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Welcome to the National Cipher Challenge 2015. Registration begins on 5th October 2015, though you might find you can register from 30th September while we warm up the website!

This year's competition centres on my adventures in post war Berlin. Join me in tracking down the Reichsdoktor by helping me to crack the enemy's codes and ciphers.

The first Challenge will go live at 3.15pm on Thursday 15th October, so you have plenty of time to prepare. You can find some notes on codebreaking somewhere on the website (you thought we would make this easy for you?) and there is a forum where you can discuss the Challenge.

Looking forward to hearing from you,

Harrys signature

On substitution ciphers

Graham A. Niblo

These notes form a brief introduction to substitution ciphers, to accompany the lesson plans provided with the University of Southampton National Cipher Challenge, 2005. We would like to thank Hugh Evans of Sholing City Technology College for his assistance in the design of these teaching materials.

Caesar shift ciphers

The easiest method of enciphering a text message is to replace each character by another using a fixed rule, so for example every letter a may be replaced by D, and every letter b by the letter E and so on.

Applying this rule to the previous paragraph produces the text

WKH HDVLHVW PHWKRG RI HQFLSKHULQJ D WHAW PHVVDJH LV WR UHSODFH HDFK FKDUDFWHU EB DQRWKHU XVLQJ D ILAHG UXOH, VR IRU HADPSOH HYHUB OHWWHU D PDB EH UHSODFHG EB G, DQG HYHUB OHWWHU E EB WKH OHWWHU H DQG VR RQ.

(Note the convention in these notes that ciphertext is written in capital letters, while plaintext is usually lowercase.)

Such a cipher is known as a shift cipher since the letters of the alphabet are shifted round by a fixed amount, and as a Caesar shift since such ciphers were used by Julius Caesar. To decode a Caesar shift it is enough to work out the amount of shift, which can be done, for example, by discovering which character has replaced the letter e. In the example above we might guess that the three-letter word starting the sentence is the and therefore that the letter e has been replaced by H. A quick check shows that the Caesar shift by 3 does indeed encode the word the as WKH, and it is easy to complete the decryption.

In fact there are only 26 Caesar shift ciphers (and one of them does nothing to the text) so it is not too hard to decipher the text by brute force. We can try each of the shifts in turn on the first word of the cipher text until we discover the correct shift. This process can be simplified by using a cipher wheel, a simple mechanical device that allows one to generate each of the Caesar shift ciphers, and to encode or decode messages using it. At the front of this pack you will find a sheet, which can be photocopied onto thin card in order to make a cipher wheel. Cut out the two discs, and fasten through their centres with a paper fastener to make the wheel. Use the convention that you read plaintext on the outer rim of the wheel and cipher text from the smaller wheel.

Keyword substitution ciphers

To increase the difficulty of deciphering the text we need a richer family of ciphers. A good example is furnished by the keyword cipher. In this we design an encryption table by choosing a keyword or phrase, which is used to jumble the alphabet as follows:

Write down the phrase, with no spaces between the letters, and omitting any repeated character. So if the phrase is “The Simpsons” we write down THESIMPON. Now we continue to go round the alphabet until every letter appears exactly once, and write the list under the standard alphabet:

abcdefghijklmnopqrstuvwxyz

THESIMPONQRUVWXYZABCDFGJKL

Of course if the key phrase is carefully chosen (for example “The quick brown fox jumps over the lazy dog”) we may not need to complete the list as above, but such a choice is not necessary. The number of such ciphers is 26!, or approximately 4x10²⁶, and brute force cannot be used to attack the problem. However an attack is possible.

Consider the text

VEP HYXHLVHTP MO AWFJYFLT H RFNEPS HJNEHAPV FL VEFU ZHC FU VEHV FV FU PHUC VM KPKMSFUP VEP IPCZMSY MS IPCNESHUP, HLY EPLRP VEP RFNEPS HJNEHAPV. VEFU FU FKNMSVHLV, APRHWUP FO VEP UPLYPS EHU VM IPPN VEP RFNEPS HJNEHAPV ML H NFPRP MO NHNPS, VEP PLPKC RHL RHNVWSP VEP NHNPS, YFURMXPS VEP IPC, HLY SPHY HLC RMKKWLFRHVFMLU VEHV EHXP APPL PLRSCNVPY ZFVE FV. EMZPXPS FO VEP IPC RHL AP RMKKFVVPY VM KPKMSC FV FU JPUU JFIPJC VM OHJJ FLVM PLPKC EHLYU.

