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d31cd1ce1ff977f57a20305b59d1496d174b4662506fb3ca62...

Tạo vào: 31 tháng 10, 2026

Trả lời bằng GPT-6 Thinking bởi Chat01

Câu hỏi

image.png

d31cd1ce1ff977f57a20305b59d1496d174b4662506fb3ca6266be8a7898868344c0e70192576030edcbfa16c6e50907b3a2611d1ace65974f30cb23613ec157cbf37ad5d70cc51dbad6c3985a6bcb30f76b97c0bfee16680bf09a151e5ae7f9a5977a2d34ac7a62174e1122b2f565a5624f44b8ab73fdabe154a2f93ccbef78c1742f8f24e42e80e0fbe5b4cca7280ebb1088b163f11b73fe203ecab5c95722bc443d511fe497265f34f411ce37478387cde48f6f6bea10a36efd8a4476118b2ea2d9ceed6c4b5866c0a131e03ddb3f15帮我解密

Trả lời

我先按你截图里已经拿到的线索继续推：先验证那 16 字节 keystream 是否能正确还原开头，再检查密文长度、重复结构和可能的帧偏移/状态模式。

Python

python
ct_hex = """d31cd1ce1ff977f57a20305b59d1496d174b4662506fb3ca6266be8a7898868344c0e70192576030edcbfa16c6e50907b3a2611d1ace65974f30cb23613ec157cbf37ad5d70cc51dbad6c3985a6bcb30f76b97c0bfee16680bf09a151e5ae7f9a5977a2d34ac7a62174e1122b2f565a5624f44b8ab73fdabe154a2f93ccbef78c1742f8f24e42e80e0fbe5b4cca7280ebb1088b163f11b73fe203ecab5c95722bc443d511fe497265f34f411ce37478387cde48f6f6bea10a36efd8a4476118b2ea2d9ceed6c4b5866c0a131e03ddb3f15"""
ct = bytes.fromhex(ct_hex)
len(ct), ct[:16]


(209, b'\xd3\x1c\xd1\xce\x1f\xf9w\xf5z 0[Y\xd1Im')

Python

python
ks0 = bytes.fromhex("b2f86c6ef756c312e0a4d5f4e03ef5e1")
pt0 = bytes(a^b for a,b in zip(ct[:16], ks0))
pt0, pt0.decode('utf-8', errors='replace')


(b'a\xe4\xbd\xa0\xe8\xaf\xb4\xe7\x9a\x84\xe5\xaf\xb9\xef\xbc\x8c', 'a你说的对，')

Python

python
s="你说的对，".encode()
len(s), s.hex()


(15, 'e4bda0e8afb4e79a84e5afb9efbc8c')

Suy nghĩ

Decrypted ciphertext and calculated UTF-8 plaintext alignment

Python

python
candidate = "a你说的对，但是《原神》是由米哈游自主研发的一款全新开放世界冒险游戏。游戏发生在一个被称作「提瓦特」的幻想世界，在这里，被神选中的人将被授予「神之眼」，导引元素之力。你将扮演一位名为「旅行者」的神秘角色，在自由的旅行中邂逅性格各异、能力独特的同伴们，和他们一起击败强敌，找回失散的亲人——同时，逐步发掘「原神」的真相。"
cb = candidate.encode()
len(cb), cb[:80]


(475,
 b'a\xe4\xbd\xa0\xe8\xaf\xb4\xe7\x9a\x84\xe5\xaf\xb9\xef\xbc\x8c\xe4\xbd\x86\xe6\x98\xaf\xe3\x80\x8a\xe5\x8e\x9f\xe7\xa5\x9e\xe3\x80\x8b\xe6\x98\xaf\xe7\x94\xb1\xe7\xb1\xb3\xe5\x93\x88\xe6\xb8\xb8\xe8\x87\xaa\xe4\xb8\xbb\xe7\xa0\x94\xe5\x8f\x91\xe7\x9a\x84\xe4\xb8\x80\xe6\xac\xbe\xe5\x85\xa8\xe6\x96\xb0\xe5\xbc\x80\xe6')

Python

python
guess = cb[:len(ct)]
ks_guess = bytes(a^b for a,b in zip(ct,guess))
ks_guess[:64].hex(), len(ks_guess)


('b2f86c6ef756c312e0a4d5f4e03ef5e1f3f6c084c8c0504ae88330159f3d1860c44b01993db0f4810a7a49f3556defbf0b4ae6b7fe76de70efa42eacf0d95bd3',
 209)

Python

python
def berlekamp_massey(bits):
    n = len(bits)
    c = [0]*n
    b = [0]*n
    c[0]=b[0]=1
    L=0;m=-1
    for N in range(n):
        d=bits[N]
        for i in range(1,L+1):
            d ^= c[i] & bits[N-i]
        if d:
            t=c[:]
            for j in range(N-m,n):
                c[j] ^= b[j-(N-m)]
            if 2*L <= N:
                L = N+1-L
                m=N
                b=t
    return L,c[:L+1]
for bitorder in ('msb','lsb'):
    bits=[]
    for byte in ks0:
        rng=range(7,-1,-1) if bitorder=='msb' else range(8)
        bits.extend((byte>>i)&1 for i in rng)
    print(bitorder, berlekamp_massey(bits)[0])

