2.mdx

author	contributors	adapted_from
0xB49bf876BE26435b6fae1Ef42C3c82c5867Fa149	0xB49bf876BE26435b6fae1Ef42C3c82c5867Fa149	https://github.com/chainlight-io/publications/blob/417d1475d3b7057db6e0da2cbaba6104576f3d20/ctf-writeups/curta-cup/AddressGame/writeup.md

Overview

This challenge was inspired by [The Password Game](https://neal.fun/password-game/).

For this challenge, you need to generate a vanity contract address that satisfies:

• The address contains box(seed, 0) % 2 vowel hexadecimal digits (0xA and 0xE).
• The address contains box(seed, 1) % 3 consonant hexadecimal digits (0xB, 0xC, 0xD, and 0xF).
• All number digits in the address ([0x0, 0x9]) sum to 25 + seed % 50.

Solving the puzzle

To generate the vanity address, you can write your own miner that satisfies the conditions above.

For example, we made the following changes to profanity2:

diff --git a/profanity.cl b/profanity.cl
index fe187fd..ddd865c 100644
--- a/profanity.cl
+++ b/profanity.cl
@@ -768,18 +768,35 @@ __kernel void profanity_score_range(__global mp_number * const pInverse, __globa
        __global const uchar * const hash = pInverse[id].d;
        int score = 0;

+    int c[16];
+    for (int i = 0; i < 16; i++) {
+        c[i] = 0;
+    }
+
+    score = 40;
+
        for (int i = 0; i < 20; ++i) {
                const uchar first = (hash[i] & 0xF0) >> 4;
                const uchar second = (hash[i] & 0x0F);
+        c[first]++;
+        c[second]++;
+       }

-               if (first >= data1[0] && first <= data2[0]) {
-                       ++score;
+    {
+        if (((c[0xa] + c[0xe]) & 1) != 0) score = 0;
+        if (((c[0xb] + c[0xc] + c[0xd] + c[0xf]) % 3) != 0) score = 0;
     }

-               if (second >= data1[0] && second <= data2[0]) {
-                       ++score;
+    {
+        int _sum = 0;
+        for (int i = 1; i < 10; i++) {
+            _sum += c[i] * i;
         }
+        if (_sum != 25 + 35) {
+            score = 0;
         }
+    }
+

While `box` returns `uint256`, you can mine for `uint64` instead since the modulo fits within 64 bits: $\log_{2}15203777108537674021<64$.

Solve script

Check out our solve test below for more details.

The solution below is for [our address](https://etherscan.io/address/0xB49bf876BE26435b6fae1Ef42C3c82c5867Fa149).

// SPDX-License-Identifier: UNLICENSED
pragma solidity ^0.8.13;

import {Test, console2} from "forge-std/Test.sol";
import {IBox, AddressGame} from "../src/Challenge.sol";

contract MyBox is IBox {
    function isSolved() external view returns (bool) {
        return true;
    }
}

contract ChallengeTest is Test {
    AddressGame public challenge;

    function setUp() public {
        challenge = new AddressGame();
    }

    function test_Solve() public {
        // Solution: see profanity2.patch and https://github.com/1inch/profanity2
        address ATTACKER = 0xB49bf876BE26435b6fae1Ef42C3c82c5867Fa149;
        address DEPLOYER = vm.createWallet(0x000039ecc4b345efae348b37e89fef3678e6ed97715a2a3b053be81f528bdad2).addr;

        uint256 seed = challenge.generate(ATTACKER);
        console2.log(seed, challenge.box(seed, 0) % 2, challenge.box(seed, 1) % 3);

        vm.startPrank(DEPLOYER);
        uint256 solution = uint256(uint160(address(new MyBox())));
        assertEq(challenge.verify(seed, solution), true);
    }
}