This is challenge 12, the Catalyst challenge. Those who are keen of mind will find their way to a unique reward, and a place of glory. When you find your solution, enter the Passcode at https://intel.ingress.com/intel
C: ptslashsofourtwominushyphenzrotwohyphenaltwoothree
A: rTLLlqZAgNgnYxlFakrMkvDxugedRGbjUAFtyEgeZywqVXtzgEnRxanrAWfJqtKHxeUcwmTgyIaeVONgeAGlHgrTtpMxtyOmoPksyfxCQgaIZFbiQwpRialUMqseQtIkpheBNsNfcTNnMFmeUcgAwsPhkgDrSNQhoGOXjEwsxOQrgMVFrLAomFLSzDnlnOchGbfijvOTcePWIerXvhBsjvGDngeZyXybzsEPyXxzMFdAepIzdlvguWuneBnrCfxrRPtLuGQTeOksaNmmFgzstgUTyiGGVrbRbuTgzrUMcniWsGgnxrLJuAfqKCu:fz
B: StmjkLSrryIIZUEJDuUfsBFBCuEeqiBGwgHNqntTGYPuuECVcpdhNzVGIeCTRTmCXfzNsKrBsnnSutMELFqIqRsczUXBsLZVcxVsAoGCeAsjEtvYdIzuSNtnLOIdsjraPtBeaacvPAgVbNGiwtUgoJBMAdcenTtvYYunbnuIXestUIbNWKrEdeYiCpdYacBuatLXeSqczAQfTPjrrUduYGWqtBaytxAsQdhralsYewqplTAApKRmPIFBRIAgrnXKqbRZRMvCTbHTssJXGWrDuPUhVQlcDEidJXxYqRSrnRwbfqRsMTbnkmKdvjt:jk
Read the puzzle, it says
pt slash so four two minus hyphen zro two hyphen al two o three
According to slash = /, four = 4, two = 2, minus = -, hyphen = —, three = 3to convert strings C, you can get:
pt/so42-—zro2—al2o3
Clearly it is a catalyst.
Pt / SO42− – ZrO2 – Al2O3
Look closely at the string, it has uppercase and lowercase letters.
We replace any lowercase letter = A and any uppercase letter = B then we get:
ABBBAABBABAABAABAAABAABAAAAABBAABBBAABAABAAABBAAABABAAAABBABAABBAABAAABAABAABBBAABBABAABAABAAABAABAAAAABBAABBBAABAABAAABBAAABABAAAABBABAABBABBAABAABAABAAABABBBAABBBABAAABBAABBBABBAABBBABAAABAABAAAAABBAABBBAABAABAAABBAAABABAAAABBABAABBABAABAAAAAABAAABAABAAABBABABBBABAAABAABAAAAABBAABBBAABAABAAABBAAABABAAAABBABAABBA
Then put this in Baconian puzzle solver.
How to decode Baconian Puzzle?
Put the long string-like AAAAABBBB to this site https://www.dcode.fr/bacon-cipher
From https://www.dcode.fr/bacon-cipher, putting the long string, we get:
oneredherringisoneredherringzeroortworedherringsarenoredherring
Split a word and get a sentence:
One red herring is one red herring, zero or two red herrings are no red herring.
It reminds us that odd numbers are useful, even numbers are useless.
Or there is another hint: 1^0=1, 0^0=0, 1^1=0 which is XOR calculation.
