There are six of them but some of these could be just added strings outputted upon completion of a stage. Lo and behold, when we dump the contents of the memory address we get "%d", which tells us that the . This looks just like phase 1. I start stepping by single instructions until I get to the point where I am about to hit the function strings_not_equal. I know that due to x86-64 calling conventions on programs compiled with GCC that %rdi and %rsi may contain pointers to the words to compare. DrEvil This command lists out all the values that each of the registers hold. 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14 by hand by running their custom bomb against their solution: For both Option 1 and Option 2, the makebomb.pl script randomly, chooses the variant ("a", "b", or "c") for each phase. The second input had to be a 11, because the the phase_4 code did a simple compare, nothing special. It's provided only for completeness. Next it takes the address of the memory location within the array indexed by the third user input and places in the empty adjacent element designated by the second user input. Let's start with when it calls sym.read_six_numbers. Some of the pass phrases could be integers, or a random set of characters if that is the case then the only way to figure things out is through dynamic analysis and disassembling the code. angelshark.ics.cs.cmu.edu So far from my understanding, two conditions need to be met: edx must equal 0xf, meaning the first input has to be 5, 21, 37, etc. CSO1 - Bomb lab. Configure the Bomb Lab by editing the following file: ./Bomblab.pm - This is the main configuration file. In this part, we are given two functions phase_4() and func4(). Each offering of the Bomb Lab starts with a clean new ./bomblab. Keep going! Is it true that the first input has to be 5, 21, 37, etc? BOOM!!! This series will focus on CMU's Binary Bomb challenge. So you think you can stop the bomb with ctrl-c, do you?' From here, we have two ways to solve this phase, a dumb way and a smart way. Pretty confident its looking for 3 inputs this time. makoshark.ics.cs.cmu.edu, Dunno, lets just get a static printout of the disassembled code and see what comes out. Next, as we scan through each operation, we see that a register is being incremented at , followed by a jump-less-than statement right afterwards that takes us back up to . The first argument must be less than 7, right? phase_defused I found various strings of interest. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. lesson and forces them to learn to use a debugger. GDB then stopped at the break before entering into the phase_1 function call. I know there has to be 6 numbers, with the range of 1-6, and there can't be any repeats. However, you do need to handle recursion actually. Now lets get started with Phase 1! correctly, else you and your students won't be able to run your bombs. phase_3 I hope it's helpful. We multiply the number by 2 each step, so we guess the sequence to be 1, 2, 4, 8, 16, 32, which is the answer. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. You continue to bounce through the array. So you got that one. Control-l can be used to refresh the UI whenever it inevitably becomes distorted. You get to know that the input sequence must be an arbitary combination of number 1,2,3,4,5,6. The source code for the different phase variants is in ./src/phases/. You have 6 phases with Solved this is binary bomb lab phase 5.I didn't solve phase - Chegg Now you can see there are a few loops. mov a b moves data from a to b as opposed to b to a). Each bomb phase tests a different aspect of machine language programs: Phase 1: string comparison. Each phase has a password/key that is solved through the hints found within the assembly code. The "report daemon" periodically, scans the scoreboard log file. And, as you can see at structure, the loop iterates 6 times. For each bomb, it tallies the number, of explosions, the last defused phase, validates each last defused, phase using a quiet copy of the bomb, and computes a score for each, student in a tab delimited text file called "scores.txt." A binary bomb is a program that consists of a . Actually in this part, the answer isn't unique. Otherwise, the bomb explodes by printing " fun7 ??? A tag already exists with the provided branch name. Phase 1 defused. How does loop address alignment affect the speed on Intel x86_64? To begin we first edit our gdbCfg file. VASPKIT and SeeK-path recommend different paths. Phase 1 defused. 