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Go through the code below by hand, statement by statement, and calculate the numbers that will be printed. n = 3 for i in range(-1, n): if i != 0: print i for i in range(1, 13, 2*n): for j in range(n): print i, j for i in range(1, n+1): for j in range(i): if j: print i, j for i in range(1, 13, 2*n): for j in range(0, i, 2): for k in range(2, j, 1): b = i > j > k if b: print i, j, k You may use a debugger, see Appendix F.1, to step through the code to see what happens.

Short Answer

Expert verified
The printed numbers are: -1, 1, 2, (1,0), (1,1), (1,2), (7,0), (7,1), (7,2), (2,1), (3,1), (3,2), (7,4,2), (7,4,3).

Step by step solution

01

Initialize Variables

First, we define the variable \( n \) and set it to \( 3 \). This variable will be used to control loops and calculate step sizes.
02

First Loop Iteration

The first loop iterates over the range from \(-1\) to \( n \), which results in values \(-1, 0, 1, 2\). The print condition is for \( i eq 0 \). Therefore, the numbers \(-1, 1, 2\) will be printed.
03

Second Loop Iteration

The second for-loop iterates over \( i \) in the range from \( 1 \) to \( 13 \) with a step of \( 2n \), or \( 6 \). This results in values \( (1, 7, 13) \). In each iteration, a nested loop over \( j \) runs for \( n \) times (0 to 2).This results in the print statements:- \( (1, 0), (1, 1), (1, 2) \)- \( (7, 0), (7, 1), (7, 2) \)(Note: it doesn't execute for \( i = 13 \) because it's beyond the range).
04

Third Loop Iteration

This loop iterates over \( i \) from \( 1 \) to \( n+1 \) (i.e., \( i = 1, 2, 3 \)) and the nested loop runs from 0 to \( i-1 \). A condition \( j eq 0 \) is checked before printing, to ensure \( j \) is not zero.The result will be:- For \( i = 2, j = 1 \) print \( (2, 1) \)- For \( i = 3, j = 1 \) and \( j = 2 \) printing \( (3, 1), (3, 2) \).
05

Fourth Loop Iteration

The fourth loop uses \( i \) with the same step of 6. It iterates \( j \) from 0 to \( i \) in steps of 2, and inside that, it checks for a condition involving \( k \) over its range. The condition \( i > j > k \) must be true to print. - For \( i = 1 \), no \( j \) value will fit.- For \( i = 7 \), valid \( j = 4 \) allows for \( k = 2, 3 \). - Results are: \( (7, 4, 2), (7, 4, 3) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Python for Loops
In Python, 'for loops' are essential for iterating over a sequence of numbers or items in a list. In the given exercise, the `for` loop is utilized in multiple scenarios to iterate over ranges dictated by conditions. The initial loop in the exercise, `for i in range(-1, n):`, generates numbers from -1 to just below 3. This range includes -1, 0, 1, and 2. However, due to the condition `if i != 0:`, the loop skips the print statement when `i` is 0 and prints -1, 1, and 2 instead.
This understanding of iterating with a loop over a range is foundational when it comes to looping in Python. The key syntax here is `range(start, stop)`, where `start` is inclusive, and `stop` is exclusive. By default, a step value is `1`, but programmers can specify different step values if needed. Additional loops in the exercise exploit different ranges and conditions to explore these programmable capabilities of Python.
Conditional Statements
Conditional statements are crucial in controlling the flow of a program. In Python, common conditional statements include `if`, `elif`, and `else`. These statements help to determine which blocks of code execute based on certain conditions. In the exercise, we see the conditional `if j:` used, which checks if `j` is a truthy value (not zero). This condition filters out integers where `j` equals zero, thus preventing zero from being printed alongside `i` and `k` values.
Another example is `if i != 0:`, which ensures that the print occurs only when `i` is not zero. This prompt filtering allows specific numbers or sequences to be processed or ignored, enhancing program logic. Understanding how these conditions affect the flow can be pivotal in constructing functional and efficient loops.
Nested Loops
Nested loops, which involve placing one loop inside another, are often used to navigate through multi-dimensional data structures like matrices or grids. In this exercise, nested loops are employed several times. One example is the loop `for j in range(n):` inside another `for i in range(1, 13, 2*n):`. Here, for each iteration of `i`, the inner loop fully iterates through its range for `j`, demonstrating how nested loops can create multiple layers of iteration.
This construct is powerful as it allows comprehensive exploration of all combinations of the loop variables. However, it also necessitates mindful design to prevent excessive computation or infinite loops. The use of nested loops is clear in extracting combinations of `i, j, k` in the final set of loops. Developers must ensure that each loop has a valid endpoint and that the operations inside are efficient.
Code Step-Through
A code step-through technique is invaluable in understanding program behavior and debugging. By stepping through each line, developers can observe changes in variable states and ensure logic flows as expected. In this exercise, manually or through a debugger, stepping through each line fosters understanding of how loops and conditions interact, ensuring comprehension of the underlying logic. Appendix F.1 suggests using a debugger, which can visually highlight which part of the code is currently active, making the learning and debugging process interactive.
By methodically assessing each iteration's state, developers can verify expected outcomes against actual variable states, spot errors, and optimize logic. Practicing code step-through, particularly with complex nested structures and multiple conditions, becomes essential in mastering precise and bug-free coding.

