Read chapter 3 in Blown To Bits.
def print_sum(matrix): sum = 0 for row in range(0,len(matrix)): for col in range(0,len(matrix[row])): sum = sum + matrix[row][col] print sum
Show how to modify this function so that it prints the sum of each row of the two-dimensional list. (HINT: Draw a flowchart of this function for yourself and then move two operations in the flowchart to get the desired behavior.)
Algorithm 1: Songs are stored in the computer's memory in arbitrary order. Each song has a code that indicates the location in memory of the song that plays next. The player keeps track of the location of the first song in the playback sequence only.Algorithm 2: Songs are stored in the computer's memory in the order of playback starting at a specific fixed location in computer memory which cannot be changed.
rpn = [23, 3, "-", 4, 6, "+", "/"]
Recall the algorithm to compute the value of a RPN expression using a stack:
1. Set i equal to 0. 2. Set x equal to rpn[i]. 3. Set s equal to an empty stack. 4. While i is not equal to the length of the rpn array, do the following: a. If x is a number, then push x on stack s. b. If x is a string (i.e. operator), then do the following: i. Pop stack s and store number in b. ii. Pop stack s and store number in a. iii. If operator is "+", push a+b on stack s. iv. If operator is "-", push a-b on stack s. v. If operator is "*", push a*b on stack s. vi. If operator is "/", push a//b on stack s. c. Add 1 to i. 5. Pop stack s and return this number.
Trace how this algorithm computes the value of the following RPN expression stored as an array:
rpn = [9, 3, "+", 7, 5, "-", "*", 6, 2, "/", 1, "+", "/"]
(Draw a new stack whenever a number is pushed or popped to show how the stack progresses throughout the computation.)
Suppose we represent a queue using an list named q such that the first element in the list (at index 0) is the front of the queue and the last element in the list (at index len(q)-1) is the rear of the queue.
def h(string, table_size): k = 0 for i in range(0,len(string)): k = ord(string[i]) + k * 256 return k % table_size
In the function above, ord(string[i]) returns the ASCII code of the ith character of string. Here are the ASCII codes for the lowercase letters:
a b c d e f g h i j k l m 97 98 99 100 101 102 103 104 105 106 107 108 109 n o p q r s t u v w x y z 110 111 112 113 114 115 116 117 118 119 120 121 122
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