6.1 graph theory · so now, the original question becomes: can you ﬁnd a path that goes through...

CHAPTER 6. OPTIONAL FUN TOPICS 150

6.1 Graph TheoryExample 1. [Seven Bridges of Koenigsburg] Here’s the problem. Can you find a path that:

• starts anywhere you like (on land)

• finishes anywhere you like (on land)

• crosses each bridge exactly once.

(Fine print: You should assume that the land areas are only connected by these bridges, you cannot walk half way across a bridge andthen turn back, you can not swim across the water, etc.)

(water) (water)

(water)

(land)

(land)

(land)

(land)


Definition. A graph is a collection of two kinds of things: (1) vertices and (2) edges such that each edge connectsexactly two vertices. We picture vertices as round dots or small circles. We picture edges as straight lines or simplecurves that connect vertices. The only information we consider in a graph is this: which vertices are connected byedges, and by how many edges. In other words, a graph is a simple representation of things that are connected alongwith the connections themselves.

Example 2. [7 Bridges of Koenigsburg continued]Here’s the graph theory representation of the 7 bridges problem:

A

B

C

D

The only thing that matters is that A is connected to B by two edges (representing the two bridges), A is connected toC by two edges, and D is connected to each of A, B and C.

So now, the original question becomes: can you find a path that goes through each edge exactly once?

Definition. Let G be a graph. A path that goes through every edge of G exactly once is call an Euler path.

Theorem 1 (Euler’s Theorem).Let G be a graph. There is an Euler path through all the vertices of G if and only if all the vertices, or all of themexcept two, are connected to an even number of edges.

If exactly two vertices have an odd number of edges, then an Euler path must start at one of these vertices and endat the other one.

Proof. Suppose we have an Euler path and let A be a vertex which is not the start or stopping point of this path. SinceA is not the start or stop, any time the path uses an edge coming in to A, it must also use a different edge going outfrom A. So, we have one of the following cases:

the path uses 1 edge coming in to A and 1 edge going out,or the path uses 2 edges coming in to A and 2 edges going out,or the path uses 3 edges coming in to A and 3 edges going out,

etc.

In each case, the path uses an even number edges connected to A. Since the path is an Euler path, we know that it useseach edge exactly once. Therefore it uses each edge connected to A exactly once, and therefore A has an even numberof edges connected to it.

This shows that any point that is not the start or stop must have an even number of edges. Therefore, if a vertexhas an odd number of edges, it must be the start or stop.

Example. [continued]Explain why it is impossible to find a path in the seven bridges graph that goes through every edge exactly once:

Solution. Note that every vertex has an odd number of edges. To find a path the way we want, we would need exactlytwo vertexes with an odd number.

Therefore, there is no path through this graph that uses every edge exactly once.


Example 3. Given the graph below:

(a) Add one edge from the white vertex to the red vertex.

(b) Find a path that starts at the blue vertex, ends at the orange vertex, and crosses each edge exactly once.

(c) Show that it’s not possible to do the same thing starting from the red vertex.

Solution. We add an edge from white to red:

and then label the edges to show which order we put them in the path, and which direction we take them:

6

578

2 1

4

3

It’s not possible to do the same thing from the red vertex because a start/stop point needs to have an odd number ofedges. Since the red vertex has an even number of edges, it cannot be a start point of a path.



Find a way to add one more edge, so that (1) it’s possible to start at the red vertex, end at the orange vertex, cross eachedge exactly once, and (2) it’s not possible to do the same thing starting from the blue vertex.

Solution. We add the following edge, and indicate the order we take the edges in our path:

9

321

6 7

4

8

5

It’s not possible anymore to find such a path starting from the blue vertex because the blue vertex has an even numberof edges.



Find a way to add one more edge, so that it’s possible to start at any vertex, make a path through each edge exactlyonce, and finish at the starting vertex again.

Solution. We add the edge shown below

since every vertex has an even number of edges it’s possible to start at any location, make a path that goes throughevery edge once, and come back to end at the same vertex as the start.


Example 6. “Koenigsburg” was the name for the town that had the seven bridges discussed above. In World WarII the town was heavily damaged through bombing, after the war it became part of the Soviet Union, and in 1946 itchanged it’s name to Kalingrad. It’s part of Russia today, and the bridges have now changed.

