Part 10 (1/2)
254--THE MOTOR-CAR TOUR
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In the above diagraood roads In just howfrom London ( every town once, and only once, on a tour, and always co back to London on the last ride? The exact reverse of any route is not counted as different
255--THE LEVEL PUZZLE
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This is a si puzzle In how many different ways can you spell out the word LEVEL by placing the point of your pencil on an L and then passing along the lines froo in any direction, backwards or forwards Of course you are not allowed to miss letters--that is to say, if you come to a letter you must use it
256--THE DIAMOND PUZZLE
IN how eo up or down, backwards or forwards, in and out, in any direction you like, so long as you always pass from one letter to another that adjoins it How many ways are there?
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257--THE DEIFIED PUZZLE
In how many different ways ement under the same conditions as in the last puzzle, with the addition that you can use any letters twice in the sa?
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258--THE VOTERS' PUZZLE
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Here we have, perhaps, theform of the puzzle In how many different ways can you read the political injunction, ”RISE TO VOTE, SIR,” under the sa of the palindrome requires the use of the central V as the middle letter
259--HANNAH'S PUZZLE
Alady whose Christian name was Hannah When he asked her to be his wife she wrote down the letters of her name in this manner:-- H H H H H H H A A A A H H A N N A H H A N N A H H A A A A H H H H H H H and promised that she would be his if he could tell her correctly in how many different ways it was possible to spell out her na froonal steps are here allowed Whether she did this merely to tease him or to test his cleverness is not recorded, but it is satisfactory to know that he succeeded Would you have been equally successful? Take your pencil and try You o backwards or forwards and in any direction, so long as all the letters in a spelling are adjoining one another How many ways are there, no two exactly alike?
260--THE HONEYCOMB PUZZLE
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Here is a little puzzle with the simplest possible conditions Place the point of your pencil on a letter in one of the cells of the honeyco always froht route you will have visited every cell once, and only once The puzzle is much easier than it looks
261--THE MONK AND THE BRIDGES
In this case I give a rough plan of a river with an island and five bridges On one side of the river is a round Now, the e once, and only once, on his return to the monastery This is, of course, quite easy to do, but on the way he thought to himself, ”I wonder how ht have selected” Could you have told him? That is the puzzle Take your pencil and trace out a route that will take you once over all the five bridges Then trace out a second route, then a third, and see if you can count all the variations You will find that the difficulty is twofold: you have to avoid dropping routes on the one hand and counting the same routes more than once on the other
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COMBINATION AND GROUP PROBLEMS
”A combination and a form indeed” Haht be tereometry of situation,” but their solution really depends on the theory of combinations which, in its turn, is derived directly from the theory of perroup puzzles and enuht, perhaps, with equal reason have been placed elsewhere; but readers are again asked not to be too critical about the classification, which is very difficult and arbitrary As I have included my problem of ”The Round Table” (No 273), perhaps a few remarks on another well-known problem of the saes,If n married ladies are seated at a round table in any determined order, in how many different ways may their n husbands be placed so that every man is between two ladies but never next to his oife?
