Minimum Spanning Trees Problem A town has a set of houses and a set of roads A road connects 2 and only 2 houses A road connecting houses u and v has a repair cost w(u, v) Goal: Repair enough (and no more) roads such that: Everyone stays connected: can reach every house from all other houses, and Total repair cost is minimum a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 CS 477/677 - Lecture 20

Minimum Spanning Trees A connected, undirected graph: Vertices = houses, Edges = roads A weight w(u, v) on each edge (u, v)  E Find T  E such that: T connects all vertices w(T) = Σ(u,v)T w(u, v) is minimized a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 CS 477/677 - Lecture 20

Minimum Spanning Trees T forms a tree = spanning tree A spanning tree whose weight is minimum over all spanning trees is called a minimum spanning tree, or MST. a b c d e g f i 4 8 7 11 1 2 14 9 10 6 CS 477/677 - Lecture 20

Properties of Minimum Spanning Trees Minimum spanning trees are not unique Can replace (b, c) with (a, h) to obtain a different spanning tree with the same cost MST have no cycles We can take out an edge of a cycle, and still have the vertices connected while reducing the cost # of edges in a MST: |V| - 1 a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 CS 477/677 - Lecture 20

Growing a MST Minimum-spanning-tree problem: find a MST for a connected, undirected graph, with a weight function associated with its edges A generic solution: Build a set A of edges (initially empty) Incrementally add edges to A such that they would belong to a MST An edge (u, v) is safe for A if and only if A  {(u, v)} is also a subset of some MST a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 We will add only safe edges CS 477/677 - Lecture 20

Generic MST algorithm A ←  while A is not a spanning tree do find an edge (u, v) that is safe for A A ← A  {(u, v)} return A How do we find safe edges? a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 CS 477/677 - Lecture 20

The Algorithm of Kruskal b c d e h g f i 4 8 7 11 1 2 14 9 10 6 Start with each vertex being its own component Repeatedly merge two components into one by choosing the light edge that connects them We would add edge (c, f) Scan the set of edges in monotonically increasing order by weight Uses a disjoint-set data structure to determine whether an edge connects vertices in different components CS 477/677 - Lecture 20

Operations on Disjoint Data Sets MAKE-SET(u) – creates a new set whose only member is u FIND-SET(u) – returns a representative element from the set that contains u May be any of the elements of the set that has a particular property E.g.: Su = {r, s, t, u}, the property is that the element be the first one alphabetically FIND-SET(u) = r FIND-SET(s) = r FIND-SET has to return the same value for a given set CS 477/677 - Lecture 20

Operations on Disjoint Data Sets UNION(u, v) – unites the dynamic sets that contain u and v, say Su and Sv E.g.: Su = {r, s, t, u}, Sv = {v, x, y} UNION (u, v) = {r, s, t, u, v, x, y} CS 477/677 - Lecture 20

KRUSKAL(V, E, w) A ←  for each vertex v  V do MAKE-SET(v) sort E into non-decreasing order by weight w for each (u, v) taken from the sorted list do if FIND-SET(u)  FIND-SET(v) then A ← A  {(u, v)} UNION(u, v) return A Running time: O(E lgV) – dependent on the implementation of the disjoint-set data structure CS 477/677 - Lecture 20

Example Add (h, g) Add (c, i) Add (g, f) Add (a, b) Add (c, f) Ignore (i, g) Add (c, d) Ignore (i, h) Add (a, h) Ignore (b, c) Add (d, e) Ignore (e, f) Ignore (b, h) Ignore (d, f) {g, h}, {a}, {b}, {c}, {d}, {e}, {f}, {i} {g, h}, {c, i}, {a}, {b}, {d}, {e}, {f} {g, h, f}, {c, i}, {a}, {b}, {d}, {e} {g, h, f}, {c, i}, {a, b}, {d}, {e} {g, h, f, c, i}, {a, b}, {d}, {e} {g, h, f, c, i, d}, {a, b}, {e} {g, h, f, c, i, d, a, b}, {e} {g, h, f, c, i, d, a, b, e} a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 1: (h, g) 2: (c, i), (g, f) 4: (a, b), (c, f) 6: (i, g) 7: (c, d), (i, h) 8: (a, h), (b, c) 9: (d, e) 10: (e, f) 11: (b, h) 14: (d, f) {a}, {b}, {c}, {d}, {e}, {f}, {g}, {h}, {i} CS 477/677 - Lecture 20