The picture below shows a satellite photo of Kalingrad today.

(a) There are 8 bridges now in Kalingrad. Identify them, and turn the land masses and bridges into a new graph.

(b) In the modern Kalingrad, is there a path that goes across each bridge exactly once? Why or why not?


Solution.

A

B

C

D


6.2 CryptologyExample 7. In 53 BC, Julius Caesar had not yet become the first Roman Emperor. He had conquered most of Gaul(France and Belgium) but was still contending with major uprisings, and in 53, the Belgaen uprising was underway.It had killed thousands of Roman soldiers and was threatening a Roman garrison in Nervii territory, commanded byCicero, that had a few thousand men. Cicero’s outpost was surrounded and outnumbered by the Gauls, cut off fromsupplies and reinforcements. Cicero manages to sneak out a message to Caesar asking for rescue. Other messengersare sent back and forth, but most are captured and tortured to death. The Gaulic forces are parleying with Cicero,asking for surrender.

Caesar sends a message, essentially containing the following1:

L FDHVDU DP DSSURDFKLQJ ZLWK WZR OHJLRQV RI URPDQ VROGLHUV L

ZLOO EH ZLWK BRX VRRQ VKRZ FRXUDJH DQG GR QRW JLYH XS

Do you think Cicero surrendered as demanded? To help decide, figure out what message says. (You may wantto know that Caesar used what’s called a “shift” cipher: the letters of his original text were all shifted, by the sameamount, down the alphabet, to produce the code.)

Solution. No doubt you figured out what the message said:

i caesar am approaching with two legions of roman soldiers i will be with you soon show courage and donot give up

or, in prettier formatting

I, Caesar, am approaching with two legions of Roman soldiers. I will be with you soon. Show courageand do not give up.

In the end, Cicero and his soldiers were heartened. They did not give up. Caesar arrived with reinforcements, Cicero’smen joined with him, and they eventually defeated the Gauls.

Here’s how the message would have been encoded in Caesar’s day. Make two alphabets, one above the other, butwith the bottom one shifted to the left three spots, moving the left end back around to the right:

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 zD 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

To encrypt, Caesar would have taken each of the original letters, looked it up the top row, and replaced it with thecorresponding letter from the bottom. Thus, i becomes L and c becomes F, etc. To decrypt, Cicero would have lookedup each letter in the bottom row, and replaced it with the letter above it.

Actually, if you are going to do it by hand, it’s best not to look up one letter at a time this way. It’s faster, becauseof the way our memory works, to look up the first letter, L , and then change all the L’s to i, all the way down themessage. Then change all the F’s, then all the D’s, etc.

Example 8. The following is the ciphertext from a shift cipher (Note: the code below does not show you the real wordbreaks. In other words, the spaces you see below are not where the spaces occur in the original message. I’ve writtenthe spaces here just because it’s easier to work with letters in groups of 5 instead of all of them written side by side).

Attack and decrypt this message.1Almost all of the above information, and the message below, comes from Caesar’s own writings, The Gaulic Wars.


SQZQD MXIME TUZSF AZIUX XODAE EFTQP QXMIM DQAZF TQZUS TFARF

TQFIQ ZFKEU JFTTQ IUXXM FFMOW FDQZF AZ

Solution. There are two ways to attack this: brute force, and frequency count. We’ll learn both in this example.The brute force method is to try each shift, for at least the first few words, until we find something that makes

sense. The fastest way to do this by hand is to fill it in one column at a time, working downwards, shifting down thealphabet by one position at each row.

SQZQD MXIME TUZSF AZIUXtrare nyjnf uvatg bajvyusbsf ozkog vwbuh cbkwzvtctg palph wxcvi dclxawuduh qbmqi xydwj edmybxvevi rcnrj yzexk fenzcywfwj sdosk zafyl gfoadzxgxk teptl abgzm hgpbeayhyl ufqum bchan ihqcfbzizm vgrvn cdibo jirdgcajan whswo dejcp kjsehdbkbo xitxp efkdq lktfieclcp yjuyq fgler mlugjfdmdq zkvzr ghmfs nmvhkgener alwas hingt onwilhfofs bmxbt ijohu poxjmigpgt cnycu jkmiv qmykn

Of course, you don’t know ahead of time how far down you should go, or how far to the right you need to go. But, youcould always start with a medium size, say 10 rows and 10 columns, and then add a few more rows and a few morecolumns, until you find what makes sense.

