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The Flooble Code (Posted on 2003-03-18) Difficulty: 4 of 5
A book titled The Bible Code introduced the topic of equidistant letter sequences (ELS), described below, for finding words “hidden” in text. That book referenced the Hebrew Bible, but prompts a question about finding any given word in any, say, English-language text.

For simplicity, and to better match the Hebrew, spaces and punctuation are removed. A particular text that I have in mind, thus crunched, has 284,939 characters remaining (letters and digits). How many times would you expect to find the word FLOOBLE as an equidistant letter sequence in the text? Ignore case. The word can start at any of the 284,939 characters and proceed by skipping any constant number of letters forward or backward. So, for example, if the 11,000th character were an F and the 10,000th an L, and the 9,000th an O, etc. that would be one occurrence. Of course we don’t expect always to find such decimally round spacings. The question again, How many do we expect to find?

The absolute and relative frequencies of the relevant letters in the text are:

B  4771 0.016744
E 36232 0.127157
F  7167 0.025153
L  9563 0.033562
O 22486 0.078915
that is, for each letter is shown the number of occurrences in the text and that number divided by the total of characters in the text.

See The Solution Submitted by Charlie    
Rating: 3.5000 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): incorrect solution? | Comment 10 of 12 |
(In reply to re: incorrect solution? by Charlie)

I never said the effects of dependency would be large or small, simply that the method you apply is incorrect. I dont know how much deviation to expect. In a binomial distribution, the expectation E = pn, where p = probability, n = number of trials. But a binomial distribution requires independent trials, which are not shown to be the case here.

As for the 9 labels 9 jars, if you read my solution to this
problem youll notice that it is precisely the demonstration that each event is isomorphic to a bernoulli trial that allows binomial distribution treatment. But this is a special case that you cannot invoke in general!

As for cory's post, he method you mention is precisely
what i had in mind to solve this exactly, and yes it requires more processing power than i can think of.
  Posted by Cheradenine on 2003-03-25 22:55:01

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