Python

python
def bits_from_bytes(bs, bitorder):
    bits=[]
    for byte in bs:
        if bitorder=='msb':
            bits += [(byte>>i)&1 for i in range(7,-1,-1)]
        else:
            bits += [(byte>>i)&1 for i in range(8)]
    return bits
def predict_with_conn(init_bits, coeffs, total):
    # coeffs c[0]=1; s[n] = sum c[i] s[n-i]
    seq=init_bits[:]
    L=len(coeffs)-1
    while len(seq)<total:
        n=len(seq)
        d=0
        for i in range(1,L+1):
            d ^= coeffs[i]&seq[n-i]
        seq.append(d)
    return seq
def bytes_from_bits(bits, bitorder):
    out=[]
    for j in range(0,len(bits)//8*8,8):
        chunk=bits[j:j+8]
        if bitorder=='msb':
            v=0
            for b in chunk: v=(v<<1)|b
        else:
            v=sum(b<<i for i,b in enumerate(chunk))
        out.append(v)
    return bytes(out)

for bo in ['msb','lsb']:
    bits=bits_from_bytes(ks0,bo)
    L,c=berlekamp_massey(bits)
    pred=bytes_from_bits(predict_with_conn(bits,c,len(ct)*8),bo)
    # compare with guessed ks after first16
    matches=sum(a==b for a,b in zip(pred[16:],ks_guess[16:]))
    bitmatches=sum(x==y for x,y in zip(bits_from_bytes(pred[16:],bo), bits_from_bytes(ks_guess[16:],bo)))
    print(bo,L,c, pred[:32].hex(), 'byte matches',matches,'/',len(ct)-16,'bitmatches',bitmatches)

Python

python
for bo in ['msb','lsb']:
    bits=bits_from_bytes(ks_guess,bo)
    # BM prefix lengths
    for nbytes in [16,32,64,128,209]:
        L,_=berlekamp_massey(bits[:nbytes*8])
        print(bo,nbytes,L)

Suy nghĩ

Computed and evaluated keystream patterns, LFSR predictions, and linear complexity

Suy nghĩ

Assessing the ciphertext

I’m treating this as likely encrypted data rather than readable text. Without the cipher mode, key, nonce, or FPGA configuration, direct recovery seems unlikely, though searching for the exact ciphertext might identify a public challenge or known solution.

Assessed the ciphertext

Searching "a你说的对" "Artix-7" CTF

Python

python
repks = (ks0 * ((len(ct)+15)//16))[:len(ct)]
pt_rep = bytes(a^b for a,b in zip(ct,repks))
pt_rep[:200], pt_rep.decode('utf-8', errors='replace')


(b'a\xe4\xbd\xa0\xe8\xaf\xb4\xe7\x9a\x84\xe5\xaf\xb9\xef\xbc\x8c\xa5\xb3*\x0c\xa79p\xd8\x82\xc2k~\x98\xa6sb\xf68\x8boe\x01\xa3"\ro/\xe2&\xdb\xfc\xe6\x01Z\rs\xed\x98\xa6\x85\xaf\x94\x1e\xd7\x81\x004\xb6y\x0b\x16\xbb Z\x06\x0fZr\x16l\xbaU>\xd1E\x93\xfb\xaeH\xb8\xd5z\xebTO\xe1\xfed\x12\x18\x17o\x16C\xc3\xfa\xb9p\xf7\xea\xc4\xd6R\xcb\x90D\xd0\xb7(\xd6\\%>\xb9\x01\xf0w\r\xdc\xf5\x1a\x99s\x8cC\xe1\xd3\xb2\xed\x92\x00_0@,\x99\xdd\xef\t\xe8\xe4\xdf\x94\xa7\xd8a\x1e\x84\xeb>U\xf7\xa2\xc3\x0e\xbcQ?\xe8\xb2T4\xbf\x90!\xe5.\t\xb2b55\x88\xe1\x98=)\x02C\xca(~\xa4H\xe4j\x9cZ\xb5\xa0\x1a:\x88J',
 'a你说的对，��*\x0c�9p\u0602�k~��sb�8�oe\x01�"\ro/�&���\x01Z\rs혦���\x1eׁ\x004�y\x0b\x16� Z\x06\x0fZr\x16l�U>�E���H��z�TO��d\x12\x18\x17o\x16C���p����RːDз(�\\%>�\x01�w\r��\x1a�s�C�Ӳ�\x00_0@,���\t��ߔ��a\x1e��>U���\x0e�Q?�T4��!�.\t�b55��=)\x02C�(~�H�j�Z��\x1a:�J�dt�\x00\x03.ާ')

Python

python
# Show first ~209 bytes of candidate safely trim to utf8 boundary
prefix=cb[:209]
for k in range(0,4):
    try:
        print(209-k, prefix[:209-k].decode())
        break
    except UnicodeDecodeError: pass

Suy nghĩ

Clarifying recoverability

I’m testing whether the known prefix reveals more than the initial keystream block. It doesn’t uniquely determine the remaining bytes: without the encryption state, key, nonce, or bitstream, many plaintexts remain mathematically possible.

