# Computer Archtiecture HW3
## 1

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## 2

| iteration | Step | Multiplier | Multiplicand | Product |
|:--------- |:--------------------------- |:----------:|:------------:|:--------:|
| 0 | Initial Value | 0110 | 00000101 | 00000000 |
| 1 | 1 : 0->no operation | 0110 | 00000101 | 00000000 |
| 1 | 2 : Shift left Multiplicand | 0110 | 00001010 | 00000000 |
| 1 | 3 : Shift right Multiplier | 0011 | 00001010 | 00000000 |
| 2 | 1a: 1->P+=Multiplicand | 0011 | 00001010 | 00001010 |
| 2 | 2 : Shift left Multiplicand | 0011 | 00010100 | 00001010 |
| 2 | 3 : Shift right Multiplier | 0001 | 00010100 | 00001010 |
| 3 | 1a: 1->P+=Multiplicand | 0001 | 00010100 | 00011110 |
| 3 | 2 : Shift left Multiplicand | 0001 | 00101000 | 00011110 |
| 3 | 3 : Shift right Multiplier | 0000 | 00101000 | 00011110 |
| 4 | 1 : 0->no operation | 0000 | 00101000 | 00011110 |
| 4 | 2 : Shift left Multiplicand | 0000 | 01010000 | 00011110 |
| 4 | 3 : Shift right Multiplier | 0000 | 01010000 | 00011110 |

0101 x 0110 = 00011110

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## 3
### a

| Hex | Binary | Decimal(2’s complement) | Decimal(unsigned) |
|:-------- |:---------------------------------------:|:-------------------------:|:-----------------:|
| 013462FD | 0000 0001 0011 0100 0110 0010 1111 1101 | 20210429 | 20210429 |

The result is same as the two’s complement.

### b

| Hex | Binary | Decimal(2’s complement) | Decimal(unsigned) |
|:-------- |:---------------------------------------:|:-----------------------:|:-----------------:|
| FECB9D03 | 1111 1110 1100 1011 1001 1101 0000 0011 | -20210429 | 4274756867 |

The result is not same as the two’s complement.
Because in this case the most left bit is 1, which is the sign bit in two’s complement.

### c

| Hex | Binary | S | Exponent | Fraction |
|:-------- |:----------------------------------:|:---:|:--------:|:-----------------------:|
| 013462FD | 0 00000010 01101000110001011111101 | 0 | 00000010 | 01101000110001011111101 |
| FECB9D03 | 1 11111101 10010111001110100000011 | 1 | 11111101 | 10010111001110100000011 |

Bias = 127
| Hex | Decimal(IEEE 754 float) |
|:-------- |:-----------------------:|
| 013462FD | 3.31317992256e-38 |
| FECB9D03 | -1.35324267299e+38 |
//Excat decimal nunber is caculated by IEEE 764 floating converter

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## 4
### a

| Decimal | Binary |
|:-------:|:----------------------------------:|
| 68.32 | 0 10000101 00010001010001111010111 |
S = 0
bias = 127
Exponent = 6+127 = 133 = 10000101
Fraction = 1.00010001010..... -1 = 00010001010001111010111
//Exact Fraction is caculated by IEEE 764 floating converter

### b
| Decimal | Binary |
|:--------:|:------------------------------------------:|
| -126.751 | 0 10000000101 11111011000000010000011..... |
S = 1
bias = 1023
Exponent = 6+1023 = 1029 = 10000000101
Fraction = 1.1111101100000..... -1 = 11111011000000010000011....(52 bits)

## 5
### a

a0 = (-1)^0 x 1.00000000 x 2^(-62)
0 0000001 00000000

### b
The largest denormalized number:
a1 = (-1)^0 x 0.11111111 x 2^(-62)
0 0000000 11111111

The second largest denormalized number:
a2 = (-1)^0 x 0.11111110 x 2^(-62)
0 0000000 11111110
### c
a0 和 a1在[8:0]是呈現完全相反的，或者從另一個角度來看a0 - 1 = a1
a0 = 0 0000001 00000000
a1 = 0 0000000 11111111

a1 和 a2只在最後一位一個是一，一個是零，從另一個角度來看a1 - 1 = a2
a1 = 0 0000000 11111111
a2 = 0 0000000 11111110

### d
1 1111111 00000000 = -infinity
0 1111111 01010011 = NAN(*Not a Number*)

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## 6
### a
不可以，因為1/6是無限循環小數，所以用浮點數表示法，一定會有些表示誤差
用浮點數表示法會表示成
0 01111100 01010101010101010101011
實際值為0.16666667163372039794921875
與理論值0.166666666666666...有些許的誤差
### b
實際作加法時

   101010101010101010101011  
+  101010101010101010101011  
+ 1010101010101010101010110  
=10000000000000000000000001
fraction = 0000 0000 0000 0000 0000 0000 1<-這個1會捨去因為超出24bit的長度

最後結果為0 01111110 00000000000000000000000
正確答案為0 01111110 00000000000000000000000
兩者並沒有差