In this case, you should be able to spot that the row starting with g is the only one that makes sense. This means todecrypt we should use a shift of S to g. The easiest way to turn this into the whole decryption is to start with g in theright spot:

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 Cg

and then start to fill in the rest of the alphabet, starting with g:

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 Cg h i j

and eventually get the whole alphabet

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 Cr 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


We’ll get the whole message below.There’s another way to attack this code. We all know that e is the most common letter in ordinary English. Go

through the message and count how many times each letter shows up. If you’re doing it by hand, here’s the fastestway. Start by writing out the alphabet. Now work on one letter at a time, and count them up. So, start by counting allthe A’s, then all the B’s, etc. Here are the totals

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 Z5 0 0 4 4 12 0 0 5 1 1 0 6 0 2 1 10 1 3 7 5 0 1 6 0 8

The simplest guess here is that e corresponds to F. If this is right, then decription would be to shift each letter backwardsin the alphabet by one spot. However, this clearly produces nonsense:

Cipher: SQZQDplain?: rpypc

Of course, not every English text has more e’s than any other letter. Sometimes e might be the second most commonletter. In this case, that would mean e corresponds to Q. As we know from our brute force attack above, this is correct.

You should easily be able to finish the rest of the decryption now and get

“gener alwas hingt onwil lcros sthed elawa reont henig htoft hetwe ntysi xthhe willa ttack trent on”Or, in prettier formatting,

“General Washington will cross the Delaware on the night of the twenty sixth. He will attack Trenton.”Note: this message is based on real events: “George Washington’s crossing of the Delaware River, which occurred onthe night of December 25–26, 1776, during the American Revolutionary War, was the first move in a surprise attack. . . They defeated British reinforcements under Lord Cornwallis at Trenton on January 2, 1777, and defeated his rearguard at Princeton on January 3.” This was first major success of the revolutionary war.


Example 9. [Very, very hard!] The following is a general substitution cipher. It may help to know that the text isformed from the extracts of two actual letters, between Mary Queen of Scots, while she was imprisoned in the Towerof London, and one of her supporters, Babington. They discuss plans to rescue her.

The letters were intercepted, decoded, and this was what lead to Mary being executed for treason to Queen Eliza-beth.

NOQMA TCDIP NXOMP FDHYI KTPWF OXPFD QIQMK DMXMO AOBDH YDJWD

OXUFO ABNDS DWQPY JEFDM UDIFT ZZBDA THDXM DDPFD MDBDI YSJOB

ZDEDJ PZDAD JTZZA NKMYR TPDXM YDJHI UFOBN PFDVD TZPFD NBDTM

POPFD WTPFO ZYWWT QIDTJ HNOQM ATCDI PNIID MRYWD UYZZQ JHDMP

TGDPF DPMTE YWTZD SDWQP YOJNO QMIDM RTJPB TBYJE POJPM QIPNI

DMRTJ PNOQA QIPDS TAYJD HQZNF OUPFD IYSED JPZDA DJDSK DWPPO

WTMMN OQPPF DYMKZ TJPFD TXXTY MIBDY JEPFQ IKMDK TMDHP FDJIF

TZZYP BDPYA DPOID PPFDI YSEDJ PZDAD JOJUO MGBDU TMDPF TPJOJ

DOXNO QMADI IDJED MIWTM MNTJN ZDPPD MIQKO JPFDA IDZRD IPTGD

FDDHO XIKYD ITJHX TZIDB MDPFM DJPFT PTMDT AOJEI PNOQG DDKJD

RDMTK TKDMT BOQPN OQPFT PATNY JTJNI OMPHO FTMAF DMDYJ YFTRD

WOAAT JHDHT AOMDT AKZDT ZKFTB DPPOB DATHD XOMNO QUFYW FFDMD

UYPFN OQIFT ZZMDW DYRDA TMN

The frequency counts for single letters and letter pairs are attached. Crack it and provide a translation.Hint: you may want to know the following:

• The most common English letters are (in order): e, t, r, n, s, o, i, a

• A single letter which appears next to a variety of other letters is probably one of: e, a, o, i, u, t, n.

