2024网鼎杯青龙组初赛 CRYPTO01 bd

1、题目信息

这题目和2023领航杯的密码题是一样的（https://www.cnblogs.com/mumuhhh/p/17789591.html）

from Crypto.Util.number import *
from secret import flag

p = getPrime(512)
q = getPrime(512)
n = p * q
d = getPrime(299)
e = inverse(d,(p-1)*(q-1))
m = bytes_to_long(flag)
c = pow(m,e,n)
hint1 = p >> (512-70)
hint2 = q >> (512-70)


print(f"n = {n}")
print(f"e = {e}")
print(f"c = {c}")
print(f"hint1 = {hint1}")
print(f"hint2 = {hint2}")

"""
n = 73337798113265277242402875272164983073482378701520700321577706460042584510776095519204866950129951930143711572581533177043149866218358557626070702546982947219557280493881836314492046745063916644418320245218549690820002504737756133747743286301499039227014032044403571945455215839074583290324966069724343874361
e = 42681919079074901709680276679968298324860328305878264036188155781983964226653746568102282190906458519960811259171162918944726137301701132135900454469634110653076655027353831375989861927565774719655974876907429954299669710134188543166679161864800926130527741511760447090995444554722545165685959110788876766283
c = 35616516401097721876690503261383371143934066789806504179229622323608172352486702183654617788750099596415052166506074598646146147151595929618406796332682042252530491640781065608144381326123387506000855818316664510273026302748073274714692374375426255513608075674924804166600192903250052744024508330641045908599
hint1 = 740477612377832718425
hint2 = 767891335159501447918
"""

2、解题方法

根据大佬师傅WP，需要利用上高位的boneh_durfee攻击

参考论文：https://eprint.iacr.org/2023/367.pdf

他们实验时的脚本：https://pastebin.com/zpUkrfDh

然后我们直接sage解密，我们只用替换我们直接的数据就可以

import time
time.clock = time.time
 
debug = True
 
strict = False
 
helpful_only = True
dimension_min = 7 # 如果晶格达到该尺寸，则停止移除
# 显示有用矢量的统计数据
def helpful_vectors(BB, modulus):
    nothelpful = 0
    for ii in range(BB.dimensions()[0]):
        if BB[ii,ii] >= modulus:
            nothelpful += 1
 #    print (nothelpful, "/", BB.dimensions()[0], " vectors are not helpful")

# 显示带有 0 和 X 的矩阵
def matrix_overview(BB, bound):
    for ii in range(BB.dimensions()[0]):
        a = ('%02d ' % ii)
        for jj in range(BB.dimensions()[1]):
            a += '0' if BB[ii,jj] == 0 else 'X'
            if BB.dimensions()[0] < 60: 
                a += ' '
        if BB[ii, ii] >= bound:
            a += '~'
        #print (a)

# 尝试删除无用的向量
# 从当前 = n-1（最后一个向量）开始
def remove_unhelpful(BB, monomials, bound, current):
    # 我们从当前 = n-1（最后一个向量）开始
    if current == -1 or BB.dimensions()[0] <= dimension_min:
        return BB
 
    # 开始从后面检查
    for ii in range(current, -1, -1):
        #  如果它没有用
        if BB[ii, ii] >= bound:
            affected_vectors = 0
            affected_vector_index = 0
             # 让我们检查它是否影响其他向量
            for jj in range(ii + 1, BB.dimensions()[0]):
                # 如果另一个向量受到影响：
                # 我们增加计数
                if BB[jj, ii] != 0:
                    affected_vectors += 1
                    affected_vector_index = jj
 
            # 等级：0
            # 如果没有其他载体最终受到影响
            # 我们删除它
            if affected_vectors == 0:
                #print ("* removing unhelpful vector", ii)
                BB = BB.delete_columns([ii])
                BB = BB.delete_rows([ii])
                monomials.pop(ii)
                BB = remove_unhelpful(BB, monomials, bound, ii-1)
                return BB
 
