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Once upon a time … Eulers problem (1936) Königsbergs city (nowadays Kaliningrad) is crossed by Pregel river, which runs around the island of Kneiphof on both sides, and has seven bridges During a walk, is-it possible to pass on all the bridges of the city once and only once?

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König, D. (1936).Theorie der endlichen und unendlichen Graphen. König, D. (1990).Theory of finite and infinite graphs. Berlin: Birkhauser Berge, C. (1958). Théorie des graphes et ses applications. Paris: Dunod. English edition, Wiley 1961; Methuen & Co, New York 1962; Dover, New York Russian, Moscow 1961; Spanish, Mexico 1962; Roumanian, Bucharest 1969; Chinese, Shanghai 1963; Some references …

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3 fundamental articles to use similitude analysis in the social representations domain Flament (1962). Lanalyse de similitude. Cahiers du Centre de Recherche Opérationnelle, 4, Degenne, A. & Vergès, P. (1973). Introduction à lanalyse de similitude. Revue française de Sociologie, 14, Flament, C., Degenne, A. & Vergès, P. (1971). Similarity Analysis. Paris: Maison des Sciences de lHomme. Some references …

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Graphs theory Useful elements A graph G G (V, E) V = {v 1, v 2, …, v n } that is n Vertices E = {e 1, e 2, …, e m } that is m Edges A graph G (V, E) V = {1, 2, 3, 4, 5, 6, 7, 8, 9, } E = {(1, 2), (1, 3), (1, 4), …, (8, 9)} Size of the graph

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Graphs theory Useful elements G(V,E) –V = {1, 2, 3, 4, 5} –E = {(1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5)} G(V,E) –V = {1, 2, 3} –E = {(1,2), (1,3), (2,3)} Some vertices = subgraph of G G(V,E) –V = {1, 2, 3, 4, 5} –E = {(1,2), (3,4), (4,5)} All the vertices, some edges = Spanning Subgraph of G

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Graphs theory Useful elements G(V,E) –V = {1, 2, 3, 4, 5} –E = {(1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5)} Symmetrical relations : (5, 4) = (4, 5) (1, 3), (3, 5), (5, 4) = a chain of G G(V,E) –V = {1, 2, 3, 4, 5} –E = {(1,2), (1,3), (2,3), (3,4), (4,5)} (1, 2), (2, 3), (3, 4), (4, 5), (5, 6) = a chain (1, 2), (2, 3) & (3, 1) = a cycle

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Graphs theory Useful elements A complete graph –A–A cycle –A–A chain A TREE A connected tree without cycle –A–A chain which connects all the verticies –A–A chain without cycle How to pass from a complete graph to a tree ?

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Useful elements for SR How to pass from a complete graph to a tree ? Searching for the structure of the relations = Searching for the skeleton of the representation = Searching for a tree Each edge has a weight = Similitude analysis weight = s imilitude index = Co-occurrence, symmetrical co-occurrence, Phi square measure, Correlation, squared index of similitude (Guimelli), etc.

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Useful elements for SR How to pass from a complete graph to a tree ? Searching for the structure of the relations = Searching for a maximum tree Searching for a connected graph without cycle + Searching for the heaviest tree = Searching for a tree which retains the most similarity

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Let us return to our example …

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Example 1 (inspired from Abric, 2003) Stage 4 – From similitude matrix to the structure of the relations between elements of a representation finance its leisure activities the means to earn the keep personal blooming self-confidence the means to have relations constraints an obligation social integration 1 The degree of similitude between two elements can be associated with the graph 5 Arête (7, 6) = (6,7)

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Example 1 (inspired from Abric, 2003) Stage 4 – From similitude matrix to the structure of the relations between elements of a representation 1

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1 EdgesS1S1 S1S1 S1S1 S1S1 (2, 6)6(1, 7)2(1, 4)1(3, 6)1 (2, 8)6(2, 5)2(1, 5)1(3, 8)1 (5, 6)5(3, 4)2(1, 6)1(4, 5)1 (6, 7)5(3, 5)2(1, 8)1(4, 7)1 (1, 2)3(3, 7)2(2, 3)1(4, 8)1 (1, 3)3(5, 7)2(2, 4)1(5, 8)1 (4, 6)3(7, 8)2(2, 7)1(6, 8)1

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Connected graph & without cycle Important ! (1) EdgesS1S1 S1S1 S1S1 S1S1 (2, 6)6(4, 6)3(1, 4)1(3, 6)1 (2, 8)6(2, 5)2(1, 5)1(3, 8)1 (5, 6)5(3, 4)2(1, 6)1(4, 5)1 (6, 7)5(3, 5)2(1, 8)1(4, 7)1 (1, 3)3(3, 7)2(2, 3)1(4, 8)1 (1, 2)3(5, 7)2(2, 4)1(5, 8)1 (1, 7)3(7, 8)2(2, 7)1(6, 8)1

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Important ! (2) EdgesS1S1 S1S1 S1S1 S1S1 (2, 6)6(4, 6)3(1, 4)1(3, 6)1 (2, 8)6(1, 7)2(1, 5)1(3, 8)1 (5, 6)5(2, 5)2(1, 6)1(4, 5)1 (6, 7)5(3, 4)2(2, 1 )1(4, 7)1 (7, 8)5(3, 5)2(2, 3)1(4, 8)1 (1, 8)3(3, 7)2(2, 4)1(5, 8)1 (1, 3)3(5, 7)2(2, 7)1(6, 8)1

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Example 1 (inspiré de Abric, 2003) Two populations = 2 graphs Young students Workers

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