Clarified recoverability

Analyzed repeated-keystream decryption and FPGA cipher structure

Trả lời

已验证：截图里的 16 字节 keystream 是对的，C[0:16] XOR keystream = "a你说的对，"。我也试了“把这 16 字节循环当密钥流”的最直接情况，第 17 字节起就立刻变成无效 UTF-8，说明不是简单循环 XOR。

Python

python
# derive guessed keystream for first 208 candidate bytes
kg = ks_guess[:208]
for p in [16,24,32,40,48,64,80,96,128]:
    matches=sum(a==b for a,b in zip(kg[p:],kg[:-p]))
    print(p,matches,len(kg)-p)

Python

python
from collections import Counter
import math
cnt=Counter(ct)
H=-sum((n/len(ct))*math.log2(n/len(ct)) for n in cnt.values())
H


6.908823950312551

Python

python
parts=[
"a你说的对，但是《原神》是由米哈游自主研发的一款全新开放世界冒险游戏。",
"游戏发生在一个被称作「提瓦特」的幻想世界，",
"在这里，被神选中的人将被授予「神之眼」，导引元素之力。",
]
for p in parts:
    print(len(p.encode()),p)
print(sum(len(p.encode()) for p in parts))

Suy nghĩ

Tested plaintext periodicity, analyzed ciphertext structure, and estimated FPGA recovery

Suy nghĩ

Searched 12 websites

Python

python
known_prefix = "a你说的对，但是《原神》是由米哈游自主研发的一款全新开放世界冒险游戏。游戏发生在一个被称作「提瓦特」的幻想世界，".encode()
len(known_prefix), known_prefix.decode()


(166, 'a你说的对，但是《原神》是由米哈游自主研发的一款全新开放世界冒险游戏。游戏发生在一个被称作「提瓦特」的幻想世界，')

Python

python
kp = bytes(a^b for a,b in zip(ct[:len(known_prefix)], known_prefix))
len(kp), kp[:32].hex(), kp[-16:].hex()


(166,
 'b2f86c6ef756c312e0a4d5f4e03ef5e1f3f6c084c8c0504ae88330159f3d1860',
 '9f96479bd849062defb45bd1b1bea368')

Python

python
import struct, math, itertools, collections
for endian in ['<','>']:
    words=[struct.unpack(endian+'Q',kp[i:i+8])[0] for i in range(0,160,8)]
    print(endian, [hex(x) for x in words[:5]])

Python

python
MASK=(1<<64)-1
def xs64(x):
    x ^= (x<<13)&MASK
    x ^= x>>7
    x ^= (x<<17)&MASK
    return x&MASK
def xs64_alt(x):
    # xorshift64* state transition Marsaglia/Vigna often 12,25,27
    x ^= x>>12
    x ^= (x<<25)&MASK
    x ^= x>>27
    return x&MASK
for endian in ['<','>']:
    words=[struct.unpack(endian+'Q',kp[i:i+8])[0] for i in range(0,160,8)]
    print('end',endian,'xs64 matches',sum(xs64(words[i])==words[i+1] for i in range(len(words)-1)))
    print('alt matches',sum(xs64_alt(words[i])==words[i+1] for i in range(len(words)-1)))

Suy nghĩ

Analyzed PRNG output relations and tested xorshift64 recurrence matches

Trả lời

我发现一个很关键的长度吻合：如果明文继续用常见的“原神”段子，那么前两段到“幻想世界，”正好是 166 字节；再接 flag{36位UUID}\n 正好总计 209 字节。这很像题目刻意设计的明文结构，我接下来按这个结构继续反推密钥流/状态。

Suy nghĩ

Continued cryptanalysis and bitstream analysis

Python

python
from math import gcd
def inv_mod_2n(a,n=32):
    mod=1<<n
    if a%2==0: return None
    return pow(a,-1,mod)
def test_lcg_words(data, wordsize=4):
    mod=1<<(8*wordsize)
    results=[]
    for off in range(wordsize):
        for endian in ['little','big']:
            words=[int.from_bytes(data[i:i+wordsize],endian) for i in range(off,len(data)-wordsize+1,wordsize)]
            if len(words)<4: continue
            # solve from first possible odd difference
            found=None
            for j in range(len(words)-2):
                d=(words[j+1]-words[j])%mod
                inv=pow(d,-1,mod) if gcd(d,mod)==1 else None
                if inv is not None:
                    a=((words[j+2]-words[j+1])*inv)%mod
                    c=(words[j+1]-a*words[j])%mod
                    matches=sum((a*words[i]+c)%mod==words[i+1] for i in range(len(words)-1))
                    found=(matches,len(words)-1,a,c,off,endian)
                    break
            if found: results.append(found)
    return sorted(results, reverse=True)
test_lcg_words(kp,4)[:10], test_lcg_words(kp,8)[:10]


([(2, 40, 3638071808, 1997681888, 1, 'big'),
  (2, 40, 2614331326, 957874691, 1, 'little'),
  (2, 40, 1504021845, 920697309, 0, 'little'),
  (2, 40, 1469451454, 1406462467, 2, 'big'),
  (2, 40, 821338831, 1057066389, 0, 'big'),
  (2, 40, 521827134, 593547419, 2, 'little'),
  (2, 39, 2225652687, 497365653, 3, 'little'),
  (2, 39, 1211730608, 3108740176, 3, 'big')],
 [(2, 19, 18109414980381024324, 12596848029656013829, 3, 'little'),
  (2, 19, 17217577938201533928, 5717537764471600019, 1, 'little'),
  (2, 19, 16754190117231553496, 1665551116145147032, 6, 'little'),
  (2, 19, 16546384507289278800, 13881936821850646320, 3, 'big'),
  (2, 19, 13605475366938356968, 9298066791957236115, 2, 'big'),
  (2, 19, 12628081763747188039, 16130533220511412707, 0, 'big'),
  (2, 19, 11205386127536930628, 3083573573766451461, 4, 'big'),
  (2, 19, 9274996692584220119, 3547359163905679059, 0, 'little'),
  (2, 19, 8668348609095215528, 5798686326067666376, 4, 'little'),
  (2, 19, 7745082455754030994, 2434365387225637775, 5, 'big')])

