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Archive-name: puzzles/archive/logic/part5
Last-modified: 17 Aug 1993
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```==> logic/smullyan/priest.p <==
In a small town there are N married couples in which one of the pair
dalliance a secret from their spouse.  Since it is a small town,
everyone knows about everyone else's infidelity.  In other words, each
spouse of an adulterer thinks there are N - 1 adulterers, but everyone
else thinks there are N adulterers.

People of this town have a tradition of denouncing their spouse in
church if they are guilty of adultery.  So far, of course, no one has
been denounced.  In addition, people of this town are all amateur
logicians of sorts, and can be expected to figure out the implications
of anything they know.

A priest has heard the confession of all the people in the town, and is
troubled by the state of moral turpitude.  He cannot break the
confessional, but knowing of his flock's logical turn of mind, he hits
upon a plan to do God's work.  He announces in Mass one Sunday that
there is adultery in the town.

Is the priest correct?  Will this result in every adulterer being denounced?

==> logic/smullyan/priest.s <==
Yes.  Let's start with the simple case that N = 1.  The offended spouse
reasons as follows: the priest knows there is at least one adulterer,
but I don't know who this person is, and I would if it were anyone
other than me, so it must be me.  What happens if N = 2?  On the first
Sunday, the two offended spouses each calmly wait for the other to get
up and condemn their spouses.  When the other doesn't stand, they
think:  They do not think that they are a victim.  But if they do not
think they are victims, then they must think there are no adulterers,
contrary to what the priest said.  But everyone knows the priest speaks
with the authority of God, so it is unthinkable that he is mistaken.
The only remaining possibility is that they think there WAS another
adulterer, and the only possibility is: MY SPOUSE!  So, they know that
they too must be victims.  So on the next Sunday, they will get up.
What if N = 3?  On the first Sunday, each victim believes that the other
two will not get up because they did not know about the other person
(in other words, they believe that each of the two other victims thought
there was only one adulterer).  However, each victim reasons, the two
will now realize that there must be two victims, for the reasons given
under the N = 2 case above.  So they will get up next Sunday.  This
excuse lasts until the next Sunday, when still no one gets up, and now
each victim realizes that either the priest was mistaken (unthinkable!)
or there are really three victims, and I am ONE!  So, on the third
Sunday, all three get up.  This reasoning can be repeated inductively
to show that no one will do anything (except use up N - 1 excuses as to
why no one got up) until the Nth Sunday, when all N victims will arise
in unison.

The induction can also be run "top down" on the priest's statement.  What
must everyone believe about what everyone else believes?
N victims of adultery believe there are N - 1 victims, and that all of these
N - 1 victims believe that there are N - 2 victims, and that all of these
N - 2 victims believe that there are N - 3 victims, and that all of these
...
2 victims believe that there is 1 victim, and that this
1 victim believes there are no victims.
Suppose the priest says, "There are N adulterers in this town."  Then
all the adulterers will know immediately that their spouses have been
unfaithful, and will denounce them the next Sunday.  Now, suppose the
priest says, "There are at least N - 1 adulterers in this town."  On
the first Sunday, the offended spouses all wait for each other to stand
up.  When none do, they all know that they have all been horribly
mistaken, and they stand up on the following Sunday.  But what if the
priest says, "There are at least N - 2 adulterers in this town."  On
the first Sunday, the victims reason, those N - 1 victims are going to
be surprised when no one stands up, and they'll stand up next Sunday.
On the second Sunday, the victims reason, wait a minute, since they
didn't stand up, I must be one of the deluded victims.  And everyone
stands up on the third Sunday.  This resoning applies inductively,
adding one Sunday at each step, until the priest's original statement
is reached, which takes N Sundays for everyone to unravel.

By the way, the rest of the town, which thinks there are N adulterers,
is about to conclude that their perfectly innocent spouses have been
conclude that the door swings both ways.  So the priest is playing a
dangerous game.  A movie plot in there somewhere?

This problem is an analogue of the "dirty children" problem discussed in
the seminal paper on common knowledge by Halpern and Moses (JACM 1990).

If the information of each victim is less than perfect, the problem is
related to the "frame" problem of AI, cf. J. M. McCarthy & P. J. Hayes,
"Some philosophical problems from the standpoint of artificial intelligence"
(1968) in _Readings in Artificial Intelligence_ (pp. 431-450),
Tioga Publishing Co., Palo Alto, CA (1981).

==> logic/smullyan/stamps.p <==
The moderator takes a set of 8 stamps, 4 red and 4 green, known to the
logicians, and loosely affixes two to the forehead of each logician so that
each logician can see all the other stamps except those 2 in the moderator's
pocket and the two on her own head.  He asks them in turn
if they know the colors of their own stamps:
A: "No"
B: "No"
C: "No"
A: "No
B: "Yes"
What are the colors of her stamps, and what is the situation?

==> logic/smullyan/stamps.s <==
B says: "Suppose I have red-red. A would have said on her
second turn: 'I see that B has red-red. If I also have red-red, then all
four reds would be used, and C would have realized that she had green-green.
But C didn't, so I don't have red-red.  Suppose I have green-green. In that
case, C would have realized that if she had red-red, I would have seen
four reds and I would have answered that I had green-green on my first
turn.  On the other hand, if she also has green-green [we assume that
A can see C; this line is only for completeness], then B would have seen
four greens and she would have answered that she had two reds.  So C would
have realized that, if I have green-green and B has red-red, and if
neither of us answered on our first turn, then she must have green-red.
"'But she didn't. So I can't have green-green either, and if I can't have
green-green or red-red, then I must have green-red.'
So B continues: "But she (A) didn't say that she had green-red, so
the supposition that I have red-red must be wrong.  And as my logic applies
to green-green as well, then I must have green-red."
So B had green-red, and we don't know the distribution of the others
certainly.
(Actually, it is possible to take the last step first, and deduce
that the person who answered YES must have a solution which would work
if the greens and reds were switched -- red-green.)

You have an empty urn, and an infinite number of labeled balls.  Each
has a number written on it corresponding to when it will go in.  At a
minute to the hour, you take the first ten balls and put them in the
urn, and remove the last ball.  At the next half interval, you put in
the next ten balls, and remove ball number 20.  At the next half
interval, you put in ten more balls and remove ball 30.  This continues
for the whole minute.... how many balls are in the urn at this point?
(infinite)

You have the same urn, and the same set of balls.  This time, you put
in 10 balls and remove ball number 1.  Then you put in another ten
balls and remove ball number 2.  Then you put in another ten balls and
remove ball number 3.  After the minute is over, how many balls are
left in the urn now? (zero)

Are the above answers correct, and why or why not?

Almost all people will intuitively feel that the first experiment
(where only balls labeled with multiples of 10 are removed) results in
an urn with an infinite number of balls.

