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TCTF/0CTF 2020 Quals - emmm (crypto)
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| #!/usr/bin/python2 | |
| import random | |
| from struct import pack,unpack | |
| from flag import flag | |
| P = 247359019496198933 | |
| C = 223805275076627807 | |
| M = 2**60 | |
| K0 = random.randint(1, P-1) | |
| K1 = random.randint(1, P-1) | |
| # not a bijection? can be adjusted but I'm lazy | |
| def encrypt_block(x): | |
| tmp = x * K0 % P | |
| tmp = tmp * C % M | |
| tmp = tmp * K1 % P | |
| return tmp | |
| for i in range(2**24): | |
| pt = random.randint(1, P-1) | |
| ct = encrypt_block(pt) | |
| print pt, ct | |
| pt = flag[5:-1] | |
| assert flag.startswith('flag{') and flag.endswith('}') and len(pt)%8==0 | |
| fmt = '%dQ' % (len(pt)/8) | |
| ct = pack(fmt, *map(encrypt_block, unpack(fmt, pt))) | |
| print ct.encode('hex') |
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| { | |
| "cells": [ | |
| { | |
| "cell_type": "markdown", | |
| "metadata": {}, | |
| "source": [ | |
| "# 0CTF/TCTF 2020 Quals - emmm (crypto)" | |
| ] | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "metadata": {}, | |
| "source": [ | |
| "In this challenge, we are given a simple block cipher (though not fully invertible), based on 3 modular multiplications." | |
| ] | |
| }, | |
| { | |
| "cell_type": "code", | |
| "execution_count": 1, | |
| "metadata": { | |
| "execution": { | |
| "iopub.execute_input": "2020-06-29T13:34:47.753110Z", | |
| "iopub.status.busy": "2020-06-29T13:34:47.751152Z", | |
| "iopub.status.idle": "2020-06-29T13:34:47.779799Z", | |
| "shell.execute_reply": "2020-06-29T13:34:47.776453Z", | |
| "shell.execute_reply.started": "2020-06-29T13:34:47.752968Z" | |
| } | |
| }, | |
| "outputs": [], | |
| "source": [ | |
| "# Sage mode\n", | |
| "P = 247359019496198933 # 2**57.78\n", | |
| "C = 223805275076627807 # 2**57.64\n", | |
| "M = 2**60\n", | |
| "K0 = random.randint(1, P-1)\n", | |
| "K1 = random.randint(1, P-1)\n", | |
| "\n", | |
| "# not a bijection? can be adjusted but I'm lazy\n", | |
| "def encrypt_block(x):\n", | |
| " tmp = x * K0 % P\n", | |
| " tmp = tmp * C % M\n", | |
| " tmp = tmp * K1 % P\n", | |
| " return tmp" | |
| ] | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "metadata": { | |
| "execution": { | |
| "iopub.execute_input": "2020-06-29T12:21:50.379449Z", | |
| "iopub.status.busy": "2020-06-29T12:21:50.378297Z", | |
| "iopub.status.idle": "2020-06-29T12:21:50.391850Z", | |
| "shell.execute_reply": "2020-06-29T12:21:50.389108Z", | |
| "shell.execute_reply.started": "2020-06-29T12:21:50.379278Z" | |
| } | |
| }, | |
| "source": [ | |
| "We are also given $2^{24}$ random plaintext-ciphertext pairs and encrypted flag.\n", | |
| "\n", | |
| "Let's build linearized relation between a plaintext and the ciphertext, by introducing variables for modular reductions." | |
| ] | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "metadata": { | |
| "execution": { | |
