AOH :: PUZZLE11.FAQ
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Article 3069 of news.answers:
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From: uunet!questrel!chris (Chris Cole)
Subject: rec.puzzles FAQ, part 11 of 15
Message-ID: <puzzles-faq-11_717034101@questrel.com>
Followup-To: rec.puzzles
Summary: This posting contains a list of
Frequently Asked Questions (and their answers).
It should be read by anyone who wishes to
post to the rec.puzzles newsgroup.
Sender: chris@questrel.com (Chris Cole)
Reply-To: uunet!questrel!faql-comment
Organization: Questrel, Inc.
References: <puzzles-faq-1_717034101@questrel.com>
Date: Mon, 21 Sep 1992 00:09:38 GMT
Approved: news-answers-request@MIT.Edu
Expires: Sat, 3 Apr 1993 00:08:21 GMT
Lines: 1889
Archive-name: puzzles-faq/part11
Last-modified: 1992/09/20
Version: 3
==> induction/hanoi.p <==
Is there an algorithom for solving the hanoi tower puzzle for any number
of towers? Is there an equation for determining the minimum number of
moves required to solve it, given a variable number of disks and towers?
==> induction/hanoi.s <==
The best way of thinking of the Towers of Hanoi problem is inductively.
To move n disks from post 1 to post 2, first move (n-1) disks
from post 1 to post 3, then move disk n from post 1 to post 2, then move
(n-1) disks from post 3 to post 2 (same procedure as moving (n-1) disks
from post 1 to post 3). In order to figure out how to move (n-1) disks
from post 1 to post 3, first move (n-2) disks . . . .
As far as an algorithm which straightens out any legal position is
concerned, the algorithm would go something like this:
1. Find the smallest disk. Call the post that it's on post 1.
2. Find the smallest disk which is not on post 1. This disk is, say,
size k. (I am calling the smallest disk size 1 here.) Call the post
that disk k is on post 2. Disks 1 through (k-1) are then all stacked up
correctly on post 1 disk k is on top of post 2. This follow from the
fact that the disks are in a legal postition.
3. Move disks 1 through (k-1) from post 1 to post 2, ignoring the other
disks. This is just the standard Tower of Hanoi problem for (k-1)
disks.
4. If the disks are not yet correctly arranged, repeat from step 1.
In fact, this gives the straightening with the fewest number of moves.
With 3 towers, for N number of disks, the formula for the minimum number of
moves to complete the puzzle correctly is:
(2^N) - 1
This bit of ancient folklore was invented by De Parville in 1884.
``In the great temple at Benares, says he, beneath the dome which
marks the centre of the world, rests a brass plate in which are
fixed three diamond needles, each a cubit high and as thick as
the body of a bee. On one of these needles, at the creation,
God placed sixty-four discs of pure gold, the largest disc resting
on the brass plate, and the others getting smaller and smaller
up to the top one. This is the Tower of Bramah. Day and night
unceasingly the priests transfer the discs from one diamond needle
to another according to the fixed and immutable laws of Bramah,
which require that the priest on duty must not move more than one
disc at a time and that he must place this disc on a needle so
that there is no smaller disc below it. When the sixty-four
discs shall have been thus transferred from the needle on which
at the creation God placed them to one of the other needles,
tower, temple, and Brahmins alike will crumble into dust, and
with a thunderclap the world will vanish.'' (W W R Ball,
MATHEMATICAL RECREATIONS AND ESSAYS, p. 304)
This has been discussed by several authors, e.g.
Er, Info Sci 42 (1987) 137-141.
Graham, Knuth and Patashnik, _Concrete_Mathematics_.
There are many papers claiming to solve this, and they are probably
all correct but they rely on the unproven "Frame's conjecture".
In particular, for the 4 peg case the conjecture states that an optimal
solution begins by forming a substack of the k smallest discs, then moving
the rest, and then moving those k again; k to be determined.
Here is a extensible bc program that does the same work. The output
format is not that great. We get 300 numbers as output. The first
hundred represent N, the next 100 represent f(N) and the last hundred
represent i, which is the number of discs to move to tmp1 using f(N).
For convenience, I have here some values for N <= 48. Enjoy.
Sharma
N 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
f(N) 1 3 5 9 13 17 25 33 41 49 65 81 97 113 129 161 193 225 257
i 0 1 1 2 2 3 3 4 5 6 6 7 8 9 10 10 11 12 13
N 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
f(N) 289 321 385 449 513 577 641 705 769 897 1025 1153 1281 1409 1537 1665
i 14 15 15 16 17 18 19 20 21 21 22 23 24 25 26 27
N 36 37 38 39 40 41 42 43 44 45 46 47 48
f(N) 1793 2049 2305 2561 2817 3073 3329 3585 3841 4097 4609 5121 5633
i 28 28 29 30 31 32 33 34 35 36 36 37 38
/* This is the bc program that gives f(N) for 4 peg case */
w = 101; /* This represents the number of disks */
m[0] = 0;
m[1] = 1;
m[2] = 3;
m[3] = 5;
m[4] = 9;
m[5] = 13;
m[6] = 17;
/* f(n) is the function that gives the min # of moves for 4 peg case */
define f(n) {
return (m[n]);
}
/* g(n) is the function that fives the min # of moves for 3 peg case */
define g(n) {
return (2^n - 1);
}
/* x(n) is the Optimization Routine */
define x(n) {
auto j
auto r
auto i
if(n == 1) return (1);
j = f(1) + g(n-1);
for(i = 2; i < n; i++) {
r = f(i) + g(n-i);
if(r < j) { j = r; d = i; }
}
return (j);
}
/* main program */
for(q = 4; q < w; q++) {
t = x(q-1);
m[q] = 2 * t + 1;
d[q] = d;
};
/*This for loop prints the number of discs from 1 <= n <= w*/
for(q = 1; q < w; q++) {
q;
}
/*This for loop prints f(n) for 1 <= n <= w */
for(q = 1; q < w; q++) {
m[q];
}
/*This for loop prints i for 1 <= n <= w
i represents the number of disks to be moved to tmp location using f(n)
N-i-1 will be moved using g(n) */
for(q = 1; q < w; q++) {
d[q];
}
--
sharma@sharma.warren.mentorg.com
==> induction/n-sphere.p <==
With what odds do three random points on an n-sphere form an acute triangle?
