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Nice to see you having a go at some of the questions that are posted with my challenges.
All posts are correct so far by those who have posted in response to them, remember some are open questions I have no answers to, and probably not easy or too time consuming to give answers.
There’s still others awaiting a solve though.@Mad_mathsman, I used Shannon’s definition. Since there are 11 ciphertext symbols, I calculated the entropy with log base 11. The result is very close to 1, meaning that the entropy is as if each character in the ciphertext was randomly selected from the set of symbols.
@Kford: night and day
@Madness:
Ah, so what you’ve really done is computed the entropy of the process where you pick one random value from the given ciphertext, and observed that this entropy is very close to the entropy of the uniform random process with 11 possible outputs. What that this really means is that each letter appears with roughly equal frequency in the given ciphertext. This is actually a property I would expect to be present (or nearly present) in many more secure ciphers. I think your ‘entropy’ only relates to the entropy of the cipher itself if each plaintext character determines a single ciphertext character, and there is no correlation between these ciphertext characters (in which case it your entropy is the entropy of the cipher divided by the length of the cipher).A bit of trivia: In the Enigma cipher, some characters were (on average) less frequent in the ciphertext than others. This is because no character could be encoded as itself, but I think the other 25 characters were roughly equally likely to occur (I haven’t got a rigourous proof of this). So the most common letters in the plaintext would be less common overall in the ciphertext.
Sorry, my fault! (I suppose you could work it out by doing the integral of the integral of g, with respect to t…)
The KA Answers – KA-3 (Initial Thoughts)
As @The-letter-wriggler correctly found out, the next term in the sequence could have been anything from Sitting In X to Stationary Indian X-Rays. But there is a pattern. Let’s remind ourselves of the first five terms that I created:
Observing New Elephants
The Witty Ones
Tribe Hannibal Reaches Extraordinary Eras
Fire Obviously Utilises Rain
Fined In Venus’s EarholesThe title was ‘Initial Thoughts’, and this could lead you to getting the initals of each word. What you get is ONE, TWO, THREE, FOUR, FIVE. So the next term in the sequence must have had the initals of SIX, then SEVEN for the next one, and so on. Here is the remaining terms I thought of whilst posting this, to get from ONE to TWENTY:
(The first five are earlier in this post.)
Successful Iraqi Xylophones
Several Egyptian Violins Eat Nothing (I put this in post #48695, page 9 of this topic)
Eating Italian Grapes Has Triumphs
No Indian Needs Everything
To Every Norwegian!
Exceptional Levitating European Violins Eat Noodles
Twenty Weird Exoplanets Lately Viewed Earth
The Highest Independent Russian To Ever Eat Nuts
Fixing Omani’s Underwater Rivers Takes Every Excited Nigerian
Finding Italian Violins Excitedly Troubles Egyptians Eating Nectarines
Succint Individual X-Rays Trouble Every Eating Namibian
Several Egyptian Violins Eat Nothing To Exterminate Ecuadorian Nuts
Every Indian Grape Has Triumphs Eating Every Nougat
No Italian Needs Terrible Egyptian Eggs Now
To Winston’s Ethiopian Nachos, Try Yemen!The KA Answers – KA-1
Original cipher:
“Well done. You have solved my trikid cipher.” (Sorry about the typo. It was meant to be ‘trifid’!)
(Updated version: “Well done. You have solved my triliteral cipher.”)
This was all done with the triliteral cipher, with 0s, 1s and 2s instead of As, Bs and Cs. See https://www.dcode.fr/triliteral-cipher for more info!Hint One:
“Solved this cipher? If so, use a similar cipher to solve the original one, but in three dimensions…”
Polybius square, with key ‘polybius’ and labels reading down the across, not across the down. I used the numbers 0-4, instead of the numbers 1-5.Hint Two:
“Hint one describes the original cipher. All other hints provide hints on other hints. Hint three will be a transposition cipher.”
Vigenère cipher, with key ‘vigenere’.Hint Three:
“Do you spot a pattern in the keys? The key is the same as the cipher used! (For numbered keys it relates to the first letter or word.)”
Columnar transposition cipher, with key ‘columnar’.Hint Four:
“The keyword for the original cipher is ‘trifidcipher’.”
Note: I accidentally did this with a columnar transposition cipher instead of a permutation cipher! The keyword is still ‘permutation’, however.Hint Five:
“Hint One is a fractionating cipher, with the original cipher being a three-dimenstional version of this. Hint two is maybe the most famous polyalphabetic substitution cipher around. Hints Three and Four are both transposition ciphers. (All these ciphers can be found at ‘crypto.interactive-maths.com’.)”