• The most common English letter pairs are (in order): th, he, in, er, re, on, an, en, at, es, ed, te, ti, or

• The most common English letter pairs that are also commonly reversed are: er, es, ed


Table 6.1: Single letter frequencies for Example 9A B C D E F G H I J K L M N O P Q R S T U V W X Y Z23 16 2 99 9 36 4 14 32 35 14 0 45 23 39 56 22 8 7 51 9 1 15 12 27 24

Table 6.2: Letter pair frequencies for Example 9A 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 1 1 0 5 0 1 0 0 1 0 1 0 0 1 3 0 1 0 0 7 0 0 0 0 1 0B 0 0 0 9 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 1 0 0 0 0 1 1C 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0D 7 2 0 3 1 1 0 6 9 11 2 0 13 1 2 10 1 1 4 7 3 1 6 3 4 1E 0 0 0 4 0 1 0 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 1 0F 0 0 0 17 0 1 0 0 0 0 0 0 1 1 5 0 1 0 0 9 0 0 0 0 1 0G 0 1 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0H 0 0 0 4 0 0 0 0 1 0 0 0 0 1 2 1 1 0 0 1 0 0 0 1 2 0I 0 1 0 8 0 3 0 0 2 0 3 0 0 0 1 6 2 0 0 1 1 0 1 0 3 0J 0 0 0 4 5 0 0 5 1 0 0 0 0 3 3 9 0 0 0 2 1 0 1 0 1 0K 0 0 0 3 0 1 0 0 0 1 0 0 2 0 1 0 0 0 0 3 0 0 0 0 1 2L 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0M 4 0 0 12 0 0 1 0 4 0 2 0 2 4 1 4 1 3 0 3 1 0 0 1 2 0N 0 1 0 1 0 1 0 0 3 0 1 0 0 0 10 1 0 0 0 1 0 0 0 1 1 1O 3 4 0 0 0 1 0 0 1 6 0 0 5 0 0 1 11 0 0 0 1 0 1 4 0 1P 1 2 0 3 0 20 0 1 0 1 0 0 2 6 5 5 0 0 0 3 0 0 1 0 3 3Q 1 0 0 0 0 0 1 0 6 1 1 0 5 0 0 5 0 0 0 0 1 0 0 0 0 1R 0 0 0 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 1 0S 0 0 0 2 2 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0T 4 3 2 0 1 0 2 2 0 8 2 0 8 1 0 6 1 1 0 0 0 0 0 1 1 8U 0 0 0 1 0 3 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 2 0V 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0W 0 0 0 3 0 2 0 0 0 0 0 0 0 0 1 1 2 0 0 5 0 0 1 0 0 0X 0 0 0 0 0 0 0 0 1 0 0 0 3 1 2 1 0 0 0 2 1 0 0 1 0 0Y 1 0 0 3 0 1 0 0 1 6 0 0 2 0 1 2 0 2 3 0 0 0 4 0 0 1Z 1 1 0 7 0 0 0 0 1 0 1 0 1 1 0 1 1 1 0 1 0 0 0 0 2 5


Solution. We start by comparing the most common cipherletters with the most common plaintext letters:

D P T M O F J Ie t n r o i s a

Since D is such an outlier, we are pretty confident that D = e. Maybe P = t, but this is less clear.Now we list the most common ciphertext and plaintext digraphs:

PF FD DM MD DJ OQ DP NO DI FTth he in er re on an en at es

Earlier we wondered if maybe P = t. This is consistant with the first digraphs. It suggests too that F = h. The next twomost common ciphertext digraph are DM and MD, a reversal. The most common reversal is er and re, so we guess thatM = r. To recap our letters:

plain 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 zcipher D F M P

So far we are pretty confident about all our guesses. The next couple are more speculative. We list the remainingciphertext and plaintext letters and digraphs:

T O J In o i s a

eJ OQ NO eI hTin on an en at es

Looking at the digraphs, we see that two of the ciphertex digraphs begin with e. The next two plaintext digraphs thatbegin with e are en and es. Looking at the single letter frequency leaders we see that J and I both show up in theciphertext and n and s both show up in the plaintext. Thus, we think we have (J,I) = (n,s) or (J,I) = (s,n). Wedon’t have a lot of reason to prefer one of these guesses more than the other. We’ll try both in a moment, but for nowlet’s move on to the next letter.

