Computer Science Exercise: depth [★★] Write a function depth : 'a tree -> int that returns the number of nodes in any longest path from the root to a leaf. For example, the depth of an empty tree (simply Leaf) is 0, and the depth of tree t above is 3. Hint: there is a library function max : 'a -> 'a -> 'a that returns the maximum of any two values of the same type. please use Ocaml for the coding

In Python3, write the code for a function mult_3 that returns a list of all non-negative multiples of 3 that are less than or equal to a given argument n.

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Scrabble Scrabble is a game where players get points by spelling words. Words are scored by adding together the point values of each individual letter. Define a function scrabble_score (word:str) -> int that takes a string word as input and returns the equivalent scrabble score for that word. score = {"a": 1, "c": 3, "f": 4, "i": 1, "1": 1, "o": 1, "r": 1, "u": 1, "t": 1, "x": 8, "z": 10} For example, "b": 3, "h": 4, "n": 1, Your Answer: == 20 assert scrabble_score("quick") assert scrabble_score("code") == 7 1 # Put your answer here 2 Submit "e": 1, "d": 2, "g": 2, "k": 5, "j": 8, "m": 3, "q": 10, "p": 3, "s": 1, "w": 4, "v": 4, "y": 4,

Recamán's sequence def recaman(n): Compute and return the first n terms of the Recamán's sequence, starting from the term a1 = 1.Note how the definition for the current next element depends on whether a particular number is already part of the previously generated part of the sequence. To let your function execute in a speedy fashion even when generating a sequence that contains millions of elements, you should use a set to keep track of which values are already part of the previously generated sequence. This way you can generate each element in constant time, instead of having to iterate through the entire previously generated list. The better technique that saves time by using more memory can create this list arbitrarily far in linear time, and should therefore be fast even for millions of elements, at least until it runs out of memory. n  Expected result 10 [1, 3, 6, 2, 7, 13, 20, 12, 21, 11] 1000000 (a list of million elements whose last five elements are [2057162,1057165,…

Links Bing Remaining Time: 1 hour, 24 minutes, 17 seconds. * Question Completion Status: Р QUESTION 8 bbhosted.cuny.edu/webapps/assessment/take/launch.jsp?course_assessment_id=_2271631_1&course_id=_2 Implement the following: 1) Define a function evenList() with an arbitrary parameter a. P 2) This function accepts any number of integer arguments. 3) evenList() stores all even numbers into a list and returns the list. 4) Call the function with 5, 6, 7, 8, 9, 10 as arguments. 5) Print the result of the function call. Make sure to precisely match the output format below. Write your code in the ANSWER area provided below (must include comments if using code is not covered in the course). Example Output [6, 8, 10] For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). BIUS Paragraph Arial QUESTION 9 Programs s 10pt Click Save and Submit to save and submit. Click Save All Answers to save all answers. く Ev Av

a Write a function cutF2 that returns its argument with the first two items removed. It should not change the argument, but should return a copy without the first two items. >>> testList = [0,1, 2,3,4,5,6,7]+ >>> cut1(testList) >>> print (testList) #Note: returns modified copy [2, 3, 4,5,6,7]- #Note: original unchanged [0, 1, 2, 3, 4, 5, 6, 7]

Here is a function that is supposed to return True if word (any string) ends with "s" and False otherwise: def ends_with_s(word):if len(word) < 1:return False    return word[-1] == "s" Here is the test function I am using to test ends_with_s: def test_ends_with_s():    assert(ends_with_s("cats")) # option 1    assert(ends_with_s("") is False) # option 2    assert(ends_with_s("apple") is False) # option 3    assert(ends_with_s("S")) # option 4   Rewrite ends_with_s so that the test passes

def find_root4(x, epsilon): ''' IN PYTHON Assume: x, epsilon are floating point numbers and epsilon > 0 Use bisection search to find the following root of x such that If x >=0, return y such that x - epsilon <= y ** 2 <= x + epsilon Else, return y such that x - epsilon <= y ** 7 <= x + epsilon Note: You must use bisection search to implement the function. '''     pass