Discussion :

[my.name]

Peace!
I want ask about the number of bits shoud I use in an fixed_point number to obtein the good precision, precision of distance eclidean.
some detail:
I want to determine an vector of foor element VECT[MOY1,MOY2,MOY3,MOY4], each element is the moyen of one quart of matrix of (64*64), the matrix' element value varies (0-255)'''it's pixel valor'''
--> the result of division (MOY1=SUM/Nbits) has 8 bits in the integer part;
Question: how many bits of fractional part shout I use to get the good result (result of euclidean distance between two vector VECT1[MOY11,MOY12,MOY13,MOY14] and VECT[MOY21,MOY22,MOY23,MOY24])..
thank you....peace:

[mith06]

I can't tell what is "the good precision"


It really depend on your application and what you want to do...
If the by the Euclidian distance you mean: cumsum(vect.*vect) and vect is 8 bits width and contains 64 elements
(8+8)+6 => 24 bits , if you don't want to loose information

[my.name]

Thank you mith06

I can't tell what is "the good precision"


It really depend on your application and what you want to do...
If the by the Euclidian distance you mean: cumsum(vect.*vect) and vect is 8 bits width and contains 64 elements

actualy, the vector has foor of fixed point values :
vect1(MOY1,MOY2,MOY3,MOY4) ,of matrix1--> MOY1= ( sum(matrix1(1:16,1:64))) / (16 * 64); /*moyen of first quart*/ , MOY1= ( sum(matrix1(17:32,1:64))) / (16 * 64); /*moyen of secound quart*/and so the others of the 3 quart rest
vect2(MOY1,MOY2,MOY3,MOY4) ,of matrix2--> MOY1= ( sum(matrix2(1:16,1:64))) / (16 * 64); /*moyen of first quart*/ ,and so the others of the 3 quart rest ,and so the others

Euclidean distance in matlab: euc.matlab = norm(vect1 - vect2)

(8+8)+6 => 24 bits , if you don't want to loose information

so the integer part can wrote in 8bits:::My question is how bits shoud I use for the fractional part Knowing that I want keep a big part of the
information.

[mith06]

norm(vect1 - vect2)

= sqrt(sum((vect1-vect2)^2))

Start with 8 bits
substraction => +1 bits
multiplcation => *2 bits
Sum over N element => + log2(N) bits
sqrt => +0 bits (you will loose information what ever the number of bits you use)

((8+1)*2)+log2(64) = 24 bits if you d'ont want to loose information