# precision of calculation with Matlab

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DRIDI Fethi on 10 Feb 2020
Commented: DRIDI Fethi on 11 Feb 2020
Hello, I have a problem of precision of calculation with Matlab when I make a multiplication of two positive numbers of 32 bits I had the results of the calculator on the other hand Matlab give us the last two bits in hex of zeros.for example:
result with calculator: 5A0C83 DE64F998F9
results with Matlab: 5A0C83 DE64F99800

James Tursa on 10 Feb 2020
You need to provide us with a complete example. Give us the exact inputs and then describe the output issues you are seeing.
DRIDI Fethi on 11 Feb 2020
clc;
clear all;
E1=31;
E2=16;
E3=(2^32)-E1-E2;
x1 = 1510769636;
x2 = 757236075;
x3 = 1466008126;
out = uint64((E3 * x1) + (E1 * x2) + (E2 * x3));
y = uint32(mod(out,2^32));
Hello,
So, normally the result equal to 5A0C83 DE64F998F9 for multiplication, I need an exact result since I only use 32-bit LSB (modular calculation).

John D'Errico on 11 Feb 2020
Edited: John D'Errico on 11 Feb 2020
This must of course fail as you are doing it, since out must generally exceed 2^53-1, the limits of flintmax for a double precision number. Remember that E1,E2,E3,x1,x2,x3 are all doubles. So when we compute
(E3 * x1) + (E1 * x2) + (E2 * x3)
It is also a double. Just because you then push it into a uint64 is not sufficient. You have already tossed too many bits into the bit bucket.
But, suppose that we force x1,x2,x3 to be uint64 in advance? That is, what if x1,x2,x3 all are created as uint64? Now products like E1*x1 will now be created directly as uint64 numbers, not temporarily as doubles. And the sum in out will now already be uint64.
E1=31;
E2=16;
E3=(2^32)-E1-E2;
x1 = uint64(1510769636);
x2 = uint64(757236075);
x3 = uint64(1466008126);
out = (E3 * x1) + (E1 * x2) + (E2 * x3);
y = mod(out,2^32)
y =
uint64
1694079225
uint32(y)
ans =
uint32
1694079225
y is the correct integer value we would expect it to have. There is no need to convert y into a uint32 though, as we effectively achieved that using the mod operation. As long as out will never overflow uint64, we have no problems.
intmax('uint64')
ans =
uint64
18446744073709551615
out =
uint64
6488706154334099705
As you can see, it does not. But can there ever be a problem? No. Since the sum E1+E2+E3 == 2^32, we need not worry about that being a problem as long as x1,x2,x3 all live inside the 32 bit limit themselves, out is just a convex linear combination of the x's, so no problem will exist.
Essentially, there is no problem, as long as you are careful in how you do the computation. Do NOT let MATLAB use doubles here though, as that will fail.

#### 1 Comment

DRIDI Fethi on 11 Feb 2020
absolutely, the problem is solved.
thank you so much

James Tursa on 10 Feb 2020
Calculators may not use the same floating point representations or arithmetic routines that MATLAB uses, so differences in the trailing bits should not be unexpected in these cases.