Breaking Civil War Codes and Encryption

by Lee A. Taylor

 

I will deal with 2 issues here, one is recovering the correspondence of flag waves in wig-wag encoding, as used by both sides during the War for Southern Independence (sometimes known as the American Civil War) and the other is much more complicated, dealing with the reading of encrypted messages.

 

FIGURING OUT THE ENCODING OF LETTERS USING FLAG WAVES

 

The recovery of the correspondence between flag waves and the underlying letters is pretty straightforward and could be easily done by an experienced signals man, even over 140 years ago. All that it requires is that you have the ability to translate the 3 flag waves with corresponding pauses to numbers, to get sequences from each message sent such as:

 

2-pause-1-1-1-pause-1-3-1-pause-2-1-pause-1-pause-1-1-1-1-pause-1-2-2-1-3-1-1-2-pause-2-2-3-1-1-pause-2-pause-2-pause-1-1-pause-1-2-1-2-pause-2-1-2-2-pause-1-1-2-pause-2-1-pause-2-1-1-2-3-3-3

 

NOTE: I purposefully did not use any preconcerted codes or short endings, as that complicates the discussion slightly. This message can be parsed into "letter groups" by using the pauses and 3's as the separators between "letter groups". Doing this, we see that

 

1 occurs 3 times

2 occurs 3 times

11 occurs 2 times

12 occurs 0 times

and so on…

 

After collecting enough messages, you could use the monographic distribution of letters in the English language (i.e., the probability of an E occurring in text, and the probability of T occurring in text, and so on and compare it to the counts to find out which letter is which, see table below for an example distribution from a corpus of English text).

 

LETTER

% time letter occurs among all letters in sample text

E

12.350%

T

9.116%

N

8.445%

O

8.327%

A

7.986%

I

7.904%

R

6.845%

S

5.505%

H

5.281%

L

4.364%

D

3.823%

C

3.093%

F

2.646%

U

2.329%

M

2.235%

G

1.964%

Y

1.952%

P

1.788%

B

1.282%

V

1.176%

W

0.882%

K

0.212%

Q

0.188%

J

0.176%

X

0.082%

Z

0.047%

 

A much more familiar way to solve this puzzle would be to assign letters to each group, such as A for 1, B for 2, C for 11, D for 12, E for 21, F for 22, G for 111, H for 112, I for 121, J for 122, K for 211, L for 212, M for 221, N for 222, O for 1111, P for 1112, Q for 1121, R for 1122, S for 1211, T for 1212, U for 1221, V for 1222, W for 2111, X for 2112, Y for 2121, Z for 2122, 0 (since I ran out of letters) for 2211, 1 for 2212, 2 for 2221 and 3 for 2222 and then the above message translates to

 

BGA AEAOU HF CBBCTZHEX

 

After you've collected enough of these messages, the problem becomes EXACTLY like solving Cryptograms in the newspaper or in a puzzle book.

BREAKING THE ENCRYPTION USED BY THE SOUTH DURING THE WAR FOR SOUTHERN INDEPENDENCE

 

First, I need to discuss the method of encryption used by The South and warn you that this section (after the description of the method of encryption) gets pretty heavy into the mathematics. The encryption scheme uses the following array of 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

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 A

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Z 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

 

Once you have this array in hand, you next need a code phrase (the above table was the only table that appears to have been used during The War). Let's use the key phrase

 

COME RETRIBUTION

 

Next, remove the spaces to get:

 

COMERETRIBUTION

 

Now, if you want to encode the message

 

THE ENEMY IS ATTACKING

 

you would first remove all of the spaces from the message to get

 

THEENEMYISATTACKING

 

and then pair the letters of your message (the "plain text") up with the code phrase (the "key") to get

 

COMERETRIBUTION

THEENEMYISATTACKING

 

            Notice that there are not enough letters in our key phrase to match up to all of the letters of the message we wish to send. So, the procedure is to repeat the key phrase as many time as necessary. For this example, you get:

 