As before we notice that the first word has three letters and, since it occurs several times, may well be the word the. This gives a strong hint that the letter e is enciphered as the letter P in the keyphrase cipher. Of course other three letter words are possible, e.g., and or but. Nonetheless a quick check shows us that the letter P is the most common letter in the enciphered text, just as e is the most common letter in English so it is reasonable to assume that the correct decryption translates P to e. This also suggests that V stands for t and E for h, allowing us to begin to decipher the text. We will use the convention that uppercase letters denote enciphered letters and lowercase denotes plaintext characters:

the HYXHLtHTe MO AWFJYFLT H RFNheS HJNhHAet FL thFU ZHC FU thHt Ft FU eHUC tM KeKMSFUe the IeCZMSY MS IeCNhSHUe, HLY heLRe the RFNheS HJNhHAet. thFU FU FKNMStHLt, AeRHWUe FO the UeLYeS hHU tM IeeN the RFNheS HJNhHAet ML H NFeRe MO NHNeS, the eLeKC RHL RHNtWSe the NHNeS, YFURMXeS the IeC, HLY SeHY HLC RMKKWLFRHtFMLU thHt hHXe AeeL eLRSCNteY ZFth Ft. hMZeXeS FO the IeC RHL Ae RMKKFtteY tM KeKMSC Ft FU JeUU JFIeJC tM OHJJ FLtM eLeKC hHLYU.

Reading carefully we see the single letter word H, and the four letter word thHt in the first line, and guess that H enciphers the letter a. Making that replacement we get:

the aYXaLtaTe MO AWFJYFLT a RFNheS aJNhaAet FL thFU ZaC FU that Ft FU eaUC tM KeKMSFUe the IeCZMSY MS IeCNhSaUe, aLY heLRe the RFNheS aJNhaAet. thFU FU FKNMStaLt, AeRaWUe FO the UeLYeS haU tM IeeN the RFNheS aJNhaAet ML a NFeRe MO NaNeS, the eLeKC RaL RaNtWSe the NaNeS, YFURMXeS the IeC, aLY SeaY aLC RMKKWLFRatFMLU that haXe AeeL eLRSCNteY ZFth Ft. hMZeXeS FO the IeC RaL Ae RMKKFtteY tM KeKMSC Ft FU JeUU JFIeJC tM OaJJ FLtM eLeKC haLYU.

Now the two 2 letter words Ft FU are probably “it is” meaning that F enciphers “i” and U enciphers “s”. Hence we get:

the aYXaLtaTe MO AWiJYiLT a RiNheS aJNhaAet iL this ZaC is that it is easC tM KeKMSise the IeCZMSY MS IeCNhSase, aLY heLRe the RiNheS aJNhaAet. this is iKNMStaLt, AeRaWse iO the seLYeS has tM IeeN the RiNheS aJNhaAet ML a NieRe MO NaNeS, the eLeKC RaL RaNtWSe teh NaNeS, YisRMXeS the IeC, aLY SeaY aLC RMKKWLiRatiMLs that haXe AeeL eLRSCNteY Zith it. hMZeXeS iO the IeC RaL Ae RMKKitteY tM KeKMSC it is Jess JiIeJC tM OaJJ iLtM eLeKC haLYs.

Continuing with appropriate guesses (haXe = have, easC = easy and so on) we decipher the text to get the following extract from Simon Singh’s excellent history of codes and ciphers, The Code Book:

“The advantage of building a cipher alphabet in this way is that it is easy to memorise the keyword or keyphrase, and hence the cipher alphabet. This is important, because if the sender has to keep the cipher alphabet on a piece of paper, the enemy can capture the paper, discover the key, and read any communications that have been encrypted with it. However if the key can be committed to memory it is less likely to fall into enemy hands."

