word puzzle AI JaneStreet Python

JaneStreet:Regional Artwork的解谜笔记

Posted by NoPanic on Wed, Jun 3, 2026

Regional Artwork

Regional Artwork 题图

这道题的标题是 Regional Artwork。题面给了 14 条线索，并说：

The answers to these clues need sorting! Move them into their proper locations, then use those moves to populate the final clue, below.

也就是说，我们需要先解出 14 个答案，再把它们“排序”到正确位置，最后用“移动”过程来提取底部的最终线索。

1. 第一个突破口：AI 不只是答案，也是提示

第三条线索是：

arguably, the story of our time (and a hint to answering all the other clues…)

答案是 AI。这既是 “the story of our time” 的答案，也是在提示其它线索的答案形态：

所有答案都应该是 A 开头、I 结尾。

因此，解每一条线索时，A…I 就成了一个很强的校验条件。很多答案本来可能有多种说法，但只有符合 A 开头、I 结尾的答案才会被保留下来。

2. 逐条解出答案

题面共有 14 条线索。因为所有答案去掉空格、标点后，长度恰好是 2 到 15 各一个，所以我把原始线索顺序标成位置 2, 3, 4, …, 15。第一条线索在原位置 2，第二条在原位置 3，以此类推。

原位置 线索 答案 解释
2 frequent Frazier foe ALI Frazier 指拳王 Joe Frazier；他的著名宿敌是 Muhammad Ali。
3 type of magazine you might start getting in your 20’s ALUMNI 大学毕业后二十多岁开始收到母校寄来的 alumni magazine，即校友杂志。
4 arguably, the story of our time AI 人工智能 AI；同时提示其它答案都是 A…I。
5 stage name of a famous skeptic (with “The”) AMAZING RANDI James Randi 的艺名是 The Amazing Randi；填答案时不取 The。
6 Oscar-winning role from the 80’s ANTONIO SALIERI 电影《Amadeus》中 F. Murray Abraham 饰演 Antonio Salieri 并获奥斯卡最佳男主角。
7 often-inset pair, for short AK/HI 美国地图上 Alaska 和 Hawaii 常以 inset 小图显示；州缩写是 AK 和 HI。
8 microwaveable snack brand that replaced a refrigerated product ACT II Act II 是微波爆米花品牌，接替早期需要冷藏的 Act I 产品。
9 where to find “take thou this vial” and “double, double toil and trouble” ACT IV, SCENE I 两句分别出自《罗密欧与朱丽叶》和《麦克白》，共同位置是第四幕第一场。
10 you might break it in Kabul AFGHANI Kabul 是阿富汗首都；afghani 是阿富汗货币。英语里 “break a bill” 是把钞票换开。
11 eponym of multiple continents AMERIGO VESPUCCI America 得名于 Amerigo Vespucci；North America 和 South America 都可算以他为名。
12 onetime Outkast output AQUEMINI Aquemini 是 OutKast 的专辑。
13 notable 2026 launch ARTEMIS II 按题面设定，这指 NASA 的 Artemis II 任务。
14 the present, in the past ANNO DOMINI Anno Domini，即 AD，是过去常用来表示当前纪元年份的写法。
15 our nearest neighbors, in some sense ALPHA CENTAURI Alpha Centauri 是离太阳系最近的恒星系统之一，可说是“某种意义上的最近邻居”。

到这里，每个答案都满足 A 开头、I 结尾。

3. 排序规则：按答案长度放回正确位置

接下来检查答案长度。计数时忽略空格、逗号、斜杠等，只数英文字母；罗马数字中的 I、V 等照常算字母。

答案 计数形式 长度，也就是正确位置
AI AI 2
ALI ALI 3
AK/HI AKHI 4
ACT II ACTII 5
ALUMNI ALUMNI 6
AFGHANI AFGHANI 7
AQUEMINI AQUEMINI 8
ARTEMIS II ARTEMISII 9
ANNO DOMINI ANNODOMINI 10
ACT IV, SCENE I ACTIVSCENEI 11
AMAZING RANDI AMAZINGRANDI 12
ALPHA CENTAURI ALPHACENTAURI 13
ANTONIO SALIERI ANTONIOSALIERI 14
AMERIGO VESPUCCI AMERIGOVESPUCCI 15

