Subj:.....The
Fighting Fishes Of Siam (S631)
From the book
"Mathematical Puzzles of Sam Loyd"
Edited by Martin Gardner
From: Dover Publications in 1959
How
long will it take one species
of
fish to vanquish the other?
The people of Siam
are natural born gamblers who would bet
their last vestige
of clothing upon any event which offers
a chance to win or
lose. They are not especially belli-
gerent themselves,
but they love to witness a fight between
any other creatures
from toads to elephants. Dog-fights,
or cocking mains
are of daily occurence and are conducted
pretty much according
to the recognized lines of civilized
countries, but in
no other land upon the globe is it possible
to witness a fish
fight!
They have two kinds
of fish which, despite their being very
choice food, are
raised and valued solely for their fighting
qualities.
The one is a large white perch known as the king
fish, and the other
is the little black carp or devil fish.
Such antipathy exists
between these two species that they
attack each other
on sight and battle to the death.
A king fish could
readily dispose of one or two of the little
fish in just a few
seconds, but the devil fish are so agile
and work together
so harmoniously that three of the little
fellows would just
equal one of the big ones, and they would
battle for hours
without any results. So cleverly and
scientifically do
they carry on their line of attack that
four of the little
fellows would kill a large one in just
three minutes and
five would administer the coup de grace
proportionately quicker.
(E.g., five would kill one king
fish in two minutes
and 24 seconds, six in two minutes,
and so on.)
These combinations
of adverse forces are so accurate and
reliable that when
a fish tournament is arranged, one can
calculate the exact
time it will take a given number of one
kind to vanquish
a certain number of the enemy.
By way of illustration
a problem is presented in wiich four
of the king fish
oppose thirteen of the little fighters.
Who should win?
And how long should it take one side to
annihilate the other?
[To avoid an ambiguity
in Loyd's statement of the problem,
it should be made
clear that the devil fish always attack
single king fish
in groups of three or more, and stay
with the large fish
until he is disposed of. We cannot,
for example, assume
that while the twelve little fish hold
the four large fish
at bay, the thirteenth devil fish darts
back and forth to
finish off the large fish by attacking
all of them simultaneously.
If we permit fractions, so to
speak, of devil fish
to be effective then we can reason that
if four devils kill
a king in three minutes, thirteen devils
will finish a king
in 12/13 minutes, or four kings in 48/13
minutes (3 minutes,
41 and 7/13 seconds). But this same
line of reasoning
would lead to conclude that twelve devils
would kill one king
in one minute, or four kings in four
minutes, even without
the aid of the thirteenth little fish
- a conclusion that
clearly violates Loyd's assumption that
three little fish
are unable to kill one devil fish. - M.G.]
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Finger pointing down
from darrell94590 on 1/2/2006 |
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...
THE SOLUTION
There would certainly
have been a battle royal in the
Siamese aquariun
had there been as many fishes in that
fight as I have received
answers to this problem, and
all maintaining such
different views!
For clearness and
simplicity, I am inclined to accept
the following decision
of the time-keeper as being
correct:
Three of the little
fish paired off with each of three
big fish, engaging
their attention while the other four
little fighters polish
off the fourth big fish in just
three minutes.
Then five little fellows tackled one big
fish and kill him
in 2 minutes and 24 seconds, while the
other little ones
were battling with the other big ones.
It it evident that
if the remaining two groups had been
assisted by one more
fighter they would all have finished
in the same time,
so there is only sufficient resistance
left in each of the
big ones to call for the attention of
a little fish for
2 minutes and 24 seconds. Therefore if
seven now attack
instead of one, they would do it in one-
seventh of that time,
or 20 and 4/7 of a second.
In dividing the little
fish forces against the remaining
two big ones - one
would be attacked by seven and the
other by six - the
last fish at the end of 20 and 4/7
seconds would still
require the punishment which one little
one could adninister
in that time. The whole thirteen
little fellows, concentrating
their attack, would give the
fish his quietus
in one-thirteenth that time, or 1 and
53/91 seconds.
Adding up the totals
of the time given in the several
rounds - 3 minutes,
2 minutes and 24 seconds, 20 and
4/7 seconds, and
1 and 53/91 seconds, we have 5 minutes,
46 and 2/13 seconds
as the entire time consumed in the
battle. |
