bushindo
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Everything posted by bushindo
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A conjugate descent algorithm allows me to improve the odd a bit further, now it's 21:36. It doesn't seem possible to get the lowest odd up much further.
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I can get 20:16 odds, but even better odds might be possible.
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If only I get paid for working on puzzles...
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Excellent puzzle. I resisted the urge to brute force this puzzle. The paper-and-pencil method, surprisingly, is faster and a lot more satisfying.
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The poet is right, it is such a joy.
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Well, I tried writing a matrix, but logic is so much easier.
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Thanks for the clarification. So, my understanding is that 1) The sequence of 26 alphabet letters are scrambled. We call this the key. A plaintext is chosen 2) The 1st letter of the plaintext is encoded as the right-hand neighbor of the same letter in the key. 3) The key is advanced 1 position. 4) Repeat step 2 and 3 with the 2nd letter of the plain text, then the 3rd, and so forth. I just need one last clarification. If the key is a scramble of the alphabet, then the last letter of the this sequence does not have a right-hand neighbor. Does the wrap-around rule, i.e. right-hand neighbor of last letter is the first letter in sequence, apply here?
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I'm not sure about the instructions. My impression is that 1) There is a certain sentence of unknown length (Maybe this key is the same as the plaintext, but scrambled. I'm not sure about this). Spaces are not present. Call this the key. 2) Each letter in the plain-text is replaced with the right-hand neighbor of the same letter in the key. 3) After encryption once, the letters of the keys are advanced one position. I'm not too clear about step 3. Does it mean that step 1-3 are repeated several times on the same plain-text?
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I've still yet to be right at one of these, so if I'm wrong AGAIN, I've always appreciated someone pointing out where I went wrong! You have the right idea, except that you started with the wrong value for the change of landing on a 6, which isn't 1/2. The post above described one way to get that value. Here's another way to get that value
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The brute force method is conceptually clear, but is a devil to program...
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Is it weird of me to say that I start to experience withdrawal symptoms if I go too long without seeing a new Walter Penney puzzle?
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I just reviewed this, and I found that the construct proof can be simplified greatly.
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Here's your second number, and a third, and a four, and so on...
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Let me refine bonanova's approach a bit
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Thanks for the detailed description. It's good to see a fellow Python user on the forum. And a better welcome to the forum, by the way.
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AWAKE and CYCLE would have worked. And so would BADGE, CABOB, and CABLE.
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Nice work. I'd like to know your approach, if you'd like to share.
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I'm not sure that I understand the instruction in this puzzle. Is the following interpretation of the problem correct? 1) A number N is chosen, where N is larger than the largest number in the cipher code. 2) The fibonnaci series F(i) modulo N has a period of 26, meaning that for any index i, F(i) mod N == F(i + 26) mod N 3) The list of 26 values from within 1 period is sorted in ascending order, and each number assigned a letter according to its ordinal index (smallest number is A, second smallest is B, and so on). 4) This transformation is then applied to "Gardening is just a soil sport" to get ( 3 0 89 1 2 34 5 34 3/ 5 144 / 8 377 144 233 / 0 / 144 34 5 13 /144 55 34 89 233 ) 5) Find N PS. Also, does this sequence start at 1 or 0? Many references for the fibonnacci series start at 0.
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This may be the reason for your results
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There is a mistake in your reasoning.
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I need my day's allowance of Walter Penney puzzle!
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That's a wonderful approach. Thanks for the description, I learned a lot from it.
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Any day that I can solve a Walter Penney puzzle without having a crash course in group theory or berlekamp factorization algorithm is a good day.
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bonanova, I'd like to ask you the same question I asked tpaxatb. Is there a non-computer particular algorithm or approach that you use to construct/deconstruct anagrams?
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Good teamwork. I have never been good at anagrams, so I'd like to know how you did it so quick. Is there a particular algorithm or approach that you use to construct/deconstruct anagrams?
