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第三届realworld 区块链wp

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前言

  • 第三届 realworld, blockchains 的 WP
  • 总共有三题，队伍总共做出两道，后续会持续更新
  • 随时欢迎大家交流，随便写写
  • 点击这里 -> 题目附件

Montagy

  • 题目目标清空余额
  • 和队友一起做出来的题目，应该是非预期做出来的，这里会将我们的解法和预期解法都稍微写一下

非预期解法

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#include <cstdio>
#include <ctime>
#include <cstdint>
#include <cstdlib>
#include <vector>
#include <algorithm>
using namespace std;

uint64_t forw(uint32_t t1, uint32_t t2, uint32_t m1, uint32_t m2, uint32_t m3, uint32_t m4)
{
        uint32_t p1, p2, p3;
        uint32_t s = 0x6644498b;
        for (int i = 0; i < 16; ++i) {
        s = s + 0x68696e74;
        p1 = (t1<<4) - m1;
        p2 = t1 + s;
        p3 = (t1>>5) + m2;
        t2 = t2 + (p1^(p2^p3));
        p1 = (t2<<4) + m3;
        p2 = t2 + s;
        p3 = (t2>>5) - m4;
        t1 = t1 + (p1^(p2^p3));
        }
        return ((uint64_t)t1 << 32) | t2;
}

uint64_t back(uint32_t t1, uint32_t t2, uint32_t m1, uint32_t m2, uint32_t m3, uint32_t m4)
{
        uint32_t p1, p2, p3;
    uint32_t s = (0x6644498b+0x68696e74ull*16) & 0xffffffff;
        for (int i = 0; i < 16; ++i) {
        p1 = (t2<<4) + m3;
        p2 = t2 + s;
        p3 = (t2>>5) - m4;
        t1 = t1 - (p1^(p2^p3));
        p1 = (t1<<4) - m1;
        p2 = t1 + s;
        p3 = (t1>>5) + m2;
        t2 = t2 - (p1^(p2^p3));
        s = s - 0x68696e74;
        }
        return ((uint64_t)t1 << 32) | t2;
}

int main()
{
        srand(time(0));
        vector<uint64_t> data;
        data.reserve(1ull<<31);
        uint32_t tb1 = 443551905, tb2 = 186746487;
        uint32_t te1 = 1418013151, te2 = 3531071330;
        for (uint64_t i = 0; i < (1ull<<31); ++i) {
                uint64_t v = forw(tb1, tb2, 0, 0, 0, i);
                data.push_back(v);
                if (i % 10000000 == 0)
                        printf("i=%ld\n",i);
        }
        printf("sort\n");
        sort(data.begin(), data.end());
        printf("unique\n");
        data.erase(unique(data.begin(), data.end()), data.end());
        uint64_t gv = 0;
        for (uint64_t m3 = 0;; ++m3) {
                vector<uint64_t> data2;
                data2.reserve(1ull<<30);
                for (uint64_t m4 = 0; m4 < (1ull<<30); ++m4) {
                        uint64_t v = back(te1, te2, 0, 0, m3, m4);
                        data2.push_back(v);
                        if (m4 % 10000000 == 0)
                                printf("i=%ld\n", m4);
                }
                printf("sort\n");
                sort(data2.begin(), data2.end());
                uint64_t i = 0, j = 0;
                while (i < data.size() && j < data2.size()) {
                        if (data[i] == data2[j]) {
                                gv = data[i];
                                break;
                        }
                        else if (data[i] < data2[j])
                                ++i;
                        else
                                ++j;
                }
                if (gv != 0) {
                        printf("gv=%lx\n", gv);
                        printf("m3=%lx\n", m3);
                        for (uint64_t m4 = 0; m4 < (1ull<<30); ++m4) {
                                uint64_t v = back(te1, te2, 0, 0, m3, m4);
                                if (v == gv) {
                                        printf("m4=%lx\n", m4);
                                        break;
                                }
                        }
                        break;
                }
        }
        for (uint64_t i = 0; i < (1ull<<31); ++i) {
                uint64_t v = forw(tb1, tb2, 0, 0, 0, i);
                if (v == gv) {
                        printf("m4=%lx\n", i);
                        break;
                }
        }
        return 0;
}
  • 这是我们部署的 newPuzzle ，我们是直接改的题目中的 newPuzzle ，把 0x403 字节位置的 0x14（等于） 改为了 0x10（小于），这样随便找一个 seed 满足条件即可，当然也可以直接写一个攻击合约调用题目的 solve 也行（填充到指定字节，然后碰撞也可以）
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
  • 因为我们是直接改的题目中的 puzzle ，所以我们需要绕过 loose ，这里是 1ph0n 师傅提供的
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# loose() writeup

