Data Structures and Algorithms
13 Hard or Intractable Problems
If a problem has an O(nk) time algorithm (where k is a constant), then we
class it as having polynomial time complexity and as being efficiently
solvable.
If there is no known polynomial time algorithm, then the problem is classed
as intractable.
The dividing line is not always obvious. Consider two apparently similar
problems:
Euler's problem asks whether there is a
(often characterized as the path through a graph which
Bridges of Königsberg - a traverses each edge only
popular 18th C puzzle) once.
Hamilton's problem asks whether there is a
path through a graph which
visits each vertex exactly
once.
Euler's problem
The 18th century German city of Königsberg was situated on the
river Pregel. Within a park built on the banks of the river, there
[Image]were two islands joined by seven bridges.
The puzzle asks whether it is possible to take a tour through the
park, crossing each bridge only once.
An exhaustive search requires starting at every possible point and
traversing all the possible paths from that point - an O(n!) problem.
However Euler showed that an Eulerian path existed iff
* it is possible to go from any vertex to any other by following the
edges (the graph must be connected) and
* every vertex must have an even number of edges connected to it, with at
most two exceptions (which constitute the starting and ending points).
It is easy to see that these are necessary conditions: to complete the tour,
one needs to enter and leave every point except the start and end points.
The proof that these are sufficient conditions may be found in the
literature . Thus we now have a O(n) problem to determine whether a path
exists.
Transform the map into a
graph in which We can now easily see that the Bridges of
the nodes represent the "dry Königsberg does not have a solution.
land" points
and the arcs represent the A quick inspection shows that it does have a
bridges. Hamiltonian path.
[Image]
However there is no known efficient algorithm for determining
whether a Hamiltonian path exists.
But if a path was found, then it can be verified to be a solution in
polynomial time: we simply verify that each edge in the path is actually an
edge (O(e) if the edges are stored in an adjacency matrix) and that each
vertex is visited only once (O(n2) in the worst case).
Classes P and NP
What does NP mean?
At each step in the algorithm, you guess
which possibility to try next. This is the
non-deterministic part: it doesn't matter
Euler's problem lies in the which possibility you try next. There is no
class P: problems solvable in information used from previous attempts
Polynomial time. Hamilton's (other than not trying something that you've
problem is believed to lie in already tried) to determine which
class NP (Non-deterministic alternative should be tried next. However,
Polynomial). having made a guess, you can determine in
polynomial time whether it is a solution or
Note that I wrote "believed" not.
in the previous sentence.
No-one has succeeded in Since nothing from previous trials helps you
proving that efficient (ie to determine which alternative should be
polynomial time) algorithms tried next, you are forced to investigate
don't exist yet! all possibilities to find a solution. So the
only systematic thing you can do is use some
strategy for systematically working through
all possibilities, eg setting out all
permutations of the cities for the
travelling salesman's tour.
Many other problems lie in class NP. Some examples follow.
Composite Numbers
Determining whether a number can be written as the product of two other
numbers is the composite numbers problem. If a solution is found, it is
simple to verify it, but no efficient method of finding the solution exists.
Assignment
Assignment of compatible room-mates: assume we have a number of students to
be assigned to rooms in a college. They can be represented as the vertices
on a graph with edges linking compatible pairs. If we have two per room, a
class P algorithm exists, but if three are to be fitted in a room, we have a
class NP problem.
Boolean satisfiability
Given an arbitrary boolean expression in n variables:
a1 op a2 op ... op an
where op are boolean operators, and, or, ..
Can we find an assignment of (true,false) to the ai so that the expression
is true? This problem is equivalent to the circuit-satisfiability problem
which asks can we find a set of inputs which will produce a true at the
output of a circuit composed of arbitrary logic gates.
A solution can only be found by trying all 2n possible assignments.
Map colouring
The three-colour map colouring problem asks if we can colour a map
so that no adjoining countries have the same colour. Once a
solution has been guessed, then it is readily proved.
[This problem is easily answered if there are only 2 colours - [Image]
there must be no point at which an odd number of countries meet -
or 4 colours - there is a proof that 4 colours suffice for any
map.]
This problem has a graph equivalent: each vertex represents a country and an
edge is drawn between two vertices if they share a common border.
