Overall

You can use the template from one of the tests at https://toebes.com/codebusters/ and just replace the questions with your own, or use the Test Builder tool to generate an appropriate balance of question types. It might be helpful to know that there is a YouTube channel full of videos that cover various aspects of test creation, printing, score balancing, and so on. While the accompanying instructional video for each question type is linked below under the appropriate heading, it also contains videos regarding more holistic and overarching elements of test creation.

For all questions, it is important to carefully check the encoded text for any inappropriate or potentially offensive words that may be generated by the encoding.

When writing the question text, all cipher or plain decoded text should be indicated by using a bold courier font.

It is best practice to avoid making questions with point values that are rounded numbers (200, 350, etc.); instead, use a variety of values in the ones and tens places (153, 287, etc.) in order to reduce the likelihood of ties. For each question, there is a "Suggest Points" button that will offer a range of values that you can use as a starting point for the point value of a question, and make changes if needed to account for the difficulty. There is a helpful YouTube video that shows how to use the features of the Test Generation tool to balance scores and reduce ties.

Timed Question

This should be a random Aristocrat with no errors and without a hint (see Aristocrats/Patristocrats). Usually Medium difficulty is appropriate, as it should be designed to be solvable by approximately 1/3 of the teams within the 10 minutes given. Ensuring that the Timed Question is solvable in 10 minutes will help create variation in the scores and reduce the likelihood of ties.

Atbash Cipher

The Atbash is one of the easiest Ciphers for students to encode or decode because the alphabet is fixed. The letter v will always stand for the letter e and vice-versa. There should only be one Atbash Cipher question on a test. The accompanying YouTube video can be found here.

The only factor for difficulty with this question is in the number of characters in the phrase. The choice of letters/words/word length has no impact on the difficulty. In general the question should be between 45 and 80 characters. It is only slightly harder to have the students Encode because it isn't obvious when they have the correct answer. An Encode problem is also slightly harder to grade as it requires carefully checking each letter instead of simply reading the answer.

An Atbash Cipher Decode should be worth around 100 points (approximately 1.2 points per character plus 2.5 points per unique character in the Plain Text). An Atbash Cipher Encode could be worth 120-150 points depending on the length of the encode string (approximately 1.2 points per character rounded to the nearest 10).

The question should clearly indicate that cipher has been encoded using the Atbash Cipher as well as the origin of the phrase or quote. It should not include a hint. Some examples:

Solve this quote from <person> which has been encoded with the Atbash Cipher.

Encode this quote by <person> using the Atbash Cipher.

Caesar Cipher

The Caesar Cipher is a fairly simple cipher where once you know a single letter mapping, the remainder of the problem is a simple transposition. There should be no more than one or two Caesar Cipher questions on a test. The accompanying YouTube video can be found here.

There are two factors which affect the difficulty. First is the length of the cipher which in general should be approximately 80-90 characters. Having a single or two character word in the Plain Text also makes it easier to solve. Secondarily is the Shift of the letters. Shifts of 1,2,3 or 13 are the easiest because once the shift is known, it can generally be done without even looking it up in a table. 13 is a special case because it operates similar to the Atbash in that letters map in pairs. It is only slightly harder to have the students Encode because it isn't obvious when they have the correct answer. An Encode problem is also slightly harder to grade as it requires carefully checking each letter instead of simply reading the answer.

A Caesar Cipher Decode should be worth 100-120 points (approximately 1 point per character in the Plain Text plus 2.5 points per unique character) with the 120 point being associated with a Caesar with a larger shift and no short words. A Caesar Cipher Encode should be worth 120-160 points with absence of short words, larger shift values and longer phrases.

The question should clearly indicate the origin of the phrase and that it has been encoded using a Caesar Cipher. It should not indicate the shift amount (for a Decode) or include a hint. For example:

Solve this quote from <person> which has been encoded with the Caesar Cipher.

A quote from <person> has been encoded with the Caesar Cipher. What does it say?

Encode this quote from <person> using a Caesar Cipher with a shift of 9 (e.g. S encodes as J).

Aristocrats/Patristocrats

These will take up the bulk of the questions. It is helpful to search for a variety of quotes and phrases to use in order to pick the ones which best meet your needs. A good quote will have around 20 words and about 100-120 characters (including spaces) for Aristocrats and 110-130 characters for Patristocrats. Note that the only difference between an Aristocrat and a Patristocrat is whether you take the spaces out or not, so the same quotes work for both. However, you generally want slightly longer phrases for the Patristocrats to give more patterns for the team to find. There are accompanying YouTube videos that demonstrate how to generate an Aristocrat, how to generate a K1/K2/K3 Aristocrat, how to generate Aristocrats with errors and misspellings, and how to generate keyword recovery Aristocrat Ciphers using the tool.

There are multiple factors which affect the difficulty of an Aristocrat. First and foremost is the match of the phrase to the frequency of distribution of letters relative to the standard distribution of English letters:

Table 1 - Frequency of English Letters

Fortunately the tool will calculate the Chi-Square Value to give an indication of how close the phrase is to a standard distribution. The lower the value, the easier the phrase.

Secondarily, the choice of words matters. A phrase using obtuse/archaic language or atypical words will be more difficult than one which uses simple words. The presence of single and double letter words helps to make the problem easier.

Third, phrases that start with it is, have multiple occurrences of the or contain the words these, there, little or people tend to be easier. You will also want some samples which have repeated words to use for test questions providing hints. In general, it is good to avoid quotes which are unattributed or by anonymous to allow the author of the quote to serve as an extra hint.

When spelling and grammar errors are introduced, the problem becomes much harder. Likewise, eliminating all the spaces for a Patristocrat makes it the hardest to solve.

In order to make the harder problems solvable, it is good to consider using a K1 or K2 alphabet to provide some additional hints. A K2 alphabet is slightly harder than a K1 alphabet, but both are easier than a random alphabet.

Aristocrats/Patristocrats have a wide range of scores, but there are some general guidance to follow in setting the score. Each of the types below provide a nominal range to start with. When encoding the sample text, the tool will report a Chi-Square Value. The range of the value provides a general guide to the difficulty of the problem (0-20 = Easy, 20-30=Medium, 30-40=Medium Hard, 40-50=Difficult, >50=Extremely Difficult). Taking the Chi-Square Value indication as a starting point, also look for short words and pattern words in the phrase as an adjustment to the difficulty up or down. A problem that is perceived to be Easy should not be worth more than 300 points while a problem that is Difficult should not be worth less than 500 points. As a rule of thumb, an Easy problem should start with a base of 200 points while a Difficult one should be 750 points. For this reason it is often good to start with a number of phrases to determine their difficulty and then assign them to the type of Aristocrat/Patristocrat problem. It is also worth noting that with a phrase that has a high Chi-Square Value, while it may not be ideal for an Aristocrat/Patristocrat, it can easily be used for just about all the other Cipher types including Baconian, Affine, Atbash, Caesar, Porta, Nihilist and Vigenère

A problem can be made less difficult in two ways. Adding a hint that reveals characters makes it easier to solve. With a hint, each letter beyond 3 letters for an Aristocrat (or 5 letters for a Patristocrat) revealed by the hint should drop 50 points from the score. By using a K1 or K2 alphabet, the score can also be dropped by 100 and 75 points respectively.

Eliminating the spaces to make a phrase into a Patristocrat automatically adds 500 points to the difficulty and eliminates any benefit from short words.

For example starting with a phrase:

Is there any knowledge in the world which is so certain that no reasonable man could doubt it?

with a Chi-Square of 11.50, we see multiple short words (it, is, in), and a hint of three letters (man) this problem would be one of the easiest with a score of 200. It would also make for a good timed question. By changing it to a Patristocrat, we would add 500 to get a base score of 700. If we encode it with a K1 alphabet we drop 100 points for a score of 600. At this range we should also be providing a hint that it starts as “isthe” revealing only 5 characters leaving this as a 600 point question.

Aristocrat with a hint

A standard Aristocrat with a hint will be about 200 points. See the General Aristocrat/Patristocrat point guidance for more details.

The question should clearly indicate that it is an Aristocrat, provide the source of the phrase and a hint of what word(s) may be found in the decoded phrase. For example:

Solve this Aristocrat which is a quote by <person> which has the word MAN in it.

<person> was heard to say the following phrase which starts with: NEVER.

Aristocrat without a hint

A standard Aristocrat without a hint will be between 250 and 350 points. See the General Aristocrat/Patristocrat point guidance for more details.

The question should clearly indicate that it is an Aristocrat and provide the source of the phrase. For example:

Solve this Aristocrat which is a quote by <person> when they <event>.

During <event>, <person> was heard to say the following phrase.

The following quote by <person> has been encoded as an Aristocrat Cipher. What does it say?

