Notes on computational linguistics.pdf - UCLA Department of ...

Stabler - Lx 185/209 2003
Two harder extra credit problems for the go-getters
(7) More extra credit, part 1. The previous extra credit problem can be solved in lots of ways. Here is a
simple way to do part a of that problem: to encode N, we count up to N with our binary sequences. But
since the front of the list is the easiest to access, we use count with the order most-significant-digit to
least-significant-digit, and then reverse the result. Here is a prolog program that does this:
e(N,L) :countupReverse(N,R),
reverse(R,L).
countupReverse(0,[]).
countupReverse(N,L) :-
N>0,
N1 is N-1,
countupReverse(N1,L1),
addone(L1,L).
addone([],[0]).
addone([0|R],[1|R]).
addone([1|R0],[0|R]) :- addone(R0,R).
This is good, but suppose that we want to communicate two numbers in sequence. For this purpose,
our binary representations are still no good, because you cannot tell where one number ends and the
next one begins.
One way to solve this problem is to decide, in advance, that every number will be represented by a
certain number of bits – say 7. This is what is done in standard ascii codes for example. But blocks
of n bits limit you in advance to encoding no more than 2n elements, and they are inefficient if some
symbols are more common than others.
For many purposes, a better strategy is to use a coding scheme where no symbol (represented by a
sequence of bits) is the prefix of any other one. That means, we would never get confused about where
one symbol ends and the next one begins. One extremely simple way to encode numbers in this way
is this. To represent a number like 5, we put in front of [1,0] an (unambiguous) representation of the
length n of [1,0] – namely, we use n 1’s followed by a 0. So then, to represent 5, we use [1,1,0,1,0]. The
first 3 bits indicate that the number we have encoded is two bits long.
So in this notation, we can unambiguously determine what sequence of numbers is represented by
[1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1].
This is a binary code for the number sequence [1, 2, 6]. Define a predicate e1(NumberSequence,BinaryCode)
that transforms any sequence of numbers into this binary code. (We will improve on this code later.)
(8) More extra credit, part 2 (hard!). While the definition of e(N, L) given above works, it involves counting
from 0=[] all the way up to the number you want. Can you find a simpler way?
Hint: The empty sequence ɛ represents 0, and any other sequence of binary digits [an,an−1,...,a0]
represents
n
(ai + 1)2 i .
i=0
So for example, [1,0] represents (0 + 1)20 + (1 + 1)21 = 1 + 4 = 5. Equivalently, [an,an−1,...,a0]
represents
2 n+1 n
− 1 + ai2 i .
So for example, [1,0] represents 2 1+1 − 1 + (0 · 2 0 ) + (1 · 2 1 ) = 4 − 1 + 0 + 2 = 5.
(Believe it or not, some students in the class already almost figured this out, instead of using a simple
counting strategy like the one I used in the definition of e above.)
25
i=0