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PUZZLE
GAMES.
"He that is beaten may be said
To lie in honour's truckle
bed."
HUDIBRAS.
It may be said generally that a game is a contest of skill for two or more
persons, into which we enter either for amusement or to win a prize. A puzzle is
something to be done or solved by the individual. For example, if it were
possible for us so to master the complexities of the game of chess that we could
be assured of always winning with the first or second move, as the case might
be, or of always drawing, then it would cease to be a game and would become a
puzzle. Of course among the young and uninformed, when the correct winning play
is not understood, a puzzle may well make a very good game. Thus there is no
doubt children will continue to play "Noughts and Crosses," though I have shown
(No. 109, "Canterbury Puzzles") that between two players who both
thoroughly understand the play, every game should be drawn. Neither player could
ever win except through the blundering of his opponent. But I am writing from
the point of view of the student of these things.
The examples that I give in this class are apparently games, but, since I
show in every case how one player may win if he only play correctly, they are in
reality puzzles. Their interest, therefore, lies in attempting to discover the
leading method of play.
392.—THE
PEBBLE GAME.
Here is an interesting little puzzle game that I used to play with an
acquaintance on the beach at Slocomb-on-Sea. Two players place an odd number of
pebbles, we will say fifteen, between them. Then each takes in turn one, two, or
three pebbles (as he chooses), and the winner is the one who gets the odd
number. Thus, if you get seven and your opponent eight, you win. If you get six
and he gets nine, he wins. Ought the first or second player to win, and how?
When you have settled the question with fifteen pebbles try again with, say,
thirteen.
Solution 393.—THE
TWO ROOKS.
This is a puzzle game for two players. Each player has a single rook. The
first player places his rook on any square of the board that he may choose to
select, and then the second player does the same. They now play in turn, the
point of each play being to capture the opponent's rook. But in this game you
cannot play through a line of attack without being captured. That is to say, if
in the diagram it is Black's turn to play, he cannot move his rook to his king's knight's
square, or to his king's rook's square, because he would enter the "line of
fire" when passing his king's bishop's square. For the same reason he cannot
move to his queen's rook's seventh or eighth squares. Now, the game can never
end in a draw. Sooner or later one of the rooks must fall, unless, of course,
both players commit the absurdity of not trying to win. The trick of winning is
ridiculously simple when you know it. Can you solve the puzzle?
Solution 394.—PUSS
IN THE CORNER.
This variation of the last puzzle is also played by two persons. One puts a
counter on No. 6, and the other puts one on No. 55, and they play alternately by
removing the counter to any other number in a line. If your opponent moves at
any time on to one of the lines you occupy, or even crosses one of your lines,
you immediately capture him and win. We will take an illustrative game.
A moves from 55 to 52; B moves from 6 to 13; A advances to 23; B goes to 15;
A retreats to 26; B retreats to 13; A advances to 21; B retreats to 2; A
advances to 7; B goes to 3; A moves to 6; B must now go to 4; A establishes
himself at 11, and B must be captured next move because he is compelled to cross
a line on which A stands. Play this over and you will understand the game
directly. Now, the puzzle part of the game is this: Which player should win, and
how many moves are necessary?
Solution 395.—A
WAR PUZZLE GAME.
Here is another puzzle game. One player, representing the British general,
places a counter at B, and the other player, representing the enemy, places his
counter at E. The Britisher makes the first advance along one of the roads to
the next town, then the enemy moves to one of his nearest towns, and so on in
turns, until the British general gets into the same town as the enemy and
captures him. Although each must always move along a road to the next town only,
and the second player may do his utmost to avoid capture, the British general
(as we should suppose, from the analogy of real life) must infallibly win. But
how? That is the question.
Solution 396.—A
MATCH MYSTERY.
Here is a little game that is childishly simple in its conditions. But it is
well worth investigation.
Mr. Stubbs pulled a small table between himself and his friend, Mr. Wilson,
and took a box of matches, from which he counted out thirty.
"Here are thirty matches," he said. "I divide them into three unequal heaps. Let
me see. We have 14, 11, and 5, as it happens. Now, the two players draw
alternately any number from any one heap, and he who draws the last match loses
the game. That's all! I will play with you, Wilson. I have formed the heaps, so
you have the first draw."
"As I can draw any number," Mr. Wilson said, "suppose I exhibit my usual
moderation and take all the 14 heap."
"That is the worst you could do, for it loses right away. I take 6 from the
11, leaving two equal heaps of 5, and to leave two equal heaps is a certain win
(with the single exception of 1, 1), because whatever you do in one heap I can
repeat in the other. If you leave 4 in one heap, I leave 4 in the other. If you
then leave 2 in one heap, I leave 2 in the other. If you leave only 1 in one
heap, then I take all the other heap. If you take all one heap, I take all but
one in the other. No, you must never leave two heaps, unless they are equal
heaps and more than 1, 1. Let's begin again."
"Very well, then," said Mr. Wilson. "I will take 6 from the 14, and leave you
8, 11, 5."
Mr. Stubbs then left 8, 11, 3; Mr. Wilson, 8, 5, 3; Mr. Stubbs, 6, 5, 3; Mr.
Wilson,4, 5, 3; Mr. Stubbs, 4, 5, 1; Mr. Wilson, 4, 3, 1; Mr. Stubbs, 2, 3, 1;
Mr. Wilson, 2, 1, 1; which Mr. Stubbs reduced to 1, 1, 1.
"It is now quite clear that I must win," said Mr. Stubbs, because you must
take 1, and then I take 1, leaving you the last match. You never had a chance.
There are just thirteen different ways in which the matches may be grouped at
the start for a certain win. In fact, the groups selected, 14, 11, 5, are a
certain win, because for whatever your opponent may play there is another
winning group you can secure, and so on and on down to the last match."
Solution 397.—THE
MONTENEGRIN DICE GAME.
It is said that the inhabitants of Montenegro have a little dice game that is
both ingenious and well worth investigation. The two players first select two
different pairs of odd numbers (always higher than 3) and then alternately toss
three dice. Whichever first throws the dice so that they add up to one of his
selected numbers wins. If they are both successful in two successive throws it
is a draw and they try again. For example, one player may select 7 and 15 and
the other 5 and 13. Then if the first player throws so that the three dice add
up 7 or 15 he wins, unless the second man gets either 5 or 13 on his throw.
The puzzle is to discover which two pairs of numbers should be selected in
order to give both players an exactly even chance.
Solution 398.—THE
CIGAR PUZZLE.
I once propounded the following puzzle in a London club, and for a
considerable period it absorbed the attention of the members. They could make
nothing of it, and considered it quite impossible of solution. And yet, as I
shall show, the answer is remarkably simple.
Two men are seated at a square-topped table. One places an ordinary cigar
(flat at one end, pointed at the other) on the table, then the other does the
same, and so on alternately, a condition being that no cigar shall touch
another. Which player should succeed in placing the last cigar, assuming that
they each will play in the best possible manner? The size of the table top and
the size of the cigar are not given, but in order to exclude the ridiculous
answer that the table might be so diminutive as only to take one cigar, we will
say that the table must not be less than 2 feet square and the cigar not more
than 4˝ inches long. With those restrictions you may take any dimensions you
like. Of course we assume that all the cigars are exactly alike in every
respect. Should the first player, or the second player, win?
Solution
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