# Simple Electromagnetic Analysis vs symmetric cryptography

I've asked this question here and was advised to post this question on this forum in hopes of an answer.

I'm currently taking a security class and i came into this issue when writing a report about side channel attacks. When reading about electromagnetic attacks, more specifically about Simple Electromagnetic Analysis in Wikipedia (https://en.wikipedia.org/wiki/Electromagnetic_attack), it states:

It is very effective against asymmetric cryptography implementations...

I understand why it is effective agaisnt an algorithm such as RSA, which is asymmetric, but doesn't the reason for its effectiveness resides on the way the algorithm is implemented and not if its symmetric or asymmetric cryptography?

For example, a common attack on asymmetric RSA relies on the fact that the encryption steps rely on the value of the key bits. Every bit is processed with a square operation and then a multiplication operation if and only if the bit is equal to 1. An attacker with a clear trace can deduce the key simply by observing where the multiplication operations are performed.

So i don't really understand why it says its effective against asymmetric cryptography.

Thanks for your attention!

It would be wrong to say that only asymmetric algorithms are vulnerable to SEMA. For example, the Pohlig-Hellman exponentiation cipher is a symmetric encryption algorithm where encryption goes $x\mapsto x^k\bmod p$ for public prime $p$ with $(p-1)/2$ prime, and $k$ a secret exponent. A naive implementation would be vulnerable to SEMA just like a naive implementation of RSA is. And it is all too frequent that a very naive implementation of DES is vulnerable to SPA/SEMA.
• The other thing that complicates this answer is that the operations that assymetric crypto uses often has identities that make it easier to add blinding factors; for example, if we need to compute $xP$ (where $x$ is a scalar and $P$ is an EC point), we can pick a random $r$ and compute $(x + rq)P$, where $q$ is the curve order. The result is the same, and this effectively randomizes each bit of the multiplier if $r$ is large enough – poncho May 7 '18 at 19:18