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Ingress 13 Archetypes 12: Catalyst

还有这种操作?

About 13 Archetypes

13 ARCHETYPES CHALLENGE

这是一个长达 13 周的解谜挑战,北京时间 12 月 10 日凌晨 01:00 起, Niantic 会在其论坛上发布一个谜题,然后让玩家解谜,结果会是一个 Passcode ,兑换后可获得 Media 一个。

Catalyst

C: ptslashsofourtwominushyphenzrotwohyphenaltwoothree
A: rTLLlqZAgNgnYxlFakrMkvDxugedRGbjUAFtyEgeZywqVXtzgEnRxanrAWfJqtKHxeUcwmTgyIaeVONgeAGlHgrTtpMxtyOmoPksyfxCQgaIZFbiQwpRialUMqseQtIkpheBNsNfcTNnMFmeUcgAwsPhkgDrSNQhoGOXjEwsxOQrgMVFrLAomFLSzDnlnOchGbfijvOTcePWIerXvhBsjvGDngeZyXybzsEPyXxzMFdAepIzdlvguWuneBnrCfxrRPtLuGQTeOksaNmmFgzstgUTyiGGVrbRbuTgzrUMcniWsGgnxrLJuAfqKCu:fz
B: StmjkLSrryIIZUEJDuUfsBFBCuEeqiBGwgHNqntTGYPuuECVcpdhNzVGIeCTRTmCXfzNsKrBsnnSutMELFqIqRsczUXBsLZVcxVsAoGCeAsjEtvYdIzuSNtnLOIdsjraPtBeaacvPAgVbNGiwtUgoJBMAdcenTtvYYunbnuIXestUIbNWKrEdeYiCpdYacBuatLXeSqczAQfTPjrrUduYGWqtBaytxAsQdhralsYewqplTAApKRmPIFBRIAgrnXKqbRZRMvCTbHTssJXGWrDuPUhVQlcDEidJXxYqRSrnRwbfqRsMTbnkmKdvjt:jk

Decode

C

按照 slash = /, four = 4, two = 2, minus = -, hyphen = —, three = 3来转换字符串 C ,可得:

pt/so42-—zro2—al2o3

Google 搜索了一下,发现是一种催化剂。

Pt / SO42− - ZrO2 - Al2O3

A

如果把字符串 A 的大小写分别转换成 AB ,即为:

ABBBAABBABAABAABAAABAABAAAAABBAABBBAABAABAAABBAAABABAAAABBABAABBAABAAABAABAABBBAABBABAABAABAAABAABAAAAABBAABBBAABAABAAABBAAABABAAAABBABAABBABBAABAABAABAAABABBBAABBBABAAABBAABBBABBAABBBABAAABAABAAAAABBAABBBAABAABAAABBAAABABAAAABBABAABBABAABAAAAAABAAABAABAAABBABABBBABAAABAABAAAAABBAABBBAABAABAAABBAAABABAAAABBABAABBA

再以此来做 Bacon 解谜,即为:

oneredherringisoneredherringzeroortworedherringsarenoredherring

分一下词得到一句话:

One red herring is one red herring, zero or two red herrings are no red herring.

在提醒我们,奇数是有用的、偶数是没用的。
或还有另一种提示:1^0=1, 0^0=0, 1^1=0,即 XOR 计算。

以字符串 C 为 key ,将字符串 A 做 Vigenere 解谜(忽略大小写),可得:

catalysisistheprocessofincreasingtherateofachemicalreactionbyaddingasubstanceknownasacatalystwhichisnotconsumedinthecatalyzedreactionandcancontinuetoactrepeatedlybecauseofthisonlyverysmallamountsofcatalystarerequiredtoalterthereactionrateinprincipleingeneralchemicalreactionsoccurfasterinthepresenceofacatalystbecau:ui

这一段出自英文维基百科 Catalysis词条的第一、二段话,只是把空格去掉了。
Catalysis is the process of increasing the rate of a chemical reaction by adding a substance known as a catalyst, which is not consumed in the catalyzed reaction and can continue to act repeatedly. Because of this, only very small amounts of catalyst are required to alter the reaction rate in principle.
In general, chemical reactions occur faster in the presence of a catalyst becau……

