MATLAB

The Language of Technical Computing

Single Precision Math

This example shows how to perform arithmetic and linear algebra with single precision data. It also shows how the results are computed appropriately in single-precision or double-precision, depending on the input.

Create Double Precision Data

Let's first create some data, which is double precision by default.

Ad = [1 2 0; 2 5 -1; 4 10 -1]
Ad =

     1     2     0
     2     5    -1
     4    10    -1

Convert to Single Precision

We can convert data to single precision with the single function.

A = single(Ad); % or A = cast(Ad,'single');

Create Single Precision Zeros and Ones

We can also create single precision zeros and ones with their respective functions.

n = 1000;
Z = zeros(n,1,'single');
O = ones(n,1,'single');

Let's look at the variables in the workspace.

whos A Ad O Z n
  Name         Size            Bytes  Class     Attributes

  A            3x3                36  single              
  Ad           3x3                72  double              
  O         1000x1              4000  single              
  Z         1000x1              4000  single              
  n            1x1                 8  double              

We can see that some of the variables are of type single and that the variable A (the single precision version of Ad) takes half the number of bytes of memory to store because singles require just four bytes (32-bits), whereas doubles require 8 bytes (64-bits).

Arithmetic and Linear Algebra

We can perform standard arithmetic and linear algebra on singles.

B = A'    % Matrix Transpose
B =

     1     2     4
     2     5    10
     0    -1    -1

whos B
  Name      Size            Bytes  Class     Attributes

  B         3x3                36  single              

We see the result of this operation, B, is a single.

C = A * B % Matrix multiplication
C =

     5    12    24
    12    30    59
    24    59   117

C = A .* B % Elementwise arithmetic
C =

     1     4     0
     4    25   -10
     0   -10     1

X = inv(A) % Matrix inverse
X =

     5     2    -2
    -2    -1     1
     0    -2     1

I = inv(A) * A % Confirm result is identity matrix
I =

     1     0     0
     0     1     0
     0     0     1

I = A \ A  % Better way to do matrix division than inv
I =

     1     0     0
     0     1     0
     0     0     1

E = eig(A) % Eigenvalues
E =

    3.7321
    0.2679
    1.0000

F = fft(A(:,1)) % FFT
F =

   7.0000 + 0.0000i
  -2.0000 + 1.7321i
  -2.0000 - 1.7321i

S = svd(A) % Singular value decomposition
S =

   12.3171
    0.5149
    0.1577

P = round(poly(A)) % The characteristic polynomial of a matrix
P =

     1    -5     5    -1

R = roots(P) % Roots of a polynomial
R =

    3.7321
    1.0000
    0.2679

Q = conv(P,P) % Convolve two vectors
R = conv(P,Q)
Q =

     1   -10    35   -52    35   -10     1


R =

     1   -15    90  -278   480  -480   278   -90    15    -1

stem(R); % Plot the result

A Program that Works for Either Single or Double Precision

Now let's look at a function to compute enough terms in the Fibonacci sequence so the ratio is less than the correct machine epsilon (eps) for datatype single or double.

% How many terms needed to get single precision results?
fibodemo('single')

% How many terms needed to get double precision results?
fibodemo('double')

% Now let's look at the working code.
type fibodemo

% Notice that we initialize several of our variables, |fcurrent|,
% |fnext|, and |goldenMean|, with values that are dependent on the
% input datatype, and the tolerance |tol| depends on that type as
% well.  Single precision requires that we calculate fewer terms than
% the equivalent double precision calculation.
ans =

    19


ans =

    41


function nterms = fibodemo(dtype)
%FIBODEMO Used by SINGLEMATH demo.
% Calculate number of terms in Fibonacci sequence.

% Copyright 1984-2014 The MathWorks, Inc.

fcurrent = ones(dtype);
fnext = fcurrent;
goldenMean = (ones(dtype)+sqrt(5))/2;
tol = eps(goldenMean);
nterms = 2;
while abs(fnext/fcurrent - goldenMean) >= tol
   nterms = nterms + 1;
   temp  = fnext;
   fnext = fnext + fcurrent;
   fcurrent = temp;
end