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Most popular questions from this chapter

Maybe you have tried to hit the square root key on a calculator multiple times and then squared the number again an equal number of times. These set of inverse mathematical operations should of course bring you back to the starting value for the computations, but this does not always happen. To avoid tedious pressing of calculator keys we can let a computer automate the process. Here is an appropriate program: from math import sqrt for n in range(1, 60): r = 2.0 for i in range(n): r = sqrt(r) for i in range(n): r = r**2 print ’%d times sqrt and **2: %.16f’ % (n, r) Explain with words what the program does. Then run the program. Round-off errors are here completely destroying the calculations when \(\mathrm{n}\) is large enough! Investigate the case when we come back to 1 instead of 2 by fixing the \(\mathrm{n}\) value and printing out \(\mathrm{r}\) in both for loops over i. Can you now explain why we come back to 1 and not 2 ? Name of program file: repeated_sqrt.py.

Type in the following program in a file and check carefully that you have exactly the same spaces: C = -60; dC = 2 while C <= 60: F = (9.0/5)*C + 32 print C, F C = C + dC Run the program. What is the first problem? Correct that error. What is the next problem? What is the cause of that problem? (See Exercise \(2.12\) for how to stop a hanging program.) The lesson learned from this exercise is that one has to be very careful with indentation in Python programs! Other computer languages usually enclose blocks belonging to loops in curly braces, parentheses, or BEGIN-END marks. Python's convention with using solely indentation contributes to visually attractive, easy-to-read code, at the cost of requiring a pedantic attitude to blanks from the programmer.

Set a variable primes to a list containing the numbers \(2,3,5,7\) 11 , and \(13 .\) Write out each list element in a for loop. Assign 17 to a variable \(\mathrm{p}\) and add \(\mathrm{p}\) to the end of the list. Print out the whole new list. Name of program file: primes.py.

Write a program that generates all odd numbers from 1 to n. Set \(n\) in the beginning of the program and use a while loop to compute the numbers. (Make sure that if \(n\) is an even number, the largest generated odd number is n-1.) Name of program file: odd.py.

Explain the outcome of each of the following boolean expressions: C = 41 C == 40 C != 40 and C < 41 C != 40 or C < 41 not C == 40 not C > 40 C <= 41 not False True and False False or True False or False or False True and True and False False == 0 True == 0 True == 1 Note: It makes sense to compare True and False to the integers 0 and 1 , but not other integers (e.g., True \(==12\) is False although the integer 12 evaluates to True in a boolean context, as in bool(12) or if 12).

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