==> induction/n-sphere.s <==
Select three points a, b, and c, randomly with respect to the surface of an
n-sphere. These three points determine a fourth, x, which is the intersection
of the sphere with the axis perpendicular to the abc plane. (Choose the pole
nearest the plane.) I could have, just as easily, selected x, a distance d
from x, and three points d units away from x. The distribution of d is not
uniform, but that's ok. For every x and d, the three points abc form an acute
triangle with probability p[n-1]. By induction, p[n] = 1/4.
==> induction/paradox.p <==
What simple property holds for the first 10,000 integers, then fails?
==> induction/paradox.s <==
Consider the sequences defined by:
s(1) = a; s(2) = b; s(n) = least integer such that s(n)/s(n-1) > s(n-1)/s(n-2).
In other words, s(n) = 1+floor(s(n-1)^2/s(n-2)) for n >= 3. These
sequences are similar in some ways to the classically-studied Pisot
sequences. For example, if a = 1, b = 2, then we get the odd-indexed
Fibonacci numbers.
D. Boyd of UBC, an expert in Pisot sequences, pointed out the following.
If we let a = 8, b = 55 in the definition above, then the resulting
sequence s(n) appears to satisfy the following linear recurrence
of order 4:
s(n) = 6s(n-1) + 7s(n-2) - 5s(n-3) - 6s(n-4)
Indeed, it does satisfy this linear recurrence for the
first 11,056 terms. However, it fails at the 11,057th term!
And s(11057) is a 9270 digit number.
(The reason for this coincidence depends on a remarkable fact
about the absolute values of the roots of the polynomial
x^4 - 6x^3 - 7x^2 + 5x + 6.)
==> induction/party.p <==
You're at a party. Any two (different) people at the party have exactly one
friend in common (the friend is also at the party). Prove that there is at
least one person at the party who is a friend of everyone else. Assume that
the friendship relation is symmetric and not reflexive.
==> induction/party.s <==
Here is an easy solution by induction. Let P be the set of people in the
party, and n the size of P. If n=2 or 3, the result is trivial. Suppose now
that n>3 and that the result is true for n-1.
For any two distinct x,y in P, write x & y to mean that `x is a friend of y',
and x ~& y to mean that `x is not a friend of y'.
Take q in P. The hypothesis on the relation & is still satisfied on P-{q}; by
induction, the result is thus true for P-{q}, and there is some p in P-{q}
such that p & x for any x in P-{p,q}. We have two cases:
a) p & q. Then the result holds for P with p.
b) p ~& q. By hypothesis, there is a unique r in P-{p,q} such that p & r & q.
For any x in P-{p,q}, if x & q, then p & x & q, and so x=r. Thus r is the
unique friend of q. Now for any s in P-{q,r} there exists some x such that s &
x & q, and so x=r. This means that r & s for any s in P-{q,r}, and as r & q,
the results holds in P with r.
The problem can also be solved by applying the spectral theory of graphs
(see for instance Bollobas' excellent book, _Extremal Graph Theory_).
The problem's condition is vacuous if there is only N=1 person at the "party",
impossible if N=2 (If you aren't your own friend, nor I mine, somebody *else*
must be our mutual friend), and trivial if N=3 (everybody must be everyone
else's friend). Henceforth assume N>3.
Let A,B be two friends, and C their mutual friend. Let a be the number
of A's friends other than B and C, and likewise b,c. Each of A's friends
is also friendly with exactly one other of A's friends, and with none of
B and C's other friends (if A1,B1 are friends of A,B resp. and of each other
then A1 and B have more than one mutual friend); likewise for B and C.
Let M=N-(a+b+c+3) be the number of people not friendly with any of A,B,C.
Each of them is friendly with exactly one of A's and one of B's friends;
and each pair of a friend of A and a friend of B must have exactly
one of them as a mutual friend. Thus M=ab; likewise M=ab=ac=bc. Thus
either M and two of a,b,c vanish, or a=b=c=k (say), M=k^2, and N=k^3+3k+3.
In the first case, say b=c=0; necessarily a is even, and A is a friend of
everybody else at the party, each of whom is friendly with exactly one other
person; clearly any such configuration (a graph of k/2+1 triangles with a
common vertex) satisfies the problem's conditions).
