Chapter IX.

Figure 140

(5) 1 (2) (1) 2 (3)

YT A EL A CSNKZYCC B IUKLQIKT B OO C LE QK C FB DD D DD ZZUNZDZGDZ D BGAGYAGABB E RZPCJMAMZY E GAOFNIVUUS HJN F MVS F HUPSCCGCCWN G QDPKZOPQWKU G CK H SP H J I Q BB I CN LQBQ J PXER J QTPWOQ K KCUTKK KGKBGBK K WETMYTW L RABC L WQQA SQ M EE VF M JM O N D N R O U G O M O P E HGGJ P MBMB OT Q SC YGBGI Q VBVWV OD R EL JXX R XXS S FQH S MGM T BKX T YBK QQ U UD YGOUBK U FIFFVG SU V MV FVY V NVU W G W B RRK X TRR YJ X GM QRNK* Y VEXUQ Y JUTTECP Z EDDDDED G Z Y

(2) 3 (4) (3) 4 (5) (4) 5 (1)

LDDDE A YYKMB YNTY A DOOD A DDPWTCPQD B GDYYKIHJO A B L L B J I C F C RZKKKKKGK C ZGBBGGGBH D MMBMG D PQQTQ OAA D DDR K E K E F E Z UEUUC F KKZVV VXYCX F WENRL F UDDSDXE G DKJGJRK GBN G KNC Z G G H B H U LQ H GF EU I ZZ SBWMV I OKTOO I M J V BMQGG J RWUTO SVS J IFZ T K V GAMBMEFWGF K QCCCCQCCPU INYG K YHXB L UMM L BHR FB L J* SXPKOMPS M DLJIKDLD A M T M IEV N GKA O N K ZZFG N GFBY E O N B O D IJIAAYI O KCQCNPR P P KVD P ZKG LL Q JY T Q H ZZDKYDKD Q KJMUJKUY R G R C LFYJ R XXOY RE S IZ UW S JJ VZVV S MVBG KK T AQ T MIDJY T QAKZZ EEV U ZLS U KHJ U VGZ QUEQVQ V IVZVZF KVVFFJ V ZSSPJS V KQLK W SKIZ W JF W GK RR X FF X X DKZT Y AFAZ QBBAA Y KTOQR ZZ Y BA Z SWVYUVIFI Z YGYNSCNQQ V Z B

Figure 141

Consideration of Alphabet No. 1

Letter: Frequency: LOW-FREQUENCY CONTACTS VARIETY OF CONTACT Left Right Total* Left Right Total* B (v) 7 2 3 5 6 6 12 G (v)(= E?) 11 5 3 8 8 8 16 K (v) 6 2 1 3 5 4 9 Z (c) 7 4 - 4 6 2 8

(*) These observations are not absolute, as in simple substitution.

In Fig. 141 we may see some data and probable conclusions concerning alphabet 1. By frequency alone, the four letters _B_, _G_, _K_, _Z_, of this alphabet might all be vowels. When variety of contact is considered in conjunction with frequency, it is noted that _Z_ shows no variety on its right. And when contacts with low-frequency letters are also considered (from information present on sheets 5 and 2; in the figure, frequencies of 1, 2, or 3 were considered to be low), it is found that in this respect, too, the letter _Z_ stands apart from the others. These observations, usually, are mental, and conclusions for any one alphabet must often be modified by what is seen in other alphabets. It may be found satisfactory to begin by selecting only the most _obvious vowel_, or vowels, in each alphabet, and to circle these, or otherwise indicate them, not only on their own sheets, but also on the two adjacent sheets where they are found again as contact letters. When this has been done, the less obvious vowels may be considered again with an additional “pointer,” whether or not they show too much contact with the more obvious vowels. Fig. 142 shows, for each alphabet, the probable conclusions which would be reached after examining the contact sheets of Fig. 140, and before any confirmation is attempted. The next step in order is that of indicating them on the cryptogram itself, and the examination of long segments in which no vowels have been marked. At this stage, too, the total number of spotted vowels may be computed to find out how much of the expected 40% is still missing. Up to this point we have nothing new, and nothing particularly difficult. Whether or not the subsequent work is to become difficult depends chiefly upon the amount of material per alphabet, though granting that the presence of probable words materially alters the case of the shorter cryptogram.

Figure 142

Conclusions for the Five Alphabets

Alphabet No. Vowels Consonants

1 B G K * Z 2 E D K Q U 3 B G M * V 4 V Z K 5 O C Q

(*) When B and G appear as vowels in two different alphabets, the graphic appearance of these two alphabets (1 and 3) should be given another inspection. It happens they are not the same.

