Internetwache – Crypto 50

In this challenge, we are given the ciphertext

♆♀♇♀♆ ♇♇♀♆⊕ ♇♀♇♀♆ ♇♆♇♆⊕ ♆♇♆♇♇ ♀♆♇♆⊕ ♆♇♆♇♆ ♇♆♇♆⊕ ♆♇♇♀♇ ♀♆⊕♇♀ ♆⊕♇♀♆ ⊕♆♇♆♇ ♇♀♆♇♆ ⊕♇♀♇♀ ♆⊕♆♇♆ ♇♆♇♇♀ ♆⊕♆♇♆ ♇♆♇♆⊕ ♆♇♆♇♆ ♇♆⊕♇♀ ♆♇♇♀♆ ♇♆⊕♇♀ ♆♇♆♇♇ ♀♆⊕♆♇ ♆♇♇♀♇ ♀♇♀♆⊕ ♆♇♆♇♇ ♀♆⊕♇♀ ♇♀♆♇♆ ⊕♆♇♇♀ ♆⊕♇♀♆ ♇♇♀♇♀ ♆⊕♆♇♆ ♇♆♇♆♇ ♆⊕♇♀♆ ♇♇♀♆♇ ♆⊕♇♀♆ ♇♆♇♇♀ ♆⊕♆♇♆ ♇♇♀♇♀ ♇♀♆⊕♇ ♀♆♇♆♇ ♆⊕♇♀♇ ♀♇♀♆⊕ ♇♀♇♀♆ ♇♆♇♆⊕ ♆♇♆♇♆ ♇♆⊕♆♇ ♇♀♇♀♆ ⊕♆♇♆♇ ♆♇♆♇♇ ♀♆⊕♇♀ ♇♀♆♇♆ ♇♆⊕♆♇ ♆♇♆♇♇ ♀♇♀♆⊕ ♆♇♆♇♆ ♇♆⊕♆♇ ♇♀♆♇♆ ♇♆⊕♆♇ ♆♇♆♇♆ ♇♆♇♆⊕ ♆♇♇♀♇ ♀♆⊕♇♀ ♇♀♆♇♆ ⊕♆♇♆⊕ ♆♇♆♇♇ ♀♇♀♇♀ ♆⊕♇♀♆ ♇♇♀♆♇ ♆⊕♇♀♇ ♀♆♇♆♇ ♆♇♆⊕♇ ♀♆♇♆⊕ ♇♀♇♀♆ ♇♆♇♆⊕ ♆♇♆♇♆ ♇♆⊕♇♀ ♆♇♆♇♇ ♀♆⊕♆♇ ♆⊕♇♀♇ ♀♇♀♆⊕ ♆♇♇♀♆ ♇♆⊕♆♇ ♇♀♇♀♇ ♀♆⊕♆♇ ♇♀♇♀♆ ♇♆⊕♆♇ ♆♇♇♀♆ ⊕♇♀♆♇ ♆♇♆♇♇ ♀♆⊕♇♀ ♆♇♆♇♆ ♇♇♀♆⊕ ♇♀♆♇♆ ♇♆♇♇♀ ♆⊕♇♀♆ ♇♆♇♆♇ ♇♀♆⊕♇ ♀♆♇♆♇ ♆♇♇♀♆ ⊕♇♀♆♇ ♆♇♆♇♇ ♀

These symbols are astronomical symbols.

☿ = 1
♀ = 2
🜨 = 3
♂ = 4
♃ = 5
♄ = 6
⛢ = 7
♆ = 8
♇ = 9

By substitution, we get

82928 99283 92928 98983 89899 28983 89898 98983 89929 28392 83928 38989 92898 39292 83898 98992 83898 98983 89898 98392 89928 98392 89899 28389 89929 29283 89899 28392 92898 38992 83928 99292 83898 98989 83928 99289 83928 98992 83898 99292 92839 28989 83929 29283 92928 98983 89898 98389 92928 38989 89899 28392 92898 98389 89899 29283 89898 98389 92898 98389 89898 98983 89929 28392 92898 38983 89899 29292 83928 99289 83929 28989 89839 28983 92928 98983 89898 98392 89899 28389 83929 29283 89928 98389 92929 28389 92928 98389 89928 39289 89899 28392 89898 99283 92898 98992 83928 98989 92839 28989 89928 39289 89899 2

which is a TAPIR cipher.

tapir_table = {0 : 'a', 1 : 'e', 2 : 'i', 3 : 'n', 4 : 'r', 50 : 'b', 51 : 'be', 52 : 'c', 53 : 'ch', 54 : 'd', 55 : 'de', 56 : 'f', 57 : 'g', 58 : 'be', 59 : 'h', 60 : 'j', 61 : 'k', 62 : 'l', 63 : 'm', 64 : 'o', 65 : ' ', 67 : 'p', 68 : 'q', 69 : 's', 70 : 't', 71 : 'te', 72 : 'u', 73 : 'un', 74 : 'v', 76 : 'w', 77 : 'x', 78 : 'y', 79 : 'z', 80 : 'wr', 81 : 'Bu', 82 : 'Zi', 83 : ' ', 84 : 'code', 85 : 'rpt', 89 : '.', 90 : ':', 91 : ',', 92 : '-', 93 : '/', 94 : '(', 95 : ')', 96 : '+', 97 : '=', 98 : '\"'}
inv_tapir_table = dict([[v,k] for k,v in tapir_table.items()])

With no OTP-key (or e.g. key = ‘a’), we obtain a morse code which translates to

KZFHWTTFMVSSCX2UGAYHCX2DOZSW4Z3SL5WGE2C7NZSXEORJPU======

This is base32 and translates to

VJ{Neee!_T00q_Cvengr_lbh_ner:)}

Obviously, this is a Caesar cipher giving

IW{Arrr!_G00d_Pirate_you_are:)}

which is the flag.

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