Aristocrat with spelling/grammar errors and a hint

These can be a lot of fun to generate and can be something that the students relate to. In general the thought process here is something that their phone mistyped when texting using voice. Generating these takes a couple of tries to get something that works out well. It is best to start out with phrases that contain homophones (like you/ewe/hue) and then either generate variations by hand or use a homophone generation tool (like http://homophonemachine.allaboutlearningpress.com/ or https://evashort.com/homophone/). The Test Generator also has built-in functionality to help create Aristocrats with homophones and misspellings, and there's an accompanying instructional YouTube video. Sometimes you can get lucky trying to send it as a voice text message with Siri or Google Voice to get a phrase which has been slightly twisted. You may want to try a couple of times to get something that is appealing. The phones have gotten a lot smarter lately and don't make as many mistakes as they used to. By using a K2 or K1 alphabet, the difficulty of solving it goes down.

An Aristocrat with spelling/grammar errors and a hint should be worth between 350 and 500 points, based on the difficulty of the phrase, number of letters revealed with the hint and whether or not a K1/K2 alphabet is used. See the General Aristocrat/Patristocrat point guidance for more details.

The question should clearly indicate that it is an Aristocrat with misspellings or grammar errors and provide the source of the phrase as well as a hint. For example:

Alexa severely misheard a phrase from <person> which has the word THE twice and then encoded it as an Aristocrat. What did it come out as?

When one of <person's> manuscripts was automatically scanned and OCRed before converted to an Aristocrat, there were quite a few mistakes but the word piece appears twice. What was the final result?

Aristocrat with spelling/grammar errors with no hint

Approach these exactly the same as Aristocrat with spelling/grammar errors and a hint except that you don't provide a hint. This is one of the hardest questions that can be found on the test. In general, these should be encoded with a K1 or K2 alphabet.

An Aristocrat with spelling/grammar errors and no hint should be worth between 450 and 700 points. See the General Aristocrat/Patristocrat point guidance for more details.

The question should clearly indicate that it is an Aristocrat with misspellings or grammar errors and provide the source of the phrase. For example:

Alexa severely misheard a phrase from <person> and then encoded it as an Aristocrat using a K1 alphabet. What did it come out as?

When one of <person's> manuscripts was automatically scanned and OCRed before converted to an Aristocrat using a K2 alphabet, there were quite a few mistakes. What was the final result?

Aristocrat K1-K3 Keyword Recovery

Recovering a keyword/phrase from an Aristocrat first requires competitors to decode the Aristocrat, so all of the guidance in the General Aristocrat/Patristocrat guidance applies. Recovering a K1 keyword (keyword in the plain text) is typically very straightforward. Recovering a K2 keyword (keyword in the cipher text) is also very doable once the concept is understood. Recovering a K3 keyword (keyword in both the cipher text and the plain text, with the alphabets shifted) requires a completely different process and merits special consideration (described below) to ensure that a plaintext phrase/quote is chosen which is a suitable candidate for this type of problem.

For K1-K3 keyword/phrase recovery problems, it is helpful to be mindful of the fact that, as is typical with keywords, duplicate letters are removed when the word or phrase is used as a key, and competitors are provided with blanks corresponding to the length of each word in the key phrase to aid them in recovering it. For example, when they recover the characters TOHESAR and are provided with an appropriate number of blanks (2,3,5), they can work out that the key phrase here is TO THE STARS. Therefore, it is good to be especially mindful to avoid using a keyword that would result as indistinguishable from another when reconstructed: for example, the two different five-letter words DOSES and DOSED would both result in a key of DOSE, so either of these would be a poor keyword choice as competitors would not be able to reliably recover which one was used.

The Aristocrat encoder prefers that the plaintext be quotes or phrases that contain at least 19 unique characters. For K1 and K2 keyword recovery, this is not strictly required. For K3 keyword recovery, 19 unique characters is an absolute minimum in selecting an appropriate phrase, as the less of the alphabet is used, the more difficult the problem becomes as it results in a higher number of unknown mappings. Letter frequencies in the plaintext still need to be distributed enough to make it solvable (i.e., if many letters are only used once, this creates its own difficulty factor for solving the Aristocrat before beginning the recovery process). A plaintext with 22 or 23 unique characters is more reliably solvable for K3 keyword recovery than one with fewer and is the easiest way to make sure that the problem isn't inadvertently too difficult without having to get into the nuts and bolts of solving these as described in the next paragraph. It is also recommended to keep the Offset at 1, 2, or 3.

The process of K3 keyword recovery is described in detail here:

• The Black Chamber: Recovering K1 to K4 Keys
• The Black Chamber: Finding K3 Alphabets Again: The Nerd Version
• The Black Chamber: Recovering K3 Alphabets: A paper and pencil approach

K1 and K2 keyword recovery are significantly easier than K3 keyword recovery. Difficulty for all types is somewhat increased when the keyword contains duplicate letters. For K3 keyword recovery, difficulty is also increased when fewer unique characters are used in the plaintext; when the keyword draws from low-frequency letters; when the reconstruction results in two or more short loops than in one long loop; and when the decimation step requires alternating interleave spacing.

See the General Aristocrat/Patristocrat point guidance for more details. Finding a K1 or K2 keyword/phrase should add 100 points to the total; finding a K3 keyword/phrase should add a minimum of 150 points to the total and could increase based on the factors for difficulty described above.

The question should clearly indicate that they are to find the keyword/key phrase and provide the source of the phrase. For example:

A famous phrase from <person> has been encoded as an Aristocrat using a K1 alphabet. What was the key phrase used to encode it?

<person> was often heard to say the following phrase which has been encoded as an Aristocrat using a K2 alphabet. Please provide the keyword used to encode it.

Patristocrats with a hint

Patristocrats should always be encoded with a K1 or K2 alphabet in order to keep them from being too difficult. It is also good to pick phrases with a very low Chi-Square Value to keep the problem from being too difficult. Note that, as of this writing, Division B requires a K1 keyword and Division C requires a K1 or K2 keyword for these.

A Patristocrat with a hint should be worth between 450 and 650 points. See the General Aristocrat/Patristocrat point guidance for more details.

The question should clearly indicate that it is a Patristocrat and provide the source of the phrase as well as a hint that reveals no more than 6 unique characters. For example:

A famous phrase from <person> has been encoded as a Patristocrat using a K1 alphabet. It starts with the letters ‘tomyf'. What did they say?

<person> was often heard to say the following phrase which has been encoded as a Patristocrat using a K2 alphabet. The sequence ‘you' appears four times in the original text. What did they say?

Patristocrats with no hint

Patristocrats with no hint must be encoded with a K1 or K2 alphabet in order to keep them from being too difficult. It is also good to pick phrases with a very low Chi-Square Value to keep the problem from being too difficult. Note that, as of this writing, Division B requires a K1 keyword and Division C requires a K1 or K2 keyword for these.

A Patristocrat without a hint should be worth between 500 and 700 points. See the General Aristocrat/Patristocrat point guidance for more details.

The question should clearly indicate that it is a Patristocrat and provide the source of the phrase. For example:

A famous phrase from <person> has been encoded as a Patristocrat using a K1 alphabet. What did they say?

<person> was often heard to say the following phrase which has been encoded as a Patristocrat using a K2 alphabet. What did they say?

Spanish Xenocrypt

These ciphers should always be encoded with a K1 or K2 alphabet.

Pick a Spanish phrase which primarily consists of words which a second-year Spanish class would cover. Phrases which have both la and las present are good choices as well as phrases which contain y or cognates (Spanish words which are substantially like their English equivalent words such as ciencia, composición and básico) are also good. For a good source see https://www.realfastspanish.com/vocabulary/spanish-cognates. Although it isn't strictly necessary, try to avoid phrases which depend on accented characters. As with the approach for the English Aristocrats, pay attention to the frequency of letters. The tool automatically calculates the Chi-Square Value to verify the match:

If the encoded string uses both N and Ñ, it is best to re-encode until you don't get them both to avoid confusion on the part of the teams. Although it is possible to get an encoding that doesn't use Ñ at all, it is perfectly fine to generate a question which has one. Having both both N and Ñ, though, increases the difficulty of the problem.

The presence of cognates greatly reduces the difficulty. It is also expected that a K1 alphabet with an English key should be used. A K2 alphabet can be used, but it doesn't reduce the difficulty as much. In Division B, only a K1 alphabet can be used. In Division C, you might expect to find one of each.

600-700.

The question should clearly indicate that it is a Spanish Xenocrypt and provide the source of the phrase (if known). It should also indicate the use of a K1 (or K2) alphabet with an English keyword. For example:

Solve this Xenocrypt which is a translation of a quote by <person> into Spanish and has been encoded with a K1 alphabet using an English keyword.

Decrypt the following cipher text which was encoded using the Affine Cipher with a=5 and b=9.

Affine Cipher encrypt

There should be at most a single one of these on the test. It should use a phrase about 25 characters long that doesn't have too many occurrences of the letter A in it, preferably with as large a variety of letters as possible… (e.g. The quick brown fox jumps over the lazy dog isn't actually bad).