If the most frequent of the spotted vowels in each alphabet can be safely assumed as _e_, the establishment of other vowel-identities can follow the rules of Chapter IX: The high-frequency vowel which practically never touches _e_ is _o_; and the one which follows it is _a_; vowels of lower frequency may precede or follow _e_, but no vowel should touch it very often. And if, in addition, the most frequent of the spotted consonants in a given alphabet can be safely assumed as _t_, then _h_ of the next alphabet will seldom be out of reach. A very material aid here is found in those _repeated digrams_ (and trigrams) whose letters are already labeled as vowels or consonants. We find, for instance, _ZD_, alphabets 1-2, occurring five times, and with both letters already spotted as consonants. This is very likely to represent _th_, especially when further examination shows it continued as a repeated _ZDB_, with _B_ already quite likely to represent the _e_ of alphabet 3. Then the contacts of _D_, alphabet 2, supposed to represent _h_, have also pointed out a probable new vowel, _A_, in alphabet 3. Again, we find _KC_, alphabets 4-5, occurring six times, and with both letters already spotted as consonants — another probable _th_ — followed three times by _G_, alphabet 1, already likely to represent _e_, and twice by _B_, which could thus represent _a_ (the famous English _the_, _tha_). And similarly we might continue with a long demonstration.

* * *

Returning, now, to our mechanical operations: Dealing, as we are, with five different alphabets, it becomes imperative that we keep track of substitutes; otherwise, with all of our numerous trials and erasures, it is almost impossible to know what substitutes have been identified and what substitutes are still available for identification. Also, totally apart from this matter of convenience, we shall probably want these five lists of substitutes for use on future cryptograms. This applies to any series of cipher alphabets, whether or not they are in any way related to one another. But it is very seldom indeed that a series of cipher alphabets used in the same cryptogram will be unrelated alphabets. Nearly always, they will have resulted from the use of a slide, and when this is true, the recovery of alphabets and parts of alphabets enables us to reconstruct the slide. The usual plan for recording substitutes is to lay out a plaintext alphabet in _A B C_ order and then, directly below it, to rule off several rows of cells, one row for each cipher alphabet. Thus, any substitute, identified in any alphabet, may be written directly below its presumed original and on the row which corresponds to its particular cipher alphabet. We sometimes speak of such a set-up as a “key-frame” or “key-skeleton,” though a better name, probably, would be “partial tableau.” (Every row, if completed, will show one cipher alphabet of the kind we saw in the partial tableaux of Fig. 138.) Such a “key-frame” for our present cryptogram can be seen in Fig. 143. At (a) of this figure we have the first tentative identifications. The most frequent vowel in each of the first four cipher alphabets has been assumed as _e_ (in practice, the _O_ of alphabet 5 would also be assumed as _e_). The _ZD_ of alphabets 1-2 and the _KC_ of alphabets 4-5 have both been assumed as _th_, and after each _th_, we are trying one letter as _a_. The five rows of this set-up we may now speak of as “alphabets.” At (b) we are beginning to speculate as to what kind of slide has been used.

Suppose that the cryptogram has been enciphered with a _Type II_ slide. If so, our plaintext alphabet, in the key-frame, is already arranged like the one on the slide; and when this is true, as may be seen by glancing back at Fig. 138, the recovered cipher alphabets will also build up with their letters in exactly the same order as that of the slide, and, in the end, if fully completed, will show a picture of the original sliding alphabet taking five of its possible positions.

Examining the first cipher alphabet of (a), we note that the _lineal_ distance from _B_ to _G_ is 4 positions. If our hypothesis is correct, then the lineal distance _BG_ will have to be 4 positions in all of the other alphabets. The third alphabet contains a _B_; measuring 4 positions to the right of this letter, we find that _G_ of the third alphabet would fall below _i_, and thus would be the substitute for _i_ in the third alphabet. To see whether or not this is likely, we return to the contact sheet, where we find that _G_ has already been spotted as a vowel (see the list in Fig. 142). So far, so good. Then, the first cipher alphabet of (a) shows the lineal distance _BZ_ as 19 positions. Returning to the third alphabet, and measuring 19 positions to the right of _B_, we find that _Z_, in this alphabet, would fall below _x_. Examination of the contact chart shows that _Z_ has not been used in the third alphabet, making it satisfactory as the substitute for _x_. Still good. Again, the third alphabet shows the distance _AB_ as 4 positions. Still pursuing our hypothesis, the first alphabet must also contain an _A_ standing 4 positions to the left of _B_. If so, it will fall below _w_, and the frequency of _A_, in the first alphabet is found to be 2, which is satisfactory as that of _w_. With _G_ and _Z_ added to alphabet 3, and with _A_ added to alphabet 1, we may now turn our attention to alphabet 4, which contains a _Z_, and, by making similar observations there, we may add to the 4th cipher alphabet the letters _A B G_, and, to the 1st and 3d alphabets, the letter _K_. Thus we arrive at (b) through what is ordinarily referred to as the “symmetry of position” existing among the several cipher alphabets.

