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 ?

I wish I didn't suck at math but I have no clue as to how you'd approach this. 256^X ?

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.

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.

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.

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.

ouch that balloons exponentially as I thought it would :)

byte = 2^8

Number of combination = (2^8)^X = 2^8X

Number of combination = (2^8)^X = 2^8X

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

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

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.

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.

No positions, *is* a position, hence 2^0 = 1

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.

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

Nice day or night.

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)

note about weights:

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

for the second digit (position 1), the maximum value (for that position alone *) is (base-1) * base^1 (weight * base)

note about weights:

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

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.

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.