Python

python
MASK32=(1<<32)-1
def xs128_next(x,y,z,w):
    t=(x ^ ((x<<11)&MASK32))&MASK32
    return (w ^ (w>>19) ^ t ^ (t>>8)) & MASK32
for off in range(4):
    for endian in ['little','big']:
        words=[int.from_bytes(kp[i:i+4],endian) for i in range(off,len(kp)-3,4)]
        if len(words)>8:
            matches=0
            for i in range(len(words)-4):
                if xs128_next(*words[i:i+4])==words[i+4]:
                    matches+=1
            if matches:
                print(off,endian,matches,len(words)-4)

Python

python
import importlib.util, sys, os, math, statistics
print(importlib.util.find_spec("z3"))

Python

python
MASK64=(1<<64)-1
def unxorshift_right(y, shift, bits=64):
    x=0
    # reconstruct from MSB to LSB
    for i in range(bits-1,-1,-1):
        bit=(y>>i)&1
        if i+shift<bits:
            bit ^= (x>>(i+shift))&1
        x |= bit<<i
    return x
def splitmix_unmix(z):
    # splitmix64: z=(x^(x>>30))*C1; z=(z^(z>>27))*C2; z ^= z>>31
    z = unxorshift_right(z,31)
    z = (z * pow(0x94D049BB133111EB,-1,1<<64)) & MASK64
    z = unxorshift_right(z,27)
    z = (z * pow(0xBF58476D1CE4E5B9,-1,1<<64)) & MASK64
    z = unxorshift_right(z,30)
    return z
for off in range(8):
    for endian in ['little','big']:
        words=[int.from_bytes(kp[i:i+8],endian) for i in range(off,len(kp)-7,8)]
        states=[splitmix_unmix(w) for w in words]
        diffs=[(states[i+1]-states[i])&MASK64 for i in range(len(states)-1)]
        c=collections.Counter(diffs)
        val,count=c.most_common(1)[0]
        if count>=2:
            print('off',off,endian,'most diff count',count,'of',len(diffs),hex(val))

Python

python
tail = ct[166:]
len(tail), tail.hex()


(43,
 '97265f34f411ce37478387cde48f6f6bea10a36efd8a4476118b2ea2d9ceed6c4b5866c0a131e03ddb3f15')

Python

python
template = b"flag{" + b"?"*36 + b"}\n"
# set hyphens positions in UUID 8,13,18,23
temp = bytearray(template)
for p in [8,13,18,23]:
    temp[5+p]=ord('-')
known_positions={}
for i,b in enumerate(temp):
    if b!=ord('?'):
        known_positions[i]=b
[(i, chr(b), hex(tail[i]^b)) for i,b in known_positions.items()]


[(0, 'f', '0xf1'),
 (1, 'l', '0x4a'),
 (2, 'a', '0x3e'),
 (3, 'g', '0x53'),
 (4, '{', '0x8f'),
 (13, '-', '0xa2'),
 (18, '-', '0x8e'),
 (23, '-', '0x5b'),
 (28, '-', '0xf4'),
 (41, '}', '0x42'),
 (42, '\n', '0x1f')]

Python

python
periods=[]
for p in range(1,166):
    if kp[p:]==kp[:-p]:
        periods.append(p)
periods[:10], len(periods)


([], 0)

Python

python
blocks=[kp[i:i+16] for i in range(0,len(kp),16)]
for bi in range(len(blocks)-1):
    a,b=blocks[bi],blocks[bi+1]
    best=[]
    for sh in range(-15,16):
        # compare a shifted with b
        pairs=[]
        for i in range(16):
            j=i+sh
            if 0<=j<16:
                pairs.append(a[i]==b[j])
        if pairs:
            m=sum(pairs)
            if m>=2:
                best.append((m,len(pairs),sh))
    if best:
        print(bi,sorted(best,reverse=True)[:3])

Python

python
full_blocks=[kp[i:i+16] for i in range(0,(len(kp)//16)*16,16)]
for bi in range(len(full_blocks)-1):
    a,b=full_blocks[bi],full_blocks[bi+1]
    best=[]
    for sh in range(-15,16):
        pairs=[]
        for i in range(16):
            j=i+sh
            if 0<=j<16:
                pairs.append(a[i]==b[j])
        if pairs:
            m=sum(pairs)
            if m>=2:
                best.append((m,len(pairs),sh))
    if best:
        print(bi,sorted(best,reverse=True)[:5])

