# Implementation of garbled circuits using RSA

I was just reading these notes on garbled (Yao) circuits and I just stuck trying to figure out how an implementation of Table 1 would work using RSA encryption. In RSA the public key is $(n,e)$. So for the first row, for example, I guess that ${\rm Enc}_{k_1^0}(k_2^0)$ means: "encrypt the "message" $k_2^0$ with the key $k_1^0$". So does the value of $k_1^0$ stand for $n$ or $e$ of RSA? What am I missing here?

• Well for starters, Yao GC uses symmetric cryptography for the gates. Does that help or did I misunderstand your question? – Thomas M. DuBuisson Oct 17 '14 at 20:15
• RSA typically isn't used for Yao's method. They define what they use for enc a little lower in the document – mikeazo Oct 17 '14 at 20:15
• Well, I didn't know that! Thanks for the heads up guys! – Cobe Oct 17 '14 at 20:25
• @ThomasM.DuBuisson Maybe change that into an answer? Negative answers are answers just as well. – Maarten Bodewes Oct 19 '14 at 16:29
• As mentioned garbled circuits usually uses symmetric cryptography. However, it also seems you are a little confused about notation. I.e., even if RSA were used, when we talk about encrypting with the key $k^0_1$, then $k^0_1$ would not be either $n$ or $e$ it would be both. In other words we would have $k^0_1 = (n, e)$. – Guut Boy Apr 2 '17 at 19:30