(e) By the end of the 4th day, the `First-prize-man' is known (by the very same process of elimination used in the existing method): and the remaining Players are paired by the same rules as before, for the 2 contests on the 5th day. If, in section (a), the Tournament was begun in the morning, the two men named in section (d) being still allowed a whole-day match, nothing would be gained in time, as the Tournament would take 4½ days, while much would be lost in interest, as the first prize would be settled in 3 days.

To illustrate these rules, I will give the complete history of a Tournament of 32 competitors, with 3 prizes. If the reader will draw out the following Tables, in blank, and fill them up for himself, referring, if necessary, to the accompanying directions, he will easily understand the workings of system.

Let the Players be arranged alphabetically, and let the relative skill with which they play in this Tournament, be --

These numbers (`1' meaning `best') will enable the reader to name the victor in any contest: but of course they are not supposed to be known to the Tournament-Committee, who have nothing to guide them but the results of actual contests. In the following Tables, `I(e)' means `first day, evening', and so on: also a Player, who is virtually proved superior to another, is entered thus `(A)'. The victor in each contest is marked<sup>*</sup>: and means `struck out'.

Directions for filling in the Tables --

Tab. I Day I (e). The names are written out alphabetically, and paired as they stand. The victors are marked with asterisks.

Tab. II. Day I (e). As B has been beaten by A, A is entered as his `superior'; C as D's superior; and so on.

Tab. I. Day II (m). We first pair together all the unbeaten, A,C,E,G, &c. Then those who have one superior, B,D,F,H, &c.

Tab. II. Day II (m). We first enter the actual superiors, C,G, &c. Then, since A has a superior C, and B has a superior A, we see that B has a virtual superior C; and so on. We then see that D has 3 superiors, and must be struck out; and so with H, &c.

Tab. I. Day II (e). We first pair together all the unbeaten, C,G, &c. Then all with one superior, A,E, &c.; but when we come to J,L, we find we have a common superior; so we pair J with P, and L with Q. This series ends with an odd one, g, who must therefore be paired with the first of those who have two superiors each, F,T, &c.

Tab. I. Day III (m). Here, in pairing those with one superior, we again end with an odd one, g, who must therefore be paired with the first of those with two superiors, viz. T. We end with an `odd man', c.

Tab. II. Day III (m). The unbeaten are now reduced to one pair, M, f, who therefore will do nothing this afternoon, but will have a whole-day contest to-morrow.

Tab. I. Day III (e). Those who have one superior are C,J,L,R, all with a common superior M; and then V,a,g, all with a common superior f. We therefore pair C with V, and so on, leaving an odd one R, who must be paired with the only one who has two superiors, viz. c.

Tab. II. Day III (e). Enter as usual.

Tab. I. Day IV (m). We pair the 2 unbeaten, M,f, for their whole-day contest. Then those with one superior.

Tab. II. Day IV (m). M and f are still contending. V and g are struck out.