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From: Jennifer Faggi Renault on 21 Apr 2010 17:49
Hi all,
i have a variabla B= '2A66311C'
so to calculate mod(B*(2*B + 1),2^32)
i did Bc=hex2dec(B,8) then i tried:
t= mod(Bc*(2*Bc + 1),2^32) but, i get after conversion into hex
t='E1DDA700' instead of t='E1DDA73C' wich i should find !!

So, can anyone help me to find my error and fix it please??
From: Roger Stafford on 21 Apr 2010 18:17
"Jennifer Faggi Renault" <asma_192(a)live.fr> wrote in message <hqnrsh$qjg$1(a)fred.mathworks.com>...
> Hi all,
> i have a variabla B= '2A66311C'
> so to calculate mod(B*(2*B + 1),2^32)
> i did Bc=hex2dec(B,8) then i tried:
> t= mod(Bc*(2*Bc + 1),2^32) but, i get after conversion into hex
> t='E1DDA700' instead of t='E1DDA73C' wich i should find !!
>
> So, can anyone help me to find my error and fix it please??

You are expecting too much of double precision floating point's 53 bits of significand accuracy. Notice that your error lies in the least six bits of the desired answer. You need to somehow have about 61 or so bits of accuracy for your calculations or to divide up the hex number into two parts and carry out the desired operations on it in an appropriate piecemeal fashion. There is a way to do the latter, even though it is rather annoying. I sometimes have to do it on my hand calculator in hex mode with its limit of eight hex characters.

Roger Stafford
From: Roger Stafford on 21 Apr 2010 19:41
"Roger Stafford" <ellieandrogerxyzzy(a)mindspring.com.invalid> wrote in message <hqnth0$o0d$1(a)fred.mathworks.com>...
> You are expecting too much of double precision floating point's 53 bits of significand accuracy. Notice that your error lies in the least six bits of the desired answer. You need to somehow have about 61 or so bits of accuracy for your calculations or to divide up the hex number into two parts and carry out the desired operations on it in an appropriate piecemeal fashion. There is a way to do the latter, even though it is rather annoying. I sometimes have to do it on my hand calculator in hex mode with its limit of eight hex characters.
>
> Roger Stafford

Here is an example of doing that "piecemeal" method I mentioned. If you had to do many multiplications like this, you could write this as a function.

B = '2A66311C';
Bc = hex2dec(B);
B1 = mod(Bc,2^16);
B2 = (Bc-B1)/2^16;
C = 2*Bc+1;
C1 = mod(C,2^16);
C2 = (C-C1)/2^16;
D = mod(B1*C2+B2*C1,2^16);
E = mod(D*2^16+B1*C1,2^32);
e = dec2hex(E)
ans = 'E1DDA73C'

It doesn't demand much more than 32-bit accuracy of the double calculations.

Roger Stafford
From: Walter Roberson on 21 Apr 2010 20:28
Roger Stafford wrote:
> "Jennifer Faggi Renault" <asma_192(a)live.fr> wrote in message
> <hqnrsh$qjg$1(a)fred.mathworks.com>...

>> i have a variabla B= '2A66311C'
>> so to calculate mod(B*(2*B + 1),2^32)
>> i did Bc=hex2dec(B,8) then i tried:
>> t= mod(Bc*(2*Bc + 1),2^32) but, i get after conversion into hex
>> t='E1DDA700' instead of t='E1DDA73C' wich i should find !!

> You are expecting too much of double precision floating point's 53 bits
> of significand accuracy. Notice that your error lies in the least six
> bits of the desired answer. You need to somehow have about 61 or so
> bits of accuracy for your calculations or to divide up the hex number
> into two parts and carry out the desired operations on it in an
> appropriate piecemeal fashion.

In addition to Roger's entirely correct analysis, note that there is a bug in
fprintf that cause the lower bits to be lost when %lx format is used with uint64:

>> num2hex(.9)
ans =
3feccccccccccccd
>> fprintf('%lx\n', typecast(.9,'uint64'))
3feccccccccccc00

This bug does not, however, affect any of the routines that have 'hex' in
their name, such as hex2dec or num2hex.
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