COMERETRIBUTIONCOME

THEENEMYISATTACKING

 

The letter in the top row tells you which COLUMN to go to in the encryption array and the corresponding letter in the bottom row tells you which ROW to go to in the encryption array. So, for example, the first C of COMERETRIBUTION and the T of THE line up, so you would look up the character in column C of the array and row T, which is the letter V, which is the first letter of the encrypted message that you will send (the encrypted message is called the "cipher text"). The first O of COMERETRIBUTION and the H of THE line up, so the next letter in the encrypted message comes from COLUMN O and ROW H, which is the letter V. Continuing in this way, we get the following letters:

 

COLUMN

ROW

CIPHER TEXT

C

T

V

O

H

V

M

E

Q

E

E

I

R

N

E

E

E

I

T

M

F

R

Y

P

I

I

Q

B

S

T

U

A

U

T

T

M

I

T

B

O

A

O

N

C

P

C

K

M

O

I

W

M

N

Z

E

G

K

 

Which gives the encrypted message:

 

VVQIEIFPQTUMBOPMWZK

 

which you could put the original spacing back in to get

 

VVQ IEIFP PQ TUMBOPMWZK

 

Now, I know this looks like it is probably a pretty good method of encryption and, if the code phrase is changed often enough (like daily!) and messages are kept very short, it can be (if you have no computers).

 

The first HUGE flaw in this scheme is in how it was used by the Confederates. For common usage, there were only 3 key phrases ever used! They were

 

COMPLETE VICTORY

COME RETRIBUTION

MANCHESTER BLUFF

 

I’ve read numerous references that hint at the use of private keys used by various spies, such as Belle Boyd and Rose O’Neal Greenhow. I’ve never seen what these keys actually were, they each supposedly only ever had one, and only one or two other people actually knew them (e.g., Major/Lt. Col. William Norris, Chief of the Confederate Signal and Secret Service in Richmond, would have been one of the people who knew the codes). Anyhow, with only 3 key phrases, it wasn’t too hard to figure out how to break the code, once you had solved a few messages. But the Confederates made it even easier! Before sending text enciphered with one of the 3 keys above, they would send a plaintext message with the key phrase in it! For example, “With the outcome of this battle, complete victory is ours”. The signal officer at the other end, knowing what the 3 possible keys were, could easily pick out which one was used and decipher the message. It is important to note that only the 30-40 signal officers in the Confederate Signal and Secret Service (ranging in rank from Sergeant to Captain, except for Norris) were actually taught the encryption method.

 

The technique I’m going to describe in this article is actually not a 21st century or even a 20th century technique. During the Civil War, if one of the Yankees had been up on his mathematical literature from Prussia, he would have found the following article:

 

Friedrich Kasiski, Die Geheimschriften und die Dechiffrierkunst ("Secret writing and the Art of Deciphering"), 1863.

 

that described a complete attack on the Vigenčre cipher as used by the Confederates. If you search for “Kasiski test”, on the Internet, you can find a number of readable versions of his algorithm that (at their core) use a higher-order language model to solve the cipher. It turned out that Charles Babbage had actually developed this attack in 1854. However, due to the need for Information Security in Great Britain, it was decided not to publish his results (the Russians also used the Vigenčre cipher, and the English didn’t want them to stop).

 

TACTICAL VS. STRATEGIC

 

Before proceeding, let’s stop for a reality check. If you are a field commander and you would like an artillery unit a mile away to open fire, do you encrypt the message? Probably not (at least not back then, encrypted radios make this a triviality today). First, it would take too long to encrypt the message; you want the artillery to open fire NOW! Second, the lifetime of the message is very short, within a few moments after sending the message; the Yankees are going to know what the message was, so it doesn’t really matter if they are able to read it five minutes after intercepting it. Encryption was primarily reserved for messages of strategic importance (e.g., troop movements over the next few days or supply line information). These messages tended to be longer and the enemy would be willing to spend more time decrypting them. It is ratio of the length of the message to the length of the key that is going to be the main approach in breaking this code.