| "iopub.execute_input": "2020-06-29T12:21:50.379449Z", | |
| "iopub.status.busy": "2020-06-29T12:21:50.378297Z", | |
| "iopub.status.idle": "2020-06-29T12:21:50.391850Z", | |
| "shell.execute_reply": "2020-06-29T12:21:50.389108Z", | |
| "shell.execute_reply.started": "2020-06-29T12:21:50.379278Z" | |
| } | |
| }, | |
| "source": [ | |
| "Consider an encryption $(x, y)$.\n", | |
| "Let \n", | |
| "$$\n", | |
| "\\begin{align}\n", | |
| "t_1 &= \\lfloor K_0x / P \\rfloor < x, \\\\\n", | |
| "t_2 &= \\lfloor (K_0x\\mod{P}) / M \\rfloor < PC/M, \\\\\n", | |
| "t_3 &= \\lfloor K_1^{-1}y / P \\rfloor + \\epsilon < y + \\epsilon, \\\\\n", | |
| "\\end{align}\n", | |
| "$$\n", | |
| "where $0 \\le \\epsilon \\le \\lfloor M / P \\rfloor = 4$ is such that $yK_1^{-1} -t_3P$ matches the plaintext side second step encryption (`tmp = tmp * C % M`). This little difference happens since the value after the second step can be large up to $M=2^{60}$ and then it is reduced modulo $P\\approx 2^{57.64}$, so a couple bits of information are lost." | |
| ] | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "metadata": {}, | |
| "source": [ | |
| "Then, following two steps of encryption of $x$ and one step of decryption of $y$ we get:\n", | |
| "$$\n", | |
| "\\begin{align}\n", | |
| "& (K_0x - t_1P)C - t_2M = K_1^{-1}y - t_3P, \\\\\n", | |
| "\\Rightarrow &\n", | |
| " xC\\cdot K_0 - y\\cdot K_1^{-1} - PC\\cdot t_1 - M\\cdot t_2 + P\\cdot t_3 = 0,\n", | |
| "\\end{align}\n", | |
| "$$\n", | |
| "where we have unknowns\n", | |
| "$$\n", | |
| "\\begin{align}\n", | |
| "0 \\le~ & K_0 < P,\\\\\n", | |
| "0 \\le~ & K_1^{-1} < P,\\\\\n", | |
| "0 \\le~ & t_1 < x,\\\\\n", | |
| "0 \\le~ & t_2 < PC/M,\\\\\n", | |
| "0 \\le~ & t_3 < y+\\epsilon.\n", | |
| "\\end{align}\n", | |
| "$$" | |
| ] | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "metadata": { | |
| "execution": { | |
| "iopub.execute_input": "2020-06-29T12:41:35.934568Z", | |
| "iopub.status.busy": "2020-06-29T12:41:35.934010Z", | |
| "iopub.status.idle": "2020-06-29T12:41:35.939785Z", | |
| "shell.execute_reply": "2020-06-29T12:41:35.938713Z", | |
| "shell.execute_reply.started": "2020-06-29T12:41:35.934501Z" | |
| } | |
| }, | |
| "source": [ | |
| "Note that $x,y,t_1,t_2,t_3$ are different for each known data pair." | |
| ] | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "metadata": {}, | |
| "source": [ | |
| "We can now use LLL to solve this constraint system. As an example, consider the following lattice for $n=3$ data pairs (rows as vectors):\n", | |
| "$$\n", | |
| "\\begin{matrix}\n", | |
| "&~~~~ eq_0 ~~~~~~ eq_1 ~~~~~~ eq_2 ~~~ K_0 ~~~ K_1 ~~~ . ~~~~ t_{1,i} ~~~ . ~~~~ . ~~~~ t_{2,i} ~~~ . ~~~~ . ~~~~ t_{3,i} ~~~ . \\hfill \\\\\n", | |
| "\\begin{matrix}\n", | |
| "K_0 \\\\ K_1 \\\\ . \\\\ t_{1,i} \\\\ . \\\\ . \\\\ t_{2,i} \\\\ . \\\\ . \\\\ t_{3,i} \\\\ .\n", | |
| "\\end{matrix} \\hspace{-1em}&\n", | |
| "\\begin{pmatrix}\n", | |
| "x_0C & x_1C & x_2C & 1 & & & & & & & & & & \\\\\n", | |