Python

python
def test_lcg_block(data, block=16):
    mod=1<<(8*block)
    res=[]
    for endian in ['little','big']:
        words=[int.from_bytes(data[i:i+block],endian) for i in range(0,(len(data)//block)*block,block)]
        for j in range(len(words)-2):
            d=(words[j+1]-words[j])%mod
            if math.gcd(d,mod)==1:
                a=((words[j+2]-words[j+1])*pow(d,-1,mod))%mod
                c=(words[j+1]-a*words[j])%mod
                matches=sum((a*words[i]+c)%mod==words[i+1] for i in range(len(words)-1))
                res.append((matches,len(words)-1,endian,a,c))
                break
    return res
test_lcg_block(kp,16)


[(2,
  9,
  'little',
  91288842069987980991399754873454776977,
  122842234262662000880282836157372175905),
 (2,
  9,
  'big',
  235489363464735757863418223273713146657,
  227708887199194692290987717446738881375)]

Suy nghĩ

Tested PRNG models, keystream structure, alignments, and recurrence patterns

Worked for 2m 48s

Trả lời

可以继续推，但仅凭你现在给的 209 字节密文 + 前 16 字节已知密钥流，还不能唯一解出 UUID。我已经把能确定的部分算出来了。

已验证：

text
cipher[0:16]
= d31cd1ce1ff977f57a20305b59d1496d

keystream[0:16]
= b2f86c6ef756c312e0a4d5f4e03ef5e1

XOR 后：
= a你说的对，

而且把这 16 字节 keystream 简单循环使用是错误的，第 17 字节开始就无法形成正常 UTF-8，所以不是普通 repeating-XOR。

有一个非常强的线索：如果后面接常见的这段文本：

text
a你说的对，但是《原神》是由米哈游自主研发的一款全新开放世界冒险游戏。游戏发生在一个被称作「提瓦特」的幻想世界，

它的 UTF-8 长度恰好是 166 字节。而：

text
flag{xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx}\n

恰好是 43 字节。

所以：

text
166 + 43 = 209

和你的密文长度完全一致。

因此我认为明文结构很可能就是：

text
a你说的对，但是《原神》是由米哈游自主研发的一款全新开放世界冒险游戏。游戏发生在一个被称作「提瓦特」的幻想世界，flag{????????-????-????-????-????????????}

也就是说 flag 从 offset 0xA6（166）开始。对应的最后 43 字节密文是：

text
97265f34f411ce37478387cde48f6f6b
ea10a36efd8a4476118b2ea2d9ceed6c
4b5866c0a131e03ddb3f15

假设 flag{UUID}\n 结构成立，我们还能得到尾部一些确定的 keystream，例如开头：

text
97 XOR 'f' = f1
26 XOR 'l' = 4a
5f XOR 'a' = 3e
34 XOR 'g' = 53
f4 XOR '{' = 8f

所以 offset 166 开始的密钥流为：

text
f1 4a 3e 53 8f ?? ?? ?? ?? ?? ?? ?? ?? a2 ...

UUID 的四个 -、最后的 } 和换行也能提供若干已知字节。

问题在于：剩余密钥流不是周期性的，而且从目前数据看也不像简单 LFSR/LCG/xorshift 的直接输出。 如果这是截图中那个 Artix-7 bitstream 里的自定义加密状态机，那么必须恢复它的状态更新逻辑；只知道前 16 字节 keystream，在密码学上不足以推出后面的 193 字节。

如果你把原始的 .bit / .bin FPGA bitstream 发上来，我可以直接沿着现在这个 166-byte known plaintext + flag{UUID} 的结构继续逆，目标就很明确：只需要生成 offset 166–208 的 keystream，就能把 UUID flag 解出来。

Câu hỏi

Cryptarithm.bit

继续解出flag

Câu hỏi

d43a680b895eb60b699fc0186d3761793e693ff0da5597bbc0eddf142185a5fd (1)

继续解出flag

Trả lời

这次有 FPGA bitstream 和另一份附件了。我会先确认它们的格式、是否包含加密逻辑或可读数据，再尝试从密钥流生成电路反推出完整明文和 flag。

Suy nghĩ

Inspected FPGA bitstream, scripts, metadata, frames, and LUTs

Suy nghĩ

Searched 4 websites

Suy nghĩ

Analyzing the cipher

I’m testing whether the ciphertext uses a simple stream generator, such as an LFSR or homemade PRNG. I’ll verify the known plaintext, derive the keystream, then examine patterns and possible recurrence structures.

Analyzed the cipher

Tested internet and FPGA tool availability

Trả lời

两份新附件已确认是同一个 bitstream（SHA-256 完全一致）。文件头显示目标器件是 xc7a35tcpg236，设计名为 crp_v3_top。现在重点是分清密钥流到底来自可重建的寄存器状态，还是需要完整恢复 FPGA 的 LUT 和布线。

Suy nghĩ

Inspected bitstream structure and analyzed keystream for LFSR patterns

Searching the web

Suy nghĩ

Searched 8 websites

Suy nghĩ

Inspected FPGA reverse-engineering resources

Suy nghĩ

Searched 7 websites

Suy nghĩ

Mapping frame layout

I’m inferring the device’s configuration geometry from repeated frame gaps and cluster sizes. The key question is whether the data represents two clock regions or truncated rows, so I’m comparing counts against the device’s documented frame organization.

Mapped frame layout

Suy nghĩ

Inferring circuit behavior

I’m separating configuration artifacts from meaningful logic. The sparse, repeated regions suggest duplicated top-and-bottom implementations, while simple lookup values may indicate a compact nonlinear pseudorandom generator rather than stored data.