It remains to show that the second case is impossible. Since N=k^2+3k+3
does not depend on A,B,C, neither does k, and it quickly follows that the
party's friendship graph is regular with reduced matrix
[ 0 k+2 0 ]
[ 1 1 k ]
[ 0 1 k+1 ]
and eigenvalues k+2 and +-sqrt(k+1) and multiplicities 1,m1,m2 for some
*integers* m1 and m2 such that (m1-m2)*sqrt(k+1)=-(k+2) (because the graph's
matrix has trace zero). Thus sqrt(k+1) divides k+2 and k+1 divides
(k+2)^2=(k+1)(k+3)+1
which is only possible if k=0, Q.E.D.
==> induction/roll.p <==
An ordinary die is thrown until the running total of the throws first
exceeds 12. What is the most likely final total that will be obtained?
==> induction/roll.s <==
Claim: If you throw a die until the running total exceeds n>=5, a final
outcome of n+1 is more likely than any other.
Assume we throw an m for a total n+k>n+1, and assume m-k>=0. Now, it
is just as likely to throw an m as an m-k+1, which means that the sum
n+1 is just as likely as any other. Now consider the series of throws
consisting of n-5 1's followed by a 6 and note that we cannot achieve
more than an n+1 by changing the last die roll. Hence, a total of n+1
is more likely than any other.
==> induction/takeover.p <==
After graduating from college, you have taken an important managing position
in the prestigious financial firm of "Mary and Lee".
You are responsable for all the decisions concerning take-over bids.
Your immediate concern is whether to take over "Financial Data".
There is no doubt that you will be successful if you are the first to
bid and that this will be profitable for the firm and you in the long
run. However, you know that there exist another n financial firms,
similar to "Mary and Lee", that are also considering the possibility.
Although you are likely to be the first one to move, you know that
just after a take-over there is a lot of adjustment that needs to be
done. In fact, for a period of time following any take-over the
successful firm becomes a prime candidate for a take-over which will
cost the job of whoever is responsable for take-overs. Among all
financial firms it is common knowledge that the managers responsable
for take-overs are rational and intelligent. What is your best response?
==> induction/takeover.s <==
Assume the takeover is wise for n. The takeover is then unwise for
n+1, as the other companies now find themselves in the same situation
as you for n. If the decision is unwise for n, by similar reasoning
it is wise to takeover FD for n+1. Now note that for n=1 the takeover
decision is clearly unwise, hence by induction you should takeover
FD iff n is even.
==> logic/29.p <==
Three people check into a hotel. They pay $30 to the manager and go
to their room. The manager finds out that the room rate is $25 and
gives $5 to the bellboy to return. On the way to the room the bellboy
reasons that $5 would be difficult to share among three people so
he pockets $2 and gives $1 to each person.
Now each person paid $10 and got back $1. So they paid $9 each,
totalling $27. The bellboy has $2, totalling $29.
Where is the remaining dollar?
==> logic/29.s <==
Each person paid $9, totalling $27. The manager has $25 and the bellboy $2.
The bellboy's $2 should be added to the manager's $25 or subtracted from
the tenants' $27, not added to the tenants' $27.
==> logic/ages.p <==
1) Ten years from now Tim will be twice as old as Jane was when Mary was
nine times as old as Tim.
2) Eight years ago, Mary was half as old as Jane will be when Jane is one year
older than Tim will be at the time when Mary will be five times as old as
Tim will be two years from now.
3) When Tim was one year old, Mary was three years older than Tim will be when
Jane is three times as old as Mary was six years before the time when Jane
was half as old as Tim will be when Mary will be ten years older than Mary
was when Jane was one-third as old as Tim will be when Mary will be three
times as old as she was when Jane was born.
HOW OLD ARE THEY NOW?
==> logic/ages.s <==
The solution: Tim is 3, Jane is 8, and Mary is 15. A little grumbling
is in order here, as clue number 1 leads to the situation a year and a
half ago, when Tim was 1 1/2, Jane was 6 1/2, and Mary was 13 1/2.
This sort of problem is easy if you write down a set of equations. Let
t be the year that Tim was born, j be the year that Jane was born, m be
the year that Mary was born, and y be the current year. As indefinite
years come up, let y1, y2, ... be the indefinite years. Then the
equations are
y + 10 - t = 2 (y1 - j)
y1 - m = 9 (y1 - t)
y - 8 - m = 1/2 (y2 - j)
y2 - j = 1 + y3 - t
y3 - m = 5 (y + 2 - t)
t + 1 - m = 3 + y4 - t
y4 - j = 3 (y5 - 6 - m)
y5 - j = 1/2 (y6 - t)
y6 - m = 10 + y7 - m
y7 - j = 1/3 (y8 - t)
y8 - m = 3 (j - m)
t = y - 3
j = y - 8
m = y - 15
==> logic/bookworm.p <==
A bookworm eats from the first page of an encyclopedia to the last page.
The bookworm eats in a straight line. The encyclopedia consists of ten
1000-page volumes. Not counting covers, title pages, etc., how many pages
does the bookworm eat through?
==> logic/bookworm.s <==
On a book shelf the first page of the first volume is on the "inside"
__ __
B| | | |F
A|1 |...........................|10|R
C| | | |O
K| | | |N
| | | |T
----------------------------------
so the bookworm eats only through the cover of the first volume, then 8 times
1000 pages of Volumes 2 - 9, then through the cover to the 1st page of Vol 10.
He eats 8,000 pages.
==> logic/boxes.p <==
Which Box Contains the Gold?
