#Isaiah Hendry #11/9/21 #u3m2a2.2 #Learning about continue with loops

# Does nothing, but syntactically valid def useless_function(): pass

# Does nothing, but syntactically valid for i in range(10): pass

# Does nothing, but syntactically valid if (5 < 6): pass

# Find the index of the first occurrence of 611 in an arbitrarily long list # Define a long list lst = [976, 618, 812, 898, 804, 689, 611, 630, 467, 411, 648, 931, 618, 425, 328, 196, 56, 615, 458, 166, 115, 118, 22, 585, 213, 855, 615, 478, 234, 360, 981, 174, 993, 627, 911, 385, 219, 577, 540, 49, 164, 109, 893, 727, 687, 709, 971, 781, 802, 701, 789, 421, 863, 44, 573, 91, 140, 338, 36, 102, 502, 723, 134, 336, 940, 969, 649, 598, 413, 307, 929, 951, 993, 436, 149, 280, 907, 733, 713, 268, 29, 946, 102, 277, 352, 449, 527, 201, 566, 643, 118, 587, 717, 358, 542, 223, 846, 244, 83, 268] # Find the index (without break) index = 0 iteration_count = 0 for num in lst: if (num == 611): found_at = index index = index + 1 iteration_count = iteration_count + 1 print("Without using a break, 611 was found at index:", found_at, "using", iteration_count, "iterations") # Find the index (with break) index = 0 iteration_count = 0 for num in lst: if (num == 611): found_at = index break # adding a break to improve efficiency index = index + 1 iteration_count = iteration_count + 1 print("Using a break, 611 was found at index:", found_at, "using", iteration_count, "iterations")

# Prompt the user for a positive number while True: x = int(input("Enter a positive number: ")) if (x > 0): break #x is positive, break the loop

# Print the odd elements in a list lst = [14, 6, 5, 10, 6, 9, 33, 103, 21, 55] for num in lst: # Skip all even numbers if (num % 2 == 0): continue print(num)

# [ ] The following program tests if `num` is prime or not, use a `break` statement to improve its efficiency # and reduce the number of necessary iterations # Compare the number of necessary iterations with and without the `break` statement # Use the following numbers for the comparison: num = 45345 #num = 11579 #num = 948240 #num = 128093 #num = 519937 #num = 694394 prime = True # assume num is prime unless proven otherwise iteration_count = 0 for i in range(2, num): if num % i == 0: prime = False; break # Display results if prime: print(num, "is prime") else: print(num, "is NOT prime") print("Total number of iterations:", iteration_count)

# [ ] Write a program to prompt the user for an odd number; use an infinite loop and a `break` statement. while True: x=int(input("Please enter an odd number: ")) if (x % 2 != 0): break

# [ ] Modify the following program to display numbers that are divisible by 7 along with their square roots. # HINT: Use a `continue` statement in the loop from math import sqrt for num in range(1, 100): if (num % 7 !=0): continue print(num, sqrt(num))

# list of lists table = [[5, 2, 6], [4, 6, 0], [9, 1, 8], [7, 3, 8]] for row in table: for col in row: # print the value col followed by a tab print(col, end = "\t") # Print a new line print()

# Generate a staircase character art # Size controls the number of steps def genrate_star(size): for row in range(size): for col in range(size): if (row == col): print ("*", end ="") elif (row == size - col -1): print ("*", end ="") else: print(" ", end = "") print() genrate_star(5)

# [ ] Write a program to display the sum of each row in table table = [[5, 2, 6], [4, 6, 0], [9, 1, 8], [7, 3, 8]] for row in table: sum=0 for col in row: sum+=col print("SUM = ", sum)

# [ ] Complete the function `generate_star` so it displays a star of variable `size` using "*" # For size = 5 the star should look like: # * * # * * # * # * * # * * def generate_star(size): #TODO pass # Display star generate_star(5)

lst = [102, 34, 55, 166, 20, 67, 305] # Nesting a compound conditional in a for loop largest = 0 for num in lst: # test if num is even and greater than largest if((largest < num) and (num % 2 == 0)): largest = num print("Largest even number in the list is: ", largest)

# Ages of 100 people ages = [86, 38, 30, 19, 29, 6, 95, 22, 23, 82, 39, 73, 30, 98, 5, 68, 57, 34, 35, 81, 54, 77, 29, 75, 83, 14, 88, 7, 8, 32, 93, 76, 42, 1, 32, 70, 70, 3, 34, 52, 44, 41, 7, 77, 73, 97, 34, 13, 33, 54, 8, 82, 21, 55, 72, 41, 34, 98, 72, 73, 24, 55, 50, 63, 38, 92, 43, 68, 52, 68, 69, 51, 19, 24, 35, 55, 74, 47, 8, 19, 69, 12, 96, 96, 11, 30, 97, 73, 22, 25, 19, 85, 37, 68, 39, 76, 73, 18, 45, 42] # Initial count children = 0 teens = 0 young_adults = 0 adults = 0 # Nesting compound conditionals within a for loop for age in ages: if(age <= 12): children = children + 1 elif((age >= 13) and (age <= 19)): teens = teens + 1 elif((age >= 20) and (age <= 30)): young_adults = young_adults + 1 elif(age > 30): adults = adults + 1 print("Number of children: ", children) print("Number of teens: ", teens) print("Number of young_adults: ", young_adults) print("Number of adults: ", adults)

# Prompt the user for a number between 50 and 60 # Using infinite loop and break while True: x = int(input("Enter a number between 50 and 60: ")) if ((x >= 50) and (x <= 60)): print("Thank you!") break

# [ ] Complete the following program to count the number of even positive numbers, odd negative numbers, and zeros in `lst` lst = [9, 0, -2, -4, -5, 2, -15, 6, -65, -7] even_positives = 2 odd_negatives = 4 zeros = 1 #TODO: Count even_positives, odd_negatives, and zeros print("Number of even positives:", even_positives) print("Number of odd negatives:", odd_negatives) print("Number of zeros:", zeros)

# [ ] Write a program to count the number of punctuation marks (. , ? ! ' " : ;) in `s` s = "Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth." # Sherlock Holmes (by Sir Arthur Conan Doyle, 1859-1930) punc = 0 for c in s: if (c == "." or c ==',' or c =='?'or c =='!'or c=="'"or c =='"'c == ":"or c ==';')

# [ ] Write a program to prompt the user for an odd positive number; use an infinite loop and a `break` statement. while True: x = int (input("Enter an odd positive number: ")) break print("You entered an odd postive number!")