Unicursal Problems

Excerpts

• Unicursal Problems

  in the determination of a route along the edges of a regular dodecahedron which will pass once and only once through every angular point.

• Unicursal Problems

  On a board containing an even number of cells the path may or may not be re-entrant, but on a board containing an odd number of cells it cannot be re-entrant.

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  His rule is that the knight must be moved always to one of the cells from which it will command the fewest squares not already traversed.

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  if we approach a point by one edge, the only routes open to us are one to the right, denoted by $r$, and one to the left, denoted by $l$.

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  The rule has not been proved to be true, but no exception to it is known

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  It is convenient to make a mark or to put down a counter at each corner as soon as it is reached, and this will prevent our passing through the same town twice.

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  Thus the cells $(x, y)$ and $(9-x, 9-y)$ are complementary, where $x$ and $y$ denote respectively the column and row occupied by the cell.

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  who invented this game---if game is the right term for it---denoted the twenty angular points on the solid by letters which stand for various towns.

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  The annulus may be divided into four closed circuits, each containing $12$ cells: these are marked respectively with the letters $a$, $b$, $c$, $d$.

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  Thus if the initial cell is on $a$, we might take either of the cycles $a\ D\ b\ C\ d\ A\ c\ B$, or $a\ D\ c\ B\ d\ A\ b\ C$.

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  By following these rules we always can connect the routes into one path, but in general it will not be re-entrant.

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  It is convenient to take the cells in each circuit in one and the same direction, but a circuit in the outer annulus must not end in a corner cell, and to avoid this we may have to alter the direction in which a circuit is taken.

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  and on the other hand is greater than $31,054144$---since this latter number is the number of re-entrant paths of a particular type

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  It leads to eight forms, similar to that in the diagram printed Jaenisch, in which the sum of the numbers in every column and every row is $260$; but although symmetrical it is not in my opinion so easy to reproduce as that given by Roget.

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  It is as yet impossible to say how many solutions of the problem exist.

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  It is evident that the question will not be affected if we suppose the islands to diminish to points and the bridges to lengthen out. In this way we ultimately obtain a geometrical figure or network.

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  Since a node of the $n$th order is one at which $n$ branches meet, there are $n$ hooks there. Also since the figure is closed, $n$ cannot be less than $2$.

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  The number of hooks at each node is even, and if they are unfastened they can be re-coupled together in pairs, the arrangement of the pairs being immaterial.

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  The number of trees with $n$ given nodes is $n^{n-2}$.

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  Of course to walk twice over every path in a labyrinth is not the shortest way of arriving at the centre, but, if it is performed correctly, the whole maze is traversed, the arrival at the centre at some point in the course of the route is certain, and it is impossible to lose one’s way

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  said to have been originally traced in the sand by the point of his scimetar without taking the scimetar off the ground or retracing any part of the figure

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  a chess-board, divided as usual by straight lines into $64$ cells, has $28$ odd nodes and $53$ even nodes: hence it would require $14$ separate pen-strokes to trace out all the boundaries without going over any more than once.

Equations

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  n^{n-2}

  The number of trees with n given nodes is n raised to the power n-2.

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  (1-x)^{-1} (1-x^2)^{-A_1} (1-x^3)^{-A_2} \dotsm & = 1 + A_1 x + A_2 x^2 + A_3 x^3 + \dotsb\, ,

  A generating-function identity: an infinite product in x equals the series whose coefficients are the numbers A_n of trees with n branches.

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  (1-x)^{-1} (1-x^2)^{-B_2} (1-x^3)^{-B_3} \dotsm & = 1 + x + 2B_2 x^2 + 2B_3 x^3 + \dotsb\,.

  A generating-function identity: an infinite product in x equals the series whose coefficients involve the numbers B_n of trees with n free branches that are bifurcations at least.

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  lr^2l = rlr

  Making the move left, then two right moves, then left has the same total effect as right, left, right on a dodecahedron's edges.

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  rl^2r = lrl

  Making the move right, then two left moves, then right has the same total effect as left, right, left on a dodecahedron's edges.

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  lr^3l=r^2

  Making the move left, three right moves, then left has the same total effect as two right moves.

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  rl^3r=l^2

  Making the move right, three left moves, then right has the same total effect as two left moves.

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  l^5=1

  Five successive left moves bring the traveller back to the starting point.

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  r^5=1

  Five successive right moves bring the traveller back to the starting point.

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  \{r^3l^3(rl)^2\}^2=1

  The twenty-step operation r r r l l l r l r l r r r l l l r l r l, read cyclically, returns to the start, so it traces a route through every town on the dodecahedron (condition (i)).

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  \{l^3r^3(lr)^2\}^2=1

  The twenty-step operation l l l r r r l r l r l l l r r r l r l r, read cyclically, returns to the start, so it traces a route through every town on the dodecahedron (condition (ii)).

Problems

No exercises in this chapter.