我在设计一个密码叫达芙妮它接受一个字节的明文返回一个加密的字节和一个修改过的达芙妮当应用于列表时，它应该返回加密字节的列表和处理完所有字节后的Daphne。我不知道如何在不占用O（n）内存的情况下做到这一点。代码如下：

byteEncrypt :: Daphne -> Word8 -> (Daphne,Word8)
byteEncrypt (Daphne key sreg acc) plain = ((Daphne key newsreg newacc),crypt) where
  left = computeLeft key sreg acc
  right = computeRight key sreg acc
  crypt = step plain left right
  newacc = acc+plain
  newsreg = Seq.drop 1 (sreg |> crypt)

byteDecrypt :: Daphne -> Word8 -> (Daphne,Word8)
byteDecrypt (Daphne key sreg acc) crypt = ((Daphne key newsreg newacc),plain) where
  left = computeLeft key sreg acc
  right = computeRight key sreg acc
  plain = invStep crypt left right
  newacc = acc+plain
  newsreg = Seq.drop 1 (sreg |> crypt)

seqEncrypt :: Seq.Seq Word8 -> (Daphne,Seq.Seq Word8) -> (Daphne,Seq.Seq Word8)
seqEncrypt Seq.Empty a = a
seqEncrypt (bs:|>b) (daph,acc) = (daph2,acc2) where
  (daph1,acc1) = seqEncrypt bs (daph,acc)
  (daph2,c) = byteEncrypt daph1 b
  acc2 = acc1 |> c

listEncrypt :: Daphne -> [Word8] -> (Daphne,[Word8])
listEncrypt daph bs = (daph1,toList seq1) where
  (daph1,seq1) = seqEncrypt (Seq.fromList bs) (daph,Seq.Empty)

seqDecrypt :: Seq.Seq Word8 -> (Daphne,Seq.Seq Word8) -> (Daphne,Seq.Seq Word8)
seqDecrypt Seq.Empty a = a
seqDecrypt (bs:|>b) (daph,acc) = (daph2,acc2) where
  (daph1,acc1) = seqDecrypt bs (daph,acc)
  (daph2,c) = byteDecrypt daph1 b
  acc2 = acc1 |> c

listDecrypt :: Daphne -> [Word8] -> (Daphne,[Word8])
listDecrypt daph bs = (daph1,toList seq1) where
  (daph1,seq1) = seqDecrypt (Seq.fromList bs) (daph,Seq.Empty)

字符串
listEncrypt在RAM中有一个列表的副本作为序列，这占用了O（n）RAM，这应该是不必要的。
完整代码在https://github.com/phma/daphne。它还没有准备好生产;我还没破译过更别说别人了我可以通过它运行千兆字节的数据并计算统计数据。