这说明 “proper locations” 就是：

长度为 2 的答案放到 2 号位，长度为 3 的答案放到 3 号位，……长度为 15 的答案放到 15 号位。

题面还特别在 “eponym of multiple continents” 后面用红字说：

and the answer that should come last after sorting

这也正好验证了 AMERIGO VESPUCCI：去掉空格后是 15 个字母，所以排序后应当排在最后。

4. “Use those moves”：用移动距离提取字母

现在每个答案都有两个位置：

  • 原位置：它在题面中的位置，标为 2 到 15。
  • 正确位置：它的字母长度，也就是 2 到 15。

“Use those moves” 的读法是：

一个答案从原位置移动到正确位置，移动了几格，就取这个答案的第几个字母。

例如 AFGHANI：

  • 原位置：10
  • 正确位置：7
  • 移动距离：3
  • AFGHANI 的第 3 个字母是 G

完整提取表如下：

正确位置 移来的答案 原位置 移动距离 取出的字母
2 AI 4 2 I
3 ALI 2 1 A
4 AK/HI 7 3 H
5 ACT II 8 3 T
6 ALUMNI 3 3 U
7 AFGHANI 10 3 G
8 AQUEMINI 12 4 E
9 ARTEMIS II 13 4 E
10 ANNO DOMINI 14 4 O
11 ACT IV, SCENE I 9 2 C
12 AMAZING RANDI 5 7 G
13 ALPHA CENTAURI 15 2 L
14 ANTONIO SALIERI 6 8 S
15 AMERIGO VESPUCCI 11 4 R

