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Burgruine-Wieladinge Gruppe

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Code Busters


Codebusters is an event in Division C for the 2023 season and was held as trial event at both the 2016 National Tournament and the 2018 National Tournament. It was going to be held as a trial event for Division B at the 2020 National Tournament before its cancellation. In this event, up to 3 participants must decode encrypted messages, or they may be required to encode messages with certain advanced ciphers. Competitors are not allowed to bring any resources to this event, but can bring a 4 or 5 function calculator - no scientific or graphing calculators allowed.




Code Busters



Tests are composed of a variety of questions where teams will encrypt or decrypt various code types. The number of questions on a given test is variable, but tests often contain anywhere from 6-24 questions, depending on the difficulty of the questions. The first question of a test is timed, and a time bonus is given for the question based on how quickly it is completed. Points may be deducted from questions based on the number of mistakes found in the answer given. If the answer differs from the solution by only one or two letters, then no points are removed, despite the answer having errors. Each additional error after two errors will result in a deduction of 100 points. The number of points removed from a question through deductions will not exceed the value of the question (meaning that deductions will never result in a question score below 0).


The very first question is timed - when solved, a team member should signal the event supervisor that they have finished the question, such as raising their hand, shouting "bingo!", or another method determined by the supervisor before the event begins. The first question will be an Aristocrat with or without a hint. Teams will receive bonus points depending on their time, and they may make as many attempts as they want to break the code without any penalties. The timing bonus will be calculated from the start of the event until the question is answered successfully, or until 10 minutes have elapsed, and is calculated with the formula: 4*(600-number of seconds taken). Teams may still answer the question after 10 minutes; however, the timing bonus is 0. The first cryptogram may also be used as a tiebreaker. Solutions are given full points only if the solution is an exact match or differs by only one or two letters. Deductions for inconsistencies in the answer to the first question are treated the same as deductions for a non-timed question.


For this example, we encode the message CODEBUSTERS using [math]\displaystyle a=9 [/math] and [math]\displaystyle b=42 [/math]. To slightly speed up the process, use the fact from modular arithmetic that the function [math]\displaystyle ax+b [/math] does not change if we also take [math]\displaystyle a, b [/math] modulo 26: the function [math]\displaystyle 9x+42 [/math] is equivalent to the function [math]\displaystyle 9x+16 [/math] or [math]\displaystyle 9x-10 [/math].


Using the alphabet square, encode the plaintext. The first message letter is S and the first key letter is S, therefore, look at the table to see where row S and column S intersect. It is clear that they intersect at K, so write down K as the first letter of the ciphertext. Repeat this with the rest of the message. Thus, the encrypted text is "KEQSYAW QTMXNACL WD AGQT"


If the key and ciphertext are both given, decoding is also possible. This is done by taking a letter of the key and finding its row, finding the corresponding letter of ciphertext in that row, and seeing what column that letter falls in. The letter in the column is the letter of the plaintext. In the previous example, the first letter is decoded by going to row S and finding K, which is located in column S. Thus, the plaintext letter is S.


To decode a message using this method, subtract instead of add the key's corresponding number: The cipher text K with the corresponding key letter S gives plain text [math]\displaystyle 10-18=-8\equiv18\pmod26\to \textS [/math].


The Baconian cipher replaces each letter of the plaintext with a 5 letter combination of 'A' and 'B'. This replacement is a binary form of encoding, in which 'A' may be considered as 0 and 'B' as 1. One variant of the Baconian Cipher uses a 24 letter alphabet with the letters 'I' and 'J' having the same code, as well as 'U' and 'V'.


Another variant of this cipher uses a unique code for each letter (26 letter alphabet). However, in a rule clarification, only the 24 letter variation is to be used on Codebusters tests for the 2019 season.


When solving a question encoded with the Baconian Cipher, it is very likely that they won't explicitly give you "A" and "B" to use to find the corresponding letters. Most of the time, they'll have certain symbols or letters that are meant to represent "A" and "B". One example of this would be using all of of the even letters in the alphabet to represent "A" while all the odd letters represent "B". There are many different variants of this, including symbols on a keyboard and vowel/consonants. To solve a Baconian, try to group the different symbols/letters into two groups based on their properties, and assign one group "A" and the other group "B". If it doesn't work the first time, then switch the groups so that the previous "A" group is now the "B" group. If even then it doesn't seem to be working, then start over again and find a different characteristic to base the groups off of. For this reason, the Baconian Cipher can take longer than some of the other ciphers.


The Morse ciphers are a pair of ciphers where you are given a long text of numbers, where each number corresponds to one or two Morse characters (dot, dash, or a separator). Some of these number values will be given while some will not. Using the properties of Morse code and logical thinking, you must find the decryption for all of the missing numbers.


In a Morbit cipher, each number corresponds to a pair of two Morse characters. Each pair of Morse character can only correspond to one number, meaning there are only nine possible decryptions. Whereas Pollux quotes cannot end with a separator, Morbit ciphers may or may not end with a separator. This is because each number has to correspond to two Morse characters, so an extra separator is added at the end if the quote in Morse code is an odd number of characters.


At this point, with very little left, one may often be able to guess what the remaining letters are without even working the Morse code out using basic decryption logic as with Aristocrats and Patristocrats. However, this example will continue using the Morse logic.


In the event that a question encoded with the running key cipher asks you to decode the phrase but does not give the key for it, then one has to use one of the documents given on the reference sheet (if applicable) and plug them all in to determine which of the documents is the key. If they give the key but do not give the text of the key (ex. "a famous quote by George Washington"), then it is probably best to skip the question.


Cody, Quinn, Luke, and M.E. may be really different, but they all share one thing in common: they love playing around with codes. In fact, they love codes so much, they have their own private club, with a super-secret hideout and passwords that change every single day.


When Cody and Quinn notice what could be a code on the window of a nearby house, the one owned by their strange neighbor, the guy they call Skeleton Man, the club gets to work. And it is a cry for help!


This exciting interactive mystery offers more than fifteen codes for you to decipher, including the Consonant code, Morse code, and American Sign Language. Test your brain with the Code Busters and solve the mystery along with them. Answers are in the back, if you ever get stuck.


This interactive mystery features more than fifteen codes and puzzles for you to decipher along with the Code Busters, including the Orienteering code, Morse code, and the Trail Signs code. Answers are in the back, if you ever get stuck.


The Code Busters are excited for their class field trip to Angel Island, known as the Ellis Island of the West. One of Mika's ancestors passed through the island's immigration station in the early twentieth century, and Mika thinks he may have left behind some secret messages . . . plus a very special box. But as the Code Busters search for hints from the past, they get caught up in an even bigger secret. Can you crack the code? Test your brain with the Code Busters to see if you have the right stuff to be an ace detective. Answers are in the back, in case you get stuck.


For several years, Cryptograms have served as the nucleus for an exciting Science Olympiad event,called Codebusters. Then Codebusters also adds several mathematical ciphers, culminatingin the RSA algorithm---the most common cipher on the internet today.In fact, I have run the Codebusters event for Wisconsin Science Olympiad since Novemberof 2013.


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