首先观察一下，单个字母里U N Q都出现过，其中尤其是N最多。这个密文应该是篇英文文章吧。在英语句子里以单个字母出现在文章里的，通常是a（做不定冠词），i（主语），这里e，o，u如果有机会出现，必定是属于引用的那种，比如说E is the most common letter in the English alphabet。所以，U N Q代表的多是a i ，既然有3个单独出现的字母，那么其中一个必定代表着（e，o，u）中的任何一个。另外，值得注意的是，文中出现了QSS这个特殊的单词。我们所知道这样构成的单词并不多，如ill，ass，all，too，egg。假如Q是t，那么Q就不能单独出现了，这样不符合英语构词规律。辅音是无法构成任何音节的。所以剩下的可能，Q只能是a或者i或者e。既然是密文，那么QSS中的S不可能是s，所以QSS只能是代表ill，all，egg。而根据字母e的构词法，e很难和一个辅音字母构成单词，如果QSS代表egg，则QY不成立。而根据文中出现的UY来看，Y绝对不是元音字母，也因此，Q不代表e，所以只代表着a和i，QSS只能是代表ill，all。继续观察，文中出现了YW，EW，GW，甚至出现了EW和GW，EW和YW连用的情况，在英语里能够这样连用而且只有2个字母且末字母相同构成的词组，我只想到了to do。显然AN EW不能代表to do，因为N和W不代表同一个字母。而EW应该就是to无疑了，毕竟这是在英语里出现几率很高的词。暂且按这个思路继续下去，可初步确定W=o，E=t。那么WH代表的不是of就是or，其中GW和YW必定有一个是do。另AN EW和EW AN可以换位使用，则AN必定代表一个代词（to的用词习惯有be+adj.+for+sb.+to do sth.,另to sb.充当介词的习惯），这样的词有me，us和it。而N可以单独构词，所以这里淘汰掉us和it，则AN=me。如此可知，A=m，N=e。这样可知，Q和U代表的就只有剩下的a和i了。既然GW和YW必定有一个是do，那么代入AWYE中，得moYt，代入YWAN中得Yome，代入YW得Yo，则Y=s。则YW=so，剩下的GW则=do，推出G=d。将Y=s，N=e，A=m代入YQAN得sQme,而Q又单独出现多次，且不能是sime，只有是a或者i，所以确定Q=a。验证一下，QY=as成立，则QSS=all（QSS只能是代表ill，all。），得S=l。现确定N=e，Q=a，则U必定=i。代入相应结果到QOG得aOd，代入QO得aO，代入WON得oOe,联立可得，O=n（唯一结果）。代入U=i到UE得it，代入到UY得is，可见U=i唯一且成立。将相应结果代入到EMN得tMe，代入到EMUY得tMis，代入到EMQE得tMat，得到M=h。代入到QDN得aDe，代入到AWDN得moDe，代入到QUD得到aiD，联立得D=r（唯一解）。代入到HUOG得Hind，代入到HDWA得Hrom，代入到HWD得Hor，代入到HUDYE得Hirst，联立得唯一解H=f。代入到FWDGY得Fords，代入到FUEM得Fith，代入到FQY得Fas，得唯一解F=w。代入到ANYYQZN得messaZe，可知此处message，因此得Z=g。代入到MQPUOZ得到haPing，代入到NPNO得ePen，联立得P=v。代入到NQYX得easX，代入到MXIWEMNYUY得hXpothesis，得X=y。代入到TNQDUOZ得Tearing，代入到TWEM得Toth，代入到TX得Ty，联立得唯一解T=b。代入到CYNG得Csed，代入到TCE得bCt，代入到EDCN得trCe，联立得唯一解C=u。代入到YNOENOLN得sentenLe，代入FMULM得whiLh，解出L=c。代入到DNLWZOUVNG得recogniVed，得V=z。代入IDNGWAUOQENY得Iredominates，代入QLLNIENG得acceIted，解出I=p。代入NKINLE得eKpect，代入NKENOE得eKtent，解出K=x。代入AQDBNG得marBed，代入YEULBAQO得sticBman，解得B=k。
密文中的24个字母明文对应如下：W=o，E=t，A=m，N=e，Y=s，G=d，Q=a，S=l，O=n，U=i，M=h，D=r，H=f，F=w，Z=g，P=v，X=y，T=b，C=u，L=c，V=z，I=p，K=x，B=k。大写是密文，小写是对应的明文。
对应译文：Having once recognized, however, that the symbols stood for letters, and having applied the rules which guide us in all forms of secret writings, the solution was easy enough. The first message submitted to me was so short that it was impossible for me to do more than to say, with some confidence, that the symbol of "the stickman with both arms extended up in the air "stood for E. As you are aware, E is the most common letter in the English alphabet, and it predominates to so marked an extent that even in a short sentence one would expect to find it most often. Out of fifteen symbols in the first message, four were the same, so it was reasonable to set this down as E. It is true that in some cases the figure was bearing a flag, and in some cases not but it was probable, from the way in which the flags were distributed, that
they were used to break the sentence up into words. I accepted this as a hypothesis, and noted that E was represented by "the stickman with
both arms extended up in the air"。
密文中的24个字母明文对应如下：W=o，E=t，A=m，N=e，Y=s，G=d，Q=a，S=l，O=n，U=i，M=h，D=r，H=f，F=w，Z=g，P=v，X=y，T=b，C=u，L=c，V=z，I=p，K=x，B=k。大写是密文，小写是对应的明文。
对应译文：Having once recognized, however, that the symbols stood for letters, and having applied the rules which guide us in all forms of secret writings, the solution was easy enough. The first message submitted to me was so short that it was impossible for me to do more than to say, with some confidence, that the symbol of "the stickman with both arms extended up in the air "stood for E. As you are aware, E is the most common letter in the English alphabet, and it predominates to so marked an extent that even in a short sentence one would expect to find it most often. Out of fifteen symbols in the first message, four were the same, so it was reasonable to set this down as E. It is true that in some cases the figure was bearing a flag, and in some cases not but it was probable, from the way in which the flags were distributed, that
they were used to break the sentence up into words. I accepted this as a hypothesis, and noted that E was represented by "the stickman with
both arms extended up in the air"。