Pick a value for a which is coprime with 26 (1,3,5,7,9,11,15,17,19,21,23 or 25). The actual value doesn't matter, but larger ones tend to be slightly harder. If you are generating tests for multiple regions, pick numbers that are near each other. I.e. 7, 9 and 11 would be good to have as equivalent a values.

Pick a value for b between 1 and 25 inclusive. Unlike a where the larger values become slightly harder, the value of b can truly be any number and be the same level of difficulty.

There is very little variability in the difficulty other than the length of the string. Larger values of a are only slightly harder while the value of b has no real impact on the difficulty.

100.

The question should state to use the Affine Cipher and include the phrase to encode along with the values of a and b for them to use (note that a and b are italicized). For example:

Encrypt the common phrase The quick brown fox jumps over the lazy dog using the Affine Cipher with a=5 and b=9.

Porta Cipher encrypt/decrypt

Depending on the rules for a given year, there can be one or two decrypts on the test. (Note that encrypts are not allowed by the current rules.) There are no restrictions on the phrase, although try to avoid a phrase with a lot of a's in it. I.e. An amazing aardvark allows all answers would be a poor choice because the letter a is trivial to encode. As this question is nominally worth two points per letter, a 50 letter phrase is ideal.

Additionally there needs to be a Key to encode it with. It should be 5 or 6 characters with no repeating letters and avoid the letter a as it causes a letter to map to itself. By setting the Block Size to the same as the length of the Key, the problem is much easier than with the default Block Size of 0 that keeps the original spacing. Setting the Block Size to a size other than the length of the Key increases the difficulty somewhat.

Note that for the Regional/Invitational events, the Key must be given as part of the question.

The Block Size is the major contributor to the difficulty followed by the length of the phrase.

100 for a Decode (approximately 2 points per letter), 120 for an Encode. If the Block Size is the same as the Key length subtract 20 points. If the Block Size is non-zero and different from the Key length, add 25 points.

The question should indicate that the Porta Cipher is being used, whether they are to Encode or Decode and the Key to use for it. It is generally nice to give the origin of the phrase if it is known. For example:

A phrase by <person> has been encoded using the Porta cipher with a code word of SLEPT. What does it say?

Using a keyword of HORSE, encode this famous quote by <person> using the Porta cipher.

Vigenère Cipher encrypt/decrypt

If the rules allow the Vigenère Cipher on a Division B/C test in a given year, there can be a couple of decryption questions. Division A allows only decryption. There are no restrictions on the phrase, although try to avoid a phrase with a lot of a's in it. I.e. An amazing aardvark allows all answers would be a poor choice because the letter a is trivial to encode. As this question is nominally worth two points per letter, a 50 letter phrase is ideal.

Additionally there needs to be a Key to encode it with. It should be 5 or 6 characters with no repeating letters and avoid the letter a as it causes a letter to map to itself. By setting the Block Size to the same as the length of the Key, the problem is much easier than with the default Block Size of 0 that keeps the original spacing. Setting the Block Size to a size other than the length of the Key increases the difficulty somewhat. There is a YouTube video demonstrating the creation of a Vigenère Cipher here.

Note that for the Regional/Invitational events, the Key must be given as part of the question.

The Block Size is the major contributor to the difficulty followed by the length of the phrase.

100 for a Decode (approximately 2 points per letter), 120 for an Encode. If the Block Size is the same as the Key length subtract 20 points. If the Block Size is non-zero and different from the Key length, add 25 points.

The question should indicate that the Vigenère Cipher is being used (don't forget the accented è), whether they are to Encode or Decode and the Key to use for it. It is generally nice to give the origin of the phrase if it is known. For example:

A phrase by <person> has been encoded using the Vigenère cipher with a code word of SLEPT. What does it say?

Using a keyword of HORSE, encode this famous quote by <person> using the Vigenère cipher.

Nihilist cipher

The Nihilist cipher is a substitution cipher. It uses two keywords, the polybius key and normal keyword, to encode the plaintext. The polybius key is used to generate a Polybius Square, which is a 5x5 table that maps every letter to a number based on its row and column. The normal keyword is used to encode the plaintext, applying each letter in the keyword in sequence in just the same way that the Vigenère cipher works. There should be no more than 2 Nihilist decryption problems on a test.

In general, more room for error makes it more difficult to get the right answer. There is more room for error when the key length and the block size are different. A block size of 0 may provide clues to the encoded words, so reduces difficulty. A longer polybius key or one with duplicate letters makes it more difficult because there is increased chance of creating the polybius square incorrectly. A longer encoding key also means there is more work to do. needed. Putting 'Z' in the polybius key means it will not have a numeric value of 55, which is a common assumption, so this slightly increases the difficulty.

Roughly 3-4 points per letter in the plaintext.

For decryption, the question text should include the polybius key and normal keyword used to encrypt the plain text. Quote length should be at most around 50 characters.

A phrase by <person> was encoded using a Nihilist Substitution Cipher with a Polybius Key of SCIENCE OLYMPIAD and a key of CODE. What does it say?

Nihilist cipher (cryptanalysis)

The Nihilist cipher is a substitution cipher. It uses two keywords to encode the plaintext. The first is used to construction a Polybius Square which is used to map all of the letters to numbers. The second keyword is used to encode the plaintext, applying each letter in the keyword in sequence in just the same way that the Vigenère cipher works. Doing cryptanalysis, some of the mapping of plain text to cipher text is given and the polybius square and keyword need to be derived from that mapping. This is an iterative process that substantially increases the difficulty.

In addition to the notes above, cryptanalysis makes Nihilist ciphers significantly more difficult due to the iteration and deduction required to solve it. As a result, these will be some of the most time-consuming problems on the test, and care should be taken to avoid making them too difficult. One factor for difficulty is the Block Size. A Block Size of 0 will make these easier to solve, while a Block Size that is nonzero and different from the keyword length will make these more difficult. Another factor for difficulty is the choice of crib. A longer crib makes the problem easier than a shorter crib, but note that too many duplicate letters within the crib will reduce the information that solvers have given to them. A third factor for difficulty is the relationship between the Keyword, the Polybius Key, and the Polybius table. If, after the Polybius table is constructed, the keyword draws primarily from the center values in the table, the problem will be more difficult. It is helpful to solvers if the keyword contains letters from a wider variety of locations in the Polybius table - both low values (letters which are also contained in the Polybius key) and high values (letters from the last row in the Polybius table), to help them analyze patterns in the cipher values. Lastly, the AutoSolver provides guidance on whether it is able to recover the keyword given the choice of keyword and crib provided, which can help guide the construction of these problems. Unlike most other kinds of ciphers, it is much easier to inadvertently make these unsolvable in competition time than it is to make them too easy, so don't hesitate to set the Block Size at 0 and to have a substantial crib. The current rules require the crib to have a length at least double the length of the keyword..

400-650. (On the lower end of this range if it is a Special Bonus Question; on the higher end if it is not.) It begins with a similar base to the Nihilist above, in the 125-150 range, and then bonuses are given for cryptanalysis (200), for the AutoSolver not being able to find the keyword or keyword positions (200-300), if the AutoSolver can find the keyword but not enough characters in the answer to solve the cipher (100-200), and if the Block Size doesn't match the keyword length (25).

For cryptanalysis, the question text should include a hint of plaintext and the location of that plaintext.

The following quote by <person> has been encoded with the Nihilist Substitution Cipher using a very common word for the key. The 35th through 39th cipher units (58 93 66 54 45) decode to be HAPPY.

Checkerboard cipher

The Checkerboard cipher is a substitution cipher. It uses three keywords: the polybius key, a row keyword, and a column keyword, to encode the plaintext. The polybius key is used to generate a Polybius Square, which is a 5x5 table that maps every letter to a pair of letters based on its row and column. To decode these ciphers, competitors are provided the polybius key but would generally determine the letters that can appear in the row and column keywords and anagram them in order to find possible keywords that can be made from those letters. It is worth noting here that the quote chosen should have a high chi-square value, as the row/column ciphertext coordinates always decode to the same letter -- with a block size of 0, these are functionally very similar to Aristocrats; and with a block size of 5, these are functionally very similar to Patristocrats. Choosing a high chi-square value for the quote will ensure that competitors try to recover the row/column keywords instead of just solving these using their Aristocrat/Patristocrat skills. There should be no more than 2 Checkerboard decryption problems on a test.

In general, more room for error makes it more difficult to get the right answer. A block size of 0 may provide clues to the encoded words, so reduces difficulty (dramatically, since the row-column coordinates will allow these to be solved like Aristocrats). For this reason, it is recommended to use a fixed block size (5 is standard, but any value is fine). A longer polybius key or one with duplicate letters will only slightly increase the difficulty because there is increased chance of creating the polybius square incorrectly. Note that the expectation is that competitors should be able to recover the word/columns by anagramming, so it would be best practice to select recognizable row and column keywords. In addition, the difficulty is affected to some degree by the choice of plaintext: a plaintext of an adequate length and variety where each of the 5 letters in the row keyword and each of the 5 letters in the column keyword can be seen in the ciphertext will be easier than ciphertext where not all of the keyword letters can be recovered. Lastly, a row or column keyword with repeated letters (such as AWARE or SWIMS); or one with multiple valid anagrams (such as STARE/RATES/TEARS) would increase the difficulty considerably but generally would still be solvable in competition time.