于是每个位置 2 到 15 都对应了一个字母。

5. 底部数字读出最终线索

底部空格下方的数字是：

17 8 10 15 12 9 / 13 6 11 3 14 / 4 2 5

把上一步得到的字母按这些位置读出：

17  8  10 15 12 9    13 6  11 3  14    4  2 5
2G  E  O  R  G  E    L  U  C  A  S     H  I T

得到最终线索：

GEORGE LUCAS HIT

6. 最终答案：AMERICAN GRAFFITI

“George Lucas hit” 指向 George Lucas 的电影：

AMERICAN GRAFFITI

它还和题目标题 Regional Artwork 对上：

  • American 可以理解为一种 regional / 地域性的描述。
  • Graffiti 是一种 artwork / 艺术作品。

更漂亮的是，它也延续了整题的结构：

1AMERICAN GRAFFITI -> AMERICANGRAFFITI

去掉空格后长度是 16，并且仍然是 A 开头、I 结尾。前面 14 个答案正好覆盖长度 2 到 15，最终答案成为自然的第 16 项。

7. 用 Python 复核提取过程

下面这个脚本把答案、长度、移动距离和最终提取全部跑一遍。它不依赖第三方库，可以直接用 Python 3 运行。

  1#!/usr/bin/env python3
  2"""Reproduce the extraction for the puzzle 'Regional Artwork'."""
  3
  4from __future__ import annotations
  5
  6from dataclasses import dataclass
  7import re
  8
  9
 10@dataclass(frozen=True)
 11class Entry:
 12    original_position: int
 13    clue: str
 14    answer: str
 15
 16
 17def clean(text: str) -> str:
 18    """Count only letters; ignore spaces, commas, slashes, apostrophes, etc."""
 19    return re.sub(r"[^A-Z]", "", text.upper())
 20
 21
 22entries = [
 23    Entry(2, "frequent Frazier foe", "ALI"),
 24    Entry(3, "type of magazine you might start getting in your 20's", "ALUMNI"),
 25    Entry(4, "arguably, the story of our time", "AI"),
 26    Entry(5, "stage name of a famous skeptic (with 'The')", "AMAZING RANDI"),
 27    Entry(6, "Oscar-winning role from the 80's", "ANTONIO SALIERI"),
 28    Entry(7, "often-inset pair, for short", "AK/HI"),
 29    Entry(8, "microwaveable snack brand that replaced a refrigerated product", "ACT II"),
 30    Entry(9, "where to find 'take thou this vial' and 'double, double toil and trouble'", "ACT IV, SCENE I"),
 31    Entry(10, "you might break it in Kabul", "AFGHANI"),
 32    Entry(11, "eponym of multiple continents", "AMERIGO VESPUCCI"),
 33    Entry(12, "onetime Outkast output", "AQUEMINI"),
 34    Entry(13, "notable 2026 launch", "ARTEMIS II"),
 35    Entry(14, "the present, in the past", "ANNO DOMINI"),
 36    Entry(15, "our nearest neighbors, in some sense", "ALPHA CENTAURI"),
 37]
 38
 39# Numbers printed under the final blanks in the image.
 40extraction_order = [7, 8, 10, 15, 12, 9, 13, 6, 11, 3, 14, 4, 2, 5]
 41final_group_lengths = [6, 5, 3]
 42
 43rows: list[dict[str, object]] = []
 44for entry in entries:
 45    normalized = clean(entry.answer)
 46    target_position = len(normalized)
 47    move_count = abs(entry.original_position - target_position)
 48    if move_count == 0:
 49        raise ValueError(f"{entry.answer!r} did not move; extraction would be ambiguous")
 50    extracted_letter = normalized[move_count - 1]
 51    rows.append(
 52        {
 53            "target": target_position,
 54            "original": entry.original_position,
 55            "answer": entry.answer,
 56            "normalized": normalized,
 57            "move": move_count,
 58            "letter": extracted_letter,
 59        }
 60    )
 61
 62lengths = sorted(row["target"] for row in rows)
 63assert lengths == list(range(2, 16)), lengths
 64assert all(clean(entry.answer).startswith("A") and clean(entry.answer).endswith("I") for entry in entries)
 65
 66by_target = {int(row["target"]): row for row in rows}
 67raw_final_clue = "".join(str(by_target[number]["letter"]) for number in extraction_order)
 68
 69chunks: list[str] = []
 70cursor = 0
 71for group_length in final_group_lengths:
 72    chunks.append(raw_final_clue[cursor : cursor + group_length])
 73    cursor += group_length
 74final_clue = " ".join(chunks)
 75
 76sorted_rows = sorted(rows, key=lambda item: int(item["target"]))
 77columns = [
 78    ("target", "target", ">"),
 79    ("original", "original", ">"),
 80    ("answer", "answer", "<"),
 81    ("normalized", "normalized", "<"),
 82    ("move", "move", ">"),
 83    ("letter", "letter", "<"),
 84]
 85widths = {
 86    key: max(len(label), *(len(str(row[key])) for row in sorted_rows))
 87    for key, label, _ in columns
 88}
 89
 90
 91def format_table_row(values: dict[str, object] | None = None) -> str:
 92    cells: list[str] = []
 93    for key, label, alignment in columns:
 94        value = label if values is None else values[key]
 95        direction = "<" if values is None else alignment
 96        cells.append(f"{str(value):{direction}{widths[key]}}")
 97    return " | ".join(cells).rstrip()
 98
 99
100print("Sorted extraction table:")
101print(format_table_row())
102print(" | ".join("-" * widths[key] for key, _, _ in columns))
103for row in sorted_rows:
104    print(format_table_row(row))
105
106print()
107print("Bottom-number order:", extraction_order)
108print("Final clue:", final_clue)
109
110final_answer = "AMERICAN GRAFFITI"
111final_answer_normalized = clean(final_answer)
112print(
113    "Final answer check:",
114    final_answer,
115    f"=> {final_answer_normalized}, length={len(final_answer_normalized)},",
116    f"A...I={final_answer_normalized.startswith('A') and final_answer_normalized.endswith('I')}",
117)

运行结果的关键部分是：

 1Sorted extraction table:
 2target | original | answer           | normalized      | move | letter
 3------ | -------- | ---------------- | --------------- | ---- | ------
 4     2 |        4 | AI               | AI              |    2 | I
 5     3 |        2 | ALI              | ALI             |    1 | A
 6     4 |        7 | AK/HI            | AKHI            |    3 | H
 7     5 |        8 | ACT II           | ACTII           |    3 | T
 8     6 |        3 | ALUMNI           | ALUMNI          |    3 | U
 9     7 |       10 | AFGHANI          | AFGHANI         |    3 | G
10     8 |       12 | AQUEMINI         | AQUEMINI        |    4 | E
11     9 |       13 | ARTEMIS II       | ARTEMISII       |    4 | E
12    10 |       14 | ANNO DOMINI      | ANNODOMINI      |    4 | O
13    11 |        9 | ACT IV, SCENE I  | ACTIVSCENEI     |    2 | C
14    12 |        5 | AMAZING RANDI    | AMAZINGRANDI    |    7 | G
15    13 |       15 | ALPHA CENTAURI   | ALPHACENTAURI   |    2 | L
16    14 |        6 | ANTONIO SALIERI  | ANTONIOSALIERI  |    8 | S
17    15 |       11 | AMERIGO VESPUCCI | AMERIGOVESPUCCI |    4 | R
18
19Bottom-number order: [7, 8, 10, 15, 12, 9, 13, 6, 11, 3, 14, 4, 2, 5]
20Final clue: GEORGE LUCAS HIT
21Final answer check: AMERICAN GRAFFITI => AMERICANGRAFFITI, length=16, A...I=True

8. 总结

这题的解题链条是：

 1AI 提示所有答案都是 A...I
 2        ↓
 3解出 14 个 A 开头、I 结尾的答案
 4        ↓
 5发现答案长度正好是 2 到 15
 6        ↓
 7按长度排序到正确位置
 8        ↓
 9用每个答案移动的距离提取字母
10        ↓
11读出 GEORGE LUCAS HIT
12        ↓
13得到 AMERICAN GRAFFITI

最终答案：

AMERICAN GRAFFITI

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