0x01 题目背景及介绍

本篇题目来自于PWNHUB五月公开赛的一道Crypto，题目叫做mylcg。

CTF的技术有一部分提示来自于题目，题目中带lcg，可能和LCG相关。

from Crypto.Util.number import * from hashlib import sha256 f = open(‘flag.txt’) flag = f.read() f.close() assert flag[:5] == ‘flag{‘ seed = getPrime(256)

打开文件flag.txt，assert断言flag格式应该为“flag{}”

class my_LCG: def __init__(self): self.p = getPrime(512) self.a = getRandomNBitInteger(256) self.b = getRandomNBitInteger(256) self.c = getRandomNBitInteger(256) self.seed = seed

定义类my_LCG

def next(self): self.seed = (self.a * self.seed * self.seed + self.b * self.seed + self.c) % self.p return self.seed >> 64 def output(self): print(“a =”, self.a) print(“b =”, self.b) print(“c =”, self.c) print(“p =”, self.p) lcg = my_LCG() lcg.output() print(“state1 =”, lcg.next()) print(“state2 =”, lcg.next()) cipher = [] for i in flag: tmp = int(sha256(i.encode()).hexdigest(), 16) cipher.append(pow(tmp, 3, seed) ^ (lcg.next() >> 192)) print(cipher) # a = 68823589009570846705623113091430161477799561031575612135516351257937127579444 # b = 76549169049080319489163980188997771079750670038002598167961495813084486794567 # c = 99215492498952976642031024510129092308042391285395704888838178561670205468882 # p = 12775129699668988473759438271274836254349225413222075887429387682336494103348583050672280757042383792640084197832605411237644937815012509935794275313643031 # state1 = 664406108637237317109372377546194289077117372665932150107678757303243619963182733408311739663233548725195723884930183442927293177078507 # state2 = 269639459994260392015105629865659210391196381140872041084737976688017858324308345528702533444116282512066085580909718347778743354754352 # [104612880565354761070267231258959974017713136013091552542621012009434344223091, 75867315122609665862412321569592448016548595925558860261941234225747875658355, 106901597652387441116797426235259979167755542628252028460970918870852084813985, 22282265919236743389726915885827434932651977730848876965702592028838506934802, 113663688553403244683128957504190135685760214545373190153400188368329624810824, 16387987427743919688629484112748920627537972533590712542304182257404148095432, 42660781848852863873070244368035358673436082124276647936063403560054320824947, 31162780219880567100421404229662317404195331726681121575505050872412728752846, 4387263209760294127327273998365291139136756172378440265726369942465661809226, 9486548512905762310741629559650376499881990751841396019470492813536817972788, 48175096956218645396129797881617184816956038092368660196837813749510220209020, 44832999096333655680734980423005605365074861714291594434259782779054037518357, 91949111038768792853022977677056688517414191501656202462769670180011454736134, 84362160451115508526234487916725793664590119636637539810168371139175756722218, 37991151261861388606427583658548923882501719664397171054526120281654768250027, 76502569176155441262877857713986290804857527394931115027371407239085302418015, 69004590108972326779999503375035935629144187797049165946644383062664912505793, 4882509736578426220928083701456224217596136409513934115208162112269124095036, 43923135793747042362291047632803354256532285967402459802384563909906286571537, 60099118217689452210082521471087211926434931778809056323284225098458442535387, 41772186654628607473153061053630877342481864865985327695842922897386493117840, 79556569085671906788310296380605892417410665990711208999213184300355584627493, 113368724525353223298893326950045576507359981373213697703045622403941359594816, 59568498228965811827206375159844085215141410061444029744235874177744506595963, 5813871055160958965491186951684884799390692165774935048726620511285107065171, 101481471612354001038968281423076176339698400003836978206153754002016784653117, 96713934756979037100616702408060142461624534046186352589457317850269578096978, 50360544372106359495068645870868114000049436655822775857718387298804448386259, 74343018392522018803983491192044402773734077434586754153316411931268932053849, 85064649783728020457864819723803995069342386266325766064354175704654032312470, 35444692928369969109775411420972206587166544062593932622816977149910920025593, 88364976585873094910882074779610461514853793492269703066264952338524869329015, 57171789241051208558861663926443006215165105080980300229007074324952708746414, 53398702677161874182117493992247921340773468735882139055903326524685751323223, 22084022599630141597766375557874162908749682780683228274573161676243137075985, 67811375554342446963928136891489894009025011339123405492350978811178609371520, 54279152372926754185972096972067394007518618911313269898203907795172124806761, 71337965908378379886587542454943435140109212437982933800558803691194797247858, 91631795365993438211094618411930046881699597378889861280702841447103987549686, 55725715889928044127700572038972949057067144434626788557169321659022938298621, 68379770126152524442093415542514227284857840135092144556912468864495685413179, 40664875356285535027116296315063139118439996515430069978781752466656941376294]

 

0x02 题目分析

cipher.append(pow(tmp, 3, seed) ^ (lcg.next() >> 192))

