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CS1022 Computer Programming & Principles Lecture 8.1 Digraphs (1)

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Plan of lecture Digraphs (definition and terminology) Simple digraphs Paths and cycles PERT charts Topological sort algorithm 2 CS1022

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Digraphs, again Directed graphs = digraphs We have used digraphs to represent relations – We did not define them formally Model partial ordering – A before B, A before C, – B before C? C before B? Networks of dependences useful for – Data flow analysis – Task scheduling Edges are directed – Finding paths require following a direction 3 CS1022 1 5 6 2 3 4

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A digraph is a pair G (V, E) where – V is a finite set of vertices – E is a relation on V Visually, a digraph is – A set of labelled vertices with – Directed edges linking pairs of vertices Directed edges are elements of E – Pairs of vertices, where the order is important – Also called arcs If u, v V are vertices and (u, v) E is an arc – We write simply uv Directed graphs 4 CS1022 ab

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Simple digraphs (1) A simple digraph has no loops or multiple arcs There is at most one arc uv from u to v and There is at most one arc vu from v to u If uv is an arc then we say u is an antecedent of v 5 CS1022

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Simple digraphs (2) Example: digraph G (V, E) where Vertex set V a, b, c, d Arc set E ab, bd, cb, db, dc Graphically: 6 CS1022 a c b d

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abcd a b c d Simple digraphs (3) Adjacency matrix (set E ab, bd, cb, db, dc ): 7 CS1022 abcd aF b c d abcd aFTFF b c d abcd aFTFF bFFFT c d abcd aFTFF bFFFT cFTFF d abcd aFTFF bFFFT cFTFF dFTTF

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Paths and cycles in digraphs A path of length k is a – Sequence of vertices v 0, v 1, , v k – Such that v i – 1 v i is an arc, 1 i k – Example: a, b, d, c is a path A cycle is a – Sequence of vertices v 0, v 1, , v k – Such that v i – 1 v i is an arc, 1 i k – v 0 v k (first and last vertices are the same) – v i v j, 0 i, j k, i 0 or j k (no other repetition) – Example: b, d, c, b is a cycle; a, b, d, c, b, a is not a cycle A graph with no cycles in it is an acyclic graph 8 CS1022 a c b d

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PERT chart (1) Acyclic graphs useful to model situations in which tasks have to be carried out in a certain order – A cycle means that a task had to precede itself! In task-scheduling problems the corresponding acyclic digraph is known as PERT chart – Project Evaluation and Review Technique (PERT) 9 CS1022

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PERT chart (2) Suppose (partial) degree programme below – Pre-requisites, so order is important 10 CS1022 ModulePre-requisites Advanced biotechnologyB BiotechnologyC Cell biologyH DNA structuresC Enzyme activitiesD, G Food scienceE Genetic engineeringC Human biologyNone

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PERT chart (3) PERT chart shows interdependence of modules 11 CS1022 ModulePre-requisites Advanced biotechnologyB BiotechnologyC Cell biologyH DNA structuresC Enzyme activitiesD, G Food scienceE Genetic engineeringC Human biologyNone A B C DE H G F

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Topological sort algorithm (1) We want to help students find an order of modules – Consistent with pre-requisites Classic solution: topological sort algorithm – Consistent labelling for vertices of acyclic digraphs Labelling 1, 2, 3, , n of vertices such that – If uv is an arc and – Vertex u has label i, and – Vertex v has label j, then – i j 12 CS1022

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Topological sort algorithm (2) Gives consistent labelling of acyclic digraph G (V, E) – Antecedents of each vertex stored in A(v) 13 CS1022 begin for v V do calculate A(v); label := 0; while unlabelled vertices v remain for which A(v) do begin label := label + 1; u := a vertex with A(u) ; assign label to u; for each unlabelled vertex v V do A(v) := A(v) {u}; % delete u from remaining vs end