for a in range(1, 16):
    for b in range(1, 16):
        for c in range(1, 16):
            if a + b + c == 7 and a ^ b ^ c == 3:
                print(a, b, c) 
# output:
# 2 2 3 
# 2 3 2 
# 3 2 2
# match them to:
# a b c
# d e f
# g h i

a = 0x2000000000000000000000000000000000000000000000000000000000000000
b = 0x2000000000000000000000000000000000000000000000000000000000000000
c = 0x3000000000000000000000000000000000000000000000000000000000000000
d = 0x2000000000000000000000000000000000000000000000000000000000000000
e = 0x3000000000000000000000000000000000000000000000000000000000000000
f = 0x2000000000000000000000000000000000000000000000000000000000000000
g = 0x3000000000000000000000000000000000000000000000000000000000000000
h = 0x2000000000000000000000000000000000000000000000000000000000000000
i = 0x2000000000000000000000000000000000000000000000000000000000000000

# test: 
t1 = (a^b^c)+(d^e^f)+(g^h^i)
t2 = (a+d+g)^(b+e+h)^(c+f+i)

print(hex(t1))
print(hex(t2))
print((t1 + t2) % 0x10000000000000000000000000000000000000000000000000000000000000000 < 0xaabbccdd )
print(t1 > 0x8261e26b90505061031256e5afb60721cb)
print(0xf35b6080614321368282376084810151606401816080016143855161051756 >= (t1*t2) % 0x10000000000000000000000000000000000000000000000000000000000000000)
print(t1 - t2 >= 0x65e670d9bd540cea22fdab97e36840e2)
  • 然后就可以找一个 seed 符合我们改过之后的条件（条件改为了小于）就行啦
  • 顺便这里插嘴一句，由于我拥有强大的暗黑之力，🤓我找到了所有队伍做这道题目的地址，吼吼哈嘿，快夸我🐶（只标注了一些队伍）
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0x05751749a85D2149C76BC035af4826208Dd9118b  AAA
0x0F4f4c78cE13b69239A676fE65C72bf974b3fCa3
0x6D2BD358b7788E27C271eFE4d40b7A82F6E6c783
0x6BA08F6731509c63026Bc431023F35e49fF4f8fC
0x8FA83b9e152120F11c13d678c7879C3eC2f14DB7
0xA54c712054dEDD5587B5c39F1DF0Aab8BD782B63
0xC66005B6295f396b4289219486aC62aCFC0Ff4d4
0xbc16Cf28E491Ae88C36638642e7673Af4570A1e4
0x3E8C7FEc0166bd59Ca1FcD0672bF8E3915c56Bdf
0x4E8d7823046b3CE7eBd99F7282B18c3DaDD879ED
0x44cE28ad66C38dD334C518EA6f07248a1B3EA08C
0x6B217cbB59e6d0edb104D815db74943670843463
0x60a05Ada89061221e1602957097248FC8c373fE4  0ops
0xDA93b922A68d42Edd7E377f4E06fD574A70B6123
0x726a02a9176e8E458CE8bD75305B92e72a8b4eA2  Sauercloud
0x00E457CD4A251d58D9Af7F674857c78a6Dc73e8F
0x11674249BdC62045947e3022808B6dfD009573B6

预期解法

EasyDefi

  • 这是 sissel 师傅做出来的题目，看看师傅会不会放 wp 啦，我就先不写啦
  • 找到一份 syang 师傅的 👉 wp 👈

billboard

  • 未解出，虽然赛后咨询了出题人正确解法，知道了大概原因，但还是准备等待官方 WP 后再更新
  • 👉 官方WP 👈 来啦 🤓
  • 下面是本地复现成功的结果，太感谢 iczc 师傅了

---------------- The End ----------------
谢谢大爷～

Author：pikachu
Link：https://hitcxy.com/2021/rw2021/
Contact：hitcxy.cn@gmail.com
本文基于 知识共享署名-相同方式共享 4.0 国际许可协议发布
转载请注明出处，谢谢！