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the computing competition for schools and colleges.
We are proud to present BIO 2004, sponsored by Lionhead Studios.
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BIO2004 Round One question 3 - worked solution

Question 3
Morse Code

The BIO has been receiving telegrams congratulating it on reaching its 10th anniversary. At least, we think it has. The telegrams have been sent in Morse code and, unfortunately, the gaps between letters have been left out.

In Morse code, each letter of the alphabet is replaced by a sequence of dots and dashes as follows:

a.-   h....   o---   v...-
b-...   i..   p.--.   w.--
c-.-.   j.---   q--.-   x-..-
d-..   k-.-   r.-.   y-.--
e.   l.-..   s...   z--..
f..-.   m--   t-      
g--.   n-.   u..-      

Every combination of between 1 and 4 dots and dashes is used, except for:


Traditionally, dots were transmitted by a short note and dashes by a longer note, with pauses between different letters. This is why some mobile phones make the sound ... -- ... when receiving a message, since this is the Morse code for SMS.

If the gaps between letters are missed out, messages can be ambiguous. For example, even if we know the message
-..-----. is made up of three letters, it might mean njg, dog, xmg or xon.

3 (a)
[26 marks]
Write a program which reads in a message (between 1 and 10 letters inclusive) and determines how many messages, with the same number of letters as the input, it might represent.   Sample run
3 (b)
[5 marks]
How many messages might ----- represent, if we do not know the number of letters in the message? How about
3 (c)
[3 marks]

It is possible to come up with new ways of encoding the alphabet so that, even when the gaps between letters are missing, messages are unambiguous. The size of such an unambiguous encoding is the total number of dots and dashes in a message containing each letter once.

For example, we could encode each letter by some dots (indicating its position in the alphabet) followed by a dash; so .- would be a, ..- would be b, and 26 dots followed by a dash would be z. This encoding has a size of 377 (2 + 3 + ... + 27).

What is the smallest size an unambiguous encoding can have?

A worked solution to the problem is in preparation.

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