Frequency analysis

A more methodical attack is frequency analysis. One counts the number of occurrences of each character in the cipher text and compares it with an expected frequency for the standard English alphabet. In the cipher text above a character count gives us the following table of occurrences:

a	b	c	d	e	f	g	h	i	j	k	l	m
7	0	12	0	26	3	0	32	6	9	11	20	18

n	o	p	q	r	s	t	u	v	w	x	y	z
16	5	55	0	14	17	2	17	35	4	4	11	4

Compare this to a table of expected frequencies, taken from Simon Singh’s “The Code Book”:

a	b	c	d	e	f	g	h	i	j	k	l	m
8.2	1.5	2.8	4.3	12.7	2.2	2.0	6.1	7.0	0.2	0.8	4.0	2.4

n	o	p	q	r	s	t	u	v	w	x	y	z
6.7	7.5	1.9	0.1	6.0	6.3	9.1	2.8	1.0	2.4	0.2	2.0	0.1

Using this and information about common one, two and three letter words we have enough to begin to tackle the cipher.

Disguising the word structure

The chink in the armour of our ciphers so far has been the preservation of word structure. This allows one to spot common words. In order to avoid such weakness cryptographers usually remove punctuation and block the characters together in groups of four or five, so our previous cipher text looks like

VEPHY XHLVH TPMOA WFJYF LTHRF NEPSH JNEHA PVFLV EFUZH CFUVE HVFVF UPHUC VMKPK MSFUP VEPIP CZMSY MSIPC NESHU PHLYE PLRPV EPRFN EPSHJ NEHAP VVEFU FUFKN MSVHL VAPRH WUPFO VEPUP LYPSE HUVMI PPNVE PRFNE PSHJN EHAPV MLHNF PRPMO NHNPS VEPPL PKCRH LRHNV WSPVE PNHNP SYFUR MXPSV EPIPC HLYSP HYHLC RMKKW LFRHV FMLUV EHVEH XPAPP LPLRS CNVPY ZFVEF VEMZP XPSFO VEPIP CRHLA PRMKK FVVPY VMKPK MSCFV FUJPU UJFIP JCVMO HJJFL VMPLP KCEHL YU

Usually the length of the text groups doesn’t matter, however, in analysing a Vigenère cipher (see below) a carelessly chosen block length may make the length of the keyword more apparent, since it can reveal the repetitions more easily.

To attack cipher text that has been grouped in this way we have to work with letters not words. To do so we use the frequency analysis described above, together with a little judgement (or luck!). The process can be hard, but wars have been won or lost on the back of it, and so have fortunes.

“It was hard going, but Jericho didn’t mind. He was taking action, that was the point. It was the same as code-breaking. However hopeless the situation, the rule was always to do something. No cryptogram, Alan Turing used to say, was ever solved by simply staring at it.” From Enigma, by Robert Harris.

Affine shift ciphers

Despite the advantages for an agent in using keyword substitution ciphers most modern ciphers are automated and rely on a mathematical encryption algorithm. Indeed the Caesar shift cipher can be viewed as just such a cipher:

We start by encoding each letter by its numerical position in the alphabet:

a	b	c	d	e	f	g	h	i	j	k	l	m
1	2	3	4	5	6	7	8	9	10	11	12	13

n	o	p	q	r	s	t	u	v	w	x	y	z
14	15	16	17	18	19	20	21	22	23	24	25	26

Next we shift the alphabet by adding 3 to each position:

a	b	c	d	e	f	g	h	i	j	k	l	m
4	5	6	7	8	9	10	11	12	13	14	15	16

n	o	p	q	r	s	t	u	v	w	x	y	z
17	18	19	20	21	22	23	24	25	26	1	2	3

Of course 24+3 = 27 ≠ 1, but here we are carrying out modular arithmetic, familiar as clock arithmetic, so that when we reach 26 we continue from 1.

Finally we replace the numbers with the letters they stand for:

This recovers the cipher table constructed in lesson plan 1 for the Caesar shift by 3.

There is a convenient shorthand for the Caesar shift by n, given by xà x+n. It is confusing since here we are using x to stand for the position of a letter, and n to stand for the shift amount, i.e., x and n are each one of the values 1 … 26. It is clear that since the shift is defined by the integer n there are only 26 Caesar shift ciphers.

There is a bigger class of shift ciphers which can be written in these terms known as the affine shift ciphers, and they exploit the fact that we can multiply as well as add integers in modular arithmetic. It is slightly complicated to set up formally but rather easy to do in practice so we will work through an example.

