Breadth First Search, Dijkstras Algorithm for Shortest Paths - PowerPoint PPT Presentation

Shortest Path Problems 3 Fall 2017 18 CS374 Sariel Har-Peled (UIUC) Many applications! Find shortest paths for all pairs of nodes. Given node s fjnd shortest path from s to all other nodes. Shortest Path Problems 2 1 18 / 42 Input A (undirected or directed) graph G = ( V , E ) with edge lengths (or costs). For edge e = ( u , v ) , ℓ ( e ) = ℓ ( u , v ) is its length. Given nodes s , t fjnd shortest path from s to t .

Shortest Path Problems 3 Fall 2017 18 CS374 Sariel Har-Peled (UIUC) Many applications! Find shortest paths for all pairs of nodes. Given node s fjnd shortest path from s to all other nodes. Shortest Path Problems 2 1 18 / 42 Input A (undirected or directed) graph G = ( V , E ) with edge lengths (or costs). For edge e = ( u , v ) , ℓ ( e ) = ℓ ( u , v ) is its length. Given nodes s , t fjnd shortest path from s to t .

Single-Source Shortest Paths: Restrict attention to directed graphs Fall 2017 19 CS374 Sariel Har-Peled (UIUC) Exercise: show reduction works. Relies on non-negativity! 3 2 1 problem - how? Undirected graph problem can be reduced to directed graph Non-Negative Edge Lengths 2 1 2 Given node s fjnd shortest path from s to all other nodes. 3 2 1 Single-Source Shortest Path Problems 1 19 / 42 Input: A (undirected or directed) graph G = ( V , E ) with non-negative edge lengths. For edge e = ( u , v ) , ℓ ( e ) = ℓ ( u , v ) is its length. Given nodes s , t fjnd shortest path from s to t . Given undirected graph G , create a new directed graph G ′ by replacing each edge { u , v } in G by ( u , v ) and ( v , u ) in G ′ . set ℓ ( u , v ) = ℓ ( v , u ) = ℓ ( { u , v } )

Single-Source Shortest Paths: Restrict attention to directed graphs Fall 2017 19 CS374 Sariel Har-Peled (UIUC) Exercise: show reduction works. Relies on non-negativity! 3 2 1 problem - how? Undirected graph problem can be reduced to directed graph Non-Negative Edge Lengths 2 1 2 Given node s fjnd shortest path from s to all other nodes. 3 2 1 Single-Source Shortest Path Problems 1 19 / 42 Input: A (undirected or directed) graph G = ( V , E ) with non-negative edge lengths. For edge e = ( u , v ) , ℓ ( e ) = ℓ ( u , v ) is its length. Given nodes s , t fjnd shortest path from s to t . Given undirected graph G , create a new directed graph G ′ by replacing each edge { u , v } in G by ( u , v ) and ( v , u ) in G ′ . set ℓ ( u , v ) = ℓ ( v , u ) = ℓ ( { u , v } )

Single-Source Shortest Paths: Restrict attention to directed graphs Fall 2017 19 CS374 Sariel Har-Peled (UIUC) Exercise: show reduction works. Relies on non-negativity! 3 2 1 problem - how? Undirected graph problem can be reduced to directed graph Non-Negative Edge Lengths 2 1 2 Given node s fjnd shortest path from s to all other nodes. 3 2 1 Single-Source Shortest Path Problems 1 19 / 42 Input: A (undirected or directed) graph G = ( V , E ) with non-negative edge lengths. For edge e = ( u , v ) , ℓ ( e ) = ℓ ( u , v ) is its length. Given nodes s , t fjnd shortest path from s to t . Given undirected graph G , create a new directed graph G ′ by replacing each edge { u , v } in G by ( u , v ) and ( v , u ) in G ′ . set ℓ ( u , v ) = ℓ ( v , u ) = ℓ ( { u , v } )

Single-Source Shortest Paths via BFS 1 Fall 2017 20 CS374 Sariel Har-Peled (UIUC) if L is large. 3 20 / 42 2 2 nodes. Run BFS ( s ) to get shortest path distances from s to all other 1 Special case: All edge lengths are 1 . O ( m + n ) time algorithm. Special case: Suppose ℓ ( e ) is an integer for all e ? Can we use BFS ? Reduce to unit edge-length problem by placing ℓ ( e ) − 1 dummy nodes on e . Let L = max e ℓ ( e ) . New graph has O ( mL ) edges and O ( mL + n ) nodes. BFS takes O ( mL + n ) time. Not effjcient

