3,5 ;initiliazation checkEven: mov dl,0002h mov ax,[data+bx] mov cx,[data+bx] xor ah,ah xor ch,ch div dl cmp ah,00h ;checking remainder je fact add bx,2 jmp checkEven fact: mov ax,0001 mov dx,0000 mult: mul cx loop mult ret start: mov si,00h mov bx,si call checkEven mov ax,0x4c00 int 0x21 Q2. Write a code in assembly language (using appropriate jumps) equivalent to this given code in C. (10 Marks) #include <stdio.h> int main() { int n1=1, n2=2; int largest; if( n1>n2) largest=n1; else largest = n2; return 0; } Solution: Assembly code: [org 0x100] jmp start n1 db 1 n2 db 2 largest db 0 start: mov ax, [n1] xor ah,ah mov bx, [n2] cmp ax,bx ja large mov [largest],bx large: mov [largest],ax mov ax, 0x4c00 int 21h
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    zareenZ
    [ORG 0X0100] start: mov ax, 0x000D int 0x10 mov ax, 0x0C02 xor bx, bx mov cx, 100 mov dx, 60 l1: int 0x10 inc cx cmp cx, 200 jne l1 mov cx, 100 mov dx, 60 l2: int 0x10 mov ax, 0x0C05 inc dx cmp dx, 120 jne l2 mov cx, 200 mov dx, 60 l3: int 0x10 mov ax, 0x0C08 inc dx cmp dx, 120 jne l3 mov cx, 100 mov dx, 120 l4: int 10h mov ax, 0x0C09 inc cx cmp cx, 200 jne l4 mov ah, 0 int 0x16 mov ax, 0x0003 int 0x10 mov ax, 0x4C00 int 0x21
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    zaasmiZ
    @aleem1990 said in CS205 Assignment 3 Solution and Discussion: Assignment No. 03 Semester: Fall 2019 CS205: Information Security Total Marks: 25 Due Date: 22/01/2020 Instructions: Please read the following instructions carefully before submitting assignment: You need to use MS word document to prepare and submit the assignment on VU-LMS. It should be clear that your assignment will not get any credit if:  The assignment is submitted after due date.  The assignment is not in the required format (doc or docx)  The submitted assignment does not open or file is corrupt.  Assignment is copied (partial or full) from any source (websites, forums, students, etc) Objectives: To enhance the learning capabilities of the students about cryptography and crypto algorithms. Assignment Question 1: You are required to encrypt the string “Security Professional” and decrypt the string “MdgfjVhjfbteut” using following tables with “Substitution Cipher Method”. [image: 2jpMsDs.png] Question 2: You are required to encrypt the plain text “ACCOUNTABILITY” using Vigenere Square Cipher method? Suppose that keyword is “SILENT”. Question 3: You are required to encrypt the plain text “wearesecurityprofessionalsinuk” into cipher text using columnar transposition method with a Key: “231645”. Best of luck Any Frnd have CS205 Solved 3rd Assignment. Kindly Share it urgent Here are the step-by-step solutions for your cryptography questions. Question 1: Substitution Cipher To solve this, we map each letter of the plaintext to its corresponding character in the provided substitution table. (Note: Since you didn’t provide the specific mapping tables, I have demonstrated the method using a standard Caesar-style shift or a placeholder logic. Please apply your specific table values to this process). Encryption (“Security Professional”): Take each letter (S-e-c-u-r-i-t-y…). Look up ‘S’ in the “Plaintext” row of your table. Write down the character directly below it in the “Cipher” row. Decryption (“MdgfjVhjfbteut”): Look up ‘M’ in the “Cipher” row. Write down the character directly above it in the “Plaintext” row. Question 2: Vigenere Square Cipher Plaintext: ACCOUNTABILITY Keyword: SILENT Step 1: Align the Key Repeat the keyword until it matches the length of the plaintext: Plaintext: A C C O U N T A B I L I T Y Keyword: S I L E N T S I L E N T S I Step 2: Intersection Logic Using a Vigenere Tableau, find the intersection of the Plaintext letter (column) and Keyword letter (row). Plain (P) Key (K) Calculation Ciphertext A (0) S (18) S C (2) I (8) K C (2) L (11) N O (14) E (4) S U (20) N (13) H N (13) T (19) G T (19) S (18) L A (0) I (8) I B (1) L (11) M I (8) E (4) M L (11) N (13) Y I (8) T (19) B T (19) S (18) L Y (24) I (8) G Final Ciphertext: SKNSHGLIMMYBLG Question 3: Columnar Transposition Plaintext: wearesecurityprofessionalsinuk Key: 231645 Step 1: Arrange into Grid The key has 6 digits, so we arrange the text in rows of 6. 2 3 1 6 4 5 w e a r e s e c u r i t y p r o f e s s i o n a l s i n u k Step 2: Read Columns by Key Order Read the columns based on the numerical order of the key (1, 2, 3, 4, 5, 6). Column 1 (Key 1): AURII Column 2 (Key 2): WEYSL Column 3 (Key 3): ECPSS Column 4 (Key 4): EIFNU Column 5 (Key 5): STEAK Column 6 (Key 6): RROON Final Ciphertext: AURIW EY SLE CPSS EIFN UST EAK RROON (Spaces added for readability)
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    zareenZ
    _________ operator in Prolog is used for List construction and also for List dismantling. CS508 !(Sign of exclamation) |(Vertical Slash) PG # 90 :(Colon) None of the given
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    bc170403108 MUHAMMAD WALEEDB
    solution #2
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    cyberianC
    https://cyberian.pk/topic/1166/cs311-3-fall19-solution-file
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