Codebusters C

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Re: Codebusters C

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TheChiScientist wrote:
jimmy-bond wrote:At a recent competition, there was a Baconian alphabet problem. Instead of A's and B's, they used the symbols <, >, [, ], {, }, (, and ). No hint was provided, and there were 52 letters in total. There were no repeating sets of symbols. How would someone tackle this problem?
I would attempt to see how grouping would create a logical response. On a test I took they used a Baconian with odd and even numbers as A and B. I suspect that in this case, A is most likely <, [, { and B is most likely >, ], }. If I could see how the test was formatted I could test this theory out but in a nutshell, you need to find patterns in Baconian ciphers. Also, don't forget to bunch everything into fives.
Adding on: with bacon the first thing to always do is to separate the groups of 5 by drawing a line between them or something. After that look for a group of 5 that are similar that fit in A or E
So like )>}]} are all similar and fit in the AAAAA for the letter A and those 5 would likely be A
Or (<}([ fit in the AABAA for E
I guess they could be annoying and set them randomly to A and B where there is no pattern. In this case you could count the frequency of each symbol for part of the cipher. Assuming they distribute each A and B equally among the symbols, A appears more frequently then B and you can assume the symbols that appear frequently are the A
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Re: Codebusters C

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Name wrote:
TheChiScientist wrote:
jimmy-bond wrote:At a recent competition, there was a Baconian alphabet problem. Instead of A's and B's, they used the symbols <, >, [, ], {, }, (, and ). No hint was provided, and there were 52 letters in total. There were no repeating sets of symbols. How would someone tackle this problem?
I would attempt to see how grouping would create a logical response. On a test I took they used a Baconian with odd and even numbers as A and B. I suspect that in this case, A is most likely <, [, { and B is most likely >, ], }. If I could see how the test was formatted I could test this theory out but in a nutshell, you need to find patterns in Baconian ciphers. Also, don't forget to bunch everything into fives.
Adding on: with bacon the first thing to always do is to separate the groups of 5 by drawing a line between them or something. After that look for a group of 5 that are similar that fit in A or E
So like )>}]} are all similar and fit in the AAAAA for the letter A and those 5 would likely be A
Or (<}([ fit in the AABAA for E
I guess they could be annoying and set them randomly to A and B where there is no pattern. In this case you could count the frequency of each symbol for part of the cipher. Assuming they distribute each A and B equally among the symbols, A appears more frequently then B and you can assume the symbols that appear frequently are the A
That was the first thing I tried, but I got 4 consonants for the first 4 letters so I stopped there. I also tried the possible combinations of groups designated as certain letters (ex. < and > are A, ( and ) are A, [ and ] are B, and { and } are B). I tried a few combinations but every time, I would get three consecutive consonants at the beginning. So my guess is they were being annoying.

I thought about the second strategy you mentioned (frequency) but quickly dismissed it. All symbols were represented relatively equally (at least it seemed so, as I did not bother to count it) so it would be strenuous to guess and test.
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Re: Codebusters C

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jimmy-bond wrote: That was the first thing I tried, but I got 4 consonants for the first 4 letters so I stopped there. I also tried the possible combinations of groups designated as certain letters (ex. < and > are A, ( and ) are A, [ and ] are B, and { and } are B). I tried a few combinations but every time, I would get three consecutive consonants at the beginning. So my guess is they were being annoying.

I thought about the second strategy you mentioned (frequency) but quickly dismissed it. All symbols were represented relatively equally (at least it seemed so, as I did not bother to count it) so it would be strenuous to guess and test.
Ok so I just took a completly random quote and encrypted it with Bacon cipher (24 letter). Almost exactly 2/3 of the letters were As, so there should be a noticable difference between frequencies where a symbol that represents A appears twice as often. You probably don't need to count too many letters, maybe 5ish would be sufficent to get a good idea.
Also what compitition is this? Im kinda interested in what the actual question is.
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Re: Codebusters C

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Name wrote:
jimmy-bond wrote: That was the first thing I tried, but I got 4 consonants for the first 4 letters so I stopped there. I also tried the possible combinations of groups designated as certain letters (ex. < and > are A, ( and ) are A, [ and ] are B, and { and } are B). I tried a few combinations but every time, I would get three consecutive consonants at the beginning. So my guess is they were being annoying.

I thought about the second strategy you mentioned (frequency) but quickly dismissed it. All symbols were represented relatively equally (at least it seemed so, as I did not bother to count it) so it would be strenuous to guess and test.
Ok so I just took a completly random quote and encrypted it with Bacon cipher (24 letter). Almost exactly 2/3 of the letters were As, so there should be a noticable difference between frequencies where a symbol that represents A appears twice as often. You probably don't need to count too many letters, maybe 5ish would be sufficent to get a good idea.
Also what compitition is this? Im kinda interested in what the actual question is.
Pearl City High School Invitational, took place around 4 hours ago. I'll be getting my test on Tuesday and will post the question then if I remember, and if I get hold of the answer key, I'll post the answer
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Re: Codebusters C

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I got my test back and realized that they encoded it using a 24-letter alphabet instead of 26 and I am now kicking myself, because the left symbols were A and right symbols were B, which is what I tried initially.
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Re: Codebusters C

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jimmy-bond wrote:I got my test back and realized that they encoded it using a 24-letter alphabet instead of 26 and I am now kicking myself, because the left symbols were A and right symbols were B, which is what I tried initially.
Oh nooooo :cry:

PS: I think this might help you :)
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Re: Codebusters C

Post by jimmy-bond »

dragonfruit35 wrote:
jimmy-bond wrote:I got my test back and realized that they encoded it using a 24-letter alphabet instead of 26 and I am now kicking myself, because the left symbols were A and right symbols were B, which is what I tried initially.
Oh nooooo :cry:

PS: I think this might help you :)
Thank you. I didn't know that. That might be the reason I mess up on most practice tests.
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Re: Codebusters C

Post by Carrot »

For solving the timed question, can you get it checked when you still have 1 or 2 errors (as often I can't think of the last word with two missing letters in it) or do you have to fully solve the problem?
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Re: Codebusters C

Post by Riptide »

Carrot wrote:For solving the timed question, can you get it checked when you still have 1 or 2 errors (as often I can't think of the last word with two missing letters in it) or do you have to fully solve the problem?
Yes you can have it checked with up to 2 errors.
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Re: Codebusters C

Post by mpnobivucyxtz »

Is there a standard format for this event? Last year's nationals test provided frequency tables and the Vigenere table as well, but some invitationals I've seen provide neither. What can I expect from regionals and states?
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