Karan's Puzzle 1

To find if a give figure can be drawn or not , just mark all the intersections in the figure and highlight them as dots and label each of them as vertex(v) from i = 1,2,3 .... ,n Now try to find the number of total lines emerging from each of the vertex and label them as number of lines L(i) where L(i) is the number of lines emerging from the vertex labelled as i. Now count the number of L(i)'s which are odd numbers for each of the i = 1,2,3 .... ,n and store it in the variable x. Now if x = 0 or 2 then the given figure can be drawn else the given figure can't be drawn.

I will be updating the proof of my solution in a week. Try to challenge me by giving some other solution or try to prove my solution.

The problem is equivalent to asking whether there exists an Eulerian circuit in each graph. A connected graph has an Eulerian circuit if and only if all of its vertices have an even degree. Since both graphs have a vertex with an odd degree, then both have no Eulerian circuit.

See: http://en.wikipedia.org/wiki/Eulerian_path

its simple .try to draw atleast once both the diagram ,you will find that it gets overwritten

This is question related to graph theory,

The basic rule of graph theory is that a figure can be drawn without lifting a pencil when it has either even number of intersections at its vertices or the same condition with maximum two number of veritce with odd intersections is also allowed.

None of the figure satifies the above conditions.

It's elementary graph theory quiz, shape can be drawn by 1 line only if it contain an eulerian path. On eulerian path, every node must has even edge pass through it. Those 2 shape is not satisfied.

Whenever the lines from the given vertex (which is L(i) in the given problem) is even, it means that there is an edge incoming to that vertex and an outgoing edge corresponding to it. So when all the vertices are having there L(i) even its simple case in which we can draw a figure. Now next comes the case in which there is a vertex with L(i) as odd. When L(i) is odd it signifies that it is having either an extra incoming line or an outgoing line, so to compensate this extra line there must be another vertex having L(i) value to be odd. Infact we can say that if there are two vertices with odd value of L(i), one of them corresponds to the vertex from which we need to start drawing the figure and the other one corresponds to the vertex where we end our figure. Next comes the case where we have more than 2 vertices with odd value of L(i), this is not possible, the explanation to this is already done in the second case, it we have more than 2 vertices which are having their L(i) value as odd, that particular vertex will have an edge which cannot be covered while drawing the figure,because there is no other edge corresponding to that particular vertex which can be drawn. Hence the solution.