從 2010 年開始，東京大学情報理工学系研究科在招生海報都會放上一段藏了訊息的 binary。去年看到這個新聞之後就多留了一份心。前幾個月研究所都在招生，就想到今年的小遊戲也差不多該出了。果然今年他們又延續了海報用和服櫻花妹附上 binary 謎題的這個優良傳統。
拿到海報圖檔之後就開工啦！
首先用 OCR 和人眼把圖檔裡的 01 抽出來，得到
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0010101100101101001011010010110100101101001010110000101001111100010010010111000001110111011001110111110000 0010100111110001101111001011100010111001110101011111000100010101000110010010000100100100100000010010110100 1110010011110101001000001010011111000110000101100011011101000110111101111100011111000111110001111100011111 0000100000011111000111110001111100011111000000101001111100011101010111001101110111001000010111110001010110 0101011001010110010101100010000001010110010101100101011001010110000010100010101100101101001011010010110100 1011010010101100101101001011010010110100101101001010110010110100101101001011010010110100101011001011010010 1101001011010010110100101011000010100111110000100000011101010110100101101100011111000110111100100000001000 0001110100011111000110100001110111001110100010000001111100011000010110101100101110011110010111110000101101 0011111001010011000010100111110000100000011001100111010000100000011111000101001100101101011101000110111101 1111000110111001110111001011110111001001111100010111110010111101100001011011110111110000101101001111100101 0100000010100111110001110011011000110111100101110011011111000110100101100001011010010010000001111100011011 1100101110011011110010000001111100010111110111010001110100011011000111110000101101001111100101010100001010 0111110001110100011101010111010101100101011111000110111001101110011011100010111001111100011011110110101001 1010110110100101111100011011000110101100101110001000000111110000101101001111100101011000101100010110010000 1010001010110010110100101101001011010010110100101011001011010010110100101101001011010010101100101101001011 0100101101001011010010101100101101001011010010110100101101001010110000101001111100010001110101010001100001 0110010001111100000010100111110001101000001011110111001101100101011111000000101001111100011011000110100000 1000000111001001111100000010100111110001000011010100110111000001000110011111000000101000101011001011010010 1101001011010010110100101011000010100000101001010010011011110111010001100001011101000110010100100000011101 0001101000011010010111001100100000011000110111010101100010011001010010000001100010011110010010000001110010 0110010101110000011001010110000101110100011010010110111001100111000010100101010000101100010010000010110001 0001010010110001010101001011000100111000101100010010010010110001010110001011000100010100101100010100100010 1100010100110010110001001001001011000101010000101100010110010010110001001111001011000100011000101100010101 0000101100010011110010110001001011001011000101100100101100010011110000101000110010001100000011000101011110 0101111000110011001000000111010001101001011011010110010101110011001011100000101000001010011000010101111001 0111100110001000100000011010010111001100100000011101010111000000101101011000010111001001110010011011110111 01110010000001101110011011110111010001100001011101000110100101101111011011100010111000001010 |
