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School
Mindanao State University - Iligan Institute of Technology *
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Course
3867
Subject
English
Date
Nov 24, 2024
Type
Pages
8
Uploaded by PrivateRose23317
NED University of Engineering & Technology
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Quick Submit
Snanja
Jsjajja Snna
Document Details
Submission ID
trn:oid:::1:2761760387
Submission Date
Nov 25, 2023, 7:04 PM GMT+5
Download Date
Dec 1, 2023, 9:00 AM GMT+5
File Name
Report-1.docx
File Size
456.7 KB
6 Pages
815 Words
3,453 Characters
Page 1 of 8 - Cover Page
Submission ID trn:oid:::1:2761760387
Page 1 of 8 - Cover Page
Submission ID trn:oid:::1:2761760387
How much of this submission has been generated by AI?
0%
of qualifying text in this submission has been determined to be
generated by AI.
Caution: Percentage may not indicate academic misconduct. Review required.
It is essential to understand the limitations of AI detection before making decisions
about a student's work.
We encourage you to learn more about Turnitin's
AI detection
capabilities before using the tool.
Frequently Asked Questions
What does the percentage mean?
The percentage shown in the AI writing detection indicator and in the AI writing report is the amount of qualifying text within the
submission that Turnitin's AI writing detection model determines was generated by AI.
Our testing has found that there is a higher incidence of false positives when the percentage is less than 20. In order to reduce the
likelihood of misinterpretation, the AI indicator will display an asterisk for percentages less than 20 to call attention to the fact that
the score is less reliable.
However, the final decision on whether any misconduct has occurred rests with the reviewer/instructor. They should use the
percentage as a means to start a formative conversation with their student and/or use it to examine the submitted assignment in
greater detail according to their school's policies.
How does Turnitin's indicator address false positives?
Our model only processes qualifying text in the form of long-form writing. Long-form writing means individual sentences contained in paragraphs that make up a
longer piece of written work, such as an essay, a dissertation, or an article, etc. Qualifying text that has been determined to be AI-generated will be highlighted blue
on the submission text.
Non-qualifying text, such as bullet points, annotated bibliographies, etc., will not be processed and can create disparity between the submission highlights and the
percentage shown.
What does 'qualifying text' mean?
Sometimes false positives (incorrectly flagging human-written text as AI-generated), can include lists without a lot of structural variation, text that literally repeats
itself, or text that has been paraphrased without developing new ideas. If our indicator shows a higher amount of AI writing in such text, we advise you to take that
into consideration when looking at the percentage indicated.
In a longer document with a mix of authentic writing and AI generated text, it can be difficult to exactly determine where the AI writing begins and original writing
ends, but our model should give you a reliable guide to start conversations with the submitting student.
Disclaimer
Our AI writing assessment is designed to help educators identify text that might be prepared by a generative AI tool. Our AI writing assessment may not always be accurate (it may misidentify
both human and AI-generated text) so it should not be used as the sole basis for adverse actions against a student. It takes further scrutiny and human judgment in conjunction with an
organization's application of its specific academic policies to determine whether any academic misconduct has occurred.
Page 2 of 8 - AI Writing Overview
Submission ID trn:oid:::1:2761760387
Page 2 of 8 - AI Writing Overview
Submission ID trn:oid:::1:2761760387
Problem 1
Instructions:
- Use "ORG" instructions to start your program at address equivalent to 1400
8
.
- Use your last university ID number to input the new values of x and y respectively. For example, if
your ID is 2415161678235, then you will use the number 3 as the value of x and 5 as the value of y.
Do not forget to change the Input and output boxes to decimal!
Proper labels and directives should be included at the end of your program.
a) Write MARIE code to perform the following pseudocode excerpt.