Testing generator families

I’m comparing the recovered stream against common pseudorandom designs, including word-based generators and state-transition variants. If those fail, I’ll need to infer the custom feedback structure rather than assume a standard algorithm.

Inferred circuit behavior and tested generator families

Calculated LUT bit mappings and analyzed cryptarithm bitstream configuration

Suy nghĩ

Correcting frame assumptions

I’m reconsidering the frame geometry because the earlier two-row model doesn’t fit the total count. The data likely combines multiple configuration regions and memory-related frames, so alignment must be fixed before decoding.

Corrected frame assumptions

Searched 1 website

Suy nghĩ

Calculated exact FPGA frame row totals

Trả lời

我已经把 bitstream 的配置帧拆出来，并定位到两片明显有逻辑活动的区域。之前“209 字节 = 166 字节原神文本 + 43 字节 flag”的长度吻合仍然只是候选结构，不能把它当作已证明的明文；我正在检查帧地址是否存在偏移，避免用错误的电路数据推一个假 flag。

Suy nghĩ

Fixing frame alignment

I’ve identified the likely source of the 12-frame discrepancy: configuration regions appear separated by two padding frames each. I’m updating the row boundaries accordingly before interpreting logic, routing, or any recovered state.

Interpreting circuit placement

I’m treating the separated logic clusters as one design spanning a clock-region boundary, not as duplicated implementations. The corrected alignment supports this layout and should guide the next decoding step.

Assessing next steps

I’m considering whether the recovered stream is linear, nonlinear, or based on an incorrect plaintext assumption. The circuit’s sparse structure suggests a compact custom generator, so I’m prioritizing functional reconstruction over standard cipher matching.

Fixed frame alignment and assessed circuit placement next steps

Searched 1 website

Suy nghĩ

Used container tool

Planning circuit recovery

I’m prioritizing direct circuit reconstruction from the sparse configuration data, especially lookup tables and routing. The goal is to recover the generator’s state transitions rather than rely on generic cipher guesses.

Recovering initial state

I’m examining flip-flop initialization bits as a possible embedded seed. If those values align with the known keystream, they could reveal the generator’s state directly; otherwise, I’ll continue with routing and transition analysis.

Exploring seed recovery

I’m checking whether initialization bits encode the generator’s starting state. Matching them against the known keystream could expose the design quickly; otherwise, I’ll rely on reconstructing its logic and routing.

Planned circuit recovery and explored recovering the initial state from seeds

Searching the web

Suy nghĩ

Explored FPGA PRNG seed reconstruction and set up SMT validation

Trả lời

关键的 12 帧偏差已经找到原因：Xilinx 配置数据在各配置区域之间有填充帧，不能直接把所有 101-word 帧连续编号当成真实 FAR 地址。我也用已知密钥流测试了几种常见 128 位 PRNG，目前没有匹配；还不能据此认定后续明文内容或 flag。

Suy nghĩ

Tested PRNG key recovery and simulated xorshift128p decryption

Suy nghĩ

Correcting LUT mapping

I’m checking whether the LUT extraction used the correct slice modes and initialization-bit positions. A mapping error there could make valid logic appear random, so I’m revisiting column-specific configuration rules before interpreting the generator.

Corrected LUT mapping

Searched 1 website

Suy nghĩ

Recomputed x0 LUT mappings with mode M

Suy nghĩ

Rechecking LUT decoding

I’m verifying slice-specific LUT bit mappings because an incorrect mode assignment could make structured cryptographic logic appear random. Correcting that mapping is essential before drawing conclusions about the circuit or its embedded state.

Identifying the cipher

I’m comparing the circuit’s size and structure with compact stream-cipher designs. The combination of many state-holding elements and nonlinear lookup logic may indicate a custom or lightweight generator rather than a standard software PRNG.

Rechecked LUT decoding and identified the cipher

Searched 1 website

Suy nghĩ

Compared FF bit patterns

Suy nghĩ

Searched 1 website

Summarizing progress

I’m treating the data as a custom stream-cipher circuit. The recovered prefix supports known-plaintext analysis, while corrected frame alignment reveals padding between configuration regions. I’m not yet able to derive the hidden state or flag reliably.

Summarized progress

Suy nghĩ

Analyzed FPGA encryption, fixed key derivation, and tested database access

Trả lời

另外确认了一点：bitstream 的 Block RAM 配置区没有发现非零初始化数据，不能直接从 BRAM 里读出一个明文密钥或 flag。现在正在把已知的 16 字节输出与更完整的轻量 PRNG 模型对照，检查是否能跳过完整布线还原。

Suy nghĩ

Created an SMT PRNG state recovery script

Trả lời

更完整的模型测试仍未找到能稳定生成已知输出的 PRNG。当前的瓶颈不是密文解析，而是缺少可验证的 FPGA 状态转移关系；仅凭已知的 16 字节无法反推出后续随机密钥流。我会把已经验证的帧结构和密钥流约束整理成可复现的结果，避免把未验证的 UUID 当成 flag。

Suy nghĩ

Analyzed bitstream frames and created a reproducible cryptarithm script

Worked for 6m 8s

Trả lời

我继续分析了你上传的 Cryptarithm.bit，但目前仍未恢复出可以验证的完整 flag。不能凭猜测给你一个 UUID。

已确认的结果

  • 两份附件内容完全相同，SHA-256 一致。
  • FPGA 型号：xc7a35tcpg236
  • 设计名称：crp_v3_top
  • 密文长度：209 字节。
  • 已验证的前 16 字节 keystream：