Its solution has a more general application. If we are scheduling work in a
factory: each vertex can represent a task to be performed - they are linked
by an edge if they share a common resource, eg require a particular machine.
A colouring of the vertices with 3 colours then provides a 3-shift schedule
for the factory.
Many problems are reducible to others: map colouring can be reduced to graph
colouring. A solution to a graph colouring problem is effectively a solution
to the equivalent map colouring or scheduling problem. The map or
graph-colouring problem may be reduced to the boolean satisfiability
problem. To give an informal description of this process, assume the three
colours are red, blue and green. Denote the partial solution, "A is red" by
ar so that we have a set of boolean variables:
ar A is red
ab A is blue
ag A is green
br B is red
bb B is blue
bg B is green
cr C is red
... ...
Now a solution to the problem may be found by finding values for ar, ab, etc
which make the expression true:
((ar and not ab and not ag) and ( (bb and (cb and (dg ....
Thus solving the map colouring problem is equivalent to finding an
assignment to the variables which results in a true value for the expression
- the boolean satisfiability problem.
There is a special class of problems in NP: the NP-complete problems. All
the problems in NP are efficiently reducible to them. By efficiently, we
mean in polynomial time, so the term polynomially reducible provides a more
precise definition.
In 1971, Cook was able to prove that the boolean satisfiability problem was
NP-complete. Proofs now exist showing that many problems in NP are
efficiently reducible to the satisfiability problem. Thus we have a large
class of problems which will are all related to each other: finding an
efficient solution to one will result in an efficient solution for them all.
An efficient solution has so far eluded a very large number of researchers
but there is also no proof that these problems cannot be solved in
polynomial time, so the search continues.
Class NP problems are solvable by non-deterministic algorithms: these
algorithms consist of deterministic steps alternating with non-deterministic
steps in which a random choice (a guess) must be made. A deterministic
algorithm must, given a possible solution,
* have at least one set of guessing steps which lead to the acceptance of
that solution, and
* always reject an invalid solution.
We can also view this from the other aspect: that of trying to determine a
solution. At each guessing stage, the algorithm randomly selects another
element to add to the solution set: this is basically building up a "game"
tree. Various techniques exist for pruning the tree - backtracking when an
invalid solution is found and trying another branch, but this is where the
exponential time complexity starts to enter!
Travelling salesman
It's possible to cast this problem - which is basically an optimality one,
we're looking for the best tour - into a yes-no one also by simply asking:
Can we find a tour with a cost less than x?
By asking this question until we find a tour with a cost x for which the
answer is provably no, we have found the optimal tour. This problem can also
be proved to be in NP. (It is reducible to the Hamiltonian circuit problem.)
Various heuristics have been developed to find near optimal solutions with
efficient algorithms.
One simple approach is the find the minimum spanning tree. One possible tour
simple traverses the MST twice. So we can find a tour which is at most twice
as long as the optimum tour in polynomial time. Various heuristics can now
be applied to reduce this tour, eg by taking shortcuts.
An algorithm due to Christofides can be shown to produce a tour which is no
more than 50% longer than the optimal tour.
It starts with the MST and singles out all cities which are linked
[Image]to an odd number of cities.
These are linked in pairs by a variant of the procedure used to
find compatible room-mates.
[Image]This can then be improved by taking shortcuts.
Another strategy which works well in practice is to divide the "map" into
many small regions and to generate the optimum tour by exhaustive search
within those small regions. A greedy algorithm can then be used to link the
regions. While this algorithm will produce tours as little as 5% longer than
the optimum tour in acceptable times, it is still not guaranteed to produce
the optimal solution.
Key terms
Polynomial Time Complexity
Problems which have solutions with time complexity O(nk) where k is a
constant are said to have polynomial time complexity.
Class P
Set of problems which have solutions with polynomial time complexity.
Non-deterministic Polynomial (NP)
A problem which can be solved by a series of guessing
(non-deterministic) steps but whose solution can be verified as correct
in polynomial time is said to lie in class NP.
Eulerian Path
Path which traverses each arc of a graph exactly once.
Hamiltonian Path
Path which passes through each node of a graph exactly once.
NP-Complete Problems
Set of problems which are all related to each other in the sense that
if any one of them can be shown to be in class P, all the others are
also in class P.
Continue on to Games Back to the Table of Contents
© John Morris, 1998