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unique 48 bits number to unique 32 bits number

Posted on 2006-07-20
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Last Modified: 2010-04-01
Hi,
as stated in the header, does anyone know an algorithm or something like that which enables me to use a 48 bit unique number and store it as a unique 32 bit number??
Regards
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Question by:TError104
8 Comments
 
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by:
sunnycoder earned 20 total points
ID: 17145091
Hi TError104,

It is not possible to store 2^48 unique numbers are 2^32 unique numbers ... there are bound to be collisions and there will be several 48 bit numbers which will map to common 32 bit numbers .... this is unavoidable ....

Cheers!
sunnycoder
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LVL 12

Assisted Solution

by:rajeev_devin
rajeev_devin earned 20 total points
ID: 17145127
If you want to store a 48 bit number then use two data members instead

32 bit integer
16 bit short integer

which is 48 bit in size.
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by:nafis_devlpr
nafis_devlpr earned 20 total points
ID: 17145213
you can use the segment:offset method, like save the first 16bit in a 16bit short int as the segment and the next 32bit in a 32bit int as the offset, by this i mean the bit mask of the 48bit in a 16+32 bit split

you can try union

union INT
      {
            unsigned short int first16;
            unsigned int next32;
      };
or a struct or union of 6 char data

union INT
      {
            unsigned char s[6];
      };

Nafis
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LVL 30

Assisted Solution

by:Axter
Axter earned 20 total points
ID: 17145532
Why not use a 64 bit integer to store the 48bit number?

__int64 My48_bitNum;
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Expert Comment

by:itsmeandnobodyelse
ID: 17145538
Do you know the whole set of 48-bit numbers where the uniqueness needs to keep guaranteed ? If yes, you might take the index of each number as a unique 32 bit equivalence (I assume the size of the set is less than 4 billions).

If no, you could simply count the 4bit numbers you have to convert and and return the count value as a unique id:

typedef unsigned short UI48[3];

unsigned int I48ToI32(UI48 ui48)
{
     static unsigned int count = 0;
     return ++count;
}

In case you could get the same input twice  you would need to store all numbers converted to a dictionary and check for existence before incrementing count.

struct UI48
{
    unsigned short us3[3];
    bool operator<(const UI48& u) const { return us3[0] < u.us3[0] ||
                                                ( us3[0] == u.us3[0] && us3[1] < u.us3[1]) ||
                                                ( us3[0] == u.us3[0] && us3[1] == u.us3[1]  && us3[2] < u.us3[2]);
                                                         }
    bool operator==(const UI48& u) const { return ( us3[0] == u.us3[0] && us3[1] == u.us3[1] && us3[2] == u.us3[2]); }
};

unsigned int I48ToI32(UI48 ui48)
{
     static unsigned int count = 0;
     static std::map<UI48, unsigned int> uimap;
     if (uimap.find(ui48) == uimap.end())
           uimap[ui48] = ++count;
     return  uimap[ui48];
}

Regards, Alex
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LVL 53

Expert Comment

by:Infinity08
ID: 17145566
First of all : what do you want to achieve ?

1) lossless compression : impossible as stated by sunnycoder (unless there are max. 2^32 valid values possible for the 48bit integer)

2) conversion of a 48bit value to a 32bit value : only possible if the maximum value of the 48bit integer is 2^32-1

3) something else : please specify
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