This tutorial is an add-on to [[Tutorial A5 Breaking AES-256 Bootloader]]. It continues working on the same firmware, showing how to obtain the hidden IV and signature in the bootloader. '''It is not possible to do this bonus tutorial without first completing the regular tutorial''', so please finish Tutorial A5 first.
''This tutorial is under construction! Check back in a few days.'' = Background === AES in CBC Mode ==* Repeat of theory from tutorial == The IV ==* Suggest some ideas== The Signature ==* Timing attack* Show firmware = Exploring the Bootloader ==
In this tutorial, we have the luxury of seeing the source code of the bootloader. This is generally not something we would have access to in the real world, so we'll try not to use it to cheat. (Peeking at <code>supersecret.h</code> counts as cheating.) Instead, we'll use the source to help us identify important parts of the power traces.
=== Bootloader Source Code ===
Inside the bootloader's main loop, it does three tasks that we're interested in:
* it decrypts the incoming ciphertext;
We'll use both of the source files throughout the tutorial.
=== Power Traces ===
After the bootloader is finished the decryption process, it executes a couple of distinct pieces of code:
* To apply the IV, it uses an XOR operation;
Let's grab a lot of traces so that we don't have to come back later. Save the project somewhere memorable, set up the capture routine to record 1000 traces, hit ''Capture Many'', and grab a coffee.
== Attacking the IV ==
We need to find the IV before we can look at the signature, so the first half of the attack will look at the IV bytes.
=== Attack Theory ===
The bootloader applies the IV to the AES decryption result by calculating
This is effectively a DPA attack on a single bit of the IV. We can repeat this attack 128 times to recover the entire IV.
=== A 1-Bit Attack ===
Unfortunately, we can't use the ChipWhisperer Analyzer to attack this XOR function. Instead, we'll write our own Python code. One thing that we ''don't'' need to do is write our own AES-256 implementation: there's some perfectly fine code in the PyCrypto library. [https://pypi.python.org/pypi/pycrypto Install PyCrypto] and make sure you can use its functions:
<pre>
for i in range(numTraces):
ct = str(bytearray(textin[i]))
d pt = aes.decrypt(ct)
d = [bytearray(pt)[i] for i in range(16)]
dr.append(d)
[[File:Tutorial-A5-Bonus-Diff-0.PNG]]
However, one of these spikes is meaningless to us. The spike around sample 1600 is caused by the signature check, which we aren't attacking yet. You can get a better feel for this by plotting an overlay of the power trace itself: <pre>plt.hold(True)plt.plot(traces[0], 'r')plt.plot(diff, 'b') plt.grid()plt.show()</pre> You can notice the input data is often "pegging" to the 0.5 limit, indicating that spike may be caused by data overload! At any rate, it's clear the actual XOR is occurring where the group of 16 identical operations happens. Let's ignore this large peak and zoom in on the smaller spikes at the start of the trace:
[[File:A5-Bonus-Diff-0-Zoom.PNG]]
These peaks are about 60 samples later (or 15 cycles, since we're using an ADC clock that's 4 times faster than the microcontroller). Also, most of these peaks are upside down! This is a pretty clear indicator that we can find the IV bits from these differential traces. The only thing that we can't tell is the polarity of these signals; there's no way to tell if right-side-up peaks indicate a bit that's set or cleared. However, that means we can narrow down the IV to two possibilities, which is a lot better than <math>2^{128}</math>.
=== The Other 127 ===
The best way to attack the IV would be to repeat the 1-bit conceptual attack for each of the bits. Try to do this yourself! (Really!) If you're stuck, here are a few hints to get you going:
* One easy way of looping through the bits is by using two nested loops, like this:
checkIfBitSet = (byteToCheck >> bit) & 0x01
</pre>
If you're ''really, really'' stuck, there's a working attack in [[#Appendix ?A: IV Attack Script]]. You should find that the secret IV is <code>C1 25 68 DF E7 D3 19 DA 10 E2 41 71 33 B0 EB 3C</code>.
= Appendix D AES-256 IV Attack Script =Attacking the Signature ==The last thing we can do with this bootloader is attack the signature. This final section will show how one byte of the signature could be recovered. If you want more of this kind of analysis, a more complete timing attack is shown in [[Tutorial B3-1 Timing Analysis with Power for Password Bypass]].
