real number representation (lecture 25 of the introduction to computer programming series) dr damian...
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Real Number Representation
(Lecture 25 of the Introduction to Computer Programming series)
Dr Damian ConwayRoom 132
Building 26
Some Terminology
• All digits in a number following any leading zeros are significant digits:
12.345 -0.12345 0.00012345
Some Terminology
• The scientific notation for real numbers is:
mantissa base exponent
Some Terminology
• The mantissa is always normalized between 1 and the base (i.e. exactly one significant figure before the point):
Normalized Unnormalized
2.9979 108
2997.9 105
B.139FC 1612
B1.39FC 1611
1.0110110101 2-3
0.010110110101 2-1
Some Terminology
• The precision of a number is how many digits (or bits) we use to represent it.
• For example:
33.143.14159263.1415926535897932384626433832795028
Representing numbers
• A real number n is represented by a floating-point approximation n*
• The computer uses 32 bits (or more) to store each approximation.
• It needs to store the mantissa, the sign of the mantissa, and the exponent (with its sign).
Representing numbers
• So it has to allocate some of its 32 bits to each task.
• The standard way to do this (specified by IEEE standard 754) is:
313002223
Representing numbers
• 23 bits for the mantissa;
• 1 bit for the mantissa's sign (i.e. the mantissa is signed magnitude);
• The remaining 8 bits for the exponent.
313002223
Representing numbers
• 23 bits for the mantissa;
• 1 bit for the mantissa's sign (i.e. the mantissa is signed magnitude);
• The remaining 8 bits for the exponent.
313002223
Representing numbers
• 23 bits for the mantissa;
• 1 bit for the mantissa's sign (i.e. the mantissa is signed magnitude);
• The remaining 8 bits for the exponent.
313002223
Representing numbers
• 23 bits for the mantissa;
• 1 bit for the mantissa's sign (i.e. the mantissa is signed magnitude);
• The remaining 8 bits for the exponent.
Representing the mantissa
• Since the mantissa has to be in the range 1 ≤ mantissa < base, if we use base 2 the digit before the decimal has to be a 1.
• So we don't have to worry about storing it!
• That way we get 24 bits of precision using only 23 bits.
Representing the mantissa
• Those 24 bits of precision are equivalent to a little over 7 decimal digits:
24
log2 10≈7.2
Representing the mantissa
• Suppose we want to represent :
3.1415926535897932384626433832795.....
• That means that we can only represent it as:
3.141592 (if we truncate)3.141593 (if we round)
Representing the mantissa
• Even if the computer appears to give you more than seven decimal places, only the first seven are meaningful.
• For example:
#include <math.h>
main(){
float pi = 2 * asin(1);printf("%.35f\n", pi);
}
Representing the mantissa
• On my machine this prints out:
3.1415927419125732000000000000000000
Representing the mantissa
• On my machine this prints out:
3.1415927419125732000000000000000000
Representing the exponent
• The exponent is represented as an excess-127 number.
