nction isPrime(n) { if (n < 2 || n % 1 ! return false; } for (let i = 2; i < %3D
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- INT_MIN = -32767 def cut_rod(price): """ Returns the best obtainable price for a rod of length n and price[] as prices of different pieces """ n = len(price) val = [0]*(n+1) # Build the table val[] in bottom up manner and return # the last entry from the table for i in range(1, n+1): max_val = INT_MIN for j in range(i): max_val = max(max_val, price[j] + val[i-j-1]) val[i] = max_val return val[n] # Driver program to test above functionsarr = [1, 5, 8, 9, 10, 17, 17, 20].What type of recursion is used in the following function? int f(int n){ if (n==1) return 1; else return n+f(n-1); } Tail recursion Multiple recursion Indirect recursion Non-tail recursionUsing a pseudo random number generation function (e.g., rand() in C or other equivalentfunctions in other languages) that generates uniformly distributed random numbers,write functions that generate the following:(a) uniformly distributed integers between 0 and 99. (b) uniformly distributed floating numbers between 0.25 and 0.5. (c) the number 1 with probability 0.5, the number 2 with probability 0.2, otherwise a floatuniformly distributed between 3 and 4.
- 19. Let f(n) be defined recursively by f(0) = 3, and f(n + 1) = 3f(n)/3, for n = 0, 1, 2, . . . Then, f(10) = _____________.In C programming Mathematically, given a function f, we recursively define fk(n) as follows: if k = 1, f1(n) = f(n). Otherwise, for k > 1, fk(n) = f(fk-1(n)). Assume that there is an existing function f, which takes in a single integer and returns an integer. Write a recursive function fcomp, which takes in both n and k (k > 0), and returns fk(n). int f(int n);int fcomp(int n, int k){Construct a DFA A so that L(A) = L(N) where N is the following NFA:
- The following function f uses recursion: def f(n): if n <= 1 return n else return f(n-1) + f(n-2) Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion? a) def f(n): a <- 0 b <- 1 if n = 0 return a elsif n = 1 return b else for i in 1..n c <- a + b a <- b b <- c return b b) def f(n): a <- 0 i <- n while i > 0 a <- a + i + (i-1) return a c) def f(n): arr[0] <- 0 arr[1] <- 1 if n <= 1 return arr[n] else for i in 2..n arr[i] <- arr[i-1] + arr[i-2] return arr[n] d) def f(n): arr[0..n] <- [0, ..., n] if n <= 1 return arr[n] else a <- 0 for i in 0..n a <- a + arr[i] return aWrite a program that prints an mxn matrix whose dimensions are specified by the user. Let the matrix values be random variables. You must use it within the repetition cycle. Example format: Enter dimension of matrix mxn: 2 4 The 2x4 matrix is : 1 2 3 4 5 6 7 8calculate number of operations in this algorithm void my_dgemv(int n, double* A, double* x, double* y) { double alpha=1.0, beta=1.0; int lda=n, incx=1, incy=1; cblas_dgemv(CblasRowMajor, CblasNoTrans, n, n, alpha, A, lda, x, incx, beta, y, incy); }
- Determine how many additions are done in the worst case scenario of the following code. Assume that all variables are properly declared. y = 1; for (i=1; i<=n;i++) { for (j = 1; j<=n;j++) { x = i + j; x = x * y; } x = y + x; }Write a function that takes in an integer n and computes n!. Do this without recursion. In [ ]: deffactorial_iter(n):"""Takes in an integer n>0 and returns the product of all integers from 1 to n."""# YOUR CODE HEREraiseNotImplementedError() In [ ]: In [ ]: assert factorial_iter(6) == 720 assert factorial_iter(7) == 5040 assert factorial_iter(10) == 3628800The following function f uses recursion: def f(n): if n <= 1 return n else return f(n-1) + f(n-2) 5 Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion? a) def f(n): a <- 0 b <- 1 if n = 0 return a elsif n = 1 return b else for i in 1..n c <- a + b a <- b b <- c return b f(n): a <- 0 i <- n while i > 0 a <- a + i + (i-1) return a f(n): arr[0] <- 0 arr[1] <- 1 if n <= 1 return arr[n] else for i in 2..n arr[i] <- arr[i-1] + arr[i-2] return arr[n] f(n): arr[0..n] <- [0, ..., n] if n <= 1 return arr[n] else a <- 0 for i in 0..n a <- a + arr[i] return a