math question "n(n-1)/2"

In symmetric cryptography, the key must be "n(n-1)/2"...
does anybody knows why the formula must be "n(n-1)/2"...
anyone can explain bout the formula?

In symmetric cryptography, the key must be "n(n-1)/2"...
does anybody knows why the formula must be "n(n-1)/2"...
anyone can explain bout the formula?

Ya know, I remember that formula from geometry class...... A triangular pattern?

Sorry Im still young and not too "in-tune" with math yet.

Ya know, I remember that formula from geometry class...... A triangular pattern?

Sorry Im still young and not too "in-tune" with math yet.

well i don't know what it has to do with crypto, but that formula is used to find the sum of all numbers prior to n.

ex: if you want the sum of all numbers leading up to 10, then
10(9) / 2 = 45

is the same as
1+2+3+4+5+6+7+8+9 = 45

i hope that sheds some light on something.

well i don't know what it has to do with crypto, but that formula is used to find the sum of all numbers prior to n.

ex: if you want the sum of all numbers leading up to 10, then
10(9) / 2 = 45

is the same as
1+2+3+4+5+6+7+8+9 = 45

i hope that sheds some light on something.

http://www.maths.mq.edu.au/~steffen/...ort/node7.html

i just found this on google, it gives good insight into anti/symmetric cryptography.

from what i know, coupled with this, i still can't think why the key must be n(n-1)/2.

should it be a function? like p = n(n-1)/2 . could it be used to transform some ascii string into some larger bit number?

in short, i don't know.

http://www.maths.mq.edu.au/~steffen/...ort/node7.html

i just found this on google, it gives good insight into anti/symmetric cryptography.

from what i know, coupled with this, i still can't think why the key must be n(n-1)/2.

should it be a function? like p = n(n-1)/2 . could it be used to transform some ascii string into some larger bit number?

in short, i don't know.

I might try to explain this but first check this out:
http://www.iam.unibe.ch/~mseeberg/theory.html
http://www.informatik.uni-bremen.de/~hs/Lehre/L11.pdf

Basically, the stated formula on top is used to estimate the entropy of the encryption key, ie how easily the key used can be reversed .... somehow it's an approximation
hope that help

I might try to explain this but first check this out:
http://www.iam.unibe.ch/~mseeberg/theory.html
http://www.informatik.uni-bremen.de/~hs/Lehre/L11.pdf

Basically, the stated formula on top is used to estimate the entropy of the encryption key, ie how easily the key used can be reversed .... somehow it's an approximation
hope that help

I could be wrong, but doesn't this formula ensure it can't be a prime number?

I could be wrong, but doesn't this formula ensure it can't be a prime number?

Uggh, been a really long time since I thought about this, but I am wanting to say the n(n-1)/2 has to do with permutations (all possible combinations of possibilities):

http://www.cut-the-knot.org/do_you_k...mutation.shtml

Number theory is a very large part of cryptography and permutation is a way to describe all the possibilties.

Suggest looking into permutations and combinatorics (but be ready for some rather dry reading and some heavy math thinking). I think encryption relies heavily on this stuff (set theory, number theory).

/nebulus

Uggh, been a really long time since I thought about this, but I am wanting to say the n(n-1)/2 has to do with permutations (all possible combinations of possibilities):

http://www.cut-the-knot.org/do_you_k...mutation.shtml

Number theory is a very large part of cryptography and permutation is a way to describe all the possibilties.

Suggest looking into permutations and combinatorics (but be ready for some rather dry reading and some heavy math thinking). I think encryption relies heavily on this stuff (set theory, number theory).

/nebulus

( 3 * (3 - 1) ) / 2 = 3
for n = 3, this function demonstrates a property Karma - you get what you give - and a prime to boot.

( 3 * (3 - 1) ) / 2 = 3
for n = 3, this function demonstrates a property Karma - you get what you give - and a prime to boot.

Now i know why...

Actually it came from arithmetic series where we could assume that

t1 = 0 = 0
t2 = 1 = 0 + 1
t3 = 3 = 0 + 1 + 2
t4 = 6 = 0 + 1 + 2 + 3

where a(initial number) a=0
and the d(differences) d = 1

we put it in arithmetic series formula Sn = (n/2)(2a+(n-1)d)

= (n/2)(2(0)+(n-1)(1))

=(n/2)(n-1)

tn = (n/2)(n-1) :D thanks!