Two boxes are labeled "A" and "B". A sign on box A says "The sign
on box B is true and the gold is in box A". A sign on box B says
"The sign on box A is false and the gold is in box A". Assuming there
is gold in one of the boxes, which box contains the gold?
==> logic/boxes.s <==
The problem cannot be solved with the information given.
The sign on box A says "The sign on box B is true and the gold is in box A".
The sign on box B says "The sign on box A is false and the gold is in box A".
The following argument can be made: If the statement on box A is true, then
the statement on box B is true, since that is what the statement on box A
says. But the statement on box B states that the statement on box A is false,
which contradicts the original assumption. Therefore, the statement on box A
must be false. This implies that either the statement on box B is false or
that the gold is in box B. If the statement on box B is false, then either
the statement on box A is true (which it cannot be) or the gold is in box B.
Either way, the gold is in box B.
However, there is a hidden assumption in this argument: namely, that
each statement must be either true or false. This assumption leads to
paradoxes, for example, consider the statement: "This statement is
false." If it is true, it is false; if it is false, it is true. The
only way out of the paradox is to deny that the statement is either true
or false and label it meaningless instead. Both of the statements on the
boxes are therefore meaningless and nothing can be concluded from them.
In general, statements about the truth of other statements lead to
contradictions. Tarski invented metalanguages to avoid this problem.
To avoid paradox, a statement about the truth of a statement in a language
must be made in the metalanguage of the language.
Common sense dictates that this problem cannot be solved with the information
given. After all, how can we deduce which box contains the gold simply by
reading statements written on the outside of the box? Suppose we deduce that
the gold is in box B by whatever line of reasoning we choose. What is to stop
us from simply putting the gold in box A, regardless of what we deduced?
(cf. Smullyan, "What Is the Name of This Book?", Prentice-Hall, 1978, #70)
==> logic/calibans.will.p <==
----------------------------------------------
| Caliban's Will by M.H. Newman |
----------------------------------------------
When Caliban's will was opened it was found to contain the following
clause:
"I leave ten of my books to each of Low, Y.Y., and 'Critic,' who are
to choose in a certain order.
No person who has seen me in a green tie is to choose before Low.
If Y.Y. was not in Oxford in 1920 the first chooser never lent me
an umbrella.
If Y.Y. or 'Critic' has second choice, 'Critic' comes before the one
who first fell in love."
Unfortunately Low, Y.Y., and 'Critic' could not remember any of the
relevant facts; but the family solicitor pointed out that, assuming the
problem to be properly constructed (i.e. assuming it to contain no
statement superfluous to its solution) the relevant data and order
could be inferred.
What was the prescribed order of choosing; and who lent Caliban an
umbrella?
==> logic/calibans.will.s <==
Let T be "person who saw Caliban in a green tie."
Let U be "person who lent Caliban an umbrella."
Then the data are:
(1) No T chooses before Low.
(2) Either Y.Y. was in Oxford in March 1920 or the first chooser is not
a U.
(3) Either Low is second or Critic is not last.
Consider first (3)
If it could be shown that Low is first, then from (3), Critic is not
last and therefore is second; i.e. the order is Low, Critic, Y.Y.
Next (1)
If both Critic and Y.Y. were T's would require Low first and (3) then
gives the order Low-Critic-Y.Y., ie. (2) would be superfluous. Hence
Critic and Y.Y. are not both T's.
If neither Critic nor Y.Y. were a T, (1) would be trivially true for
any ordering and therefore would give no information, i.e. would be
superfluous. Hence just one of Y.Y. and Critic is a T. It follows
that the only possible order in which Low is not first is:
Not T, Low, T
Now (2)
First if Y.Y was in Oxford in March 1920, nothing follows from (2)
about the order and (2) is superfluous. Hence Y.Y. was not in Oxford.
If Low were a U he would not, by (2) come first, and so by (1) the
order would be:
Not T, Low, T
i.e. (1) and (2) alone would fix an order, and (3) would be superfluous.
Hence Low is not a U.
It now follows, by the arguments just given for T's under (1) that just
one of Y.Y. and Critic is a U. If the same one is the T and the U (2)
follows from (1) (since Low is not a U); i.e (2) is superfluous. The
situation is therefore:
T's: just one of Y.Y. and Critic; not Low
U's: the other one of Y.Y; not Low
It now follows that "not T, Low, T" is impossible, for the "not T" is
the "U" and therefore, by (2), is not first. Hence Low is first, and
(3) gives the order:
Low, Critic, Y.Y.
Finally, Y.Y. is a T, and Critic is a U. For if Critic is a T, then
by (1) Low precedes Critic and hence (3) allows only "Low, Critic, Y.Y";
(2) is superfluous. I.e. Critic (only) lent Caliban an umbrella.
The problem is from _Problems Omnibus_ by Hubert Phillips,
Arco Publications, London, 1960. Hubert Phillips was a noted puzzelist
who contributed under his own name and the pseudonyms of "Caliban",
"T.O. Hare", and "The Doc".
==> logic/camel.p <==
An Arab sheikh tells his two sons that are to race their camels to a
distant city to see who will inherit his fortune. The one whose camel
is slower will win. The brothers, after wandering aimlessly for days,
ask a wiseman for advise. After hearing the advice they jump on the
camels and race as fast as they can to the city. What did the wiseman
say?
==> logic/camel.s <==
The wiseman tells them to switch camels.
