Tutorial B11 Breaking RSA
RSA Attack Theory
We won't go into what RSA is, see the RSA Wikipedia article for a quick background. What we really care about is the following pieces of sudocode from that article used in decrypting a message:
/**
* Decrypt
*
* @param {c} int / bigInt: the 'message' to be decoded (encoded with RSA.encrypt())
* @param {d} int / bigInt: d value returned from RSA.generate() aka private key
* @param {n} int / bigInt: n value returned from RSA.generate() aka public key (part I)
* @returns {bigInt} decrypted message
*/
RSA.decrypt = function(c, d, n){
return bigInt(c).modPow(d, n);
};
The most critical piece of information is that value d. It contains the private key, which if leaked would mean a compromise of the entire system. So let's assume we can monitor a target device while it decrypts any message (we don't even care what the message is). Our objective is to recover d.
Let's consider our actual target code, which will be the RSA implementation in avr-crypto-lib. This has been copied over to be part of the ChipWhisperer repository, and you can see the implementation rsa_basic.c of rsa_dec(). The function in question looks like this:
uint8_t rsa_dec(bigint_t* data, const rsa_privatekey_t* key){
if(key->n == 1){
bigint_expmod_u(data, data, &(key->components[0]), &key->modulus);
return 0;
}
if(key->n == 5){
if (rsa_dec_crt_mono(data, key)){
return 3;
}
return 0;
}
if(key->n<8 || (key->n-5)%3 != 0){
return 1;
}
//rsa_dec_crt_multi(data, key, (key->n-5)/3);
return 2;
}
We'll consider the case where key->n == 5, so we have the rsa_dec_crt_mono() to attack. You can see that function Line 53 of that same file. I've removed all the debug code in the following so you can better see the program flow:
uint8_t rsa_dec_crt_mono(bigint_t* data, const rsa_privatekey_t* key){
bigint_t m1, m2;
m1.wordv = malloc((key->components[0].length_B /* + 1 */) * sizeof(bigint_word_t));
m2.wordv = malloc((key->components[1].length_B /* + 1 */) * sizeof(bigint_word_t));
if(!m1.wordv || !m2.wordv){
//Out of memory error
free(m1.wordv);
free(m2.wordv);
return 1;
}
bigint_expmod_u(&m1, data, &(key->components[2]), &(key->components[0]));
bigint_expmod_u(&m2, data, &(key->components[3]), &(key->components[1]));
bigint_sub_s(&m1, &m1, &m2);
while(BIGINT_NEG_MASK & m1.info){
bigint_add_s(&m1, &m1, &(key->components[0]));
}
bigint_reduce(&m1, &(key->components[0]));
bigint_mul_u(data, &m1, &(key->components[4]));
bigint_reduce(data, &(key->components[0]));
bigint_mul_u(data, data, &(key->components[1]));
bigint_add_u(data, data, &m2);
free(m2.wordv);
free(m1.wordv);
return 0;
}
Note all the calls to bigint_expmod_u() with the private key material. If we could attack that function, all would be lost. These functions are elsewhere - it's in the bigint.c file at Line 812. Again we can see the source code here:
oid bigint_expmod_u(bigint_t* dest, const bigint_t* a, const bigint_t* exp, const bigint_t* r){
if(a->length_B==0 || r->length_B==0){
return;
}
bigint_t res, base;
bigint_word_t t, base_b[MAX(a->length_B,r->length_B)], res_b[r->length_B*2];
uint16_t i;
uint8_t j;
res.wordv = res_b;
base.wordv = base_b;
bigint_copy(&base, a);
bigint_reduce(&base, r);
res.wordv[0]=1;
res.length_B=1;
res.info = 0;
bigint_adjust(&res);
if(exp->length_B == 0){
bigint_copy(dest, &res);
return;
}
uint8_t flag = 0;
t=exp->wordv[exp->length_B - 1];
for(i=exp->length_B; i > 0; --i){
t = exp->wordv[i - 1];
for(j=BIGINT_WORD_SIZE; j > 0; --j){
if(!flag){
if(t & (1<<(BIGINT_WORD_SIZE-1))){
flag = 1;
}
}
if(flag){
bigint_square(&res, &res);
bigint_reduce(&res, r);
if(t & (1<<(BIGINT_WORD_SIZE-1))){
bigint_mul_u(&res, &res, &base);
bigint_reduce(&res, r);
}
}
t<<=1;
}
}
SET_POS(&res);
bigint_copy(dest, &res);
}
Within that file, the part is the loop at the end. This is actually going through and doing the required a**exp % r function. If you look closely into that loop, you can see there is a variable t, which is set to the value t = exp->wordv[i - 1];. After each run through the loop it is shifted left one. That t variable contains the private key, and the reason it is shifted left is the following piece of code is checking if the MSB is '0' or '1':
bigint_square(&res, &res);
bigint_reduce(&res, r);
if(t & (1<<(BIGINT_WORD_SIZE-1))){
bigint_mul_u(&res, &res, &base);
bigint_reduce(&res, r);
}
What does this mean? While there is data-dependent code execution! If we could determine the program flow, we could simply read the private key off one bit at a time. This will be our attack on RSA that we perform in this tutorial.