200-450. A straightforward one where the row and column letters result in only one anagrammable keyword for each would be on the lower end; the points would raise for multiple of the same letter within a single row or column keyword or for a row/column keyword with several valid anagrams.

For decryption, the question text should include the polybius key. Quote length should be around 55-75 characters with a high chi-square value.

A phrase by <person> was encoded using a Checkerboard Cipher with a Polybius Key of SCIENCE OLYMPIAD. What does it say?

Checkerboard cipher (cryptanalysis)

The Checkerboard cipher is a substitution cipher. It uses three keywords to encode the plaintext. The first is used to construct a Polybius Square which is used to create a 5x5 array of all of the letters in the alphabet (with I and J sharing a cell). The other keywords are for the row and column headers to map the letters. Doing cryptanalysis, some of the mapping of ciphertext to plaintext is given, and the polybius square and polybius keyword need to be derived from that mapping. This is an iterative process that substantially increases the difficulty. Note that it is possible to treat the Checkerboard cipher as an Aristocrat/Patristocrat if you read the pairs of letters as a symbol.

In addition to the notes above, cryptanalysis makes Checkerboard ciphers significantly more difficult due to the iteration and deduction required to solve it. While Checkerboard cryptanalysis and Nihilist cryptanalysis have some commonalities, a Checkerboard cryptanalysis is significantly easier due to the fact that a ciphertext pair always represents the same plaintext character. One factor for difficulty is the Block Size. A Block Size of 0 will make these easier to solve, while a Block Size that is nonzero will make these more difficult. Another factor for difficulty is the choice of crib. A longer crib makes the problem easier than a shorter crib, but note that too many duplicate letters within the crib will reduce the information that solvers have given to them. A third factor for difficulty is whether the row/column keywords are words that have several valid anagrams, or whether they have letters duplicated within them, which will be a significant factor. Lastly, the AutoSolver provides guidance on whether it is able to recover the keyword given the choice of keyword and crib provided, which can help guide the construction of these problems. The current rules require a "crib" of at least 5 characters.

300-550. It begins with a similar base to the Checkerboard above, in the 200ish range, and then bonuses are given for cryptanalysis, for the AutoSolver not being able to find the keyword or keyword positions, if the AutoSolver can find the keyword but not enough characters in the answer to solve the cipher, and if the row/column keywords contain duplicate letters or have more than just two possibilities for anagrams.).

For cryptanalysis, the question text should include a hint of plaintext and the location of that plaintext.

The following quote by <person> has been encoded with the Checkerboard Cipher. The 35th through 39th cipher units (YN OT RT RT LN) decode to be HAPPY.

Homophonic Cipher cryptanalysis

The Homophonic cipher is a monoalphabetic substitution cipher. It uses one keyword (with four unique letters) to map each letter of the alphabet onto one of four values. The first letter of the keyword is mapped to the number 1, and the numbers wrap around the alphabet: with I/J sharing a value, the numbers 1-25 are assigned to the letters. The second letter of the keyword is mapped to 26; the third is mapped to 51; and the fourth is mapped to 76. Each time a letter is encoded in ciphertext, one of the four corresponding values is chosen to represent it. In a cryptanalysis problem, be sure to carefully follow the rules as to how many of the alphabets must be revealed by the crib that is selected, but in general the idea is that a partial decryption (a "crib" mapping of ciphertext to plaintext) is provided to competitors. There should be no more than 2 Homophonic cryptanalysis problems on a test.

A block size of 0 may provide clues to the encoded words, so reduces difficulty (dramatically, especially if there are single-letter words that would reveal an alphabet). For this reason, it is recommended to use a fixed block size (5 is standard, but any value is fine). A rarer keyword or one with a more unusual letter pattern would make this problem more difficuilt. Note that the expectation is that competitors should be able to recover the keyword by anagramming, so it would be best practice to select a recognizable keyword. In addition, the difficulty is affected to some degree by the choice of plaintext: a plaintext with a lower chi-square value containing more common words would be easier for competitors to solve. Lastly, the best way to evaluate the solvability of this cipher will be to look at the Auto-Solver's initial work, and see whether it appears to give competitors a solid place to start working from; and then look at its end state and see if it is able to completely decrypt the ciphertext. Does it manage to completely solve the cipher? How many attempts does it have to make? When it begins its work by filling in the provided plaintext, does it fill in enough pieces of one or more words that competitors can make educated guesses as to other letters?

250-350. A straightforward one where the provided crib gives competitors a solid place to make educated guesses as to one or more words would be on the lower end; the points would raise if the crib given only shows competitors plaintext letters that are scattered about and requires them to make multiple attempts in guessing some of the alphabets.

For cryptanalysis, the question text should include a hint of plaintext and the location of that plaintext.

A phrase by <person> was encoded using a Homophonic Cipher. The 8th through 12th cipher units (62 34 15 77 16) decode to be PIZZA. What does it say?

Homophonic cipher (decode)

The Homophonic cipher, described above, uses a four-letter keyword to map each letter of the alphabet to four potential numeric values. For a Homophonic decode, no mapping of ciphertext to plaintext is given. Instead, a number of letters in the keyword are provided in accordance with the rules for a given division. This typically means that competitors need to think about the most likely positions where the given letters can appear in a four-letter word, and make and test hypotheses in order to determine where in the keyword those letters appear, and what the rest of the keyword is then likely to be.

In addition to the notes above, a decode using a partial set of keyword letters makes Homophonic ciphers significantly more difficult due to the iteration and deduction required to solve it. One factor for difficulty is the Block Size. A Block Size of 0 will make these easier to solve, while a Block Size that is nonzero will make these more difficult. Another factor for difficulty is the choice of keyword letters that are provided. For example, a keyword that contains Q is then likely to also contain U right after it; an E is more likely to appear word-final than word-initial, and so on. A third factor for difficulty is whether the letter pattern in the keyword is unusual. Lastly, it's always helpful to look at the AutoSolver's work and see how it might show the amount of trial and error a human competitor might have to do to solve the cipher.

300-400. It begins with a similar base to the Homophonic above, in the 200ish range, and then bonuses are given for Decode, and then from there there are bonuses based on the work the AutoSolver is able to complete. The point value given is less if the position of the letters in the keyword is provided, and greater if they are provided without their accompanying positions.

For a decode, the question text can provide either the position of the provided letters in the Keyword, or just the letters themselves without revealing their positions. Again, ensure that you are providing the number of letters as appropriate for the rules in a given Division.

The following quote by <person> has been encoded with the Homophonic Cipher. You are told that the keyword contains the following letters in the given positions: _OL_.

<Person> has been known to say the following phrase. It has been encoded using the Homophonic Cipher. You are told that the keyword contains the following letters: C,O.

Baconian cipher

This is where the test writer can have a lot of fun. The phrase should be approximately 20-35 characters long and ideally should be one for which the ending can't necessarily be guessed once they have decoded the first half; it is important to make sure that they have to work the entire problem. There is little impact on the difficulty.

With the Words Baconian cipher which is the harder of the three options, it is important to pick a pattern for the letters mapping to A and B. Obviously the default with the tool of ABABAB… is the easiest and it is not recommended to use this for the test. There are lots of patterns which work (first half of the alphabet is A, second half is B). But there are some that don't work such as BABABA… which eliminates all vowels from being an A. Fortunately the tool tells you when the pattern picked don't have any words to match. Note that with the Words Baconian, a crib and crib location should both be provided. The tool will highlight the letters that are given with the crib, helping the test writer to evaluate how much information is available to the competitors. The accompanying YouTube video can be found here.

For the Letter for letter and Sequence Baconian ciphers it is possible to have multiple letters stand for A and B, but it is recommended to use symbols (which can be pasted into the fields). Having a variety of symbols which have potential overlap can make it harder. For example on the 2018 North Carolina State test, the key to the solution was that 'A' was represented by any arrow pointing down (e.g. ) and 'B' was represented by any arrow pointing up (e.g. ); so that the students had to figure out whether the direction of the arrow or the line was what indicated the difference. As in this example, it is best practice for there to be some way for competitors to deduce at least two different groups of symbols they can assign to A and B: while they may have to test a couple of different hypotheses, it should not be completely random.

It is worth noting that the biggest difference between a Letter for letter Baconian and a Sequence Baconian is how the groups of symbols representing A and B are used. In a Letter for Letter Baconian, each instance of A will be represented by one of the symbols denoted for A; and each instance of B will be represented by one of the symbols denoted for B. However, in a Sequence Baconian, the position of the letters matters. This type of cipher is typically used with a visibly repeating pattern. The idea is to look at the pattern and determine, for each position in the pattern, whether a symbol corresponds to A or B. So, unlike the Letter for letter Baconian, in a Symbol Baconian, the same symbol could theoretically represesent A in one position, but B in a different position. The key is to look at where in the repeating sequence the symbol appears.