           # 等级：1
            #如果只有一个受到影响，我们会检查
            # 如果它正在影响别的向量
            elif affected_vectors == 1:
                affected_deeper = True
                for kk in range(affected_vector_index + 1, BB.dimensions()[0]):
                    # 如果它影响哪怕一个向量
                    # 我们放弃这个
                    if BB[kk, affected_vector_index] != 0:
                        affected_deeper = False
                # 如果没有其他向量受到影响，则将其删除，并且
                # 这个有用的向量不够有用
                #与我们无用的相比
                if affected_deeper and abs(bound - BB[affected_vector_index, affected_vector_index]) < abs(bound - BB[ii, ii]):
                    #print ("* removing unhelpful vectors", ii, "and", affected_vector_index)
                    BB = BB.delete_columns([affected_vector_index, ii])
                    BB = BB.delete_rows([affected_vector_index, ii])
                    monomials.pop(affected_vector_index)
                    monomials.pop(ii)
                    BB = remove_unhelpful(BB, monomials, bound, ii-1)
                    return BB
    # nothing happened
    return BB
 
""" 
Returns:
* 0,0   if it fails
* -1，-1 如果 "strict=true"，并且行列式不受约束
* x0,y0 the solutions of `pol`
"""
def boneh_durfee(pol, modulus, mm, tt, XX, YY):
    """
    Boneh and Durfee revisited by Herrmann and May
 
 在以下情况下找到解决方案：
* d < N^delta
* |x|< e^delta
* |y|< e^0.5
每当 delta < 1 - sqrt（2）/2 ~ 0.292
    """
 
    # substitution (Herrman and May)
    PR.<u, x, y> = PolynomialRing(ZZ)   #多项式环
    Q = PR.quotient(x*y + 1 - u)        #  u = xy + 1
    polZ = Q(pol).lift()
 
    UU = XX*YY + 1
 
    # x-移位
    gg = []
    for kk in range(mm + 1):
        for ii in range(mm - kk + 1):
            xshift = x^ii * modulus^(mm - kk) * polZ(u, x, y)^kk
            gg.append(xshift)
    gg.sort()
 
    # 单项式 x 移位列表
    monomials = []
    for polynomial in gg:
        for monomial in polynomial.monomials(): #对于多项式中的单项式。单项式（）：
            if monomial not in monomials:  # 如果单项不在单项中
                monomials.append(monomial)
    monomials.sort()
 
    # y-移位
    for jj in range(1, tt + 1):
        for kk in range(floor(mm/tt) * jj, mm + 1):
            yshift = y^jj * polZ(u, x, y)^kk * modulus^(mm - kk)
            yshift = Q(yshift).lift()
            gg.append(yshift) # substitution
 
    # 单项式 y 移位列表
    for jj in range(1, tt + 1):
        for kk in range(floor(mm/tt) * jj, mm + 1):
            monomials.append(u^kk * y^jj)
 
    # 构造格 B
    nn = len(monomials)
    BB = Matrix(ZZ, nn)
    for ii in range(nn):
        BB[ii, 0] = gg[ii](0, 0, 0)
        for jj in range(1, ii + 1):
            if monomials[jj] in gg[ii].monomials():
                BB[ii, jj] = gg[ii].monomial_coefficient(monomials[jj]) * monomials[jj](UU,XX,YY)
 
    #约化格的原型
    if helpful_only:
        #  #自动删除
        BB = remove_unhelpful(BB, monomials, modulus^mm, nn-1)
        # 重置维度
        nn = BB.dimensions()[0]
        if nn == 0:
            print ("failure")
            return 0,0
 
