# Lab 7
Name: G Jaswanth
Roll No.: CS22B020

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## Question 1

To convert 0.1 (in base ten) to binary, we can use the method of multiplying 
by 2 and noting the integer part of the result at each step.
Here are the steps:

1. Multiply 0.1 by 2: 0.1 * 2 = 0.2, integer part is 0
2. Multiply 0.2 by 2: 0.2 * 2 = 0.4, integer part is 0
3. Multiply 0.4 by 2: 0.4 * 2 = 0.8, integer part is 0
4. Multiply 0.8 by 2: 0.8 * 2 = 1.6, integer part is 1
5. Multiply 0.6 by 2: 0.6 * 2 = 1.2, integer part is 1
6. Multiply 0.2 by 2: 0.2 * 2 = 0.4, integer part is 0
7. Multiply 0.4 by 2: 0.4 * 2 = 0.8, integer part is 0
8. Multiply 0.8 by 2: 0.8 * 2 = 1.6, integer part is 1
9. Multiply 0.6 by 2: 0.6 * 2 = 1.2, integer part is 1

We continue this process until we either reach the desired precision or the 
fractional part becomes zero. The binary representation of 0.1 is therefore
0.0001100110011... (the digits 1100 repeat indefinitely).

### Output

0.0001100110011... 

## Question 2

-> sign bit of 0.1 is 0
-> now it's exponent in scientific form is -4 so it's exponent is 123 (01111011)
since it's bias for precesion is 127 

now 0.1/0.0625 is equal to 1.6

now representing 0.6 in binary form:

1. Start with 0.6.
2. Multiply 0.6 by 2:
   - 0.6 * 2 = 1.2 ⟶ Bit: 1
3. Take the fractional part (0.2) and repeat:
   - 0.2 * 2 = 0.4 ⟶ Bit: 0
   - 0.4 * 2 = 0.8 ⟶ Bit: 0
   - 0.8 * 2 = 1.6 ⟶ Bit: 1
   - 0.6 * 2 = 1.2 ⟶ Bit: 1
   - 0.2 * 2 = 0.4 ⟶ Bit: 0
   - 0.4 * 2 = 0.8 ⟶ Bit: 0
   - 0.8 * 2 = 1.6 ⟶ Bit: 1
   - 0.6 * 2 = 1.2 ⟶ Bit: 1
   - ... and so on.

The pattern repeats: 1001. Note that this pattern repeats every 4 bits after the binary point.
To get 23 bits after the binary point, we can extend this pattern:

0.6 in binary with 23 bits after the binary point:
0.10011001100110011001100

### Output:

0 01111011  10011001100110011001100

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## Question 3

**Part1**

The exponent of 0.2 in scientific notation is -3 so it's value is 124

representation of 124 in binary is 01111100

when we divide 0.2/0.125 we get 1.6 , represent 0.6 in 23 bit binary number:

now representing 0.6 in binary form:

1. Start with 0.6.
2. Multiply 0.6 by 2:
   - 0.6 * 2 = 1.2 ⟶ Bit: 1
3. Take the fractional part (0.2) and repeat:
   - 0.2 * 2 = 0.4 ⟶ Bit: 0
   - 0.4 * 2 = 0.8 ⟶ Bit: 0
   - 0.8 * 2 = 1.6 ⟶ Bit: 1
   - 0.6 * 2 = 1.2 ⟶ Bit: 1
   - 0.2 * 2 = 0.4 ⟶ Bit: 0
   - 0.4 * 2 = 0.8 ⟶ Bit: 0
   - 0.8 * 2 = 1.6 ⟶ Bit: 1
   - 0.6 * 2 = 1.2 ⟶ Bit: 1
   - ... and so on.

The pattern repeats: 1001. Note that this pattern repeats every 4 bits after the binary point. To get 23 bits after the binary point, we can extend this pattern:

0.6 in binary with 23 bits after the binary point:
0.10011001100110011001100

now entire representation:
0 01111100 10011001100110011001100

**Part2**

now reverting back

sign bit 0

exponent is 124 so the exponent in scientific notation is -3

now we can see in binary form it is 

$$
1 \times 2^{-1} + 0 \times 2^{-2} + 0 \times 2^{-3}+\ldots
$$

so we approximately get 0.6

so answer is 1.6*(0.125)

so value is 0.2

## Question 4

we can use a simple scheme where each floating-point number is represented by two registers:
one for the integer part and one for the fractional part.
We will assume a fixed number of decimal places for simplicity.

