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Alice and Bob communicate using the ECC public-key cryptosystem. They choose the elliptic curve y2 = x + and the finite field 2z3 Bob also chooses P B , ILB and cal...

Question

Alice and Bob communicate using the ECC public-key cryptosystem. They choose the elliptic curve y2 = x + and the finite field 2z3 Bob also chooses P B , ILB and calculates Bob sends his public key to Alice:Alice encrypts numerical message M, and sends her public key containing the encrypted message to Bob_You are BobllDecrypt the numerical message_You may use the "point addition" chart:(15,1)(15,22),(11,21)

Alice and Bob communicate using the ECC public-key cryptosystem. They choose the elliptic curve y2 = x + and the finite field 2z3 Bob also chooses P B , ILB and calculates Bob sends his public key to Alice: Alice encrypts numerical message M, and sends her public key containing the encrypted message to Bob_ You are Bobll Decrypt the numerical message_ You may use the "point addition" chart: (15,1) (15,22),(11,21)



Answers

Describe the steps that Alice and Bob follow when they use the Diffie-Hellman key exchange protocol to generate a shared key. Assume that they use the prime $p=101$ and take $a=2,$ which is a primitive root of $101,$ and that Alice selects $k_{1}=7$ and $\mathrm{Bob}$ selects $k_{2}=9 .$ (You may want to use some computational aid.)

All right. So Alice and Bob are gonna try to share a key here, and what we're given is P is 23. A is equal to five. Okay, so that's what they have chosen to share. So, step one is those items were chosen. Step to Alice is gonna choose a secret integer K one equal to eight, and we'll send a to the power of K one mod p two Bob. Okay, well, so this is going to be a is five. So we have two power eight. Wouldn't take that mod 23. This is gonna be equal to 390 1006 125 mod 23 which is 16. Okay, so Alice tells Bob 16. Step three bothers choosing secret into jure que tu, which is equal 25 and is going to send a to the power of K two mod pea to Alice. Okay, so then we get five to the power of five mod pea, which is 3125 mon 23. It is going to be twenties, and Bob tells Alice 20. Okay, women. Step three, run Step four. So, no, Alice will compute. Uh, a to the power of K two. Okay, one model P. Okay, so this is equal to into the power of K two Monde p the power of eight, ma 23. Okay. Which is going to be equal to 20 to the power of AIDS. Mod 23. Uh, it's just gonna be equal to six. 11 or 12 digit numbers. I didn't feel like writing it out, but, uh, feel free to use the power of technology to calculate that if you desire Okay, this is involved is going to do a similar computation A to the K one to the cage to Mod pea. And so this is going to be equal to a to the K one, my p to the power of five. My God, 23. Okay, which is 16 t power of five mile 23. This is 10 for 8576 mod 23 which is equal to six. What would you look at that? Both in step five and Instant four, we get a six. Right? So the the end of the protocol, they've each computer Their shared key of six

So in this problem we're working with some protocol photography. And it makes the note in here that the original message could be retrieved by multiplying the encrypted matrix by the inverse of the key matrix. Okay, so the first part they give us the key matrix K. So we need to find K inverse. Well, an easy way to do this is to go to matrix calculator. I'll go over here too. Dismas dot com matt. And then the map tools when you pull up that matrix calculated right there and you get this matrix calculator. So at the new matrix here And our Matrix is a three x 3. And so the entries are too one one, one, one zero. And then 111 one one one. Those are matrix. So then I'll have to do is go A and a inverse with that key right there and there's the inverse of it right there. 10 -1. So this is 10 -1 minus one, 11 & 0 -1. 10 -11. Okay, that gave us the inverse. Now R B says you as a result from our part A to decode the encrypted matrix E. So, what we're looking for is m which is E times K inverse. All right. So E is 47. 33. 44 36. 27. 47. 41 20 there's E times RK inverse up there. So let's go return matrix calculator then. And let's put in another new matrix, which will be This three x 3. Matrix e here I'll call it be in the in the matrix calculator. But it's E. In our problem statement. So 47 34 33. 44 36 87 47 41 and 20. Okay, so there's that matrix so now I have to do is take B times A. And it's the inverse of a. Right? And there is what I get. So I get 13 120 13 1 20 89 19 89 19 06:21 14 six 21 14. Okay, so then part C. Each entry and the resulting matrix that we got right represents the position of a letter in the english alphabet where A is one. These two C. is three and so on. Okay. So what was the original message? Well, let's see, 13 So ABCD. E. Let's just do this. Let's write some more of this table out here for ourselves. Days for is five F. Six G. is seven h. is eight. I is nine jay is 10. Okay. Is 11 L. Is 12, mm is 13. Hence 140. Is 15. P. 16 Q 17 are 18. S 19 T is 20 You is 21. That's the highest number I have on the on the listing there. Right? So 13 his M one is a 20 is mhm. Eight is H. Nine is I 19 is S and six is F. 21 is you And 14 is in. And so what do we have? We have. M a t h Math is fine. Here we go.


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