midterm_1_uofzona

School

University Of Arizona *

*We aren’t endorsed by this school

Course

244

Subject

Computer Science

Date

Dec 6, 2023

Type

Pages

Uploaded by MajorTankBarracuda45

#ODD ALL THE ODD NUMBER FROM A LIST def add_odd(list): sum = 0 for i in list: if i % 2 != 0: sum += i return sum list = [2,3,4,17,6] print(add_odd(list)) #-------------------------------------------------- #ODD THE EVEN DIGITS OF A NUMBER def even_sum(n): sum = 0 for i in str(n): if int(i) % 2 == 0: sum += int(i) return sum print(even_sum(4456)) #-------------------------------------------------- #RETURN THE LARGEST NUMBER AMONG 3 NUMBERS def find_max(a,b,c): max = 0 if a >= b and a >= c: max = a elif b > c: # b > a and c > a max = b else: max = c return max print(find_max(0,0,1)) #--------------------------------------------------- #COMPUTE AND RETURN THE SUM OF SQUARE UP TO THE GIVEN NUMBER N def sum_of_squares(n): sum = 0

for i in range(n): sum += (i**2) return sum print(sum_of_squares(100)) #---------------------------------------------------- #GIVEN A STRING, RETURN TRUE IF ITS PURE, OTHERWISE RETURN FALSE def is_pure(str1): alpha = "abcdefghijklmnopqrstuvwxyz" str2 = str1.lower() print(str2) for i in str2: if (i in alpha) == False: print(i) return False break for i in str2: if (i in alpha) == True: print(i) return True print(is_pure("GhJifJ")) #----------------------------------------------------- #COMPUTE THE AVERAGE OF NUMBERS IN A STRING def average(str1): sum = 0 count = 0 for i in str1: if i.isnumeric(): sum += int(i) count += 1 num = sum / count return num print(average("2gos6nds4")) #------------------------------------------------------ #RETURN THE GREATEST COMMON DEVISOR AMONG 3 NUMBERS def gcd(a,b,c): lowest = 0 if (a <= b) and (a <= c):

lowest = a elif b <= c: lowest = b else: lowest = c divisors = [] for x in range(1,lowest+1): if ((a % x == 0) and (b % x == 0) and (c % x == 0)): divisors.append(x) return max(divisors) print(gcd(12,16,14)) #------------------------------------------------------- #RETURN BASE 12 NUMBER OF A DECIMAL NUMBER def base12(x): result = " " while (x > 0): r = x % 12 if (x == 10): result = "A" + result elif (x == 11): result = "B" + result else: result = str(r) + result x = x // 12 return result print(base12(912)) #-------------------------------------------------------- #SEPERATE THE NUMBERS IN THE STRING AND ODD THEM def sum_of_num(str1): sum = 0 for i in str(str1): if i.isnumeric(): sum += int(i) return sum print(sum_of_num("asdk5ksd8dsk3"))

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

#-------------------------------------------------- # SUM OF ALTERNATIVE NUMBERS def num_sum(str1): sum = 0 list = str1.split(",") count = 0 for i in list: count += 1 if (count % 2 != 0): sum = float(i) + sum elif (count % 2 == 0): sum = sum - float(i) return sum print(num_sum("65.21,7.3,44,12.1,33.2")) print(num_sum("65.21,7.3,44,12.1,33.2,1,0")) print(num_sum("65.21,7.3,44,12.1,33.2,1,0,2.0")) print(num_sum("65.21,7.3,44,12.1,33.2,1,0,2.0,5.0")) # -------------------------------------------------- # SMALLEST NUMBER def smallest(a, b, c): min = 0 if a < b and a < c: min = a elif b < c: min = b else: min = c return min print(smallest(8, 8, 10)) # -------------------------------------------------- # REMOVE THE NUMBERS FROM THE STRING def number(str1): str2 = "" for i in str1:

