While CS2103T is about technical skills and CS2101 is imbued with insights around the topics of communication, both aimed to teach me how to work with others effectively. The former opened my eyes to working in a technical environment and the latter taught me how to interact with others to strive towards excellence in that environment. My reflection will be deeply involved in how the theme of persuasive presentation connects the dots in the module CS2101.

The first part of the module introduced me to persuasion stemming from being audience-centered. Before taking this module, I have never believe in the thinking that in order to persuade someone, one has to be perform strategic needs analysis or conduct pre-presentation surveys. I valued more on being interesting and informational in a presentation or a speech. That led to a rather average performance during my OP1 presentation which I focused on squeezing in all the content about conflict management styles and disregarded the audience profile. It was not effective as I did not put a premium on connecting with the audience. I learned from this experience that persuasive presentation should start with identifying the target audience and also identifying with the target audience. It would be better if I had analyzed the audience who were my fellow computing students, and constructed my content based on their needs and the benefits that they might get out of from learning about the various management styles.

Following from keeping the target audience in mind, persuasion also requires careful consideration of the purpose that the speaker is trying to achieve. After learning about the audience's preconceived notions and their expectations, a convincing presentation should focus on aspects of the completing values framework (CVF) and employ rhetorical appeals. During OP2 product demo, I tried to strike a balance between the informational and promotional aspects of CVF. On top of presenting the features of the product, I empathized on the design considerations and how they were catered to solve problems that potential users might have. I also utilized "Pathos" by opening my presentation with a relatable story that the group of computing students who were the audience could understand. In my explanation, I also included a reference to a famous design principle named "KISS" to establish credibility in my claims. I ripped the benefits of applying what I learned in the module to organize my presentation to be context and purpose driven.

The last part of persuasion in a presentation comes from having an unified and cohesive delivery with strong teamwork. The difficulty in a team presentation context is with transitioning of ideas and the entire flow of presentation. While I understood the importance of coding styles that were taught in CS2103T to make our code appear unified in the technical context, I realized that I lacked that consideration when preparing for a team presentation. Often times the team members decide on what they want to say before a proper discussion on the flow and purpose of the presentation. This often results in minor issues such as the lack of sign posting and unnatural transition between the team members and sometimes serious drawbacks when team members repeat what was communicated earlier or angled their sharing on a separate front. In this aspect, I think while following a structure such as motivated sequence pattern (MSP) is important to achieving soundness in logic and coherence, the planning of talking points among the team members and constant review of the integration between different teammates is crucial.

In conclusion, I was able to identify and practice many of the said ideas in the opportunities provided by the module. Keeping in mind the devices that help build a persuasive presentation, there are also other important aspects of communication such as participating in audience interaction and providing constructive feedback. Communication skills require practice and I am grateful to be armed with the relevant knowledge to further practice what was taught in the module in the future.

P.S. The following information is based on various resources from the course CS2100 NUS, certain parts are adapted to include my comments and explanations for my better understanding & review. I will try to include source in relevant section as much as possible to honor the authors' work (Mainly Prof Aaron & Prof Colin & TA Alvin).

P.P.S Half way through the course, I decided to change the way I take notes and therefore the information contained here only cover the first few chapters of the module.

C

• Read Array

// while-loop
int readArray(int arr[], int limit) {
    int index, input;
    printf("Enter up to %d integers, terminating with a negative integer.\n", limit);
    index = 0; // Must remember to initialize to zero
    scanf("%d", &input);
    while (input >= 0) {
        arr[index] = input;
        index++;
        scanf("%d", &input);
    }
    return index; // Which is the actual array size
}

// for-loop
int readArray2(int arr[], int limit) {
    int index, input;
    printf("Enter up to %d integers, terminating with a negative integer.\n", limit);
    for (index = 0; index < limit; index ++) {
        if (arr[index] < 0) {
            break;
        }
        scanf("%d", &arr[index]);
    }
    return index - 1; // Which is the actual array size
}

• Reverse Array

// iterative while-loop
void reverseArray(int arr[], int size) {
    int left = 0, right = size - 1, temp;
    while (left < right) {
        temp = arr[left];
        arr[left] = arr[right];
        arr[right] = temp;
        left++;
        right--;
    }
}

// iteractive for-loop
void reverseArray2(int arr[], int size) {
    int start, temp, end = size - 1;
    for (start = 0; start < size / 2; start++) {
        temp = arr[start];
        arr[start] = arr[end];
        arr[end] = temp;
        end--;
    }
}

// recursive
void reverseArray3(int arr[], int size) {
    reverseArrayRec(arr, 0, size - 1);
}

void reverseArrayRec(int arr[], int left, int right) {
    int temp;
    if (left < right) {
        temp = arr[left];
        arr[left] = arr[right];
        arr[right] = temp;
        reverseArrayRec(arr, left + 1, right - 1);
    }
}

// recursive C-style : from TA Alvin CS2100
void reverseArray4(int arr[], int size) {
    if (size <= 1) {
        return;
    } else {
        temp = arr[0];
        arr[0] = arr[size - 1];
        arr[size - 1] = temp;
        // Add 1 to array pointer => shift pointer by 4 bytes
        // => move to next item in array, that is the new base
        // => treat it as removing the front and back items,
        // hence size - 2
        reverseArray4(arr + 1, size - 2)
    }
}

Sign extension

When we want to represent an n-bit signed integer as an m bit signed integer, where m > n. We do this by copying the sign-bit of the n-bit signed m - n times to the left of the n-bit number to create an m-bit number.