 

For a mathematical diversion (which could lead to a more sophisticated attack), let me point out something of the mathematical structure of this code. If you’ve studied modular arithmetic, you might have recognized the structure in the array of letters used to encrypt in the Vigenčre cipher. Except for a handful of mathematicians, modular arithmetic was not commonly taught at the time (it’s main use was in Number Theory, see, for example, Carl Friedrich Gauss’ 1801 text, Disquitiones Arithmeticć). Notice that if we assign the number 0 to A, 1 to B, and so on up to 25 for Z, then the "letter" (encoded using the numbers 0, 1, …, 25) in row i and column j of the array, can be figured out by the simple formula

 

(i + j) (mod 26)

 

For example, in encoding the first letter of our message, we went to ROW T (the T from the word THE in the plain text) and COLUMN C (C from COMERETRIBUTION). T is the 20th letter of the alphabet, so it has the code 19 and C is the 3rd letter, so it has the code 2 (remember, we started with A=0 NOT A=1). So the letter at position (T,H)=(20,2)=19+2(mod 26)=21(mod 26)=21, which corresponds to the letter V, which is exactly what we got before!

 

Now, I'm going to assume that both the code phrase and the message were both in American English as spoken during The War (the method gets a little more complicated when shorthand, or the preconcerted, codes are introduced, but not too much and the general technique will still work, so I'll stick with the easy case). This means that we can use the monographic distribution table found earlier in this article (for different languages, you can find appropriate distributions on the Internet or use a large body of text in the language and compute estimates for the probabilities yourself).

 

Before getting to the point, though, where we can use the distribution of letters in English, we’re going to need to figure out how long the key phrase is. Kasiski’s approach simply notes that in Indo-European languages, there are a lot of common trigraphs (3 letters occurring in a row). In English, “THE” is the most frequent trigraph (not only is it a word, but it occurs in common words such as THEre, THEir, farTHEr, norTHErn, moTHEr, souTHErn, etc.). Kasiski’s approach is to find how far apart repeated trigraphs are in the encrypted message and stare at these to figure out how long the key phrase is. Let’s use the following encrypted message as our example:

 

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

 

The text of this message is a message sent to General Stuart by General Lee just before the Battle of Gettysburg, as we shall see shortly. Anyhow, we get the following distribution of distances between common trigraphs in the enciphered text (COMERETRIBUTION is 15 letters long):

 

DISTANCE

# common trigraphs DISTANCE letters apart

1

0

2

0

3

0

4

0

5

0

6

0

7

0

8

0

9

0

10

0

11

0

12

0

13

2

14

0

15

4

16

0

17

1

18

1

19

1

20

0

21

0

22

0

23

0

24

0

25

0

26

0

27

1

28

0

29

0

30

9

31-40

0

41

1

42-44

0

45

12

46

1

47-59

0

60

4

61-70

0

71

1

72

0

73

1

74

0

75

10

 

If I were to continue, you would see that the count is much higher at multiples of 15 (15, 30, 45, 60, 75, 90, 105, 120, 135, …) than anywhere else. This leads us to guess that the key phrase is 15 letters long (REMEMBER: we don’t know what the key phrase is yet if we’re the Yankees).

 

The next step is to divide the encrypted message into 15 groups, one for each letter of the (as yet unknown) key. The first group will consist of the first letter, the sixteenth letter, the 31st letter, the 46th letter, etc., as each letter in this group will have been encrypted with the same letter of the key phrase. Similarly, the second group will consist of the second letter, the 17th letter, the 32nd letter, the 47th letter, etc. and so on. The 15 groups are, then, as follows:

 

