re-entrant knight's tour may be made on each part. Cuts along the dotted lines will not do, as the four central squares of the board would be either detached or hanging on by a mere thread. (2/2)
339. THE FOUR KNIGHTS' TOURS. I will repeat that if a chessboard be cut into four equal parts, as indicated by the dark lines in the illustration, it is not possible to perform a knight's tour, either re-entrant or not, on one of the parts. The best re-entrant attempt is shown, in which each knight has to trespass twice on other parts. The puzzle is to cut the board differently into four parts, each of the same size and shape, so that a (1/2)
SOLUTION TO 375. FIVE JEALOUS HUSBANDS. (5/6) Show more
similar opportunity occurs in crossings 10 and 11, where the party again had the option of sending over two ladies or one only.
To those who think they have solved the puzzle in nine crossings I would say that in every case they will find that they are wrong. No such jealous husband would, in the circumstances, send his wife over to the other bank to a man or men, even if she assured him that she was coming
SOLUTION TO 375. FIVE JEALOUS HUSBANDS. (4/6) Show more
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
SOLUTION TO 375. FIVE JEALOUS HUSBANDS. (3/6) Show more
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."
SOLUTION TO 375. FIVE JEALOUS HUSBANDS. (2/6) Show more
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
SOLUTION TO 375. FIVE JEALOUS HUSBANDS. (1/6) Show more
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
Call the men A, B, C, D, E, and their respective wives a, b, c, d, e. To go over and return counts as two crossings. No tricks such as ropes,
swimming, currents, etc., are permitted. (2/2)
375. FIVE JEALOUS HUSBANDS. During certain local floods five married couples found themselves surrounded by water, and had to escape from their unpleasant position in a boat that would only hold three persons at a time. Every husband was so jealous that he would not allow his wife to be in the boat or on either bank with another man (or with other men) unless he was himself present. Show the quickest way of getting these five men and their wives across into safety.
266. A TENNIS TOURNAMENT. Four married couples played a "mixed double" tennis tournament, a man and a lady always playing against a man and a lady. But no person ever played with or against any other person more than once. Can you show how they all could have played together in the two courts on three successive days? This is a little puzzle of a quite practical kind, and it is just perplexing enough to be interesting.
SOLUTION TO 269. THREE MEN IN A BOAT. (7/7) Show more
and showed what that general law is and how the groups should be posed for any number of girls. I gave actual arrangements for numbers that had previously baffled all attempts to manipulate, and the problem may now be considered generally solved. Readers will find an excellent full account of the puzzle in W.W. Rouse Ball's _Mathematical Recreations_,
SOLUTION TO 269. THREE MEN IN A BOAT. (6/7) Show more
solution of this puzzle had exercised the ingenuity of mathematicians since 1850, when the question was first propounded, until recently. In 1908 and the two following years I indicated (see _Educational Times Reprints_, Vols. XIV., XV., and XVII.) that all our trouble had arisen from a failure to discover that 15 is a special case (too small to enter into the general law for all higher numbers of girls of the form 6n+3),
SOLUTION TO 269. THREE MEN IN A BOAT. (5/7) Show more
It will be found that no two men ever go out twice together, and that no man ever goes out twice in the same boat.
This is an extension of the well-known problem of the "Fifteen Schoolgirls," by Kirkman. The original conditions were simply that fifteen girls walked out on seven days in triplets without any girl ever walking twice in a triplet with another girl. Attempts at a general
SOLUTION TO 269. THREE MEN IN A BOAT. (4/7) Show more
3 5 4 1 2
3rd Day (AJM) (BEH) (CFI) (DKO) (GNL)
7 6 8 9 1
4th Day (AEK) (CGM) (BOI) (DHL) (JNF)
4 5 3 10 2
5th Day (AHN) (CDJ) (BFL) (GEO) (MKI)
6 7 8 10 1
6th Day (AFO) (BGJ) (CKH) (DNI) (MEL)
5 4 3 9 2
7th Day (AIL) (BDM) (CEN) (GKF) (JHO)
SOLUTION TO 269. THREE MEN IN A BOAT. (3/7) Show more
Of course only a certain number of all these arrangements are available when we have that other condition of using the smallest possible number of boats. As a matter of fact we need employ only ten different boats. Here is one the arrangements:--
1 2 3 4 5
1st Day (ABC) (DBF) (GHI) (JKL) (MNO)
8 6 7 9 10
2nd Day (ADG) (BKN) (COL) (JEI) (MHF)
SOLUTION TO 269. THREE MEN IN A BOAT. (2/7) Show more
the reader will realize why this must be, for although, as an example, A must go out once with B and once with C, it does not necessarily follow that he must go out with C on the same occasion that he goes with B. He might take any other letter with him on that occasion, though the fact of his taking other than B would have its effect on the arrangement of the other triplets.
SOLUTION TO 269. THREE MEN IN A BOAT. (1/7) Show more
If there were no conditions whatever, except that the men were all to go out together, in threes, they could row in an immense number of different ways. If the reader wishes to know how many, the number is 455^7. And with the condition that no two may ever be together more than once, there are no fewer than 15,567,552,000 different solutions--that is, different ways of arranging the men. With one solution before him,
The men within each pair of brackets are here seen to be in the same boat, and therefore A can never go out with B or with C again, and C can never go out again with B. The same applies to the other four boats. The figures show the number on the boat, so that A, B, or C, for example,
can never go out in boat No. 1 again. (5/5)
readers to discover how the men should have been placed in the boats. As their names happen to have been Andrews, Baker, Carter, Danby, Edwards,
Frith, Gay, Hart, Isaacs, Jackson, Kent, Lang, Mason, Napper, and Onslow, we can call them by their initials and write out the five groups for each of the seven days in the following simple way:
1 2 3 4 5
First Day: (ABC) (DEF) (GHI) (JKL) (MNO).