Problem Set: Binary Floating-Point Representation

Tags: #◇Computer Science #◇CSARCH2 #◇Computer Architecture #◇Problem Sets #◇IEEE-754 #◇Floating-Point Representation #◇Single Precision #◇Double Precision #◇Subnormal Numbers #◇Binary Conversion

◇Overview & Theoretical Foundations

In digital computing, fixed-point representations suffer from a severe dynamic range limitation: allocating a fixed number of integer and fractional bits restricts the range of numbers that can be represented. The IEEE-754 Standard for Floating-Point Arithmetic resolves this by representing real numbers in scientific notation:

$\text{Value} = (-1)^{\text{Sign}} \times \text{Significand} \times 2^{\text{Exponent}}$

IEEE-754 Single Precision (32-bit):
 1 bit       8 bits                  23 bits
┌──────┬──────────────────┬───────────────────────────────────────┐
│ Sign │  Biased Exponent │         Fraction (Mantissa)           │
│ (S)  │       (E)        │                 (F)                   │
└──────┴──────────────────┴───────────────────────────────────────┘
 31     30              23 22                                    0

IEEE-754 Double Precision (64-bit):
 1 bit       11 bits                 52 bits
┌──────┬──────────────────┬───────────────────────────────────────┐
│ Sign │  Biased Exponent │         Fraction (Mantissa)           │
│ (S)  │       (E)        │                 (F)                   │
└──────┴──────────────────┴───────────────────────────────────────┘
 63     62              52 51                                    0

1. Structural Parameters of IEEE-754 Formats

2. Classification of Representable Numbers

IEEE-754 partitions the exponent field $E$ into special regimes:

                  ┌─────────────────┬──────────────────┐
                  │      E = 0      │   1 <= E <= Emax │
                  ├────────┬────────┼──────────────────┤
                  │ F = 0  │ F != 0 │                  │
                  ├────────┼────────┼──────────────────┤
                  │ Zero   │Subnorm │ Normalized Number│
                  │ (+/-0) │(Denorm)│                  │
                  └────────┴────────┴──────────────────┘
                                    ┌──────────────────┐
                                    │     E = Emax     │
                                    ├────────┬─────────┤
                                    │ F = 0  │  F != 0 │
                                    ├────────┼─────────┤
                                    │Infinity│   NaN   │
                                    │ (+/-oo)│(Quiet/  │
                                    │        │Signal)  │
                                    └────────┴─────────┘

1. Normalized Numbers ($1 \le E \le 2^k - 2$):
   • Significand has an implicit leading bit of $1$: $M = 1.F = 1 + \sum_{i=1}^p f_i 2^{-i}$.
   • Unbiased exponent: $e = E - \text{Bias}$.
   • $\text{Value} = (-1)^S \times (1.F)_2 \times 2^{E - \text{Bias}}$.
2. Subnormal / Denormalized Numbers ($E = 0, F \neq 0$):
   • Used for gradual underflow to fill the gap between $0$ and the smallest normalized number.
   • No implicit leading 1 (leading bit is $0$): $M = 0.F$.
   • The unbiased exponent is fixed at: $e = 1 - \text{Bias}$ (i.e., $-126$ for single, $-1022$ for double).
   • $\text{Value} = (-1)^S \times (0.F)_2 \times 2^{1 - \text{Bias}}$.
3. Signed Zero ($E = 0, F = 0$):
   • Represents $+0$ ($S=0$) or $-0$ ($S=1$).
4. Infinity ($E = 2^k - 1, F = 0$):
   • Represents $+\infty$ ($S=0$) or $-\infty$ ($S=1$), produced by overflow or division by zero.
5. Not a Number / NaN ($E = 2^k - 1, F \neq 0$):
   • Represents invalid operations (e.g., $0/0$, $\sqrt{-1}$).
   • Quiet NaN (QNaN): MSB of fraction is $1$ ($f_1 = 1$).
   • Signaling NaN (SNaN): MSB of fraction is $0$ ($f_1 = 0, F \neq 0$).