    # 检查向量是否有帮助
    if debug:
        helpful_vectors(BB, modulus^mm)
 
    # 检查行列式是否正确界定
    det = BB.det()
    bound = modulus^(mm*nn)
    if det >= bound:
        print ("We do not have det < bound. Solutions might not be found.")
        print ("Try with highers m and t.")
        if debug:
            diff = (log(det) - log(bound)) / log(2)
            print ("size det(L) - size e^(m*n) = ", floor(diff))
        if strict:
            return -1, -1
    else:
        print ("det(L) < e^(m*n) (good! If a solution exists < N^delta, it will be found)")
 
    # display the lattice basis
    if debug:
        matrix_overview(BB, modulus^mm)
 
    # LLL
    if debug:
        print ("optimizing basis of the lattice via LLL, this can take a long time")
 
    #BB = BB.BKZ(block_size=25)
    BB = BB.LLL()
 
    if debug:
        print ("LLL is done!")
 
    # 替换向量 i 和 j ->多项式 1 和 2
    if debug:
        print ("在格中寻找线性无关向量")
    found_polynomials = False
 
    for pol1_idx in range(nn - 1):
        for pol2_idx in range(pol1_idx + 1, nn):
 
            # 对于i and j, 构造两个多项式
 
            PR.<w,z> = PolynomialRing(ZZ)
            pol1 = pol2 = 0
            for jj in range(nn):
                pol1 += monomials[jj](w*z+1,w,z) * BB[pol1_idx, jj] / monomials[jj](UU,XX,YY)
                pol2 += monomials[jj](w*z+1,w,z) * BB[pol2_idx, jj] / monomials[jj](UU,XX,YY)
 
            # 结果
            PR.<q> = PolynomialRing(ZZ)
            rr = pol1.resultant(pol2)
 
 
            if rr.is_zero() or rr.monomials() == [1]:
                continue
            else:
                print ("found them, using vectors", pol1_idx, "and", pol2_idx)
                found_polynomials = True
                break
        if found_polynomials:
            break
 
    if not found_polynomials:
        print ("no independant vectors could be found. This should very rarely happen...")
        return 0, 0
 
    rr = rr(q, q)
 
    # solutions
    soly = rr.roots()
 
    if len(soly) == 0:
        print ("Your prediction (delta) is too small")
        return 0, 0
 
    soly = soly[0][0]
    ss = pol1(q, soly)
    solx = ss.roots()[0][0]
    return solx, soly
 
def example():
    ############################################
    # 随机生成数据
    ##########################################
    #start_time =time.perf_counter
    start =time.clock()
    size=512
    length_N = 2*size;
    ss=0
    s=70;
    M=1   # the number of experiments
    delta = 299/1024
    # p =  random_prime(2^512,2^511)
    for i in range(M):
#         p =  random_prime(2^size,None,2^(size-1))
#         q =  random_prime(2^size,None,2^(size-1))
#         if(p<q):
#             temp=p
#             p=q
#             q=temp
        N = 104769059324906604819374246969389472089736482039584780304698351288134425847574721209477631552050746222528061242850563906415558000954816414452571907898376586538455570846715727736834959625908944488834642926192746728574287181536549647851644625185864257557629579686099455733892320222578364826099212655146530976379
        e = 12337109880409970018293646110440488264982341274846829641219533345965373708872641944832903882339212178067485766669515688243675673212167726028183775964215646348775048640061665951311218967384639999950950042290221189659835294938061099700246737365693200129282703765155456889082133763568539014092220899267025682857
        c = 31744736423783628269884009616541129531740686983212218114995065554639252322714403985771782435353721009653250709135160293375136413735234647281736871541268953447552855923299477737849706638177219571453513142214997506075291749228813720600113175989090030091204440975462838480365583907951185017109681679559591532826
        hint1 = 864467081468962738290   # p高位
        hint2 =  939654974954806345061 # q高位
#         print ("p真实高",s,"比特：", int(p/2^(512-s)))
#         print ("q真实高",s,"比特：", int(q/2^(512-s)))
 