Single-Source Shortest Paths via BFS 1 Fall 2017 20 CS374 Sariel Har-Peled (UIUC) if L is large. 3 20 / 42 2 2 nodes. Run BFS ( s ) to get shortest path distances from s to all other 1 Special case: All edge lengths are 1 . O ( m + n ) time algorithm. Special case: Suppose ℓ ( e ) is an integer for all e ? Can we use BFS ? Reduce to unit edge-length problem by placing ℓ ( e ) − 1 dummy nodes on e . Let L = max e ℓ ( e ) . New graph has O ( mL ) edges and O ( mL + n ) nodes. BFS takes O ( mL + n ) time. Not effjcient

Single-Source Shortest Paths via BFS 1 Fall 2017 20 CS374 Sariel Har-Peled (UIUC) if L is large. 3 20 / 42 2 2 nodes. Run BFS ( s ) to get shortest path distances from s to all other 1 Special case: All edge lengths are 1 . O ( m + n ) time algorithm. Special case: Suppose ℓ ( e ) is an integer for all e ? Can we use BFS ? Reduce to unit edge-length problem by placing ℓ ( e ) − 1 dummy nodes on e . Let L = max e ℓ ( e ) . New graph has O ( mL ) edges and O ( mL + n ) nodes. BFS takes O ( mL + n ) time. Not effjcient

Single-Source Shortest Paths via BFS 1 Fall 2017 20 CS374 Sariel Har-Peled (UIUC) if L is large. 3 20 / 42 2 2 nodes. Run BFS ( s ) to get shortest path distances from s to all other 1 Special case: All edge lengths are 1 . O ( m + n ) time algorithm. Special case: Suppose ℓ ( e ) is an integer for all e ? Can we use BFS ? Reduce to unit edge-length problem by placing ℓ ( e ) − 1 dummy nodes on e . Let L = max e ℓ ( e ) . New graph has O ( mL ) edges and O ( mL + n ) nodes. BFS takes O ( mL + n ) time. Not effjcient

Single-Source Shortest Paths via BFS 1 Fall 2017 20 CS374 Sariel Har-Peled (UIUC) if L is large. 3 20 / 42 2 2 nodes. Run BFS ( s ) to get shortest path distances from s to all other 1 Special case: All edge lengths are 1 . O ( m + n ) time algorithm. Special case: Suppose ℓ ( e ) is an integer for all e ? Can we use BFS ? Reduce to unit edge-length problem by placing ℓ ( e ) − 1 dummy nodes on e . Let L = max e ℓ ( e ) . New graph has O ( mL ) edges and O ( mL + n ) nodes. BFS takes O ( mL + n ) time. Not effjcient

Towards an algorithm 2 Fall 2017 21 CS374 Sariel Har-Peled (UIUC) observe that edge lengths are non-negative. Proof. Why does BFS work? 21 / 42 1 Let G be a directed graph with non-negative edge lengths. Let Lemma BFS (s) explores nodes in increasing distance from s dist ( s , v ) denote the shortest path length from s to v . If s = v 0 → v 1 → v 2 → . . . → v k shortest path from s to v k then for 1 ≤ i < k : s = v 0 → v 1 → v 2 → . . . → v i is shortest path from s to v i dist ( s , v i ) ≤ dist ( s , v k ) . Relies on non-neg edge lengths. Suppose not. Then for some i < k there is a path P ′ from s to v i of length strictly less than that of s = v 0 → v 1 → . . . → v i . Then P ′ concatenated with v i → v i + 1 . . . → v k contains a strictly shorter path to v k than s = v 0 → v 1 . . . → v k . For the second part,

Towards an algorithm 2 Fall 2017 21 CS374 Sariel Har-Peled (UIUC) observe that edge lengths are non-negative. Proof. Why does BFS work? 21 / 42 1 Let G be a directed graph with non-negative edge lengths. Let Lemma BFS (s) explores nodes in increasing distance from s dist ( s , v ) denote the shortest path length from s to v . If s = v 0 → v 1 → v 2 → . . . → v k shortest path from s to v k then for 1 ≤ i < k : s = v 0 → v 1 → v 2 → . . . → v i is shortest path from s to v i dist ( s , v i ) ≤ dist ( s , v k ) . Relies on non-neg edge lengths. Suppose not. Then for some i < k there is a path P ′ from s to v i of length strictly less than that of s = v 0 → v 1 → . . . → v i . Then P ′ concatenated with v i → v i + 1 . . . → v k contains a strictly shorter path to v k than s = v 0 → v 1 . . . → v k . For the second part,