if not (i.isnumeric()): str2 = str2 + str(i) return str2 print(number("h3y2jda")) # -------------------------------------------------- # LARGEST CUBE BEFORE THE GIVEN NUMBER def cube(n): largest = 0 x = 1 while (x ** 3) <= n: if (x ** 3) > largest: largest = (x ** 3) x += 1 return largest print(cube(100)) # -------------------------------------------------- # GIVE THE NO. OF PALINDROME NUMBERS SMALLER THAN 'N' def palindrome(n): count = 0 for i in range(n): # going from 0 to n-1 counting 0 as a palindrome if str(i)[0] == str(i)[len(str(i)) - 1]: count += 1 return count print(palindrome(44)) # -------------------------------------------------- # GIVEN A NUMBER PRINT TRUE IF THE NUMBER IS PERFECT AS # N FOR WHICH THE SUM OF DIVISORS IS 2N, N = 6 IS A PERFECT NUMBER SINCE WE HAVE # 1 + 2 + 3 + 6 = 12.

def num(n): sum = 0 for i in range(1, n + 1): if (n % i) == 0: sum += i if sum == (2 * n): return True else: return False print(num(6)) # -------------------------------------------------- #Write a program that examines three variables x , y , and z, and prints the largest odd number #among them. If none of them are odd, it should print a message to that effect. x = 888 y = 653 z = 34 max = 0 if (x >= 0) and (x % 2 == 1): max = x elif (y >= 0) and (y % 2 == 1): max = y elif (z >= 0) and (z % 2 == 1): max = z else: print("No numbers are odd") print(max) #-------------------------------------------------- #Write a program that asks the user to input 10 integers, and then prints the largest odd number #that was entered. If no odd number was entered, it should print a message to that effect. This is a #more general version of Question 1. For example of the integers were: 10, 9, 7, 12, 2, 5, 15, 100, 90, 60, #then the program would print 15 as the largest odd number.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

x = 0 max = 0 while x < 10: number = int(input("Define Integer")) if (number > max) and (number % 2 == 1): max = number x = x + 1 if (max != 0): print(max, 'is the greatest number') else: print('No number is odd') #-------------------------------------------------- #Write a program that adds the amounts of money in a list. An example input is: #23 CAD, 345 CAD, 55 CAD. The output for this input is 423 CAD amount=[] a= int(input("Input number of Items: ")) for i in range(a): b=float(input("Input Amount: ")) amount.append(b) total=sum(amount) print("Sum of amount = ", total, 'CAD') #-------------------------------------------------- #Write a program that converts a given positive integer to its binary representation. #Hint: write the program using a for loop. To get the number of bits in a given integer, use the #function bit_length() in python. For example, for a = 983265982365, a.bit_length() will be 40. a=3 print(bin(a)) y =a.bit_length() for i in range(y): if a>= 2**(y-i-1): print(1, end="") a= a-2**(y-i-1) else: print(0, end="") print()

#-------------------------------------------------- #Let f(x) = x**5 + 52*x**3 - 29 be a function. This function has only one root in the real #numbers. Write a bisection search algorithm to find that root up to two digits of accuracy, i.e., #for epsilon = 0.01 as in the class. def ComputeFunction(x): return x ** 5 + 52 * x ** 3 - 29 a = 0 c = 2 b = a + (c - a)/2 epsilon = b - a count = 0 while epsilon > 0.01: if ComputeFunction(a) * ComputeFunction(b) < 0: a = a c = b b = a + (c - a)/2 else: a = b c = c b = a + (c - a)/2 epsilon = b - a count += 1 print(a) print("The number of iterations is: ", count) #-------------------------------------------------- #Solve question 2 using the Newton-Raphson method. Compare the number of iterations with #the bisection search. def ComputeFuntion(x): return x ** 5 + 52 * x ** 3 - 29 def ComputeDerivative(x): return 5 * x ** 4 + 156 * x ** 2 a = 0.1 b = 2 epsilon = b -a count = 0 while epsilon > 0.01:

