The Side Channel - 2015-01-11

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10:36 AM

hello im working on a little cryptosystem .... just wondering what relation between b and c verifies a unique solution of : a+b modulo c = A knowin thet a , b are inferior to c and A desired to be uique

hence /// between backets ///// i regret the french murdering incident and condemn the committers of such barbaric actee .... religion is entirely independant and a more convenient appelation of this would be score settling ..... thanks

2 hours later…

12:24 PM

3

Q: Is masking effective for thwarting side channel attacks?

I'm working with some bigint public-key cryptography code. Is it safe to use bitwise masking to ensure that the calculation timing and memory addresses accessed are independent of the data values? Is this technique vulnerable to side-channel attacks based on instruction timing, power, RF emissio...

c++ c cryptography

Just spotted this on SO, thought I'd share it here, on the remote chance that someone follows this chat but not [tag:cryptography] on SO.

side channel means bruteforcing ?

it is not a differential methode is it ?

Side channel attack

In cryptography, a side channel attack is any attack based on information gained from the physical implementation of a cryptosystem, rather than brute force or theoretical weaknesses in the algorithms (compare cryptanalysis). For example, timing information, power consumption, electromagnetic leaks or even sound can provide an extra source of information which can be exploited to break the system. Some side-channel attacks require technical knowledge of the internal operation of the system on which the cryptography is implemented, although others such as differential power analysis are effective...

any attack would be fruitful if the insider system is revealed

@AbdouAbdou Anyway... do I understand you right that you want to solve a + b ≡ A (mod c) for a, given b, c and A? Or something else?

a isnt given ... it is variable and inferior to c

but what relation between c and b makes a unique solution

in the case of existence

12:34 PM

How much do you know about modular arithmetic?

4 + 3 mod 5 =2 .... this accepts two solutions 2 and 7

for exple

do u mean congruence ?

Because in general, the solution is simply a ≡ A - b (mod c), and will be "unique modulo c" -- that is, if a is a solution, then so is a + kc for any integer k.

Which, in particular, means that there will be a unique solution in the range [0, c-1].

i know but somtimes a solution could be wrong like 2

What is "wrong" about 4 + 3 ≡ 2 (mod 5)?

the number we seek to find is 7 not 2

so we can reverse the operation

and be back to 4

orelse it wud be asymetric

12:42 PM

In arithmetic modulo 5 (for example), 7 ≡ 2, so you cannot tell those solutions apart using just modular arithmetic.

But that's fine, since 2 - 3 ≡ 4 (mod 5).

(and, of course, 2 - 3 (mod 5) also equals -6, -1, 9, 14, etc., since those numbers are all equivalent modulo 5.)

(but 4 is generally considered the "canonical" solution, since it's the smallest non-negative one.)

look

my idea is

taken an encrypted ascci number

add it to some variable

so we have a sum

we do modulo (length of ascii code)=128

here we have two solutions

not just one

im stucked just here

which solution leads to original ascii code ?

The one that is a valid ASCII code, i.e. between 0 and 127.

(But you might as well work with bytes modulo 256, so you can encrypt UTF-8, too.)

some text editors dont support unicode

So? Most text editors will also barf on most ASCII codes below 32 (= space), except for tabs and newlines.

(Ps. Most stream ciphers use XOR instead of addition modulo 128 or 256, since that's even easier to invert.)

xor is easy to be reversed

and cryptanalysed

12:55 PM

So is modular addition, too.

In fact, XOR is really just addition modulo 2.

yes .... vnt noticed it before hahaha

The trick is that you need a cryptographically strong (pseudo)random stream for your key; then the cipher cannot be broken unless one can somehow distinguish the keystream from true randomness.

i thought of that

i used datetime variant

Datetime?

new date().gettime()

12:59 PM

That's not really very random, if I know when you encrypted the message.

Or if you encrypt several messages in close succession.

is there alternatives ?

Depends on exactly what you want to do, but see e.g. en.wikipedia.org/wiki/…

truecrypt uses this standard

in encryption

And, since you seem to be interested in these kinds of ciphers, en.wikipedia.org/wiki/One-time_pad

a random num is excerpted from mouse consecutive positions

so ...... i have another had example

taken a = 4 and b=1

1:05 PM

Yeah, that's one reasonable way to collect randomness on desktop computers, if you don't trust the OS randomness collector.

(Although, if you don't trust the OS to give you good random numbers, why do you trust it to give you correct mouse positions?)

Anyway, OK, I have a = 4 and b = 1. What should I do with them?

Add them modulo 5? That gives 5 ≡ 0 (mod 5).

If I add them modulo 128, I get 5 ≡ 133 ≡ 261 ≡ 389 ≡ 517 ≡ ... (mod 128).

(And of course there are negative solutions, too.)

ok ok

so either 5 or 0

we take which within bounds

sp b must be independant of a

orelse we should have collisions

but in xor operation the addition is performed between a and a dependant variable

As long as you know b, you can always recover a (modulo whatever) from a + b, simply by a = (a + b) - b. But yes, if you can't figure out b without knowing a first, then just having a + b doesn't help.

i cant deal with such collisions apart from exhaustive essays

yes thats what i came up with

(In fact, that's exactly why a stream cipher / one-time pad can be decoded with the key, but not without it.)

Anyway, what do you mean by "in xor operation the addition is performed between a and a dependant variable"?

1:21 PM

a + a modulo 2 where b=a is relevant

in some cases a dependant variable works fine

Yeah, you can't recover a from a + a = 2 a ≡ 0 a = 0 (mod 2), because 2 doesn't have a multiplicative inverse modulo 2.

In fact, you can't uniquely recover a from 2 a (mod m) for any even modulus m.

Similarly, you cannot uniquely solve for a given a + a + a = 3 a (mod m), if m is a multiple of 3.

yes yes correct its always zero

1:51 PM

those things must be dealed using congruence

dealt*

2:06 PM

tring all combinations and detecting collisions

but right the moment i avoid any side effects by manipulating totally independant vars

Honestly, I'm not entirely sure what, exactly, you're trying to do, so I can't really help much. I do recommend reading the Wikipedia articles I linked to above.

there is a condition when a*b mod c has unique solutions and it s dealt with congruence

thats my point

but a+b mod c is always unique as long as we expulse the option out of boundary

its easy.... wonder why it took for me that much of time to get in mind

look the multiplication table ... the multiplication by 2 modulo 10 has many redundancies but multiplication by 9 is different .... well thats the line where on we should move from

thanks . gdby

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Jan '1511

The Side Channel

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