Figure 143

(a) 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 1 - B G Z 2 - E D 3 - A B 4 - Z K 5 - C

(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 1 - B G K Z A 2 - E D 3 - A B G K Z 4 - Z A B G K 5 - C

But the second and fifth alphabets cannot yet be combined with the other three, since neither of these contains any letter in common with them, and thus we have no point from which to measure lineal distances. We know, however, that if our hypothesis is correct, the letters _A B G K Z_, in these alphabets also, will be found at exactly the same lineal distances as before. It would be possible to prepare a sort of slide on which these letters, written twice in succession, are spaced as in the other three alphabets, and use this in experimenting with alphabets 2 and 5.

The cryptogram, as we first saw it, showed all substitutions which are possible from (b) of Fig. 143, together with a few _v_-_c_ notations which were listed in Fig. 142 but not further investigated. In case the student cares to complete solution, he might refer to certain precautions mentioned at the beginning of the chapter. Notice, in the last figure, the lineal distance from _G_ to _K_; what letters would you feel inclined to try in the three intervening positions? Or notice the distance _BG_. What letter is very likely to have been taken here for use in the key-word, and where is it likely to stand in that word? If the index-letter was _A_, does it seem possible that the _a_-substitutes could all be selected in advance directly from the contact sheets? Would this be possible if the encipherment process were varied so that an index, selected in the sliding alphabet, were brought to stand below keys in the stationary one? The cryptogram is known to contain the word SUPPOSE, and the period is 5. Is there any room here for _pattern_ methods?

* * *

Our _Type II_ slide, then, unlike the remaining three, builds up automatically in the key-frame, _owing to the simple fact that we are able to set down the plaintext alphabet in the encipherer’s original order_. The method of solution, so far as we know, was first published (1883) by Auguste Kerckhoffs, who seems to have originated the term “symmetry of position.” The invention of the cipher is credited to “a member of the (French) Commission on Military Telegraphy.”

If these parallel cipher alphabets are to be avoided in the key-frame, but still using a _Type II_ slide, General Sacco has suggested that the encipherment process be altered as follows: Let the index-letter and the key-letter both be found in the upper alphabet. Slide the plaintext letter to stand below the index-letter, and use the substitute which will then be standing below the key-letter. This, of course, would have to be letter-by-letter encipherment, and represents one of those rare cases in which a slide is less convenient and rapid than its equivalent tableau. If this tableau be laid out in full, as explained for Beaufort alphabets, it shows, on its 26 rows, 26 cipher alphabets not one of which appears to be at all related to the others. One of these (the one in which index-letter and key-letter are the same) will be the normal alphabet. We may find the original sliding alphabet, however, by looking at _columns_. Such a tableau is exactly equivalent to the _Delastelle tableau_ if the _Z_-alphabet be made the normal one. Delastelle’s tableau was described as follows: Using the mixed alphabet, fill in the tableau by columns, beginning each column with whatever letter, in the mixed alphabet, follows the plaintext letter shown above the column. This causes the final alphabet to come out in _A B C_ order. The Delastelle tableau is not nearly so easy to reconstruct as that of the ordinary _Type II_ slide; the method, however, will be plain enough when we have understood the reconstruction of _Types III_ and _IV_.

* * *

The _Type I_ slide, as pointed out in the beginning, is somewhat out of place in the present chapter; every frequency count will follow the graph of the mixed plaintext alphabet, so that all can be “lined up” by their common pattern. Having letters, and not numbers, the “top” of a frequency count may be anywhere; it is usually best to prepare at least one of the frequency counts of double length in order to effect the alignment. Granting, however, that for some reason the common pattern of the frequency counts has not been recognized, then the method of decryptment would be exactly the same as for any other case of mixed alphabets.

