Aoccdrnig to a rscheearch at Cmabrigde Uinervtisy, it deosn't mttaer in waht oredr the
ltteers in a wrod are,
the olny iprmoetnt tihng is taht the frist and lsat ltteer be at the rghit pclae. The rset
can be a toatl mses
and you can sitll raed it wouthit porbelm. Tihs is bcuseae the huamn mnid deos not raed
ervey lteter by istlef,
but the wrod as a wlohe.
-- Unknown Author.
"Ennyn Durin Aran Moria. Pedo mellon a Minno. Im Narvi hain echant. Celebrimbor o Eregion
tethant. I thiw
hin"
"The Doors of Durin, Lord of Moria. Speak friend and enter. I Narvi made them. Celebrimbor
of Hollin drew these
signs."
-- "The Fellowship of the Ring" J. R. Tolkien
The Doors of Durin were opened by Gandalf and the Fellowship entered in
Moria
.
As they found, this underground kingdom has gates on every passageway. Each gate has a written
message which
contains the key word. Luckily, Gimli knows how to recognize the key in these messages. It's the
most "like" word,
which has the greatest average "likeness" coefficient with other words in the message.
You are given a message. You need to pick out all words (a letter or group of letters that has meaning when spoken or written and typically shown with a space or other punctuation sign on either side when written or printed),
calculate the "likeness" coefficients with other words, take an average of them and choose the
greatest. Count
"likeness" coefficient even for the same words (of course it's 100). If several words have the
same resulting value,
then choose the word closest to the end of the message. Words in the message can be separated by
whitespaces and
punctuation. There are no numbers.
"Likeness" coefficient for two words is measured in percentages using the following rules:
- Letter case does not matter ("A" == "a");
- If the first letters of the words are equal, then add 10%;
- If the last letters of the words are equal, then add 10%;
- Add length comparison as
(length_of_word1 / length_of_word2) * 30%, if
length_of_word1 ≤ length_of_word2;
, else
(length_of_word2 / length_of_word1) * 30%
- Take all unique letters from the both words in one set and count how many letters appear in
the both words. Add to
the coefficient (common_letter_number / unique_letters_numbers) * 50;
So the maximum coefficient of likeness is 100%. For example: for the words "Bread" and "Beard".
The result should be given in the lower case.
Let's look at an example. The message "Friend Fred and friend Ted."
First, pick out words - ("friend", "fred", "and", "friend", "ted").
Next we calculate "likeness" for the first word with other.
"friend" and "fred" have the same first and last letters, so add 20.
Then length comparison: the length of "fred" is lesser than "friend", so add (4/6)*30=20.
The last rule: for these words unique letters are "definr" and the intersected letters are
"defr".
So add (4/6)*50≈33.333. And the summary is 73.333.
This way we will count all other coefficients and get the following table
(results are rounded just for simplicity). The greatest average is 62.976 and the result is
"friend".
| friend | fred | and | friend | ted |
--------|---------|---------|---------|---------|---------|
friend | ------ | 73.333 | 39.286 | 100.0 | 39.286 |
fred | 73.333 | ------ | 40.833 | 73.333 | 52.5 |
and | 39.286 | 40.833 | ------ | 39.286 | 50.0 |
friend | 100.0 | 73.333 | 39.286 | ------ | 39.286 |
ted | 39.286 | 52.5 | 50.0 | 39.286 | ------ |
--------|---------|---------|---------|---------|---------|
sum | 251.905 | 239.999 | 169.405 | 251.905 | 181.072 |
average | 62.976 | 60 | 42.351 | 62.976 | 45.268 |
Input:
A message as a string (unicode)
Output:
The keyword as a string.
Example:
find_word("Speak friend and enter.") == "friend"
How it is used:
This is a fabricated algorithm which can be modified and be used for linguistic research and
text pattern
recognition.
Precondition:
0 < len(message)
all(x in (string.ascii_letters + string.punctuation + " ") for x in message)