To show that sign extension is value-preserving, we can simply check two cases:

• If the signed bit is 0, adding 0s to the left is similar to adding 0 to the right of a decimal point. It does not change the value.
• If the signed bit is 1, adding 1s to the left seems to change the value, but not really. This is because the newly added 1s to the right of the newly added signed bit increases the value while the signed bit decreases the value of the number. This effectively cancels out the changes in value. E.g. Focusing on the value that signed bit and the extended signed bit portion, original: 1001 => -2**3 = -8, extended: 111001 => -2**5 + 2**4 + 2**3 = -8

Addition of two 1's complement number with decimal

• Sign extend or pad zero to the end so that they are of equal length.
• Invert all the bits for negative number to use addition.
• If there is a carry out, add 1 to the result.
• Check for overflow if the result is opposite sign of A and B.

Converting decimal numbers to fixed-point binary

When doing the rounding

• 0.0000 is rounded to 0.000.
• 0.0001 is rounded to 0.001.

Represent decimal value in IEEE 754 format

• When converting the 32 bits to Hexadecimal, make sure that the signed bit is also included.

Bitwise operations

a = 001010
b = 101011
a | b = 101011
a & b = 001010
a ^ b = 100001
~a = 110101
a << 2 = 101000 # equivalent to a *= 4
a >> 2 = 001010 # equivalent to floor(b /= 4)

Special thing about XOR

x ^ x = 0
x ^ 0 = x

a = 00110
a = 00110
a ^ a = 00000

b = 00000
c = 00110
c ^ b = 00110

Swap without temporary variable, with bitwise operator

void swap(int *a, int *b) {
    *a = *a ^ *b
    *b = *a ^ *b # equivalent to *a ^ *b ^ *b = *a
    *a = *a ^ *b # equivalent to *a ^ *b ^ *a = *b
}

MIPS Masking

Summary​

• andi/ori/xori use zero-extension to fill in upper 32 bits. The operation acts on all 32 bits.

andi x, 1 = x
andi x, 0 = 0
ori x, 0 = x
ori x, 1 = 1
nor x, 0 = ~x
xor x, 1 = ~x
xor x, 0 = x

Set bits 2, 8, 16 to 1 in b, a -> $s0, b -> $s1, c -> $s2​

# 16th bit is outside the range of immediate, which is only 0 - 15th bit
# use intermediate registers for results
lui $t0, 0b1                # this sets the 16th bit
ori $t0, 0b0000000100000100 # this sets 0 - 15th bit
or $s1, $s1, $t0

Copy over bits 1, 3 of b into a​

lui $t1, 0b1111111111111111
ori $t1, t1, 0b1111111111110101
and $s0, $s0, $t1                 # clear the two bits from a
andi $t0, $s1, 0b0000000000001010 # get the two bits from b
or $s0, $s0, $t0                  # copy the bits over

Make bits 2, 4 of c inverse of bits 1, 3 of b​

xori $t0, $s1, 0b0000000000001010 # this invert and copy
andi $t0, $t0, 0b0000000000001010 # clear everything else
sll $t0, $t0, 1                   # shift left to 2, 4
lui $t1, $t1, 0b1111111111111111  # clear out 2, 4 of c, 3 steps, lui, ori, and
ori $t1, $t1, 0b1111111111101011
and $s2, $s2, $t1
or $s2, $s2, $t0                  # copy over to c

MIPS arithmetic

a -> $s0 b -> $s1 c -> $s2 d -> $s3

d = 6a + 3(b - 2c) d = 6a + 3b - 6c d = 3(2a + b - 2c) d = 3(2(a - c) + b)

sub $t0, $s0, $s2   # a - c
sll $t0, $t0, 1     # 2(a - c)
add $t0, $t0, $s1   # 2(a - c) + b
sll $s3, $t0, 2     # 4(2(a - c) + b)
sub $s3, $s3, $t0   # 3(2(a - c) + b)

MIPS tracing

add $t0, $s0, $ zero    # make a copy
lui $t1, 0x8000         # set t1 to 1 at MSB and 0 else where
lp: beq $t0, $zero, e   # if t0 == zero, exit
    andi $t2, $t0, 1    # t2 left with LSB
    beq $t2, $zero, s   # if t2 LSB == 0, s
    xor $s0, $s0, $t1   # invert MSB
s:  srl $t0, $t0, 1     # discard LSB
    j   lp              # loop back
e:                      # exit

What happens when integer overflow in C

• int is 4 bytes (in sunfire), which is 4 x 8 = 32 bits
• the range is from -2,147,483,648 (-2^31) through +2,147,483,647(2^31 - 1)
• This range comes from 32 bits 2s complement representation,
• from 10000000000000000000000000000000 (smallest negative number)
• to 01111111111111111111111111111111 (largest positive number)
• when adding 1 to the largest positive number, it becomes the smallest negative number
• adding another 1 will make it 10000000000000000000000000000001,
• which is -2147483647, (literally add 1 to the decimal number)

For an n-bit sign-and-magnitude representation, what is the range of values that can be represented?

• -(2^(n-1) - 1)
• 2^(n-1) - 1

1s Complement

• -x = 2^n - 1 - x (in decimal)
• -x = 2^n - 1 - x + 1 (in decimal)
• minus 1 because you want the largest possible values in that base

2s Complement

• -2^(n - 1)
• 2^(n-1) - 1
• Adding A + B
• Ignore the carry out of the MSB
• Check for overflow. Overflow occurs if the carry in and the carry out of the MSB are different, or if result is opposite sign of A and B

P.S. The following information on CS2030/S Programming Methodology II is based on past experience and subject to changes. The purpose is to provide a rough idea of what is to come for prospective students so that one can prepare early if possible.

In the recent run of the module, "A module about abstractions" is Prof Henry's description for CS2030. Having the opportunity to be a lab tutor for this module, I think it is a very fair way to describe it.

Updated as of: 28/4/2021