Take the string C as the key and solve the string A for Vigenere cipher from this site https://www.dcode.fr/vigenere-cipher (ignoring the case), you can get:
catalysisistheprocessofincreasingtherateofachemicalreactionbyaddingasubstanceknownasacatalystwhichisnotconsumedinthecatalyzedreactionandcancontinuetoactrepeatedlybecauseofthisonlyverysmallamountsofcatalystarerequiredtoalterthereactionrateinprincipleingeneralchemicalreactionsoccurfasterinthepresenceofacatalystbecau:ui
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……
Take the above string as the key, and perform the Beaufort puzzle from this site dcode.fr/beaufort-cipher (ignoring the case) to obtain the string B:
khhrbnarbkklikliliknanahlinakkhhknarbnalihlinaknalikrbhnaklihhrblihnakkrbnakkrbklikkklirbrbrblinaknananaknalilikklikknanakrblinanahknalinahhnanarbknarbhrbnanalinahrbnakhknanarbrbhrbnakklinaknanahrbkkrblinalinaknaklinananahrbrbknarbkkrblilinahrbnakknananahhkklinanahkklikklirblinakkkhrbnakkkhrbnanalinaklihhklihrbhrb:ly
Excluding the trailing colon and the following parts, you can find that the previous parts are all Group 1 (hydrogen and alkali metal) elements, namely:
| 1 2 3 4 5 | Hydrogen, H lithium, Li sodium, Na potassium, K Rubidium, Rb |
If you treat uppercase and lowercase letters as 1 and 0, respectively, A is:
011100110100100100010010000011001110010010001100010100001101001100100010010011100110100100100010010000011001110010010001100010100001101001101100100100100010111001110100011001110110011101000100100000110011100100100011000101000011010011010010000001000100100011010111010001001000001100111001001000110001010000110100110
B is:
100001100011111110100111101000110011000111100111000010111011110110010101000100111101010001110111001010110100100101001100111000001010000011010110001001111000010011000001100011011101001010010010001101000110110001001110010000101000000100000111011011111110001100111101101100111101011011001100110101100100001011000010000
Calculation A XOR ??? = Bcan be obtained when:
111101010111011010110101101011111101010101101011010110110110111010110111010111011011110101010101011010101101010111011101011010101011101010111010101101011010101010110101111010101011010111010110101101110101010101101101010101101011010111010101011010111010101111101010111101110101010111110101111101010101011011110110110
After restoring the case to decoded B, it becomes:
KHHRbNaRbKKLiKLiLiKNaNaHLiNaKKHHKNaRbNaLiHLiNaKNaLiKRbHNaKLiHHRbLiHNaKKRbNaKKRbKLiKKKLiRbRbRbLiNaKNaNaNaKNaLiLiKKLiKKNaNaKRbLiNaNaHKNaLiNaHHNaNaRbKNaRbHRbNaNaLiNaHRbNaKHKNaNaRbRbHRbNaKKLiNaKNaNaHRbKKRbLiNaLiNaKNaKLiNaNaNaHRbRbKNaRbKKRbLiLiNaHRbNaKKNaNaNaHHKKLiNaNaHKKLiKKLiRbLiNaKKKHRbNaKKKHRbNaNaLiNaKLiHHKLiHRbHRb
That is, all elements are written correctly.
So in order H = 1, ... Rb = 5to convert, that is:
41153544242243312344114353212343245134211521344534454244425552343334322442443345233143231133543515332315341433551534423433154452323434233315543544522315344333114423314424425234441534441533234211421515
Do this Polybius puzzle: ( https://www.dcode.fr/polybius-cipher )
| 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 | FROM |
Available:
QEPTIGSLHTASXFHSIVOFEFOUOURTRZWONOMIRTNUHLSHANYPENHEODNZEORONETWMOOHNEYPTWHEOSNATHLTIRWOTEOTENHRAREE
A 100-character string. After every two strings are grouped, the odd and even arrays are respectively formed into a string, and we can get
QEIGHTXFIVEFOURZNORTHLANENODEONEMONETWOSTHIRTEENARPTSLASHSOFOURTWOMINUSHYPHENZROTWOHYPHENALTWOOTHREE
You can see that the second string is the same as the string C, so even lines are really useless.
The odd-numbered lines according EIGHT = 8, FIVE = 5, FOUR = 4, ONE = 1, TWO = 2, THIRTEEN = 13to the conversion, can be obtained:
Q8X54ZNORTHLANENODE1M12S13AR
Divide the words according to the passcode format of the 13AR series.
Q8X54Z Northlane Node 1m12s 13AR
A Google search revealed that Node is an album by the Australian band team Northlane and the title of their third song.
And the 1′12 ″ lyrics of this music are:
You can be the catalyst
After several attempts, I found that the keyword is bethecatalyst, that is, the passcode is:
q8x54zbethecatalyst13ar
So Ingress 13 Archetypes 12 Catalyst solution is
Finally, go to https://intel.ingress.com/intel and put this passcode to get the media.
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