The algorithm of Prim The edges in set A always form a single tree Starts from an arbitrary “root”: VA = {a} At each step: Find a light edge crossing cut (VA, V - VA) Add this edge to A Repeat until the tree spans all vertices Greedy strategy At each step the edge added contributes the minimum amount possible to the weight of the tree a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 CS 477/677 - Lecture 20

How to Find Light Edges Quickly? Use a priority queue Q: Contains all vertices not yet included in the tree (V – VA) V = {a}, Q = {b, c, d, e, f, g, h, i} With each vertex we associate a key: key[v] = minimum weight of any edge (u, v) connecting v to a vertex in the tree Key of v is  if v is not adjacent to any vertices in VA After adding a new node to VA we update the weights of all the nodes adjacent to it We added node a  key[b] = 4, key[h] = 8 a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 CS 477/677 - Lecture 20

PRIM(V, E, w, r) 0         Q ←  for each u  V do key[u] ← ∞ π[u] ← NIL INSERT(Q, u) DECREASE-KEY(Q, r, 0) while Q   do u ← EXTRACT-MIN(Q) for each v  Adj[u] do if v  Q and w(u, v) < key[v] then π[v] ← u DECREASE-KEY(Q, v, w(u, v)) a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 0         Q = {a, b, c, d, e, f, g, h, i} VA =  Extract-MIN(Q)  a CS 477/677 - Lecture 20

Example 0         Q = {a, b, c, d, e, f, g, h, i} VA =  4 8 7 11 1 2 14 9 10 6 0         Q = {a, b, c, d, e, f, g, h, i} VA =  Extract-MIN(Q)  a 4  8 key [b] = 4  [b] = a key [h] = 8  [h] = a 4      8  Q = {b, c, d, e, f, g, h, i} VA = {a} Extract-MIN(Q)  b a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 CS 477/677 - Lecture 20

Example key [c] = 8  [c] = b key [h] = 8  [h] = a - unchanged 8     8  Q = {c, d, e, f, g, h, i} VA = {a, b} Extract-MIN(Q)  c 4  8  8 a b c d e h g f i 4 8 7 11 1 2 14 9 10 6  key [d] = 7  [d] = c key [f] = 4  [f] = c key [i] = 2  [i] = c 7  4  8 2 Q = {d, e, f, g, h, i} VA = {a, b, c} Extract-MIN(Q)  i 4  8 7 a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 2 4 CS 477/677 - Lecture 20

Example key [h] = 7  [h] = i key [g] = 6  [g] = i 7  4  8 7  4  8 Q = {d, e, f, g, h} VA = {a, b, c, i} Extract-MIN(Q)  f 4 7  8 2 a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 key [g] = 2  [g] = f key [d] = 7  [d] = c unchanged key [e] = 10  [e] = f 7 10 2 8 Q = {d, e, g, h} VA = {a, b, c, i, f} Extract-MIN(Q)  g 7 6 4 7  6 8 2 a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 10 2 CS 477/677 - Lecture 20

Example key [h] = 1  [h] = g 7 10 1 4 7 10 2 8 key [h] = 1  [h] = g 7 10 1 Q = {d, e, h} VA = {a, b, c, i, f, g} Extract-MIN(Q)  h a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 7 10 Q = {d, e} VA = {a, b, c, i, f, g, h} Extract-MIN(Q)  d 1 4 7 10 1 2 8 a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 CS 477/677 - Lecture 20

Example key [e] = 9  [e] = f 9 Q = {e} VA = {a, b, c, i, f, g, h, d} 4 7 10 1 2 8 a b c d e h g f i 4 8 7 11 1 2 14 9 10 6 key [e] = 9  [e] = f 9 Q = {e} VA = {a, b, c, i, f, g, h, d} Extract-MIN(Q)  e Q =  VA = {a, b, c, i, f, g, h, d, e} 9 CS 477/677 - Lecture 20