The affine shift x -> 3x+5

We start as before with the position table, but this time instead of replacing a position x with the number x+3 we will replace it by the number 3x+5, where this number is interpreted appropriately. So for example 2 à 3.2+5 = 11, while 8 à 3.8+5 = 29 which is interpreted as 3 (29=26+3). Whenever the result of the computation is larger than 26 we keep subtracting 26 until it becomes smaller. More formally we compute 3x+5 and then take the remainder after division by 26. This yields the table:

a	b	c	d	e	f	g	h	i	j	k	l	m
8	11	14	17	20	23	26	3	6	9	12	15	18

n	o	p	q	r	s	t	u	v	w	x	y	z
21	24	1	4	7	10	13	16	19	22	25	2	5

And from this we recover the encryption table as given on the handout for lesson 3:

The affine shift ciphers can also be written in a shorthand form x à ax+b and the Caesar shift ciphers are special cases of the affine shift ciphers with a=1.

Now notice that in both the Caesar shift x -> x+3 and the affine shift x à 3x+5 the letter y is enciphered as B, since 25+3 = 28 = 26+2, and 3.25+5 = 80 = 3.26+2. It follows that two different affine shift ciphers can encrypt a letter in the same way, so it is no longer sufficient to discover the letter substituting for e in order to decipher the message. Since there are two degrees of freedom in our choice of cipher we might hope that deciphering two letters is sufficient, and it is, since, if we know two values of the expression ax+b we can solve the two corresponding simultaneous equations to find a and b.

We may be more familiar with this exercise when solving the equations over the real numbers, but the same method works for modular arithmetic, with the caveat that in general we cannot divide. This caveat has an interpretation in cryptography. In order for the rule x -> ax+b to define a cipher it had better be the case that each of the numbers 1 … 26 appears exactly once in the list of numbers ax+b as x ranges from 1 to 26. If we choose a carelessly (so that we can’t divide by a mod 26) this might not be the case.

For example the rule x -> 2x tries to encipher both m and z as Z, since 2.13=26 and 2.26 = 52 both of which are equal to 26 modulo 26. Such an encryption cannot easily be deciphered since the recipient of the message is unable to determine whether the sender intended Z to be read as m or z.

From a mathematician’s point of view the enciphering rule defines a function from the alphabet to itself, and this needs an inverse if the cipher is to be decipherable in a deterministic way. In other words the number theory function x -> ax+b needs to have an inverse in mod 26 arithmetic. It is a fact from elementary number theory that it will have such an inverse if and only if a is coprime to 26, that is, their only common divisor is 1.

There are 12 numbers less than 26 and coprime to it (those odd numbers not divisible by 13) so we have 12 possible choices of the number a, and 26 choices for the number b, yielding 312 affine shift ciphers. This makes a brute force attack, without frequency analysis, less practical than the much simpler situation for Caesar shift ciphers.

Polyalphabetic ciphers

The main weakness allowing us to tackle a substitution cipher is the irregularity in the distribution of letters in English text. Other languages demonstrate similar (though language specific) irregularities and you can find frequency tables for them on the web.

In order to remove this weakness from a cipher it is necessary to disguise the frequencies of letters in the plaintext and the easiest way to do this is by using a polyalphabetic cipher. In such a cipher each plaintext letter may be encoded in more than one way so that, for example, the letter e may be enciphered as both X and G within the ciphertext. One problem with this approach is that if X and G both encode for e we don’t have enough letters left to encode the other 25 letters. One elegant solution to this problem is the famous French cipher known as the Vigenère cipher.

In a Vigenère cipher ANY letter might be encoded by any other; a given Vigenère cipher uses a subset of the 26 possible Caesar shift ciphers. Of course for a genuine recipient to have any hope of deciphering the message there has to be a way to determine for each cipher character which of the shifts has been used. The answer to this tricky problem is to choose a sequence of them known to both parties but to no-one else.

So the two parties might agree to use shifts of 22, 9, 7, 5, 14, 5 18, and 5 in that order and to continue repeating the pattern for the entire text: 22, 9, 7, 5, 14, 5 18, 5, 22, 9, 7, 5, 14, 5 18, 5, 22, 9 etc..

In order to decode the cipher text the recipient shifts the first cipher character back by 22, the second back by 9 and so on to recover the cipher text. Of course the question remains how one can memorise the correct sequence, but here we borrow an idea from the keyword cipher. The shift numbers 1, …, 26 are taken to stand for the alphabet a, …, z, and then the pattern 22, 9, 7, 5, 14, 5 18, 5 spells the word vigenere.