◇Problem Solutions

Problem 1: Convert $1.1101_2 \times 2^{135}$ to IEEE-754 Single Precision

Problem Statement

Represent the floating-point value $+1.1101_2 \times 2^{135}$ using 32-bit single-precision binary format. Specify the sign bit, 8-bit exponent representation, 23-bit fraction, and hexadecimal representation.

Step-by-Step Derivation

1. Sign Bit ($S$):

   • The value is positive $\implies S = \mathbf{0}$.

2. Exponent Analysis & Calculation:

   • Unbiased exponent: $e = 135$.
   • Single-precision bias: $\text{Bias} = 127$.
   • Biased exponent:
     $E = e + \text{Bias} = 135 + 127 = 262$
   • In single precision, the 8-bit exponent field $E$ can only represent integers from $0$ to $255$ ($2^8-1$).
   • Maximum finite normalized exponent is $E_{\max} = 254$ (which corresponds to $e_{\max} = 254 - 127 = +127$).
   • Since $e = 135 > 127$ (and $E = 262 > 255$), this value exceeds the maximum representable finite number in IEEE-754 single precision.

3. IEEE-754 Standard Overflow Handling:

   • Because of Exponent Overflow, the IEEE-754 standard maps this value to Positive Infinity ($+\infty$):

• $\text{Sign } S = 0$
• $\text{Exponent } E = 255 = 11111111_2$
• $\text{Fraction } F = 00000000000000000000000_2$ (23 zeros)
• 32-bit pattern: 0 11111111 00000000000000000000000
• Hexadecimal: 0x7F800000

4. Theoretical / Unbounded 9-Bit Exponent Derivation (Academic Context):
   • If evaluated without clamping to standard single-precision overflow saturation:

• $262_{10} = 256 + 6 = 100000110_2$ (9 bits).
• Truncated to 8 bits modulo 256: $262 \pmod{256} = 6 = 00000110_2$.
• Fraction (23 bits): $F = 1101\,0000\,0000\,0000\,0000\,000_2$.

Result Summary Table

[!warning] Architecture Note
The maximum representable finite float in 32-bit single precision is $(2 - 2^{-23}) \times 2^{127} \approx 3.4028235 \times 10^{38}$. Since $1.1101_2 \times 2^{135} \approx 7.89 \times 10^{40}$, it strictly overflows single precision.

Problem 2: Convert $1.1101_2 \times 2^{135}$ to IEEE-754 Double Precision

Problem Statement

Represent $+1.1101_2 \times 2^{135}$ using 64-bit double-precision binary format.

Step-by-Step Derivation

1. Sign Bit ($S$):

   • Positive number $\implies S = \mathbf{0}$.

2. Exponent Calculation:

   • Unbiased exponent: $e = 135$.
   • Double-precision bias: $\text{Bias} = 1023$.
   • Biased exponent:
     $E = e + 1023 = 135 + 1023 = 1158$
   • Convert $1158_{10}$ to 11-bit binary:
     $1158 = 1024 + 128 + 4 + 2 = 2^{10} + 2^7 + 2^2 + 2^1$
     $E = \mathbf{10000000000}_2 + \mathbf{00010000000}_2 + \mathbf{00000000110}_2 = \mathbf{100\,1000\,0110}_2$
   • Check range: $1 \le 1158 \le 2046 \implies$ Valid normalized exponent.