PRIM(V, E, w, r) Q ←  for each u  V do key[u] ← ∞ π[u] ← NIL INSERT(Q, u) DECREASE-KEY(Q, r, 0) ► key[r] ← 0 while Q   do u ← EXTRACT-MIN(Q) for each v  Adj[u] do if v  Q and w(u, v) < key[v] then π[v] ← u DECREASE-KEY(Q, v, w(u, v)) Total time: O(VlgV + ElgV) = O(ElgV) O(V) if Q is implemented as a min-heap Min-heap operations: O(VlgV) Executed |V| times Takes O(lgV) Executed O(E) times Constant O(ElgV) Takes O(lgV) CS 477/677 - Lecture 20

Shortest Path Problems How can we find the shortest route between two points on a map? Model the problem as a graph problem: Road map is a weighted graph: vertices = cities edges = road segments between cities edge weights = road distances Goal: find a shortest path between two vertices (cities) CS 477/677 - Lecture 20

Shortest Path Problems Input: Directed graph G = (V, E) Weight function w : E → R Weight of path p = v0, v1, . . . , vk Shortest-path weight from u to v: δ(u, v) = min w(p) : u v if there exists a path from u to v ∞ otherwise Shortest path u to v is any path p such that w(p) = δ(u, v) 3 9 5 11 6 7 s t x y z 2 1 4 p CS 477/677 - Lecture 20

Variants of Shortest Paths Single-source shortest path G = (V, E)  find a shortest path from a given source vertex s to each vertex v  V Single-destination shortest path Find a shortest path to a given destination vertex t from each vertex v Reverse the direction of each edge  single-source Single-pair shortest path Find a shortest path from u to v for given vertices u and v Solve the single-source problem All-pairs shortest-paths Find a shortest path from u to v for every pair of vertices u and v CS 477/677 - Lecture 20

Optimal Substructure of Shortest Paths Given: A weighted, directed graph G = (V, E) A weight function w: E  R, A shortest path p = v1, v2, . . . , vk from v1 to vk A subpath of p: pij = vi, vi+1, . . . , vj, with 1  i  j  k Then: pij is a shortest path from vi to vj Proof: p = v1 vi vj vk w(p) = w(p1i) + w(pij) + w(pjk) Assume  pij’ from vi to vj with w(pij’) < w(pij)  w(p’) = w(p1i) + w(pij’) + w(pjk) < w(p) contradiction! vj pjk v1 pij p1i pij’ vk vi p1i pij pjk CS 477/677 - Lecture 20

Negative-Weight Edges s  a: only one path (s, a) = w(s, a) = 3 s  b: only one path (s, b) = w(s, a) + w(a, b) = -1 s  c: infinitely many paths s, c, s, c, d, c, s, c, d, c, d, c cycle has positive weight (6 - 3 = 3) s, c is shortest path with weight (s, b) = w(s, c) = 5 What if we have negative-weight edges? 3 -1 - -4 2 8 -6 s a b e f 5 11 -3 y 6 4 7 c d g CS 477/677 - Lecture 20

Negative-Weight Edges s  e: infinitely many paths: s, e, s, e, f, e, s, e, f, e, f, e cycle e, f, e has negative weight: 3 + (- 6) = -3 can find paths from s to e with arbitrarily large negative weights (s, e) = -   no shortest path exists between s and e Similarly: (s, f) = - , (s, g) = -  3 -1 - -4 2 8 -6 s a b e f 5 11 -3 y 6 4 7 c d g  j h i 2 3 -8 h, i, j not reachable from s (s, h) = (s, i) = (s, j) =  CS 477/677 - Lecture 20

Negative-Weight Edges Negative-weight edges may form negative-weight cycles If such cycles are reachable from the source: (s, v) is not properly defined Keep going around the cycle, and get w(s, v) = -  for all v on the cycle 3 -4 2 8 -6 s a b e f -3 y 5 6 4 7 c d g CS 477/677 - Lecture 20

Cycles Can shortest paths contain cycles? Negative-weight cycles Positive-weight cycles: By removing the cycle we can get a shorter path Zero-weight cycles No reason to use them Can remove them to obtain a path with similar weight We will assume that when we are finding shortest paths, the paths will have no cycles No! No! CS 477/677 - Lecture 20

