shortest path by SanjuDudeja


More Info
									                      Midterm page 1
    The worst-case complexity of an algorithm is ...
      a function from N (or size of instance) to N.
    (n lg n) is ... a set of functions.
    A random variable is ...
       a function from elementary event to real number.
    Polyhash is ... a set of (hash) functions.
    T(n) = 3 T(n/2) + n         T(n) is (nlg3)
    T(n) = 3 T(n/3) + 5         T(n) is (n)
    T(n) = 3 T(n/4) + n lg n T(n) is (n lg n)
    False A heap of height n can have 2n+1 –1 nodes
    False A tree of height k might only have k+1 nodes
    True A red-black can implement a priority queue efficiently
1                                                 CSE 202 - Shortest Paths
                     Midterm page 2
Yes, you can implement “Decrease-Key(S,x,k)” in a max
 heap efficiently. Go to node x (which you can do since
 you have a pointer to it), change its priority to k, apply
 Heapify (which “bubbles downward”).
Given a red-black tree ... in black subtree
    – Every non-leaf has 2, 3, or 4 children (by P3).
    – Every leaf has depth k (by P4).
    – Thus # black nodes  2k+1 – 1 (obvious, or by induction)
    In full red-black tree, N  # black nodes  2k+1 – 1
    – So lg N  lg (2k+1 – 1)  lg 2k = k; that is , k < lg N.
    – Longest path  2 k (by P4).
    – So height tree = longest path  2 lg N.
NOTE: Just saying “Tree is balanced” is insufficient – we
 never even really defined what “balanced” means.
2                                                   CSE 202 - Shortest Paths
                  Midterm page 3
If keys are more or less uniformly randomly distributed
 (so Bucket Sort take O(N) time) and K is large (or grows
 with N, so KN isn’t O(N)), then Bucket Sort will be
 faster than Radix Sort.
If the keys are very skewed, and K is small, Radix Sort
 will be faster (actually, all you really need is that K is
 very small, that is, K << lg N).
Greedy change-making can be arbitrarily bad. Given any c,
 let D2 = 2c, D3 = 2c+1, and N = 4c. Then greedy will give
 one D3 and 2c-1 D1’s, or 2c coins. But optimal is to give
 two D2’s, or 2 coins. So greedy is c times worse.

3                                             CSE 202 - Shortest Paths
                     Midterm page 5
Let A(i) = best sum of numbers in the first i columns that
 doesn’t use any entry in column i.
B(i) = best sum [of ...] that uses only the top entry in column i
C(i) = best sum that uses middle entry in column i
D(i) = best sum that uses only the bottom entry in column i
E(i) = best sum that uses top and bottom entry of column i.
I claim, given A(i), ..., E(i), we can compute A(i+1), ..., E(i+1) in
  the “obvious” way. This will take (N) time.
E.g. A(i+1) = max(A(i), B(i), C(i), D(i), E(i))
     B(i+1) = T(1,i+1) + max(A(i), C(i), D(i)),
4                                                    CSE 202 - Shortest Paths
                  Midterm, page 6
Basic idea: use divide an conquer.
ktile(A,N,K) {
   /* print the K-tiles of A[1], ..., A[N] */
   Let M = select(A, N, K/2);           /* takes O(N) time */
  Split A by M into Big and Small (each of size N/2);
  if (K>1) call ktile (Big, N/2, K/2);
  print M;
  if (K>1) call ktile (Small, N/2, K/2);
Complexity: T(N,K) = 2T(N/2, K/2) + cN
Recursion tree has lg K levels, each requires cN time.
So complexity is c N lg K.
5                                             CSE 202 - Shortest Paths
           CSE 202 - Algorithms
            Shortest Paths Problems

11/12/02           CSE 202 - Shortest Paths
              Shortest Paths Problems
                v                        t
          5                     4                9
                    8               13
     u         3        x                    4       y
                    2               10
          1                 1                    2
                w                        z

Given a weighted directed graph,
    <u,v,t,x,z> is a path of weight 29 from u to z.
    <u,v,w,x,y,z> is another path from u to z; it has
      weight 16 and is the shortest path from u to z.

7                                                    CSE 202 - Shortest Paths
     Variants of Shortest Paths Problems
     A. Single pair shortest path problem
       •   Given s and d, find shortest path from s to d.
     B. Single source shortest paths problem
       •   Given s, for each d find shortest path from s to d.
     C. All-pairs shortest paths problem
       •   For each ordered pair s,d, find shortest path.
(1) and (2) seem to have same asymptotic complexity.
(3) takes longer, but not as long as repeating (2) for each s.

