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WGU Foundations-of-Computer-Science Exam Syllabus Topics:

Section	Weight	Objectives
Algorithms & Complexity	25%	- Algorithm design and analysis - Big O notation, time and space complexity - Recursion and iterative structures - Sorting and searching algorithms
Software Engineering & Programming Basics	15%	- Testing and debugging fundamentals - Basic syntax and control structures - Software development lifecycle - Programming paradigms
Data Structures	20%	- Trees, graphs, hash tables - Data storage and retrieval principles - Primitive and composite data types - Arrays, linked lists, stacks, queues
Computer Architecture & Organization	15%	- Instruction sets and execution cycles - Von Neumann architecture - Memory hierarchy and performance - CPU, memory, I/O systems
Discrete Mathematics & Logic	25%	- Proof techniques and mathematical induction - Set theory, relations, functions - Propositional and predicate logic - Boolean algebra and digital logic

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問題 #49
What stores the location of the next node in a linked list?

A. The header
B. The index
C. The pointer
D. The value

答案：C

解題說明：
A linked list is a dynamic data structure made up of nodes, where each node typically contains two components: a data field (the value being stored) and a link field (commonly called a pointer or reference).
The pointer's role is to store the memory address (or reference) of the next node in the sequence, thereby maintaining the logical order of the list even though nodes may be scattered throughout memory. This is a key contrast with arrays, which store elements contiguously and rely on index arithmetic to locate the next element.
Because each node explicitly points to the next node, linked lists support efficient insertion and deletion operations compared with arrays. To insert a node, you allocate it and then adjust pointers so it fits into the chain. To delete a node, you redirect the pointer of the previous node to skip over the removed node.
Traversal is performed by starting at the head node and repeatedly following the pointer until a null reference indicates the end of the list.
The other options do not correctly describe what stores the location of the next node. An index is used in array-like structures, not in a standard linked list node. The value is the payload data, not the link.
The "header" (often called the head pointer) is an external reference to the first node, not the field inside each node that links to the next. Therefore, the correct answer is the pointer.

問題 #50
What is the expected result of running the following code: list1[0] = "California"?

A. The first value in the list will be replaced with "California".
B. The list will be extended by adding "California" at the end.
C. A second element will be added to the line "California".
D. A new list will be created with the value "California".

答案：A

解題說明：
Python lists are mutable sequences, which means elements can be changed in place after the list has been created. The expression list1[0] = "California" uses indexing to target the element at position 0 (the first element, because Python uses zero-based indexing) and assignment (=) to replace that element with a new value. As a result, the list keeps the same length, but its first entry becomes "California".
This operation does not create a new list (so option A is incorrect); it modifies the existing list object referenced by list1. It also does not append to the end of the list (so option C is incorrect). Appending would use methods like list1.append("California"). Option D is not meaningful in Python list semantics; assignment to a single index replaces exactly one element rather than "adding a second element to the line." Textbooks highlight this difference between mutable and immutable sequence types. For example, strings are immutable, so you cannot assign to some_string[0]. Lists, however, are designed for collections that change over time, supporting updates, insertions, deletions, and reordering. Index assignment is fundamental for many algorithms: updating an array-like buffer, modifying a dataset row, replacing incorrect values, or implementing in-place transformations efficiently.

問題 #51
How is a NumPy array named data with 6 elements reshaped into a 2x3 array?

A. np.reshape(data, (2, 3))
B. data_reshape[2, 3]
C. data.set_shape(2, 3)
D. np_reshape(list, (2, 3))

答案：A

解題說明：
Reshaping is the operation of changing the "view" of an array so that the same elements are arranged with new dimensions. In NumPy, reshaping is possible when the total number of elements stays the same. A 2x3 array contains 6 elements, so a 1D array data of length 6 can be reshaped into shape (2, 3) without adding or removing values. Textbooks stress this invariant: the product of the dimensions must equal the original size.
NumPy provides two standard reshaping interfaces: the function np.reshape(data, (2, 3)) and the method data.
reshape(2, 3) (or data.reshape((2, 3))). Option A is correct because it uses the official NumPy function with the proper arguments: the original array and the target shape. The shape is passed as a tuple describing rows and columns.
Option B is incorrect because np_reshape is not the correct NumPy function name, and it references an unrelated identifier list. Option C is incorrect because NumPy arrays do not provide a set_shape method like that. Option D is not valid NumPy syntax for reshaping.
Reshaping is fundamental in data analysis and machine learning: it converts flat vectors into matrices, prepares batches of samples, and aligns dimensions for matrix multiplication and broadcasting.

問題 #52
m = 30
n = 30
What will be the output of print(id(m), id(n)) after executing the following code?

A. Error
B. 0 0
C. Two identical numbers
D. Two different numbers

答案：C

解題說明：
In Python, id(x) returns the "identity" of an object, which in CPython (the most common implementation) is typically the object's memory address. When you write m = 30 and n = 30, both names may refer to thesame integer objectbecause CPython caches a range of small integer objects for efficiency. This optimization means that commonly used small integers are pre-created and reused, so repeated occurrences of the same small integer literal often point to the same object, producing identical id() values. As a result, print(id(m), id (n)) will most likely displaytwo identical numbersin standard CPython builds when 30 falls within the cached range. (Real Python) This behavior is an implementation detail, but it is widely discussed in Python education because it illustrates the difference between object identity (whether two variables reference the same object) and value equality (whether two objects have the same value). Even if id(m) and id(n) were different in some edge environment, m == n would still be True because the values are equal; id() is about identity, not value. The options "0 0" and "Error" are not consistent with how id() works for valid objects.

問題 #53
Given the following code, what is the expected output?

A. array([10, 20, 30, 40])
B. [1, 2, 3, 4]
C. [1, 10]
D. [10, 20, 30, 40]

答案：B

解題說明：
In NumPy, a 2D array can be visualized as a table of rows and columns. When you write np_2d[0], you are usingzero-based indexingto select thefirst rowof that 2D array. This is a standard convention in Python and many other programming languages: index 0 refers to the first element, index 1 to the second, and so on.
Therefore, np_2d[0] returns all the elements in row 0.
With a typical construction such as np_2d = np.array([[1, 2, 3, 4], [10, 20, 30, 40]]), the first row is [1, 2, 3,
4], so printing np_2d[0] displays that row. NumPy returns the row as a 1D NumPy array, and when printed it often appears in bracket form like [1 2 3 4] (spaces rather than commas are common in NumPy's display).
Conceptually, however, the contents are exactly the first row values, matching option C.
Option A and D show the second row (index 1), not the first. Option B incorrectly suggests a column extraction rather than a row selection.

問題 #54
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Foundations-of-Computer-Science權威考題 &最新Foundations-of-Computer-Science考題

WGU Foundations-of-Computer-Science Exam Syllabus Topics:

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