There are accompanying YouTube videos for creating Letter for Letter and Sequence Baconians.

There are three factors which affect the difficulty. First is the choice of symbols to represent the Baconian. The harder to distinguish, the more points the question should be worth. Second is how much of a clue is provided. Lastly is the size of the grouping. By picking a Line Width which is not a multiple of 5, it ensures that the students have to carefully manage wrapping a group of 5 across a line boundary.

200-400. A Letter for letter with a single substitution and Line Width which is multiple of 5 should be worth 200 points. A Words Baconian with a complex pattern and a couple letters of clues would be worth 400. The basic calculation is roughly 6 points per letter, plus various bonuses based on the type of Baconian, the patterns it uses, and the other factors for difficulty mentioned above.

The question should indicate that the Baconian Cipher is being used and the origin of the phrase if known and any clue. For example:

The following symbols encodes a phrase by <person> using a Baconian alphabet. What does it say?

The following strange headlines encodes a phrase by <person> using a Baconian alphabet. You have been told that it starts out with “SOME.” What does it say?

The following odd symbols were found when a tomb was opened, but you recognized it as a prankster who scratched it on the wall using the Baconian alphabet. What does it say?

Cryptarithm

The Cryptarithm is a cipher that maps each digit 0-9 to a different letter and provides an entire mathematical equation where the digits are substituted with the corresponding letters. Competitors can use mathematical deduction to determine which digit each letter corresponds to, and then they are asked to decode a phrase where the digits are provided and they are to use the information they've gathered to decode it. There is an accompanying YouTube video that covers how to put a cryptarithm into a test.

A primary factor for difficulty in these is the choice of math problem. Competitors will be accustomed to working with basic addition or subtraction problems, so providing any type of multiplication or long division problem will make it significantly more difficult. These problems also become more difficult when more than one value can map to a given letter. One way to make sure competitors have a place to approach these problems is through providing an addition or subtraction problem where the largest value has more digits than the other values (as demonstrated in the sample question below), so they can deduce that it is likely to begin with a 1. Note that if you use the tool to generate the cryptarithm, it will estimate the difficulty of the cipher on a scale of 1-5. Aside from the math problem itself, competitors typically work it alongside the "phrase to decode," so this phrase also affects the difficulty: longer phrases become somewhat easier, as competitors can begin deducing letter frequencies and patterns as they do for Aristocrat ciphers, and short phrases containing short, uncommon words are more difficult.

150-600. A good starting point for a standard cryptarithm would be about 250-300 points, or 150 points plus the product of 50 times the difficulty level. This can be adjusted (usually bumped up) using the factors for difficulty above.

The question should indicate that they are to solve the cryptarithm, and it should contain the values that competitors are to decode. For example:

After solving the cryptarithm SEND+MORE=MONEY decode the phrase 9015 10992.

The test generator tool can generate cryptarithms given a list of words separated by spaces or commas, as described in this YouTube video. There are several other sites that allow for generation of cryptarithms including:

Complete Columnar

The Complete Columnar Cipher is a transposition cipher. Plain text is written across a table with a pre-defined number of columns. The column order is mixed up by providing a column ordering string. For example, if the column count is 7 and the column order string is SCIENCE, the column order will be rearranged to be in the order: 2, 6, 4, 7, 3, 5, 1. The column order string gets alphabetically sorted to CCEEINS and we find the first character C in position 2 of the column sorting string, the next C in position 6 of the column sorting string, the first E in position 4, etc. The columns of the plain text table are rearranged using this order and the cipher text is generated by reading down each column, starting with the first column on the left and working across.

Note: Single letter and/or number characters can be used in the column order string. Numbers 0-9 and uppercase characters A-Z are used and lower case letters are automatically converted to upper case. The order is determined by sorting the characters using their ASCII value (numbers come before letters). Duplicate characters are taken in order, left to right. For greater than 10 columns, double digit numbers are not honored. However, letters can be used to avoid duplicate values in the column order string, although it is not necessary.

To put these columns back in the correct order (mentally renumbering them 1-7), take them in numeric order. I.e., the column labeled 1 is in the seventh position, the column labeled 2 is in the first position, the column labeled 3 is in the fifth position, etc. So the transposed table would be rewritten with column ordering: 7, 1, 5, 3, 6, 2, 4 to reveal the original plain text in a seven column table.

A crib is given for cryptanalysis to provide a starting point for deducing the column order and decoding the ciphertext. The location of the crib text is not given, but is deduced by the competitors. There is a YouTube video that demonstrates generating a Complete Columnar cipher with the Test Generator that can be found here.

Difficulty is increased when the crib is split over two rows; when the crib appears in more than one location in the column block; when there are duplicate letters in the crib; when there are more possible columns (i.e. more factors of the cipher text length) that must be analyzed; when there are no pad characters (i.e. cipher text is a multiple of the number of encode columns).

150-275. The point value should be higher when more columns are used in the encoding; when the crib is split over two rows; when there is more than one instance of the crib; when there are duplicate letters in the crib; when there are more possible arrays of columns and rows that could have been used to encode the cipher text; when no pad characters are used in the cipher text to make it fit in the grid. Also, the longer the cipher text, the more difficult because it just takes more time to analyze.

The question should indicate that they are to use the crib text (given in the question text) to help determine the number of columns and the order of the columns to decode the cipher text. For example:

Decode a quote by Baden-Powell which has been encoded using the Complete Columnar Cipher. You are told that the quote has BADGE somewhere in it.

The crib should contain a sequence of plain text characters which is no shorter than one less than the number of columns used to encode at the Invitational/Regional level, and no shorter than three less than the number of columns used to encode at the State/National level. In choosing a crib, review the section on "Setting Difficulty" above, as the choice of crib is a primary factor in setting difficulty for these. Reviewing the work done by the Auto-Solver can also help guide the choice of crib.

Decryption matrix for Hill Cipher

For Regional and Invitational competitions, only the Compute Decryption option is used for the Hill Cipher. For this there needs to be a 4-character phrase for a 2x2 Hill or a 9-character phrase for a 3x3 Hill which corresponds to an invertible matrix. (https://en.wikipedia.org/wiki/Invertible_matrix). Fortunately, the tool will tell you if it is not invertible. There is also a list of known valid word keys at https://toebes.com/codebusters/HillKeys.html (which is linked to at the top of the tool) for the 2x2 and 3x3 Hill Ciphers?. Note that there are other keys that can be used which are not words. It is more likely to be invertible if you use the odd letters B, D, F, H, L, N, R, T, X and Z. as they are odd and non-prime, but you can mix in some other letters. Just make sure that the keyword is not an inappropriate phrase. A total nonsense phrase is perfectly acceptable, but it helps the style of the test if it looks like a word.

For both 2x2 and 3x3 matrices, we can improve the likelihood that the determinant is invertible if half to three-quarters of the letters are odd.

There is little variability in the difficulty although letters at the end of the alphabet generate larger numbers.

100.

The question should indicate that they are to compute the decryption matrix using the Hill Cipher and provide the key. For example:

Using a key of TEST, compute the decryption matrix for a 2x2 Hill with a 26-character alphabet.

For a 2x2 matrix we can be explicit. Since it is very likely that we will want either the 2nd or 3rd letter to be a vowel, the more useful advice would be that one or both of the 2nd and 3rd letters should be vowels, while both the first and last letters come from the set of odd letters listed above. This would guarantee that the determinant is odd, and then we only need to check to make sure it isn't 0 (mod 13), which means we would succeed with probability 12/13.

For a 3x3 matrix it isn't as easy, but if we ask for about half the letters to be odd (same set as before), then there is a good chance that the determinant will be odd. In fact, assuming random placement of the odd letters, the exact odds for each possible number of odd letters is as follows:

Basically, we want 5, 6, or 7 odd letters in the mix in order to have a decent shot at being invertible mod 2. If we are lucky, we may get by with 4 odd letters; with only 3 odd letters we need to be extremely lucky; and for any other number of odd letters, it is simply impossible.

Morbit Cipher Decode

The Morbit Cipher encodes text by first converting the plain text into morse code with the space between characters represented by an × and the space between words represented by two × characters. The resulting morse code is then broken into pairs of characters (adding an × at the end if necessary). The numbers 1-9 are randomly assigned to the unique pairs of characters (●●, ●–, ●×, –●, ––, –×, ×●, ×–, ××). The pairs of morse characters are then replaced with the corresponding number to generate the final cipher string. Decoding is done by reversing the process.

The difficulty is driven by the number of letters and the variety of morse code combined with which letters are chosen. Looking at the generated solving guide can help guide the question writing. Not telling them which digit corresponds to ×× increases the difficulty.

100-150.

The question should be around 40 characters.

Solve this quote from <person> which has been encoded using the Morbit Cipher. You are told that 2=●●, 4=●–, 6=●×, 9=–●, 1=––, and 7=–×.