This difficult problem was first solved by Laisant, and thetable is due to Moreau:-- 4 0 2 5 3 13 6 13 80 7 83 579 8 592 4738 9 4821 43387 10 43979 439792 The first column shows the number of married couples The numbers in the second column are obtained in this way: 5 3 + 0 - 2 = 13; 6 13 + 3 + 2 = 83; 7 83 + 13 - 2 = 592; 8 592 + 83 + 2 = 4821; and so on Find all the numbers, except 2, in the table, and the method will be evident It will be noted that the 2 is subtracted when the first number (the number of couples) is odd, and added when that number is even The numbers in the third column are obtained thus: 13 - 0 = 13; 83 - 3 = 80; 592 - 13 = 579; 4821 - 83 = 4738; and so on The nuive the required solutions Thus, four husbands may be seated in tays, five husbands hty ways
The following method, by Lucas, will show the remarkable way in which chessboard analysis may be applied to the solution of a circular problem of this kind Divide a square into thirty-six cells, six by six, and strike out all the cells in the long diagonal froht-hand corner, also the five cells in the diagonal next above it and the cell in the bottoht-hand corner The answer for six couples will be the same as the nu the cancelled cells) so that no rook shall ever attack another rook It will be found that the six rooks rees with the above table
262--THOSE FIFTEEN SHEEP
A certain cyclopaedia has the following curious problem, I am told: ”Place fifteen sheep in four pens so that there shall be the same number of sheep in each pen” No anshatever is vouchsafed, so I thought I would investigate thewould appear to be quite impossible, since four times any number must be an even nuht, therefore, that there enerally known So I decided to interview some farmers on the subject The first one pointed out that if we put one pen inside another, like the rings of a target, and placed all sheep in the sht But I objected to this, because you admittedly place all the sheep in one pen, not in four pens The second man said that if I placed four sheep in each of three pens and three sheep in the last pen (that is fifteen sheep in all), and one of the ewes in the last pen had a laht, there would be the sa This also failed to satisfy me
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The third farot four hurdle pens down in one of my fields, and a small flock of wethers, so if you will just step doith me I will show you how it is done” The illustration depicts my friend as he is about to demonstrate the matter to me His lucid explanation was evidently that which was in the mind of the writer of the article in the cyclopaedia What was it? Can you place those fifteen sheep?
263--KING ARTHUR'S KNIGHTS
King Arthur sat at the Round Table on three successive evenings with his knights--Beleobus, Caradoc, Driam, Eric, Floll, and Galahad--but on no occasion did any person have as his neighbour one who had before sat next to hi they sat in alphabetical order round the table But afterwards King Arthur arranged the two next sittings so that he ht have Beleobus as near to him as possible and Galahad as far away frohts to the best advantage, rehbour twice?
264--THE CITY LUNCHEONS
Twelve e firether every day in the same room The tables are small ones that only accommodate two persons at the saether on eleven days in pairs, so that no two of theether? We will represent the men by the first twelve letters of the alphabet, and suppose the first day's pairing to be as follows-- (A B) (C D) (E F) (G H) (I J) (K L)
Then give any pairing you like for the next day, say-- (A C) (B D) (E G) (F H) (I K) (J L), and so on, until you have co twice There are a good ements possible Try to find one of them
265--A PUZZLE FOR CARD-PLAYERS
Twelve ether on eleven evenings, but no player was ever to have the same partner more than once, or the sa how they ? Call the twelve players by the first twelve letters of the alphabet and try to group them
266--A TENNIS TOURNAMENT
Four married couples played a ”mixed double” tennis tournaainst a ainst any other person more than once Can you sho 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
267--THE WRONG HATS
”One of the s I have cohtnot wisely but too well at a certain London restaurant They were the last to leave, but not one man was in a condition to identify his own hat Now, considering that they took their hats at random, what are the chances that everyto hi,” said Mr Waterson, ”is to see in how ht hats could be taken”
”That is quite easy,” Mr Stubbs explained ”Multiply together the numbers, 1, 2, 3, 4, 5, 6, 7, and 8 Let me see--half a minute--yes; there are 40,320 different ways”
”Now all you've got to do is to see in how many of these cases no man has his own hat,” said Mr Waterson
”Thank you, I' any,” said Mr Packhurst ”I don't envy theout all those forty-thousand-odd cases and then picking out the ones he wants”
They all agreed that life is not long enough for that sort of a at the answer, the matter was postponed indefinitely Can you solve the puzzle?
268--THE PEAL OF BELLS
A correspondent, who is apparently y, asks me how he is to construct what he calls a ”true and correct” peal for four bells He says that every possible per once, and once only He adds that no bell must move more than one place at a time, that no bell must make more than two successive strokes in either the first or the last place, and that the last change must be able to pass into the first These fantastic conditions will be found to be observed in the little peal for three bells, as follows:-- 1 2 3 2 1 3 2 3 1 3 2 1 3 1 2 1 3 2 How are we to give him a correct solution for his four bells?