其中

pow(tmp, 3, seed)为，tmp的3次方对seed取余

pow(tmp, 3, seed) ^ (lcg.next() >> 192)为pow(tmp, 3, seed)与(lcg.next() >> 192)的二进制异或

getPrime(512)

getRandomNBitInteger(256)

知识点

思路应该是先推导出seed再进行flag的判断，尝试下的结果：

尝试了一下发现

state1不是state2的输入，只是输出，原因在于对>>符号的理解

比特右移（>>）运算符可以是算术（左端补最高有效位）或是逻辑（左端补 0）位移。例如，将 11100011 右移 3 比特，算术右移后成为 11111100

如果再移回来就是

a = 0b10001

a>>2 = 0b100

(a>>2)<<2 = 0b10000

知识点

“(a ^ b)^b = a”

a = 0b10000110

b = 0b110100101

a ^ b = 0b100100011

(a ^ b)^b = 0b10000110

 

0x03 错误思路

一开始本想通过爆破逆推，尝试后发现，较短的数据可以做到，较长的效果极其不好，所以换为官方的思路

from Crypto.Util.number import * from hashlib import sha256 a = 68823589009570846705623113091430161477799561031575612135516351257937127579444 b = 76549169049080319489163980188997771079750670038002598167961495813084486794567 c = 99215492498952976642031024510129092308042391285395704888838178561670205468882 p = 12775129699668988473759438271274836254349225413222075887429387682336494103348583050672280757042383792640084197832605411237644937815012509935794275313643031 state1 = 664406108637237317109372377546194289077117372665932150107678757303243619963182733408311739663233548725195723884930183442927293177078507 state2 = 269639459994260392015105629865659210391196381140872041084737976688017858324308345528702533444116282512066085580909718347778743354754352 for i in range(2**64): s = state1<<64 s += i if ((a*s*s+b*s+c)%p)>>64 == state2: s2 = (a*s*s+b*s+c)%p print(s2)

 

0x04 题目思路

这道题的关键在于Coppersmith一元、二元攻击

import itertools def small_roots(f, bounds, m=1, d=None): if not d: d = f.degree() R = f.base_ring() N = R.cardinality() f /= f.coefficients().pop(0) f = f.change_ring(ZZ) G = Sequence([], f.parent()) for i in range(m + 1): base = N ^ (m – i) * f ^ i for shifts in itertools.product(range(d), repeat=f.nvariables()): g = base * prod(map(power, f.variables(), shifts)) G.append(g) B, monomials = G.coefficient_matrix() monomials = vector(monomials) factors = [monomial(*bounds) for monomial in monomials] for i, factor in enumerate(factors): B.rescale_col(i, factor) B = B.dense_matrix().LLL() B = B.change_ring(QQ) for i, factor in enumerate(factors): B.rescale_col(i, 1 / factor) H = Sequence([], f.parent().change_ring(QQ)) for h in filter(None, B * monomials): H.append(h) I = H.ideal() if I.dimension() == -1: H.pop() elif I.dimension() == 0: roots = [] for root in I.variety(ring=ZZ): root = tuple(R(root[var]) for var in f.variables()) roots.append(root) return roots return [] a = 68823589009570846705623113091430161477799561031575612135516351257937127579444 b = 76549169049080319489163980188997771079750670038002598167961495813084486794567 c = 99215492498952976642031024510129092308042391285395704888838178561670205468882 p = 12775129699668988473759438271274836254349225413222075887429387682336494103348583050672280757042383792640084197832605411237644937815012509935794275313643031 state1 = 664406108637237317109372377546194289077117372665932150107678757303243619963182733408311739663233548725195723884930183442927293177078507 state2 = 269639459994260392015105629865659210391196381140872041084737976688017858324308345528702533444116282512066085580909718347778743354754352 PR. < s1_low, s2_low > = PolynomialRing(Zmod(p)) f = a * ((state1 << 64) + s1_low) ^ 2 + b * ((state1 << 64) + s1_low) + c – (state2 << 64) – s2_low state1 = small_roots(f, (2 ^ 64, 2 ^ 64), m=3)[0][0] + (state1 << 64) state2 = small_roots(f, (2 ^ 64, 2 ^ 64), m=3)[0][1] + (state2 << 64) print(state2)

此段代码的目的是用于输出之后的state，之前有推测过，知道flag需要知道seed

求seed方程：

P.<x> = PolynomialRing(Zmod(p)) f = a * x * x + b * x + c – state1 f = f.monic() print(f.roots())

得到seed：

seed = 931619026756858577869325713310380185951406212494138341790364440

通过seed，求解flag：

seed = 93161902675685857786932571331038018595140621249413834179036444009740278201203 hashtable = [] flag = ” lcg = my_LCG() for i in printable: tmp = int(sha256(i.encode()).hexdigest(), 16) hashtable.append(pow(tmp, 3, seed)) for i in range(len(cipher)): tmp = lcg.next() >> 192 ind = hashtable.index(tmp ^ cipher[i]) flag += printable[ind] print(flag) References: https://mp.weixin.qq.com/s/yDFu4S7fBaxI5HmWrxrduA