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accuracy problems in uint64 numbers

Asked by Ansam Osamah on 18 Sep 2019
Latest activity Commented on by Ansam Osamah on 20 Sep 2019
I encrypt uint64 numbers. Number x which encrypted as, for example, 9824265115183455531 decrypted as 9824265115183455488. This difference affects the final decrypted text. Is Matlab not gives accurate results when operates on uint64? how I overcome this problem?

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I think it must be something else, perhaps loss of precision on an intermediate step involving smaller numbers.
>> double(uint64(9824265115183455531)) - uint64(9824265115183455488)
ans =
uint64
768
>> uint64(double(uint64(9824265115183455531)))
ans =
uint64
9824265115183456256
I implemented my code on uint8 numbers to test the correctness of code. everything was OK and the decrypted text was correct. but when I implemented the same code on uint64, I gained a wrong result.
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1 Answer

Answer by Steven Lord
on 18 Sep 2019
 Accepted Answer

It really depends on how you performed your computations. Let's take a somewhat large number, 3^40.
xDouble = 3^40;
xUint64 = uint64(3)^40;
areTheseTheSame = [uint64(xDouble); xUint64]
The answer to the implied question is no. The reason the two elements of areTheseTheSame are not the same is that 3^40 is greater than flintmax.
isGreaterThanFlintmax = xDouble > flintmax % true
The key statement in the flintmax documentation is "Above this value, double-precision format does not have integer precision, and not all integers can be represented exactly." The next largest representable number above flintmax is not flintmax+1 but flintmax+2.
>> F1 = flintmax;
>> F2 = flintmax + 1;
>> F1 == F2
ans =
logical
1
>> eps(flintmax)
ans =
2
>> F3 = flintmax + 2;
>> F1 == F3
ans =
logical
0
However, 3^40 is less than the maximum value that you can store in a uint64 so it can be exactly represented in uint64.
isLessThanMaximumUint64 = xUint64 < intmax('uint64') % true
If your computations are likely to result in large (greater than flintmax) integer values, you should probably operate on values stored in an integer data type as much as you can, like I did when I computed xUint64 by computing mpower (the ^ operator) on a uint64 value.

  7 Comments

Thank you very much
But, what about dec2bin for uint64
K>> dec2bin(uint64(8417341953491579463),64)
ans =
0111010011010000011001111110100011100010110110111011010000000000
but it is actually
‭0111010011010000011001111110100011100010110110111011001001000111
dec2bin() converts its input to double before working out the binary.
u64 = uint64(8417341953491579463);
reshape(dec2bin(typecast(swapbytes(u64),'uint8'),8).',1,[])
Thank you very much to all, your advice helped me too much

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