Calcul XOR : array , times

Ruby
1

Hello ,

I have problems with my code for XOR calcul :

a = [1, 0, 1, 1, 0, 0, 1]
b = ["101101", "101100", "110011", "000111", "010110"]
# good result : 000001, 011010,101000, 001010,110000
# false result : ["000001", "000000", "011111", "101011", "111010"]

=> output with this code
# Why ? Because he spends that time with respect b[i].lenght ( = 6
generally ) that prevents take in this case the element 7 and 8 of “a”.
# works in case a.length <b [x]. length
# How to resolv this problem ?
i = j = x = 0
c = []
d = []
f = []
5.times { |x| c.push b[x].split(’’) }

5.times { |i|
        e = c[i].length
        e.times { |j| d.push(a[j % 7].to_i ^ c[i][j].to_i) }
        f.push d
        d = []
}
puts "Résultat brute: "
p f

z = f.length
z.times { |h| f[h] = f[h].join("")}

puts "Résultats :"
p f

In this case :

a = [1, 0, 1, 1, 0, 0, 1]
b = ["101101", "101100", "110011", "000111", "010110"]

After splitting

b : [["1","0","1","1","0","1"], ["1","0","1","1","0","0"],

[“1”,“1”,“0”,“0”,“1”,“1”],
[“0”,“0”,“0”,“1”,“1”,“1”], [“0”,“1”,“0”,“1”,“1”,“0”]]

But a.length => 7 and b[x].length =>6

In this code, the xor calcul is performed with only 6 elements in a on
7.

So the calculation is performed here:

a[0] ^ b[0][0] ; a[1] ^ b[0][1] ; a[2] ^ b[0][2] ; a[3] ^ b[0][3] ;

a[4] ^ b[0][4] ; a[5] ^ b[0][5] ;
a[0] ^ b[1][0] ; a[1] ^ b[1][1] ; a[2] ^ b[1][2] ; a[3] ^ b[1][3] ;
a[4] ^ b[1][4] ; a[5] ^ b[1][5] ;
a[0] ^ b[2][0] ; a[1] ^ b[2][1] ; a[2] ^ b[2][2] ; a[3] ^ b[2][3] ;
a[4] ^ b[2][4] ; a[5] ^ b[2][5] ;
a[0] ^ b[3][0] ; a[1] ^ b[3][1] ; a[2] ^ b[3][2] ; a[3] ^ b[3][3] ;
a[4] ^ b[3][4] ; a[5] ^ b[3][5] ;
a[0] ^ b[4][0] ; a[1] ^ b[4][1] ; a[2] ^ b[4][2] ; a[3] ^ b[4][3] ;
a[4] ^ b[4][4] ; a[5] ^ b[4][5] ;

a[6] is never use.

and the calculation should be done:

a[0] ^ b[0][0] ; a[1] ^ b[0][1] ; a[2] ^ b[0][2] ; a[3] ^ b[0][3] ;

a[4] ^ b[0][4] ; a[5] ^ b[0][5] ;
a[6] ^ b[1][0] ; a[0] ^ b[1][1] ; a[1] ^ b[1][2] ; a[2] ^ b[1][3] ;
a[3] ^ b[1][4] ; a[4] ^ b[1][5] ;
a[5] ^ b[2][0] ; a[6] ^ b[2][1] ; a[0] ^ b[2][2] ; a[1] ^ b[2][3] ;
a[2] ^ b[2][4] ; a[3] ^ b[2][5] ;
a[4] ^ b[3][0] ; a[5] ^ b[3][1] ; a[6] ^ b[3][2] ; a[0] ^ b[3][3] ;
a[1] ^ b[3][4] ; a[2] ^ b[3][5] ;
a[3] ^ b[4][0] ; a[4] ^ b[4][1] ; a[5] ^ b[4][2] ; a[6] ^ b[4][3] ;
a[0] ^ b[4][4] ; a[1] ^ b[4][5] ;

I want to make the right calculation of course but I do not see how to
this.

How to use a[6] as above ?

Thanks

2

On Fri, May 13, 2011 at 9:25 PM, aix aix [email protected] wrote:

Hello ,

I have problems with my code for XOR calcul :

a = [1, 0, 1, 1, 0, 0, 1]
b = [“101101”, “101100”, “110011”, “000111”, “010110”]

good result : 000001, 011010,101000, 001010,110000

false result : [“000001”, “000000”, “011111”, “101011”, “111010”]

How to use a[6] as above ?

Sorry, I did not look for a problem with your code. I did it another
way.
This is not tested so it may have some problems but maybe it will give
you some ideas.

a = [1, 0, 1, 1, 0, 0, 1]
b = [“101101”, “101100”, “110011”, “000111”, “010110”]

bmod = b.join.split(//).map{|x| x.to_i}
amod = a*(bmod.size/a.size+1)
res = amod[0…bmod.size].zip(bmod).map{|z| z[0]^z[1]}

p [].tap{|y| b.size.times{y << res.slice!(0…b[0].size)}}.map{|g|
g.join}

Harry

3

aix aix wrote in post #998497:

How to use a[6] as above ?

Thanks

You want Array#cycle.

4

7stud – wrote in post #998575:

aix aix wrote in post #998497:

How to use a[6] as above ?

Thanks

You want Array#cycle.

Here’s an example using cycle and Enumerator#next:

a = [1, 0, 1, 1, 0, 0, 1]
b = [“101101”, “101100”, “110011”, “000111”, “010110”]

enum = a.cycle

b.each do |str|
result = “”

str.each_char do |char|
result << (char.to_i ^ enum.next).to_s
end

p result
end

–output:–
“000001”
“011010”
“101000”
“001010”
“110000”