// Numbas version: exam_results_page_options {"name": "Networks 2: Dijkstra's Algorithm v2 ", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Networks 2: Dijkstra's Algorithm v2 ", "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-ShareAlike 4.0 International"}, "statement": "

In a weighted graph $G$, the weight of a path is the sum of all edge weights in that path.

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Suppose we fix a vertex $v_0$. A minimal $v_0$-spanning tree is a special spanning tree. Like all spanning trees there is a unique path from $v_0$ to any other vertex. But the special thing about this tree is that each of these paths is the shortest. The procedure to construct a minimal $v_0$-spanning tree is called Dijkstra's Algorithm:

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1. Start by defining a tree $T=(V,E)$ which contains only the vertex $v_0$, i.e. $V = \\left\\{v_0\\right\\}$ and $E=\\left\\{\\right\\}$.

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2. Consider all edges with one vertex in $V$ and the other vertex not in $V$:

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$N = \\left\\{ [v,u] : v \\in V \\text{ and } u \\not \\in V\\right\\}$.

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3. Find the edge $[u,v] \\in N$ which minimises the sum:

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$|v_0,v| + \\operatorname{weight}([v,u])$.

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4. Add $u$ to $V$ and add $[u,v]$ to $E$.

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5. Repeat steps 2, 3 and 4 until $T$ is a spanning tree for $G$.

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When constructed in this way, $(V,E)$ will be a minimal $v_0$-spanning tree.

", "advice": "

Dijkstra's algorithm is described in detail above. If the answer-checker's feedback wasn't helpful to you, you could try performing the algorithm on a smaller/simpler graph of your own design for practice. 

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i < edgenames.length; i++) {\n var e = edgenames[i];\n \n var u0 = e.split(\",\")[0];\n var u1 = e.split(\",\")[1];\n \n if (V.has(u0) && V.has(u1)) {\n if (-1 !== Numbas.exam.currentQuestion.parts[0].tree.indexOf(e) ) {\n // edge is in the tree and should already be coloured black\n } else {\n // edge is not in the tree, but both start and end vertex are.\n // this edge would create a loop so mark it as unselected.\n lines[e].setAttribute({strokeWidth:2,strokeColor:'blue'});\n lines[e].selected = false;\n } \n } else if ((!V.has(u0) && V.has(u1)) || (V.has(u0) && !V.has(u1))) {\n // edge has one vertex in V and one not in V\n lines[e].setAttribute({strokeWidth:4,strokeColor:'#008000'});\n } else {\n // edge is not connected to the tree.\n lines[e].selected = false;\n lines[e].setAttribute({strokeWidth:2,strokeColor:'blue'});\n }\n }\n \n Numbas.exam.currentQuestion.parts[0].distance = distance;\n \n // update graph display of distance \n for (var i = 0; i < points.length; i++) {\n var vertex = points[i].getAttribute(\"name\").split(\":\")[0];\n if (vertex in distance) {\n points[i].setAttribute({\"name\" : (vertex + \":\" + distance[vertex])});\n } else {\n points[i].setAttribute({\"name\" : (vertex)});\n }\n }\n \n \n });\n}\n\nreturn div;"}}, "preamble": {"js": "\n// perform dijkstra's algorithm on subgraph with given weights.\n// return dictionary of minimum distances from v0.\ncalculateDistance = function(subgraph, weight){\n \n \n // the set of vertices reached by the graph\n var V = new Set();\n for (var i=0; i < subgraph.length; i++) {\n var e = subgraph[i];\n var u0 = e.split(\",\")[0];\n var u1 = e.split(\",\")[1];\n V.add(u0);\n V.add(u1);\n }\n \n // answers are the definiatve list of shortest distances from v0\n var answers = {\"v0\":0};\n \n // calculate distances from v0 according to given subgraph \n var skip = 0;\n while (Object.keys(answers).length + skip < V.size) {\n var neighbours = {};\n var addedSomething = false;\n for (var i=0; i < subgraph.length; i++) {\n var e = subgraph[i];\n for (var j = 0; j < 2; j++) {\n var u0 = e.split(\",\")[j % 2];\n var u1 = e.split(\",\")[(j+1) % 2];\n \n if (u0 in answers && !(u1 in answers)) {\n addedSomething = true;\n if (u1 in neighbours) {\n // take the shortest path from all seen paths\n neighbours[u1] = Math.min(answers[u0] + weight[e],neighbours[u1]);\n } else {\n // only one path seen so far to u1\n neighbours[u1] = answers[u0] + weight[e];\n }\n }\n } \n }\n if (!addedSomething) {\n skip++;\n }\n \n // select minimum weight answer from all answers\n var min = 1000;\n var v;\n for (var k in neighbours) {\n if (neighbours[k] < min) {\n v = k;\n min = neighbours[k];\n }\n }\n if (min < 1000) {\n answers[v]=min;\n }\n \n } \n return answers; 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Perform Dijkstra's algorithm on the following graph.

\n\n

{graph(vertices,edges,weights,edgenames)}

\n

\u200b

\n

"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "contributors": [{"name": "Daniel Mansfield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/743/"}, {"name": "Joshua Capel", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1479/"}, {"name": "Sean Gardiner", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2443/"}]}]}], "contributors": [{"name": "Daniel Mansfield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/743/"}, {"name": "Joshua Capel", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1479/"}, {"name": "Sean Gardiner", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2443/"}]}