Fig. 144 shows the development of the key-frame in this case. At (a), some substitutes have been correctly identified in each of four cipher alphabets. But long before reaching this stage, the most careless of decryptors must have noticed that the difference between any two cipher alphabets is purely a matter of alphabetical shift. This is particularly visible as between alphabets 3 and 4, where the alphabetical interval is only 1; examination of alphabets 1 and 2 shows that wherever both substitutes are present, their alphabetical difference is 14; and further examination shows that the alphabetical distance from alphabet 2 to alphabet 3 is 17. The use, here, of a Saint-Cyr slide enables us to arrive very quickly at (b). The alphabets of (b) are, of course, secondary cipher alphabets, and the primary one obviously runs in normal order (or, at worst, in a strictly methodized order which is easily obtainable from the normal one). What we still lack, in order to reconstruct the slide, is the mixed plaintext alphabet, and this can be recovered as at (c). Write out the normal alphabet (known to be the original cipher alphabet), then, using any one of the secondary alphabets, place originals above their substitutes wherever these are known. In the given example, all missing letters can be filled in by alphabetical sequence; and even though the index-letter was one of low frequency, and thus was not used in the message, the student should have no trouble whatever in discovering the key-word which governs the four cipher alphabets.

* * *

In considering the reconstruction of the remaining two slides, we shall have to keep clearly in mind the _imaginary tableau_ on which the plaintext alphabet has exactly the order of the one on the slide, so that cipher alphabets, also, have exactly the order of the one on the slide, and are shifted one letter at a time, as in the Vigenère tableau. For one thing, we are going to call some of these alphabets by numbers, or refer to them as odd-numbered and even-numbered alphabets. Thus, with the alphabet we have been using, the _first_ alphabet in the imaginary tableau is position 1 of the slide: _C U L P E R Z Y X W V T_. . . . . . , the _second_ alphabet is position 2 of the slide: _U L P E R Z_. . . . . . . , the _third_ alphabet is position 3 of the slide: _L P E R Z Y_. . . . . , and so on to the _26th_ alphabet, which is the final position of the slide: _A C U L_. . . . . But over and above this, it must be remembered that _the columns of this imaginary tableau are duplicates of the rows_, just as they are in the Vigenère tableau. We do not, of course, recover any of these alphabets in the order mentioned, since our plaintext alphabet of the key-frame must necessarily be arranged in its _a_-_b_-_c_ order. For instance, the _fourth_ alphabet, which, in the imaginary tableau, begins with its key-letter, _P_, and runs in the order _P E R Z Y X W V T_. . . . . , comes out in one of our examples (_Type III_ slide) as _L U P C Y A B D_. . . . . , and in the other (_Type IV_ slide) as _E W Y P V T S_. . . .

Figure 144

The Alphabets from a TYPE I Slide:

(a) 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

1 - L . . J Q . H G F . . . . B A . . R Y X . . . . . . 2 - Z . . . E . . . T . . C Q . O . . . . . B . . . . . 3 - Q P . O V N . . . . I . . G F U . . D C . B A . Y . 4 - R . . P W . . M L . J . . . G . . . . D . . . . . .

(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

1 - L K . J Q I H G F . D O C B A P . R Y X N W V . T . 2 - Z Y . X E W V U T . R C Q P O D . F M L B K J . H . 3 - Q P . O V N M L K . I T H G F U . W D C S B A . Y . 4 - R Q . P W O N M L . J U I H G V . X E D T C B . Z .

(c) Alphabet No. 1:

Plaintext: o n m k . i h g f d b a . u l p e r . y . w v t s . CIPHER (Rearranged): 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

* * *

The _Type III_ slide is, in many respects, the most interesting member of its family. With every alphabet taking exactly the same order (that is, the plaintext alphabet, the key-alphabet, and all cipher alphabets in the imaginary tableau), it parallels the Vigenère in every particular except the order of the 26 letters. It has a corresponding Beaufort form, and a corresponding variant in which complementary keys are based on the order of the mixed alphabet. Its 14th alphabet, like the _N_-alphabet of the Vigenère, is reciprocal throughout. And its first alphabet, like the _A_-alphabet of the Vigenère, is a duplicate of the plaintext alphabet. This was pointed out in connection with the slide of Fig. 138, where key-letter and index-letter were both _C_. There are two ciphers, then (the _Type III_ slide and the Delastelle tableau), in which we are sometimes able to find, among a number of mixed frequency counts, a single one which follows perfectly the graph of the Vigenère _A_-alphabet. Concerning the 14th alphabet, however, we are dealing altogether, here, with a 26-letter alphabet; and some of what follows is being explained on the theory that the number 26 contains no factors other than 2 and 13. If the student will give his careful attention to _reasons_, as well as to methods, he will be able to adjust these methods to alphabets of other lengths, as, for instance, the very common 25-letter alphabet met with in foreign texts. The first alphabet, of course, duplicates the plaintext alphabet regardless of what alphabet-length is being considered, and thus, whenever a _Type III_ slide has been used, we are always in full possession of one of the cipher alphabets.

Figure 145

The Alphabets from a TYPE III Slide:

Behavior of an EVEN-NUMBERED Alphabet

1st Alphabet (Always normal): 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 4th Alphabet: L U P C Y A B D F G H R I J K Z M X N O E Q S T V W

For an EQUIVALENT SLIDE, follow the chain AL-LR-RX-XT-TO......