Towards an algorithm 2 Fall 2017 21 CS374 Sariel Har-Peled (UIUC) observe that edge lengths are non-negative. Proof. Why does BFS work? 21 / 42 1 Let G be a directed graph with non-negative edge lengths. Let Lemma BFS (s) explores nodes in increasing distance from s dist ( s , v ) denote the shortest path length from s to v . If s = v 0 → v 1 → v 2 → . . . → v k shortest path from s to v k then for 1 ≤ i < k : s = v 0 → v 1 → v 2 → . . . → v i is shortest path from s to v i dist ( s , v i ) ≤ dist ( s , v k ) . Relies on non-neg edge lengths. Suppose not. Then for some i < k there is a path P ′ from s to v i of length strictly less than that of s = v 0 → v 1 → . . . → v i . Then P ′ concatenated with v i → v i + 1 . . . → v k contains a strictly shorter path to v k than s = v 0 → v 1 . . . → v k . For the second part,

Towards an algorithm Lemma Fall 2017 21 CS374 Sariel Har-Peled (UIUC) observe that edge lengths are non-negative. Proof. 2 1 Let G be a directed graph with non-negative edge lengths. Let 21 / 42 dist ( s , v ) denote the shortest path length from s to v . If s = v 0 → v 1 → v 2 → . . . → v k shortest path from s to v k then for 1 ≤ i < k : s = v 0 → v 1 → v 2 → . . . → v i is shortest path from s to v i dist ( s , v i ) ≤ dist ( s , v k ) . Relies on non-neg edge lengths. Suppose not. Then for some i < k there is a path P ′ from s to v i of length strictly less than that of s = v 0 → v 1 → . . . → v i . Then P ′ concatenated with v i → v i + 1 . . . → v k contains a strictly shorter path to v k than s = v 0 → v 1 . . . → v k . For the second part,

A proof by picture Sariel Har-Peled (UIUC) CS374 22 Fall 2017 22 / 42 s = v 0 v 6 v 2 v 5 v 4 v 1 v 3 Shortest path from v 0 to v 6

A proof by picture Sariel Har-Peled (UIUC) Fall 2017 22 CS374 22 / 42 Shorter path from v 0 to v 4 s = v 0 v 6 v 2 v 5 v 4 v 1 v 3 Shortest path from v 0 to v 6

A proof by picture Sariel Har-Peled (UIUC) Fall 2017 22 CS374 22 / 42 A shorter path from v 0 to v 6 . A contradiction. s = v 0 v 6 v 2 v 5 v 4 v 1 v 3 Shortest path from v 0 to v 6

A Basic Strategy Explore vertices in increasing order of distance from s : Fall 2017 23 CS374 Sariel Har-Peled (UIUC) How can we implement the step in the for loop? i 'th closest to s and that no edge has zero length) 23 / 42 (For simplicity assume that nodes are at different distances from s Initialize for each node v , dist ( s , v ) = ∞ Initialize X = { s } , for i = 2 to | V | do (* Invariant: X contains the i − 1 closest nodes to s *) Among nodes in V − X , find the node v that is the Update dist ( s , v ) X = X ∪ { v }

A Basic Strategy Explore vertices in increasing order of distance from s : Fall 2017 23 CS374 Sariel Har-Peled (UIUC) How can we implement the step in the for loop? i 'th closest to s and that no edge has zero length) 23 / 42 (For simplicity assume that nodes are at different distances from s Initialize for each node v , dist ( s , v ) = ∞ Initialize X = { s } , for i = 2 to | V | do (* Invariant: X contains the i − 1 closest nodes to s *) Among nodes in V − X , find the node v that is the Update dist ( s , v ) X = X ∪ { v }

Finding the i th closest node 1 Fall 2017 24 CS374 Sariel Har-Peled (UIUC) than v . Implies v is not the i ’th closest node to s - recall that X If P had an intermediate node u not in X then u will be closer to s Proof. Let P be a shortest path from s to v where v is the i th closest node. Claim 2 24 / 42 X contains the i − 1 closest nodes to s Want to fjnd the i th closest node from V − X . What do we know about the i th closest node? Then, all intermediate nodes in P belong to X . already has the i − 1 closest nodes.

Finding the i th closest node 1 Fall 2017 24 CS374 Sariel Har-Peled (UIUC) than v . Implies v is not the i ’th closest node to s - recall that X If P had an intermediate node u not in X then u will be closer to s Proof. Let P be a shortest path from s to v where v is the i th closest node. Claim 2 24 / 42 X contains the i − 1 closest nodes to s Want to fjnd the i th closest node from V − X . What do we know about the i th closest node? Then, all intermediate nodes in P belong to X . already has the i − 1 closest nodes.

