Write codes to transform a given score to GPA following the rules shown in the right table. Submit the screenshot of your test results and check if it's what you expected.
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- (Practice) Although the total number of bytes varies from computer to computer, memory sizes of millions and billions of bytes are common. In computer language, the letter M representsthe number 1,048,576, which is 2 raised to the 20th power, and G represents 1,073,741,824, which is 2 raised to the 30th power. Therefore, a memory size of 4 MB is really 4 times 1,048,576 (4,194,304 bytes), and a memory size of 2 GB is really 2 times 1,073,741,824 (2,147,483,648 bytes). Using this information, calculate the actual number of bytes in the following: a. A memory containing 512 MB b. A memory consisting of 512 MB words, where each word consists of 2 bytes c. A memory consisting of 512 MB words, where each word consists of 4 bytes d. A thumb drive that specifies 2 GB e. A disk that specifies 4 GB f. A disk that specifies 8 GBThe factorial function f(n) = n! is defined by n! = 1 2 3... n whenever n is a positive integer, and O! = 1. Provide big-O estimates for the factorial function and the logarithm of the factorial function. As an illustration, 1! = 1, 2! = 1 2=2, 3! = 1 23 = 6, and 4! = 1 234=24.Take note of how quickly the function n! expands. Consider the following: 20! = 2,432,902,008,176,640,000.implement anyEvenBit(x) Return 1 if any even bit in x is set to 1 you are only allowed to use the following eight operators: ! ~ & ^ | + << >> “Max ops” field gives the maximum number of operators you are allowed to use to implement each function /* * anyEvenBit - return 1 if any even-numbered bit in word set to 1* Examples anyEvenBit(0xA) = 0, anyEvenBit(0xE) = 1* Legal ops: ! ~ & ^ | + << >>* Max ops: 12*/int anyEvenBit(int x) {return 2;}
- Q3) Let Σ = {a , b}. Find DFA's for: d) L = { wab: w€ {a,b}*}e) L = { abwab: w€ {a,b}*}f) L = { ab5wb4 : w € {a,b}* }Write a program that swaps 5th~11th bits in data_a with 25th~31th bits in data_bYour program must work for any data given, not just the example belowIn this question, we assmue that the positions of bits count from right to left.That is, the first bit is the least significant bit.Using a 5-bit cell to store integers in 2's complement, what is the largest integer that can be stored? Question 74 options: A. 15 B. 31 C. 3 D. 8 E. 7 Consider the following program segment, where a is an integer constant and b is an integer variable that holds a positive value.int r = 0, n = b;while (n != 0){ r += 2*a; n--; }Which of the following is a loop invariant for the while loop? r = 2*a*(b-1); r = a*b; r = 2*a*b ; r = a*(b+1);
- Give big-O estimates for the factorial function and the logarithm of the factorial function, wherethe factorial function f(n) = n! is defined byn! = 1 · 2·3· ... ·nwhenever n is a positive integer, and O! = 1 . For example,1 ! = 1, 2! = 1 · 2=2, 3 ! =1· 2·3 = 6, 4! = 1 · 2·3· 4=24.Note that the function n! grows rapidly. For instance,20! = 2,432,902,008,176,640,000.implement greatestBitPos function Compute a mask marking the most significant 1 bit. you are only allowed to use the following eight operators: ! ~ & ^ | + << >> “Max ops” field gives the maximum number of operators you are allowed to use to implement each function /* * greatestBitPos - return a mask that marks the position of the* most significant 1 bit. If x == 0, return 0* Example: greatestBitPos(96) = 0x40* Legal ops: ! ~ & ^ | + << >>* Max ops: 70* Rating: 4 */int greatestBitPos(int x) {return 2;}Suppose the economies of the world use a set of currencies C1, . . . , Cn; think of these as dollars, pounds, Bitcoin, etc. Your bank allows you to trade each currency Ci for any other currency Cj, and finds some way to charge you for this service. Suppose that for each ordered pair of currencies (Ci, Cj ), the bank charges a flat fee of fij > 0 dollars to exchange Ci for Cj (regardless of the quantity of currency being exchanged). Describe an algorithm which, given a starting currency Cs, a target currency Ct, and a list of fees fij for all i, j ∈ {1, . . . , n}, computes the cheapest way (that is, incurring the least in fees) to exchange all of our currency in Cs into currency Ct. Also, justify the its runtime. [A description or pseudocode (either is OK) of the algorithm, as well as a brief justification of its runtime.]
- implement bitcount function Count the number of 1’s in x. you are onlyallowed to use the following eight operators:! ~ & ^ | + << >> “Max ops” field gives the maximumnumber of operators you are allowed to use to implement each function /** bitCount - returns count of number of 1's in x* Examples: bitCount(5) = 2, bitCount(7) = 3* Legal ops: ! ~ & ^ | + << >>* Max ops: 40*/int bitCount(int x) {return 2;}x=[10,1,20,3,6,2,5,11,15,2,12,14,17,18,29] for i in range(0, len(x)): m = x[i] l = i # Finding the minimum value and its location for j in range(i+1, len(x)): if(m >x [j]): m= x[j] l = j # Now we will do the swapping tmpry = m x[l] = x[i] x[i] = tmpry print(x) 8.The code above is written in Python programming language. In this sample, there are many sectors where you can apply knowledge of code refactoring. Mention each criteria where this code lacks and provide appropriate suggestions to improve. (Hint: Code Smells e.g: Naming Convention, Code length, Assertion etc.)implement byteSwap(x,n,m) Swap the m and n byte of x /* * byteSwap - swaps the nth byte and the mth byte* Examples: byteSwap(0x12345678, 1, 3) = 0x56341278* byteSwap(0xDEADBEEF, 0, 2) = 0xDEEFBEAD* You may assume that 0 <= n <= 3, 0 <= m <= 3* Legal ops: ! ~ & ^ | + << >>* Max ops: 25*/ int byteSwap(int x, int n, int m) {return 2;}