JAKPPGCWCEQKCDPUOUFHJGGTATKVUCIMVHQCPXQGYUOCNPGVUVOGTKKTOIVCAHITKPWUGGQUQTQUPQHGCGVJUVGVCHNYQIQERXGTG

SCBXRFBGGVFBHSUOOSZFORFSCWAVPDBMVHFHSSKFVGMBHOFOSVCFCBGIAOCBOHHSBOTHGFBWTRTCUBHZWQHSWSHGFCGWAGKVSSGIB

MRUQUMPFDMKSQFRDEFQAHNMYGSMFGBAAPTDEJDZNQMIOTZQUMQHUARUOMPIPZTTREZRAMKSZFEFRFFTQZMTEZDFAQDRXMFFROYBXQ

HREFRPEFIWSXQESMXLXQICPEGERLXIVYVISLXXCIXVMIIHERJVIKTSSXRIEVHIIVAHMKRXXKLXLXSLIJWRISXXIJEQMPGVSYXIICV

HFAJXPDVXVLYFBIVVZKKEKYZRUUVJRKYRDNVUFFRYFKUUTJJKZFYJIEZUJKVZVZFVFTLUYYLVYVYNVSKDKDFFFIVJVIIKVDCZETPR

YVYXGSXIESVERICWVWLLSLSRRIAXLVLEASRTEJYFIYLSEVXMIZRXGQWSIPGERRVQWEMEFMITEIXIEGVMYSFRQQXAWVWISISERXXSP

TMGNTNHGKYFMXGHHLMXXMXHLELBAHMPWMNBARKPEKGHBFHHGKXTHHTVGKXAKMXYMMAXKKGOHKFPXKHBGLVNXTHAXMEMTWMKGTLYN

IYVRMIUIUKVTPZLIKFDDYDBZVKKILFRSYEXVRVZVPULERJWVTIEWCKXJFWKFYDIYFCEUZXREDFFEUDXKKFKIIIVCRVUTRNIUCZLI

BMXZIVIMAWVWEIZYWJMJQMMVIWPMTJZMQBPZVLTBWBBOOABQZGLMTQQBNBPNMGWMNMBBVKTBGDJMEUIPLCQGGZUTBBQPGQWKTITA

FSNUMPZDUCTOJNDVQBOVOOSBWXEFEFEUTBUEEFMPVIIUFUIUPPGXFPWPUCFUFMONUBQIHMMIBFSNBBEFPOUPMPPTFUWUBMXJZNMS

LHGWLNBYBUOZFQIUOWANACMWYULIBGCNMCWMGLBDWYCBSBYBMOYYWHYNBYZBPYNIBPCYCYYYMGCSLHYGQNBOUQPWXYCBHFVLIPSY

LOTHRXTBXVIXEBFKKVXBLYTMMMTMXHMXBGKMHBHNTBGXHXFXLFXEMIBAXAEXXTKNXBVIGTRKKXZFKWLHAXBVGMXHBKLXWYXVNXTE

IQRUOADDXKXLVTUBKWBKMOZQEKEPVDPZLBWWDKELVZLUCZWZQCTTQZVMJQIIVDMVAVSIOVKMMVIWMMBCIZVZLPUZVPQXTWECZZVM

FFUASCSSICCSCZWSVOWOWSAJCVKSCWWKSCGKSYSUDOSOQWIQBGHGBCGQFBBFHWHHVUSGSOZOUHRJBFCBHOYCOSSDAWCCCZOAAMRS

ZTRNASWQESFEGVFENAGAMALROUVEGAAVBZFABGIRNEEYNIANTGUGTIGBVQXZBAVNRFGFIYBENFRVGBOGUPGFSOAFLYAGAYGFBEGT

 

In each group, see which letter occurs the most often, and guess that it is the encrypted form of the letter “E”. This first attempt gives us the following key phrase:

 

COBERADEXQUTROX

 

Now, we could do one of 2 things here. We could either attempt to decrypt with this key phrase and the fix the wrong letters in the decrypted message and try again, or do something more sophisticated, like look for the entire letter distribution in each group and see which possible key letter gives the best decrypt. Let’s try the first way and see what happens:

 

