For function log, write the missing base case condition and the recursive call. This function computes the log of n to the base b. As an example: log 8 to the base 2 equals 3 since 8 = 2*2*2. We can find this by dividing 8 by 2 until we reach 1, and we count the number of divisions we make. You should assume that n is exactly b to some integer power. Examples: log(2, 4) -> 2 and  log(10, 100) -> 2   public int log(int b, int n ) {   if <> {     return 0;   } else {     return <>   } }

In kotlin, Write a recursive linear search function that takes a String and a Char and returns true if the String contains the Char.  The base cases are that a) the length of the String is 0 or b) the first character in the String is the character we are searching for. then Modify the linear search function so that both the String and the Char are nullable.  If either is null, return false.

Can you please help me solve problem 2?

Which is the base case of the following recursion function: def mult3(n):    if n == 1:        return 3    else:        return mult3(n-1) + 3     else     n == 1     mult3(n)     return mult3(n-1) + 3

Let recursive algorithm F be defined as follows: F(int n):if n=1 then    return 1else   return F(n-1)*n (a) i. Assuming x is a positive integer, what mathematical expression is returned by F(x)?ii. Repeat (i) but assume the line “return 1” is replaced by “return 0”? (b) Which of the following replacements for the last line will yield a function guaranteed to terminate for all valid inputs? i. return F(n-2)+2ii. return F( )*2iii. return F( )*2

Using recursion, write a Python function def before(k,A) which takes an integer k and an array A of integers as inputs and returns a new array consisting of all the integers in A which come before the last  occurrence of k in A, in the same order they are in A. For example, if A is [1,2,3,6,7,2,3,4] then before(3,A) will return [1,2,3,6,7,2]. If k does not occur in A, the function should return None.

Consider a recursive function, called f, that computes powers of 3 using only the + operator. Assume n > = 0. int f(int n) { if (n == 0) return 1; return f(n-1) + f(n-1) + f(n-1); } Give an optimized version of f, called g, where we save the result of the recursive call to a temporary variable t, then return t+t+t. i got int g(int n) { if (n == 0) return 1; int t = g(n - 1); return t+t+t; } so now Write a recurrence relation for T(n), the number addition operations performed by g(n) in terms of n.

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Give a recursive definition for the set of all strings of a’s and b’s that begins with an a and ends in a b. Say, S = { ab, aab, abb, aaab, aabb, abbb, abab..} Let S be the set of all strings of a’s and b’s that begins with a and ends in a b. The recursive definition is as follows – Base:... Recursion: If u ∈ S, then... Restriction: There are no elements of S other than those obtained from the base and recursion of S.

Need some help with this c++ recursive function: Write the function int countEights(int n). Inside the function, compute recursively (no loops) the count of the occurrences of 8 as a digit inside the parameter n. Except that an 8 with another 8 immediately to its left counts double, so 8818 yields 4. Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost digit (126 / 10 is 12). Here are some examples: countEights(8) returns 1 countEights(818 returns 2 countEights(8818) returns 4