Caesar shift cipher, with key of 6 and alphabet shown in Hint Six. By the way, did I put in ‘three-dimenstional’? I meant ‘three-dimensional’!Hint Six:
“The alphabet for the previous hint is a space, a full stop, a comma, a question mark, an exclamation mark, a forward slash, an opening bracket, a closing bracket, a hyphen, an apostrophe, and the normal alphabet (in that order). It is the same cipher as this one, with twice the key number.”
Another caesar shift cipher, with key of 6 and normal alphabet.The KA Answers – KA-2
Original cipher:
“Congratulations on solving the four-square cipher! I did not expect you to solve this without Hint One first, so give yourself a pat on the back if you needed no hints! (And now you know about digraph substitution ciphers!)”
This was a four-square cipher, with the two keys of ‘142onefourtwo’ and ‘857eightfiveseven’, in that order. The alphabet was also ‘abcdefghijklmonpqrstuvwxyz0123456789’.Hint One – Practice:
“This paragraph will tell you one out of six steps to get the keys for the original cipher.This paragraph tells you how to solve the next cipher. Note that in Step Six, the final step, it tells you what the original cipher has been encrypted with. The keys are what the above paragraph is for!
And this paragraph tells you how to solve the previous cipher. Note that in Step One, this is not applicable, so it is missed out.”
Caesar shift cipher, with key of 1.Hint One – Step One:
“Welcome to my mega-hint – if you like. Start with the number of words in this sentence.The next cipher is a columnar transposition cipher, with key ‘two’.”
Mixed alphabet cipher, with key ‘one’.Hint One – Step Two:
“The next step is to add ninety to your previous total.The next cipher will be a permutation cipher, with key ‘three’. This follows a sequence that will be put in Hint Two.
The previous cipher was a mixed alphabet cipher, with key ‘one’.”
Columnar transposition cipher, with key ‘two’. (Note: Did you remember to decrypt each paragraph separately???)Hint One – Step Three:
“Now, multiply your result by ten thousand, one hundred and one.The next cipher is a vigenere cipher.
The previous cipher was a columnar transposition cipher.”
Permutation cipher, with key of ‘three’.Hint One – Step Four:
“You should have a six-digit number. Now divide this number by seven.Next up is the polybius square, which is hard to crack! (Just wait until you see Step Six!)
If you haven’t already got it, Hint Three was encrypted with the permutation cipher.”
Vigenere cipher, with key of ‘four’.Hint One – Step Five:
“Well done for getting this far! You should hopefully have a 6-digit number. Split this number into two halves (for instance, if the number was 123456, split this number into 123 and 456.The final step in Hint One of KA-2 will be encrypted with the fiendish playfair cipher! (Note that that is a digraph substitution cipher. So is the original cipher!)
If you want to know the previous step, it was encrypted using the vigenere cipher – ‘the unbreakable cipher’, as they used to call it (until it was breakable…)”
Polybius square, with key of ‘five’ and alphabet of ‘abcdefghijklmnopqrstuvwxyz0123456789’.Hint One – Step Six:
“The final step is solved! Well done for finishing Hint One! Anyway, what you should do with your two numbers is this: put after each number each of its three digits in written form. Here is an example: if the numbers were 123 and 456, they would be changed to 123onetwothree and 456fourfivesix. 789 would be changed to 789seveneightnine. Now you have your two keys for the original cipher!The original cipher was encrypted with the four-square cipher, which can be found at Crypto Corner.
Hint Five was encrypted with the polybius square.”
Playfair cipher, with key of ‘six’ and alphabet of ‘abcdefghijklmnopqrstuvwxyz0123456789’.Hint Two:
“The keywords (for Hint One) are the number of the step the message is in. So the keyword for Step One is ‘one’, ‘two’ for Step Two, and so on.”
Caesar shift cipher, with key of 9.Hint Three first!
“The alphabet For Hint One, Step Five is the standard alphabet, followed by the ten numbers in numerical order (i.e. starting with zero and ending in nine).”
Caesar shift cipher used, with key 22.Hint Four:
“The alphabet for the original cipher is the same as in the latter stages of Hint One!”
Caesar shift cipher, with key 1. (Sorry, did I say KA-2 Hint Five when I posted this???)TLW NUMBER THEORY
========================
Can you answer these three Number Theory questions?
====================================================== NT 1
The largest palindrome made from the product of
two 3-digit numbers.
====================================================== NT 2
the smallest positive number that is evenly divisible
by all of the numbers from 1 to 20.