'''NB: This script works for 0.10 release or later, see local copy in doc/html directory of chipwhisperer release if you need === Attack Theory ===Recall from earlier versions'''that the signature check in C looks like:
Full attack script, copy<pre>if ((tmp32[0] == SIGNATURE1) && (tmp32[1] == SIGNATURE2) && (tmp32[2] == SIGNATURE3) && (tmp32[3] == SIGNATURE4)){</paste into a file then add as active attack scriptpre>or, in assembly,<pre> 37a: 89 89 ldd r24, Y+17 ; 0x11 37c: 88 23 and r24, r24 37e: 09 f0 breq .+2 ; 0x382 380:98 cf rjmp .-208 ; 0x2b2
<pre>#IV Attack Script 382: 8a 89 ldd r24, Y+18 ; 0x12from chipwhisperer.common.autoscript import AutoScriptBase 384: 8b 3e cpi r24, 0xEB ; 235#Imports from Preprocessingimport chipwhisperer 386: 09 f0 breq .analyzer.preprocessing as preprocessing+2 ; 0x38a#Imports from Capturefrom chipwhisperer 388: 94 cf rjmp .analyzer.attacks.CPA import CPAfrom chipwhisperer.analyzer.attacks.CPAProgressive import CPAProgressiveimport chipwhisperer.analyzer.attacks.models.AES128_8bit# Imports from utilList-216 ; 0x2b2
# Imports for AES256 Attack 38a: 8b 89 ldd r24, Y+19 ; 0x13from chipwhisperer 38c: 82 30 cpi r24, 0x02 ; 2 38e: 09 f0 breq .analyzer.attacks.models+2 ; 0x392 390: 90 cf rjmp .AES128_8bit import getHW-224 ; 0x2b2
#Imports for IV Attack 392: 8c 89 ldd r24, Y+20 ; 0x14from Crypto 394: 8d 31 cpi r24, 0x1D ; 29 396: 09 f0 breq .+2 ; 0x39a 398: 8c cf rjmp .-232 ; 0x2b2</pre>In C, boolean expressions support ''short-circuiting''. When checking multiple conditions, the program will stop evaluating these booleans as soon as it can tell what the final value will be. In this case, unless all four of the equality checks are true, the result will be false. Thus, as soon as the program finds a single false condition, it's done.Cipher import AES
class AESIVAttackThe assembly code confirms this short-circuiting operation. Each of the four assembly blocks include a comparison (object<code>and</code> or <code>cpi</code>): numSubKeys = 16, a ''branch if equal'' (<code>brqe</code>), and a relative jump (<code>rjmp</code>). All four of the relative jumps return the program to the same location (the start of the <code>while(1)</code> loop), and all four of the branches try to avoid these relative jumps. If any of the comparisons are false, the relative jumps will return the program back to the start of the loop. All four branches must succeed to get into the body of the <code>if</code> block.
@staticmethod def leakage(textinThe short-circuiting conditions are perfect for us. We can use our power traces to watch how long it takes for the signature check to fail. If the check takes longer than usual, textout, guess, bnum, setting, state): knownkey = [0x94, 0x28, 0x5D, 0x4D, 0x6D, 0xCF, 0xEC, 0x08, 0xD8, 0xAC, 0xDD, 0xF6, 0xBE, 0x25, 0xA4, 0x99, 0xC4, 0xD9, 0xD0, 0x1E, 0xC3, 0x40, 0x7E, 0xD7, 0xD5, 0x28, 0xD4, 0x09, 0xE9, 0xF0, 0x88, 0xA1] knownkey = str(bytearray(knownkey)) ct = str(bytearray(textin))then we know that the first byte of our signature was right.