Input a value for x
Input a value for y if(x > y)
max= x;
else
max=y;
Output the value of max
MARIE Code:
Code (Without Comments)
Code (With Comments)
ORG 300
Input
Store x
Input
Store y
Load x
Subt y
Skipcond 800
Jump Else
Load x
Store max
Output
Halt
Else, Load y
Store max
Output
Halt
x, DEC 0
y, DEC 0
max, DEC 0
ORG 300
// Start the program at 1400
8
(300
16
)
Input
// Input x
Store x
// Store x in memory location x
Input
// Input y
Store y
// Store y in memory location y
Load x
// Load the value of x
Subt y
// Subtract y from x
Skipcond 800 // Skip the next instruction if result is positive (x > y)
Jump Else
// Jump to Else if x is not greater than y
Load x
// Load x if x > y
Store max // Store x in max
Output
// Output the value of max
Halt
Else, Load y
// Load y if x is not greater than y
Store max // Store y in max
Output
// Output the value of max
Halt
x, DEC 0
// Memory location to store x
y, DEC 0
// Memory location to store y
max, DEC 0 // Memory location to store max
Page 3 of 8 - AI Writing Submission
Submission ID trn:oid:::1:2761760387
Page 3 of 8 - AI Writing Submission
Submission ID trn:oid:::1:2761760387
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According to the instructions, the origin of the code should be from 1400
8
(equivalent hexadecimal 300
16
).
Wiled the labeled code is also attached but for some simulator like the Online Marie Simulator may cause
an error so we have to remove the comments.
b) Provide a screenshot of the simulation result (A screenshot of the MARIE Simulator window after
running the program, showing the value at the output window).
On the left side of the assembler is shown while on right side there is the output log. While on the bottom
there are the memory allocations. Here the code has been run using step by step to observe the output and
RTL log. Whenever a step is taken the program counter PC and other registers like AC and IR, IN and
OUT are changed.
As the student ID is
212183
so we will take
x = 8
and
y = 3
. As all the steps results snaps are given as
following.
Page 4 of 8 - AI Writing Submission
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Page 4 of 8 - AI Writing Submission
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Page 5 of 8 - AI Writing Submission
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Page 6 of 8 - AI Writing Submission
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Page 6 of 8 - AI Writing Submission
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Page 7 of 8 - AI Writing Submission
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Page 7 of 8 - AI Writing Submission
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Problem 2
Suppose that you have a computer with a 16K x 24 word-addressable main memory. In this
computer, the assembly program's instruction set consists of 585 different operations. All
instructions have an operation code part (opcode) and an address part (allowing for only one
address). Each instruction is stored in one word of memory.
a.
How many bits are needed for the opcode?
The given operation is 585 so to represent these operations we have to find enough bits to find each
operation. The required bits are calculated by the following formula.
No. of Required Operations = Log
2
(585) = 9.191
As the answer is in the digits so we have to take the ceiling of the answer to find the exact number
of bits.
No. of Required Operations = Ceiling [Log
2
(585)] = ceiling [9.191] = 10
So, a total of 10 bits are needed for the opcode.
b.
How many bits are left for the address part of the instruction?
The total number of bits in each individual instruction are 16-bits (according to the word size). As
according to the previous question, the 10 bits are used for the Opcode, so the remaining bits are for
the address.
Address Bits = Instruction Bits
–
Opcode bits = 16
–
10 = 6
c.
How many additional instructions could be added to this instruction set without exceeding the
assigned number of bits? Discuss and show your calculations.
If every instruction uses the 16-bits where 10-bits are used for the Opcode the remaining 6-bits can
be used for the additional instructions as well. The number of additional instructions that could be
added in our set are given as,
No. of Additional Instructions = 2
6
= 64
d.
What is the largest unsigned binary number that the address can hold?
As already in the previous question we have calculated the instructions count now for the largest
unsigned binary instruction address will be 63 if the address started from 0.
Largest Unsigned Address = max (2
6
-1) = 63
Page 8 of 8 - AI Writing Submission
Submission ID trn:oid:::1:2761760387
Page 8 of 8 - AI Writing Submission
Submission ID trn:oid:::1:2761760387