B

以上述字符串为 key ,将字符串 B 做 Beaufort 解谜(忽略大小写),可得:

khhrbnarbkklikliliknanahlinakkhhknarbnalihlinaknalikrbhnaklihhrblihnakkrbnakkrbklikkklirbrbrblinaknananaknalilikklikknanakrblinanahknalinahhnanarbknarbhrbnanalinahrbnakhknanarbrbhrbnakklinaknanahrbkkrblinalinaknaklinananahrbrbknarbkkrblilinahrbnakknananahhkklinanahkklikklirblinakkkhrbnakkkhrbnanalinaklihhklihrbhrb:ly

除去结尾冒号及之后的部分,可以发现前面的部分都是第 1 族(氢和碱金属)元素,即:

1
2
3
4
5
氢, H
锂, Li
钠, Na
钾, K
铷, Rb

如果将大小写字母分别视为 1 和 0 ,即 A 为:

011100110100100100010010000011001110010010001100010100001101001100100010010011100110100100100010010000011001110010010001100010100001101001101100100100100010111001110100011001110110011101000100100000110011100100100011000101000011010011010010000001000100100011010111010001001000001100111001001000110001010000110100110

B 为 :

100001100011111110100111101000110011000111100111000010111011110110010101000100111101010001110111001010110100100101001100111000001010000011010110001001111000010011000001100011011101001010010010001101000110110001001110010000101000000100000111011011111110001100111101101100111101011011001100110101100100001011000010000

计算 A XOR ??? = B 时可得:

111101010111011010110101101011111101010101101011010110110110111010110111010111011011110101010101011010101101010111011101011010101011101010111010101101011010101010110101111010101011010111010110101101110101010101101101010101101011010111010101011010111010101111101010111101110101010111110101111101010101011011110110110

再把大小写还原到解码后的 B 之后,即可变成:

KHHRbNaRbKKLiKLiLiKNaNaHLiNaKKHHKNaRbNaLiHLiNaKNaLiKRbHNaKLiHHRbLiHNaKKRbNaKKRbKLiKKKLiRbRbRbLiNaKNaNaNaKNaLiLiKKLiKKNaNaKRbLiNaNaHKNaLiNaHHNaNaRbKNaRbHRbNaNaLiNaHRbNaKHKNaNaRbRbHRbNaKKLiNaKNaNaHRbKKRbLiNaLiNaKNaKLiNaNaNaHRbRbKNaRbKKRbLiLiNaHRbNaKKNaNaNaHHKKLiNaNaHKKLiKKLiRbLiNaKKKHRbNaKKKHRbNaNaLiNaKLiHHKLiHRbHRb

也就是所有元素字母大小写正确的写法。

那么按顺序 H = 1, ... Rb = 5 来转换,即为:

41153544242243312344114353212343245134211521344534454244425552343334322442443345233143231133543515332315341433551534423433154452323434233315543544522315344333114423314424425234441534441533234211421515

将这一串数字做 Polybius 解谜:

1 2 3 4 5
1 A B C D E
2 F G H I/J K
3 L M N O P
4 Q R S T U
5 V W X Y Z

可得:

QEPTIGSLHTASXFHSIVOFEFOUOURTRZWONOMIRTNUHLSHANYPENHEODNZEORONETWMOOHNEYPTWHEOSNATHLTIRWOTEOTENHRAREE

一个 100 字符的字符串。每两个字符串一组后,再将奇数组和偶数组分别组成一个字符串,可得:

QEIGHTXFIVEFOURZNORTHLANENODEONEMONETWOSTHIRTEENAR
PTSLASHSOFOURTWOMINUSHYPHENZROTWOHYPHENALTWOOTHREE

可以发现第二个字符串和字符串 C 是一样的,所以偶数行确实没用。

将奇数行按照 EIGHT = 8, FIVE = 5, FOUR = 4, ONE = 1, TWO = 2, THIRTEEN = 13 来转换,可得:

Q8X54ZNORTHLANENODE1M12S13AR

按照 13AR 系列的 passcode 格式来分一下词,可得

Q8X54Z Northlane Node 1m12s 13AR

Google 搜索后得知,Node 是 澳大利亚乐队 Nourthlane 的一个专辑,也是其中的第 3 首音乐的标题。

而这首音乐的 1′12″ 的歌词为:

You can be the catalyst

Northlane Node

多次尝试后发现, keyword 为 bethecatalyst,即 passcode 为:

q8x54zbethecatalyst13ar

兑换获得 Media 一枚。

Media

三个单词组成一个 password ,竟然还有这种操作?

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