==> logic/centrifuge.p <==
You are a biochemist, working with a 12-slot centrifuge. This is a gadget
that has 12 equally spaced slots around a central axis, in which you can
place chemical samples you want centrifuged. When the machine is turned on,
the samples whirl around the central axis and do their thing.
To ensure that the samples are evenly mixed, they must be distributed in the
12 slots such that the centrifuge is balanced evenly. For example, if you
wanted to mix 4 samples, you could place them in slots 12, 3, 6 and 9
(assuming the slots are numbered from 1 to 12 like a clock).
Problem: Can you use the centrifuge to mix 5 samples?
==> logic/centrifuge.s <==
The superposition of any two solutions is yet another solution, so given
that the factors > 1 of 12 (2, 3, 4, 6, 12) are all solutions, the
only thing to check about, for example, the proposed solution 2+3 is
that not all ways of combining 2 & 3 would have centrifuge tubes
from one subsolution occupying the slot for one of the tubes in
another solution. For the case 2+3, there is no problem: Place 3
tubes, one in every 4th position, then place the 4th and 5th
diametrically opposed (each will end up in a slot adjacent to one of
the first 3 tubes). The obvious generalization is, what are the
numbers of tubes that cannot be balanced? Observing that there are
solutions for 2,3,4,5,6 tubes and that if X has a solution, 12-X has
also one (obtained by swapping tubes and holes), it is obvious that
1 and 11 are the only cases without solutions.
Here is how this problem is often solved in practice: A dummy tube
is added to produce a total number of tubes that is easy to balance.
For example, if you had to centrifuge just one sample, you'd add a
second tube opposite it for balance.
==> logic/children.p <==
A man walks into a bar, orders a drink, and starts chatting with the
bartender. After a while, he learns that the bartender has three
children. "How old are your children?" he asks. "Well," replies the
bartender, "the product of their ages is 72." The man thinks for a
moment and then says, "that's not enough information." "All right,"
continues the bartender, "if you go outside and look at the building
number posted over the door to the bar, you'll see the sum of the
ages." The man steps outside, and after a few moments he reenters and
declares, "Still not enough!" The bartender smiles and says, "My
youngest just loves strawberry ice cream."
How old are the children?
A variant of the problem is for the sum of the ages to be 13 and the
product of the ages to be the number posted over the door. In this
case, it is the oldest that loves ice cream.
Then how old are they?
==> logic/children.s <==
First, determine all the ways that three ages can multiply together to get 72:
72 1 1 (quite a feat for the bartender)
36 2 1
24 3 1
18 4 1
18 2 2
12 6 1
12 3 2
9 4 2
9 8 1
8 3 3
6 6 2
6 4 3
As the man says, that's not enough information; there are many possibilities.
So the bartender tells him where to find the sum of the ages--the man now knows
the sum even though we don't. Yet he still insists that there isn't enough
info. This must mean that there are two permutations with the same sum;
otherwise the man could have easily deduced the ages.
The only pair of permutations with the same sum are 8 3 3 and 6 6 2, which both
add up to 14 (the bar's address). Now the bartender mentions his
"youngest"--telling us that there is one child who is younger than the other
two. This is impossible with 8 3 3--there are two 3 year olds. Therefore the
ages of the children are 6, 6, and 2.
Pedants have objected that the problem is insoluble because there could be
a youngest between two three year olds (even twins are not born exactly at
the same time). However, the word "age" is frequently used to denote the
number of years since birth. For example, I am the same age as my wife,
even though technically she is a few months older than I am. And using the
word "youngest" to mean "of lesser age" is also in keeping with common parlance.
So I think the solution is fine as stated.
In the sum-13 variant, the possibilities are:
11 1 1
10 2 1
9 3 1
9 2 2
8 4 1
8 3 2
7 5 1
7 4 2
7 3 3
6 6 1
6 5 2
6 4 3
The two that remain are 9 2 2 and 6 6 1 (both products equal 36). The
final bit of info (oldest child) indicates that there is only one
child with the highest age. This cancels out the 6 6 1 combination, leaving
the childern with ages of 9, 2, and 2.
==> logic/condoms.p <==
How can you have mutually safe sex with three women with only two condoms?
==> logic/condoms.s <==
Use both condoms on the first woman. Take off the outer condom (turning it
inside-out in the process) and set it aside. Use the inner condom alone on
the second woman. Put the outer condom back on. Use it on the third woman.
==> logic/dell.p <==
How can I solve logic puzzles (e.g., as published by Dell) automatically?