Let's represent the floating-point numbers using two registers (R1 and R2 for the integer and fractional parts, respectively).
Each register can store values from 0 to 9. 
We'll perform addition using this representation and store the result in another pair of registers (R3 and R4).

Given the numbers to add are 5.391 and 3.012, we initialize the registers as follows:

- R1: 5 (integer part of 5.391)
- R2: 391 (fractional part of 5.391, after the decimal point)
- R3: 3 (integer part of 3.012)
- R4: 012 (fractional part of 3.012, after the decimal point)

Now, let's perform the addition:

1. Add the fractional parts:
   - R2 (391) + R4 (012) = 403 (carry 0, result in R4)

2. Add the integer parts with the carry from the fractional part addition:
   - R1 (5) + R3 (3) + carry (0) = 8 (result in R3)

The result stored in registers R3 and R4 is 8.403,
which corresponds to the addition of 5.391 and 3.012. 
Note that this representation is simplistic and assumes fixed decimal places and limited range of values in each register. 

### Question 5

.data
num1: .word 0b01000001001100000000000000000000 # 11.0
num2: .word 0b11000001001010000000000000000000 # -10.5
num3: .word 0
exmask: .word 0x7f800000
mmask: .word 0b00000000011111111111111111111111
extra1: .word 0b00000000100000000000000000000000
mantissa: .word 0b00000000010000000000000000000000

.text
lw x1 num1
lw x2 num2
srli a1 x1 31 # sign
srli a2 x2 31 
lw t1 exmask 
lw t2 mmask
lw t0 extra1

# Handling the sign parities seperately, increased redundancy

beq a1 a2 same
j notsame

same:
    and a1 x1 t1
    srli s1 a1 23 # exponent
    mv s9 s1
    and a1 x1 t2 # mantissa
    or a1 a1 t0
    
    and a2 x2 t1
    srli s2 a2 23
    mv s10 s2 
    and a2 x2 t2
    or a2 a2 t0
    
expsame:
    beq s10 s9 exit
    blt s10 s9 inc
    addi s9 s9 1
    srli a1 a1 1
    j expsame 
    
inc:
    addi s10 s10 1
    srli a2 a2 1 
    j expsame
    
exit:
    add t6 a1 a2 # mantissa sum
    srli t5 t6 23 # extra mantissa
    and t6 t6 t2
    addi x1 x0 1
    
shiftexp:
    beq x1 t5 exit1
    addi s9 s9 1
    srli t5 t5 1
    srli t6 t6 1
    j shiftexp
    
exit1:
    #concatenate the answer
    
    lw x1 num1
    srli a1 x1 31
    slli a1 a1 31
    slli s9 s9 23
    add a1 a1 s9
    and t6 t6 t2
    add a1 a1 t6
    la x1 num3
    sw a1 0(x1)
    j exit3

notsame:
    # Assuming abs(a) > abs(b)
    # Otherwise need to swap'em
    
    and a1 x1 t1
    srli s1 a1 23 # exponent
    mv s9 s1
    and a1 x1 t2 # mantissa
    or a1 a1 t0
    
    and a2 x2 t1
    srli s2 a2 23
    mv s10 s2 
    and a2 x2 t2
    or a2 a2 t0
    
expnotsame:
    beq s10 s9 exit2
    blt s10 s9 inc2
    addi s9 s9 1
    srli a1 a1 1
    j expnotsame 
    
inc2:
    addi s10 s10 1
    srli a2 a2 1 
    j expnotsame
    
exit2:
    sub t6 a1 a2 # mantissa difference
    lw s7 extra1
    mv s4 t6
    and s6 s7 t6
    
shiftexp2:
    beq s6 s7 exit1
    slli t6 t6 1
    addi s9 s9 -1
    and s6 s7 t6
    j shiftexp2

exit3:
    addi a7 x0 10
    ecall

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