Pollux Cipher Decode

The Pollux Cipher encodes text by first converting the plain text into morse code with the space between characters represented by an × and the space between words represented by two × characters. The resulting morse code is then randomly assigned to digits (0-9) chosen to represent the morse characters: (●, –, ×) Each morse characters is then replaced with the corresponding number to generate the final cipher string. Decoding is done by reversing the process. Note that because one or more digits may represent a morse character piece, the mapping can be somewhat random.

The difficulty is driven by the number of letters and the variety of morse code combined with which letters are chosen. Looking at the generated solving guide can help guide the question writing. Not telling them all the Hint Digits which correspond to × increases the difficulty. Assigning more characters for one morse digit, thereby reducing another, also increases the difficulty.

100-150.

The question should be around 40 characters.

Solve this quote from <person> which has been encoded using the Pollux Cipher. You are told that 2,3=●, 4,7=–, 9,1=×.

Fractionated Morse Cryptanalysis

The Fractionated Morse Cipher encodes text by first converting the plain text into morse code with the space between characters represented by an × and the space between words represented by two × characters. The resulting morse code string is divided into groups of three characters called a morslet. The morslet is made up of combinations of these characters: (●, –, ×). There are 27 possible morselets using combinations of those characters. Each morslet is assigned to letters using a K1-like replacement table. 26 morselets are used in the replacement table; (×××) is not used. The table is always the same, and the letters in the keyword (with duplicates removed) are always the initial letters in the table, with the remaining letters following alphabetically. The morslet (sets of morse characters) are then replaced with the corresponding letter to generate the final cipher string. Decoding is done by reversing the process. The YouTube video on how to generate a Fractionated Morse cipher using the Test Generator can be found here.

The difficulty will depend on how many cipher characters are revealed by the crib. This is determined by the length of the crib and where it is placed. The difficulty is also related to the letters that are chosen for the keyword, although less so. Difficulty increases when the keyword draws letters spread throughout the alphabet, when it draws from the end of the alphabet, when the keyword is very long, and when there are duplicate letters in the keyword. Another factor for difficulty is how frequently the cipher letters that are revealed by the crib are used in the cipher.

A crib which provides 10 to 12 mappings of cipher letters to morslets should be adequate to solve the cipher in a reasonable amount of time. As stated above, the position and length of the crib determine how many cipher letters are revealed. A shorter crib may reveal more cipher letters than a longer crib, it all depends on the length of the morse characters generated by the plain text. The auto-solver output shows what cipher characters the crib reveals, so trying different cribs gives immediate feedback to the test writer. The auto-solver also makes an estimate of the length of the keyword. How close this estimate is to the actual length of the keyword will make the problem marginally easier.

Looking at the auto-solver output should give an idea of how much effort a problem will take to solve. The auto-solver iterates through logic and builds upon previous progress--techniques a student would use. Put another way, in general, the longer the output from the auto-solver, the more effort it will take a student to solve it, thus a more difficult problem.

350-420. This can be adjusted using the factors for difficulty above. The 'Suggest Points' button uses these considerations to determine a score.

The question should be around 40 characters. Note that the encoded text revealed by the 11 characters of the crib has a length of 13 characters, but only 9 are unique.

Someone passed you a piece of paper with this Fractionated Morse encoded phrase of a quote. You are told that the quote has EDSOMEMAGIC in it corresponding to the encoded text N G S N L L E L B J S V H, meaning B = ●–×; E = ●××; G = ●●×; H = –●×; J = ––●; L = ––×; N = ●×–; S = ×●●; V = ×–●. What does it say?

Vigenère Cipher Decrypt with a known plaintext

There can be one of these on a State/National test. There are no restrictions on the phrase, although try to avoid a phrase with a lot of a's in it. I.e.An amazing aardvark allows all answers would be a poor choice because the letter A is trivial to decode. As this question is nominally worth four points per letter, a 50 letter phrase is ideal.

Additionally there needs to be a Key to encode it with. It should be 5 or 6 characters with no repeating letters and avoid the letter a as it causes a letter to map to itself. By setting the Block Size to the same as the length of the Key, the problem is much easier than with the default Block Size of 0 that keeps the original spacing. Setting the Block Size to a size other than the length of the Key increases the difficulty somewhat.

For the decoding clue, pick a word in the phrase which is at least 5 characters long, carefully count to determine the position in the encoded phrase as well as the encoded character.

The length of the phrase is the major factor affecting the difficulty, as is the choice of crib: it will be easier if the letters in the crib fall aligned with the pattern of the key. Unlike many other keyword ciphers, this one actually becomes marginally easier if there are duplicate letters in the key, as it reduces the number of rows and columns that competitors have to consult.

200-250. If the Block Size is 0, add 25 points. If the Block Size is non-zero and different from the Key length, add 50 points.

The question should indicate that the Vigenère Cipher is being used (don't forget the accented è), the length of the Key and the plain text corresponding to some part of the phrase. It is generally nice to give the origin of the phrase if it is known. For example:

<person> once said this about <topic>. It has been encoded using the Vigenère cipher using a very common five letter word. You have been told that the 17th through the 22nd letters in the code (YMNCHU) actually is the word REASON. What does the message decode to?

Porta Cipher Cryptanalysis with a known plaintext B,C{3.f.iii}

There can be one of these on a State/National test. There are no restrictions on the phrase, although try to avoid a phrase with a lot of a's in it. I.e.An amazing aardvark allows all answers would be a poor choice because the letter A is trivial to decode. As this question is nominally worth four points per letter, a 50 letter phrase is ideal.

Additionally there needs to be a Key to encode it with. It should be 5 or 6 characters with no repeating letters and avoid the letter a as it causes a letter to map to itself. By setting the Block Size to the same as the length of the Key, the problem is much easier than with the default Block Size of 0 that keeps the original spacing. Setting the Block Size to a size other than the length of the Key increases the difficulty somewhat.

For the decoding clue, pick a word in the phrase which is at least 5 characters long, carefully count to determine the position in the encoded phrase as well as the encoded character.

The length of the phrase is the major factor affecting the difficulty, as is the choice of crib: it will be easier if the letters in the crib fall aligned with the pattern of the key. Unlike many other keyword ciphers, this one actually becomes marginally easier if there are duplicate letters in the key, as it reduces the number of rows and columns that competitors have to consult.

200-250. If the Block Size is 0, add 25 points. If the Block Size is non-zero and different from the Key length, add 50 points.

The question should indicate that the Porta Cipher is being used, the length of the Key and the plain text corresponding to some part of the phrase. It is generally nice to give the origin of the phrase if it is known. For example:

<person> once said this about <topic>. It has been encoded using the Porta cipher using a very common five letter word. You have been told that the 17th through the 22nd letters in the code (YMNCHU) actually is the word REASON. What does the message decode to?

RSA Cipher

There are several variants of the RSA question. The Safe Combo and Exchange Keys options test the knowledge of the RSA algorithm by presenting randomized order of RSA components with randomized values. The Quantum Computer and Compute d options test using the extended Euclidean Algorithm compute the inverse. Decode Year tests using Rapid Modular Exponentiation to compute the mod of a prime raised to a large factor. All of these dynamically generate and update the question text with all of the computed values.

For the Safe Combo and Exchange Keys choices, it is good to pick values for the Prime Digits and Safe Combination Digits/Data Digits so that the generated RSA keys and numbers provide lots of similar digits. The Randomize button will regenerate the values. Note that the student names are also randomized, picking from the names which were common around the turn of this century.

For the other options, click on the Randomize button until it generates a problem which has the desired level of difficulty.

The choice of question type and the size of the digits in the key are the major impact on difficulty.

Safe Combo: 100 points

Quantum Computer, Compute d: 15 points per step of the extended Euclidean Algorithm (shown by the tool) with a minimum of 120 points.

Decode Year: 15 points per 1 bit and 5 points per 0 bit in the binary representation of d (shown by the tool) with a minimum of 120 points.

Exchange Keys: 120 points

The question text should be automatically generated, but can be edited. Note that the tool keeps track of all the edits so that if it generates new values, the edits are kept in place.

PigPen Cipher

The PigPen/Masonic Cipher is one of the easiest Ciphers for students to decode because the alphabet is fixed. The letter E will always be represented by a square. There can be a couple of PigPen Cipher questions on a test. There is an accompanying YouTube video for creating a PigPen cipher that can be found here.

The only factor for difficulty with this question is in the number of characters in the phrase. The choice of letters/words/word length has no impact on the difficulty. In general the question should be approximately 40-50 characters.

A PigPen Cipher Decode should be worth 60-80 points (approximately 1 point per character in the Plain Text plus 1.5 points for each unique character).

The question should clearly indicate that cipher has been encoded using the PigPen Cipher as well as the origin of the phrase or quote. It should not include a hint. For example:

Solve this quote from <person> which has been encoded with the PigPen Cipher.

Knights Templar Cipher

The Knights Templar Cipher is similar to the PigPen/Masonic Cipher and is generally easy for students to decode because the alphabet is fixed. The letter N will always be represented by an X. There can be a couple of Knights Templar Cipher questions on a test.

