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TWCTF 2017 - Liar's Trap
#!/usr/bin/env ruby
require 'securerandom'
## Parameters
P = 115792089237316195423570985008687907853269984665640564039457584007913129639747
N = 100
K = 25
L = 38 # The number of liars
def apply_polynomial(coeffs, x)
r = 0
coeffs.inject(0) do |r, c|
r = (r * x + c) % P
def create_polynomial
coeff = SecureRandom.hex(64 * K)
coeff.chars.each_slice(128).map{|a|a.join.to_i(16) % P}
secret = SecureRandom.hex(64).to_i(16) % P
polynomial = create_polynomial
polynomial[-1] = secret
# Distribute Secret
user = {|i| apply_polynomial(polynomial, i + 1) }
# Some user are liars
[*0...N].shuffle[0, L].each{|i| user[i] = rand(P) }
STDOUT.puts "--- Distributed Secret ---"
N.times do |i|
STDOUT.puts "User #{i}: #{user[i]}"
STDOUT.puts "What's secret? "
input_secret = STDIN.gets.to_i
if input_secret == secret
STDOUT.puts "OK. I'll give you the flag"
STDOUT.puts "Wrong"
In the challenge we have the Shamir's Secret Sharing scheme.
The secret is splitted into 100 shares with threshold 25.
However, 38 shares are trashed (the "liars").
This is related to Reed-Solomon codes decoding.
But since this is a PPC challenge, we can go bruteforce!
Indeed, the fraction of 25-element subsets without liars is quite high:
sage: math.log( binomial(100-38, 26) / binomial(100, 26) , 2)
To verify that interpolation is good, we can simply wait until the decoded value is repeated twice.
For wrong decodings this is highly unlikely.
That is, we need to make around 2 x 2^21 = 4 millions interpolations.
The challenge server gives us 30 seconds. Maybe we won't fit, but trying a few times is not a big deal.
Additionally, let's optimize the interpolation. Instead of choosing subsets randomly each time,
we can modify a previous subset e.g. by replacing a single point in it.
This can improve the time from quadratic to linear per single interpolation.
In a few tries the script succeeds and gets the flag:
from random import randint, choice, shuffle
from libnum import invmod
P = 115792089237316195423570985008687907853269984665640564039457584007913129639747
N = 100
K = 25
L = 38 # The number of liars
class Interpol(object):
Lagrange Interpolation for the constant term only.
Allows fast replacement of a single point (in linear instead of quadratic time)
def __init__(self, lst):
self.xset = {v[0] for v in lst}
self.xs = [v[0] for v in lst]
self.ys = [v[1] for v in lst]
self.vecs = []
self.n = len(lst)
for j in xrange(self.n):
top = 1
bot = 1
for i in xrange(self.n):
if i == j:
top *= -self.xs[i]
top %= P
bot *= (self.xs[j] - self.xs[i])
bot %= P
self.vecs.append(top * invmod(bot, P) % P)
self.vecs[-1] %= P
def ans(self):
return sum(y * v % P for y, v in zip(self.ys, self.vecs)) % P
def replace(self, jj, xy):
xnew, ynew = xy
assert (self.xs[jj] == xnew) ^ (self.ys[jj] == ynew) ^ True
if self.xs[jj] == xnew:
return False
if xnew in self.xset:
return False
xold = self.xs[jj]
yold = self.ys[jj]
for j in xrange(self.n):
if j == jj:
# self.vecs[j] /= (-xold) / (self.xs[j] - xold)
# self.vecs[j] *= (-xnew) / (self.xs[j] - xnew)
top = (-xnew) * (self.xs[j] - xold) % P
bot = (-xold) * (self.xs[j] - xnew) % P
self.vecs[j] *= top * invmod(bot, P) % P
self.vecs[j] %= P
self.xs[jj] = xnew
self.ys[jj] = ynew
# vecs[jj] rebuild
top = 1
bot = 1
for i in xrange(self.n):
if i == jj:
top *= (-self.xs[i])
top %= P
bot *= (self.xs[jj] - self.xs[i])
bot %= P
self.vecs[jj] = top * invmod(bot, P) % P
return True
import ast, os
vals = ast.literal_eval(open("challenge").read())
I = Interpol(vals[:25])
open("/tmp/tags/%d" % sum(I.ys), "w").close()
open("/tmp/ans/%d" % I.ans(), "w").close()
ntried = 0
while True:
# random walk on subsets until a decoded value is repeated
# tag is hash of the subset, needed to avoid false positives
pos = randint(0, 24)
valnew = choice(vals)
if not I.replace(pos, valnew):
ntried += 1
ans = I.ans()
tag = sum(I.ys)
if os.path.isfile("/tmp/ans/%d" % ans) and not os.path.isfile("/tmp/tags/%d" % tag):
print "Decoded!", "itr", ntried, "ans", ans
open("/tmp/tags/%d" % sum(I.ys), "w").close()
open("/tmp/ans/%d" % I.ans(), "w").close()
#-*- coding:utf-8 -*-
This script gets a challenge from the server and runs workers.
Can be run repeatedly to succeed.
import gevent, os
from gevent.pool import Pool
from gevent.subprocess import check_output
from sock import *
print "cleaning ans, tags"
# shared storage for the poor
os.system("mkdir /tmp/{ans,tags}")
os.system("find /tmp/ans/ -type f -delete")
os.system("find /tmp/tags/ -type f -delete")
print "getting challenge"
f = Sock(" 42453", timeout=1000)
vals = []
for i in xrange(1, 101):
parts = f.read_line().split()
assert parts[0] == "User"
assert parts[1] == str(i-1)+":"
vals.append((i, int(parts[2])))
open("challenge", "w").write(`vals`)
# parallel computing is hard!!
def worker(i):
out = check_output(["./"])
print "got solution:", "worker", i, "outout", out
ans = int(out.split()[-1])
print "Answer:"
print `f.read_all()`
print "running workers"
p = Pool(8)
p.map_async(worker, range(8))
print "no luck.."
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