Finding the i th closest node 1 Fall 2017 24 CS374 Sariel Har-Peled (UIUC) than v . Implies v is not the i ’th closest node to s - recall that X If P had an intermediate node u not in X then u will be closer to s Proof. Let P be a shortest path from s to v where v is the i th closest node. Claim 2 24 / 42 X contains the i − 1 closest nodes to s Want to fjnd the i th closest node from V − X . What do we know about the i th closest node? Then, all intermediate nodes in P belong to X . already has the i − 1 closest nodes.

Finding the i th closest node repeatedly An example Fall 2017 25 CS374 Sariel Har-Peled (UIUC) 25 / 42 10 9 a 18 6 0 6 6 11 30 13 19 20 6 8 16 25

Finding the i th closest node repeatedly An example Fall 2017 25 CS374 Sariel Har-Peled (UIUC) 25 / 42 b f 10 9 a 18 6 0 c g 6 6 11 30 13 19 20 6 8 16 d 25 e h

Finding the i th closest node repeatedly An example Fall 2017 25 CS374 Sariel Har-Peled (UIUC) 25 / 42 b f 10 9 a 18 6 0 c g 6 6 6 11 30 13 19 20 6 8 16 d 25 e h

Finding the i th closest node repeatedly An example Fall 2017 25 CS374 Sariel Har-Peled (UIUC) 25 / 42 b f 10 9 9 a 18 6 0 c g 6 6 6 11 30 13 19 20 6 8 16 d 25 e h

Finding the i th closest node repeatedly An example Fall 2017 25 CS374 Sariel Har-Peled (UIUC) 25 / 42 b f 10 9 9 a 18 6 0 c g 6 6 6 11 30 13 19 20 6 8 16 d 25 e 13 h

Finding the i th closest node repeatedly An example Fall 2017 25 CS374 Sariel Har-Peled (UIUC) 25 / 42 b f 10 9 19 9 a 18 6 0 c g 6 6 6 11 30 13 19 20 6 8 16 d 25 e 13 h

Finding the i th closest node repeatedly An example Fall 2017 25 CS374 Sariel Har-Peled (UIUC) 25 / 42 b f 10 9 19 9 a 18 6 0 c g 6 6 6 11 30 13 25 19 20 6 8 16 d 25 e 13 h

Finding the i th closest node repeatedly An example Fall 2017 25 CS374 Sariel Har-Peled (UIUC) 25 / 42 b f 10 9 19 9 a 18 6 0 c g 6 6 6 36 11 30 13 25 19 20 6 8 16 d 25 e 13 h

Finding the i th closest node repeatedly An example Fall 2017 25 CS374 Sariel Har-Peled (UIUC) 25 / 42 b f 10 9 19 9 a 18 6 0 c g 6 6 6 36 11 30 13 25 19 20 6 8 16 d 25 e 38 13 h

Finding the i th closest node Corollary Fall 2017 26 CS374 Sariel Har-Peled (UIUC) The i th closest node is adjacent to X . 26 / 42 10 9 a 18 6 0 6 6 30 11 13 19 20 6 8 16 25

Finding the i th closest node 1 Fall 2017 27 CS374 Sariel Har-Peled (UIUC) Lemma 2 1 2 to u using only nodes in X as intermediate vertices. 1 2 27 / 42 X contains the i − 1 closest nodes to s Want to fjnd the i th closest node from V − X . For each u ∈ V − X let P ( s , u , X ) be a shortest path from s Let d ′ ( s , u ) be the length of P ( s , u , X ) Observations: for each u ∈ V − X , dist ( s , u ) ≤ d ′ ( s , u ) since we are constraining the paths d ′ ( s , u ) = min t ∈ X ( dist ( s , t ) + ℓ ( t , u )) - Why? If v is the i th closest node to s , then d ′ ( s , v ) = dist ( s , v ) .

Finding the i th closest node 1 Fall 2017 27 CS374 Sariel Har-Peled (UIUC) Lemma 2 1 2 to u using only nodes in X as intermediate vertices. 1 2 27 / 42 X contains the i − 1 closest nodes to s Want to fjnd the i th closest node from V − X . For each u ∈ V − X let P ( s , u , X ) be a shortest path from s Let d ′ ( s , u ) be the length of P ( s , u , X ) Observations: for each u ∈ V − X , dist ( s , u ) ≤ d ′ ( s , u ) since we are constraining the paths d ′ ( s , u ) = min t ∈ X ( dist ( s , t ) + ℓ ( t , u )) - Why? If v is the i th closest node to s , then d ′ ( s , v ) = dist ( s , v ) .