一共是 2848 個 0 或 1。
2848 可以被 8 整除，直接把這東西轉換成 binary data 試試，果然得到一份題目的文字檔。
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+----+ |Ipwg| |o..u|EFHI KNOR |acto||||| |||| |usw!|VVVV VVVV +----+----+----+----+ | uil|o t|hw: |ak.y|->S | ft |S-to|nw/r|_/ao|->T |scys|iai |o.o |_ttl|->U |tuue|nnn.|ojki|lk. |->V,Y +----+----+----+----+ |GTad| |h/se| |lh r| |CSpF| +----+ Rotate this cube by repeating T,H,E,U,N,I,V,E,R,S,I,T,Y,O,F,T,O,K,Y,O 201^^3 times. a^^b is up-arrow notation. |
看來是個 4x4x4 的魔術方塊（展開圖），而 E, F, H, I 等等則各是代表轉動魔術方塊的其中一層。接下來我們要做的就是依照他的指示，重複 THEUNIVERSITYOFTOKYO 這個動作 201^^3 次。問題來了，所謂的 201^^3 到底是多少呢？查了一下 Wiki，原來 up-arrow notation 是這個意思。
也就是說，201^^3 代表
\[201 \uparrow\uparrow 3 = 201^{201^{201}}\]
是個超級大的數字！
看到這麼大的數字，第一個想法就是：絕對不會需要轉這麼多次！
首先，就算 4x4x4 的魔術方塊能夠轉出所有的組合（事實上沒辦法），頂多也就
\[(4 \times 4 \times 6)! = 96!\]
這麼多的組合方式。而且，
\[96! < 201^{201} = 201 \uparrow\uparrow 2 < 201 \uparrow\uparrow 3\]
根據鴿籠原理，在不斷重複 THEUNIVERSITYOFTOKYO 這個動作的情況下，因為 \(96! < 201 \uparrow\uparrow 3\)，過程中必然會有轉到之前轉出過組合的狀況（做了多餘的事情）。所以關鍵在於，如何找出方法，轉個少少次，就有轉 \(201 \uparrow\uparrow 3\) 次的效果。
在做更進一步的事情之前，還有一個問題，那就是如何用程式來計算所謂的“旋轉魔術方塊”這件事。
假如我們把魔術方塊上的每個小面都標上編號，所謂的“旋轉魔術方塊”這件事情，就轉變成了“位置與編號的重新對應”。對程式來說，就是決定“新狀態的某個位置根據舊狀態來看，新的編號應該是多少”。舉個例子來說，如果把一個位置代碼和編號一樣的狀態叫做 identity：
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+-----------+ | 0 1 2 3| | 4 5 6 7| | 8 9 10 11| |12 13 14 15| +-----------+-----------+-----------+-----------+ |16 17 18 19|20 21 22 23|24 25 26 27|28 29 30 31| |32 33 34 35|36 37 38 39|40 41 42 43|44 45 46 47| |48 49 50 51|52 53 54 55|56 57 58 59|60 61 62 63| |64 65 66 67|68 69 70 71|72 73 74 75|76 77 78 79| +-----------+-----------+-----------+-----------+ |80 81 82 83| |84 85 86 87| |88 89 90 91| |92 93 94 95| +-----------+ |
那 identity 經過 E 這個動作之後，會變成這樣：
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+-----------+ | 0 1 2 3| | 4 5 6 7| | 8 9 10 11| |79 63 47 31| +-----------+-----------+-----------+-----------+ |64 48 32 16|12 21 22 23|24 25 26 27|28 29 30 80| |65 49 33 17|13 37 38 39|40 41 42 43|44 45 46 81| |66 50 34 18|14 53 54 55|56 57 58 59|60 61 62 82| |67 51 35 19|15 69 70 71|72 73 74 75|76 77 78 83| +-----------+-----------+-----------+-----------+ |68 52 36 20| |84 85 86 87| |88 89 90 91| |92 93 94 95| +-----------+ |
只要建立個對照表，讓程式知道如何根據舊狀態建立新狀態就好了。 轉換了問題之後，程式的撰寫就變得簡單了。為了方便起見，我是把它寫成 Python 的一個小 module。其中包含了一個 rotate 的函數，還有一個代表 THEUNIVERSITYOFTOKYO 這個動作的 M。