#         N = p*q;
 
 
    # 解密指数d的指数( 最大0.292)
 
 
 
        m = 7   # 格大小（越大越好/越慢）
        t = round(((1-2*delta) * m))  # 来自 Herrmann 和 May 的优化
        X = floor(N^delta)  # 
        Y = floor(N^(1/2)/2^s)    # 如果 p、 q 大小相同，则正确
        for l in range(int(hint1),int(hint1)+1):
            print('\n\n\n l=',l)
            pM=l;
            p0=pM*2^(size-s)+2^(size-s)-1;
            q0=N/p0;
            qM=int(q0/2^(size-s))
            A = N + 1-pM*2^(size-s)-qM*2^(size-s);
        #A = N+1
            P.<x,y> = PolynomialRing(ZZ)
            pol = 1 + x * (A + y)  #构建的方程
 
            # Checking bounds
            #if debug:
                #print ("=== 核对数据 ===")
                #print ("* delta:", delta)
                #print ("* delta < 0.292", delta < 0.292)
                #print ("* size of e:", ceil(log(e)/log(2)))  # e的bit数
                # print ("* size of N:", len(bin(N)))          # N的bit数
                #print ("* size of N:", ceil(log(N)/log(2)))  # N的bit数
                #print ("* m:", m, ", t:", t)
 
            # boneh_durfee
            if debug:
                ##print ("=== running algorithm ===")
                start_time = time.time()
 
 
            solx, soly = boneh_durfee(pol, e, m, t, X, Y)
 
 
            if solx > 0:
                #print ("=== solution found ===")
                if False:
                    print ("x:", solx)
                    print ("y:", soly)
 
                d_sol = int(pol(solx, soly) / e)
                ss=ss+1

                print ("=== solution found ===")
                print ("p的高比特为：",l)
                print ("q的高比特为：",qM)
                print ("d=",d_sol) 
 
            if debug:
                print("=== %s seconds ===" % (time.time() - start_time))
            #break
        print("ss=",ss)
                            #end=time.process_time
        end=time.clock()
        print('Running time: %s Seconds'%(end-start))
if __name__ == "__main__":
    example()

解出d

l= 864467081468962738290
19 / 47  vectors are not helpful
det(L) < e^(m*n) (good! If a solution exists < N^delta, it will be found)
optimizing basis of the lattice via LLL, this can take a long time
LLL is done!
在格中寻找线性无关向量
found them, using vectors 0 and 1
=== solution found ===
p的高比特为： 864467081468962738290
q的高比特为： 939654974954806345060
d= 739638974439475993531255172202354119076210219461888654018566065491412925302888252209575273
=== 36.80956530570984 seconds ===
ss= 1
Running time: 36.81700682640076 Seconds

然后解密即可

from Crypto.Util.number import *
n = 104769059324906604819374246969389472089736482039584780304698351288134425847574721209477631552050746222528061242850563906415558000954816414452571907898376586538455570846715727736834959625908944488834642926192746728574287181536549647851644625185864257557629579686099455733892320222578364826099212655146530976379
e = 12337109880409970018293646110440488264982341274846829641219533345965373708872641944832903882339212178067485766669515688243675673212167726028183775964215646348775048640061665951311218967384639999950950042290221189659835294938061099700246737365693200129282703765155456889082133763568539014092220899267025682857
c = 31744736423783628269884009616541129531740686983212218114995065554639252322714403985771782435353721009653250709135160293375136413735234647281736871541268953447552855923299477737849706638177219571453513142214997506075291749228813720600113175989090030091204440975462838480365583907951185017109681679559591532826=

d = 739638974439475993531255172202354119076210219461888654018566065491412925302888252209575273
m = pow(c, d, n)
print(long_to_bytes(m))

得到flag

b'wdflag{c8546072-924f-46b6-a548-1a21bf67bcfc}'

参考文章

2024网鼎杯青龙组初赛---WriteUp