b = a - (ComputeFunction(a) / ComputeDerivative(a)) epsilon = b - a a = b count += 1 print(a) print("The number of iterations is: ", count) #-------------------------------------------------- #You have written an essay that should be at least five pages long, but it is only 4.5 pages. You #ecide to use your new python skills to stretch it out by putting spaces between each letter #and the next. Write a function that takes a string and a number and prints out the string with #number of spaces between each pair of letters. #def spacedout(str, numSpaces) def spacedout(str, nmb): separator = " " * nmb return separator.join(str) s = input("input content") n = int(input("the number of of space = ")) print(spacedout(s,n)) #-------------------------------------------------- #Write a python program that returns the reverse of a given string. #'def reverse(str1)' def reverse(str): new = "" n = len(str) for i in range(n): new = new + str[n - i - 1] return new def isPalindrome(str): if reverse(str) == str: return (True) else: return(False) str = "Hello" print(reverse(str)) print(isPalindrome(str))

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

#-------------------------------------------------- #Write a recursive function that prints all the permutations of a given string. def returnPermutation(s): n = len(s) l = [] if n == 1: l.append(s) return l for i in range(n): c_0 = s[i] s_1 = s[:i] + s[i+1:] L = returnPermutation(s_1) for w in L: l.append(c_0 + w) return l print(returnPermutation("abc")) #-------------------------------------------------- #Given an integer n, write a python program that finds the digit with the highest number of #occurrences in n. For example, for n = 823778823478234768238 the program should return 8. #'def findHighestOccur(n)' def findHighestOccur(n): count = [0 for i in range(10)] for one in str(n): count[int(one)] += 1 max_occur = count.index(max(count)) return max_occur n = 983254682376587235235 print(findHighestOccur(n)) #-------------------------------------------------- #Assume that L is a list with one element. The element might be a number or it might be a list #of one element, where that element might be a number or a list of one element, and so on.

#Return the innermost number. You must use recursion. Your solution cannot use a loop. def innermost(x): #Defining a function to give the innermost element from the list if type(x[0]) == int: #Opening an 'if and else' statement return x[0] else: y = x[0] return innermost(y) l = [5] #Assigning 'l' a value print(innermost(l)) #printing the innermost element from the above element l = [[5]] #Assigning 'l' a value #-------------------------------------------------- #Assume that T is a tuple. Return a tuple which is the reverse of T. You must use recursion. #Your solution cannot use a loop or a list or a built-in reverse function. Hint: Reverse the tuple #starting at the second element, and concatenate to the end of that reversed tuple the first #element. #Test your program with the following: #(1, 2, 3, 4, 5) #('red', 'green', 'blue', 'yellow') def reverse(t): #Defining a function to reverse new = () if len(t) == 1: #Opening an 'if and else' statement new = t return new else: new = reverse(t[1:]) + t[:1] return new t = ("a", "b", "c") #Giving 't' a value print(reverse(t)) #printing the output

t = (1, 2, 3, 4, 5) #Giving 't' a value print(reverse(t)) #printing the output t = ('red', 'green', 'blue', 'yellow') #Giving 't' a value print(reverse(t)) #-------------------------------------------------- #Assume that sub and bigstring are strings. Return True if sub is part of bigstring, False #otherwise. You must use recursion. Your solution cannot use a string funtion or the 'in #operation'. Your solution cannot use a loop. Hint: check if sub is the beginning of bigstring, #and if not, consider the string obtained by removing the first letter from bigstring. def isSubstring(str, s): l1 = len(str) l2 = len(s) if l1 < l2: return False else: if str[:l2] == s: return True else: return isSubstring(str[1:], s) str = "I am a student" s = "I am" print(isSubstring(str, s)) #-------------------------------------------------- #Assume that phrase is a string. Return the number of words in phrase. Assume that words #are separated by one or more blanks. You are not allowed to use the built-in function split() def numberofwords(s): counter = 0 for c in s: if c == " ": counter += 1