1 first, so gdb is the most recent available version of GDB. How about saving the world? - Main daemon (bomblab.pl). @cinos hi, I had same problem, I couldn't understand, I must have ecx 15 too, but I couldn't figure it out. GitHub - Taylor1VT/HW-5-Binary-Bomb Connect and share knowledge within a single location that is structured and easy to search. In order to do this you must look at the various integers within the array and then place them in ascending order by the index of those integer containing elements. . Then we use strings command to find out the answer, Having a look at the code structure, you should notice that there exists a loop structure. Work fast with our official CLI. The input should be "4 2 6 3 1 5". This command sets breakpoints throughout the code. 0000000000401062 <phase_5>: 401062: 53 push % rbx 401063: 48 83 ec 20 sub $ 0x20, % rsp 401067: 48 89 fb mov % rdi, % rbx 40106a: . I dereference the string pointed to by %rdi using x/s $rdi and see that the string pointed to is 'blah'. You don't need to understand any of this to. On line <phase_4+16>, the <phase_4> function is pushing a fixed value stored at memory address 0x8049808 onto the stack right before a call to scanf is made. Can you still use Commanders Strike if the only attack available to forego is an attack against an ally? Could this mean alternative endings? Instructors and students view the scoreboard by pointing their, The online Bomb Lab is self-grading. Contribute to hengyingchou/CSE351 development by creating an account on GitHub. While layout asm is helpful, also helpful to view the complete disassembled binary. servers running. As we have learned from the past phases, fixed values are almost always important. Each phase expects you to type a particular string. Explosion and, diffusions from bombs whose LabIDs are different from the current. You signed in with another tab or window. Ok, let's get right to it and dig into the <phase_5> code: So, what have we got here? On to the next' or 'So you got that one. Work fast with our official CLI. Did the drapes in old theatres actually say "ASBESTOS" on them? Set a breakpoint on phase 3 and start the process again and you should come to the following. offer the lab. "make cleanallfiles" resets the lab from scratch, deleting all data specific to a particular instance of the lab, such, as the status log, all bombs created by the request server, and the, scoreboard log. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. ordered by the total number of accrued points. your answer turns out to be 21 115, The solution is : 5 115. phase_3 phase_2 You signed in with another tab or window. Otherwise, the bomb explodes by printing "BOOM!! The smart way of solving this phase is by actually figuring out the cypher. The student then saves the tar file to disk. For more information, you can refer to this document, which gives a handy tutorial on the phase 6. Dump of assembler code for function phase_5: 0x0000000000401002 <+0>: sub $0x18,%rsp ; rsp = rsp - 24, 0x0000000000401006 <+4>: lea 0x8(%rsp),%rcx ; rcx = *(rsp + 8) (function argument), 0x000000000040100b <+9>: lea 0xc(%rsp),%rdx ; rdx = *(rsp + 12) (function argument), 0x0000000000401010 <+14>: mov $0x401ebe,%esi ; esi = "%d %d", 0x0000000000401015 <+19>: mov $0x0,%eax ; eax = 0, 0x000000000040101a <+24>: callq 0x400ab0 <__isoc99_sscanf@plt>, 0x000000000040101f <+29>: cmp $0x1,%eax ; if (eax > 1) goto 0x401029, 0x0000000000401022 <+32>: jg 0x401029 , 0x0000000000401024 <+34>: callq 0x40163d ; if (eax <= 1) explode_bomb(), 0x0000000000401029 <+39>: mov 0xc(%rsp),%eax ; eax = *(rsp + 12) ::function parameter, 0x000000000040102d <+43>: and $0xf,%eax ; eax = eax & 0xf (last 2 bits), 0x0000000000401030 <+46>: mov %eax,0xc(%rsp) ; *(rsp + 12) = eax, 0x0000000000401034 <+50>: cmp $0xf,%eax ; if (eax == 0xf) explode_bomb(), 0x0000000000401037 <+53>: je 0x401065 , 0x0000000000401039 <+55>: mov $0x0,%ecx ; ecx = 0, 0x000000000040103e <+60>: mov $0x0,%edx ; edx = 0, 0x0000000000401043 <+65>: add $0x1,%edx ; edx = edx + 0x1, 0x0000000000401046 <+68>: cltq ; sign extend eax to quadword (rax), 0x0000000000401048 <+70>: mov 0x401ba0(,%rax,4),%eax ; eax = *(rax * 4 + 0x401ba0), 0x000000000040104f <+77>: add %eax,%ecx ; ecx = ecx + eax, 0x0000000000401051 <+79>: cmp $0xf,%eax ; if (eax != 0xf) goto 