The real excitement starts with the experiment where balls are removed
in increasing order, but 10 times slower than they are added.  Some
feel that the urn will not get empty, due to the slowness of removing.
Some others feel that the urn does get empty, since each ball is
removed at some time during the experiment.  The remaining people claim
that the experiment is not well defined, that it is not possible to do
something an infinite number of times, or something similar,
effectively dismissing the experiment.

Just to put a bit of doubt in some peoples mind, I will add a third experment:

Let us suppose that at 1 minute to 12 p.m. balls nummbered 1 through 9
are placed in the urn, and instead of withdrawing a ball we add a zero
to the label of ball number 1 so that its label becomes 10. At 1/2 minute
to 12 p.m., balls numbered 11 through 19 are placed in the urn, and we
add a zero to the label of ball number 2 so that it becomes ball number 20.
At 1/4 minute to 12 p.m., balls numbered 21 through 29 are placed in the
urn and ball number 3 becomes ball number 30, and so on.
At each instant, instead of withdrawing the ball with the smallest label
we add a zero to its label so that its number is multiplied by 10.
How many balls are in the urn at 12 p.m. and what are their labels?

If we look at this experiment, at any point in time the inside of the
urn looks exactly like the inside during the execution of the original
paradoxical experiment. However, since no balls leave the urn, it is
now impossible to conclude that the urn will be empty at 12 p.m.
Still, there is no natural number that is the label of any ball in the
urn.  Instead, each ball in the urn will have as its lable a natural
number followed by an infinite number of zero's.

A possible question is now: does this support that the outcome of the
original experiment where balls are removed in increasing order is that
there are an infinite number of balls in the urn? Possibly also with
'infinite natural numbers' as their labels, or are these experiments so
different that the answer is still a clear 'zero'?

I now come to the main points.
1. Our normal mathematical models do not cater for the COMPLETION of infinite
2. Since we intuitively feel that for many of these experiments there
are obvious outcomes, we would like to enhance our model to describe the
outcomes of these experiments.
3. In the enhancement of the model continuity should play an important role.

We include statement 3, since a model in which the conclusion of all
these experiments is that, at 12 p.m. the urn contains "exactly 7
balls, all red" is not desirable, nor useful.

It can be easily shown that general continuity is unattainable. For
instance the sentence "it is before midnight" is true during the
experiment, but is suddenly false after the experiment.

The people claiming that in the second experiment the urn will contain
an infinite number of balls, base this on the fact that the number of
balls in the urn during the experiment, is 9n at (1/2)^(n-1) minute
before 12. They thus assume that this statement is continuous. This
remains to be seen, however.

We have not come to a clear set of criteria which decide whether a
given statement is continuous with respect to performing supertasks. We
did define a "kinematical principle of continuity", which is roughly
formalised as:

If at some moment before 12 p.m. a ball comes to rest at a particular
position, which it does not leave till 12 p.m., then it is still at
that position at 12 p.m.

If we look at the three experiments mentioned, then we can see that in
each case we can come to a conclusion on the contents of the urn.

1. In the first experiment, with the 10-folds being removed, each ball
which number is a multiple of 10 comes to rest outside the urn (just
after being removed) and thus is outside the urn at 12 p.m. All other
balls come to rest inside the urn (just after being placed there), and
thus are inside the urn at 12 p.m.  Therefore the urn contains an infinite
number of balls at 12 p.m.

2. In the second experiment, with the balls being removed in increasing order,
each balls comes to rest outside the urn. Thus all balls involved are not
in the urn. Thus the urn is empty.

3. In the third experiment, all balls come to rest inside the urn and thus the
urn contains an infinite number of balls. The labels of these balls are
naturall number followed by an infinite number of zero's (since each of the
numbers is not changed, and zero's once added remain at the label, we can
draw this conclusion).

The first and third experiment are rather straightforward, while the
one way of extending our model to include super tasks.  We have only
shown that for these experiments, in our model, we come to consistent
conclusions. It does not mean that there are no other models which lead
to different, but also, within that model, consistent solutions.

A final remark: while thinking about these matters, we have wondered
whether we could create a model in which the second experiment would
lead to an urn containing an infinite number of balls. A possibility is
assuming that if a position is continuously occupied by a ball,
although the occupant ball may be swapped every now and again for
another ball, that at 12 p.m. the position is occupied by a so-called
LIMIT BALL.  For the second experiment we could than place balls 1, 10,
100 ..  2, 20, 200, .. each at its own spot in the urn. Each spot in
the urn, once occupied is than continuously occupied with a ball,

This idea of continuity is stronger than the kinematic principle
suggested above, and we have not followed these ideas up enough to
decide whether this extended principle can be made consistent. If any
of the readers have feelings whether this can or cannot be done, I
would be interested to hear their arguments.

I conclude by stating that the result of the super task depends on how
our standard models are enlarged to include the execution of
results for the supertasks suggested by Ross. Other models may lead to
different, but also consistent, conclusions.

Reference:

Victor Allis and Teunis Koetsier (1991).
On Some Paradoxes of the Infinite.
Brit. J. Phil. Sci. 42 pp. 187-194.

-- allis@cs.rulimburg.nl (Victor Allis)

I am interested in the origin of the puzzle.  As far as I know in this
form the puzzle occurs for the first time in Littlewood's "Mathematical
Miscellanea", which is an amusing little booklet from the 1950s (it may
be even older). Littlewood does not discuss the puzzle.  DOES ANYONE
KNOW OF EARLIER REFERENCES TO THIS PUZZLE?  The puzzle also occurs in
S. Ross's "A first course in probability", New York and London, 1988,
without critical comment.

-- teun@cs.vu.nl (Teun Koetsier)

==> logic/timezone.p <==
Two people are talking long distance on the phone; one is in an East-
Coast state of the US, the other is in a West-Coast state of the US.
The first asks the other "What time is it?", hears the answer, and
says, "That's funny.  It's the same time here!"

==> logic/timezone.s <==
One is in Eastern Oregon (in Mountain time), the other in
Western Florida (in Central time), and it's daylight-savings
changeover day at 1:30 AM.

==> logic/unexpected.p <==
Swedish civil defense authorities announced that a civil defense drill would
be held one day the following week, but the actual day would be a surprise.
However, we can prove by induction that the drill cannot be held.  Clearly,
they cannot wait until Friday, since everyone will know it will be held that
day.  But if it cannot be held on Friday, then by induction it cannot be held
on Thursday, Wednesday, or indeed on any day.

What is wrong with this proof?

==> logic/unexpected.s <==
This problem has generated a vast literature (see below).  Several
solutions of the paradox have been proposed, but as with most paradoxes
there is no consensus on which solution is the "right" one.