To set up a Vigenère cipher the two parties agree in advance to use the shift pattern encoded by some agreed keyword or phrase; in our previous Golden Jubilee Cipher challenge we used a Vigenère cipher based on the keyword GOLD, so characters were shifted in turn by 7, 15, 12, 4. Such a cipher is very hard to crack.

The method we recommend is due to Babbage and Kasiski who independently discovered it, and is based on the regularity of the repetition. An analysis of repeated strings of letters is used to try to determine the length of the keyword, and once this is done a standard frequency analysis is applied to each part of the ciphertext encoded by a single cipher. A very good account of Babbage-Kasiski deciphering can be read in chapter 2 of Simon Singh’s The Code Book.

On transposition ciphers

Sometimes when you carry out a frequency analysis you will find that each letter occurs with about the same frequency as you would expect in natural English text (or whichever language you are studying). This is a broad hint that the text is not enciphered using a substitution cipher, but rather by an anagram or transposition cipher, also known as an anagram cipher. In such a cipher the letters of the message are not replaced by substitutes, but rather jumbled using some rule which allows them to be untangled again to decipher the message.

Example

We will encipher the text:

The quick brown fox jumps over the lazy dog

We start with a keyword. Suppose our keyword is BAD. We write it at the head of a table with three columns, then enter the ciphertext in the boxes below. The last, empty, box is paddded with an X.

B	A	D
t	h	e
q	u	i
c	k	b
r	o	w
n	f	o
x	j	u
m	p	s
o	v	e
r	t	h
e	l	a
z	y	d
o	g	x

Next we rearrange the columns so that the letters in the keyword are now in alphabetic order

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

Giving a ciphertext of

HTEUQIKCBORWFNOJXUPMSVOETRHLEAYZDGOX

If the keyword contains repeated letters then we delete them as we would if it were the keyword for a substitution cipher before constructing the grid. Hence if the keyword was TOFFEE we would use a grid of width 4 with header TOFE and we would rearrange the grid so that its header appeared as EFOT to encipher the message.

How do we tackle such a cipher?

Clearly the length of the keyword is quite crucial. You should be able to guess this from the length of the ciphertext, which will be a multiple of it. So in our example the ciphertext has length 36 which has factors 2,3,4,6,9 and so on. Hence we could try laying out the text in grids of width 2,3,4,6,9 respectively (a keyword of length 12 or more is unlikely) and examining the rows.

Of course a grid of width 2 would leave us just switching alternate letters so we probably don’t need to lay it out that way to check it. Having checked and dismissed the idea of a keyword of length 2 the first grid we try looks like the second grid above. Having got to this point the best hope for a quick solution is to find a crib. If there is a word you think ought to appear in the cipher text then you could try looking for anagrams of that word. This is made difficult by the fact that in splitting the text into blocks (blocks of three in the example), If your crib word does not take up an entire block then even the characters from the crib that do appear will be jumbled with other nearby characters, so you need a reasonably long crib. On the other hand if it is too long only part of the word will appear in that block so you are looking for anagrams of parts of the crib.

In our example if we knew, for some reason, that the text was likely to contain the word “jumps” we could look for anagrams of “jum”, “ump”, “mps. Looking carefully you should see the anagram PMS in the text and we might guess that the first and second columns have been transposed while the third has remained fixed. Checking this we find have cracked the cipher.

Things are harder with longer keywords but the principle remains the same. Things get tougher if the plaintext is not in our own language, since it is harder to say what makes sense. Of course even in this case it may be that part of the message is in your language and the rest in another. In this case you might hope to crack the ciphertext corresponding to your native language, and apply the knowledge that gives you about the cipher to write down a decrypt of the entire message, even when the text is unfamiliar.

Other (subtle) cribs: In English the letters q and u occur together so if they are separated either you are not looking at English text or they should be brought back together by undoing the anagram.

Numbers often represent dates, so for example the letters/numbers 2, 1, s, t in proximity might represent 21st, while 2,1,t,h might represent 12th.

About the Challenge

Welcome to the National Cipher Challenge, a nationwide, online codebreaking competition, which will run from October 15th 2015 to January 7th 2016. We will open for online registration on October 5th.

