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Trevor Brown trevor.brown@uwaterloo.ca DC 2338, Office hour M3-4pm
CS 341: Algorithms Trevor Brown DC 2338, Office hour M3-4pm
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This time Depth-first search (DFS) Complexity DFS forest edge types
Application: cycle detection via DFS Application: topological sort via DFS (also without DFS)
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Depth First Search (DFS)
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Basic DFS Properties to recall
Also gets a discovery time π[π£] at this point Nodes start white A node π£ turns gray when it is discovered, which is when the first call to π·πΉππππ ππ‘(π£) happens After π£ is turned gray, we recurse on its neighbours After recursing on all neighbours, we turn π£ black Recursive calls on neighbours end first, so neighbours turn black first Also gets a finish time π[π£] at this point
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Complexity with adjacency lists is π(π+π)
Home exercise: complexity with adjacency matrix?
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Classifying edges in the DFS forest
Recall the example:
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What are the edge types in the example graph?
Can we classify edges without inspecting the DFS forest? Perhaps using π[β¦], π β¦ and/or πππππ’π[β¦]?
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Definitions Definition: we use π° π to denote (π[π’],π[π’]), which we call the interval of π Definition: π is white-reachable from π if there is a path from π’ to π£ containing only white nodes 2 3 12 7 π 4 9 11 13 6 1 8 10 5
![Definitions Definition: we use π° π to denote (π[π’],π[π’]), which we call the interval of π.](http://slideplayer.com./slide/16235604/95/images/8/Definitions+Definition%3A+we+use+%F0%9D%91%B0+%F0%9D%92%96+to+denote+%28%F0%9D%91%91%5B%F0%9D%91%A2%5D%2C%F0%9D%91%93%5B%F0%9D%91%A2%5D%29%2C+which+we+call+the+interval+of+%F0%9D%92%96..jpg "Definition: π is white-reachable from π if there is a path from π’ to π£ containing only white nodes π")
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Exploring d[], f[] and colour[]
Observe: every node π£ that is white-reachable from π’ when we call π·πΉππππ ππ‘(π’) (discover π’) becomes gray after π and black before π (so πΌ π£ is nested inside πΌ π’ ) Note: π£ is discovered during the recursive π·πΉππππ ππ‘ calls, which is equivalent to π being a descendent of π in the DFS forest Start π·πΉππππ ππ‘(π’) and colour π’ grey Perform π·πΉππππ ππ‘ calls recursivelyβ¦ π π π Colour π’ black and return from π·πΉππππ ππ‘(π’)
![Exploring d[], f[] and colour[]](http://slideplayer.com./slide/16235604/95/images/9/Exploring+d%5B%5D%2C+f%5B%5D+and+colour%5B%5D.jpg "Observe: every node π£ that is white-reachable from π’ when we call π·πΉππππ ππ‘(π’) (discover π’) becomes gray after π and black before π (so πΌ π£ is nested inside πΌ π’ ) Note: π£ is discovered during the recursive π·πΉππππ ππ‘ calls, which is equivalent to π being a descendent of π in the DFS forest. Start π·πΉππππ ππ‘(π’) and colour π’ grey. Perform π·πΉππππ ππ‘ calls recursivelyβ¦ π. π. π. Colour π’ black and return from π·πΉππππ ππ‘(π’)")
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more generally Theorem: Let π,π be any nodes. The following statements are all equivalent. (π£ is discovered during π·πΉππππ ππ‘(π’)) (π£ is white-reachable from π’ when we call π·πΉππππ ππ‘(π’)) (π£ is a descendant of π in the DFS forest) (π£ turns grey after π’ and black before π’) ( πΌ π£ nested inside πΌ π’ )
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π£ is not a descendent, and not an ancestor
Classification of edge types in DFS DFS inspects every edge in the graph. When DFS inspects an edge {π,π}, the colour of π£ and relationship between the intervals of π’ and π£ determine the edge type. π£ discovered during π«ππΊπ½ππππ(π) but not at the first level of recursion (else {π’,π£} is a tree edge). π π Q1? Q2? π£ is a child of π’ Q4? Q3? π£ is a descendent of π’ Q6? Q5? π£ is an ancestor of π’ Q8? Q7? π£ is not a descendent, and not an ancestor Break up this slide? Recall: (π£ is discovered during π·πΉππππ ππ‘(π’)) β (π£ is white-reachable from π’ when we call π·πΉππππ ππ‘(π’)) β (π£ is a descendant of π in the DFS forest) β (π£ turns grey after π’ and black before π’) β ( πΌ π£ nested inside πΌ π’ ) So the recursive call in π·πΉππππ ππ‘(π’) that discovered π£ must finish (and set π£ black) before π·πΉππππ ππ‘(π’) inspects {π’,π£}
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useful fact: Parenthesis theorem
Theorem: for each for each pair of nodes π’,π£ the intervals of π’ and π£ are either disjoint or nested Proof: Suppose the intervals are not disjoint. Then either π π£ β πΌ π’ or π[π’]β πΌ π£ WLOG suppose π π£ β πΌ π’ Then π£ is discovered during π·πΉππππ ππ‘(π’), so π£ is white-reachable from π’ when π·πΉππππ ππ‘(π’) starts So, π£ must turn black before π’, so π π£ <π[π’] So the intervals are nested. QED π«ππΊπ½ππππ(π) π[π’] π[π’] π[π£] π[π£] β π[π£] π[π’] π[π’] π[π£] π[π£] β
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Classification of edge types in DFS
So, πΌ π£ must be earlier. DFS inspects every edge in the graph. When DFS inspects an edge {π,π}, the colour of π£ and relationship between the intervals of π’ and π£ determine the edge type. If πΌ π’ were earlier, then π£ would be discovered before π finishes (because of edge {π’,π£}), so intervals would not be disjoint! π π Intervals π° π and π° π must be disjoint. But which is earlier? Q8? Q7? π£ is not a descendent, and not an ancestor Break up this slide? Recall: (π£ is discovered during π·πΉππππ ππ‘(π’)) β (π£ is white-reachable from π’ when we call π·πΉππππ ππ‘(π’)) β (π£ is a descendant of π in the DFS forest) β ( πΌ π£ nested inside πΌ π’ )
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DFS Application: Testing whether a graph is a DAG
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Back edge: points to an ancestor in the DFS forest
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β¦ Case β : Suppose β directed cycle. Show β back edge.