The earliest writers (O'Connor, Cohen, Alexander) see the announcement as
simply a statement whose utterance refutes itself.  If I tell you that I
will have a surprise birthday party for you and then tell you all the
details, including the exact time and place, then I destroy the surprise,
refuting my statement that the birthday will be a surprise.

Soon, however, it was noticed that the drill could occur (say on Wednesday),
and still be a surprise.  Thus the announcement is vindicated instead of
being refuted.  So a puzzle remains.

One school of thought (Scriven, Shaw, Medlin, Fitch, Windt) interprets
the announcement that the drill is unexpected as saying that the date
of the drill cannot be deduced in advanced.  This begs the question,
deduced from which premises?  Examination of the inductive argument
shows that one of the premises used is the announcement itself, and in
particular the fact that the drill is unexpected.  Thus the word
"unexpected" is defined circularly.  Shaw and Medlin claim that this
circularity is illegitimate and is the source of the paradox.  Fitch
uses Godelian techniques to produce a fully rigorous self-referential
announcement, and shows that the resulting proposition is
self-contradictory.  However, none of these authors explain how it can
be that this illegitimate or self-contradictory announcement
nevertheless appears to be vindicated when the drill occurs.  In other
words, what they have shown is that under one interpretation of "surprise"
the announcement is faulty, but their interpretation does not capture the
intuition that the drill really is a surprise when it occurs and thus
they are open to the charge that they have not captured the essence of

Another school of thought (Quine, Kaplan and Montague, Binkley,
Harrison, Wright and Sudbury, McClelland, Chihara, Sorenson) interprets
"surprise" in terms of "knowing" instead of "deducing."  Quine claims
that the victims of the drill cannot assert that on the eve of the last
day they will "know" that the drill will occur on the next day.  This
blocks the inductive argument from the start, but Quine is not very
explicit in showing what exactly is wrong with our strong intuition
that everybody will "know" on the eve of the last day that the drill
will occur on the following day.  Later writers formalize the paradox
using modal logic (a logic that attempts to represent propositions
knowing are at fault, e.g., the axiom that if one knows something, then
one knows that one knows it (the "KK axiom").  Sorenson, however,
formulates three ingenious variations of the paradox that are
independent of these doubtful axioms, and suggests instead that the
problem is that the announcement involves a "blindspot":  a statement
that is true but which cannot be known by certain individuals even if
they are presented with the statement.  This idea was foreshadowed by
O'Beirne and Binkley.  Unfortunately, a full discussion of how this
blocks the paradox is beyond the scope of this summary.

Finally, there are two other approaches that deserve mention.  Cargile
interprets the paradox as a game between ideally rational agents and finds
fault with the notion that ideally rational agents will arrive at the same
conclusion independently of the situation they find themselves in.  Olin
interprets the paradox as an issue about justified belief: on the eve of
the last day one cannot be justified in believing BOTH that the drill will
occur on the next day AND that the drill will be a surprise even if both
statements turn out to be true; hence the argument cannot proceed and the
drill can be a surprise even on the last day.

are Bennett-Cargile and both papers of Sorenson.  All of these provide
overviews of previous work and point out some errors, and so it's helpful to
"deducibility" side, Shaw, Medlin and Fitch are good representatives.  Other
papers that are definitely worth reading are Quine, Binkley, and Olin.

D. O'Connor, "Pragmatic Paradoxes," Mind 57:358-9, 1948.
L. Cohen, "Mr. O'Connor's 'Pragmatic Paradoxes,'" Mind 59:85-7, 1950.
P. Alexander, "Pragmatic Paradoxes," Mind 59:536-8, 1950.
M. Scriven, "Paradoxical Announcements," Mind 60:403-7, 1951.
D. O'Connor, "Pragmatic Paradoxes and Fugitive Propositions," Mind 60:536-8,
1951
P. Weiss, "The Prediction Paradox," Mind 61:265ff, 1952.
W. Quine, "On A So-Called Paradox," Mind 62:65-7, 1953.
R. Shaw, "The Paradox of the Unexpected Examination," Mind 67:382-4, 1958.
A. Lyon, "The Prediction Paradox," Mind 68:510-7, 1959.
D. Kaplan and R. Montague, "A Paradox Regained," Notre Dame J Formal Logic
1:79-90, 1960.
G. Nerlich, "Unexpected Examinations and Unprovable Statements," Mind
70:503-13, 1961.
M. Gardner, "A New Prediction Paradox," Brit J Phil Sci 13:51, 1962.
K. Popper, "A Comment on the New Prediction Paradox," Brit J Phil Sci 13:51,
1962.
B. Medlin, "The Unexpected Examination," Am Phil Q 1:66-72, 1964.
F. Fitch, "A Goedelized Formulation of the Prediction Paradox," Am Phil Q
1:161-4, 1964.
R. Sharpe, "The Unexpected Examination," Mind 74:255, 1965.
J. Chapman & R. Butler, "On Quine's So-Called 'Paradox,'" Mind 74:424-5, 1965.
J. Bennett and J. Cargile, Reviews, J Symb Logic 30:101-3, 1965.
J. Schoenberg, "A Note on the Logical Fallacy in the Paradox of the
Unexpected Examination," Mind 75:125-7, 1966.
J. Wright, "The Surprise Exam: Prediction on the Last Day Uncertain," Mind
76:115-7, 1967.
J. Cargile, "The Surprise Test Paradox," J Phil 64:550-63, 1967.
R. Binkley, "The Surprise Examination in Modal Logic," J Phil 65:127-36,
1968.
C. Harrison, "The Unanticipated Examination in View of Kripke's Semantics
for Modal Logic," in Philosophical Logic, J. Davis et al (ed.), Dordrecht,
1969.
P. Windt, "The Liar in the Prediction Paradox," Am Phil Q 10:65-8, 1973.
A. Ayer, "On a Supposed Antinomy," Mind 82:125-6, 1973.
M. Edman, "The Prediction Paradox," Theoria 40:166-75, 1974.
J. McClelland & C. Chihara, "The Surprise Examination Paradox," J Phil Logic
4:71-89, 1975.
C. Wright and A. Sudbury, "The Paradox of the Unexpected Examination,"
Aust J Phil 55:41-58, 1977.
I. Kvart, "The Paradox of the Surprise Examination," Logique et Analyse
337-344, 1978.
R. Sorenson, "Recalcitrant Versions of the Prediction Paradox," Aust J Phil
69:355-62, 1982.
D. Olin, "The Prediction Paradox Resolved," Phil Stud 44:225-33, 1983.
R. Sorenson, "Conditional Blindspots and the Knowledge Squeeze: A Solution to
the Prediction Paradox," Aust J Phil 62:126-35, 1984.
C. Chihara, "Olin, Quine and the Surprise Examination," Phil Stud 47:191-9,
1985.
R. Kirkham, "The Two Paradoxes of the Unexpected Hanging," Phil Stud
49:19-26, 1986.
D. Olin, "The Prediction Paradox: Resolving Recalcitrant Variations," Aust J
Phil 64:181-9, 1986.
C. Janaway, "Knowing About Surprises: A Supposed Antinomy Revisited," Mind
98:391-410, 1989.

