0xGame 2024 [Week 2] 报告哈基米
0xGame 2024 报告哈基米新知识：Tupper（塔珀自指公式）；Arnold Cat（猫映射）
下载文件是一个png图片，010打开查看
发现是倒着的pk
https://img2023.cnblogs.com/blog/3540423/202411/3540423-20241104223921181-1607510860.png
转一下保存为zip文件，打开后是一个txt文件
https://img2023.cnblogs.com/blog/3540423/202411/3540423-20241104223921593-332731555.png
有两个地方有提示，一个是十六进制里面的Maybe You Need To Kown Arnold Cat？另有一个是txt里面的Is This Tupper？
去搜一下Tupper是什么
代码急转弯——Tupper（塔珀自指公式）-CSDN博客
其代码运行出来的图片就是公式本身，在txt文件中第一句是倒序，所以我们也将下面的k值进行逆序，下面是在网上找的大佬的代码，应该是有三个算法在里面，我们就用Tupper一个就好了，把k值丢进去跑一下得到图片
########################################################################
#
#
# Tupper’s Self-Referential Formula
# Tupper.py
#
# MAIN
#
# Copyright (C) 2015 Ulrik Hoerlyk Hjort
#
#Tupper’s Self-Referential Formula is free software;you canredistribute it
#and/or modify it under terms of theGNU General Public License
#as publishedby the Free SoftwareFoundation;either version 2,
#or (at your option) any later version.
#Tupper’s Self-Referential Formula is distributed in the hope that it will be
#useful, but WITHOUT ANY WARRANTY;without even theimplied warranty
#of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
#See the GNU General Public License formore details.
#You should havereceiveda copy of the GNU General
#Public Licensedistributed with Yolk.If not, writetotheFree
#Software Foundation,51FranklinStreet,FifthFloor, Boston,
#MA 02110 - 1301, USA.
########################################################################
from PIL import Image
# Tupper
k1=9489414856877039590479997730148554425666925984049232945604842888420596111937489062065081199094002132087091572191187170308560128611026043144427876131133135794969867759108490917632153891963456295991713868378392769549376070709924497237322046334486274987407067993824142187115870972520417207510521083293280152434558803258138899515603807505064799735152359900010019631133734298562293682916239050320580346316026460860919542540955914826806059123630945216006606268974979135253968165822806241305783300650874506602000048154282039485531804337171305656252
# Pacman
# k2=144520248970897582847942537337194567481277782215150702479718813968549088735682987348888251320905766438178883231976923440016667764749242125128995265907053708020473915320841631792025549005418004768657201699730466383394901601374319715520996181145249781945019068359500510657804325640801197867556863142280259694206254096081665642417367403946384170774537427319606443899923010379398938675025786929455234476319291860957618345432248004921728033349419816206749854472038193939738513848960476759782673313437697051994580681869819330446336774047268864
# Euler
# k3=2352035939949658122140829649197960929306974813625028263292934781954073595495544614140648457342461564887325223455620804204796011434955111022376601635853210476633318991990462192687999109308209472315419713652238185967518731354596984676698288025582563654632501009155760415054499960
# Assign k1,k2, k3 to k to get desired image
k = k1
width = 106
height = 17
scale = 5
fname = "foo"
image= Image.new("RGB", (width, height),(255, 255, 255))
for x in range (width):
for y in range (height):
if ((k+y)//17//2**(17*int(x)+int(y)%17))%2 > 0.5:
# Image need to be flipped vertically - therefore y = height-y-1
image.putpixel((x, height-y-1), (0,0,0))
#scale up image
image = image.resize((width*scale,height*scale))
image.save(fname+".png")https://img2023.cnblogs.com/blog/3540423/202411/3540423-20241104223921903-1802122767.png
看大括号的位置，这应该是flag的后半段
找Arnold Cat是个什么东西
在数学中，阿诺德猫映射是一种从环面到自身的混沌映射，以弗拉基米尔·阿诺德的名字定名，他于 20 世纪 60 年代使用猫的图像展示了它的效果，因此得名。该图的一个特点是图像在变换过程中看似随机化，但经过多少步骤后又规复到原始状态。猫的原始图像在变换的第一次迭代中被剪切然后重新包裹。经过几次迭代后，天生的图像看起来相称随机或无序，但经过进一步的迭代后，图像似乎有了进一步的秩序——猫的幽灵般的图像、以重复结构分列的多个较小的副本，甚至是原始图像的颠倒副本——并最终规复到原始图像。
发现最初的图片看起来非常无序，应该是经过了猫映射（https://github.com/zhanxw/cat github代码）（https://www.jasondavies.com/catmap/ 在线网站）
不知道咋回事网站和代码运行后都得不到原图，看起来还是很无序
卡死在这了
末了看了大佬的wp，太强了，脚本一运行直接出来了，放一下脚本
from PIL import Image
img = Image.open('mijiha.png')
if img.mode == "P":
img = img.convert("RGB")
assert img.size == img.size
dim = width, height = img.size
st = 1
a = 35
b = 7
for _ in range(st):
with Image.new(img.mode, dim) as canvas:
for nx in range(img.size):
for ny in range(img.size):
y = (ny - nx * a) % width
x = (nx - y * b) % height
canvas.putpixel((y, x), img.getpixel((ny, nx)))
canvas.show()
canvas.save('result.png')得到图片
https://img2023.cnblogs.com/blog/3540423/202411/3540423-20241104223922456-1268296898.png
得到flag前半段，题解
总结：遇到了两个从来没有见过的算法，算是长了见识
免责声明：如果侵犯了您的权益，请联系站长，我们会及时删除侵权内容，谢谢合作！更多信息从访问主页：qidao123.com:ToB企服之家，中国第一个企服评测及商务社交产业平台。
页:
[1]