Let π£ 1 , π£ 2 ,β¦, π£ π , π£ 1 be a directed cycle WLOG let π£ 1 be earliest discovered node in the cycle π π π π = = π π π[ π£ π ] π[ π£ 1 ] π[ π£ 1 ] π[ π£ π ] π[ π£ π ] π[ π£ 1 ] π[ π£ 1 ] π[ π£ π ] π[ π£ 1 ] π[ π£ π ] π[ π£ π ] π[ π£ 1 ] π[ π£ 1 ] π[ π£ 1 ] π[ π£ π ] π[ π£ π ] Discovered before π£ 2 ,β¦, π£ π π£ 2 π£ 2 π£ 3 π£ 3 π£ 1 π£ 1 Recall: every node π£ π that is white-reachable from π£ 1 when we discover π£ 1 (call π·πΉππππ ππ‘( π£ 1 )) turns black before π π (π π£ π <π[ π£ 1 ]) π£ 4 π£ 4 Consider edge { π π , π π } Recall: nodes become gray when discovered Since π π£ 1 <π π£ π , { π π , π π } must be a back or cross edge. Why? π£ π π£ π So π£ π must turn black before π£ 1 , and we have π π£ π <π[ π£ 1 ]. π£ 5 π£ 5 β¦ π£ 6 π£ 6 So when π π is discovered, π£ 2 ,β¦, π£ π are all white Thus, { π π , π π } must be a back edge. QED
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Turning the lemma into an algorithm
Search for back edges How to identify a back-edge? Back edge When we observe an edge from π’ to π£, check if π is gray π π π
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DFS: Testing whether a graph is a DAG
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Home exercise Try running the algorithm on this graph:
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Topological sort Finding node orderings that satisfy given constraints
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Example problem: getting dressed in the morning
Dependency graph Edge {π’,π£} means π’ must be completed before π£ Example problem: getting dressed in the morning Watch any time Pants before belt Socks before shoes Could do various things first. Which ones are possible? What do they have in common?
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Topological sort IFF there is a (directed) cycle!
Might not be possible! How can this happen? Try to order nodes linearly so there are only pointers from left to right! Topological sort
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more realistic Use cases (1/2)
Compiler design: ordering program compilation steps Not just compilers⦠Linkers, build systems (makefiles, etc.) Can do parse.y, then parse.h, then parse.c, then main.c, then parse.o, then main.o, then prog⦠Cell evaluation-order in spreadsheets⦠Can verify pointers only go left-to-right. Other orders are possible⦠Manufacturing processes, chemical reaction processes, ⦠all sorts of planning and scheduling
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More realistic use cases (2/2): task scheduling
Goal: output a serial order of tasks that can be run without worrying about dependencies (because all dependencies are already satisfied by the time a task is run). (Nodes are numbered according to one such order.) Can even schedule tasks to run in parallel! Can do 1||2 then 3||13 then 4| 5 |15 etc. Similar to instruction scheduling and pipelining in modern processors!
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Formal definitions
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One of these must be repeated. So there is a cycle!
Useful fact One of these must be repeated. So there is a cycle! β¦ π π+π π π π π π π π π
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Existence of topological sort
Not a very efficient algorithm We can build a more efficient algorithm by using DFS
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Recall this fact from the DAG-testing application
13 To see why we get a topological ordering, suppose D is a DAG and we order nodes from largest to smallest finish time, so π π π > π π π >β¦> π π πβπ > π π π So edges are left-to-right, hence this is a topological order! β¦ β¦ β¦ π π π π π π π π π π But this contradicts the node ordering! Suppose a right-to-left edge { π£ π , π£ π } exists Since edge { π£ π , π£ π } exists, the lemma implies π π£ π < π π£ π
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Topological ordering via DFS
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π(π+π) with adjacency lists
Running time? π(π+π) with adjacency lists We are pushing smallest finishing times first into a stack, so when we pop them out, we will get largest finishing time first Save each node when it finishes Save each node when it finishes
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Example On the boardβ¦
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A different algorithm for topological sort
Running time is π(π+π) with adjacency lists
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Home exercise (Kahnβs algorithm)
Run on this graphβ¦
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