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# identity element identity = range(4*4*6) rotate_maps = dict() rotate_maps['E'] = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 79, 63, 47, 31, 64, 48, 32, 16, 12, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 80, 65, 49, 33, 17, 13, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 81, 66, 50, 34, 18, 14, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 82, 67, 51, 35, 19, 15, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 83, 68, 52, 36, 20, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, ] rotate_maps['F'] = [ 0, 1, 2, 3, 4, 5, 6, 7, 78, 62, 46, 30, 12, 13, 14, 15, 16, 17, 18, 19, 20, 8, 22, 23, 24, 25, 26, 27, 28, 29, 84, 31, 32, 33, 34, 35, 36, 9, 38, 39, 40, 41, 42, 43, 44, 45, 85, 47, 48, 49, 50, 51, 52, 10, 54, 55, 56, 57, 58, 59, 60, 61, 86, 63, 64, 65, 66, 67, 68, 11, 70, 71, 72, 73, 74, 75, 76, 77, 87, 79, 80, 81, 82, 83, 69, 53, 37, 21, 88, 89, 90, 91, 92, 93, 94, 95, ] rotate_maps['H'] = [ 0, 1, 2, 3, 77, 61, 45, 29, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 4, 23, 24, 25, 26, 27, 28, 88, 30, 31, 32, 33, 34, 35, 36, 37, 5, 39, 40, 41, 42, 43, 44, 89, 46, 47, 48, 49, 50, 51, 52, 53, 6, 55, 56, 57, 58, 59, 60, 90, 62, 63, 64, 65, 66, 67, 68, 69, 7, 71, 72, 73, 74, 75, 76, 91, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 70, 54, 38, 22, 92, 93, 94, 95, ] rotate_maps['I'] = [ 76, 60, 44, 28, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 0, 27, 43, 59, 75, 92, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 1, 26, 42, 58, 74, 93, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 2, 25, 41, 57, 73, 94, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 3, 24, 40, 56, 72, 95, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 71, 55, 39, 23, ] rotate_maps['K'] = [ 0, 1, 2, 19, 4, 5, 6, 35, 8, 9, 10, 51, 12, 13, 14, 67, 16, 17, 18, 83, 68, 52, 36, 20, 15, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 87, 69, 53, 37, 21, 11, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 91, 70, 54, 38, 22, 7, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 95, 71, 55, 39, 23, 3, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 72, 84, 85, 86, 56, 88, 89, 90, 40, 92, 93, 94, 24, ] rotate_maps['N'] = [ 0, 1, 18, 3, 4, 5, 34, 7, 8, 9, 50, 11, 12, 13, 66, 15, 16, 17, 82, 19, 20, 21, 22, 23, 24, 14, 26, 27, 28, 29, 30, 31, 32, 33, 86, 35, 36, 37, 38, 39, 40, 10, 42, 43, 44, 45, 46, 47, 48, 49, 90, 51, 52, 53, 54, 55, 56, 6, 58, 59, 60, 61, 62, 63, 64, 65, 94, 67, 68, 69, 70, 71, 72, 2, 74, 75, 76, 77, 78, 79, 80, 81, 73, 83, 84, 85, 57, 87, 88, 89, 41, 91, 92, 93, 25, 95, ] rotate_maps['O'] = [ 0, 17, 2, 3, 4, 33, 6, 7, 8, 49, 10, 11, 12, 65, 14, 15, 16, 81, 18, 19, 20, 21, 22, 23, 24, 25, 13, 27, 28, 29, 30, 31, 32, 85, 34, 35, 36, 37, 38, 39, 40, 41, 9, 43, 44, 45, 46, 47, 48, 89, 50, 51, 52, 53, 54, 55, 56, 57, 5, 59, 60, 61, 62, 63, 64, 93, 66, 67, 68, 69, 70, 71, 72, 73, 1, 75, 76, 77, 78, 79, 80, 74, 82, 83, 84, 58, 86, 87, 88, 42, 90, 91, 92, 26, 94, 95, ] rotate_maps['R'] = [ 16, 1, 2, 3, 32, 5, 6, 7, 48, 9, 10, 11, 64, 13, 14, 15, 80, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 12, 31, 47, 63, 79, 84, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 8, 30, 46, 62, 78, 88, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 4, 29, 45, 61, 77, 92, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 0, 28, 44, 60, 76, 75, 81, 82, 83, 59, 85, 86, 87, 43, 89, 90, 91, 27, 93, 94, 95, ] rotate_maps['S'] = [ 3, 7, 11, 15, 2, 6, 10, 14, 1, 5, 9, 13, 0, 4, 8, 12, 28, 29, 30, 31, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, ] rotate_maps['T'] = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 44, 45, 46, 47, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, ] rotate_maps['U'] = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 60, 61, 62, 63, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, ] rotate_maps['V'] = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 76, 77, 78, 79, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 92, 88, 84, 80, 93, 89, 85, 81, 94, 90, 86, 82, 95, 91, 87, 83, ] rotate_maps['Y'] = rotate_maps['V'] def rotate(vector, map): return [vector[map[i]] for i in identity] M = identity for c in 'THEUNIVERSITYOFTOKYO': M = rotate(M, rotate_maps[c]) |
回到一開始的問題，那我麼要怎麼樣簡化那 \(201 \uparrow\uparrow 3\) 次的轉動呢？這裡就必須考慮魔術方塊的兩個特性：
- 狀態數為有限
- 每個動作都是可逆的（就反轉而已）
根據 finite state machine 的理論，我們可以找得到一個最小的正整數 \(n < 96!\) 使得 \[(\text{operation THEUNIVERSITYOFTOKYO})^n \equiv M^n \equiv \text{identity}\] 二話不說，讓我們寫支程式來找看看
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import rubiks v = rubiks.identity counter = 0 done = False while not done: v = rubiks.rotate(v, rubiks.M) counter += 1 if v == rubiks.identity: done = True print 'n =', counter |
跑完得到 \(n = 1008\)。
也就是說，
\[M^{201 \uparrow\uparrow 3} \equiv M^{(201 \uparrow\uparrow 3)\mod{1008}}\]
計算 \((201 \uparrow\uparrow 3)\mod{1008}\) 的部分我是用 Mathematica 的 PowerMod （拿來做 RSA 也很方便 XD），如果你電腦上沒有 Mathematica，也可以用 Wolfram Alpha 來算。所以我們現在知道
\[M^{201 \uparrow\uparrow 3} \equiv M^{153}\]
只差最後臨門一腳啦！櫻花妹我來了！！！抽出原題目被轉亂的字串，重新轉正。
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import rubiks ciphertext = 'Ipwgo..uactousw! uilo thw: ak.y ft S-tonw/r_/aoscysiai o.o _ttltuuennn.ojkilk. GTadh/selh rCSpF' # PowerMod[201, 201^201, 1008] = 153 v = [c for c in ciphertext] for i in range(153): v = rubiks.rotate(v, rubiks.M) print ''.join(v) |
BINGO!
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Congratulations!Follow the link.http://www.is.s.u-tokyo.ac.jp/ist_ueno_yuuka GradSch of IST |
跟著這個連結進去可以聽到一首歌。
最後引用一句 up-arrow notation 發明人的話
Beware of bugs in the above code; I have only proved it correct, not tried it.
— Donald Knuth
我也要說，上面的程式碼雖然在我這裡可以跑，但不保證到其他地方能用。 XD
後記
其實這是一篇測試文。 XD
當初就在想說到底要用什麼樣的主題，才能夠用一篇文章把所有的東西都測過一遍。看來這個主題是挑得不錯，幾乎所有的排版元素和功能都測到了。哈！
喔對了！還有一件事情我一定要補充一下，就是海報和連結裡面的那個女生根本是高中生，請不要以為東大裡面會有這種年輕可愛的美眉。（東大你騙得我好苦啊 QQ）
你好,我是做完題目之後,用答案去搜索到貴站的^^
那個153我沒算出來直接跑答案的,我想請教一下計算那個1008的過程,感謝啦~
沒事打擾啦,我突然想到怎麼算了,打擾了
No worries! 有些地方我的確是跳得比較快，有任何問題都可以提出來討論。而且我也蠻好奇你所謂直接跑答案的方法是什麼，是一直做 THEUNIVERSITYOFTOKYO 然後用人眼看的意思嗎？
是的,我把輸出的格式弄成每一個側面的字母鏈接成一行字符串,然後用逗號隔開,把結果複製到csv文件裡面看哪條包含有意思單詞的OTL
之後參考了您的方法之後,就把第一行生成的字符串和之後的每一行做判斷,的確會得出n=1008會重複的現象,現在想起來真是太笨了(黑線
了解，直接看也不失為一個方法。話說壓根沒想到會有人研究這篇文章，看到讀者有收穫真的很開心。 ^_^