A L R X T O K H D C P Z W S N J G B U E Y V Q M I F L R X T O K H D C P Z W S N J G B U E Y V Q M I F A (1)........... ═ ═ ═ (2)................. ═══ ═══ ═══ ═══ ═══

To find the ORIGINAL SLIDE from the EQUIVALENT one:

(1) Either take letters constantly at interval 9, which is the interval V-W-X:

R Z Y X W V T S Q O N M K J I H G F D B A C U L P E (R) X W V T S Q O N M K J I H G F D B A C U L P E R Z Y (X)

(2) Or: Spread the letters apart so that the alphabetical sequences K(JI)H, Z(YX)W, etc. are standing at the right interval, always maintaining the alphabet-length, 26, and intertwine. Both alphabets are the same in this slide:

(The interbals A . . L . . R . . X . . T . . O . . K . . H . . D . are always odd, . C . . P . . Z . . W . . S . . N . . J . . G . . B 3, 5, 7, etc.) . . U . . E . . Y . . V . . Q . . M . . I . . F . .

Behavior of an ODD-NUMBERED Alphabet

1st Alphabet: 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 3d Alphabet: U C L A Z B D F G H I E J K M R N Y O Q P S T V W X

1st HALF-CHAIN: 2d HALF-CHAIN: A U P R Y W T Q N K I G D (A) B C L E Z X V S O M J H F (B)

Spread the letters of each half, trying interval 2, interval 4, interval 6, and so on, treating both halves alike, until the intertwining of the two will set up some alphabetical sequences:

. A . U . P . R . Y . W . T . Q . N . K . I . G . D B . C . L . E . Z . X . V . S . O . M . J . H . F .

Now, granting that we have completed the decryptment of a message, we have before us a key-frame in which several cipher alphabets are at least partially recovered. With one alphabet fully known in advance, the recovery of another full alphabet usually enables us very quickly to restore the original slide, or an equivalent slide. The ideal case is that in which we recover one of the even-numbered alphabets (except No. 14). The recovery of an odd-numbered alphabet will, at times, leave us with thirteen possibilities; while the recovery of the 14th alphabet could be useless, provided we have no other information. The method of reconstruction can be followed in Fig. 145.

First, we have the perfect case, one in which an even-numbered alphabet (the 4th) has been recovered in full. We begin by writing this recovered cipher alphabet letter for letter below (or above) the one which is always known to us; thus the two substitutes for _a_ are in a same column, the two substitutes for _b_ are in a same column, the two substitutes for _c_ are in a same column, and so on. The columns themselves are not in their original order, but the two alphabets, throughout, are running parallel, just as they would in the imaginary tableau, and thus the _columnar distance is uniform_ which separates each pair of substitutes; that is, the _vertical_ distances _AL_, _BU_, _CP_, _DC_, _EY_, etc., are all equal in the imaginary tableau. If these be rearranged in such a way that the last letter of each pair is the beginning letter of the next, we have a chain _AL_-_LR_-_RX_-_XT_-_TO_-_OK_. . . . . made up entirely of equal vertical intervals, from which the repeated letters may be dropped: _A L R X T_. . . . . , leaving us a series of 26 letters known to be equally spaced in the columns of the tableau. Then, remembering that the columns of this tableau are duplicates of its rows, we have also a series of 26 letters known to be equally spaced on the rows. That this series of letters, _A L R X T_. . . . , _sliding against itself_, produces exactly the results of the original series, the student may ascertain for himself; also that a number of other _equivalent slides_ can be formed by taking letters of this series at any constant interval which is not divisible by 2 or 13. The total number possible is eleven, of which one was the original. An equivalent slide, of course, is all that we actually need for enciphering and deciphering cryptograms. But where alphabetical sequences existed in the original alphabet, two methods are shown for obtaining it without writing out the entire eleven possibilities: (1) Find, at some constant interval, the letters of an alphabetical, or nearly alphabetical, sequence, as the (reversed) _V W X_ of the figure, standing at interval 9; the taking of all letters at this interval brings back the original order. (2) Find pairs of consecutive letters, as the (reversed) _HK_, _WZ_, _GJ_, which, if all spread apart to the same extent (some odd interval, as 3 of the figure), would then be standing at their normal alphabetical intervals, or nearly so. Lay out the 26 positions, and spread the entire alphabet, maintaining the common interval even after the 26th position is reached. The figure shows this on several rows; in practice, there is only one.

