accuracy problems in uint64 numbers
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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?
4 个评论
James Tursa
2019-9-18
Please show us your code. Perhaps you are doing intermediate calculations in double and that is causing your loss of trailing digits problem.
Walter Roberson
2019-9-18
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
Ansam Osamah
2019-9-18
Walter Roberson
2019-9-18
Sorry, it is a cloudy night here and my telescope cannot see your computer screen.
采纳的回答
Steven Lord
2019-9-18
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 个评论
Ansam Osamah
2019-9-19
James Tursa
2019-9-19
This is what everyone has been warning you about. All that stuff inside the uint64( ) conversion are intermediate calculations that are not done in uint64 arithmetic. So force it to use uint64 arithmetic in your example:
>> plain_uint64=uint64(bin2dec(plain_binary(1:32))*uint64(2)^32+ bin2dec(plain_binary(33:64)))
plain_uint64 =
6082227239949792032
Ansam Osamah
2019-9-19
Walter Roberson
2019-9-19
Unfortunately bin2dec returns a double, not a uint64.
Another approach is to bin2dec 8 bits at a time, uint8() that, and then typecast to uint64. You can do the same thing up to uint32 as the carrier type. But be careful because the bit order of dec2bin is big-endian (msb is to the left and bit significance strictly decreases to the right) but the bit order of all current matlab implementations is little-endian so swapping bytes might be needed depending what you do with the results.
Ansam Osamah
2019-9-20
编辑：Ansam Osamah
2019-9-20
Walter Roberson
2019-9-20
dec2bin() converts its input to double before working out the binary.
u64 = uint64(8417341953491579463);
reshape(dec2bin(typecast(swapbytes(u64),'uint8'),8).',1,[])
Ansam Osamah
2019-9-20
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