Sections on this page: — Collini's Method — Tours of type IIIOOOIO — Tours of type OOOIIOII — Tours of type IIIOOIOO and Tours of type IIOOIIOO — Tours of type IIOOIOIO, IIOIOOIO and IOIOIOIO — Symmetric Tours by Double Linking

As reported in the history pages, Collini (1772-3) published a method for constructing tours beginning from the pseudotour consisting of eight concentric circuits (A0), and gave 20 examples. Here we look into Collini's method in more detail and derive new results. If we label the circuits 'a' to 'h' as in (A1) then it may be noted that no moves are possible connecting ae, bf, cg, dh, ac, bd, eg, fh. The transitions possible form the network (A2). Thus, if for example we wish to construct a tour from a cell in circuit 'a' to a cell in circuit 'c' by a simply -linked open tour, we first find a tour of (A2) from node 'a' to node 'c'. Not all tours of the network lead to board tours however; for example moves across the principal diagonal such as bad, bcd, hgf, hef are impossible.

Using the sequence ahbgdefc (A3) I formed the corner-to-corner linkage polygon (A4) to generate the corner-to -corner tour (A5). This is an improvement on Collini's corner-to-corner example (17) which employed nine deletions.

The followng four tours are also improvements on Collini's examples. The further four with ends on adjacent cells are cases not shown by Collini. Some of the linkage polygons in these tours are interesting patterns in themselves. That for the a2-a3 tour below, resembling a ‘flying pelican’, has only one self-intersection. A non-intersecting linkage polygon seems to be impossible.

Theorem: __To show all possible sequences of inner and outer circuits requires 33 open tours, or 7 closed__.
*Proof:* (a) The pattern of alternating inner and outer circuits can be represented ...IOIOIO...
The number of arrangements of 4 Is and 4 Os in sequence is 8!/4!4! = 70, but from this we must remove the 8 that
contain IIII or OOOO which are impossible in tours. There are thus 62 possible sequences of 4 Is and 4 Os in simply
linked open tours. (b) We must now take account of symmetry. The number of arrangements of 2 Is and 2 Os
in one half of the tour is 4!/2!2! = 6, but two of these are the impossible OOIIIIOO and IIOOOOII. Thus of the
62 sequences only four are symmetric; 58 are asymmetric. (c) The reversal of a tour has the reverse 'IO' code, so
to show all possible codes we need to construct 58/2 + 4 = 33 open tours. (d) A single closed tour, by removal of
any one of its eight linking moves, can illustrate up to 16 cases (8 choices of break point, two choices of
direction of description). For closed tours we need only count the number of different cyclic arrangements of
4 Is and 4 Os which is 8, from which we exclude the impossible case IIIIOOOO, giving 7. It remains to show by
examples that all the other cases are possible.

Here is a list of the seven cyclic sequences. The first number in brackets after the code is the number of
non-cyclic arrangements formed by breaking the cycle or reversing its sequence, or both (these total 62), the
second is the number of derivable open cases not counting reversals as different (total 33):

IIIOOOIO (16, 8)

OOOIIOII (8, 4)

IIIOOIOO (8, 4)

IIOOIIOO (4, 3)

IIOOIOIO (16, 8)

IIOIOOIO (8, 5)

IOIOIOIO (2, 1)

There are 24 such tours. All contain the OOO pattern (B1). The single inner circuit can go in two ways (B2), and for the triple inner circuits there are four configurations (B3), their orientation being determined by that of the single inner circuit. Finally there are three ways of connecting the single outer circuit to the two loose ends as in (B4). For example with the loose ends at c3, c4 we have (1) c3-b5...a3-c4 or (2) c3-b1...a3-c4 or (3) c3-b1...d2-c4 (where ... indicates the deleted move in the outer circuit). Total therefore 2×4×3 = 24. The diagram (B5) shows one of the 24.

The OOO must be as in (C1) and the other O from corner to corner of the central 4×4 (as in C2), which can be rotated into four different positions. The two IIs must each connect a terminal of OOO to one of O as in the four 4×4 half tour diagrams (C3). For each of the two IIs there is a choice of two routes. Total therefore 4×2×2 = 16. The example (C4) is a tour with an axially symmetric centre.

A tour of type IIIOOIOO was given by E. M. Laquière (1880). The III connections are the four three-quarter tours of the 4×4 as given in (B3) above. If the terminals of these are c3 and f3 then the single I must be d3...f4. There are three ways of connecting a pair of adjacent terminals of the inner circuits by OO circuits, as in (D1), (D2) and (D3), in which the cells visited are paired in a ‘domino’ pattern (pairs of adjacent cells). These can be combined in pairs in four ways: D1 with D2 or D3, arranged with the dominoes horizontal or vertical. Note that combinations D1-D1, D2-D2, D3-D3 or D2-D3 are impossible because the dominoes must be oriented in the same way. The total is therefore 4×4 = 16 tours. Diagram (D4) shows Laquière's example.

A closed tour in which each circuit is linked to one of the same type must be of pattern IIOOIIOO. All contain the OO pattern (E1). For the other OO we have four choices, the two shown in (E2) and (E3) or their half-turns, each connecting one pair of adjacent corners of the 4×4. For each case there is a choice of two ways of connecting the terminals within the 4×4 and for each connection there is a choice of two paths within the 4×4, see (C3) above. Thus there are 4×2×2 = 16 such tours. Diagram (E4) shows one example.

There are more tours in these classes than in those above. The enumeration remains to be completed. Examples
of these three types are shown in (F1) (F2), (G1), (G2) and (H1). Tour (H1) is one of the earliest tours I
constructed myself, since I have recorded the date as March 1970, though it was not published until 1985 in
*Chessics*. Collini's (15) and (25) shown in the history section also provide examples of the alternating type.

By the nature of Collini's method none of the tours it produces by simple-linking can be symmetric. However, symmetric tours can be produced by double-linking, as explained in the page on the simple-linking method. We can derive four symmetric doubly-linked tours from any singly-linked closed tour by using the same 16-move linkage polygon rotated to the other side of the board (i.e by a half-turn) and then joining up the pseudotour thus formed (consisting of two 32-move circuits) by reversing one of the central pairs of linkages (i.e. instead of connecting a-b, c-d we connect a-c, b-d).

Here are some distinctive tours found by this method.

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