T Oo i a

OQ NO hTin on an at

We think that T and O are both probably vowels (this is confirmed by looking at the digraph table where each of themshows up pretty frequetly on both sides of digraphs). Looking at plaintext digraphs, there is only one more commondigraph that starts with h, and it’s ha. So, we make a guess that T = a. Looking at OQ and NO we scan down the plaintexdigraphs that haven’t been used and find the next ones that begin and end with the same vowel, and that aren’t alreadymatched up with the ciphertext. We see in and the next on with an i is ti, but we already know what t is, and it’s notin OQ or NO. There are no other common plaintext digraphs of the form ?i. So, probably O 6=i. This leaves O = o.

We add our last guesses to our list

plain 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 zcipher T D F J O M I P

and now start trying to translate the ciphertext.

NoQrA aCest NXort heHYs KatWh oXthe QsQrK erXro AoBeH YenWe oXUho ABNeS eWQtYnEher Uesha ZZBeA aHeXr eethe reBes YSnoB ZeEen tZeAe naZZA NKrYR ateXr YenHsUhoBN theVe aZthe NBear tothe Watho ZYWWa Qsean HNoQr AaCes tNsse rRYWe UYZZQ


nHert aGeth etraE YWaZe SeWQt YonNo Qrser RantB aBYnE tontr QstNs erRan tNoQAQsteS aAYne HQZNh oUthe sYSEe ntZeA eneSK eWtto WarrN oQtth eYrKZ anthe aXXaYrsBeY nEthQ sKreK areHt hensh aZZYt BetYA etose tthes YSEen tZeAe nonUo rGBeUareth atnon eoXNo QrAes senEe rsWar rNanN Zette rsQKo ntheA seZRe staGe heeHoXsKYe sanHX aZseB rethr entha tarea AonEs tNoQG eeKne ReraK aKera BoQtN oQthatAaNY nanNs ortHo harAh ereYn YhaRe WoAAa nHeHa Aorea AKZea ZKhaB ettoB eAaHeXorNo QUhYW hhere UYthN oQsha ZZreW eYReA arN

At this point there is no formula: simply read through the passage and see if your brain recognizes any word fragments.See if you can fill in missing letters to make these fragments fit. In line 1, group 3, we see NXort he which could befor the. In line 3, group 1 we see Xr eethe which could be free the. So we proceed with guessing X = f.

NoQrA aCest Nfort heHYs KatWh ofthe QsQrK erfro AoBeH YenWe ofUho ABNeS eWQtYnEher Uesha ZZBeA aHefr eethe reBes YSnoB ZeEen tZeAe naZZA NKrYR atefr YenHsUhoBN theVe aZthe NBear tothe Watho ZYWWa Qsean HNoQr AaCes tNsse rRYWe UYZZQnHert aGeth etraE YWaZe SeWQt YonNo Qrser RantB aBYnE tontr QstNs erRan tNoQAQsteS aAYne HQZNh oUthe sYSEe ntZeA eneSK eWtto WarrN oQtth eYrKZ anthe affaYrsBeY nEthQ sKreK areHt hensh aZZYt BetYA etose tthes YSEen tZeAe nonUo rGBeUareth atnon eofNo QrAes senEe rsWar rNanN Zette rsQKo ntheA seZRe staGe heeHofsKYe sanHf aZseB rethr entha tarea AonEs tNoQG eeKne ReraK aKera BoQtN oQthatAaNY nanNs ortHo harAh ereYn YhaRe WoAAa nHeHa Aorea AKZea ZKhaB ettoB eAaHeforNo QUhYW hhere UYthN oQsha ZZreW eYReA arN

Line 9, group 1 has affaY rs which certainly looks like affairs. Thus, we guess that Y = i. Line 9, starting atgroup 5 has t hensh aZZ looks like then shall. Thus, we guess taht Z = l.