-- tycchow@math.mit.edu.

==> logic/verger.p <==
A very bright and sunny Day
The Priest did to the Verger say:
"Last Monday met I strangers three
None of which were known to Thee.
I ask'd Them of Their Age combin'd
which amounted twice to Thine!
A Riddle now will I give Thee:
Tell Me what Their Ages be!"

So the Verger ask'd the Priest:
"Give to Me a Clue at least!"
"Keep Thy Mind and Ears awake,
And see what Thou of this can make.
Their Ages multiplied make plenty,
Fifty and Ten Dozens Twenty."

The Verger had a sleepless Night
To try to get Their Ages right.
"I almost found the Answer right.
Please shed on it a little Light."
"A little Clue I give to Thee,
I'm older than all Strangers three."
After but a little While
The Verger answered with a Smile:
"Inside my Head has rung a Bell.
Now I know the answer well!"

Now, the question is:

How old is the PRIEST??
======

==> logic/verger.s <==
The puzzler tried to take the test;
Intriguing rhymes he wished to best.
But "Fifty and ten dozens twenty"
When he finally found some leisure,
He took to task this witty treasure.

"The product of the age must be
Twenty-Four Hundred Fifty!"
Knowing that, he took its primes,
permuted them as many times
as needed, til he found amounts
equal to, by all accounts,
twice the Verger's age, so that
He would have that next day's spat.

The reason for the lad's confusion
was due to multiple solution!
Hence he needed one more clue
to give the answer back to you!
Since only one could fit the bill,
and then confirm the priest's age still,
the eldest age of each solution
by one could differ, with no coercion.  <=(Sorry)

Else, that last clue's revelation
would not have brought information!
With two, two, five, seven, and seven,
construct three ages, another set of seven.
Two sets of three yield sixty-four,
Examine them, yet one time more.
The eldest age of each would be
forty-nine, and then, fifty!

With lack of proper rhyme and meter,
I've tried to be the first completor
of this poem and a puzzle;
my poetry, you'd try to muzzle!
And lest you think my wit is thrifty,
The answer, of course, must be fifty!
If dispute, you wish to tender,
note my addresss, as the sender!

--
Kevin Nechodom <knechod@stacc.med.utah.edu>

==> logic/weighing/balance.p <==
You are given 12 identical-looking coins, one of which is counterfeit
and weighs slightly more or less (you don't know which) than the
others.  You are given a balance scale which lets you put the same
number of coins on each side and observe which side (if either) is
heavier.  How can you identify the counterfeit and tell whether it
is heavy or light, in 3 weighings?

More generally, you are given N coins, one of which is heavy or light.

==> logic/weighing/balance.s <==
Martin Gardner gave a neat solution to this problem.

Assume that you are allowed W weighings.  Write down the 3^W possible
length W strings of the symbols '0', '1', and '2'.  Eliminate the three
such strings that consist of only one symbol repeated W times.

For each string, find the first symbol that is different from the symbol
preceeding it.  Consider that pair of symbols.  If that pair is not 01,
12, or 20, cross out that string.  In other words, we only allow strings
of the forms 0*01.*, 1*12.*, or 2*20.* ( using ed(1) regular expressions ).

You will have (3^W-3)/2 strings left.  This is how many coins you can
handle in W weighings.

Perform W weighings as follows:

For weighing I, take all the coins that have a 0 in string
position I, and weigh them against all the coins that have
a 2 in string position I.

If the side with the 0's in position I goes down, write down
a 0.  If the other side goes down, write down a 2.  Otherwise,
write down a 1.

After the W weighings, you have written down an W symbol string.  If
your string matches the string on one of the coins, then that is the
odd coin, and it is heavy.  If none of them match, than change every
2 to a 0 in your string, and every 0 to a 2.  You will then have a
string that matches one of the coins, and that coin is lighter than
the others.

Note that if you only have to identify the odd coin, but don't have to
determine if it is heavy or light, you can handle (3^W-3)/2+1 coins.
Label the extra coin with a string of all 1's, and use the above
method.

Note also that you can handle (3^W-3)/2+1 coins if you *do* have to
determine whether it is heavy or light, provided you have a single reference
coin available, which you know has the correct weight. You do this by
labelling the extra coin with a string of all 2s. This results in it being
placed on the same side of the scales each time, and in that side of the
scales having one more coin than the other each time. So you put the
reference coin on the other side of the scales to the "all 2s" coin on each
weighing.

Proving that this works is straightforward, once you notice that the
method of string construction makes sure that in each position, 1/3
of the strings have 0, 1/3 have 1, and 1/3 have 2, and that if a
string occurs, then the string obtained by replacing each 0 with a
2 and each 2 with a 0 does not occur.

If you already know the odd coin is heavy (or light), you can handle
3^W coins.  Given W weighings, there can only be 3^W possible
combinations of balances, left pan heavy, and right pan heavy.

The algorithm in this case:

Divide the coins into three equal groups... A, B, and C. Weigh A
against B. If a pan sinks, it contains the heavy coin, otherwise, the
heavy coin is in group C. If your group size is 1, you've found the coin,
otherwise recurse on the group containing the heavy coin.

==> logic/weighing/box.p <==
You have ten boxes; each contains nine balls.  The balls in one box
weigh 0.9 kg; the rest weigh 1.0 kg.  You have one weighing on an
accurate scale to find the box containing the light balls.  How do you
do it?

==> logic/weighing/box.s <==
Number the boxes 0-9.  Take 0 balls from box 0, 1 ball from box 1, 2
balls from box 2, etc.  Now weigh all those balls and follow this
table:

If odd box is	Weight is
0		45 kg
1		44.9 kg
2		44.8 kg
3		44.7 kg
4		44.6 kg
5		44.5 kg
6		44.4 kg
7		44.3 kg
8		44.2 kg
9		44.1 kg

==> logic/weighing/find.median.p <==
What is the least number of pairwise comparisons needed to find the median of
2n+1 distinct real numbers?

==> logic/weighing/find.median.s <==
Consider median of three numbers {a,b,c}.
Let G[a,b]=max(a,b) and L[a,b]=min(a,b)
If we assume that median of {a,b,c} can be computed by two
comparisons, then the solution must be one of the following:
G[c,G[a,b]], G[c,L[a,b]], L[c,G[a,b]], L[c,L[a,b]].
However, it is easily seen that none of these provides
a solution. Therefore, we need more than two comparisons to
get median of three numbers.

