It would appear that no general rule can be given for solving these river-crossing puzzles.  A formula can be found for a particular case (say on No. 375 or 376) that would apply to any number of individuals under the restricted conditions; but it is not of much use, for some little added stipulation will entirely upset it.  As in the case of the measuring puzzles, we generally have to rely on individual ingenuity.

374.—­CROSSING THE RIVER AXE.

Here is the solution:—­

| {J 5) | G T8 3
5  | ( J } | G T8 3
5  | {G 3) |  JT8
53 | ( G } |  JT8
53 | {J T) | G  8
J  5  | (T 3} | G  8
J  5  | {G 8) |   T  3
G  8   | (J 5} |   T
G  8   | {J T) |     53
JT8   | ( G } |     53
JT8   | {G 3) |     5
G T8 3 | ( J } |     5
G T8 3 | {J 5) |

G, J, and T stand for Giles, Jasper, and Timothy; and 8, 5, 3, for L800, L500, and L300 respectively.  The two side columns represent the left bank and the right bank, and the middle column the river.  Thirteen crossings are necessary, and each line shows the position when the boat is in mid-stream during a crossing, the point of the bracket indicating the direction.

It will be found that not only is no person left alone on the land or in the boat with more than his share of the spoil, but that also no two persons are left with more than their joint shares, though this last point was not insisted upon in the conditions.

375.—­FIVE JEALOUS HUSBANDS.

It is obvious that there must be an odd number of crossings, and that if the five husbands had not been jealous of one another the party might have all got over in nine crossings.  But no wife was to be in the company of a man or men unless her husband was present.  This entails two more crossings, eleven in all.

The following shows how it might have been done.  The capital letters stand for the husbands, and the small letters for their respective wives.  The position of affairs is shown at the start, and after each crossing between the left bank and the right, and the boat is represented by the asterisk.  So you can see at a glance that a, b, and c went over at the first crossing, that b and c returned at the second crossing, and so on.

ABCDE abcde *|..|
|  |
1.  ABCDE    de  |..|*       abc
2.  ABCDE  bcde *|..|        a
3.  ABCDE     e  |..|*       abcd
4.  ABCDE    de *|..|        abc
5.     DE    de  |,,|* ABC   abc
6.    CDE   cde *|..|  AB    ab
7.         cde  |..|* ABCDE ab
8.        bcde *|..|  ABCDE a
9.           e  |..|* ABCDE abcd
10.       bc e *|..|  ABCDE a  d
11.             |..|* ABCDE abcde

There is a little subtlety concealed in the words “show the quickest way.”

Everybody correctly assumes that, as we are told nothing of the rowing capabilities of the party, we must take it that they all row equally well.  But it is obvious that two such persons should row more quickly than one.

Therefore in the second and third crossings two of the ladies should take back the boat to fetch d, not one of them only.  This does not affect the number of landings, so no time is lost on that account.  A similar opportunity occurs in crossings 10 and 11, where the party again had the option of sending over two ladies or one only.