Shortest-Path Representation For each vertex v  V: d[v] = δ(s, v): a shortest-path estimate Initially, d[v]=∞ Reduces as algorithms progress [v] = predecessor of v on a shortest path from s If no predecessor, [v] = NIL  induces a tree—shortest-path tree Shortest paths & shortest path trees are not unique 3 9 5 11 6 7 s t x y z 2 1 4 CS 477/677 - Lecture 20

Initialization Alg.: INITIALIZE-SINGLE-SOURCE(V, s) for each v  V do d[v] ←  [v] ← NIL d[s] ← 0 All the shortest-paths algorithms start with INITIALIZE-SINGLE-SOURCE CS 477/677 - Lecture 20

Relaxation Relaxing an edge (u, v) = testing whether we can improve the shortest path to v found so far by going through u If d[v] > d[u] + w(u, v) we can improve the shortest path to v  update d[v] and [v] s s 5 9 2 u v 5 6 2 u v After relaxation: d[v]  d[u] + w(u, v) RELAX(u, v, w) RELAX(u, v, w) 5 7 2 u v 5 6 2 u v CS 477/677 - Lecture 20

RELAX(u, v, w) if d[v] > d[u] + w(u, v) then d[v] ← d[u] + w(u, v) [v] ← u All the single-source shortest-paths algorithms start by calling INIT-SINGLE-SOURCE then relax edges The algorithms differ in the order and how many times they relax each edge CS 477/677 - Lecture 20

Bellman-Ford Algorithm Single-source shortest paths problem Computes d[v] and [v] for all v  V Allows negative edge weights Returns: TRUE if no negative-weight cycles are reachable from the source s FALSE otherwise  no solution exists Idea: Traverse all the edges |V – 1| times, every time performing a relaxation step of each edge CS 477/677 - Lecture 20

BELLMAN-FORD(V, E, w, s) INITIALIZE-SINGLE-SOURCE(V, s) for i ← 1 to |V| - 1 do for each edge (u, v)  E do RELAX(u, v, w) for each edge (u, v)  E do if d[v] > d[u] + w(u, v) then return FALSE return TRUE  6 5 7 9 s t x y z 8 -3 2 -4 -2  6 5 7 9 s t x y z 8 -3 2 -4 -2 6 7 E: (t, x), (t, y), (t, z), (x, t), (y, x), (y, z), (z, x), (z, s), (s, t), (s, y) CS 477/677 - Lecture 20

Example (t, x), (t, y), (t, z), (x, t), (y, x), (y, z), (z, x), (z, s), (s, t), (s, y) 6  7 5 9 s t x y z 8 -3 2 -4 -2 6  7 5 9 s t x y z 8 -3 2 -4 -2 Pass 1 Pass 2 4 11 2 Pass 3 6  7 5 9 s t x y z 8 -3 2 -4 -2 11 4 6  7 5 9 s t x y z 8 -3 2 -4 -2 11 4 Pass 4 2 -2 CS 477/677 - Lecture 20

Detecting Negative Cycles for each edge (u, v)  E do if d[v] > d[u] + w(u, v) then return FALSE return TRUE  c s b 2 3 -8  c s b 2 3 -8 -3 2 5 c s b 3 -8 Look at edge (s, b): d[b] = -1 d[s] + w(s, b) = -4  d[b] > d[s] + w(s, b) -3 2 -6 -1 5 2 CS 477/677 - Lecture 20

BELLMAN-FORD(V, E, w, s) INITIALIZE-SINGLE-SOURCE(V, s) for i ← 1 to |V| - 1 do for each edge (u, v)  E do RELAX(u, v, w) for each edge (u, v)  E do if d[v] > d[u] + w(u, v) then return FALSE return TRUE Running time: O(VE) (V) O(V) O(E) O(E) CS 477/677 - Lecture 20

Shortest Path Properties Triangle inequality For all (u, v)  E, we have: δ(s, v) ≤ δ(s, u) + w(u, v) If u is on the shortest path to v we have the equality sign s 5 7 2 u v s 5 6 2 u v CS 477/677 - Lecture 20