HELDQYQEEPRSRRXYOQNOVJUPCNVZRRINTAJYDRAXMAAGPNJPBSXKNCECODMLNDTNGGQILWRYXEYERLLYSKEYZTEJOQANOAMXEQLJHAMEUUSEBEIDEPNEIMEOASCEGEHQDEHEGUCCHLSESVGZMACTOQORJOUVCRYDUPGODINRTHEJPZYFEUECATPMORULHTLLEOEBEEAKIDVLXWICLTNGQORCEHCNOMDIDSACIEWEEBFARKECMADTEVIGZAURTHLSEEHIWJBMLCCFAYDLPTTLUZPYGEKIEFRZMTLUZMFTITAYHAGENSJUTYGSVIKEDMYTLUZPYIFXEYERLLHSEXPCSAIMJREXAIRIVYLCTZVPYOFCARBRLGETNOMRIRADIIGZHATTHSIMLNDAYGSORANWTTHEHEXXEPPOTYECSBFTSLEHWOHEEOEAPAEAVJBMPMOMIYGNZRTLMNCOITYIYKYZUHETOPETEIWTTHORAAJUTDSIUEZFTSEMSKAELINKOXORCOWRYTSECRFSDATDHETXRCOSTFWYNEITDEONYOMOMEZVECTOJHROPRITKEOWYYOYMVWWHONEGERMEAFBREZJUUGPWHPTHIHLZFCAEPLSSLROYDQESEIIACMYHITLEHESINUECANNEDSYAREHEDAWLTSEDECNRPYOLCLNAYDCVEFDEHEIIGERPASXESESEMFUYTATNSMDRTEHEICLSELFTIHPCZSSZNRTHPRIZUEJZUMLSEMOGEORQAOQEECTSERTGHXESPHELCSEROZPSGEYWPCTZNRINQORQQGTZNPIOGISTONWSTTGEIESERUNTISDFEZTHVCZMMLNDIHBQEHESRTGAOESPUSEMEHZNOTOHATGXGSPFLRNVANOREEHBQEHERRXYAYDIRJUPPVEETZFTSEERUZJWEAMIYGTSEIVVEZYTRVTTREQROQJUPXOUETLINDWEWJBQEHEJHPNAYDOEXYPLVIEGDUFQICMUAEAICBEESTZGUEHQESEPRSDESLNDFHVYRINXEGERJTHMDTNWEAEAWONRTHILNWWEYTLZSIYGUTEAESERVACOFEHEEHZJLSRVGLRDDTHICBGPMEETDOFEHEXMBMCIGRDPSOQTHIUAPXYMFVTNGEOWEHQHLRRVNEONEHEGEZXLNDVRZFTSEBVYTLOESKOMELPFTMDGSPMOLNEAIYSMYIGOZWHRTSECLNTSSBFYTEIANTTSEMFKGTEHIEKEHEDOORUEJZUCIODSIYTOQQEJWANUAQTECTOQEECZWTYEMETEERXXRXZVEDEYTSZFEAUYWDCOIPDARPASWJNEPDIEMJFOCMEVBREEERYIWLSQIRWJQTGISZOYWIWLRIQPSEHEGOEOMLCTSTNJLNDCOYGSEREIJJTWLFFLWOWEOMSHEZHBENAECHQULEDQNTRCLMDPENTIRQYWJOUIMZVEXENXIVLXVEIYCESAECXVHWWYAEDERUWYYSKEDCELVERENPRAP

 

This isn’t quite so easy to figure out. So now, let’s go back and try the second way, where we declare a key letter to be good if, when decrypting its corresponding group, we get something that looks like a distribution where “E” is the most common letter, “T”, the second, and so on. Using this technique, we get the following guess at a key phrase:

 

COMERETRIBUTION

 

To see this entire attack at work, here is a C program (for Microsoft .NET) that will take you through the entire process described in this article (no comments in the code, see if you can figure it out):

 

#include "stdafx.h"

#include "stdio.h"

#include "string.h"

#include "conio.h"

#include "dos.h"

 

#using <mscorlib.dll>

#include <tchar.h>

#include <ctype.h>

 

double monoprobs[26] = { 0.07986, 0.01282, 0.03093, 0.03823, 0.12350, 0.02646, 0.01964, 0.05281, 0.07904, 0.00176, 0.00212, 0.04364,

