Algorithm Binary Search Input: L = a1, a2, ... , an list of distinct integers; x integer Output: p integer i = 1; j = n; found = F; p = 0; while i ≤ j and found = F m = ⌊i+j2⌋; if am = x then found = T and p = m else if am < x then i = m + 1 else j = m - 1; return p; When performing Algorithm Binary Search with input L = 1,3,5,7,9,11,13 and x = 9, how many times is x compared to an element ai of the list L?
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Input: L = a1, a2, ... , an list of distinct integers; x integer
Output: p integer
i = 1;
j = n;
found = F;
p = 0;
while i ≤ j and found = F
m = ⌊i+j2⌋;
if am = x then found = T and p = m
else if am < x then i = m + 1
else j = m - 1;
return p;
When performing Algorithm Binary Search with input L = 1,3,5,7,9,11,13 and x = 9, how many times is x compared to an element ai of the list L?
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- int functionC (int n) { int i, j, sumC = 0; for (i=n; i > 0; i=i-5) for (j=1; j 0) { if (functionC(n) % 2 == 0) { for (i=m; i > 0; i=i/3) sumE++; } else (10) Asymptotic runtime of functionE { for (i=m; i > 0; i=i-3) sumE--; } n--; } return sumE;int n = 1; int k - 2; int r = n; if (k < n) { r - kc++ computer language randomly generates integers in the interval [1,10] a.1+rand()%11 b.rand()%11 c.rand()%10 d.1+rand()%10
- CodeW For fun X C Solved https://codeworkou... 臺亂 CodeWorkout X272: Recursion Programming Exercise: Is Reverse For function isReverse, write the two missing base case conditions. Given two strings, this function returns true if the two strings are identical, but are in reverse order. Otherwise it returns false. For example, if the inputs are "tac" and "cat", then the function should return true. Examples: isReverse("tac", "cat") -> true Your Answer: 1 public boolean isReverse(String s1, String s2) { 2. if > 3. 4. else if > return true; return false; 5. 6. else { String s1first = String s2last return s1first.equals (s2last) && 51. substring(0, 1); s2, substring(s2.length() 1); 7. 8. 6. isReverse(s1.substring(1), s2.substring(0, s2.length() 1)); { 12} 1:11AM 50°F Clear 12/4/2021calculate 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); }Algorithm problem w/ recurrence: Frying pancakes: a small pan can only hold two pancakes at a time. Each pancake needs to be fried on both sides. Frying one side takes 1 minute, no matter how many pancakes are on the pan. Consider this recursive algorithm: If n <= 2, fry the pancakes or the two pancakes together on each side. If n > 2, fry any two pancakes together on each side and then apply the same process recursively to the remaining n-2 pancakes. a. Set up and solve the recurrence for the amount of time this algorithm needs to fry n pancakes. b. Explain why this algorithm does not fry the pancakes in the minimum time for all n > 0. c. Give a correct recursive algorithm that executes the task in the minimum amount of time. > I was not sure how to start this. I have had trouble with recurrence in the past. Also from the work that I did do, I didn't know how there could be a better algorithm. Thanks in advance
- - .is Complete history of everything that the agent has ever perceived. * Percept percept sequence Agent functions NoneMemoization technique is one of the popular techniques that improve the performance of the recursive algorithms. When applied to a recursive problem how does it affect overall performance? Time complexity decreases and the space complexity increases Time complexity decreases and the space complexity decreases Time complexity increases and the space complexity decreases Time complexity increases and the space complexity increasesAlgorithm Binary Search Input: L = a₁, a2, ..., an list of distinct integers; x integer Output: p integer i=1; j= n; found = F; p = 0; while iNuts and bolts You are given a collection of n bolts of different widths and n corresponding nuts. You are allowed to try a nut and bolt together, from which you can determine whether the nut is larger than the bolt, smaller than the bolt, or matches the bolt exactly. However, there is no way to compare two nuts together or two bolts together. The problem is to match each bolt to its nut. Design an algorithm for this problem with average-case efficiency in (n log n).Outputs of the following: void main () { int M[3]; int "ptr; ptr=M; *ptr=100; ptr++; "ptr=200; ptr=&M[2]; *ptr=300; for (int n=0; n<3; n++) cout <1: |R|←|P| Reserve space for |P| = 13 values.2: x ← n3: for i ← 0 ...(|P| − 1) do4: c ← x div Pi Number of multiplicands Pi in x.5: Ri ← c6: x ← x − c · Pi7: end for8: return RA Java programmer could implement Algorithm by first modelling the primitivenumbers with the enumeration type RomanNumeral. Each enum constant (I, IV, ..., M) is declared with its decimal value, which can be accessed with the function getValue().SEE MORE QUESTIONSRecommended textbooks for youC++ Programming: From Problem Analysis to Program…Computer ScienceISBN:9781337102087Author:D. S. MalikPublisher:Cengage LearningC++ Programming: From Problem Analysis to Program…Computer ScienceISBN:9781337102087Author:D. S. MalikPublisher:Cengage Learning