Now, consider median of 5 numbers {a,b,c,d,e}.
There are two possible ways to start the computation.
Let Ci[a,b] denote the ith comparison between {a} and {b}.
We ignore other trivialities such as C1[a,a] or C2[a,C1[a,b]].

In the first case, the next operation is to find
median of S={e,C1[a,b],C2[c,d]} which requires at least three
So the total cost of the first approach is at least 5 comparisons.
However, if the median is not equal to {e} then we can always
choose C1 and C2 such that the median is not in S.

In the second case, the next step is to find median of S={C2[a,C1[b,c]],d,e}.
Again, the total cost of this approach is at least five comparisons.
If median is not equal to {d} or {e}, we can again always
choose C1 and C2 such that the median is not in S.

Other starting sets, such as {e,d,c,b,a}, can always be ordered
as {a,b,c,d,e}. This shows that the argument covers all possible cases.

Navid,

==> logic/weighing/gummy.bears.p <==
Real gummy drop bears have a mass of 10 grams, while imitation gummy
drop bears have a mass of 9 grams.  Spike has 7 cartons of gummy drop bears,
4 of which contain real gummy drop bears, the others imitation.
Using a scale only once and the minimum number of gummy drop bears, how
can Spike determine which cartons contain real gummy drop bears?

==> logic/weighing/gummy.bears.s <==
Spike uses 51 gummy drop bears:  from the 7 boxes he takes respectively
0, 1, 2, 4, 7, 13, and 24 bears.

The notion is that each box of imitation bears will subtract its
number of bears from the total "ideal" weight of 510 grams (1 gram of
missing weight per bear), so Spike weighs the bears, subtracts the
result from 510 to obtain a number N, and finds the unique combination
of 3 numbers from the above list (since there are 3 "imitation" boxes)
that sum to N.

The trick is for the sums of all triples selected from the set S of
numbers of bears to be unique.  To accomplish this, I put numbers into
S one at a time in ascending order, starting with the obvious choice,
0.  (Why is this obvious?  If I'd started with k > 0, then I could
have improved on the resulting solution by subtracting k from each
number) Each new number obviously had to be greater than any previous,
because otherwise sums are not unique, but also the sums it made when
paired with any previous number had to be distinct from all previous
pairs (otherwise when this pair is combined with a third number you
can't distinguish it from the other pair)--except for the last box,
where we can ignore this point.  And most obviously all the new
triples had to be distinct from any old triples; it was easy to find
what the new triples were by adding the newest number to each old sum
of pairs.

Now, in case you're curious, the possible weight deficits and their
unique decompositions are:

3       = 0 + 1 + 2
5       = 0 + 1 + 4
6       = 0 + 2 + 4
7       = 1 + 2 + 4
8       = 0 + 1 + 7
9       = 0 + 2 + 7
10      = 1 + 2 + 7
11      = 0 + 4 + 7
12      = 1 + 4 + 7
13      = 2 + 4 + 7
14      = 0 + 1 + 13
15      = 0 + 2 + 13
16      = 1 + 2 + 13
17      = 0 + 4 + 13
18      = 1 + 4 + 13
19      = 2 + 4 + 13
20      = 0 + 7 + 13
21      = 1 + 7 + 13
22      = 2 + 7 + 13
24      = 4 + 7 + 13
25      = 0 + 1 + 24
26      = 0 + 2 + 24
27      = 1 + 2 + 24
28      = 0 + 4 + 24
29      = 1 + 4 + 24
30      = 2 + 4 + 24
31      = 0 + 7 + 24
32      = 1 + 7 + 24
33      = 2 + 7 + 24
35      = 4 + 7 + 24
37      = 0 + 13 + 24
38      = 1 + 13 + 24
39      = 2 + 13 + 24
41      = 4 + 13 + 24
44      = 7 + 13 + 24

Note that there had to be (7 choose 3) distinct values; they end up
ranging from 3 to 44 inclusive with 7 numbers missing:  4, 23, 34, 36,
40, 42, and 43.

-- David Karr (karr@cs.cornell.edu)

==> logic/weighing/optimal.weights.p <==
What is the smallest set of weights that allow you to weigh on a
balance scale every integer number of kilograms up to some number N?

==> logic/weighing/optimal.weights.s <==
a) EQUATION
-----------
Obviously (I hate this word! :-) any weight Y that can be weighed
by X1, X2, ... Xn can be written as:

Y = A1*X1 + A2*X2 + .... + An*Xn

where Ai is equal to -1, or 0, or 1.

b) UPPER BOUND FOR Y(n)
-----------------------
Each Ai can have one of three values, so the total number of
combinations of values for A1,A2, ... An is 3^n. At least one of these
combinations gives Y=0 (A1=A2=...=An=0). Out of the remaining 3^n-1
combinations, some give a negative Y (for example A1=A2=...=An=-1).
and some a positive Y (and some might also give zero values, e.g. if
X1=X2, and A1=1, A2=-1).
Because of symmetry it's easy to see that the combinations that give Y>0
are at most half i.e. (3^n-1)/2. It is also possible that different
combinations might give the same value of Y, and it is also possible
that these values of Y are not successive.
However, to obtain an upper bound, lets assume the best case i.e.
all (3^n-1)/2 combinations give different, positive, and successive
values, so:
Y(n) <= (3^n-1)/2

c) AN OPTIMAL ALGORITHM FOR CHOOSING Xn
---------------------------------------
I will present an algorithm for choosing the weights X1,X2,...Xn.
Then we will prove that it is optimal.

For n=1, we choose X1=1, and of course Y(1) = 1.

Let's assume that we have already chosen n weights X1, X2 ... Xn,
and that we can weigh all Y where 1<=Y<=Y(n).
We are allowed to get an extra new weight Xn+1.
If we choose:
Xn+1 = 2*Y(n)+1
then we get
Y(n+1) = Y(n) + Xn+1 = 3*Y(n)+1

Proof:
for 1<= Y <= Y(n):
use the weights X1...Xn (ignoring the new one)
for Y(n)+1 <= Y < Xn+1 = 2*Y(n)+1
Put Xn+1 on one side of the scale, and on the other side put
the unknown weight, and a combination of X1...Xn so that
this combination weighs "Xn+1 - Y" (which is a number
in the range 0...Y(n), so *there exists* such a combination)
for 2*Y(n)+1 <= Y <= 3*Y(n)+1:
put the unknown weight on one side, and on the other side
put Xn+1, and combination of X1...Xn with a weight equal to
"Y - Xn+1" (which again is a number in the range 0...Y(n),
so there exists such a combination)

So, to summarize, if we use such an algorithm, we have:
X1 = 1;
Y(1) =1;

Xn+1 = 2*Y(n)+1
Y(n+1) = Y(n) + Xn+1 = 3*Y(n) + 1

It's easy to prove (e.g. by induction) that:
Y(n) = (3^n-1)/2
X(n) = 3^n

So, Y(n) is equal to the upper bound we found before, so this algorithm is
optimal, and the weights you must choose are powers of 3.

