T HERE is an old proverb to the effect that walls have ears. To which— in view of cipher No. 58 published three weeks ago—it might be well to add that they also have tongues.
The above cipher, you may remember, was apparently only the picture of a stone wall. And yet it had very definite information to convey. In fact, it expressed no less than two distinct messages, either of which could be deciphered independently of the other.
One of these messages was in triliteral cipher, the three-unit symbols of which were represented by groups of three stones. The other message was in biliteral cipher, groups of five stones replacing the five-unit symbols of Bacon's biliteral alphabet.
To translate the triliteral cipher, first substitute a's for stones, whose sides are approximately parallel; b's for those having a right taper; and c's for those with a left taper. Then group the resultant series by threes, and substitute from the triliteral alphabet, and you will have the message: SEARCH DILIGENTLY AND YOU WILL FIND.
This does not look so encouraging for one seeking the miser's gold. But let's proceed. The wall has at least said something.
The triliteral cipher depended upon the shapes of the stones. In a somewhat similar manner the biliteral cipher is dependent upon their sizes. In other words, by substituting a's for long stones and b's for short ones, and grouping by fives, the resultant symbols convey the message: REMOVE CENTRAL STONE.
Armed with this information we may well suppose that the decipherer will immediately proceed to loosen the central stone in the wall. And that he will find back of it papers or instructions telling just where to look for the hidden gold.
At any rate, the following decipherment of the first range of stones is appended to illustrate the method of translating the remaining six. Observe that the first stone stands for a c in the triliteral cipher, and a h in the biliteral; that the second stone signifies a in both ciphers; and so on, each stone contributing its share to both messages.
S E A R (C) c a a-a b b-a a a-b c c-a b a a a a-a a b a a-a b a R E (M)
When the present alphabets are used, unit substitutes can often be determined by their frequencies. For example, in the Bacon biliteral cipher, calculated on a basis of ten thousand letters of ordinary English text, the a's average 64.1%, and the b's 35.9%. On a similar basis in the present triliteral cipher the a's average 36%; the b's, 36.9%; and the c's, 27.1%. In examples like No. 58, however, either of these ciphers could be solved by simple substitution methods without knowing the alphabetical order.
One object of this cipher was to see how many readers would find both messages. Did some of you get one of them, and thinking you had exhausted the possibilities, leave the other unread? Results on this point should be interesting; and a list of the solutions submitted will accordingly lie published in an early issue.
In embodying two messages in the wall, based on the shapes and sizes of the stones, the limit was by no means reached. By the introduction of further differences in the stones, other ciphers conveying additional messages could be included.
Such a cipher is No. 66, below, where the seven lines of letters represent the colors in the seven courses of stone in the same wall pictured in cipher No. 58; B being used for brown, D for drab, G for gray, P for pink, and Y for yellow. Can you decipher this third message? Try to get it; it may further elucidate the mystery of the hidden fortune!
Last week's ciphers are next in order. Here is the solution to No. 62:
Question: The Berlillon system of anthropometrical identification is based chiefly upon what measurements of the human body ?
Answer: Head length; head width; length of left middle finger; length of left foot; and length of left forearm from elbow to extremity of middle finger.
Each number in cipher No. 63 stood for a letter in the following alphabet; the.first figure indicating the row in being odd or even, and the sum of the two figures indicating the column. Thus the first number of the cryptogram, 41, indicates the EVEN row, and column five, or J. Similarly, 12 is E; 57 is W, and so on. Notice that the ODD row contains only the odd numbered letters of the alphabet, and the EVEN row the even numbered letters:
1 2 3 4 5 6 7 8 9 10 11 12 13 ODD: A C E G I K M O Q S U W Y EVEN: B D F H J L N P R T V X Z
Message: Jewel thieves have been located in down town hotel with loot valued at five thousand dollars.
Now try your hand at this week's ciphers. No. 65 uses two simple substitution alphabets; one for the question, and the other for the answer. If you are looking for some curious and interesting information, dig into this one!
No. 66 is the additional "stone wall" cipher already mentioned. In No. 67 Mr. Walker uses the following four alphabets: (a) the linotype keyboard—an old form of frequency alphabet; (b) the ordinary alphabet; (c) based on the key phrase, "Pack my box with five dozen liquor jugs"; and (d) the standard office typewriter keyboard:
(a) E T A O I N S H R D L U C M F W Y P V B G K Q J X Z (b) A B C D E F G H I J K L M N O P Q R S T U V W X Y Z (c) P A C K M Y B O X W I T H F V E D Z N L Q U R J G S (d) Q W E R T Y U I O P A S D F G H J K L Z X C V B N M
To solve this cipher you must determine the- method of using these alphabets. Can you turn the trick?
CIPHER No. 65.
Question: MTXW MXR WTN CFKPIT "IXVZPNW PDZF?" Answer: UP QZZSWY WTYUCYX XFT- SPN CLY TYSNP QZ AQFSE GH JLYTY CLY AYCCYTE QZ EFEBYWCYX BYT- EQPE JYTY QBYPYX UPX TYUX VK BFVASW QZZSWSUAE VYZQTY VYSPN EYPC QP CQ CLYST XYECSPUCSQP.
CIPHER No. 66.
G Y G D P Y B Y Y P P G B Y Y P B D D Y G B D D P D D P B G P Y G D B Y Y G Y Y Y P B B D P D Y G Y Y B D B B Y P D D P Y G Y Y G B D P G B G Y D Y D Y D P B B B P D D B P D P Y G
CIPHER No. 67 (M. Walker, Akron, Ohio).
AOQJT ZELSL AWTSE UBLMO RHDYN AZDCA UDCID SADEV UWADB LDDGL AUOSG PJBEO THUSW MEMCS DOOZX KIYBE PVJUA VUUMP YZTYD ZNZVS AOHAA AZUQL I