0.02235, 0.08445, 0.08327, 0.01788, 0.00188, 0.06845, 0.05505, 0.09116, 0.02329, 0.01176, 0.00882, 0.00082, 0.01952, 0.00047 };

 

using namespace System;

 

char message[5000] = "Headquarters Army of Northern VirginiaJune 23, 1863--5 p.mMaj. Gen. J. E. B. STUART, Commanding Cavalry: General,    Your notes of 9 and 10.30 a.m. to-day have just been received. As regards the purchase of tobacco for your men, supposing that Confederate money will not be taken, I am willing for your commissaries or quartermasters to purchase this tobacco and let the men get it from them, but I can have nothing seized by the men. If General Hooker's army remains inactive, you can leave two brigades to watch him, and withdraw with the three others, but should he not appear to be moving northward, I think you had better withdraw this side of the mountain to-morrow night, cross at Shepherdstown next day, and move over to Fredericktown.You will, however, be able to judge whether you can pass around their army without hinderance, doing them all the damage you can, and cross the river east of the mountains. In either case, after crossing the river, you must move on and feel the right of Ewell's troops, collecting information, provisions, &c.Give instructions to the commander of the brigades left behind, to watch the flank and rear of the army, and (in the event of the enemy leaving their front) retire from the mountains west of the Shenandoah, leaving sufficient pickets to guard the passes, and bringing everything clean along the Valley, closing upon the rear of the army.As regards the movements of the two brigades of the enemy moving toward Warrenton, the commander of the brigades to be left in the mountains must do what he can to counteract them, but I think the sooner you cross into Maryland, after to-morrow, the better.The movements of Ewell's corps are as stated in my former letter. Hill's first division will reach the Potomac to-day, and Longstreet will follow to-morrow.Be watchful and circumspect in all your movements.                   I am, very respectfully and truly, yours,R. E. Lee                     General.";

int delta[5000];

 

char key[16] = "COMERETRIBUTION";

char encrypted[5000];

 

char atoffsets[15][1000];

 

void removespaces (char *);

void encryptit (char *, char *);

void decryptit (char *s, char *k);

char mostfreq (char *s);

char tryaletter (char *s);

 

int _tmain(void)

{

       char x;

       char guess, keyguess[100];

       int i, j, k;

       FILE *fout;

 

       fout = fopen ("C:\\junkout.txt", "w");

 

       removespaces (message);

       fprintf (fout, "Message with spaces removed: %s\n", message);

       encryptit (message, key);

       fprintf (fout, "Encrypted with '%s': %s\n", key, message);

 

       strcpy (encrypted, message);

      

       for (i = 1; i < ((int) strlen (encrypted)) - 1; i++)

       {

              delta[i - 1] = 0;

              for (j = 0; j < ((int)strlen (encrypted)) - i - 3; j++)

                     if (strncmp (&(encrypted[j]), &(encrypted[j+i]), 3) == 0)

                            delta[i - 1]++;

       }

 

       for (i = 1; i < ((int)strlen (encrypted)) - 1; i++)

              fprintf (fout, "delta[%d] = %d\n", i, delta[i - 1]);

 

       for (i = 0; i < 15; i++)

       {

              for (j = 0, k = 0; (i + j) < strlen (encrypted); j += 15, k++)

                     atoffsets[i][k] = encrypted[i + j];

 

              atoffsets[i][k] = '\0';

 

              fprintf (fout, "Group %d = %s\n", i, atoffsets[i]);

 

              x = mostfreq (atoffsets[i]);

              guess = ((char)(((int)(x - 'E' + 26)) % 26)) + 'A';

              keyguess[i] = guess;

 

              fprintf (fout, "Most frequent letter = %c, so guess at key letter is %c\n", x, guess);

       }

 

       keyguess[15] = '\0';

 

       strcpy (message, encrypted);

 

       decryptit (message, keyguess);