aes = AES.new(knownkey== Finding a Single Byte ===Okay, AESwe know that our power trace will look a lot different for one of our choices of signatures.MODE_ECB) pt = aesLet's figure out which one.decrypt(ct) return getHW(bytearray(pt)[bnum] ^ guess)We'll start by finding the average over all of our 1000 traces:
class userScript(AutoScriptBase):<pre> preProcessingList # Find the average over all of the tracesmean = [] def initProjectnp.average(selftraces, axis=0): pass</pre>
def initPreprocessing(self): self.preProcessingResyncSAD0 = preprocessing.ResyncSAD.ResyncSAD(self.parent) self.preProcessingResyncSAD0.setEnabled(True) self.preProcessingResyncSAD0.setReference(rtraceno=0Then, refpoints=we'll split our traces into 256 different groups (6300,6800one for each plaintext), inputwindow=(6000,7200)) self.preProcessingResyncSAD1 = preprocessing.ResyncSAD.ResyncSAD(self.parent) self.preProcessingResyncSAD1.setEnabled(True) self.preProcessingResyncSAD1.setReference(rtraceno=0Since we know the IV, refpoints=(4800,5100), inputwindow=(4700,5200)) self.preProcessingList = [self.preProcessingResyncSAD0,self.preProcessingResyncSAD1,] return self.preProcessingListwe can now use it to recover the actual plaintext that the bootloader checks:
def initAnalysis(self):<pre> self.attack # Split the traces into groupsgroupedTraces = CPA[[] for _ in range(self.parent, console=self.console, showScriptParameter=self.showScriptParameter256)] self.attack.setAnalysisAlgorithmfor i in range(CPAProgressive, AESIVAttack, NonenumTraces): self.attack.setTraceStart( group = dr[i][0)] ^ 0xC1 self.attack.setTracesPerAttack(100) self.attack.setIterations(1) self.attack.setReportingInterval(25) self.attack groupedTraces[group].setTargetBytesappend(traces[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15i]) self.attack.setTraceManager(self.traceManager()) self.attack.setProject(self.project()) self.attack.setPointRange((4800,6500)) return self.attack</pre>
def initReportingNext, we can find the mean for each group and see how much they differ from the overall mean: <pre># Find the mean for each groupmeans = np.zeros(self[256, resultstraceLen])for i in range(256): if len(groupedTraces[i]) > 0: resultsmeans[i] = np.setAttackaverage(selfgroupedTraces[i], axis=0) plt.plot(means[0] - mean)plt.plot(means[1] - mean)plt.plot(means[2] - mean)plt.plot(means[3] - mean)plt.grid()plt.show()</pre> The plot that comes out of this should look a bit like: [[File:Tutorial-A5-Bonus-Signature.PNG]] Wow - looks like we found it! However, let's clean this up with some statistics. We can use the correlation coefficient to see which bytes are the furthest away from the average: <pre>corr = []for i in range(256): corr.append(np.corrcoef(mean[1500:1700], means[i][1500:1700])[0, 1])print np.sort(corr)print np.argsort(corr)</pre> This should print something that looks like: <pre>[ 0.67663819 0.9834704 0.98855246 0.98942869 0.98994226 0.99019698 0.99082752 0.99159262 0.99166859 0.99169598 0.99216366 0.99229359 0.99239152 0.99240231 0.99246389 0.99254908 0.99258285 0.9926239 0.99280577 0.99302107 0.99339631 0.99394492 0.99396965 0.99403114 0.99408231 0.99410649 0.99424916 0.99460312 0.99464971 0.99468856 0.99483947 0.99512576 0.99553707 0.99570373 0.99572752 0.99577311 0.99582226 0.99587666 0.99590395 0.99623462 0.99630861 0.99639056 0.99644546 0.99646228 0.99653183 0.99661089 0.9966309 0.99665822 0.9966832 0.99670105 0.99673815 0.99679397 0.99690167 0.99692316 0.9969269 0.99694459 0.99703105 0.99704228 0.99705158 0.99708642 0.99709179 0.9971024 0.99710707 0.99711091 0.99711536 0.99715928 0.99720665 0.99721363 0.99721902 0.99722437 0.99722547 0.99723478 0.99724198 0.997244 0.99724712 0.99728416 0.99728622 0.99729196 0.99734564 0.99737952 0.99739401 0.99742793 0.99745246 0.99747648 0.99750044 0.9975651 0.99760837 0.99762965 0.99763106 0.99763222 0.99765327 0.9976662 0.9976953 0.99769761 0.99771007 0.99773553 0.99775314 0.99777414 0.99782335 0.99785114 0.99786062 0.99787688 0.99788584 0.99788938 0.9978924 0.99793722 0.99797874 0.99798273 0.9980249 0.99807047 0.99807947 0.99810194 0.99813208 0.9982722 0.99838807 0.99843216 0.99856034 0.99856295 0.99863064 0.9987529 0.99878124 0.99882028 0.99884917 0.99890103 0.99890116 0.99890879 0.99891135 0.99891317 0.99893291 0.99893508 0.99894488 0.99894848 0.99897892 0.99898304 0.9989834 0.99898804 0.99901833 0.99905207 0.99905642 0.99905798 0.99908281 0.99910538 