I've been playing NetHack and digging into its internals. In the end of the /usr/lib/games/nethack/nhdat file I've found such strings:
Ujaq0rc}(dicp(inw${xnwhl0ocim0xmqz0emc(Cqw`{0um$`qwg$i0bmkd0qgp&
...
Ujaq0rc}(dicp(inw${xnwhl0ogrmb!qhmuq"jmqs"mffhqmj|d"va~f"szqhvl{>
Ujaq0rc}(dicp(inw${xnwhl0ogrmb!vvq0um$du`ta(dig$leoeag~!um|x!c$jqf"kn0fgi{>
Ujaq0rc}(dicp(inw${xnwhl0sgigfd"}ges"ez}np$jugmvm0rkp|yoe$g~!c$|xsmjm>
...
Vg$`qwg$fuv"siir"kn0egpmsukjo0upaisigvq>/,[WO^][WO^][WO^][W
Vgp(dnuadc!oecu!evmqu"smqqmj{1^][WO^][WO^][WO^][WO^][WO^][W
Zakg{hgY(Gicp(q!rm|i-"}ge!aef~nv$zu`f$ad ][WO^][WO^][WO^][W
My gut says, this is a simple encryption, maybe XOR-ing, maybe 5-byte key (you see repeating pattern of 5-character length: "[WO^]").
I'll try simulated annealing to find XOR key, again:
state/key: 00000000: 01 02 04 08 10 ..... decrypted: 00000000: 59 6F 75 20 61 72 65 20 74 68 65 20 6F 6E 65 20 You are the one 00000010: 6D 69 6C 6C 69 6F 6E 74 68 20 76 69 73 69 74 6F millionth visito 00000020: 72 20 74 6F 20 74 68 69 73 20 70 6C 61 63 65 21 r to this place! 00000030: 20 20 50 6C 65 61 73 65 20 77 61 69 74 20 32 30 Please wait 20 00000040: 30 20 74 75 72 6E 73 20 66 6F 72 20 79 6F 75 72 0 turns for your 00000050: 20 77 61 6E 64 20 6F 66 20 77 69 73 68 69 6E 67 wand of wishing 00000060: 2E 0E
Correct key found. As we supposed, this is English text. For other languages, other statistical features are to be used instead.
Of course, brute-force is not feasible.
All the files, including last version of my decrypting utility.