The competition is a great extension activity (or a fantastic maths club project) that can be tackled by students in teams or on their own. There is no charge to register or take part, and all you need to get involved is a reasonably modern web browser.

This year competitors will be challenged to help Harry track down the Reichsdoktor in postwar Berlin. His enemies are cunning and the story will unfold in a series of short encrypted messages. Competitors will work with Harry to decipher these documents in order to untangle the truth.

You can download lesson plans and notes on codebreaking from the Black Chamber on the competition website. This is the competition library and, alongside the challenges themselves, it provides a range of supporting materials that provide a good introduction to the skills needed by a successful code-breaker.

Entrants can take part alone or in teams of any size. To take part you will need to register on the website, and your account also gives you access to the forum where you can discuss a whole range of things connected to the competition, and quite a few that are totally unrelated.

Everyone involved in the challenge can have their own account to allow them access to the forum, but by default only the team captain can post entries for a team with multiple members. If you want to allow other members of the team to edit and submit entries then the team captain can use our Team Builder to invite others to join. Details are given below.

Competition Schedule

Registration will open online on 5th October 2015 and the first challenge will be published at 3.15pm on October 15th. Challenges will be set periodically on the web-site www.cipher.maths.soton.ac.uk in the Challenges section, following this schedule:

Challenge	Publication date 15:15 on	Solution deadline 23:59 on
1	15/10/2015	21/10/2015
2	22/10/2015	04/11/2015
3	05/11/2015	11/11/2015
4	12/11/2015	18/11/2015
5	19/11/2015	25/11/2015
6	26/11/2015	02/12/2015
7	03/12/2015	16/12/2015
8	17/12/2015	07/01/2016

Points are awarded for speed and accuracy (with accuracy more important) but you do not have to rush to download the first challenges immediately as you have a day or two in which you can still get top marks. In later challenges speed will become important, and the full schedule of marks will be published so you can see how quickly you need to get started in each round.

The first two challenges should be thought of as a “warm-up” exercise and will not count in the final leader board rankings or for the award of main prizes, however it is still worth tackling rounds one and two as they give excellent practice and do develop the storyline. There will also be a range of smaller prizes for those challenges and you will still be able to download certificate recording your team’s performance.

As usual we apologise in advance if your school holidays clash with the schedule, but hopefully not too many of you will be affected.

Registration

To take part you will need to register for the competition on our registration page:

https://www.cipher.maths.soton.ac.uk/register

This will be open from October 5thth, and you will need to provide the following information:

Username: This will be the name you use to log on to the site to post comments, and also to submit your entries, check feedback and to print your certificate. Choose something memorable. You can share this info among the team, but do not let anyone else have it as they can use it to pose as you!

Password: Again this is for logging on. Choose it carefully and keep it secret.

Email address: This will be used to confirm your registration so it must be an active account you can check to authorise the account. If we need to contact you this is how we will do it so add the account This email address is being protected from spambots. You need JavaScript enabled to view it. to the account address book to avoid sending our emails to your junk mail bin, make sure the account is not too full, and check it regularly.

Teacher contact: Give the name of a teacher we can write to if we need to check anything. You should get their permission first! We don't usually do this unless you win a prize. If you are home schooled give us a parent or carer's name here and write home schooled in the school name field. You will still need to give us contact details in the address fields below.

School: We have provided a drop down list, sorted alphabetically by school. If yours is missing let us know!

We also ask for some other contact info that is useful if you win a prize/

We will also ask you to say on the registration form how you would like your name to appear on the leader boards and on your certificates. The options are to show your first name, your full name or just the team name (choose “none”), and you can choose a different convention for the online leader board and the certificate.

Why would I click the “Ineligible for a prize” box on the registration form

If you are a teacher who is registering in order to keep an eye on the forum, or a Cipher Challenge alumnus who is now too old to take part but just can’t keep away, or ineligible for some other reason, then please tick this box so that the computer doesn’t award you a prize by mistake! It is embarrassing for us to have to ask for it back. Thanks.

Teams and solo entries

If you are taking part on your own you only need to register as above and this will create a team of 1 with you as Captain. The default team name will be your user ID but you can change that by editing your Profile. If you want to enter as a group you should each sign up for an account and choose one member of the team as Captain. The Captain should change his or her team name to the one you all want, and can then use the Team Profile tab to invite you all to join the team. Once you accept the invite you will be part of that team throughout the competition. You can do this at any time but you cannot transfer any score that you have gained yourself to the team you join.