If the recovered alphabet is an odd-numbered one, the same plan is followed, but results in a chain of only 13 letters; it is necessary to begin with some other letter, not included among the first 13, and form another 13-letter chain. Having absolutely no additional information, we cannot combine these two halves with certainty unless the original alphabet contained some alphabetical, or nearly alphabetical sequences. Presuming that it did, the method ordinarily described for combining the two halves is that of the figure. Spread the letters of the two halves (plan 2 of the preceding case), treating both halves exactly alike, until a point is found at which the two halves can be intercombined to show alphabetical sequences.

For this case, however, George C. Lamb, the author of Chapter X, suggests another plan which would seem to be more direct and less troublesome than the standard one. Lamb, incidentally, is to be congratulated here for his entirely new observation: If the two half-chains can be properly adjusted with reference to each other, _each pair of letters, regardless of the order, will be a digram belonging to the original mixed alphabet_. The reason for this division into halves, of the odd-numbered alphabets, is probably self-evident: One half contains only odd letters (1-3-5-7-9. . . . . . . . .) and the other contains only even letters (2-4-6-8-10. . . . . . .). If both halves were recovered in this order, and if one half were written directly below the other with letters 1-2 standing together, then the other pairs would also be standing together: 3-4, 5-6, 7-8, and so on. We seldom recover them in straight order; but whatever rearrangement has taken place in one of the halves has taken place, also, in the other half; should one be recovered with letters in the order 1-7-13-19-25-5-11. . . . (each third letter in a series 1-3-5-7. . . . .), then the other will be recovered with letters in the order 2-8-14-20. . . . . . (each third letter in a series 2-4-6. . . . . .), though neither half necessarily begins with the first letter of its series. If, then, we are able to place together letters 1-2, the other pairs will also be adjusted, perhaps in the order 1-2, 7-8, 13-14, 19-20, and so on. These pairs may then be taken at some regular interval and will bring back the original order 1-2, 3-4, 5-6, and so on.

Figure 146

Another Method for Combining the two Half-Chains of an Odd-Numbered Alphabet

(Originated by GEORGE C. LAMB)

With a Type III slide, based on the key-word EXCORIATE, the 7th alphabet, as recovered from a cryptogram would come out as shown: H K B L A M N P........

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 H K B L A M N P G Q S U V W D Y Z F E J X C O T R I

The second half-chain, started with D, must be re-adjusted so as to plece in correspondence the alphabetical sequences PQ, YZ, FG, MN, etc.

1st HALF-CHAIN: A H P Y R F M V C B K S E 2d HALF-CHAIN: (d l u x) T J Q Z I G N W O D L U X

Each corresponding pair of letters was a digram in the original cipher alphabet. Taking some two letters, as FG, which form an alphabetical sequence, look for another pair, such as HJ, which may be its continuation. HJ having been found at the interval 9, try taking pairs at the interval 9:

FG HJ KL MN PQ SU VW YZ EX CO RI AT BD (FG)

Lamb’s method, applied to an actual cipher alphabet, can be seen in Fig. 146. The first half-chain, if started at _A_, will include the letters _B_ and _C_, so that the second half-chain would probably have been started at _D_. But _AD_ will not be a correctly adjusted digram. It is necessary to look for one which forms an alphabetical sequence, as _FG_; and when the two halves are adjusted so that _F_ and _G_ are together, other alphabetical, or nearly alphabetical, sequences are also found to be adjusted, as _MN_, _VW_, _BD_, _KL_, making it likely that we have found some digrams belonging to the original mixed alphabet. With this adjustment reached, it is found that pairs can be taken at the constant interval 9, _AT_-_BD_-_FG_-_HJ_. . . . . , thus bringing back the original cipher alphabet.

Presuming that the original alphabet did not contain these alphabetical sequences, then there are thirteen possible adjustments for the two half-chains, and any one of these could be the original alphabet (or its equivalent). But it must not be forgotten that in an actual case our key-frame always contains portions of other cipher alphabets; and if the foregoing principle has been well understood, it may be readily seen how we could make use of these in order to determine which of the thirteen possible adjustments is correct. Even the recovery of the 14th cipher alphabet (which results in thirteen 2-letter chains), would not be useless with this other information present always in every key-frame. The _Type III_ slide, in fact, can often be reconstructed without possessing any fully recovered cipher alphabet. This cipher is very popular with members of the American Cryptogram Association, and is usually known, for no very good reason, as “the Quagmires cipher.”