 

       fprintf (fout, "Attempted decrypted message is: %s\n", message);

 

       for (i = 0; i < 15; i++)

       {

              x = tryaletter (atoffsets[i]);

              keyguess[i] = x;

 

              fprintf (fout, "Guess at key letter is %c\n", x);

       }

 

       fclose (fout);

}

 

void removespaces (char *s)

 

{

       int i, j, slen;

 

       slen = strlen (s);

       for (i = 0, j = 0; j < slen; j++)

       {

              if (isalpha (s[j]))

              {

                     s[i] = toupper (s[j]);

                     i++;

              }

       }

 

       s[i] = '\0';

}

 

void encryptit (char *s, char *k)

 

{

       int i, j, klen, slen;

 

       slen = strlen (s);

       klen = strlen (k);

 

       for (i = 0, j = 0; j < slen; j++)

       {

              s[j] = ((s[j] + k[i] - 2 * ((int)'A')) % 26) + (int)'A';

              i++;

              if (i >= klen)

                     i = 0;

       }

}

 

void decryptit (char *s, char *k)

 

{

       int i, j, klen, slen;

 

       slen = strlen (s);

       klen = strlen (k);

 

       for (i = 0, j = 0; j < slen; j++)

       {

              s[j] = ((s[j] - k[i] + 26) % 26) + (int)'A';

              i++;

              if (i >= klen)

                     i = 0;

       }

}

 

char mostfreq (char *s)

 

{

       char retval;

       int freqs[26], i, max;

 

       for (i = 0; i < 26; i++)

              freqs[i] = 0;

 

       for ( ; *s; s++)

              freqs[(*s) - 'A']++;

 

       max = 0;

       for (i = 0; i < 26; i++)

       {

              if (freqs[i] > max)

              {

                     max = freqs[i];

                     retval = (char) (i + 'A');

              }

       }

 

       return (retval);

}

 

char tryaletter (char *s)

 

{

       char c, retval;

       double weights[26];

       double highscore, score;

       int distro[26];

       int i, slen;

 

       slen = strlen (s);

       highscore = 0;

       for (c = 'A'; c <= 'Z'; c++)

              {

              for (i = 0; i < 26; i++)

                     distro[i] = 0;

 

              for (i = 0; i < slen; i++)

                     distro[(s[i] - c + 26) % 26]++;

 

              score = 0;

              for (i = 0; i < 26; i++)

                     score += distro[i] * monoprobs[i];

 

              if (score > highscore)

              {

                     highscore = score;

                     retval = c;

              }

       }

 

       return (retval);

}

YANKEE ENCRYPTION (THE ROUTE CIPHER)

 

The North didn’t use the Vigenčre Cipher. For the most part they used a Route Cipher, which the South never broke. Union messages were often put in newspapers in the South with rewards offered for their solution. The key components to the Route Cipher as used by the North was:

 

- common political, military and geographic terms were substituted with more common words (for example, Abraham Lincoln might be an “oak tree” and Washington might be “brick”)

- the words of the message (after the above substitution) were written out in a rectangular matrix of some prescribed shape (sometimes pad words were added which had nothing to do with the message, just filler to complete the rectangular shape)

- a prescribed path through the rectangular matrix (usually things like “up the second column”, then “across to the right”, etc.) was followed and the words copied out in this new, jumbled order

 

I think this could have been solved by 19th Century techniques, but would have required a great deal of organization and a fairly large number of methodical clerks, as counts would have to have been kept across hundreds of messages and correlated with events of the time the message was received to try to uncover what common terms had been substituted (e.g., messages coming out of Washington are more likely to mention President Lincoln or Secretary of War Stanton then a message coming out of Kentucky). Major Norris could have even caused messages to be sent by having his spies in Washington cause some sort of uproar and then see what messages came out. Had the South known about the rectangular shapes, they could have tried various spacings of words to put the messages back together again (if you go “up a column”, the words will have originally been equally spaced apart, dependent only on the number of words in each row of the matrix).