# blinder   ## Time

8 hours
Solved after the CTF ended, with @tjbecker (PPP)

# Behavior

Same as blind, but the order of GF(p) isn't smooth anymore.

# Solution

I found another solution which is more efficient.

## TL;DR

1. Search a elliptic curve with smooth order
2. Recover the index of y using Pohlig-hellman.
3. Recover the index of y in each subgroup with naive discrete log.
4. Calculate y.

After the CTF ended, there's some discussion of this challenge.

 00:29    hellmann| for blinder my guess would be to construct an elliptic curve with small factor in the order and perform dlp there,
using oracle operations to implement the ecc group
00:29      _0xbb_| you need to discuss that with y7 later^^
00:30    hellmann| looking forward :)
00:42    tjbecker| hellmann: that was my approach, but I couldn't get it down to only 0x4000 queries
00:43    tjbecker| would love to see what optimizations I was missing, if that was the intended solution
00:43    hellmann| how close could you get?
00:44    tjbecker| I needed roughly 0x20000, which is 8x too many
00:45    hellmann| still sounds cool, how many curves did you use? what were the factors
00:47    tjbecker| a,b = (242251381217038, 29739910446124)
00:47    tjbecker| gives a curve of order 2 * 3 * 11 * 13 * 17 * 67 * 71 * 131 * 173 * 179
00:48    tjbecker| (which is also nice because it's square free)
00:48    hellmann| nice, I thought about using many curves
00:48    tjbecker| maybe that's what I was missing
00:48    hellmann| maybe with montgomery arithmetic it should be good enough?
00:48    tjbecker| more liekly, I was just bad at optimizing point addition
00:50    tjbecker| yeah I also thought about using Edwards curves, but got too tired to stay up to try it
00:50    tjbecker| I was happy enough with a solution which was much better than sqrt(p)
01:01       sasdf| tjbecker: How do you construct that curve?
01:02    tjbecker| just try random (a,b) pairs, constructing the curve over GF(p) and take the one with the smoothest order
01:02    tjbecker| took like 10 minutes to find that curve


The method using Elliptic curve sounds promising. After @tjbecker gave me his script, we work on optimizing the algorithm together.

Here's the summary of our algorithm:

• Let C = EllipticCurve, T = Target y, G = Generator, o = C.cardinality().
• Calculate P = C.lift_x(T).
• Similar to what we do in blind, For each factor f in o:
• Calculate X = (o / f) G
• Calculate Y = (o / f) P
• Find i such that i X = Y, which means P.order() % f = i.
• Reconstruct P.order() using Pohlig-Hellman algorithm.
• Calculate T

## Embed integer

We can't encrypt arbitary integer directly. We need to construct them using 1 and add operation.

First, we calculate a list of power of two (using double and 1), and we can sum those needed bits to construct arbitary number.

## Offline calculation of operation on G

This is our first optimization.

Operation on the elliptic curve is much more expensive than embedding an integer, so we calculated all operation about the generator (i.e. X = (o / f) G and i X) offline, and then embed the result using previous algorithm.

For comparsion between i X and Y, we only compare their y coordinate.

## NAF form

For square-and-multiply (or double-and-add), we use NAF form instead of typical binary form to minimize the operation count.

## Merge order

When calculating double-and-add, we have to sum a list of numbers. If we sum them sequentially:

result = reduce(add, points, zero)


We (almost) won't hit the cache in add operation.

a, b, c, d, e, f, g -> (a+b), (c+d), (e+f), g


We will have a better chance that hit the cache at first round.

For Point multiplication (i.e. Y = (o / f) P), we know all the operands in advance. We use a greedy algorithm: add up most common pair first. It's not optimal, but much better then merging neighbors first.

A = 0, 1, 2
B = 0, 1, 3
C = 0, 1, 4, 5, 6
D = 5, 6
With following order:


## Constant power

There're two operation using constant power:

• inv = exp(x, p-2)
• sqrt = exp(x, (p+1)/4) p-2 is a very bad constant when using square-and-multiply, because of it's binary representation: 111111111111111111111111111111111110101101001111. We have to sum all those 1 and that is a huge amount of operations.

Fortunately, we cat factor p-2 to 123 * 2288414444759, which means we can calculate exp(exp(x, 123), 2288414444759) instead.

And we can apply similar techniques to sqrt too.

## Linear search v.s. Baby Step Giant Step

tjbecker implemented Baby Step Giant Step too. The algorithm has lower complexity, but it needs to run point addition sqrt(f) times, which needs too much operations.

A better curve with linear search works better.

## Search the curve

After we finish our algorithm, we can estimate the cost of each part:

• Y = (o / f) P takes about 2000 operations of each f.
• Comparing iX and Y takes about 24 operation of each test.
• Expectation of the needed tests to find i is f / 2.

We can use the cost for ranking curves when searching. Here's the script we use to search the curve.

## Flag

Now combining everything together, It's time for getting our flag :) Here's the script.

It uses about 13000 operations on localhost.

...

So where's the flag?

TimeoutError

It seems that I have to find a server in Germany to get the flag...