Shortest Path Properties Upper-bound property We always have d[v] ≥ δ(s, v) for all v. Once d[v] = δ(s, v), it never changes. The estimate never goes up – relaxation only lowers the estimate 6  7 5 9 s v x y z 8 -3 2 -4 -2 11 4 6  7 5 9 s v x y z 8 -3 2 -4 -2 11 4 Relax (x, v) CS 477/677 - Lecture 20

Shortest Path Properties No-path property If there is no path from s to v then d[v] = ∞ always. δ(s, h) = ∞ and d[h] ≥ δ(s, h)  d[h] = ∞ 3 -1 - -4 2 8 -6 s a b e f 5 11 -3 y 6 4 7 c d g  j h i 2 3 -8 h, i, j not reachable from s (s, h) = (s, i) = (s, j) =  CS 477/677 - Lecture 20

Shortest Path Properties Convergence property If s u → v is a shortest path, and if d[u] = δ(s, u) at any time prior to relaxing edge (u, v), then d[v] = δ(s, v) at all times afterward. If d[v] > δ(s, v)  after relaxation: d[v] = d[u] + w(u, v) d[v] = 5 + 2 = 7 Otherwise, the value remains unchanged, because it must have been the shortest path value u v 2 6 5  8 11 5 s 4 4 7 CS 477/677 - Lecture 20

Shortest Path Properties Path relaxation property Let p = v0, v1, . . . , vk be a shortest path from s = v0 to vk. If we relax, in order, (v0, v1), (v1, v2), . . . , (vk-1, vk), even intermixed with other relaxations, then d[vk ] = δ(s, vk). v1 v2 2 d[v2] = δ(s, v2) 6  5   11 7 v4  5 14 v3 3 d[v1] = δ(s, v1) 4 s  d[v4] = δ(s, v4) 11 d[v3] = δ(s, v3) CS 477/677 - Lecture 20

Single-Source Shortest Paths in DAGs Given a weighted DAG: G = (V, E) – solve the shortest path problem Idea: Topologically sort the vertices of the graph Relax the edges according to the order given by the topological sort for each vertex, we relax each edge that starts from that vertex Are shortest-paths well defined in a DAG? Yes, (negative-weight) cycles cannot exist 6  5 2 4 y u v 11 x CS 477/677 - Lecture 20

Dijkstra’s Algorithm Single-source shortest path problem: No negative-weight edges: w(u, v) > 0  (u, v)  E Maintains two sets of vertices: S = vertices whose final shortest-path weights have already been determined Q = vertices in V – S: min-priority queue Keys in Q are estimates of shortest-path weights (d[v]) Repeatedly select a vertex u  V – S, with the minimum shortest-path estimate d[v] CS 477/677 - Lecture 20

Dijkstra (G, w, s) INITIALIZE-SINGLE-SOURCE(V, s) S ←  Q ← V[G]  10 1 5 2 s t x y z 3 9 7 4 6 INITIALIZE-SINGLE-SOURCE(V, s) S ←  Q ← V[G] while Q   do u ← EXTRACT-MIN(Q) S ← S  {u} for each vertex v  Adj[u] do RELAX(u, v, w)  10 1 5 2 s t x y z 3 9 7 4 6 10 5 CS 477/677 - Lecture 20

Example 10  5 s t x y z 8 14 5 7 s t x y z 8 14 13 7 8 9 5 7 s t x y 10  5 1 2 s t x y z 3 9 7 4 6 8 14 5 7 10 1 2 s t x y z 3 9 4 6 8 14 13 7 8 9 5 7 10 1 2 s t x y z 3 4 6 8 13 5 7 10 1 2 s t x y z 3 9 4 6 9 CS 477/677 - Lecture 20

Dijkstra (G, w, s) INITIALIZE-SINGLE-SOURCE(V, s) S ←  Q ← V[G] while Q   do u ← EXTRACT-MIN(Q) S ← S  {u} for each vertex v  Adj[u] do RELAX(u, v, w) Running time: O(VlgV + ElgV) = O(ElgV) (V) O(V) build min-heap Executed O(V) times O(lgV) O(E) times; O(lgV) CS 477/677 - Lecture 20