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

counter += 1 return counter s = "A white cat and a black cow" print(numberofwords(s)) #-------------------------------------------------- #Return the first value of n such that the sum #Sum = 4/1 - 4/3 + 4/5 - 4/7 + ... + 4 * (-1)**(n+1)/(2n - 1) #is within epsilon of math.pi = 3.141592653589793. For example, if epsilon is 1, then return 1, #since 4 is within 1 of pi. If epsilon is 0.5 then return 2, since 4 is not within 0.5 of pi but 4 - 4/3 #= 2.67, and that is within 0.5 of pi epsilon = float(input("Enter Integer")) #Defining a variable b = 1 #Defining 'b' a value sum = 0 #Defining 'sum' a value pi = 3.14 #Giving the value to 'pi' while True: #Opening a while loop sum = sum + 4 * (-1) ** (b + 1) / (2 * b - 1) if abs(sum - pi) <= epsilon: #Opening an 'if' statement break b += 1 print(b) #-------------------------------------------------- #Find a given element in a list using a recursive function. The function is defined as #find(elm_list, x), where elm_list is the input list and x is the element to be found. The function #return the index of x if x is in the list, otherwise it returns -1. def find(elm_list, x): #Defining a function as find(elm_list, x) if (len(elm_list) == 1): if (elm_list[0] == x):

#Opening an 'if' statement to find x in the list return x #If 'x' is found it is returned else: #else return -1 #-1 is returned if (elm_list[0] == x): return x else: return find(elm_list[1:], x) elm_list = [1.2, 3, 555, 66, 6, 35, 345, 345, 78, 12341] x = 78 print(find(elm_list,x)) print(elm_list.index(x)) #-------------------------------------------------- #Write a function that computes the inner product of two lists. The inner product of two lists #L1 and L2 of length n is defined as L1[0] * L2[0] + L1[1] * L2[1] + ... + L1[n 1]. If #the lists are not of the same length, the function should return None. def innerProduct(l1, l2): #Defining a function as innerProduct len1 = len(l1) #Assigning the variable 'len1' a value len2 = len(l2) #Assigning the vairable 'len2' a value total = 0 counter = 0 if len1 == len2: #Opening an if statement, if the length of the two lists are equal for i in l1: #Opening a for loop total += (i * l2 [counter]) counter += 1 print(total) return total else: return None print(innerProduct([8.17, 4.99, 16.56, 7.31, 0.21, 16.18, 13.75,

11.28, 4.3, 17.79], #Giving the lists to print the product [17.68, 1.19, 2.64, 13.01, 8.5, 15.04, 0.3, 18.72, 4.58, 9.66])) #-------------------------------------------------- #Write a function that extract a number from a string. The string can contain any character. #The number you return is the concatenation of all the digits you find in the string. For #example, the extracted number from the string is 823755. str = "1ksdj4sdf>R?387BDYGrrf%44@2" #Assigning the values to the variable print("Given String : " + str) #Printing the given string num = "" for c in str: #Opening a for loop, to find a digit in the given string if c.isdigit(): #Checking digits in the stirng num = num + c print("Extracted Numbers : " + num) #-------------------------------------------------- #Write a function that computes the sum #1 + 1 / 2**2 + 1 / 3**2 + ... + 1 / n**2 #for a given integer n. For large n, this sum converges to pi**2 / 6 where pi = #3.141592653589793. How large should n be for the sum to be within distance 0.001 of pi**2 / 6? def equation(n): #Defining a function as equation sum = 0 for i in range(n): #Opening a for loop sum += 1/(i+1)**2 #If sum is equal to 1/(i+1)**2 return sum #Return sum def piEquation(): #Defining a function as 'piEquation' pi = 3.141592653589793

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

#Assigning the variable 'pi' a value total = pi**2/6 return total #Returning total def nFind(): #Defining a function as 'nFind' n = 1 #Assigning 'n' a value of 1 epsilon = 0.001 #Assigning epsilon value while True: #Opening a while loop if abs(equation(n) - piEquation()) <= epsilon: #Opening an if statement break n += 1 return n #Returning n print(nFind()) #--------------------------------------------------

midterm_1_uofzona

Related Documents