if you have X bytes, how would you calculate the number of different combinations that would give? (2 bytes or all bytes *can* be the same after all but the whole still would be different :/) :confused:

I wish I didn't suck at math but I have no clue as to how you'd approach this. 256^X ?
Posted on 2003-11-23 06:11:10 by Hiroshimator
X bytes = 8*X bits
And we know that a byte has 256 possible combinations ( 2^8 ), two bytes (a word 16 bits) has 256 * 256 possible cominations which is 65536 or if you like it in this format (2^8)^2 = 2^(8*2) = 2^16.
So when you say that 256^X is the solution you're right since:
256^X = (2^8)^X = 2^(8*X)
And if you replace X with 2 then you have the possible combinations for 2 bytes.
QED.
Posted on 2003-11-23 08:09:18 by scientica
Also when you are 'constructing a true table' and you have two prepositions, then you have 4 posible

00
01
10
11

for 3 are 16

The formula is this: 2^n where n is the number of entryes.

then take 256 and this is 2*2*2*2*2*2*2*2 or 2^8 for one byte
for two bytes 2^16
for three bytes 2^24

Ok doing some algebra:

2*2*2*2*2*2*2*2 = 2^8 =256
2^16 = 65536
2^24 = 16777216

And see that the exponents 8, 16, 24 are multiple of 2, and remembering that a exponent of a exponent is a multimplication, this is... 2^(4*2) = 2^8 = 256. Then you can do some like:
2^(8*1) = (2^8)^1 = 256^1 = 256
2^(8*2) = (2^8)^2 = 256^2 = 65536
2^(8*3) = (2^8)^3 = 256^3 = 16777216

Then te fromula is 256^n where n is the number of bytes

---------------------------------------

I find this: (you need translate) the combinations with repetitions (CR) is a litle extrange...

http://thales.cica.es/rd/Recursos/rd99/ed99-0516-02/practica/index.html

There explain combinations wit and without repetition, permutations and thigs like that.

Hope this help.

Nice day or night.
Posted on 2003-11-23 08:25:10 by rea
ouch that balloons exponentially as I thought it would :)
Posted on 2003-11-23 08:49:19 by Hiroshimator
byte = 2^8

Number of combination = (2^8)^X = 2^8X
Posted on 2003-11-23 10:36:01 by roticv
Im personally not 100% sure what your asking to find out about. But to give you tools in math, when you have exponents of a common base, their exponent adds/subtracts together:

Ex: what is the maximum unique numbers in 21 bits?

To keep it simple:

2^8 = 256 (kinda a memorized fact)

1000 = 4 * 250 therefor 1k must be 4 * 256. --> 4=2^2 and 2^8 = 256. Multiplying them together (with common bases) 2^2 * 2^8 = 2^10 = 1024 = 1k

So now we know that every 10 bits is 1024 unique numberical values.

Looking again at 21 bits we know that 2 is the base since they are bits: 2^21 =?

You can start by factoring out 1k units (2^10 = 1k):
2^21 = 2^10 * 2^(-10) * 2^(21)
2^21 = 2^10 * 2^(-10 + 21)
2^21 = 2^10 * 2^(11)
2^21 = 2^(11) * 'k'

Seing there still 11 bits we can factor out another 'k'

2^21 = 2^(10) * 2^(-10) * 2^(11) * 'k'
2^21 = 2^(10) * 2^(11-10) * 'k'
2^21 = 2^(10) * 2^(1) * 'k'
2^21 = 2^(1) * 'k' * 'k'

2^1 = 2 (one number 2 is two)
k * k = meg = (2^20)

Thus we know:

2^21 = 2 Meg (or approximately 2 Million unique combinations)

I say approximatley cause 1024 *1024 != 1,000,000 exaclty. Its actually 1,048,576.

In short we make use of the base unit '2' alot such that we can factor out the exponents into smaller more realizable values (2^10, 2^20 == 'k' and 'M') and most have all the values from 2^0 -> 2^10 memorized (1,2,4,8,16,32,64,128,256,512,1024). This is all you need to know to figure out any bit combinations when factorizing the exponents.

88 bits would be -> 2^20 * 2^20 * 2^20 * 2^20 * 2^8 = 256 Meg*Meg*Meg*Meg = 256 * Tera * Tera = 256 * Yotta !!
(According to the SI Units "Yotta" is 10^24).

Hope it helps...
:alright:
NaN
Posted on 2003-11-23 10:51:20 by NaN
I was thinking in the formula for preprosition(true/false):
2^n

Now you have no propositions, then this becomes:

2^0

that is 1

Now I question: what is that one?

ok.... n is the number of propositions (maybe you will never have 0), but what happend if you have 0?? = 1.. what is that 1?, is true, is false, is flasphemous?, is some that never happend?

Nice day or night.
Posted on 2003-11-29 14:25:17 by rea
No positions, *is* a position, hence 2^0 = 1
Posted on 2003-11-29 15:19:17 by NaN
Yes, but that position represent a value of true/false, or 0/1,

or the question need be... What is the meaning of one position?

Nice day or night.
Posted on 2003-11-29 16:00:02 by rea
Ok, the first digit has the position 0, this the maximum value for that position (alone) is: (base-1) * base^0 (weight * base^0)
for the second digit (position 1), the maximum value (for that position alone *) is (base-1) * base^1 (weight * base)

every number can be writen as: n * b^p ("potensform")
where n is teh weight, b the base and p? the position of the wieght.

in our ten base its:
n * 10^p where 0 >= n >= 9

and in binary it's:
n * 2^p where 0 >= n >= 1 (thus 0 and 1)

Thus the position 0 can have the values (for while n is an integer)
0 * 2^0 = 0
1 * 2^0 = 1
for p = 1
0 * 2^1 = 0
0 * 2^1 = 1

you can combine two (or more) values to make another value:
11 = 1 * 2^1 + 1 * 2^0
10 = 1 * 2^1 + 0 * 2^0
Posted on 2003-11-30 09:01:40 by scientica
I think for p=1
0 * 2^1 = 0
1 * 2^1 = 2

then adding (1* 2^1) + (1*2^0) = 11 = 3

Ok, what I try to show is this:
1)
gived: I am in M?xico
or
p: I am in M?xico

I only have one preposition, then I have two posible values true or false, 2) with 2 prepositions with the formula we have 4 posibilities for obtain true or false, now supose you take

3)
p: the

p is not a preposition, then this is 2^0 or only 1 posibility for obtain true or false

1)

p
0 true/false
1 true/false

2)
pq
00 true/false
01 true/false
10 true/false
11 true/false

3)
p
x true/false

Here you see with 2^n posibilities you have only true/false result(or binary result) by each posibility calculate with the formula, but the interesting is in 3) (if posible) is that with a x value you can obtain the two diferent values (true/false)? if that is posible, what is the value that will be choiced for x?

Nice day or night.
Posted on 2003-11-30 16:24:25 by rea