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Floating point numbers

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Computable reals computable numbers may be described briefly as the real numbers whose expressions as a decimal are calculable by finite means.( A. M. Turing, On Computable Numbers with an Application to the Entschiedungsproblem, Proc. London Mathematical Soc., Ser. 2, Vol 42, pages 230-265, 1936-7.)

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Look first at decimal reals A real number may be approximated by a decimal expansion with a determinate decimal point. As more digits are added to the decimal expansion the precision rises. Any effective calculation is always finite – if it were not then the calculation would go on for ever. There is thus a limit to the precision that the reals can be represented as.

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Transcendental numbers In principle, transcendental numbers such as Pi or root 2 have no finite representation We are always dealing with approximations to them. We can still treat Pi as a real rather than a rational because there is always an algorithmic step by which we can add another digit to its expansion.

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First solution Store the numbers in memory just as they are printed as a string of characters. 249.75 Would be stored as 6 bytes as shown below Note that decimal numbers are in the range 30H to 39H as ascii codes 3234392E3735 Full stop char Char for 3

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Implications The number strings can be of variable length. This allows arbitrary precision. This representation is used in systems like Mathematica which requires very high accuracy.

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Example with Mathematica 5! Out[1]=120 In[2]:=10! Out[2]=3628800 In[3]:=50! Out[3]=3041409320171337804361260816 60647688443776415689605120000000000 00

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Decimal byte arithmetic 9+ 8= 17 decimal 39H+38H=71H hexadecimal ascii 57+56=113 decimal ascii Adjust by taking 30H =48 away -> 41H =65 If greater than 9= 39H =57 take away 10= 0AH and carry 1 Thus 41H-0Ah = 65-10=55= 37H so the answer would be 31H,37H = 17

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Representing variables Variables are represented as pointers to character strings in this system A=249.75 A 3234392E3735

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Advantages Arbitrarily precise Needs no special hardware Disadvantages Slow Needs complex memory management

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Binary Coded Decimal (BCD) or Calculator style floating point Note that 249.75 can be represented as 2.4975 x 10 2 Store this 2 digits to a byte to fixed precision as follows 24975002 32 bits overall Each digit uses 4 bits exponentmantissa

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Normalise Convert N to format with one digit in front of the decimal point as follows: 1.If N>10 then Whilst N>10 divide by 10 and add 1 to the exponent 2.Else whilst N<1 multiply by 10 and decrement the exponent

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Add floating point 1.Denormalise smaller number so that exponents equal 2.Perform addition 3.Renormalise Eg 949.75 + 52.0 = 1002.75 9.49750 E02 5.20000 E01 0.52000 E02 + 10.02750 E02 1.00275 E03

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Note loss of accuracy Compare Octave which uses floating point numbers with Mathematica which uses full precision arithmetic Octave floating point gives only 5 figure accuracy Octave fact(5) ans = 120 fact(10) ans = 3628800 fact(50) ans = 3.0414e+64 Mathematica 5! Out[1]=120 10! Out[2]=3628800 50! Out[3]=3041409320171337804 3612608166064768844377641 568960512000000000000

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Loss of precison continued When there is a big difference between the numbers the addition is lost with floating point Octave 325000000 + 108 ans = 3.2500D+08 Mathematica In[1]:= 325000000 + 108 Out[1]= 325000108

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IEEE floating point numbers Institution of Electrical and Electronic Engineers

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Single Precision EF

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Definition N=-1 s x 1.F x 2 E-128 Example 1 3.25 In fixed point binary = 11.01 = 1.101 x 2 1 In IEEE format this is s=0 E=129, F=10100… thus in IEEE it is S E F 0|1000 0001|1010 0000 0000 0000 0000 000 Delete this bit

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Example 2 -0.375 = -3/8 In fixed point binary = -0.011 =-1 1 x 1.1 x 2 -2 In IEEE format this is s=1 E=126, F=1000 … thus in IEEE it is S E F 1|0111 1110|1000 0000 0000 0000 0000 000

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Range IEEE32 1.17 * 10 –38 to +3.40 * 10 38 IEEE64 2.23 * 10 –308 to +1.79 * 10 308 80bit 3.37 * 10 –4932 to +1.18 * 10 4932

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Binary numbers. Primary memory Memory = where programs and data are stored – Unit = bit “BIT” is a contraction for what two words? Either a 1 or a 0 (because.

Binary numbers. Primary memory Memory = where programs and data are stored – Unit = bit “BIT” is a contraction for what two words? Either a 1 or a 0 (because.

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