====================================================== NT 3
The difference between the sum of the squares of the
first one hundred natural numbers and the square of the sum.
======================================================The KA Answers – KA-4 (Simplification of Infinitely Long Sums)
Let’s do the anwers first and I’ll show you the method afterwards.1. x=(1+sqrt(5))/2 (otherwise known as the golden ratio)
2. x=1+sqrt(2)
3. x=2
4. x=4
5. x=(3+sqrt(33))/6
6. x=(9+sqrt(241))/10
(Note that number 3 is referring to my updated one, which is ‘[1+[1/(2+1/(2+1/(2+1…]]^2’. Thanks to @Madness for pointing this out!)And here is that method…
The first question was ‘1+1/(1+1/(1+1/(1+1…’. I put in Hint One that this can be simplified to ‘x=1+1/x’. Multiplying by x and rearranging gets ‘x^2-x-1=0’. Using the quadratic formula (which is ‘(-b±sqrt(b^2-4ac))/2’ for anyone who forgot about their GCSE maths exams at school!), we get ‘x=(1±sqrt(5))/2’. It should be easy to eliminate the ‘x=(1-sqrt(5))/2’ option just by trying something like the approximation of ‘1+1/(1+1/1)’, which is closer to ‘x=(1+sqrt(5))/2’ than ‘x=(1-sqrt(5))/2’. (The answer is 1.5, if you’re interested.) So the answer is ‘x=(1+sqrt(5))/2’. And you can use that exact same method for the rest of them!
The KA Answers – KA-5
Here is the answer:
A is 8, B is 6, C is 3, D is 1, E is 9, F is 7, G is 2 and H is 0.My method is shown below (also in post #48947, page 12). There is an interesting discussion about the redundancy of some of the clues on page 12 of this topic!
“Here is a reminder of the clues (I will refer to these throughout my method – for those who didn’t get it):
1. E-D=A
2. EH+E=A+D
3. DG=H+G
4. CG-G=G(G+H)
5. C+G=A-C
6. A=5+C
7. E^2=10A+1
8. B-A=D-C=F-E=H-GThe first step was to work out which letter H was. From Clue 1, E could not be 0 (as A would have been a negative number otherwise), so it was safe to divide by E in Clue 2. Rearranging Clue 1 makes E=A+D. So E can be substituted for ‘A+D’ in Clue 2, which comes up as EH+E=E. Dividing by E gets us H+1=1, which means H is 0. Substitute H for 0 in Clue 3 and you get DG=G. Dividing by G (as G isn’t 0) gets us D=1, our next letter. Substitute H for 0 again and in Clue 4 and you get CG-G=G(G). Divide by G and you get C-1=G. Substitute G for ‘C-1’ in Clue 5 and you get C+C-1=A-C. Adding C to both sides and simplifying gets 3C-1=A, which in turn makes Clue 6 (after substituting A for ‘3C-1’) 3C-1=5+C. This can be rearranged to make 2C=6, or C=3. Putting this into Clue 6 gets us A=5+3, or A=8. Doing the same to Clue 5 gets 3+G=8-3, or 3+G=5. This means that G is 2. In Clue 7, we can replace A for 8, giving us E^2 = 10*8+1, or E^2=81. So E=9. The only letters remaining are now B and F. In Clue 8, we are given a statement connecting all 8 letters. D-C is 1-3, or -2. So we can say that B-A=-2 and F-E=-2. Let’s start with B-A=-2 first. A is 8, so B-8=-2, or B=6. Now for F-E=-2. E is 9, so F-9=-2, or F=7. Just to summarize the answers: A is 8, B is 6, C is 3, D is 1, E is 9, F is 7, G is 2 and H is 0.”
(P.S. Sorry if the explanation was a bit long!)
@Kford-academy:
You’ve misspelt ‘fifteen’ (in “Finding Italian Violins Excitedly Troubles Egyptians Eating Nectarines”)The Third Snake Place Puzzle – My Solution
[5][6][3][4][2][3][4][5] R1
[4][1][2][5][1][6][X][6] R2 1>RURDD, 1>URRRD
[3][2][X][6][4][5][6][6] R3
[1][1][1][2][3][4][5][5] R4 1>DDDDR, 1>ULUUR, 1>RRURU
[2][5][4][1][2][3][3][4] R5 1>RRURU
[3][6][3][2][3][X][2][1] R6 1>LURUU
[4][X][2][1][4][5][6][6] R7 1>URDRR
[5][6][1][1][2][3][4][5] R8 1>UUULD, 1>RRRRU@Kford-academy
Snake Place #3 is correct.
NT #2 is correct.
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