Dynamic Programming: All-Points Shortest-Path Problem college
topic: CS577 (Algorithms)
Dynamic programming finds optimal solutions to problems by recursively figuring out which sub-solution, combined with the choices at the current level, will produce the best result
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\centerline{\bf All--Points Shortest Path Problem}
\centerline{\bf Zak Smith}

\vskip 1in

\subsection{Description of Problem}

Given a directed graph, find the shortest path from each vertex to every
other vertex in the graph.

\subsection{Relation to Dynamic Programming}

Dynamic programming finds optimal solutions to problems by recursively figuring out
which sub--solution, combined with the choices at the current level, will produce
the best result.

For example, if we are solving problem $A$, which can be broken up and solved
4 ways $a_1, a_2, a_3, a_4$, we do the following:

	\item figure out cost of $a_1, \dots, a_4$: $COST(a_i)$
	\item pick best of $a_1, \dots, a_4$ by choosing the minimum of
		$COMBINE\_COST(a_i) + COST(a_i)$: the cost to combine $a_i$
		at the current level plus the cost of $a_i$ at the lower levels

	\item we then return this solution and cost to the caller

Another feature of dynamic programming is the caching of solutions to sub--problems.  Once we
find the best solution for a sub--solution, we can remember this value and we need not
recalculate it as we take other branches down the recursion tree.

A simple example is if we
are trying to calculate $x!$ for $x = 1, 2, 3, 4, 5$.  Once we calculate $4!$, for $x=4$, we
need not recalculate $4!$ when we find $5!$, we can simple use the value we already calculated.

Often these cached values are kept in a matrix, and often the dependencies between
entries have geometric significance (like in the matrix--multiply problem).  In the example
above ($5!$), an element on depends only on the element directly to the left of it.

\subsection{Solution to Problem}

The input to this algorithm is a sparsely populated matrix $M(i,j)$, whose $(i,j)$ element
is the distance from vertex $v_i$ to vertex $v_j$.  The input matrix only contains values
for cities which are directly connected.  As the algorithm runs, it fills in the remainder
of the entries.

\[ M(i,j) = \mbox{MIN}\left\{ M(i,k) + M(k,j) \right\} \] $\dots$ for all choices of $k$
such that $v_i$ is connected to $v_k$, ie: we pick the best sub--solution for all
the vertices directly connected to the current vertex, based on the cost to combine
that vertex $v_k$ with the current vertex $v_i$.  The ``destination'' vertex is $v_j$.

After $M(i,j)$ is computed, the value is stored in the matrix position in $M$ so that
subsequent queries do not have to recompute $M(i,j)$.

I have just described how to find a singular shortest--path.  To find all the shortest--paths, 
we merely look up the next $(i,j)$ pair using the same matrix.  As more and more paths
are known, we quickly build a ``library'' of cached sub--solutions so instead of lots
of computation, we merely look up the answers to the sub--solutions in our matrix.  $M(i,j)$ is
called with all possible vertex pairs, $n^2$.  The first computed pairs take generally longer to compute, and
the later pairs take generally less time to compute because of the already--computed values
in the matrix.  


This algorithm does not fill the matrix in a geometric order because the graph is, by
nature, not geometrically regular.  Nevertheless, this algorithm does use the ideas
of Dynamic Programming to find the best solution in the least ammount of time.

An alternative algorithm would be to start at each vertex, $1 \dots n$, and
find the shortest path to each other vertex, using a method similar to
the one specified above.  Thus, when starting at the first vertex, we would
visit $n-1$ vertices.  At the end, we would only need to go to one other
vertex.  At the first vertex, it must visit $O(n^2)$ nodes, so it is
equivalent in complexity to the solution I have proposed above.


[Zak Smith] [zak@computer.org] [/~zak/documents/college/cs577-shortest-path/tex]
$Id: documents,v 1.5 2000/09/28 21:20:39 zak Exp zak $
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