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[[Image:Gleich.jpg|thumb|right|793px|CSUM in Digital Signal Processing terms]]
<pre>
<pre>
function s=csum(x)
function s=csum(x)
% CSUM Sum of elements using a compensated summation algorithm
% CSUM Sum of elements using a compensated summation algorithm
%
%
-
% For large vectors, the native sum command in Matlab does not appear to
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% For large vectors, the native sum command in Matlab does
-
% use a compensated summation algorithm which can cause significant round
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% not appear to use a compensated summation algorithm which
-
% off errors.
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% can cause significant roundoff errors.
%
%
-
% This code implements a variant of Kahan's compensated summation algorithm
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% This code implements a variant of Kahan's compensated
-
% which often takes about twice as long, but produces more accurate sums
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% summation algorithm which often takes about twice as long,
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% when the number of elements is large.
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% but produces more accurate sums when the number of
 +
% elements is large.
%
%
% See also SUM
% See also SUM
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% David Gleich, Stanford University, 2008
% David Gleich, Stanford University, 2008
-
shat=0; y=0; e=0;
+
s_hat=0; y=0; e=0;
for i=1:numel(x)
for i=1:numel(x)
-
shat_old = shat;
+
s_hat_old = s_hat;
y = x(i) + e;
y = x(i) + e;
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shat = shat_old + y;
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s_hat = s_hat_old + y;
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e = (shat_old - shat) + y;
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e = (s_hat_old - s_hat) + y;
end
end
</pre>
</pre>

Revision as of 19:49, 17 February 2009

CSUM in Digital Signal Processing terms
CSUM in Digital Signal Processing terms
function s=csum(x)
% CSUM Sum of elements using a compensated summation algorithm
%
% For large vectors, the native sum command in Matlab does 
% not appear to use a compensated summation algorithm which 
% can cause significant roundoff errors.
%
% This code implements a variant of Kahan's compensated 
% summation algorithm which often takes about twice as long, 
% but produces more accurate sums when the number of 
% elements is large.
%
% See also SUM
%
% Example:
%   v=rand(1e7,1);
%   sum1 = sum(v);
%   sum2 = csum(v);
%   fprintf('sum1 = %18.16e\nsum2 = %18.16e\n', sum1, sum2);

% David Gleich, Stanford University, 2008

s_hat=0; y=0; e=0;
for i=1:numel(x)
   s_hat_old = s_hat; 
   y = x(i) + e; 
   s_hat = s_hat_old + y; 
   e = (s_hat_old - s_hat) + y;
end
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