Finding the i th closest node 1 Fall 2017 27 CS374 Sariel Har-Peled (UIUC) Lemma 2 1 2 to u using only nodes in X as intermediate vertices. 1 2 27 / 42 X contains the i − 1 closest nodes to s Want to fjnd the i th closest node from V − X . For each u ∈ V − X let P ( s , u , X ) be a shortest path from s Let d ′ ( s , u ) be the length of P ( s , u , X ) Observations: for each u ∈ V − X , dist ( s , u ) ≤ d ′ ( s , u ) since we are constraining the paths d ′ ( s , u ) = min t ∈ X ( dist ( s , t ) + ℓ ( t , u )) - Why? If v is the i th closest node to s , then d ′ ( s , v ) = dist ( s , v ) .

Finding the i th closest node Let v be the i th closest node to s . Then there is a shortest path P Fall 2017 28 CS374 Sariel Har-Peled (UIUC) from s to v that contains only nodes in X as intermediate nodes (see Proof. Lemma 2 1 Given: 28 / 42 X : Set of i − 1 closest nodes to s . d ′ ( s , u ) = min t ∈ X ( dist ( s , t ) + ℓ ( t , u )) If v is an i th closest node to s , then d ′ ( s , v ) = dist ( s , v ) . previous claim). Therefore d ′ ( s , v ) = dist ( s , v ) .

Finding the i th closest node Lemma Corollary Proof. Sariel Har-Peled (UIUC) CS374 29 Fall 2017 29 / 42 If v is an i th closest node to s , then d ′ ( s , v ) = dist ( s , v ) . The i th closest node to s is the node v ∈ V − X such that d ′ ( s , v ) = min u ∈ V − X d ′ ( s , u ) . For every node u ∈ V − X , dist ( s , u ) ≤ d ′ ( s , u ) and for the i th closest node v , dist ( s , v ) = d ′ ( s , v ) . Moreover, dist ( s , u ) ≥ dist ( s , v ) for each u ∈ V − X .

Algorithm Correctness: By induction on i using previous lemmas. Fall 2017 30 CS374 Sariel Har-Peled (UIUC) 1 30 / 42 using only X as intermediate nodes*) Initialize for each node v : dist ( s , v ) = ∞ Initialize X = ∅ , d ′ ( s , s ) = 0 for i = 1 to | V | do (* Invariant: X contains the i − 1 closest nodes to s *) (* Invariant: d ′ ( s , u ) is shortest path distance from u to s Let v be such that d ′ ( s , v ) = min u ∈ V − X d ′ ( s , u ) dist ( s , v ) = d ′ ( s , v ) X = X ∪ { v } for each node u in V − X do � � d ′ ( s , u ) = min t ∈ X dist ( s , t ) + ℓ ( t , u ) Running time: O ( n · ( n + m )) time. n outer iterations. In each iteration, d ′ ( s , u ) for each u by scanning all edges out of nodes in X ; O ( m + n ) time/iteration.

Algorithm Correctness: By induction on i using previous lemmas. Fall 2017 30 CS374 Sariel Har-Peled (UIUC) 1 30 / 42 using only X as intermediate nodes*) Initialize for each node v : dist ( s , v ) = ∞ Initialize X = ∅ , d ′ ( s , s ) = 0 for i = 1 to | V | do (* Invariant: X contains the i − 1 closest nodes to s *) (* Invariant: d ′ ( s , u ) is shortest path distance from u to s Let v be such that d ′ ( s , v ) = min u ∈ V − X d ′ ( s , u ) dist ( s , v ) = d ′ ( s , v ) X = X ∪ { v } for each node u in V − X do � � d ′ ( s , u ) = min t ∈ X dist ( s , t ) + ℓ ( t , u ) Running time: O ( n · ( n + m )) time. n outer iterations. In each iteration, d ′ ( s , u ) for each u by scanning all edges out of nodes in X ; O ( m + n ) time/iteration.

Algorithm Correctness: By induction on i using previous lemmas. Fall 2017 30 CS374 Sariel Har-Peled (UIUC) 1 30 / 42 using only X as intermediate nodes*) Initialize for each node v : dist ( s , v ) = ∞ Initialize X = ∅ , d ′ ( s , s ) = 0 for i = 1 to | V | do (* Invariant: X contains the i − 1 closest nodes to s *) (* Invariant: d ′ ( s , u ) is shortest path distance from u to s Let v be such that d ′ ( s , v ) = min u ∈ V − X d ′ ( s , u ) dist ( s , v ) = d ′ ( s , v ) X = X ∪ { v } for each node u in V − X do � � d ′ ( s , u ) = min t ∈ X dist ( s , t ) + ℓ ( t , u ) Running time: O ( n · ( n + m )) time. n outer iterations. In each iteration, d ′ ( s , u ) for each u by scanning all edges out of nodes in X ; O ( m + n ) time/iteration.