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Find consistent labelling for digraph of modules Step 0 – Antecedent sets are: – A(A) {B} – A(B) {C} – A(C) {H} – A(D) {C} – A(E) {D, G} – A(F) {E} – A(G) {C} – A(H) begin for v V do calculate A(v); label := 0; while unlabelled vertices v remain for which A(v) do begin label := label + 1; u := a vertex with A(u) ; assign label to u; for each unlabelled vertex v V do A(v) := A(v) {u}; end Topological sort algorithm (3) 14 CS1022 for v V do calculate A(v); A B C DE H G F

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Topological sort algorithm (4) Step 1 – Enter while loop: – Assign label 1 to H – Delete H from remaining A(v) – A(A) {B} – A(B) {C} – A(C) – A(D) {C} – A(E) {D, G} – A(F) {E} – A(G) {C} 15 CS1022... label := label + 1; u := a vertex with A(u) ; assign label to u; for each unlabelled vertex v V do A(v) := A(v) {u};...

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Topological sort algorithm (5) Step 2 – second pass through while loop: – Assign label 2 to C – Delete C from remaining A(v) – A(A) {B} – A(B) – A(D) – A(E) {D, G} – A(F) {E} – A(G) 16 CS1022... label := label + 1; u := a vertex with A(u) ; assign label to u; for each unlabelled vertex v V do A(v) := A(v) {u};...

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Topological sort algorithm (6) Step 3 – third pass through while loop: – There is a choice of labels to choose from – Each choice leads to distinct consistent labelling – Assign label 3 to B and delete B from remaining A(v) – A(A) – A(D) – A(E) {D, G} – A(F) {E} – A(G) 17 CS1022... label := label + 1; u := a vertex with A(u) ; assign label to u; for each unlabelled vertex v V do A(v) := A(v) {u};...

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Topological sort algorithm (7) Step 4 – fourth pass through while loop: – There is again a choice of labels to choose from – Assign label 4 to A and delete A from remaining A(v) – A(D) – A(E) {D, G} – A(F) {E} – A(G) 18 CS1022... label := label + 1; u := a vertex with A(u) ; assign label to u; for each unlabelled vertex v V do A(v) := A(v) {u};...

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Topological sort algorithm (8) Step 5 – fifth pass through while loop: – Assign label 5 to D and delete D from remaining A(v) – A(E) {G} – A(F) {E} – A(G) 19 CS1022... label := label + 1; u := a vertex with A(u) ; assign label to u; for each unlabelled vertex v V do A(v) := A(v) {u};...

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Topological sort algorithm (9) Step 6 – sixth pass through while loop: – Assign label 6 to G and delete G from remaining A(v) – A(E) – A(F) {E} 20 CS1022... label := label + 1; u := a vertex with A(u) ; assign label to u; for each unlabelled vertex v V do A(v) := A(v) {u};...

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Topological sort algorithm (10) Step 7 – seventh pass through while loop: – Assign label 7 to E and delete E from remaining A(v) – A(F) 21 CS1022... label := label + 1; u := a vertex with A(u) ; assign label to u; for each unlabelled vertex v V do A(v) := A(v) {u};...

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Topological sort algorithm (11) Step 7 – final pass through while loop: – Assign label 8 to F a – There are no remaining vs to delete E from 22 CS1022... label := label + 1; u := a vertex with A(u) ; assign label to u; for each unlabelled vertex v V do A(v) := A(v) {u};...

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Topological sort algorithm (12) Algorithm found one possible consistent labelling: H, C, B, A, D, G, E, F This gives an order in which modules can be taken – Consistent with pre-requisites 23 CS1022 A B C DE H G F

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Some remarks Algorithm analysed a graph and ordered vertices – “Sort” vertices based on incidence of arcs Approach was exhaustive... – However, it did not try all traversals of the digraph – It relied on visiting vertices (labelling them) in some order Why should you care? – If you ever need to perform similar process you can (you should!) re-use the algorithm – Algorithm can be implemented in different languages 24 CS1022

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Further reading R. Haggarty. “Discrete Mathematics for Computing”. Pearson Education Ltd. 2002. (Chapter 8) Wikipedia’s entry on directed graphs Wikibooks entry on graph theory 25 CS1022

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