Spyros Potamianos
potamian@hpl.hp.com

==> logic/weighing/weighings.p <==
Some of the supervisors of Scandalvania's n mints are producing bogus coins.
It would be easy to determine which mints are producing bogus coins but,
alas, the only scale in the known world is located in Nastyville,
which isn't on very friendly terms with Scandalville.  In fact, Nastyville's
king will only let you use the scale twice.  Your job?  You must determine
which of the n mints are producing the bogus coins using only two weighings
and the minimum number of coins (your king is rather parsimonious, to put it
nicely).  This is a true scale, i.e. it will tell you the weight of whatever
you put on it.  Good coins are known to have a weight of 1 ounce and it is
also known that all the bogus mints (if any) produce coins that are
light or heavy by the same amount.

Some examples: if n=1 then we only need 1 coin, if n=2 then clearly 2
coins suffice, one from each mint.

What are the solutions for n=3,4,5?  What can be said for general n?

==> logic/weighing/weighings.s <==
Oh gracious and wise king, I have solved this problem by first
simplifying and then expanding.  That is, consider the problem of
being allowed only a single weighing.  Stop reading right now if you
want to think about it further.

There are three possible outcomes for each mint (light, OK, heavy)
which may be represented as (-1, 0, +1).  Now, let each mint represent
one place in base 3.  Thus, the first mint is the ones place, the
second the threes place, the third is the nines place and so on.  The
number of coins from each mint must equal the place.  That is, we'll
have 1 coin from mint 1, 3 from mint 2, 9 from mint 3, and, in
general, 3^(n-1) from mint n.

By weighing all coins at once, we will get a value between 1 + 3 + 9 +
...  and -1 + -3 + -9 + ...  In fact, we notice that that value will
be unique for any mint outcomes.  Thus, for the one weighing problem,
we need

sum for i=1 to n (3^(i-1))

which evaluates to (3^n - 1)/2

I'm fairly satisfied that this is a minimum for a single weighing.
What does a second weighing give us?  Well, we can divide the coins
into two groups and use the same method.  That is, if we have 5 mints,
one weighing will be:

1 coin from mint 1 + 3 coins from mint 2 + 9 coins from mint 3

while the other weighing will be:

1 coin from mint 4 + 3 coins from mint 5

It's pretty plain that this gives us a total coinage of:

3^(n/2) - 1 for even n and, after some arithmetic agitation:
2 * 3^((n-1)/2) - 1 for odd n

I think the flaw in this solution is that we don't know ahead of time
the amount by which the coins are off weight.  So if you weigh 1 coin
from mint 1 together with 3 coins from mint 2 and the result is heavy
by 3x units, you still don't know whether the bogus coins are from
mint 3 (heavy by x units) or from mint 1 (heavy by 3x units).  Note
that we're not given the error amount, only the fact that is is equal
for all bogus coins.

Here is my partial solution:

After considering the above, it would seem that on each of the two
weighings we must include coins from all of the mints (except for the
special cases of small n).  So let ai (a sub i) be the number of coins
from mint i on weighing 1 and bi be the number of coins from mint i on
weighing 2.  Let the error in the bogus coins have a value x, and let
ci be a the counterfeit function: ci is 0 if mint i is good, 1
otherwise.

Then
Sum   ai ci x = delta1		error on weighing 1
Sum   bi ci x = delta2		error on weighing 2

Now the ratio of delta1 to delta2 will be rational regardless of the
value of x, since x will factor out; let's call this ratio p over q (p
and q relatively prime).  We would like to choose { ai } and { bi }
such that for any set of mints J, which will be a subset of { 1 , 2 ,
... , n }, that

Sum aj	( = Sum  ai ci ) is relatively prime to Sum bj.

If this is true then we can determine the error x; it will simply be
delta1/p, which is equal to  delta2/q.

If the { ai } have been carefully chosen, we should be able to figure
out the bogus mints from one of the weighings, provided that
all subsets ( { { aj } over all J } ) have unique sums.
This was the strategy proposed above, where is was suggested
that ai = 3 ** (i-1) ; note that you can use base 2 instead
of base 3 since all the errors have the same sign.

Well, for the time being I'm stumped.

This agrees with the analysis I've been fighting with.  I actually
came up with a pair of functions that "almost" works.  So that the
rest of you can save some time (in case you think the way I did):
Weighing 1: 1 coin from each mint
Weighing 2: 2^(k-1) coins from mint k, for 1...k...n
(total 2^n - 1 coins)

Consider the n mints to be one-bit-each -- bit set -> mint makes bogus
coins.  Then we can just state that we're trying to discover "K",
where K is a number whose bit pattern _just_ describes the bogosity of
each mint.  OK - now, assuming we know 'x', and we only consider the
*difference* of the weighing from what it should be, for weighing 1,
the devaiation is just the Hamming weight of K -- that is the number
of 1-bits in it -- that is, the number of bogosifying mints.  For
weighing 2, the deviation is just K!  When the nth bit of K is set,
then that mint contributes just 2^n to the deviation, and so the total
deviation will just be K.

So that set me in search of a lemma: given H(x) is the hamming weight
of x, is f(x) = x / H(x) a 1-1 map integers into rationals?  That is,
if x/H(x) = y/H(y) can we conclude that x = y?

The answer (weep) is NO.  The lowest pair I could find are 402/603
(both give the ratio 100.5).  Boy it sure looked like a good
conjecture for a while!  Sigh.

There are two parts to the problem. First let us try to come up with a
solution to finding the answer in 2 weighings - then worry about using the
min. number of coins.
Solutions are for GENERAL n.

Let N = set of all mints, 1 to n. Card(N) = n.
Let P = set of all bogus mints. Let Card(P) = p.

Weighing I: Weigh n coins, 1 from each mint.

Since each "good" coins weighs one ounce, let delta1 be the error in weighing.
Since all bogus coins are identical, let delta1 be abs(error).
If x is the weight by which one bogus coin differs from a good coin,
delta1 = p * x.

Weighing II: The coins to be weighed are composed thusly.

Let a1 be the number of coins from mint 1, a2 # from mint2 .. and an from
mint n. All ai's are distinct integers.

Let A = Set of all ai's.

Let delta2 = (abs.) error in weighing 2 = x * k
where k is the number of coins that are bogus in weighing two.
Or more formally
k = sigma(ai)
(over all i in P)

Assuming p is not zero (from Weighing I - in that case go back and get beheaded
Let ratio = delta1/delta2 = p/k.
Let IR = delta2/delta1 = k/p = inverse-ratio (for later proof).