3. Fraction Part ($F$):

   • Significand: $M = 1.1101_2$.
   • Implicit leading 1 is dropped, leaving fraction bits: $1101_2$.
   • Pad with zeros to 52 bits:
     $F = \mathbf{1101}\underbrace{000000000000000000000000000000000000000000000000}_{48\text{ zeros}}$

4. Assemble 64-bit Binary Representation:
   $\underbrace{0}_{S\text{ (1)}} \quad \underbrace{100\,1000\,0110}_{E\text{ (11)}} \quad \underbrace{1101\,0000\,0000\,\dots\,0000}_{F\text{ (52)}}$

5. Convert to Hexadecimal (4-bit nibbles):

   • 0100 $\to$ 4
   • 1000 $\to$ 8
   • 0110 $\to$ 6
   • 1101 $\to$ D
   • 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 $\to$ 000000000000
   • Hexadecimal: 0x486D000000000000

Result Summary Table

Problem 3: Convert $1.1101_2 \times 2^{-135}$ to IEEE-754 Single Precision

Problem Statement

Represent $+1.1101_2 \times 2^{-135}$ using 32-bit single-precision format.

Step-by-Step Derivation

1. Sign Bit ($S$):

   • Positive number $\implies S = \mathbf{0}$.

2. Exponent Analysis & Subnormal Detection:

   • Unbiased exponent: $e = -135$.
   • Single-precision bias: $\text{Bias} = 127$.
   • Naive biased exponent:
     $E = -135 + 127 = -8$
   • Because $E \le 0$, this number cannot be represented as a normalized single-precision float (minimum normal exponent is $e_{\min} = -126$ with $E = 1$).
   • This value falls into the Subnormal (Denormalized) range!

3. Subnormal Representation Mechanism:

   • In IEEE-754 subnormals:

• The exponent field is set to $E = 00000000_2$ ($0$).
• The implicit leading bit is $0$ rather than $1$.
• The actual exponent is fixed to:
  $e_{\text{subnormal}} = 1 - \text{Bias} = 1 - 127 = -126$
  • We must equate:
    $(-1)^0 \times (0.F)_2 \times 2^{-126} = 1.1101_2 \times 2^{-135}$
  • Solve for $(0.F)_2$:
    $(0.F)_2 = 1.1101_2 \times 2^{-135 - (-126)} = 1.1101_2 \times 2^{-9}$
  • Shifting $1.1101_2$ right by $9$ positions:
    $(0.F)_2 = 0.\underbrace{00000000}_{8\text{ zeros}}11101_2$
  • The 23-bit fraction field $F$ is:
    $F = \underbrace{00000000}_{8\text{ zeros}}11101\underbrace{0000000000}_{10\text{ zeros}} = \mathbf{0000\,0000\,1110\,1000\,0000\,000}_2$

4. Assemble 32-bit Binary Layout:

   • Bit 31 ($S$): 0
   • Bits 30–23 ($E$): 00000000
   • Bits 22–0 ($F$): 00000000111010000000000

5. Convert to Hexadecimal:
   $\begin{aligned} \text{Bits 31–28:} &\quad 0000 \to \mathbf{0} \\ \text{Bits 27–24:} &\quad 0000 \to \mathbf{0} \\ \text{Bits 23–20:} &\quad 0000 \to \mathbf{0} \\ \text{Bits 19–16:} &\quad 0000 \to \mathbf{0} \\ \text{Bits 15–12:} &\quad 0111 \to \mathbf{7} \\ \text{Bits 11–8:} &\quad 0100 \to \mathbf{4} \\ \text{Bits 7–4:} &\quad 0000 \to \mathbf{0} \\ \text{Bits 3–0:} &\quad 0000 \to \mathbf{0} \end{aligned}$

   • Hexadecimal: 0x00007400

Result Summary Table

Problem 4: Convert $1.1101_2 \times 2^{-135}$ to IEEE-754 Double Precision

Problem Statement

Represent $+1.1101_2 \times 2^{-135}$ using 64-bit double-precision format.

Step-by-Step Derivation

1. Sign Bit ($S$):

   • Positive number $\implies S = \mathbf{0}$.