NoQrA aCest Nfort heHis KatWh ofthe QsQrK erfro AoBeH ienWe ofUho ABNeS eWQtinEher Uesha llBeA aHefr eethe reBes iSnoB leEen tleAe nallA NKriR atefr ienHsUhoBN theVe althe NBear tothe Watho liWWa Qsean HNoQr AaCes tNsse rRiWe UillQnHert aGeth etraE iWale SeWQt ionNo Qrser RantB aBinE tontr QstNs erRan tNoQAQsteS aAine HQlNh oUthe siSEe ntleA eneSK eWtto WarrN oQtth eirKl anthe affairsBei nEthQ sKreK areHt hensh allit BetiA etose tthes iSEen tleAe nonUo rGBeUareth atnon eofNo QrAes senEe rsWar rNanN lette rsQKo ntheA selRe staGe heeHofsKie sanHf alseB rethr entha tarea AonEs tNoQG eeKne ReraK aKera BoQtN oQthatAaNi nanNs ortHo harAh erein ihaRe WoAAa nHeHa Aorea AKlea lKhaB ettoB eAaHeforNo QUhiW hhere UithN oQsha llreW eiReA arN

Line 4, groups 1–2 have atefr ienHs which looks like it contains friends. Thus, we guess H = d. There area few spots where B shows up and it looks like it could be b, although not with a lot of certainty: line 2, group 1,AoBeH (maybe to be something); line 2, groups 7–8, Uesha llBeA (maybe shall be something); line 3 groups2–5, eethe reBes iSnoB leEen (maybe there be ? noble ?). We apply B = b.

NoQrA aCest Nfort hedis KatWh ofthe QsQrK erfro Aobed ienWe ofUho AbNeS eWQtinEher Uesha llbeA adefr eethe rebes iSnob leEen tleAe nallA NKriR atefr iendsUhobN theVe althe Nbear tothe Watho liWWa Qsean dNoQr AaCes tNsse rRiWe UillQndert aGeth etraE iWale SeWQt ionNo Qrser Rantb abinE tontr QstNs erRan tNoQAQsteS aAine dQlNh oUthe siSEe ntleA eneSK eWtto WarrN oQtth eirKl anthe affairsbei nEthQ sKreK aredt hensh allit betiA etose tthes iSEen tleAe nonUo rGbeUareth atnon eofNo QrAes senEe rsWar rNanN lette rsQKo ntheA selRe staGe heedofsKie sandf alseb rethr entha tarea AonEs tNoQG eeKne ReraK aKera boQtN oQthatAaNi nanNs ortdo harAh erein ihaRe WoAAa ndeda Aorea AKlea lKhab ettob eAadeforNo QUhiW hhere UithN oQsha llreW eiReA arN


The rest proceeds similarly, and we eventually get:

yourm ajest yfort hedis patch ofthe usurp erfro mobed ience ofwho mbyex ecutingher wesha llbem adefr eethe rebes ixnob legen tleme nallm ypriv atefr iendswhoby theze althe ybear tothe catho licca usean dyour majes tysse rvice willundert aketh etrag icale xecut ionyo urser vantb abing tontr ustys ervan tyoumustex amine dulyh owthe sixge ntlem enexp ectto carry outth eirpl anthe affairsbei ngthu sprep aredt hensh allit betim etose tthes ixgen tleme nonwo rkbewareth atnon eofyo urmes senge rscar ryany lette rsupo nthem selve stake heedofspie sandf alseb rethr entha tarea mongs tyouk eepne verap apera bouty outhatmayi nanys ortdo harmh erein ihave comma ndeda morea mplea lphab ettob emadeforyo uwhic hhere withy ousha llrec eivem ary

Below, I have the original passage (the salutations and signatures have been added by me)

[Your Majesty,]

For the dispatch of the Usurper, from obedience of whom by executing her we shall be made free, therebe six noble gentlemen, all my private friends, who by the zeal they bear to the Catholic cause and yourmajesty’s service will undertake the tragical execution.

[Your servant,Babington]

[Trusty servant,]

You must examine duly how the six gentlemen expect to carry out their plan. The affairs being thusprepared then shall it be time to set the Six Gentlemen on work. Beware that none of your messengerscarry any letters upon themselves. Take heed of Spies and false-brethren that are amongst you. Keepnever a paper about you that may in any sort do harm: Herein I have commanded a more ample Alphabetto be made for you, which herewith you shall receive.

[Mary]

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

6.1 graph theory · so now, the original question becomes: can you ﬁnd a path that goes through...

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