0.99911272 0.99911782 0.99912193 0.99912223 0.9991229 0.99914415 0.99914732 0.99916885 0.99917188 0.99917941 0.99918178 0.99919009 0.99921141 0.99923463 0.99924823 0.99924986 0.99925438 0.99925524 0.99926407 0.99927205 0.99927364 0.99928305 0.99928533 0.99929447 0.99929925 0.99930205 0.99930243 0.99930623 0.99931579 0.99932861 0.99933414 0.99933806 0.99933992 0.99934213 0.99935681 0.99935721 0.9993594 0.9993601 0.99936267 0.99936373 0.99936482 0.99937458 0.99937665 0.99937706 0.99938049 0.99938241 0.99938251 0.999391 0.99940622 0.9994087 0.99940929 0.9994159 0.99941886 0.99942033 0.99942274 0.99942601 0.9994279 0.99943674 0.99943796 0.99944123 0.99944152 0.99944193 0.99944859 0.9994499 0.99945661 0.9994776 0.99948316 0.99949018 0.9994928 0.99949457 0.99949475 0.99949542 0.99949547 0.99949835 0.99950941 0.99951938 0.99951941 0.99953141 0.9995379 0.99954004 0.99954337 0.99954548 0.99955606 0.9995565 0.99956179 0.99956494 0.99956494 0.99956716 0.99957014 0.99957477 0.99957663 0.99958413 0.99958574 0.99958651 0.99958795 0.99958879 0.99959042 0.99959141 0.99959237 0.99959677 0.99961313 0.99962923 0.99963177 0.9996504 0.99968832 0.99969333 0.99969583 0.99969834 0.99970998 0.99972495 0.99972646 nan nan nan][ 0 32 128 255 160 223 8 16 48 96 40 1 95 215 2 33 34 64 4 36 127 207 239 254 253 247 222 251 159 191 221 219 80 129 136 176 168 192 144 56 224 162 130 119 87 72 132 24 126 9 17 111 123 18 112 68 63 125 79 3 66 93 94 49 42 161 237 206 31 35 104 20 98 245 37 238 10 65 52 50 246 231 243 44 41 183 5 6 214 97 190 12 250 220 91 175 199 252 205 249 189 151 235 143 218 157 158 213 203 38 100 211 187 217 155 55 200 226 11 107 138 120 152 23 103 137 81 145 25 30 118 109 110 60 7 184 202 146 117 21 225 177 131 208 77 148 78 193 71 85 140 196 133 47 185 115 15 86 233 169 61 172 194 232 122 186 62 92 102 75 124 212 116 29 150 180 156 57 230 121 90 182 240 167 76 170 165 88 43 229 166 46 147 27 188 163 149 19 198 51 210 53 73 83 142 135 59 114 22 197 241 45 236 227 89 174 82 67 13 244 14 181 228 69 195 58 39 26 242 173 113 74 179 141 106 99 234 105 216 28 139 153 209 201 204 248 54 108 84 171 101 70 154 164 134 178]</pre> This output tells us two things:* The first list says that almost every trace looks very similar to the overall mean (98% correlated or higher). However, there's one trace that is totally different, with 68% correlation. This is probably our correct guess.* The second list gives the signature guess that matches each of the above correlations. The first number in the list is 0x00, which is the correct signature!Note that three numbers in this output show a correlation of <code>nan</code> because none of the captured traces had any data on them. However, this doesn't matter to us - we found our byte. To finish this attack, you could force the capture software to send more specific text. To find the next byte of the signature, you'd want to fix byte 0 at 0x00 and make byte 1 random. Then, the plaintext should be XORed with the known IV and encrypted with the known AES-256 key. This is left as an exercise for the reader. == Appendix A: IV Attack Script ==This is the author's script to automatically attack the secret IV. If you've completed [[#A 1-Bit Attack]], you can paste this snippet immediately after it: <pre># Attack!for byte in range(16): location = 51 + byte * 60 iv = 0 for bit in range(8): results# Check if the decrypted bits are 0 or 1 pt_bits = [((dr[i][byte] >> (7 - bit)) & 0x01) for i in range(numTraces)] # Split the traces into two groups groupedPoints = [[] for _ in range(2)] for i in range(numTraces): groupedPoints[pt_bits[i]].setTraceManagerappend(selftraces[i][location]) # Get the means for each bit and subtract them means = [] for i in range(2): means.traceManagerappend(np.average(groupedPoints[i])) self.results diff = resultsmeans[1] - means[0] # Look in point of interest location iv_bit = 1 if diff > 0 else 0 iv = (iv << 1) | iv_bit print iv_bit, print "%02x" % iv</pre> The output from this script is:<pre>1 1 0 0 0 0 0 1 c10 0 1 0 0 1 0 1 250 1 1 0 1 0 0 0 681 1 0 1 1 1 1 1 df1 1 1 0 0 1 1 1 e71 1 0 1 0 0 1 1 d30 0 0 1 1 0 0 1 191 1 0 1 1 0 1 0 da0 0 0 1 0 0 0 0 101 1 1 0 0 0 1 0 e20 1 0 0 0 0 0 1 410 1 1 1 0 0 0 1 710 0 1 1 0 0 1 1 331 0 1 1 0 0 0 0 b01 1 1 0 1 0 1 1 eb0 0 1 1 1 1 0 0 3c</pre>
def doAnalysis(self): self.attack.doAttack()</pre>== Links ==
= Attacking the Signature ={{Template:Tutorials}}[[Category:Tutorials]]