==> logic/dell.s <==
#include <setjmp.h>
#define EITHER if (S[1] = S[0], ! setjmp((S++)->jb)) {
#define OR } else EITHER
#define REJECT longjmp((--S)->jb, 1)
#define END_EITHER } else REJECT;
/* values in tmat: */
#define T_UNK 0
#define T_YES 1
#define T_NO 2
#define Val(t1,t2) (S->tmat[t1][t2])
#define CLASS(x) \
(((x) / NUM_ITEM) * NUM_ITEM)
#define EVERY_TOKEN(x) \
(x = 0; x < TOT_TOKEN; x++)
#define EVERY_ITEM(x, class) \
(x = CLASS(class); x < CLASS(class) + NUM_ITEM; x++)
#define BEGIN \
struct state { \
char tmat[TOT_TOKEN][TOT_TOKEN]; \
jmp_buf jb; \
} States[100], *S = States; \
\
main() \
{ \
int token; \
\
for EVERY_TOKEN(token) \
yes(token, token); \
EITHER
/* Here is the problem-specific data */
#define NUM_ITEM 5
#define NUM_CLASS 6
#define TOT_TOKEN (NUM_ITEM * NUM_CLASS)
#define HOUSE_0 0
#define HOUSE_1 1
#define HOUSE_2 2
#define HOUSE_3 3
#define HOUSE_4 4
#define ENGLISH 5
#define SPANISH 6
#define NORWEG 7
#define UKRAIN 8
#define JAPAN 9
#define GREEN 10
#define RED 11
#define IVORY 12
#define YELLOW 13
#define BLUE 14
#define COFFEE 15
#define TEA 16
#define MILK 17
#define OJUICE 18
#define WATER 19
#define DOG 20
#define SNAIL 21
#define FOX 22
#define HORSE 23
#define ZEBRA 24
#define OGOLD 25
#define PLAYER 26
#define CHESTER 27
#define LSTRIKE 28
#define PARLIA 29
char *names[] = {
"HOUSE_0", "HOUSE_1", "HOUSE_2", "HOUSE_3", "HOUSE_4",
"ENGLISH", "SPANISH", "NORWEG", "UKRAIN", "JAPAN",
"GREEN", "RED", "IVORY", "YELLOW", "BLUE",
"COFFEE", "TEA", "MILK", "OJUICE", "WATER",
"DOG", "SNAIL", "FOX", "HORSE", "ZEBRA",
"OGOLD", "PLAYER", "CHESTER", "LSTRIKE", "PARLIA",
};
BEGIN
yes(ENGLISH, RED); /* Clue 1 */
yes(SPANISH, DOG); /* Clue 2 */
yes(COFFEE, GREEN); /* Clue 3 */
yes(UKRAIN, TEA); /* Clue 4 */
EITHER /* Clue 5 */
yes(IVORY, HOUSE_0);
yes(GREEN, HOUSE_1);
OR
yes(IVORY, HOUSE_1);
yes(GREEN, HOUSE_2);
OR
yes(IVORY, HOUSE_2);
yes(GREEN, HOUSE_3);
OR
yes(IVORY, HOUSE_3);
yes(GREEN, HOUSE_4);
END_EITHER
yes(OGOLD, SNAIL); /* Clue 6 */
yes(PLAYER, YELLOW); /* Clue 7 */
yes(MILK, HOUSE_2); /* Clue 8 */
yes(NORWEG, HOUSE_0); /* Clue 9 */
EITHER /* Clue 10 */
yes(CHESTER, HOUSE_0);
yes(FOX, HOUSE_1);
OR
yes(CHESTER, HOUSE_4);
yes(FOX, HOUSE_3);
OR
yes(CHESTER, HOUSE_1);
EITHER yes(FOX, HOUSE_0);
OR yes(FOX, HOUSE_2);
END_EITHER
OR
yes(CHESTER, HOUSE_2);
EITHER yes(FOX, HOUSE_1);
OR yes(FOX, HOUSE_3);
END_EITHER
OR
yes(CHESTER, HOUSE_3);
EITHER yes(FOX, HOUSE_2);
OR yes(FOX, HOUSE_4);
END_EITHER
END_EITHER
EITHER /* Clue 11 */
yes(PLAYER, HOUSE_0);
yes(HORSE, HOUSE_1);
OR
yes(PLAYER, HOUSE_4);
yes(HORSE, HOUSE_3);
OR
yes(PLAYER, HOUSE_1);
EITHER yes(HORSE, HOUSE_0);
OR yes(HORSE, HOUSE_2);
END_EITHER
OR
yes(PLAYER, HOUSE_2);
EITHER yes(HORSE, HOUSE_1);
OR yes(HORSE, HOUSE_3);
END_EITHER
OR
yes(PLAYER, HOUSE_3);
EITHER yes(HORSE, HOUSE_2);
OR yes(HORSE, HOUSE_4);
END_EITHER
END_EITHER
yes(LSTRIKE, OJUICE); /* Clue 12 */
yes(JAPAN, PARLIA); /* Clue 13 */
EITHER /* Clue 14 */
yes(NORWEG, HOUSE_0);
yes(BLUE, HOUSE_1);
OR
yes(NORWEG, HOUSE_4);
yes(BLUE, HOUSE_3);
OR
yes(NORWEG, HOUSE_1);
EITHER yes(BLUE, HOUSE_0);
OR yes(BLUE, HOUSE_2);
END_EITHER
OR
yes(NORWEG, HOUSE_2);
EITHER yes(BLUE, HOUSE_1);
OR yes(BLUE, HOUSE_3);
END_EITHER
OR
yes(NORWEG, HOUSE_3);
EITHER yes(BLUE, HOUSE_2);
OR yes(BLUE, HOUSE_4);
END_EITHER
END_EITHER
/* End of problem-specific data */
solveit();
OR
printf("All solutions found\n");
exit(0);
END_EITHER
}
no(a1, a2)
{
int non1, non2, token;
if (Val(a1, a2) == T_YES)
REJECT;
else if (Val(a1, a2) == T_UNK) {
Val(a1, a2) = T_NO;
no(a2, a1);
non1 = non2 = -1;
for EVERY_ITEM(token, a1)
if (Val(token, a2) != T_NO)
if (non1 == -1)
non1 = token;
else
break;
if (non1 == -1)
REJECT;
else if (token == CLASS(a1) + NUM_ITEM)
yes(non1, a2);
for EVERY_TOKEN(token)
if (Val(token, a1) == T_YES)
no(a2, token);
}
}
yes(a1, a2)
{
int token;
if (Val(a1, a2) == T_NO)
REJECT;
else if (Val(a1, a2) == T_UNK) {
Val(a1, a2) = T_YES;
yes(a2, a1);
for EVERY_ITEM(token, a1)
if (token != a1)
no(token, a2);
for EVERY_TOKEN(token)
if (Val(token, a1) == T_YES)
yes(a2, token);
else if (Val(token, a1) == T_NO)
no(a2, token);
}
}
solveit()
{
int token, tok2;
for EVERY_TOKEN(token)
for (tok2 = token; tok2 < TOT_TOKEN; tok2++)
if (Val(token, tok2) == T_UNK) {
EITHER
yes(token, tok2);
OR
no(token, tok2);
END_EITHER;
return solveit();
}
printf("Solution:\n");
for EVERY_ITEM(token, 0) {
for (tok2 = NUM_ITEM; tok2 < TOT_TOKEN; tok2++)
if (Val(token, tok2) == T_YES)
printf("\t%s %s\n",names[token],names[tok2]);
printf("\n");
}
REJECT;
}
---
james@crc.ricoh.com (James Allen)
==> logic/elimination.p <==
97 baseball teams participate in an annual state tournament.