Let S(i) be the bag of all numbers generated by summing i distinct elements
from A. Clearly there will be nCi (that n comb. i) elements in S(i).

[A bag is a set that can have the same element occur more than once.]

So S(1) = A
and S(n) will have one element that is the sum of all the elements of A.

Let R(i) = {x : For-all y in S(i), x = i/y} expressed as p/q (no common
factors).
(R is a bag too).

Let R-A = Bag-Union(R(i) for 1>= i >=n). (can include same element twice)

Choose A, such that all elements of R-A are DISTINCT, i.e. Bag(R-A) = Set(R-A).

Let the sequence a1, a2, .. an, be an L-sequence if the above property is
true. Or more simply, A is in L.

**********************************************************************
CONJECTURE: The bogus mint problem is solved in two weighings if A is in L.

Sketchy proof: R(1) = all possible ratios (= delta1/delta2) when p=1.
R(i) = all possible ratio's when p=i.

Since all possible combinations of bogus mints are reflected in R, just match
the actual ratio with the generated table for n.

************************************************************************
A brief example. Say n=3. Skip to next line if you want.
Let A=(2,3,7).

p=1 possible ratios = 1/2 1/3 1/7
p=2 possible ratios = 2/5 2/9 1/5(2/10)
p=3 possible ratios = 1/4(3/12) (lots of blood in Scandalvania).

As all outcomes are distinct, and the actual ratio MUST be one of these,
match it to the answer, and start sharpening the axe.

Note that the minimum for n=3 is A=(0,1,3)
possible ratios are
p=1 infinity (delta2=0),1,1/3
p=2 2/1,2/3,1/2
p=3 3/4

************************************************************************

All those with the determination to get this far are saying OK, OK how do we
get A.

I propose a solution that will generate A, to give you the answer in two
weighings, but will not give you the optimal number of coins.

Let a1=0

For i>=2 >=n

ai = i*(a1 + a2 + ... + ai-1) + 1

*****************************************
*		i-1			*
*	ai = i* [Sigma(aj)] + 1 	*   ****Generator function G*****
*		j=1			*
*****************************************

If A is L, all RATIO's are unique. Also all inverse-ratio's (1/ratio) are
unique. I will prove that all inverse-ratio's (or IR's) are unique.

Let A(k), be the series generated by the first k elements from eqn. G.(above)

************************************************************************

PROOF BY INDUCTION.

A(1) = {0} is in L.
A(2) = {0,1} is in L.

ASSUME A(k) = {0,1, ..., ak} is in L.

T.P.T. A(k+1) = {0,1, ..., ak, D) is in L where D is generated from G.

We know that all IR's(inverse ratio's)  from A(k) are distinct.

Let K = set of all IR's of A(k).

Since A(k+1) contains A(k), all IR's of A(k) will also be IR's of A(K+1).

So for all P, such that (k+1) is not in P, we get a distinct IR.

So consider cases when (k+1) is in P.

p=1 (i.e. (k+1) = only bogus mint), IR = D

______________________________________________________________________
CONJECTURE: Highest IR for A(k) = max(K) = ak

Proof: 	Since max[A(k)] = ak,
for p'= 1, max IR = ak/1 = ak
for p'= 2, max IR (max sum of 2 ai's)/2
= (ak + ak-1)/2 < ak (as ak>ak-1).
for p'= i max IR sum of largest i elements of A(k)
--------------------------------
i
< i * ak/i = ak.
So max. IR for A(k) is ak.
______________________________________________________________________

D > ak
So for p=1 IR is distinct.

Let Xim be the IR formed by choosing i elements from A(k+1).
Note: We are choosing D and (i-1) elements from A(k).
m is just an index to denote each distinct combination of
(i-1) elemnts of A(i).

______________________________________________________________________
CONJECTURE : For p=j, all new IR's Xjm are limited to the range
D/(j-1) > Xjm > D/j.

Proof:
Xjm = (D + {j-1 elements of A(k)})/j

Clearly Xjm > D/j.

To show: max[Xjm] < D/(j-1)

Note: a1 + a2 .. + ak < D/(k+1)

max[Xjm] = (D + ak + ak-1 +  ... + a(k-j+1))/j
< (D + D/(k+1))/j
= D (k+2)/(k+1)j
= [D/(j-1)] * alpha.

alpha = (j-1)/(j)  *  (k+2)/(k+1)

Since j <= k, (j-1)/j <= (k-1)/k < (k+1)/(k+2)

IMPLIES alpha < 1.

Conjecture proved.

______________________________________________________________________
CONJECTURE : For a given p, all newly generated IR's are distinct.

Assume this is not so.

Implies
(D + (p-1) elements of A(k))/p
= (D + some other (p-1) elements of A(k))/p

Implies SUM[(p-1) elements of A(k)] = SUM[ some other (p-1) elements of A(k)]

Implies SUM[(p-1) elements of A(k)]/(p-1)
= SUM[some other (p-1) elements]/(p-1)

Implies A(k) is NOT in L.

Contra.

Hence conjecture.
______________________________________________________________________

CONJECTURE: A(k+1) is in L.

Since all newly generated IR's are distinct from each other, and all newly generated IR's are greater than previous IR's, A(k+1) is in L.

==> logic/zoo.p <==
I took some nephews and nieces to the Zoo, and we halted at a cage marked

Tovus Slithius, male and female.
Beregovus Mimsius, male and female.
Rathus Momus, male and female.
Jabberwockius Vulgaris, male and female.

The eight animals were asleep in a row, and the children began to guess
which was which.  "That one at the end is Mr Tove."  "No, no!  It's Mrs
Jabberwock," and so on.  I suggested that they should each write down
the names in order from left to right, and offered a prize to the one
who got most names right.

As the four species were easily distinguished, no mistake would arise in
pairing the animals; naturally a child who identified one animal as Mr
Tove identified the other animal of the same species as Mrs Tove.

The keeper, who consented to judge the lists, scrutinised them carefully.
"Here's a queer thing.  I take two of the lists, say, John's and Mary's.
The animal which John supposes to be the animal which Mary supposes to be
Mr Tove is the animal which Mary supposes to be the animal which John
supposes to be Mrs Tove.  It is just the same for every pair of lists,
and for all four species.

"Curiouser and curiouser!  Each boy supposes Mr Tove to be the animal
which he supposes to be Mr Tove; but each girl supposes Mr Tove to be
the animal which she supposes to be Mrs Tove.  And similarly for the oth-
er animals.  I mean, for instance, that the animal Mary calls Mr Tove
is really Mrs Rathe, but the animal she calls Mrs Rathe is really Mrs
Tove."

"It seems a little involved," I said, "but I suppose it is a remarkable
coincidence."

"Very remarkable," replied Mr Dodgson (whom I had supposed to be the
keeper) "and it could not have happened if you had brought any more
children."