Algorithm Correctness: By induction on i using previous lemmas. Fall 2017 30 CS374 Sariel Har-Peled (UIUC) 1 30 / 42 using only X as intermediate nodes*) Initialize for each node v : dist ( s , v ) = ∞ Initialize X = ∅ , d ′ ( s , s ) = 0 for i = 1 to | V | do (* Invariant: X contains the i − 1 closest nodes to s *) (* Invariant: d ′ ( s , u ) is shortest path distance from u to s Let v be such that d ′ ( s , v ) = min u ∈ V − X d ′ ( s , u ) dist ( s , v ) = d ′ ( s , v ) X = X ∪ { v } for each node u in V − X do � � d ′ ( s , u ) = min t ∈ X dist ( s , t ) + ℓ ( t , u ) Running time: O ( n · ( n + m )) time. n outer iterations. In each iteration, d ′ ( s , u ) for each u by scanning all edges out of nodes in X ; O ( m + n ) time/iteration.

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 10 a 9 18 6 0 6 6 30 11 13 19 20 6 8 16 25

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 S 10 a 9 18 6 0 0 6 6 30 11 13 19 20 6 8 16 25

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 S 10 a 9 18 6 0 0 6 6 30 11 13 19 20 6 8 16 25

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 S 10 a 9 9 18 6 0 0 6 6 6 30 11 13 19 20 6 8 16 25 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 S 10 a 9 9 18 6 0 0 6 6 6 6 30 11 13 19 20 6 8 16 25 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 S 10 a 9 9 18 6 0 0 6 6 6 6 30 11 13 19 20 6 8 16 25 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 S 10 a 9 24 9 18 6 0 0 6 6 6 6 30 11 13 36 19 20 6 8 16 25 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 10 a 9 9 24 9 18 6 0 0 6 6 6 6 30 11 13 36 19 20 6 8 16 25 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 10 a 9 9 19 24 9 18 6 0 0 6 6 6 6 30 11 13 36 19 20 6 8 16 25 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 10 a 9 9 19 24 9 18 6 0 0 6 6 6 6 30 11 13 36 19 20 6 8 16 25 13 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 10 a 9 9 19 24 9 18 6 0 0 6 6 6 6 30 11 13 36 33 19 20 6 8 16 25 38 13 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 10 a 9 9 19 19 24 9 18 6 0 0 6 6 6 6 30 11 13 36 33 19 20 6 8 16 25 38 13 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 10 a 9 9 19 19 24 9 18 6 0 0 6 6 6 6 30 11 13 33 36 36 25 19 20 6 8 16 25 38 13 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 10 a 9 9 19 19 24 9 18 6 0 0 6 6 6 6 30 11 13 36 36 33 25 25 19 20 6 8 16 25 38 13 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 10 a 9 9 19 19 24 9 18 6 0 0 6 6 6 6 36 30 11 13 36 36 33 25 25 19 20 6 8 16 25 38 13 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 10 a 9 9 19 24 19 9 18 6 0 0 6 6 6 6 36 36 36 30 11 13 25 25 36 36 33 19 20 6 8 16 25 38 13 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 10 a 9 9 19 24 19 9 18 6 0 0 6 6 6 6 36 36 36 30 11 13 25 25 36 36 33 19 20 6 8 16 25 38 13 13

Example: Dijkstra algorithm in action Sariel Har-Peled (UIUC) Fall 2017 31 CS374 31 / 42 10 a 9 9 19 24 19 9 18 6 0 0 6 6 6 6 36 36 36 30 11 13 25 25 36 36 33 19 20 6 8 16 25 38 13 38 13