You do not have to all be at the same school, we will use the Captain’s school and email address for any communications with the team. The names of the members of the team will appear on the leader boards and certificates according to the individual preferences of the team members as discussed above. Anyone in the team can post a solution to the Challenge, and since your marks can’t go down as a result there is never any harm in this. You can also all read the feedback and download individual certificates in the Black Chamber.

The structure of the competition

You will find the Challenges on the Challenge page. Each round of the competition will come in two parts, Part A and Part B. Think of them as the “easy” and the “hard” challenges (or the “hard” and “much harder” challenges if you prefer). Part A challenges will consist of communications between Harry and his friends as they try to find the Reichsdoktor. You can expect these messages to be fairly lightly encrypted, at least at first, although in the latter stages of the competition security will be tightened and you will find the Part A ciphers harder to crack. Part B consists of intercepts that Harry is using in the hunt. At the start of the challenge the encryption is not too hard to crack, but as the net closes you will find that they use much higher security. You may find that learning to use a spreadsheet or even to programme will be of particular value in tackling the later challenges.

Submitting your solutions

You can submit solutions to either part A or part B at any time during a round by typing them into the submissions page. If you need to resubmit (because you found a mistake, or because we pointed one out to you) you can use the same form. Just paste your entry as text in the appropriate box on the form. It doesn’t matter how you format your answer – with or without punctuation and spaces and whether or not you use capital letters, however you must only type or paste in the exact text of a decrypt of the message. It is a good idea to use a simple text editor to type up your solution (rather than something like Word) as the spell checker sometimes tries to change what you are typing and any “mistake” in the text might be deliberate. Don’t try to correct any errors you think we have made, always type in an exact decryption of the text. Don’t try to tell us what cipher we used, or to ask us a question, or to say how you solved the cipher in the entry form, we don’t read it and it will be marked as an error in the solution. If you need to get hold of us you can post a message on the forum or send us an email at This email address is being protected from spambots. You need JavaScript enabled to view it..

Getting help

We offer online feedback on submissions during each round to help you if you make mistakes. The feedback is delayed so you will lose points if you rely on it rather than trying to correct your own errors quickly, but it can be useful if you are on the right track (and speed doesn’t matter for part A challenges which are only scored for accuracy). The feedback consists of a score for accuracy, together with a copy of your submission with the first error highlighted. The feedback also contains a link to your certificate for the round. At the end of each round we will publish the official decrypts of part A and part B on the Solutions page.

Participants often get stuck on a Challenge but, as in real life, sometimes a good night's rest is all you need. Other times you might need more practical help and can turn to the website for clues, either hidden in earlier rounds of the competition, or posted (by us) as comments on the forum. We ask you not to post hints of your own without checking them with us first as this will spoil the Challenge for others. Anyone posting solutions or links to solutions on our site or elsewhere may be barred from the site and disqualified from the competition - we do search for them and do find them!

Scoring

Each of the two challenges in a round (part A and part B) are scored for accuracy in the same way. We strip out all the non-ascii characters, spaces and punctuation from your solution, convert it to lower case and compare that string of letters with our solution, which we have treated the same way. The more similar they are the higher the score you will get, and if they are identical you will score 100% for that challenge. If you spot a mistake in your answer you can submit again - we only ever take your most accurate answer into account and accuracy beats speed in every case, though speed is also important in the part B competition. In part B we look at all your submissions for the round and find those with the highest mark. We then take the first one of those that you submitted and award you points depending on how quickly you submitted it, according to a schedule that is published with each challenge. There are no speed points for part A, only for part B. You can find your scores for each round in the feedback section of the site, and we will publish a leader board for each round. The first two rounds are a warm-up so the points will not count for the overall leader boards but from round 3 we will publish a Championship leader board based on your total points from then in each of the competitions.

Prizes

The GCHQ prize of £1,000 will be awarded to the top individual entry this year as measured by performance in the part B competition. Other prizes will be annouced shortly.

The Prizegiving

As usual we hope to be hosting a prizegiving ceremony for selected participants at Bletchley Park. Dates will be announded as soon as possible and some tickets will be available by lottery. You can apply for them online at

https://www.cipher.maths.soton.ac.uk/tickets

How many can enter?