| "y_0 & y_1 & y_2 & & 1 & & & & & & & & & \\\\\n", | |
| "CP & & & & & 1 & & & & & & & & \\\\\n", | |
| " & CP & & & & & 1 & & & & & & & \\\\\n", | |
| " & & CP & & & & & 1 & & & & & & \\\\\n", | |
| "M & & & & & & & & 1 & & & & & \\\\\n", | |
| " & M & & & & & & & & 1 & & & & \\\\\n", | |
| " & & M & & & & & & & & 1 & & & \\\\\n", | |
| "P & & & & & & & & & & & 1 & & \\\\\n", | |
| " & P & & & & & & & & & & & 1 & \\\\\n", | |
| " & & P & & & & & & & & & & & 1 \\\\\n", | |
| "\\end{pmatrix}\n", | |
| "\\end{matrix}\n", | |
| "$$" | |
| ] | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "metadata": {}, | |
| "source": [ | |
| "We are looking for a linear combination of rows that makes the first $n$ entries zero, and the others to respect our bounds. We can achieve this by scaling the coordinates (columns): first $n$ columns should be multiplied by a very large number (forcing LLL to make it zero), the other columns should be multiplied inversely to their bounds. After applying the LLL, we need to scale back." | |
| ] | |
| }, | |
| { | |
| "cell_type": "code", | |
| "execution_count": 2, | |
| "metadata": { | |
| "execution": { | |
| "iopub.execute_input": "2020-06-29T13:34:47.783252Z", | |
| "iopub.status.busy": "2020-06-29T13:34:47.782103Z", | |
| "iopub.status.idle": "2020-06-29T13:35:16.183268Z", | |
| "shell.execute_reply": "2020-06-29T13:35:16.182586Z", | |
| "shell.execute_reply.started": "2020-06-29T13:34:47.783130Z" | |
| } | |
| }, | |
| "outputs": [], | |
| "source": [ | |
| "f = open(\"res\")\n", | |
| "data = []\n", | |
| "for line in f:\n", | |
| " try:\n", | |
| " pt, ct = map(int, line.split())\n", | |
| " except:\n", | |
| " break\n", | |
| " data.append((pt, ct))" | |
| ] | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "metadata": {}, | |
| "source": [ | |
| "Note that the *bounds* of $t_{1,i},t_{2,i}$ depend on the actual plaintexts. We are thus interested in smallest plaintexts and ciphertexts. As we shall see, 20 smallest pairs are enough!" | |
| ] | |
| }, | |
| { | |
| "cell_type": "code", | |
| "execution_count": 3, | |
| "metadata": { | |
| "execution": { | |
| "iopub.execute_input": "2020-06-29T13:35:16.184553Z", | |
| "iopub.status.busy": "2020-06-29T13:35:16.184146Z", | |
| "iopub.status.idle": "2020-06-29T13:36:02.050930Z", | |
| "shell.execute_reply": "2020-06-29T13:36:02.050459Z", | |
| "shell.execute_reply.started": "2020-06-29T13:35:16.184500Z" | |
| } | |
| }, | |
| "outputs": [], | |
| "source": [ | |
| "data.sort(key=lambda a: a[0]**2 + a[1]**2)" | |
| ] | |
| }, | |
| { | |
| "cell_type": "code", | |
| "execution_count": 4, | |
| "metadata": { | |
| "execution": { | |
| "iopub.execute_input": "2020-06-29T13:36:02.051933Z", | |
| "iopub.status.busy": "2020-06-29T13:36:02.051607Z", | |
| "iopub.status.idle": "2020-06-29T13:36:02.055496Z", | |
| "shell.execute_reply": "2020-06-29T13:36:02.054720Z", | |
| "shell.execute_reply.started": "2020-06-29T13:36:02.051880Z" | |
| } | |
| }, | |