另外，修正了之前的配置帧偏移问题：

已确认 5420 帧中有 5408 个实际配置帧和 12 个填充帧。

Bitstream 中能定位到活动的 LUT 逻辑区域，但尚未完整恢复寄存器之间的布线和状态更新规则。Block RAM 配置区域也没有发现非零初始化数据。

当前 flag 线索

如果 166 字节的《原神》文本猜测成立，后面会是：

对应的密文起始位置是 0xA6。但这仍需通过恢复出的 keystream 验证。

分析文件

我保存了可复现的解析脚本和结果：

解析脚本

分析结果

剩下的关键步骤是完整还原 FPGA 的 LUT 输入连接和状态转移，再重放加密器，生成第 16–208 字节的 keystream。 当前还没有完成这一步，因此没有可靠的最终 flag。

Câu hỏi

能解出flag吗这个？nc 47.94.231.96 20560这个是容器地址

Trả lời

有了这个 nc 容器地址，就可以尝试从服务本身入手，不一定要完整逆向 FPGA bitstream。我先测试连接、查看交互协议，再结合现有密文判断能否直接恢复 flag。

Suy nghĩ

Inspected CTF files and diagnosed network connectivity and FPGA access

Suy nghĩ

Searched 1 website

Suy nghĩ

Tested proxy settings and network connectivity

Assessing the challenge

I’m checking whether I can solve this offline, but the remote service appears necessary. I can still explain how to connect locally, capture its prompt and responses, then analyze the encryption once that output is available.

Assessed the challenge

Searching the web

Trả lời

连接测试发现了环境限制：不仅题目端口 47.94.231.96:20560 返回 Connection refused，连其他公网 TCP 测试地址也同样被拒绝，因此暂时不能判断是题目容器故障还是当前执行环境禁止外连。我会再从附件里检查是否存在不依赖远程交互的解题入口。

Suy nghĩ

Continuing offline analysis

I’m treating the 16-byte match as a clue, not proof. The remaining ciphertext doesn’t reveal a short recurrence, so I need the service’s prompts and responses to identify the intended oracle or exploit path.

Continued offline analysis

Inspected bitstream strings and ciphertext

Worked for 1m

Trả lời

Text version

有机会解出 flag，但目前还差容器的交互信息。

我已经尝试连接：

返回：

同时，当前环境对其他公网 TCP 地址的连接也被拒绝，因此更可能是执行环境的网络限制，不能据此判断你的 CTF 容器已经关闭。

目前已经有密文、FPGA bitstream，以及前 16 字节的已知密钥流。最值得尝试的是从容器寻找加密接口：如果它允许我们提交明文，并且重复使用同一密钥流，就可能通过选择明文攻击直接恢复完整的 209 字节明文，不必完全逆向 FPGA。

请你在自己的终端运行：

把连接后出现的欢迎信息、菜单和题目提示完整复制给我。如果有交互选项，也发一下。

有了这些信息，我就能判断能否构造自动化解题脚本，进一步尝试拿到真正的 flag{...}，而不是猜测 UUID。

Câu hỏi

访问不了

Trả lời

那先不依赖 nc 了。既然你这边也访问不了，可能是题目实例已停止、端口未开放或服务异常。我再检查一下已上传的 bitstream 和分析文件，看看离线还能不能推进到 flag。

Suy nghĩ

Investigating challenge options

I’m assessing whether the FPGA bitstream can be analyzed offline, whether the challenge name has public documentation, and whether the available script or embedded logic reveals the cipher. I’ll focus on legitimate CTF-oriented approaches.

Investigated challenge options

Inspected uploaded CTF artifacts offline

Suy nghĩ

Searched 5 websites

Suy nghĩ

Checked FPGA tools and databases, then ran Berlekamp–Massey analysis

Worked for 22s

Trả lời

那就先不依赖容器了。
我检查了你上传的 Cryptarithm.bit 和之前的分析结果。目前可以确认：
  • 209 字节密文完整，没有明显缺失。
  • 已知前 16 字节密钥流，但不足以解出后续全部内容。
  • 已测试简单循环 XOR 和线性反馈模型，未找到有效的密钥流生成规则。
  • FPGA bitstream 还需要进一步还原 LUT、寄存器和布线逻辑。可以使用 Project X-Ray 的 bit2fasm 工具辅助分析。
现在还不能给出真实、经过验证的 flag。
对于 47.94.231.96:20560，建议回到 CTF 平台点击「重启容器」或「重新启动实例」，确认是否生成了新端口。如果平台也启动不了，那很可能是题目服务端的问题。
如果你能拿到这道题的完整题目描述、附件列表或题目页面截图，发给我。除了逆向 FPGA，这些信息也可能提供另一条解题路径，不一定非要连接容器。
Text version

那就先不依赖容器了。

我检查了你上传的 Cryptarithm.bit 和之前的分析结果。目前可以确认：

  • 209 字节密文完整，没有明显缺失。
  • 已知前 16 字节密钥流，但不足以解出后续全部内容。
  • 已测试简单循环 XOR 和线性反馈模型，未找到有效的密钥流生成规则。
  • FPGA bitstream 还需要进一步还原 LUT、寄存器和布线逻辑。可以使用 Project X-Ray 的 bit2fasm 工具辅助分析。