The only factor for difficulty with this question is in the number of characters in the phrase. The choice of letters/words/word length has no impact on the difficulty. In general the question should be approximately 40-50 characters.

A Knights Templar Cipher Decode should be worth 60-80 points (approximately 1 point per character in the Plain Text plus 1.5 points for each unique character).

The question should clearly indicate that cipher has been encoded using the Knights Templar Cipher as well as the origin of the phrase or quote. It should not include a hint. For example:

Solve this quote from <person> which has been encoded with the Knights Templar Cipher.

Standard Galactic Alphabet Cipher

The Standard Galactic Alphabet is another type of simple cipher with a fixed alphabet. Each letter in the alphabet corresponds with a symbol, and is represented by the same symbol each time it appears. The letter Y will always be represented by a symbol that looks like two vertical lines ||. Additionally, it is worth noting that in its appearance on the resource sheet, the Standard Galactic Alphabet is not provided in alphabetic order, but instead is represented with a common pangram (a sentence that includes each letter of the alphabet). Students will need to familiarize themselves with the pangram sentence "The quick brown fox jumps over the lazy dog" in order to be able to map the Standard Galactic symbols with the corresponding plaintext letters. There can be several Standard Galactic Alphabet questions on a test.

The two factors for difficulty with this question are the number of characters in the phrase, and the number of unique characters that are used. The choice of letters/words/word length has no impact on the difficulty. In general the question should be approximately 40-50 characters.

A Standard Galactic Alphabet Cipher Decode should be worth roughly 80 points (approximately 1 point per character in the Plain Text plus 2 points for each unique character).

The question should clearly indicate that cipher has been encoded using the Standard Galactic Alphabet as well as the origin of the phrase or quote. It should not include a hint. For example:

Solve this quote from <person> which has been encoded with the Standard Galactic Alphabet.

Tap Code Cipher

The Tap Code is is a simple cipher to decode because the mapping is is fixed. A single dot followed by a single dot will always stand for the letter a. What makes it slightly more challenging is that the students need to memorize the table or just know how to recreate it on the test. There should be one or two Tap Code Cipher questions on a test. The accompanying instructional YouTube video can be found here.

The two factors for difficulty with this question is in the number of characters in the phrase and how far down in the alphabet the letters are. The further in the alphabet, the more characters that they have to count.

A Tap Code Cipher should be worth 55-75 points. This works out to about 1.5 points per letter or one point for every 3 or 4 taps generated.

The question should clearly indicate that cipher has been encoded using the Tap Code Cipher. It should not include a hint. For example:

Solve this quote from <person> which has been encoded with the Tap Code Cipher.

Dancing Men Cipher

The Dancing Men Cipher (also known as the Running Men Cipher) is one of the easiest Ciphers for students to decode because the alphabet is fixed. The letter E will always be represented by a person with both arms up in the air and both legs split. There can be a couple of Dancing Men Cipher questions on a test. There is a YouTube video showing how to implement a Dancing Men cipher here.

The only factor for difficulty with this question is in the number of characters in the phrase. The choice of letters/words/word length has no impact on the difficulty. In general the question should be approximately 40 characters.

A Dancing Men Cipher Decode should be worth 80 points (approximately 1 point per character in the Plain Text and 2 points for how many of the characters are unique).

The question should clearly indicate that cipher has been encoded using the Dancing Men Cipher as well as the origin of the phrase or quote. It should not include a hint. For example:

Solve this quote from <person> which has been encoded with the Dancing Men Cipher.

Affine Cipher decrypt

There should be a single one of these on the test. It should use a phrase about 25 characters long that doesn't have too many occurrences of the letter A in it, preferably with as large a variety of letters as possible… (e.g. The quick brown fox jumps over the lazy dog isn't actually bad).

Pick a value for a which is coprime with 26 (1,3,5,7,9,11,15,17,19,21,23 or 25). The actual value doesn't matter, but larger ones tend to be slightly harder. If you are generating tests for multiple regions, pick numbers that are near each other. I.e. 7, 9 and 11 would be good to have as equivalent a values.

Pick a value for b between 1 and 25 inclusive. Unlike a where the larger values become slightly harder, the value of b can truly be any number and be the same level of difficulty. There is an accompanying YouTube video demonstrating how to create an Affine Cipher Decrypt problem in the YouTube video linked here.

There is very little variability in the difficulty other than the length of the string. Larger values of a are only slightly harder while the value of b has no real impact on the difficulty.

100.

The question should state to use the Affine Cipher and give the values of a and b for them to use (note that a and b are italicized). For example:

Decrypt the following cipher text which was encoded using the Affine Cipher with a=5 and b=9.

Affine Cryptanalysis

There should be a single one of these on the State/National test. It should use a phrase about 25 characters long that doesn't have too many occurrences of the letter a in it, preferably with as large a variety of letters as possible. (e.g. The quick brown fox jumps over the lazy dog isn't actually bad).

Pick a value for a which is coprime with 26 (1,3,5,7,9,11,15,17,19,21,23 or 25). The actual value doesn't matter, but larger ones tend to be slightly harder. If you are generating tests for multiple regions, pick numbers that are near each other. I.e. 7, 9 and 11 would be good to have as equivalent a values.

Pick a value for b between 1 and 25 inclusive. Unlike a where the larger values become slightly harder, the value of b can truly be any number and be the same level of difficulty. There is a YouTube video demonstrating the creation of an Affine Cryptanalysis cipher here.

There is very little variability in the difficulty other than the length of the string. Larger values of a are only slightly harder while the value of b has no real impact on the difficulty. Picking the two mapping letters so that they are toward the end of the alphabet also increase the difficulty. Having phrases which use the letters G, W, Y, B, V, K, X, J, Q, and Z increase the difficulty slightly.

140-240. A good rule of thumb is 6 points per character plus 6 points for each of the letters G, W, Y, B, V, K, X, J, Q, and Z that are used.

The question should state to use the Affine Cipher and the indication of what two letters map to. It should not give the values of a or b. For example:

A message from <person> encrypted with the Affine Cipher using an alphabet of 26 characters has been received. You have been told that the first two letters are TH. With that knowledge, what does this message say?

Affine Cipher encrypt

Decryption matrix for Hill Cipher

For Regional and Invitational competitions, only the Compute Decryption option is used for the Hill Cipher. For this there needs to be a 4-character phrase for a 2x2 Hill or a 9-character phrase for a 3x3 Hill which corresponds to an invertible matrix. (https://en.wikipedia.org/wiki/Invertible_matrix). Fortunately, the tool will tell you if it is not invertible. There is also a list of known valid word keys at https://toebes.com/codebusters/HillKeys.html (which is linked to at the top of the tool) for the 2x2 and 3x3 Hill Ciphers?. Note that there are other keys that can be used which are not words. It is more likely to be invertible if you use the odd letters B, D, F, H, L, N, R, T, X and Z. as they are odd and non-prime, but you can mix in some other letters. Just make sure that the keyword is not an inappropriate phrase. A total nonsense phrase is perfectly acceptable, but it helps the style of the test if it looks like a word.

For both 2x2 and 3x3 matrices, we can improve the likelihood that the determinant is invertible if half to three-quarters of the letters are odd.

There is little variability in the difficulty although letters at the end of the alphabet generate larger numbers.

100.

The question should indicate that they are to compute the decryption matrix using the Hill Cipher and provide the key. For example:

Using a key of TEST, compute the decryption matrix for a 2x2 Hill with a 26-character alphabet.

For a 2x2 matrix we can be explicit. Since it is very likely that we will want either the 2nd or 3rd letter to be a vowel, the more useful advice would be that one or both of the 2nd and 3rd letters should be vowels, while both the first and last letters come from the set of odd letters listed above. This would guarantee that the determinant is odd, and then we only need to check to make sure it isn't 0 (mod 13), which means we would succeed with probability 12/13.

For a 3x3 matrix it isn't as easy, but if we ask for about half the letters to be odd (same set as before), then there is a good chance that the determinant will be odd. In fact, assuming random placement of the odd letters, the exact odds for each possible number of odd letters is as follows:

Basically, we want 5, 6, or 7 odd letters in the mix in order to have a decent shot at being invertible mod 2. If we are lucky, we may get by with 4 odd letters; with only 3 odd letters we need to be extremely lucky; and for any other number of odd letters, it is simply impossible.

Hill Cipher Encrypt

Pick a phrase to encode. As a rule of thumb for a 2x2 matrix, every pair of letters is worth 16 points. Ideally you want an odd length string to force them to use a padding Z. For a 3x3 matrix, every group of three letters is worth 21 points. It is important to pick a phrase which is not a multiple of 3 characters long so that they must add the appropriate number of padding characters.

