# Large Integer problem

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deniz kartan 3 Mar 2011
댓글: Walter Roberson 14 Jan 2016
Hi, I am getting these strange results:
Equations are of the form (x*y + n) - x*y which has the obvious result n. But Matlab gives strange results. These are the results I get in Matlab 2007 and 2010a:
(1812433253*19650218 + 1) - (1812433253*19650218) = 0
(1812433253*19650218 + 2) - (1812433253*19650218) = 0
(1812433253*19650218 + 3) - (1812433253*19650218) = 4
I realized that 1812433253*19650218 has more than 53 bits and these results has something to do with the way Matlab handles numbers between 53 bits and 64 bits. But I couldn't find a reliable way to carry out above operations correctly. Thanks.

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### 채택된 답변

Sean de Wolski 3 Mar 2011

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### 추가 답변(3개)

Matt Fig 3 Mar 2011
You are seeing the results of the limitations of floating point arithmetic.
Read this if you plan on using MATLAB much:
You can see what is going on by looking at this:
N = 1812433253*19650218;
eps(N)
ans =
4
This means that whenever you add a number greater than eps/2 to N, the result is represented as the next largest number available, N+eps(N). Then it will make sense that:
N + eps(N) - N % Equals eps(N)
ans =
4
Conversely, whenever you add a number smaller than eps(N)/2 to N, the result is represented as N. So it makes sense that
N + eps(N)/3 - N % eps(N)/3 is less than eps(N)/2.
ans =
0
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Matt Fig 3 Mar 2011
I recommend thoroughly understanding this because it is the same thing for non-integers.

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the cyclist 3 Mar 2011
I recommend John D'Errico's "Variable Precision Integer" code from the FEX:
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Michelle Hirsch 13 Jan 2016
I'm 5 years late to the party, but another lightweight way to address this specific behavior is to force the variables to be int64 (or uint64), which can express integer values larger than 2^53.
>>(int64(1812433253*19650218) + 1) - int64(1812433253*19650218)
ans =
1
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
Walter Roberson 14 Jan 2016
There are also some things that cannot be done with int64 or uint64. Like bin2dec() into one of them, or get a correct dec2bin()
>> dec2bin((uint64(2^33)^2) - uint64(1023))
ans =
10000000000000000000000000000000000000000000000000000000000000000
>> dec2bin((uint64(2^33)^2) - uint64(1024))
ans =
1111111111111111111111111111111111111111111111111111100000000000
dec2bin() converts to double() before it does the conversion to bits, which is a waste of computation power.

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