2. Exponent Calculation:

   • Unbiased exponent: $e = -135$.
   • Double-precision bias: $\text{Bias} = 1023$.
   • Biased exponent:
     $E = e + 1023 = -135 + 1023 = 888$
   • Convert $888_{10}$ to 11-bit binary:
     $888 = 512 + 256 + 64 + 32 + 16 + 8 = 2^9 + 2^8 + 2^6 + 2^5 + 2^4 + 2^3$
     $E = \mathbf{011\,0111\,1000}_2$
   • Check range: $1 \le 888 \le 2046 \implies$ Valid normalized double-precision number!

3. Fraction Part ($F$):

   • Significand: $M = 1.1101_2$.
   • Drop implicit leading 1: $1101_2$.
   • Pad to 52 bits:
     $F = \mathbf{1101}\underbrace{000000000000000000000000000000000000000000000000}_{48\text{ zeros}}$

4. Assemble 64-bit Binary Representation:
   $\underbrace{0}_{S\text{ (1)}} \quad \underbrace{011\,0111\,1000}_{E\text{ (11)}} \quad \underbrace{1101\,0000\,\dots\,0000}_{F\text{ (52)}}$

5. Convert to Hexadecimal:

   • 0011 $\to$ 3
   • 0111 $\to$ 7
   • 1000 $\to$ 8
   • 1101 $\to$ D
   • Trailing 48 zero bits $\to$ 000000000000
   • Hexadecimal: 0x378D000000000000

Result Summary Table

Problem 5: Decode Single-Precision 0xC1C20000 to Decimal

Problem Statement

Determine the equivalent fixed-point decimal value of the IEEE-754 single-precision floating-point bit pattern 0xC1C20000.

Step-by-Step Derivation

1. Convert Hexadecimal to 32-bit Binary:
   $\begin{aligned} \text{Hex: } &\quad \mathtt{C} & \mathtt{1} & \mathtt{C} & \mathtt{2} & \mathtt{0} & \mathtt{0} & \mathtt{0} & \mathtt{0} \\ \text{Binary: } &\quad 1100 & 0001 & 1100 & 0010 & 0000 & 0000 & 0000 & 0000 \end{aligned}$

2. Deconstruct IEEE-754 Single-Precision Fields:

   • Sign Bit ($S$, bit 31):
     $S = 1 \implies \text{Negative number } (-1)^1 = -1$
   • Biased Exponent ($E$, bits 30–23):
     $\text{Bits: } 10000011_2 = 128 + 2 + 1 = 131_{10}$
     Unbiased exponent $e$:
     $e = E - \text{Bias} = 131 - 127 = \mathbf{+4}$
   • Fraction Field ($F$, bits 22–0):
     $\text{Bits: } 100\,0010\,0000\,0000\,0000\,0000_2$
     Significand (with implicit leading 1):
     $M = 1.F = 1.100001_2$

3. Compute Value in Binary Scientific Notation:
   $\text{Value} = -1 \times 1.100001_2 \times 2^4$

4. Shift Radix Point by Exponent ($e = +4$):
   $1.100001_2 \times 2^4 = 11000.01_2$

5. Convert Fixed-Point Binary to Decimal:

   • Integer Part ($11000_2$):
     $1 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 0 \cdot 2^1 + 0 \cdot 2^0 = 16 + 8 = \mathbf{24}$
   • Fractional Part ($0.01_2$):
     $0 \cdot 2^{-1} + 1 \cdot 2^{-2} = 0 + \frac{1}{4} = \mathbf{0.25}$
   • Combining with sign:
     $\text{Value} = \mathbf{-24.25}$

Result Summary Table

Problem 6: Decode Double-Precision 0x4027800000000000 to Decimal

Problem Statement

Determine the equivalent fixed-point decimal value of the IEEE-754 double-precision bit pattern 0x4027800000000000.