• That is:
00000000 –12700000001 –12601111111 010000000 +111111111 +128
Representing the exponent
• However, the IEEE standard restricts exponents to the range:
–126 ≤ exponent ≤ +127
• The exponents –127 and +128 have special meanings (basically, zero and infinity respectively)
Floating point overflow
• Just like the integer representations in the previous lecture, floating point representations can overflow:
9.999999 10127
+ 1.111111 10127
1.1111110 10128
Floating point overflow
• Just like the integer representations in the previous lecture, floating point representations can overflow:
9.999999 10127
+ 1.111111 10127
1.1111110 10128
Floating point overflow
• Just like the integer representations in the previous lecture, floating point representations can overflow:
9.999999 10127
+ 1.111111 10127
∞
Floating point underflow
• But floating point numbers can also get too small:
1.000000 10-126
÷ 2.000000 100
5.000000 10-127
Floating point underflow
• But floating point numbers can also get too small:
1.000000 10-126
÷ 2.000000 100
5.000000 10-127
Floating point underflow
• But floating point numbers can also get too small:
1.000000 10-126
÷ 2.000000 100
0
Floating point addition
• Five steps to add two floating point numbers:– Express them with the same exponent
(denormalize)– Add the mantissas– Adjust the mantissa to one digit/bit
before the point (renormalize)– Round or truncate to required precision.– Check for overflow/underflow
Floating point addition example
x = 9.876 107
y = 1.357 106
Floating point addition example
1. Same exponents:
x = 9.876 107
y = 0.1357 107
Floating point addition example
2. Add mantissas:
x = 9.876 107
y = 0.1357 107
x+y = 10.0117 107
Floating point addition example
3. Renormalize sum:
x = 9.876 107
y = 0.1357 107
x+y = 1.00117 108
Floating point addition example
4. Trucate or round:
x = 9.876 107
y = 0.1357 107
x+y = 1.001 108
Floating point addition example
5. Check overflow and underflow:
x = 9.876 107
y = 0.1357 107
x+y = 1.001 108
Floating point addition example 2
x = 3.506 10-5
y = -3.497 10-5
Floating point addition example 2
1. Same exponents:
x = 3.506 10-5
y = -3.497 10-5
Floating point addition example 2
2. Add mantissas:
x = 3.506 10-5
y = -3.497 10-5
x+y = 0.009 10-5
Floating point addition example 2
3. Renormalize sum:
x = 3.506 10-5
y = -3.497 10-5
x+y = 9.000 10-8
Floating point addition example 2
4. Trucate or round:
x = 3.506 10-5
y = -3.497 10-5
x+y = 9.000 10-8 (no
change)
Floating point addition example 2
5. Check overflow and underflow:
x = 3.506 10-5
y = -3.497 10-5
x+y = 9.000 10-8
Floating point addition example 2
Question: should we believe these zeroes?
x = 3.506 10-5
y = -3.497 10-5
x+y = 9.000 10-8
Floating point multiplication
• Five steps to multiply two floating point numbers:– Multiply mantissas– Add exponents– Renormalize mantissa– Round or truncate to required
precision.– Check for overflow/underflow
Floating point multiplication example
x = 9.001 105
y = 8.001 10-3
1&2. Multiply mantissas/add exponents:
x = 9.001 105
y = 8.001 10-3
x y = 72.017001 102
Floating point multiplication example
3. Renormalize product:
x = 9.001 105
y = 8.001 10-3
x y = 7.2017001 103
Floating point multiplication example
4. Trucate or round:
x = 9.001 105
y = 8.001 10-3
x y = 7.201 103
Floating point multiplication example
4. Trucate or round:
x = 9.001 105
y = 8.001 10-3
x y = 7.202 103
Floating point multiplication example
5. Check overflow and underflow:
x = 9.001 105
y = 8.001 10-3
x y = 7.202 103
Floating point multiplication example
Limitations
• Float-point representations only approximate real numbers.
• The normal laws of arithmetic don't always hold (even less often than for integer representations).
• For example, associativity is not guaranteed:
Limitations
x = 3.002 103
y = -3.000 103
z = 6.531 100
Limitations
x = 3.002 103
x+y = 2.000 100
y = -3.000 103
z = 6.531 100
Limitations
x = 3.002 103
x+y = 2.000 100
y = -3.000 103
(x+y)+z = 8.531 100
z = 6.531 100
Limitations
x = 3.002 103
y = -3.000 103
z = 6.531 100
Limitations
x = 3.002 103
y = -3.000 103
y+z = -2.993 103
z = 6.531 100
Limitations
x = 3.002 103
x+(y+z) = 0.009 103
y = -3.000 103
y+z = -2.993 103
z = 6.531 100
Limitations
x = 3.002 103
x+(y+z) = 9.000 100
y = -3.000 103
y+z = -2.993 103
z = 6.531 100
Limitations
x = 3.002 103
x+(y+z) = 9.000 100
y = -3.000 103
(x+y)+z = 8.531 100
z = 6.531 100
Limitations
• Consider the other laws of arithmetic:– Commutativity (additive and multiplicative)– Associativity – Distributivity– Identity (additive and multiplicative)
• Spend some time working out which ones (if any!) always hold for floating-point numbers.
Reading (for the very keen)
Goldberg, D., What Every Computer Scientist Should Know About Floating-Point Arithmetic, ACM Computing Surveys, Vol.23, No.1, March 1991.