How many nephews and nieces were there?  Was the winner a boy or a girl?
And how many names did the winner get right?	[by Sir Arthur Eddington]

==> logic/zoo.s <==
Given that there is at least one boy and one girl (John and Mary are
mentioned) then the answer is that there were 3 nephews and 2 nieces,
the winner was a boy who got 4 right.

Number the animals 1 through 8, such that the females are even and the
males are odd, with members of the same species consecutive; i.e.  1 is
Mr. Tove, 2 Mrs. Tove, etc.

Then each childs guesses can be represented by a permutation.  I use
the standard notation of a permutation as a set of orbits.  For
example: (1 3 5)(6 8) means 1 -> 3, 3 -> 5, 5 -> 1, 6 -> 8, 8 -> 6 and
2,4,7 are unchanged.

  Let P be any childs guesses.  Then P(mate(i)) = mate(P(i)).

  If Q is another childs guesses, then [P,Q] = T, where [P,Q] is the
commutator of P and Q (P composed with Q composed with P inverse
composed with Q inverse) and T is the special permutation (1 2) (3 4)
(5 6) (7 8) that just swaps each animal with its spouse.

  If P represents a boy, then P*P = I (I use * for composition, and
I for the identity permutation: (1)(2)(3)(4)(5)(6)(7)(8)

  If P represents a girl, then P*P = T.

 and  together mean that all girl's guesses must be of the form:
(A B C D) (E F G H) where A and C are mates, as are B & D,
E & F G & H.

So without loss of generality let Mary = (1 3 2 4) (5 7 6 8)
Without to much effort we see that the only possibilities for other
girls "compatible" with Mary (I use compatible to mean the relation
expressed in ) are:
g1:  (1 5 2 6) (3 8 4 7)
g2:  (1 6 2 5) (3 7 4 8)
g3:  (1 7 2 8) (3 5 4 6)
g4:  (1 8 2 7) (3 6 4 5)

Note that g1 is incompatible with g2 and g3 is incompatible with g4.
Thus no 4 of Mary and g1-4 are mutually compatible.  Thus there are at
most three girls: Mary, g1 and g3 (without loss of generality)

By  and , each boy must be represented as a product of
transpostions and/or singletons: e.g. (1 3) (2 4) (5) (6) (7) (8) or
(1) (2) (3 4) (5 8) (6 7).

Let J represent John's guesses and consider J(1).
If J(1) = 1, then J(2) = 2 (by ) using  and Mary J(3) = 4, J(4) =
3,  and g1 & J => J(5) = 6, J(6) = 5, & g3 & J => J(8) = 7 J(7) = 8
i.e. J = (1)(2)(3 4)(5 6)(7 8).  But the [J,Mary] <> T.  In fact, we
can see that J must have no fixed points, J(i) <> i for all i, since
there is nothing special about i = 1.

If J(1) = 2, then we get from Mary that J(3) = 3.  contradiction.

If J(1) = 3, then J(2) = 4, J(3) = 1, J(4) = 2 (from Mary) =>
J(5) = 7, J(6) = 8, J(7) = 5, J(8) = 6 => J = (1 3)(2 4)(5 7)(6 8)
(from g1)
But then J is incompatible with g3.

A similar analysis shows that J(1) cannot be 4,5,6,7 or 8; i.e. no J
can be compatible with all three girls.  So without loss of generality,
throw away g3.

We have Mary = (1 3 2 4) (5 7 6 8)
g1   = (1 5 2 6) (3 8 4 7)

The following are the only possible boy guesses which are compatible
with
both of these:

B1: (1)(2)(3 4)(5 6)(7)(8)
B2: (1 2)(3)(4)(5)(6)(7 8)
B3: (1 3)(2 4)(5 7)(6 8)
B4: (1 4)(2 3)(5 8)(6 7)
B5: (1 5)(2 6)(3 8)(4 7)
B6: (1 6)(2 5)(3 7)(4 8)

Note that B1 & B2 are incombatible, as are B3 & B4, B5 & B6, so at most
three of them are mutually compatible.  In fact, Mary, g1, B1, B3 and
B5 are all mutually compatible (as are all the other possibilities you
can get by choosing either B1 or B2, B3 or B4, B5 or B6.  So if there
are 2 girls there can be 3 boys, but no more, and we have already
eliminated the case of 3 girls and 1 boy.

The only other possibility to consider is whether there can be 4 or
more boys and 1 girl.  Suppose there are Mary and 4 boys.  Each boy
must map 1 to a different digit or they would not be mutually
compatible.  For example if b1 and b2 both map 1 to 3, then they both
map 3 to 1 (since a boy's map consists of transpositions), so both
b1*b2 and b2*b1 map 1 to 1.  Furthermore, b1 and b2 cannot map 1 onto
spouses.  For example, if b1(1) = a and b is the spouse of a, then
b1(2) = b.  If b2(1) = b, then b2(2) = a.  Then b1*b2(1) = b1(b) = 2
and b2*b1(1) = b2(a) = 2 (again using the fact that boys are all
transpostions).  Thus the four boys must be:

B1: (1)(2)...    or (1 2)....
B2: (1 3)...     or (1 4) ...
B3: (1 5) ...    or (1 6) ...
B4: (1 7) ...    or (1 8) ...

Consider B4.  The only permutation of the form (1 7)... which is
compatible
with Mary ( (1 3 2 4) (5 7 6 8) ) is:

(1 7)(2 8)(3 5)(4 6)

The only (1 8)... possibility is:

(1 8)(2 7)(3 6)(4 5)

Suppose B4 = (1 7)(2 8)(3 5)(4 6)

If B3 starts (1 5), it must be (1 5)(2 6)(3 8)(4 7) to be compatible
with B4.  This is compatible with Mary also.

Assuming this and B2 starts with (1 3) we get B2 = (1 3)(2 4)(5 8)(6 7)
in order to be compatible with B4.  But then B2*B3 and B3*B2 moth map 1
to 8.  I.e. no B2 is mutually compatible with B3 & B4.

Similarly if B2 starts with (1 4) it must be (1 4)(2 3)(5 7)(6 8) to
work with B4, but this doesn't work with B3.

Likewise B3 starting with (1 6) leads to no possible B2 and the
identical reasoning eliminates B4 = (1 8)...

So no B4 is possible!

I.e at most 3 boys are mutually compatiblw with Mary, so 2 girls & 3
boys is optimal.

Thus:

Mary = (1 3 2 4) (5 7 6 8)
Sue  = (1 5 2 6) (3 8 4 7)
John = (1)(2)(3 4)(5 6)(7)(8)
Bob  = (1 3)(2 4)(5 7)(6 8)
Jim  = (1 5)(2 6)(3 8)(4 7)

is one optimal solution, with the winner being John (4 right: 1 2 7 & 8)

```

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