Improved Algorithm 1 Fall 2017 32 CS374 Sariel Har-Peled (UIUC) time values is O n s u Finding v from d 3 2 n outer iterations and in each iteration following steps 1 32 / 42 2 // Main work is to compute the d ′ ( s , u ) values in each iteration d ′ ( s , u ) changes from iteration i to i + 1 only because of the node v that is added to X in iteration i . Initialize for each node v , dist ( s , v ) = d ′ ( s , v ) = ∞ Initialize X = ∅ , d ′ ( s , s ) = 0 for i = 1 to | V | do // X contains the i − 1 closest nodes to s , and the values of d ′ ( s , u ) are current Let v be node realizing d ′ ( s , v ) = min u ∈ V − X d ′ ( s , u ) dist ( s , v ) = d ′ ( s , v ) X = X ∪ { v } Update d ′ ( s , u ) for each u in V − X as follows: � � d ′ ( s , u ) = min d ′ ( s , u ) , dist ( s , v ) + ℓ ( v , u ) Running time: O ( m + n 2 ) time. updating d ′ ( s , u ) after v is added takes O ( deg ( v )) time so total work is O ( m ) since a node enters X only once

Improved Algorithm 1 Fall 2017 32 CS374 Sariel Har-Peled (UIUC) time values is O n s u Finding v from d 3 2 n outer iterations and in each iteration following steps 1 32 / 42 2 // Main work is to compute the d ′ ( s , u ) values in each iteration d ′ ( s , u ) changes from iteration i to i + 1 only because of the node v that is added to X in iteration i . Initialize for each node v , dist ( s , v ) = d ′ ( s , v ) = ∞ Initialize X = ∅ , d ′ ( s , s ) = 0 for i = 1 to | V | do // X contains the i − 1 closest nodes to s , and the values of d ′ ( s , u ) are current Let v be node realizing d ′ ( s , v ) = min u ∈ V − X d ′ ( s , u ) dist ( s , v ) = d ′ ( s , v ) X = X ∪ { v } Update d ′ ( s , u ) for each u in V − X as follows: � � d ′ ( s , u ) = min d ′ ( s , u ) , dist ( s , v ) + ℓ ( v , u ) Running time: O ( m + n 2 ) time. updating d ′ ( s , u ) after v is added takes O ( deg ( v )) time so total work is O ( m ) since a node enters X only once

Improved Algorithm 1 Fall 2017 32 CS374 Sariel Har-Peled (UIUC) 3 2 n outer iterations and in each iteration following steps 32 / 42 // Initialize for each node v , dist ( s , v ) = d ′ ( s , v ) = ∞ Initialize X = ∅ , d ′ ( s , s ) = 0 for i = 1 to | V | do // X contains the i − 1 closest nodes to s , and the values of d ′ ( s , u ) are current Let v be node realizing d ′ ( s , v ) = min u ∈ V − X d ′ ( s , u ) dist ( s , v ) = d ′ ( s , v ) X = X ∪ { v } Update d ′ ( s , u ) for each u in V − X as follows: � � d ′ ( s , u ) = min d ′ ( s , u ) , dist ( s , v ) + ℓ ( v , u ) Running time: O ( m + n 2 ) time. updating d ′ ( s , u ) after v is added takes O ( deg ( v )) time so total work is O ( m ) since a node enters X only once Finding v from d ′ ( s , u ) values is O ( n ) time

Dijkstra’s Algorithm 1 Fall 2017 33 CS374 Sariel Har-Peled (UIUC) 2 1 Priority Queues to maintain dist values for faster running time 33 / 42 2 update dist values after adding v by scanning edges out of v eliminate d ′ ( s , u ) and let dist ( s , u ) maintain it Initialize for each node v, dist ( s , v ) = ∞ Initialize X = ∅ , dist ( s , s ) = 0 for i = 1 to | V | do Let v be such that dist ( s , v ) = min u ∈ V − X dist ( s , u ) X = X ∪ { v } for each u in Adj ( v ) do � � dist ( s , u ) = min dist ( s , u ) , dist ( s , v ) + ℓ ( v , u ) Using heaps and standard priority queues: O (( m + n ) log n ) Using Fibonacci heaps: O ( m + n log n ) .

Dijkstra’s Algorithm 1 Fall 2017 33 CS374 Sariel Har-Peled (UIUC) 2 1 Priority Queues to maintain dist values for faster running time 33 / 42 2 update dist values after adding v by scanning edges out of v eliminate d ′ ( s , u ) and let dist ( s , u ) maintain it Initialize for each node v, dist ( s , v ) = ∞ Initialize X = ∅ , dist ( s , s ) = 0 for i = 1 to | V | do Let v be such that dist ( s , v ) = min u ∈ V − X dist ( s , u ) X = X ∪ { v } for each u in Adj ( v ) do � � dist ( s , u ) = min dist ( s , u ) , dist ( s , v ) + ℓ ( v , u ) Using heaps and standard priority queues: O (( m + n ) log n ) Using Fibonacci heaps: O ( m + n log n ) .

