Be careful, it’s cheating!
nc 109.233.61.11 3126
Summary: discrete logarithm with group oracle
This challenge was in FigureCrypting category in Olympic CTF 2014. At first, there was some proof of work required (I’ve stolen this idea from some GitS ctf):
$ nc -n 109.233.61.11 3126 Proof of work, please Prefix is (hexed) 25cbcf3a964c3fc5 sha1 prefix (hexed) must be 000000 input (hexed): |
Puzzle can be solved with this C script.
... Let's play a guess game: How many 'tinsels' are there in 'therese'? You can use [mix a b] or [inv a] commands 251 times. Use [guess x] to submit your guess Good luck! |
The idea behind the challenge is simple discrete logarithm, while the group is given as an oracle. The elements are encoded as random words, and a player have access to a group operation and an element inverse operation. In this setting the lower bound for discrete logarithm is asymptotically equal to sqrt(p) operations (p is order of the group and is a prime), because you can’t exploit encoding properties of the elements like in index calculus method (as there are no properties at all).
Since the group size couldn’t be too large, players could simply guess the answer. So the server was ‘cheating’: it moved the secret while the group rules were not broken.
The intended way to solve this was to use Baby Step Giant Step algorithm (which is based just on the meet-in-the-middle technique). First, we need to estimate the group order. One can make a dozen of random guesses and see that the order should be just a bit larger than 30000. One can guess that it’s 31337 (and confirm with a bunch of mix operations) or make a table of first 100 elements and then try to compute powers in range(30000, 35000, 100) until one matches (you can use fast exponentiation here for power).
For classic BSGS you need 2 * sqrt(31337) = 2 * 177 = 354 operations, but only 251 requests are allowed. Easy to notice that inverse oracle doesn’t count:
mix jefferson jefferson bessemer (250 questions left) inv bessemer sally (250 questions left) inv sally bessemer (250 questions left) mix sally bessemer postcards (249 questions left) |
So in the first step of BSGS we can make table of the same size with twice fewer operations – just inverse each element of the table. But 177/2 + 177 = 265 is still too much! Can we use the same trick for the second step of BSGS? No, because if some element falls in our table, it’s inverse falls too! (that’s how our table is constructed). E.g. you gain no profit of computing inverses in the second stage.
So what do we have? We need to minimize x + y (x – number of operations in the first stage, y – number of operations in the second stage), while we need to cover the whole group: (2x + 2)y >= 31337 (2x+2 means that we already know two values – g and inv(g) and we gain two table items for one mix request.
Let’s check for some good values for x:
for x in range(100, 150): print x, x + math.ceil(31337 / (2*x + 2)) 100 255.0 101 254.0 102 254.0 103 253.0 104 253.0 105 252.0 106 252.0 107 252.0 108 251.0 109 251.0 110 251.0 111 250.0 112 250.0 113 250.0 114 250.0 115 250.0 116 249.0 117 249.0 118 249.0 119 249.0 120 249.0 121 249.0 122 249.0 123 249.0 124 249.0 125 249.0 126 249.0 127 249.0 128 249.0 129 249.0 130 249.0 131 249.0 132 249.0 133 249.0 134 250.0 135 250.0 136 250.0 137 250.0 138 250.0 139 250.0 140 251.0 141 251.0 142 251.0 143 251.0 144 252.0 145 252.0 146 252.0 147 252.0 148 253.0 149 253.0 |
Seems the minimum is 249 operations (and anything between 116 and 133 works)! But wait, we haven’t counted one operation of doubling our “giant step” – we need it because we have larger table). So, we can do it in 250 operations!
Bugs
That was the intended solution, but there was a flaw: when you make a request with a word from dictionary, which was not given to you in current session, it was binded to the first free number: 0, 2, 3, … (1 is given to you already). So you can ask an inverses of words from dictionary and they’ll get freezed. Using 31337/2 inverse operations you could guess the number with 0 mix oracles :)
Some teams used this partially: you can make a table of any size without using mix oracles.
I allowed using non-given words because it seems logical – for a group you should be able to pick any random element. But they should really be frozen to random elements, or otherwise picking new word should be counted as oracle request too.
Implementation
A few notes about how it was implemented:
For any group, where we are looking only at powers of some generator, we can switch to the group of indexes (powers of the generator) which is just additive group mod N. So actually “discrete logarithm” can be treated as finding v/g in additive group (or even finding v, because g=1). But since you can’t apply extended gcd as usual (which bases on a properties of element encoding), this is not easier to compute, but is easier to think as.
The game (G, V) contains two tables, matching words with numbers.
The first table hold elements which don’t depend on V (therefore are fixed (frozen) and can’t be moved). This elements could be computed as G + G, G – G, 2G + 4G, etc.
The second table hold elements dependent on V, with coefficients of V: xV + yG -> (x, y), e.g. V + V + G -> (2, 1).
It’s now easy to compute all operations – convert both operands to xV + yG form and compute the result in this form. Then decide where to store the result – in the first table or in the second.
V is binded to some value while the second table is not intersecting the first. When they intersect, new value for V is searched. If player is good enough, there can be no good place for V and then it’s frozen forever.