Teams of any size and composition may enter, and a school can enter as many teams as it wishes. Teams can be run from one or several individual accounts (see above) and inter-school teams are also allowed, indeed, encouraged.

Frequently Asked Questions

What does it cost to take part?

At the moment we are lucky to have several generous sponsors, and the costs of the competition are covered by the School of Mathematics at the University of Southampton so there is no charge to take part.

When does the competition start?

Registration opens on October 5th. The first part of the competition will be published at 3.15 on Thursday October 15th. There is no need to rush to download it as you have one or two days in which to submit to achieve full speed marks. Often the website is overloaded for the first half an hour or so, and it probably pays to wait out the rush. The part B prizes will be allocated based on performance in rounds 3-8 so if you miss the first couple of challenges it won’t matter too much.

When will the Challenges be published?

See the schedule above.

How long do we have to complete each challenge?

The deadline for each Challenge is 11.59pm on the day before the next Challenge is published, The number of points you score in part B depends on how early you submit your best attempt.

What age group is it aimed at?

Principally this is an extension activity for older pupils, but the early stages of the competition are aimed at a wide audience and there is something for everyone. We have had bright year 6 pupils do well on early rounds and this prepares them for further achievement in future competitions, though it is definitly not written or intended for the junior school age group. Many pupils return year after year as they try to improve on the number of stages they can successfully complete, and the staged certificates give everyone an incentive to keep going as long as possible. There are also some small prizes awarded at random to a few participants each week as an added incentive.

Is this appropriate for a Math Club activity?

Certainly. The entire math club could enter as a team or you could divide up into smaller groups and use the math club meetings to discuss techniques and strategies.

Can pupils enter on their own?

Yes, we get many solo entrants and teachers do not have to be involved, but we do ask for the name of a teacher contact for prize administration.

Do team members have to all come from the same school?

No, in the past we have had several teams made up of members of different schools and colleges, and this is great. We do ask for the name of at least one of the schools and a teacher contact for prize administration, and, for now at least, the team captain’s school will get all the credit!

Does everyone in a team need their own account?

No, but everyone can have one! The team is principally associated with one account holder (the team captain), but team captains can invite others to join the team at any point including when it is set up. You don’t need to have an individual account to be associated with a team, as you can be listed as a member on the team captain’s account. You do however need an account if you want to be able to post comments to the forum or to post solutions for the team. You also need one to log in to see feedback and get certificates.

Who else is taking part

Click on the map on the home page to see the schools taking part. It might be worth zooming out as sometimes we get schools overseas who enter for the fun of it even thought they can’t win a prize.

Where can I find out more?

The website is stocked with information about the competition, including the rules and regulations, some notes on codebreaking, some lesson plans and a guide to programming for codebreakers. We will also publish updates about the competition on the home page www.cipher.maths.soton.ac.uk.

It is also worth taking part in the forums, which we will monitor and where we will publish occasional updates and answers to common questions.

Click on the appropriate icon for the previous challenges.


Cipher 2002	Cipher 2003	Cipher 2004	Cipher 2005	Cipher 2006	Cipher 2007


Cipher 2008	Cipher 2009	Cipher 2010	Cipher 2011	Cipher 2012	Cipher 2013


Cipher 2014

In accepting these Terms & Conditions you are agreeing to abide by the rules of the competition as published on the competition website.

You are also agreeing that the challenge organisers are permitted to publish on the challenge website the content of those fields that are marked as visible on the registration page together with the contents of the field entitled "Leader Board", if you have chosen the options "First Name" or "Full name".
You are further agreeing that we may use your contact details to contact you in relation to your entry in the Cipher Challenge competition and that we may contact your "Teacher Contact" if this is felt necessary.
The software we use to generate our webpages sometimes places small amounts of information on your device, for example, computer or mobile phone. These include small files known as cookies. They cannot be used to identify you personally but are used to allow you to login to our site and take part in the Cipher Challenges and read/post information on the forums. Under EU law, we need your consent to store these cookies on your device. Consequently, you are further agreeing to allow our software to store these cookies on your device.
We agree that we may not use any of the information obtained from the registration page or from your entry to the competition for any other purpose except administration of the competition and information about future Challenges unless we have your explicit permission to do so.
We also agree not to divulge any information about you except where specified in item (2) to any third-party.

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Welcome

Codebreaking