6a0877384d
- Remove undeployed challenges: Phantom_Byte, Cr4cK_w1f1, Lain_Br34kC0r3 V1, Lain_VS_Knights, Lets_All_Love_UART, AETHER_NET, Last_Train_451, Web3/ - Sync 24 solve/ files from main CTF-Espilon repo - Update all READMEs with real CTFd final scores at freeze - Add git-header.png banner - Rewrite README: scoreboard top 10, edition stats (1410 users, 264 boards, 1344 solves), correct freeze date March 26 2026
64 lines
1.5 KiB
Markdown
64 lines
1.5 KiB
Markdown
# ESP Start — Solution
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**Difficulty:** Easy | **Category:** ESP | **Flag:** `ESPILON{st4rt_th3_w1r3}`
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## Overview
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Flash the provided firmware onto an ESP32. On boot, the device outputs an
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XOR-encrypted flag along with the XOR key via UART at 115200 baud.
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## Step 1 — Flash the firmware
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```bash
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esptool.py --chip esp32 --port /dev/ttyUSB0 --baud 460800 write_flash -z \
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0x1000 bootloader.bin \
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0x8000 partition-table.bin \
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0x10000 hello-espilon.bin
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```
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## Step 2 — Read the UART output
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```bash
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screen /dev/ttyUSB0 115200
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# Or:
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minicom -D /dev/ttyUSB0 -b 115200
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```
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The device prints:
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```text
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=== Hello ESP ===
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System ready.
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Encrypted flag: 09 12 19 07 00 0E 07 35 3F 35 7D 3C 38 1E 3D 26 7F 1E 3E 7F 3E 72 34
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XOR Key: 4C 41 49 4E
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```
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## Step 3 — Decrypt the flag
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XOR key is `LAIN` (`4C 41 49 4E`). Apply it cyclically:
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```python
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enc = bytes([0x09,0x12,0x19,0x07,0x00,0x0E,0x07,0x35,
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0x3F,0x35,0x7D,0x3C,0x38,0x1E,0x3D,0x26,
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0x7F,0x1E,0x3E,0x7F,0x3E,0x72,0x34])
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key = b"LAIN"
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flag = bytes(b ^ key[i % len(key)] for i, b in enumerate(enc))
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print(flag.decode())
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# ESPILON{st4rt_th3_w1r3}
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```
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## Key Concepts
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- **ESP32 flashing**: `esptool.py` writes bootloader, partition table, and application at their respective offsets
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- **UART monitoring**: ESP32 default baud rate is 115200, 8N1
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- **XOR cipher**: Simple symmetric cipher — key is broadcast in plaintext here as an intro challenge
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## Flag
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`ESPILON{st4rt_th3_w1r3}`
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## Author
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Eun0us
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