The way the champion is chosen for this tournament is by the same old
elimination schedule. That is, the 97 teams are to be divided into
pairs, and the two teams of each pair play against each other.
After a team is eliminated from each pair, the winners would
be again divided into pairs, etc. How many games must be played
to determine a champion?
==> logic/elimination.s <==
In order to determine a winner all but one team must lose.
Therefore there must be at least 96 games.
==> logic/family.p <==
Suppose that it is equally likely for a pregnancy to deliver
a baby boy as it is to deliver a baby girl. Suppose that for a
large society of people, every family continues to have children
until they have a boy, then they stop having children.
After 1,000 generations of families, what is the ratio of males
to females?
==> logic/family.s <==
The ratio will be 50-50 in both cases. We are not killing off any
fetuses or babies, and half of all conceptions will be male, half
female. When a family decides to stop does not affect this fact.
==> logic/flip.p <==
How can a toss be called over the phone (without requiring trust)?
==> logic/flip.s <==
A flips a coin. If the result is heads, A multiplies 2 90-digit prime
numbers; if the result is tails, A multiplies 3 60-digit prime
numbers. A tells B the result of the multiplication. B now calls
either heads or tails and tells A. A then supplies B with the
original numbers to verify the flip.
==> logic/friends.p <==
Any group of 6 or more contains either 3 mutual friends or 3 mutual strangers.
Prove it.
==> logic/friends.s <==
Take a person X. Of the five other people, there must either be at least three
acquaintances of X or at least three strangers of X. Assume wlog that X has
three strangers A,B,C. Unless A,B,C is the required triad of acquaintances,
they must include a pair of strangers, wlog A,B. Then X,A,B is the required
triad of strangers, QED.
==> logic/hundred.p <==
A sheet of paper has statements numbered from 1 to 100. Statement n says
"exactly n of the statements on this sheet are false." Which statements are
true and which are false? What if we replace "exactly" by "at least"?
==> logic/hundred.s <==
It is tempting to argue as follows:
Since only one statement can be true (they are mutually contradictory),
therefore 99 are false. So, all are false except for statement 99.
If replaced by "at least", and the "real" number of false statements is
x, then statements x+1 to 100 will be false (since they falsely claim
that there are more false statements than there actually are). So, 100-x are
false, ie. x=100-x, so x=50. The first 50 statements are true, and statements
51 to 100 are false.
However, there is a hidden and incorrect assumption in this argument.
To see this, suppose that there is one statement on the sheet and it
says "One statement is false" or "At least one statement is false,"
either way it implies "this statement is false," which is a familiar
paradoxical statement. We have learned that this paradox arises because
of the false assumption that all statements are either true or false.
This is the hidden assumption in the above reasoning.
If it is acknowledged that some of the statements on the page may be
neither true nor false (i.e., meaningless), then nothing whatsoever can
be concluded about which statements are true or false.
This problem has been carefully contrived to appear to be solvable (like
the vacuous statement "this statement is true"). By changing the
numbers in some statements and changing "true" to "false," various
circular forms of the liar's paradox can be constructed.
From _Litton's Problematical Recreations_
==> logic/inverter.p <==
Can a digital logic circuit with two inverters invert N independent inputs?
The circuit may contain any number of AND or OR gates.
==> logic/inverter.s <==
It can be shown that N inverters can invert 2N-1 independent inputs, given
an unlimited supply of AND and OR gates. The classic version of this
puzzle is to invert 3 independent inputs using AND gates, OR gates, and
only 2 inverters.
So, start with N inverters. Replace 3 of them with 2.
Keep doing that until you're down to 2 inverters.
I was skeptical at first, because such a design requires so much feedback
that I was sure the system would oscillate when switching between two
particular states. But after writing a program to test every possible state
change (32^2), it appears that this system settles after a maximum of
3 feedback logic iterations. I did not include gate delays in the simulation,
however, which could increase the number of iterations before the system
settles.