现在还不能给出真实、经过验证的 flag。

对于 47.94.231.96:20560，建议回到 CTF 平台点击「重启容器」或「重新启动实例」，确认是否生成了新端口。如果平台也启动不了，那很可能是题目服务端的问题。

如果你能拿到这道题的完整题目描述、附件列表或题目页面截图，发给我。除了逆向 FPGA，这些信息也可能提供另一条解题路径，不一定非要连接容器。

Câu hỏi

活动区的置位配置位都能由数据库中的特征解释，暂时没有发现漏掉的配置位。不过，电路的输入驱动和默认常量仍需复核；异或求解器在设定时限内也还没找到状态。

Trả lời

这个进展说明配置位映射可能已经比较完整，下一步更值得查的是电路模型是否正确，而不是继续扩大异或求解时间。我会对照现有分析脚本，重点检查输入驱动、未连接引脚的默认值，以及求解约束有没有把真实状态排除。

Suy nghĩ

Reviewed cryptanalysis tools, results, and image analysis

Suy nghĩ

继续诊断解码

我会把已覆盖的配置位与真正的连接、求解模型分开检查。当前结果只能说明解码范围足够，还不能证明电路语义或状态恢复正确；下一步应优先验证已知明文约束。

Continued diagnosing decoding issues

Searched 13 websites

Suy nghĩ

Inspected FPGA solver resources and tested database network access

Suy nghĩ

Refining the approach

I’m checking whether the recovered prefix corresponds to a recognizable plaintext or cipher structure before trusting the solver. The most useful next step is validating the keystream against independent constraints, then narrowing the generator model.

Testing cipher families

I’m comparing the recovered keystream with common generators and checking its linear complexity. A simple XOR recurrence appears unlikely, so I’m prioritizing nonlinear or state-dependent designs rather than extending the existing timeout.

Expanding the diagnostics

I’m treating the available ciphertext as too short to identify a general generator confidently. I’ll prioritize checking offsets, bit order, polarity, reset handling, and known-prefix alignment before attempting more sophisticated state recovery.

Refined the approach, tested cipher families, and expanded diagnostics

Searched 5 websites

Suy nghĩ

Validating alignment

I’m stopping short of claiming a recovered state. I’ll first verify bit order, polarity, reset timing, and prefix alignment; without the actual netlist or solver model, further state searches would be speculative.

Assessing the cipher

I’m treating the guessed plaintext as a useful oracle, not proof. Since the stream looks random and the state search timed out, I’m checking alignment and reset assumptions before trusting deeper cryptanalytic conclusions.

Requesting solver artifacts

I’m not treating the timeout as evidence that the state is impossible. I need the extracted netlist, configuration dump, or solver model to distinguish a wrong circuit abstraction from a genuinely difficult search.

Separating evidence

I’m not treating the timeout as failure. Only the initial known plaintext is confirmed; later guessed text must remain tentative. I need the extracted netlist or solver model to test incremental constraints and identify modeling errors.

Validated alignment, assessed cipher, and separated solver evidence

Searched 1 website

Worked for 1m 3s

Trả lời

Text version

这个进展很关键。配置位已经基本解释完毕，但异或求解器超时，不代表密钥状态无解。

目前我更怀疑是电路连接、初始状态或时钟时序建模存在问题，而不是缺少更多配置位。

在 Xilinx 7 系列 FPGA 中，还需要特别处理不直接体现在普通配置位中的固定路由（pseudo PIP）、LUT 输入的默认值，以及寄存器的初始化和复位行为。(link.springer.com github.com)

接下来建议优先做三个验证：

  1. 确认输入驱动：逐个追踪所有状态寄存器的 D、CE、CLK、SR 引脚，不能直接把未找到连线的信号全部设成 0。
  2. 确认初始状态：区分 FF 的 INIT、同步/异步复位以及电路实际开始输出密钥流的时刻。
  3. 确认求解模型：先只约束前 16 字节已知 keystream，检查是否可满足，再逐步增加约束。不要一开始就把 209 字节全部加入求解器。

我核对现有附件后的发现

目前上传的 cryptarithm_analysis.py 只包含 bitstream 帧解析，没有完整的 LUT 连线模型或异或求解器代码。因此，暂时无法独立复现你提到的“配置位全部覆盖”和“求解超时”。

真正确定的前 16 字节密钥流仍然是：

之前推测的 166 字节《原神》明文不能直接作为硬约束，否则一旦有一个字不同，整个求解模型就可能被错误排除。

建议把求解长度依次设为 1、2、4、8、16 字节，记录每次结果究竟是 SAT、UNSAT 还是 UNKNOWN（超时）。如果模型包含非线性 LUT 或进位逻辑，单纯的 GF(2) 异或线性求解可能也不够，需要完整的 SAT/SMT 建模。

现在距离真正恢复 flag 最关键的材料，是已经提取出的 .fasm 文件、Verilog 网表 .v，或者当前异或求解器的 .py 代码。 有其中任意一份，就可以具体检查哪条连接或约束阻碍状态恢复，而不必从 2.1 MB 的 bitstream 重新开始分析。

目前还没有可信的 flag{...}，但现阶段应优先验证电路模型，而不是盲目延长求解时间。

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