 8                                             CSE 202 - Shortest Paths
             More Shortest Paths Variants
     1. All weights are non-negative.
     2. Weights may be negative, but no negative cycles
       (A cycle is a path from a vertex to itself.)
     3. Negative cycles allowed.
       Algorithm reports “- ” if there is a negative cycle on
          path from source to destination
(2) and (3) seem to be harder than (1).
               v                           5                  u    v  w
        5                   u v    w                     u    0   - -
                       u    0 5    8   u       -3   2    v        - -
  u           3    2   v   -3 0    3
                                                         w       - -
                       w   -6 -1   0       6
        -6                                      w
 9                                                  CSE 202 - Shortest Paths
                   Single Source Shortest Paths
Non-negative weights (problem B-1):
  From source, construct tree of shortest paths.
               • Why is this a tree?
               • Is this a Minimum Spanning Tree?
         Basic approach: label nodes with shortest path found so far.
         Relaxation step: choose any edge (u,v) . If it gives a shorter
           path to v, update v ’s label.

                       4                                4       4       9
               4               9
                           5        5           s                   5         5
                   9                                        9
           1                   2                    1                   2
                       9                                        7
                                   this edge   improves
                                               this node        CSE 202 - Shortest Paths
         Single Source Shortest Paths
Simplest strategy:
     – Keep cycling through all edges
     – When all edges are relaxed, you’re done.
        What is upper bound on work?
     We mark nodes when we’re sure we have best path.
     Key insight: if you relax all edges out of marked nodes, the
       unmarked node that is closest to source can be marked.
     This gives Dijkstra’s algorithm
        Relax edge (u,v) only once - just after u is marked.
        Keep nodes on min-priority queue – need E Decrease-Key’s.
        Gives O(E lg V) (or O(E + V lg V) using Fibonacci heap.)
        There’s a simple O(V2) implementation – when is it fastest?

11                                                  CSE 202 - Shortest Paths
        Single Source Shortest Paths
Negative weights allowed (problem B-2 and B-3)
     Why can’t we just add a constant to each weight?
     Why doesn’t Dijkstra’s algorithm work for B-2 ?
     Bellman-Ford: Cycle through edges V-1 times.
       O(VE) time.

PERT chart: Arbitrary weights, graph is acyclic:
     Is there a better algorithm than Bellman-Ford?

12                                       CSE 202 - Shortest Paths
              All-pairs Shortest Paths
Naïve #1: Solve V single-source problems.
     – O(V2 E) for general problem.
     – O(V3) or O(V E lg V) for non-negative weights.
Naïve #2: Let dk(i,j) = shortest path from i to j
                             involving  k edges.
                    d1(i,j) = is original weight matrix.
     Compute dk+1’s from dk’s by seeing if adding edge helps:
                    dk+1(i,j) = Min { dk(i,m) + d1(m,j) }
        Hmmm ...this looks a lot like matrix multiplication
     If there are no negative cycles, dV-1(i,j) is solution.
     If there is a negative cycle, then dV-1  dV
     Complexity is O(V4)
13                                                    CSE 202 - Shortest Paths
            All-pairs Shortest Paths
No negative cycles – Divide and Conquer says:
       “Shortest path from a to b with at most 2k+1 hops
       is a shortest path from a to c with at most 2k hops
       followed by one from c to b with at most 2k hops.”
     T(k+1) = T(k) + V3 so T(lg V) is O(V3 lg V).

     This also looks like matrix multiplication, using
     d2k = dk x dk instead of dk+1= dk x d1

14                                          CSE 202 - Shortest Paths
            All-pairs Shortest Paths
No negative cycles – Dynamic Programming says:
       “Number cities 1 to V. Shortest path from a to b
        that only uses first cities 1 to k+1 as intermediate
        points is a path that goes from a to k+1 using only
        cities 1 to k, followed by a path going from k+1 to b
        (or shortest from a to b avoiding k+1 altogether).”
     T(k+1)= T(k) + cV2, and T(0) is 0.
       Thus, T(V) is O(V3).
     This is the Floyd-Warshall algorithm

15                                           CSE 202 - Shortest Paths
      Johnson’s All-Pairs Shortest Paths
Motivation: for sparse non-negative weights,
       “O(V E lg V)”
                                    Dijkstra V times
     is better than
       “O(V3)”             Floyd-Warshall

     We’ll convert “arbitrary weights” (C3) to “non-
      negative” (C1) problem via reweighting.

                                            0                0
        -3           -1
                                      -                             1
                                      3                 4
                 8                                           -9
        5            -10                    8

                                   Starting bonus
                                 or finishing penalty       CSE 202 - Shortest Paths
                Johnson’s Algorithm
Reweighting can make all edges non-negative

                                         0   5          4
        -3           -1
                                    2                           10
                 8                                0
        5            -10                 3              0

Make node-weight(x) = shortest path to x from anywhere.
Sounds like all-pairs problem            0
     Slick trick                        0    -3                  -1
     use Bellman-Ford           0
     (only O(VE) time)                       5                   -10

17                                                    CSE 202 - Shortest Paths
     Glossary (in case symbols are weird)
            
      
 subset  element of       infinity  empty
 for all  there exists  intersection 
 big theta  big omega  summation
 >=          <=        about equal
 not equal  natural numbers(N)
18                                 CSE 202 - Shortest Paths

To top