* * *

In the case of the _Type IV_ slide, we do not begin reconstruction with one complete cipher alphabet already in our possession. It becomes necessary that we recover two, the perfect case being that in which one is an odd-numbered alphabet and the other an even-numbered one. We will follow this case in Fig. 147, where the two recovered alphabets are Nos. 4 and 7. This tableau, like the preceding one, has columns which are duplicates of its rows, and to see our preceding case again (with its one modification), let us begin by looking only at the three alphabets immediately below the heading. One of these, the plaintext alphabet, shown in lower-case letters, we will disregard for a moment, giving our attention only to the two cipher alphabets.

Figure 147

The Alphabets from a TYPE IV Slide:

Plaintext letters: 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 (1) 4th Alphabet: E W Y P V T S Q O N M K R J I H G F Z D X B A C U L (2) 7th Alphabet: Y S V Z Q O N M K J I H X G F D B A W C T U L P E R

A CHAIN Started with ab

ab ys vd qx nt jp gl bi sf du xm tz pw lr io fk uh me zc wa ry ov kq hn ej cg (ab) EW UZ BP GC JD NH SK WO ZT PX CR DL HA EF OI TM XQ RV LY AE FU IB MG QJ VN YS

REARRANGEMENT of this CHAIN:

ab bi io ov vd du uh hn nt tz zc cg gl lr ry ys sf fk kq qx xm me ej jp pw wa EW WO OI IB BP PX XQ QJ JD DL LY YS SK KF FU UZ ZT TM MG GC CR RV VN NH HA AE

A Reconstructed EQUIVALENT Slide:

Plaintext: a b i o v d u h n t z c g l r y s f k q x m e j p w CIPHER: E W O I B P X Q J D L Y S K F U Z T M G C R V N H A

ORIGINAL Slide, Found by Taking Letters at the Interval 5:

b u c s m a d z y x w v t ....... W X Y Z R E P L U C A B D .......

These two alphabets, like the two from the Type _III_ slide, are running parallel in the imaginary tableau, so that we have, as before, a series of 26 _vertical_ distances, _EY_, _WS_, _YV_, and so on, all known to be equal in the columnar direction and therefore known to be equal distances on any row. A chain may be started, exactly as in the other case, _EY_, _YV_, _VQ_, _QM_, _MI_. . . . . , resulting in a series of equally-spaced letters _E Y V Q M I F A L R_. . . . . . which is either the original cipher alphabet, or the original one with letters taken at some odd interval other than 13. It is, however, only the _cipher alphabet_; the mixed _plaintext alphabet_ must still be found. This may be done, as in the case of the _Type I_ plaintext alphabet, by using either of the two cipher alphabets which were first recovered and setting originals above their substitutes. If this is done with our cipher alphabet in the order _E Y V Q M I F A L R_. . . . . . , then the plaintext alphabet comes out in the order _a c e h k o r w z m_. . . . . . , and we have an equivalent slide. If we first rearrange the sliding alphabet (each 9th letter of the series _E Y V Q_. . . .), we obtain the plaintext alphabet also rearranged.

Continuing, now, with the rest of our figure: The method we have just seen was based on a tableau, and our equal intervals were all vertical. In the figure, we are dealing purely with horizontal distances, and our method is based, not on a tableau, but on a _slide_ (as it was with the _Type II_). Our 4th and 7th (secondary) cipher alphabets, after all, are merely two different positions of the same slide. If we select any two letters, as _a_ and _b_ of the plaintext alphabet, and find that their substitutes are, respectively, _E_ and _W_, in alphabet 4, then the lineal distance _ab_ in the stationary alphabet must be exactly equal to the lineal distance _EW_ in the sliding one; if this were not true, the letters could not have coincided as they do. Then, if we find the same substitutes, _E_ and _W_, in alphabet 7, and note that, in this position of the slide, they have coincided, respectively, with plaintext letters _y_ and _s_, then the distance _EW_ in the sliding alphabet must be exactly equal to the distance _ys_ in the stationary one. It follows from this that _ys_ and _ab_ are equal in the stationary alphabet. If we begin again with the lineal interval _ys_, we find that this is equal to _UZ_ of alphabet 4, and that _UZ_, found again in alphabet 7, is equal to _vd_. Here, then, is another interval, _vd_, which is equal to both ab and _ys_. And so we may continue, forming a chain made up of these known equal intervals, _ab_, _ys_, _vd_, _qx_, etc., for the plaintext alphabet, and _EW_, _UZ_, _BP_, _GC_, for the cipher alphabet. Sometimes we return to _ab_ (_EW_) without having included all 26 letters, and in that case (unless the number of letters included is a divisor of 26), it becomes necessary to abandon _ab_, and try starting with some other interval, as _ac_.

