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<b>Data Structures and Algorithms
with Object-Oriented Design Patterns in C++</b><br>
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<H3><A NAME="SECTION0015112000000000000000">Greedy Algorithm</A></H3>
<P>
A cashier does not really consider all the possible
ways in which to count out a given sum of money.
Instead, she counts out the required amount
beginning with the largest denomination
and proceeding to the smallest denomination.
<P>
For example,
suppose we have ten coins:
five pennies, two nickels, two dimes and quarter.
I.e., <IMG WIDTH=315 HEIGHT=24 ALIGN=MIDDLE ALT="tex2html_wrap_inline68455" SRC="img1816.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img1816.gif" >.
To count out 32 cents,
we start with a quarter,
then add a nickel followed by two pennies.
This is a greedy strategy because once a coin has been counted out,
it is never taken back.
Furthermore, the solution obtained is the correct solution
because it uses the fewest number of coins.
<P>
If we assume that the pieces of money (notes and coins)
are sorted by their denomination,
the running time for the greedy algorithm is <I>O</I>(<I>n</I>).
This is significantly better than that of the brute-force
algorithm given above.
<P>
Does this greedy algorithm always produce the correct answer?
Unfortunately it does not.
Consider what happens if we introduce a 15-cent coin.
Suppose we are asked to count out 20 cents
from the following set of coins: <IMG WIDTH=151 HEIGHT=24 ALIGN=MIDDLE ALT="tex2html_wrap_inline68459" SRC="img1817.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img1817.gif" >.
The greedy algorithm selects 15 followed by five ones--six coins in total.
Of course, the correct solution requires only two coins.
The solution found by the greedy strategy is a feasible solution,
but it does not minimize the objective function.
<P>
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