Candy.

Candy. Would you like some?

Is that candy?

A prowler?

Chair. Candy. Hey.

A fast network logon cracker

Bannerman, safecracker.

This candy costs eighty cents.

Microsoft Windows password cracker

A 100 years ago, the Cracker

Is there a lock picker in the crowd?

It must've been a burglar.

The candy I had in my bag went mushy in the heat.

The burglar pointed his gun at the victim.

The burglar gained access to the house through this door.

I'm not trying to intrude, but I'm in sort of a jam.

When I woke up, there was a burglar in my room.

Are you a prowler or do you have a search warrant?

And therefore, the adversary has negligible advantage in defeating deterministic counter mode.

Candy, sugar, bitter, good, taste, tooth, nice, honey, soda, chocolate, heart, cake, tart, pie.

Adewale Taiwo, aka FeddyBB master bank account cracker from Abuja in Nigeria.

Joe Grand, Electrical Engineer amp Hardware Hacker, Grand Idea Studio, http www.grandideastudio.com

Adewale Taiwo, aka FreddyBB master bank account cracker from Abuja in Nigeria.

Sour, candy, sugar, bitter, good, taste, tooth, nice, honey, soda, chocolate, heart, cake, tart, pie.

A lock breaker may look for mechanical flaws or failing that, extract information in order to narrow down the correct combination.

And as a result, the adversary will ever only see one ciphertext encrypted using this key.

And then our goal is to say that the adversary can't distinguish between these two experiments.

The top and the bottom of the hat I work in the same fashion as in previously published instruction Lollipop hand fan.

So the adversary outputs two messages, m0 and m1, and then he gets either the encryption of m0 or the encryption of m1, these are two different experiments.

Well, this adversary when he received an encryption of the message m1 he will always output 0. and when he receives an encryption of the message m0 it will always output 1.

A PRF is indistinguishable from a truly random function, therefore when we replace the PRF on the left with a truly random function on the right, the adversary is going to behave the same.

like this beautiful chair by Dirk Vander Kooij, where a robot deposits a toothpaste like slur of recycled refrigerator parts, as if he were a big candy, and makes a chair out of it.

You said, hey, this is all about things that are related to sweet things in one way or another sweet sugar, sweet candy, sweet and sour, honey is sweet, chocolate is sweet.

He was indicted on July 6, 2010, by a U.S. Federal Court in Seattle, Washington, on charges of transporting another stolen aircraft in that state.

World . And if the two ciphertext blocks are the same then the adversary knows that he received an encryption of the message Hello Hello and as a difference he knows that he received encryption of the message Hello World . So, he just follows a simple strategy here.

But now because F is a truly random function, the pad here is a truly one time pad, and therefore no adversary can distinguish an encryption of m0 from an encryption of m1 under the one time pad.

I'll just remind you that essentially for every adversary A that's trying to attack deterministic counter mode, we prove that there's an adversary B that's trying to attack the PRF.

So an attacker who looks at the ciphertext, even though he might not know what's actually written in these blocks, we'll know that these two blocks are equal.

But most foods posing as health food raise the same question as vitamin supplemented candy does adding vitamins really make a food better for you?

And similarly again when we go back from a truly random function here to a PRF, because the PRF is secure, the adversary can't distinguish these two bottom distributions, the left from the right.

So basically, what we want to show is, when the adversary's given the encryption of the message m0, here, this is the encryption of the message, m0. m0 XOR counter applied to the PRF, versus in giving the encryption of the message, m1.

So the way we do that is basically we say, well the top distribution, if instead of a PRF, we use a truly random function, namely here f is a truly random function, then the adversary, because of the property of the PRF, the adversary cannot distinguish these two experiments, right.

I couldn't be expected to foresee that the scheme, in itself a cracker jack, would skid into the ditch as it had done but all the same I'm bound to admit that I didn't relish the idea of meeting Corky again until time, the great healer, had been able to get in a bit of soothing work.