bushindo
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Everything posted by bushindo
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No computer algebra software was used in the solving of this puzzle. Though for most practical purposes, I did the same algorithm that an algebra software would have used.
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Well, seems like nobody wants to tag team with me. I'll have to do it myself.
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If I think of how I got it...I'll tell someone. This is the approach using group theory. Many thanks to profmmv for introducing me to the basic elements of cycles in this
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Here is a 3x3 matrix, A = [ 1 9 10 ] [ 5 4 1 ] [ 4 13 9 ] How many square root matrices of A are there? That is, how many complex matrices R are there such that R*R = A? By complex matrices, I mean that the elements of R can either be real or imaginary.
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My bad, apparently I missed a '4' in my code, and that produced the result above
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I need a tag team. I got this sucker within range, someone go for the knock-down
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I got the following average number of questions, anyone got lower?
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That's a marvelous solution. Makes me regret that I didn't take group theory when I had the chance. Can you take a look at this Walter Penney also? I can solve it using the matrix method I described above, but cycles are a lot more elegant and I learned a lot from your solution here.
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I omitted some details earlier. Here's the revised part, which should convey the point better.
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Here's the logical way
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Nicely done. If you're not too busy, I'd like to know how you approached this problem.
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Everyday that you can solve a Walter Penney puzzle is a good day.
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I see you're not only 'well-versed' in mathematics, bonanova. Well done.
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A certain short phrase is encoded using simple substitution cipher. That is, every single letter of the alphabet is replaced with another letter of the alphabet. The blank spaces in the original text are already removed before encryption. The cipher code is BMDDEMGQBKGEQCWGNCDGCELMWXBJTBDKNMKXLCBTGMESBRZWMTENQWBJMXQCDKXZMTZBJTBXDCELMWX What is the plain text?
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I guess you did this in C or one of its derivatives. I did this in Python, which has a regular expression (re) package. I used the built-in function re.search( substring, parent_string) which basically tells me whether the substring is in parent_string or not (also how many times too ). If you haven't come across regular expression before, I highly recommend a look. It is one of the more useful things I learned in my undergrad years.
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I did some more work, ignore the above thought. Answers for the above.
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Do you mean that assume N is not equal to 3*n + 1, for some integer n? Because
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That's some slick calculations. Very nicely done.
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The typo is final's, not yours.
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and for the record im dumb again Ha, you beat me to it. I'd like to know your technique, though.
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I can't replicate the "to be" example neither. I write the "to be" quote horizonally in row 5, 3, 2, 6, 4, and then I take out the letters from column 5, 3, 2, 6, 4, 1. My coded text comes out as T O O O Q H T T B H T O S O T I B N R U A T I E E E N E T S By the way, if there happens to be a row of blanks in the middle of the horizonal fill-in operation (as a result of say this scrambling vector ( 1,2, 6, 3, 4, 5) ), how would that come out in the encoded text?
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think this works but the number does seem too simple
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This is a followup to Bonanova's killville problem. I'll quote his statement of the problem here "You live in Killville - a town populated by 10 killers and 10 pacifists. When a pacifist meets a pacifist, nothing happens. When a pacifist meets a killer, the pacifist is killed. When two killers meet, both die. Assume meetings always occur between exactly two persons and the pairs involved are completely random." These meetings continue until all killers are dead. Now, you are an insurance underwriter. One of the pacifists come to you before joining Killville, and ask for a 1 million dollars insurance policy. In order to set the premium, you need to compute his chance of surviving Killville. 1) Compute the pacifist's chance of survival. 2) Compute the pacifist's chance of survival exactly ( simulation and approximation methods are not allowed) .
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My my, HoustonHokie is too fast. Solving this puzzle in less than 11 minutes, that's insane. PS. These puzzles are great. Keep them coming, superprismatic.
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Marvelous work. Well done.