0x401043 (inc edx), 0x0000000000401054 <+82>: jne 0x401043 , 0x0000000000401056 <+84>: mov %eax,0xc(%rsp) ; *(rsp + 12) = eax, 0x000000000040105a <+88>: cmp $0xc,%edx ; if (edx != 12) explode_bomb(), 0x000000000040105d <+91>: jne 0x401065 , 0x000000000040105f <+93>: cmp 0x8(%rsp),%ecx ; if (ecx == *(rsp + 8)) goto 0x40106a, 0x0000000000401063 <+97>: je 0x40106a , 0x0000000000401065 <+99>: callq 0x40163d ; explode_bomb(), 0x000000000040106a <+104>: add $0x18,%rsp ; rsp = rsp + 24, 0x000000000040106e <+108>: retq ; return, --------------------------------------------------------------------------------. Use arg1 and address ebp-0x20 as arguments of function read_six_numbers. So, the value of node1 to node6 are f6, 304, b7, eb, 21f, 150. When we hit phase_1, we can see the following code: The code is annotated with comments describing each line. Lets clear all our previous breakpoints and set a new one at phase_2. From this, we can deduce that the input for phase_2 should be 1 2 4 8 16 32. Maybe you get an alternative string for the bomb blowing up if done so via the secret stage? When you fail a phase, and the bomb goes off, you probably get the string 'BOOM!!!' Essentially what is happening is, each character from our string is ANDed with 0xf, and the result is used to get the character with the corresponding index from the array. node3 Hello world. Then we can get the range of the first argument from the line. DePaul University - System I - Winter 2017, **Note: I made this repo with the intent to help others solve their own Bomb Labs. BombID: Each bomb in a given instance of the lab has a unique, non-negative integer called the "bombID. A binary bomb is a program that consists of a sequence of phases. Are you sure you want to create this branch? Are you sure you want to create this branch? Give 0 to ebp-8, which is used as loop condition. I also wanted to see groupings of strings that may have similar prefixes and so I sorted the strings program output and looked for anything interesting in that manner. Content Discovery initiative April 13 update: Related questions using a Review our technical responses for the 2023 Developer Survey. There is an accessed memory area that serves as a counter. sign in Lets enter a test string to let the program hit our break point. Thus, each student, gets a unique bomb that they must solve themselves. Please, Understanding Bomb Lab Phase 5 (two integer input), https://techiekarthik.hashnode.dev/cmu-bomblab-walkthrough?t=1676391915473#heading-phase-5. There are many things going on with shuffling of variables between registers, some bit shifting, and either a subtraction or an addition being applied to some of the hard coded constants. The following lines are annotated. Keep going! Now switch to Visual mode with v, cycle the print mode with p until you see the disassembled function, toggle your cursor with c, then finally move down to the movzx edx, byte . From phase_4, we call the four arguments of func4 to be a, b(known, 0), c(known, 14), d(known, 0). At any point in time, the, tab-delimited file (./bomblab/scores.txt) contains the most recent, scores for each student. You can tell, makebomb.pl to use a specific variant by using the "-p" option. ', It is not clear what may be the output string for solving stage 4 or 5. The variable being used in this comparison is $eax. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. phase_4() - In this phase you are dealing with a recursively called function. Please, Your answer could be improved with additional supporting information. bomblab-Angr/Phase 5 x86_64.ipynb at master - Github So you think you can stop the bomb with ctrl-c, do you? Link to Bomb Lab Instructions (pdf) in GitHub Repository. Cannot retrieve contributors at this time. Maybe function names or labels? Attack Lab Phase 1: Buffer Overflow (CS:APP) - YouTube p # Change print mode in Visual/Graph mode. CMU Bomb Lab with Radare2 Phase 1 | by Mark Higgins - Medium You've defused the secret stage! This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. Then you can solve this problem by making a table(Yeah, it may seem silly, but I think it's the most convenient way). is "defused." The code is comparing the string (presumably our input) stored in %eax to a fixed string stored at 0x804980b. Could there be a randomization of stages or two planned routes through the bomb?
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