Henry Ernest Dudeney/Modern Puzzles/157 - Crossing the Lines/Solution

by : $157$

 * Crossing the Lines

Solution
The solution is straightforward by Characteristics of Traversable Graph, but it is interesting to see how wordy could be:


 * Let us suppose that we cross the lines by bridges, represented in Figure $1$ by the little parallels.


 * Dudeney-Modern-Puzzles-157-solution.png


 * Now, in Figure $2$, I transform the diagram, reducing the spaces $A$, $B$, $C$, $D$, $E$ to mere points,
 * and representing the bridges that connect these spaces by lines or roads.
 * This transformation does not affect the conditions,
 * for there are $16$ bridges or roads in one case, and $16$ roads or lines in the other,
 * and they connect with $A$, $B$, $C$, $D$, $E$ in precisely the same way.
 * It will be seen that $9$ bridges or roads connect with the outside.
 * Obviously we are free to join these up in pairs in any way we choose, provided the roads do not cross one another.
 * The simplest way is shown in Figure $3$, where on coming out from $A$, $B$, $C$, or $E$,
 * we immediately return to the same point by the adjacent bridge, leaving one point, $X$, necessarily in the open.
 * In Figure $2$ there are $4$ odd nodes, $A$, $B$, $D$, and $X$ (if we decide on the exits and entrances, as in Figure $3$),
 * so, as I have already explained, we require $2$ strokes (half of $4$) to go over all the roads,
 * proving a perfect solution to be impossible.


 * Now, let us cancel the line $AB$.
 * Follow the line in Figure $3$ and you will see that this can be done, omitting the line from $A$ to $B$.
 * This route the reader will easily transform into Figure $4$ if he says to himself,
 * "Go from $X$ to $D$, from $D$ to $E$, from $E$ to the outside and return into $E$," and so on.
 * The route can be varied by linking up those outside bridges differently,
 * by making $X$ an outside bridge to $A$ or $B$, instead of $D$, and by taking the cancelled line either at $AB$, $AD$, $BD$, $XA$, $XB$, or $XD$.
 * In Figure $5$ I make $X$ lead to $B$.
 * We still omit $AB$, but we must start and end at $D$ and $X$.
 * Transformed in Figure $6$, this will be seen to be the precise example that I gave in the question.
 * The reader can now write out as many routes as he likes for himself,
 * but he will always find it necessary to omit one line or crossing.


 * It is thus seen how easily sometimes a little cunning, like that of the transformation shown,
 * will settle a perplexing question of this kind.