| "outputs": [], | |
| "source": [ | |
| "n = 20\n", | |
| "pairs = data[:20]" | |
| ] | |
| }, | |
| { | |
| "cell_type": "code", | |
| "execution_count": 5, | |
| "metadata": { | |
| "execution": { | |
| "iopub.execute_input": "2020-06-29T13:36:02.057363Z", | |
| "iopub.status.busy": "2020-06-29T13:36:02.056610Z", | |
| "iopub.status.idle": "2020-06-29T13:36:02.230072Z", | |
| "shell.execute_reply": "2020-06-29T13:36:02.227888Z", | |
| "shell.execute_reply.started": "2020-06-29T13:36:02.057272Z" | |
| } | |
| }, | |
| "outputs": [], | |
| "source": [ | |
| "m = matrix(QQ, 2 + 3*n, 2 + 4*n)\n", | |
| "m[:,n:] = identity_matrix(2+3*n)\n", | |
| "for i, (x, y) in enumerate(pairs):\n", | |
| " m[0,i] = C*x\n", | |
| " m[1,i] = y\n", | |
| " m[0*n+2+i,i] = C*P\n", | |
| " m[1*n+2+i,i] = M\n", | |
| " m[2*n+2+i,i] = P\n", | |
| " \n", | |
| "bounds = [1] * n + [P] * 2\n", | |
| "bounds += [pt for pt, ct in pairs]\n", | |
| "bounds += [P*C / M] * n\n", | |
| "bounds += [ct for pt, ct in pairs]\n", | |
| "assert len(bounds) == m.ncols()" | |
| ] | |
| }, | |
| { | |
| "cell_type": "code", | |
| "execution_count": 6, | |
| "metadata": { | |
| "execution": { | |
| "iopub.execute_input": "2020-06-29T13:36:02.232429Z", | |
| "iopub.status.busy": "2020-06-29T13:36:02.231585Z", | |
| "iopub.status.idle": "2020-06-29T13:36:06.011783Z", | |
| "shell.execute_reply": "2020-06-29T13:36:06.011175Z", | |
| "shell.execute_reply.started": "2020-06-29T13:36:02.232267Z" | |
| } | |
| }, | |
| "outputs": [], | |
| "source": [ | |
| "# scale\n", | |
| "for i, b in enumerate(bounds):\n", | |
| " m.set_column(i, m.column(i)/b)\n", | |
| "# LLL\n", | |
| "m = m.LLL()\n", | |
| "for i, b in enumerate(bounds):\n", | |
| " m.set_column(i, m.column(i)*b)" | |
| ] | |
| }, | |
| { | |
| "cell_type": "code", | |
| "execution_count": 7, | |
| "metadata": { | |
| "execution": { | |
| "iopub.execute_input": "2020-06-29T13:36:06.012649Z", | |
| "iopub.status.busy": "2020-06-29T13:36:06.012340Z", | |
| "iopub.status.idle": "2020-06-29T13:36:06.024758Z", | |
| "shell.execute_reply": "2020-06-29T13:36:06.023676Z", | |
| "shell.execute_reply.started": "2020-06-29T13:36:06.012615Z" | |
| } | |
| }, | |
| "outputs": [ | |
| { | |
| "name": "stdout", | |
| "output_type": "stream", | |
| "text": [ | |
| "Row 11: key recovered: 1df19a439748567 29ad0f3aac513b9\n" | |
| ] | |
| } | |
| ], | |
| "source": [ | |
| "for irow, row in enumerate(m):\n", | |
| " k0, negk1i = row[n:n+2]\n", | |
| " if gcd(negk1i, P) == 1:\n", | |
| " k1 = inverse_mod(-int(negk1i), P)\n", | |
| " for x, y in pairs:\n", | |
| " tmp = x * k0 % P\n", | |
| " tmp = tmp * C % M\n", | |
| " tmp = tmp * k1 % P\n", | |
| " if tmp != y:\n", | |
| " break\n", | |
| " else:\n", | |
| " k0 %= P\n", | |
| " k1 %= P\n", | |
| " print(\"Row %d: key recovered: %x %x\" % (irow, k0, k1))\n", | |
| " break\n" | |
| ] | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "metadata": {}, | |
| "source": [ | |