Step-by-Step Derivation

1. Convert Hexadecimal to 64-bit Binary:
   $\begin{aligned} \text{Hex: } &\quad \mathtt{4} & \mathtt{0} & \mathtt{2} & \mathtt{7} & \mathtt{8} & \mathtt{0\dots0} \\ \text{Binary: } &\quad 0100 & 0000 & 0010 & 0111 & 1000 & 0000\dots0000 \end{aligned}$

2. Deconstruct IEEE-754 Double-Precision Fields:

   • Sign Bit ($S$, bit 63):
     $S = 0 \implies \text{Positive number } (+)$
   • Biased Exponent ($E$, bits 62–52):
     $\text{Bits: } 0100\,0000\,0010_2 = 1024 + 2 = 1026_{10}$
     Unbiased exponent $e$:
     $e = E - \text{Bias} = 1026 - 1023 = \mathbf{+3}$
   • Fraction Field ($F$, bits 51–0):
     $\text{Bits: } 0111\,1000\,0000\,\dots\,0000_2$
     Significand (with implicit leading 1):
     $M = 1.F = 1.01111_2$

3. Compute Value in Binary Scientific Notation:
   $\text{Value} = + 1.01111_2 \times 2^3$

4. Shift Radix Point by Exponent ($e = +3$):
   $1.01111_2 \times 2^3 = 1011.11_2$

5. Convert Fixed-Point Binary to Decimal:

   • Integer Part ($1011_2$):
     $1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0 = 8 + 0 + 2 + 1 = \mathbf{11}$
   • Fractional Part ($0.11_2$):
     $1 \cdot 2^{-1} + 1 \cdot 2^{-2} = \frac{1}{2} + \frac{1}{4} = 0.5 + 0.25 = \mathbf{0.75}$
   • Combining:
     $\text{Value} = \mathbf{+11.75}$

Result Summary Table

◇Python Verification Script

import struct

def verify_all():
    # Problem 2
    val2 = 1.8125 * (2.0 ** 135)
    hex64_2 = f"0x{struct.unpack('>Q', struct.pack('>d', val2))[0]:016X}"
    print(f"Prob 2: Expected 0x486D000000000000 -> Got {hex64_2} (Match: {hex64_2 == '0x486D000000000000'})")

    # Problem 4
    val4 = 1.8125 * (2.0 ** -135)
    hex64_4 = f"0x{struct.unpack('>Q', struct.pack('>d', val4))[0]:016X}"
    print(f"Prob 4: Expected 0x378D000000000000 -> Got {hex64_4} (Match: {hex64_4 == '0x378D000000000000'})")

    # Problem 5
    f5 = struct.unpack('>f', struct.pack('>I', 0xC1C20000))[0]
    print(f"Prob 5: Expected -24.25 -> Got {f5} (Match: {f5 == -24.25})")

    # Problem 6
    f6 = struct.unpack('>d', struct.pack('>Q', 0x4027800000000000))[0]
    print(f"Prob 6: Expected 11.75 -> Got {f6} (Match: {f6 == 11.75})")

verify_all()

◇Key Takeaways & Exam Summary

1. Biased Exponent Formula: $E = e + \text{Bias}$ where $\text{Bias} = 2^{k-1} - 1$.
   • Single precision ($k=8$): $\text{Bias} = 127$.
   • Double precision ($k=11$): $\text{Bias} = 1023$.
2. Special Exponent Codes:
   • $E = 0, F = 0 \implies \pm 0$.
   • $E = 0, F \neq 0 \implies \text{Subnormal (Gradual Underflow, } e = 1 - \text{Bias}, M = 0.F\text{)}$.
   • $1 \le E \le 2^k - 2 \implies \text{Normalized (Implicit leading 1, } e = E - \text{Bias}, M = 1.F\text{)}$.
   • $E = 2^k - 1, F = 0 \implies \pm\infty \text{ (Overflow)}$.
   • $E = 2^k - 1, F \neq 0 \implies \text{NaN}$.
3. Double Precision Cushion: Values that overflow or underflow single precision (e.g., $e = \pm 135$) easily fit within double precision normalized range ($e \in [-1022, +1023]$).

◇References & Navigation

• Related Notes: ◇IEEE 754 Floating-Point Representation (Single & Double Precision)