Priority Queues 5 Fall 2017 34 CS374 Sariel Har-Peled (UIUC) decreaseKey is implemented via delete and insert . meld : merge two separate priority queues into one. 7 6 Data structure to store a set S of n elements where each element delete ( v ): Remove element v from S . 4 3 fjndMin : fjnd the minimum key in S . 2 makePQ : create an empty queue. 1 following operations: 34 / 42 v ∈ S has an associated real/integer key k ( v ) such that the extractMin : Remove v ∈ S with smallest key and return it. insert ( v , k ( v ) ): Add new element v with key k ( v ) to S . decreaseKey ( v , k ′ ( v ) ): decrease key of v from k ( v ) (current key) to k ′ ( v ) (new key). Assumption: k ′ ( v ) ≤ k ( v ) . All operations can be performed in O (log n ) time.

Priority Queues 5 Fall 2017 34 CS374 Sariel Har-Peled (UIUC) decreaseKey is implemented via delete and insert . meld : merge two separate priority queues into one. 7 6 Data structure to store a set S of n elements where each element delete ( v ): Remove element v from S . 4 3 fjndMin : fjnd the minimum key in S . 2 makePQ : create an empty queue. 1 following operations: 34 / 42 v ∈ S has an associated real/integer key k ( v ) such that the extractMin : Remove v ∈ S with smallest key and return it. insert ( v , k ( v ) ): Add new element v with key k ( v ) to S . decreaseKey ( v , k ′ ( v ) ): decrease key of v from k ( v ) (current key) to k ′ ( v ) (new key). Assumption: k ′ ( v ) ≤ k ( v ) . All operations can be performed in O (log n ) time.

Priority Queues 5 Fall 2017 34 CS374 Sariel Har-Peled (UIUC) decreaseKey is implemented via delete and insert . meld : merge two separate priority queues into one. 7 6 Data structure to store a set S of n elements where each element delete ( v ): Remove element v from S . 4 3 fjndMin : fjnd the minimum key in S . 2 makePQ : create an empty queue. 1 following operations: 34 / 42 v ∈ S has an associated real/integer key k ( v ) such that the extractMin : Remove v ∈ S with smallest key and return it. insert ( v , k ( v ) ): Add new element v with key k ( v ) to S . decreaseKey ( v , k ′ ( v ) ): decrease key of v from k ( v ) (current key) to k ′ ( v ) (new key). Assumption: k ′ ( v ) ≤ k ( v ) . All operations can be performed in O (log n ) time.

Dijkstra’s Algorithm using Priority Queues . Fall 2017 35 CS374 Sariel Har-Peled (UIUC) 3 2 1 Priority Queue operations: 35 / 42 decreaseKey Q ← makePQ () insert ( Q , ( s , 0 ) ) for each node u � = s do insert ( Q , ( u , ∞ )) X ← ∅ for i = 1 to | V | do ( v , dist ( s , v )) = extractMin ( Q ) X = X ∪ { v } for each u in Adj ( v ) do � ��� � � Q , u , min dist ( s , u ) , dist ( s , v ) + ℓ ( v , u ) O ( n ) insert operations O ( n ) extractMin operations O ( m ) decreaseKey operations

Implementing Priority Queues via Heaps Using Heaps Store elements in a heap based on the key value 1 Sariel Har-Peled (UIUC) CS374 36 Fall 2017 36 / 42 All operations can be done in O (log n ) time Dijkstra’s algorithm can be implemented in O (( n + m ) log n ) time.

Implementing Priority Queues via Heaps Using Heaps Store elements in a heap based on the key value 1 Sariel Har-Peled (UIUC) CS374 36 Fall 2017 36 / 42 All operations can be done in O (log n ) time Dijkstra’s algorithm can be implemented in O (( n + m ) log n ) time.

Priority Queues: Fibonacci Heaps/Relaxed Heaps 2 Fall 2017 37 CS374 Sariel Har-Peled (UIUC) (European Symposium on Algorithms, September 2009!) implement, and perform well in practice. Rank-Pairing Heaps work has obtained data structures that are easier to analyze and Data structures are complicated to analyze/implement. Recent 37 / 42 Fibonacci Heaps 1 3 2 1 extractMin , insert , delete , meld in O (log n ) time decreaseKey in O ( 1 ) amortized time: ℓ decreaseKey operations for ℓ ≥ n take together O ( ℓ ) time Relaxed Heaps: decreaseKey in O ( 1 ) worst case time but at the expense of meld (not necessary for Dijkstra’s algorithm) Dijkstra’s algorithm can be implemented in O ( n log n + m ) time. If m = Ω( n log n ) , running time is linear in input size.