| "That was fast! Now let's decrypt the flag (recall that a couple of bits is missing, so we have to check a few candidates per block). Also the first reduction modulo $P$ destroys a few bits too." | |
| ] | |
| }, | |
| { | |
| "cell_type": "code", | |
| "execution_count": 8, | |
| "metadata": { | |
| "execution": { | |
| "iopub.execute_input": "2020-06-29T13:36:06.027049Z", | |
| "iopub.status.busy": "2020-06-29T13:36:06.026321Z", | |
| "iopub.status.idle": "2020-06-29T13:36:06.049992Z", | |
| "shell.execute_reply": "2020-06-29T13:36:06.047456Z", | |
| "shell.execute_reply.started": "2020-06-29T13:36:06.026967Z" | |
| } | |
| }, | |
| "outputs": [], | |
| "source": [ | |
| "k0i = inverse_mod(k0, P)\n", | |
| "k1i = inverse_mod(k1, P)\n", | |
| "Ci = inverse_mod(C, M)\n", | |
| "\n", | |
| "def decrypt_block(y): \n", | |
| " for t in range(5):\n", | |
| " v = y * k1i % P\n", | |
| " v += t*P \n", | |
| " if v >= M: continue\n", | |
| " v = v * Ci % M\n", | |
| " if v >= P: continue\n", | |
| " v = v * k0i % P\n", | |
| " while v < 2**64:\n", | |
| " #assert encrypt_block(v) == y\n", | |
| " yield v\n", | |
| " v += P" | |
| ] | |
| }, | |
| { | |
| "cell_type": "code", | |
| "execution_count": 10, | |
| "metadata": { | |
| "execution": { | |
| "iopub.execute_input": "2020-06-29T13:36:06.890826Z", | |
| "iopub.status.busy": "2020-06-29T13:36:06.890285Z", | |
| "iopub.status.idle": "2020-06-29T13:36:07.673682Z", | |
| "shell.execute_reply": "2020-06-29T13:36:07.672780Z", | |
| "shell.execute_reply.started": "2020-06-29T13:36:06.890736Z" | |
| } | |
| }, | |
| "outputs": [ | |
| { | |
| "name": "stdout", | |
| "output_type": "stream", | |
| "text": [ | |
| "b'_p4droNe'\n", | |
| "b'_a5k3d_m'\n", | |
| "b'3_7o_br1'\n", | |
| "b'ng_tHis_'\n", | |
| "b'foR_thE_'\n", | |
| "b'5ign0Ra.'\n", | |
| "b'flag{_p4droNe_a5k3d_m3_7o_br1ng_tHis_foR_thE_5ign0Ra.}'\n" | |
| ] | |
| } | |
| ], | |
| "source": [ | |
| "from struct import unpack, pack\n", | |
| "ct = open(\"res\").read()[-200:].strip().split()[-1]\n", | |
| "ct = bytes.fromhex(ct)\n", | |
| "fmt = '%dQ' % (len(ct)/8)\n", | |
| "ct = unpack(fmt, ct)\n", | |
| "\n", | |
| "flag = b\"\"\n", | |
| "for block in ct:\n", | |
| " for dec in decrypt_block(block):\n", | |
| " dec = pack(\"<Q\", dec)\n", | |
| " if all(0x20 <= v < 127 for v in dec):\n", | |
| " print(dec)\n", | |
| " flag += dec\n", | |
| "\n", | |
| "print(b\"flag{\" + flag + b\"}\")" | |
| ] | |
| } | |
| ], | |
| "metadata": { | |
| "kernelspec": { | |
| "display_name": "SageMath 9.0", | |
| "language": "sage", | |
| "name": "sagemath3" | |
| }, | |
| "language_info": { | |
| "codemirror_mode": { | |
| "name": "ipython", | |
| "version": 3 | |
| }, | |
| "file_extension": ".py", | |
| "mimetype": "text/x-python", | |
| "name": "python", | |
| "nbconvert_exporter": "python", | |
| "pygments_lexer": "ipython3", | |
| "version": "3.7.3" | |
| } | |
| }, | |
| "nbformat": 4, | |
| "nbformat_minor": 4 | |
| } |
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