Pick an encoding key. For a 2x2 it is 4 characters long and for a 3x3 it is 9 characters long. This is probably the hardest part to making the test as the matrix has to be invertible (https://en.wikipedia.org/wiki/Invertible_matrix). Fortunately, the tool will tell you if it is not invertible. There is also a list of known valid keys at https://toebes.com/codebusters/HillKeys.html for both the 2x2 and 3x3 encodings. In general, it is more likely to be invertible if you use the letters B, D, F, H, L, N, R, T, X and Z. as they are odd and non-prime, but you can mix in some other letters. Just make sure that the keyword is not an inappropriate phrase. A total nonsense phrase is perfectly acceptable, but it helps the style of the test if it looks like a word. There is a YouTube video on generating a Hill Cipher using the tool which can be found here.

Encrypt is harder than Decrypt and 3x3 is harder than 2x2. The other factor is the number of characters for them to encode.

160-250. About 16 points per group of 2 for a 2x2, 21 points per group of 3 for a 3x3.

The question should indicate whether they are to Encode or Decode using the Hill Cipher with a 26 character alphabet along with the Key. For example:

Using a key of CARNIVALS encode the string ASTROBIOLOGIST using the Hill Cipher with a 26-character alphabet.

Decode the following quote by <person> which was encoded as a 2x2 Hill Cipher using a keyword of LOST.

Rail Fence Cipher

The rail fence cipher encodes messages by 'zig-zagging' the letters along 2 to 6 'rails'. The cipher can be decoded by brute force; trial and error; or procedurally by applying some simple math and understanding spacing of characters in the rail patterns.

The difficulty gets slightly higher with more rails because it requires more diligence when solving. It becomes slightly more difficult when there is an offset, and significantly moreso when the offset is unknown because of the additional time investment required for trial and error.

100-150. A rule of thumb: 2 rails = 100; 3 rails = 110; 4 rails = 120; 5 rails = 130; 6 rails = 140 points. 150 points for when only the range of rails is given.

The question should be around 75 characters.

Solve this quote from <person> which has been encoded with the rail fence cipher. The message was encoded with <rail count> rails.

Morbit Cipher Decode

The Morbit Cipher encodes text by first converting the plain text into morse code with the space between characters represented by an × and the space between words represented by two × characters. The resulting morse code is then broken into pairs of characters (adding an × at the end if necessary). The numbers 1-9 are randomly assigned to the unique pairs of characters (●●, ●–, ●×, –●, ––, –×, ×●, ×–, ××). The pairs of morse characters are then replaced with the corresponding number to generate the final cipher string. Decoding is done by reversing the process.

The difficulty is driven by the number of letters and the variety of morse code combined with which letters are chosen. Looking at the generated solving guide can help guide the question writing. Not telling them which digit corresponds to ×× increases the difficulty.

100-150.

The question should be around 40 characters.

Solve this quote from <person> which has been encoded using the Morbit Cipher. You are told that 2=●●, 4=●–, 6=●×, 9=–●, 1=––, and 7=–×.

Morbit Cipher Cryptanalysis

The Morbit Cipher encodes text by first converting the plain text into morse code with the space between characters represented by an × and the space between words represented by two × characters. The resulting morse code is then broken into pairs of characters (adding an × at the end if necessary). The numbers 1-9 are randomly assigned to the unique pairs of characters (●●, ●–, ●×, –●, ––, –×, ×●, ×–, ××). The pairs of morse characters are then replaced with the corresponding number to generate the final cipher string. Decoding is done by reversing the process.

The difficulty is driven by the number of letters and the variety of morse code combined with which letters are chosen. Looking at the generated solving guide can help guide the question writing. Not giving them Hint Digits which exposes the digit corresponding to ×× increases the difficulty.

150-200.

The question should be around 40 characters.

Solve this quote from <person> which has been encoded using the Morbit Cipher. You are told that the first four letters are SOME.

Pollux Cipher Decode

The Pollux Cipher encodes text by first converting the plain text into morse code with the space between characters represented by an × and the space between words represented by two × characters. The resulting morse code is then randomly assigned to digits (0-9) chosen to represent the morse characters: (●, –, ×) Each morse characters is then replaced with the corresponding number to generate the final cipher string. Decoding is done by reversing the process. Note that because one or more digits may represent a morse character piece, the mapping can be somewhat random.

The difficulty is driven by the number of letters and the variety of morse code combined with which letters are chosen. Looking at the generated solving guide can help guide the question writing. Not telling them all the Hint Digits which correspond to × increases the difficulty. Assigning more characters for one morse digit, thereby reducing another, also increases the difficulty.

100-150.

The question should be around 40 characters.

Solve this quote from <person> which has been encoded using the Pollux Cipher. You are told that 2,3=●, 4,7=–, 9,1=×.

Pollux Cipher Cryptanalysis

The Pollux Cipher encodes text by first converting the plain text into morse code with the space between characters represented by an × and the space between words represented by two × characters. The resulting morse code is then randomly assigned to digits (0-9) chosen to represent the morse characters: (●, –, ×) Each morse character is then replaced with the corresponding number to generate the final cipher string. Decoding is done by reversing the process. Note that because one or more digits may represent a morse character piece, the mapping can be somewhat random.

The difficulty is driven by the number of letters and the variety of morse code combined with which letters are chosen. Looking at the generated solving guide can help guide the question writing. Not giving them Hint Digits which exposes all the digits corresponding to × increases the difficulty. Assigning more characters for one morse digit, thereby reducing another, also increases the difficulty.

150-200.

The question should be around 40 characters.

Solve this quote from <person> which has been encoded using the Pollux Cipher. You are told that the first four letters are SOME.

Vigenère Cryptanalysis

There are no restrictions on the phrase, although try to avoid a phrase with a lot of a's in it. I.e.An amazing aardvark allows all answers would be a poor choice because the letter A is trivial to decode. As this question is nominally worth four points per letter, a 50 letter phrase is ideal.

Additionally there needs to be a Key to encode it with. It should be 5 or 6 characters with no repeating letters and avoid the letter a as it causes a letter to map to itself. By setting the Block Size to the same as the length of the Key, the problem is much easier than with the default Block Size of 0 that keeps the original spacing. Setting the Block Size to a size other than the length of the Key increases the difficulty somewhat.

For the decoding clue, pick a word in the phrase which is at least 5 characters long, carefully count to determine the position in the encoded phrase as well as the encoded character.

The length of the phrase is the major factor affecting the difficulty, as is the choice of crib: it will be easier if the letters in the crib fall aligned with the pattern of the key. Unlike many other keyword ciphers, this one actually becomes marginally easier if there are duplicate letters in the key, as it reduces the number of rows and columns that competitors have to consult.

200-250. If the Block Size is 0, add 25 points. If the Block Size is non-zero and different from the Key length, add 50 points.

The question should indicate that the Vigenère Cipher is being used (don't forget the accented è), the length of the Key and the plain text corresponding to some part of the phrase. It is generally nice to give the origin of the phrase if it is known. For example:

<person> once said this about <topic>. It has been encoded using the Vigenère cipher using a very common five letter word. You have been told that the 17th through the 22nd letters in the code (YMNCHU) actually is the word REASON. What does the message decode to?

Running Key Cipher

Pick a phrase which is approximately 80 characters long. The actual content has little impact on the difficulty. By default there are four encoding texts (Gettysburg address, Declaration of Independence, Constitution of United States of America and MAGNA CARTA (In Latin)) but they can be changed by going to https://toebes.com/codebusters/EditRunningKeys.html.

There is little variability in the difficulty other than the length. Encode is slightly harder than Decode.

200-250. Approximately 3 points per letter.

The question should indicate that the Running Key Cipher is being used and whether they are to Encode or Decode. If they are to Encode, it should indicate which encoding text to use. For example:

The following quote from <person> has been encoded using a running key cipher against a famous document. What does it say?

Encode what <person> said about <topic> using a running key cipher against the MAGNA CARTA.

RSA Cipher

There are several variants of the RSA question. The Safe Combo and Exchange Keys options test the knowledge of the RSA algorithm by presenting randomized order of RSA components with randomized values. The Quantum Computer and Compute d options test using the extended Euclidean Algorithm compute the inverse. Decode Year tests using Rapid Modular Exponentiation to compute the mod of a prime raised to a large factor. All of these dynamically generate and update the question text with all of the computed values.

For the Safe Combo and Exchange Keys choices, it is good to pick values for the Prime Digits and Safe Combination Digits/Data Digits so that the generated RSA keys and numbers provide lots of similar digits. The Randomize button will regenerate the values. Note that the student names are also randomized, picking from the names which were common around the turn of this century.

For the other options, click on the Randomize button until it generates a problem which has the desired level of difficulty.

The choice of question type and the size of the digits in the key are the major impact on difficulty.

Safe Combo: 100 points

Quantum Computer, Compute d: 15 points per step of the extended Euclidean Algorithm (shown by the tool) with a minimum of 120 points.

Decode Year: 15 points per 1 bit and 5 points per 0 bit in the binary representation of d (shown by the tool) with a minimum of 120 points.

Exchange Keys: 120 points

The question text should be automatically generated, but can be edited. Note that the tool keeps track of all the edits so that if it generates new values, the edits are kept in place.