Figure 148

Some EXERCISES in the RECONSTRUCTION OF ALPHABETS

Plaintext ....... 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

Exercise 1: Q * Z A X B O C N * E R F P V G * Y M U I * W * T L

Exercise 2: U V D W S X K Y H Z C F R J Q L I N G P T O M E A B

Exercise 3: H J G K F P E Q O R S T D M B U V W X A Y Z C L I N

Exercise 4: (1) V N U X J Y Z D Q E M P O W C K R I A T L S B F G H (2) H S G J R K L N F P Q B U I V A W C X Y T Z D E M O

Exercise 5: (1) G X Y Z M H A F T R L K E V Q U O J W I P N S B C D (2) E * G J I K * L B * * U T C V W * Q D X S * * * * *

The keywords, all selected by Mr. A. F. SEMPER, do not contain any repeated letters. The TYPES of slide, respectively, are: I, III, III, IV, and III. In the 5th exercise, the completely recovered alphabet is a "number 14." (See also practice-cryptogram No. 46).

Here, as in the case of the _Type III_ slide, we sometimes obtain two 13-letter chains; but in the fortunate case of having recovered in full both an odd-numbered and an even-numbered cipher alphabet, we end the chain with every letter represented twice, both in the stationary alphabet and in the sliding one. Pairs of letters (representing horizontal intervals) can then be rearranged as in the other case, the second letter of one becoming the first letter of the next (in both series), and the dropping out of duplicated letters gives us an equivalent slide. In the figure, starting with _ab_ (_EW_), we find a plaintext alphabet _a b i o v d u_. . . . . and a cipher alphabet _E W O I B P X_. . . . . . This, as mentioned, is one of eleven possible equivalent slides, of which one is the original. Here, the original can be found by taking letters at interval 21. In the figure, letters were taken at interval 5, a result of observing the sequence _W X Y Z_ standing at that interval in the lower alphabet, and the slide comes out in reverse order. This is still an equivalent slide, and the decryptor may or may not care to decide which alphabet runs backward.

Since the reconstruction of these mixed-alphabet slides is probably the most fascinating subject in the whole field of cryptanalysis, several problems are being appended in Fig. 148. In all of these, Semper has selected key-words or phrases to contain no repeated letters. With reference to the practice cryptograms, only one of those submitted was thought to have enough alphabet-length for purely analytical attack. The others, even with their probable words or partial translations, will still require some work. The periods of these examples are said to be, respectively, 6, 7, 8, 3, and (?).

135. By NEMO. (Type II. Partial solution: WHEN JACK BOOMER,GREEN RIVER,WYO,B..)

T E R P J Y D B N Q S A I M B X B L Y M D O B I T Z P T I H K O K A G M Y Q R X T D W U U X B O B Q Y D B W Z S V Z G C U P R Z S W V O D M T Q Z C A T S M Y Q F D B H Z Q U T I H F S V S Y N F U L Z G L B G D M T R M U C N A J M I Y N Q O F B D P Q L G X U Y W U I P C A Y N J N X S B K W I J G R L G I B.

136. By NEMO. (Type III. Partial solution: ALOIS STEPHEN,YOUNG VIENNESE CHARGED W..)

H G K S T I L O Y D B O L E G A Z N G P D U W P B D R V Z Q Y Z X F L Q S B H S L T U Q P S Z G X V A Y T G B C B X K H U R I E D M D X B T O E P S A R I N K X K J B I T P Y I X R I U Z Y O M I M H P H B E J D N E B S E F L B F D B H F J B F N L G P L J M I B O G T A W D U E Q E Z T Y U S I.

137. By THE SQUIRE. (Type III. Probable words: AMERICAN CONSTITUTION. GLADSTONE).

H T F M R S R T Y E O V P D S Z L A X B A C N T N A K X R C S Z K G O Q U O F A Z R E T D S V I W K W T E L K F R P R B I H I N A S W R R S B O H T F L A D D L U B U F M Q O J A G I L I D W T Z I M M R H L L V K W U J S.

138. By ALII KIONA. (Is this a diplomatic telegram?)

L L D R K Y C R F A S E V S U K T D U L X V K E V C A B L Y U P Y M R K B E X U B T E L W P J F P T I I U Q Q K T F C T P S K Q L W N D A P B F A E S N M P R K A P T T S H F K B Z R M G P P Y V M S A I F N P Z A L T S U S A U D N L X A A Z Y P U C H K N P Y V M S I A X K K D B E T P S A T P K P S Y V T A Y E A P B T E L W P J F P T A X N.

139. By PICCOLA. (This is a straight-alphabet cipher. Won't tell which one!)

A N D N Y L M Y X N K D L R P G C X G Q N A A R Z L D E P L G I A W Q N E I O G A G P Q G Z V D E I E Z R H A Y P L B P N A G E L N V A G T D H O K H V G T I N D O L S F C P L R T.