In any case, it appears that the world needs only 2 inverters! :-)
==> logic/josephine.p <==
The recent expedition to the lost city of Atlantis discovered scrolls
attributted to the great poet, scholar, philosopher Josephine. They
number eight in all, and here is the first.
THE KINGDOM OF MAMAJORCA, WAS RULED BY QUEEN HENRIETTA I. IN MAMAJORCA
WOMEN HAVE TO PASS AN EXTENSIVE LOGIC EXAM BEFORE THEY ARE ALLOWED TO
GET MARRIED. QUEENS DO NOT HAVE TO TAKE THIS EXAM. ALL THE WOMEN IN
MAMAJORCA ARE LOYAL TO THEIR QUEEN AND DO WHATEVER SHE TELLS THEM TO.
THE QUEENS OF MAMAJORCA ARE TRUTHFUL. ALL SHOTS FIRED IN MAMAJORCA CAN
BE HEARD IN EVERY HOUSE. ALL ABOVE FACTS ARE KNOWN TO BE COMMON
KNOWLEDGE.
HENRIETTA WAS WORRIED ABOUT THE INFIDELITY OF THE MARRIED MEN IN
MAMAJORCA. SHE SUMMONED ALL THE WIVES TO THE TOWN SQUARE, AND MADE
THE FOLLOWING ANNOUNCEMENT. "THERE IS AT LEAST ONE UNFAITHFUL HUSBAND
IN MAMAJORCA. ALL WIVES KNOW WHICH HUSBANDS ARE UNFAITHFUL, BUT HAVE
NO KNOWLEDGE ABOUT THE FIDELITY OF THEIR OWN HUSBAND. YOU ARE
FORBIDDEN TO DISCUSS YOUR HUSBAND'S FAITHFULNESS WITH ANY OTHER WOMAN.
IF YOU DISCOVER THAT YOUR HUSBAND IS UNFAITHFUL, YOU MUST SHOOT HIM AT
PRECISELY MIDNIGHT OF THE DAY YOU FIND THAT OUT."
THIRTY-NINE SILENT NIGHTS FOLLOWED THE QUEEN'S ANNOUNCEMENT. ON THE
FORTIETH NIGHT, SHOTS WERE HEARD. QUEEN HENRIETTA I IS REVERED IN
MAMAJORCAN HISTORY.
As with all philosophers Josephine doesn't provide the question, but leaves
it implicit in his document. So figure out the questions - there are two -
and answer them.
Here is Josephine's second scroll.
QUEEN HENRIETTA I WAS SUCCEEDED BY DAUGHTER QUEEN HENRIETTA II. AFTER
A WHILE HENRIETTA LIKE HER FAMOUS MOTHER BECAME WORRIED ABOUT THE
INFIDELITY PROBLEM. SHE DECIDED TO ACT, AND SENT A LETTER TO HER
SUBJECTS (WIVES) THAT CONTAINED THE EXACT WORDS OF HENRIETTA I'S
FAMOUS SPEECH. SHE ADDED THAT THE LETTERS WERE GUARENTEED TO REACH
ALL WIVES EVENTUALLY.
QUEEN HENRIETTA II IS REMEMBERED AS A FOOLISH AND UNJUST QUEEN.
What is the question and answer implied by this scroll?
==> logic/josephine.s <==
The two questions for scroll #1 were:
1. How many husbands were shot on that fateful night?
2. Why is Queen Henrietta I revered in Mamajorca?
The answers are:
If there are n unfaithful husbands (UHs), every wife of an UH knows of
n-1 UH's while every wife of a faithful husband knows of n UHs. [this
because everyone has perfect information about everything except the
fidelity of their own husband]. Now we do a simple induction: Assume
that there is only one UH. Then all the wives but one know that there
is just one UH, but the wife of the UH thinks that everyone is
faithful. Upon hearing that "there is at least one UH", the wife
realizes that the only husband it can be is her own, and so shoots
him. Now, imagine that there are just two UH's. Each wife of an UH
assumes that the situation is "only one UH in town" and so waits to
hear the other wife (she knows who it is, of course) shoot her husband
on the first night. When no one is shot, that can only be because her
OWN husband was a second UH. The wife of the second UH makes the same
deduction when no shot is fired the first night (she was waiting, and
expecting the other to shoot, too). So they both figure it out after
the first night, and shoot their husbands the second night. It is
easy to tidy up the induction to show that the n UHs will all be shot
just on the n'th midnight.
The question for scroll #2 is:
3. Why is Queen Henrietta II not?
The answer is:
The problem now is that QHII didn't realize that it is *critical* that
all of the wives, of faithful and UH's alike, to
*BEGIN*AT*THE*SAME*MOMENT*. The uncertainty of having a particular
wife's notice come a day or two late makes the whole logic path fall
apart. That's why she's foolish. She is unjust, because some wives,
honed and crack logicians all, remember, will *incorrectly* shoot
faithful husbands. Let us imagine the situation with just a SINGLE UH
in the whole country. And, wouldn't you know it, the notice to the
wife of the UH just happens to be held up a day, whereas everyone
else's arrived the first day. Now, all of the wives that got the
notice the first day know that there is just one UH in the country.
And they know that the wife of that UH will think that everyone is
faithful, and so they'll expect her to figure it out and shoot her
husband the first night. BUT SHE DIDN"T GET THE NOTICE THE FIRST
NIGHT.... BUT THE OTHER WIVES HAVE NO WAY OF KNOWING THAT. So, the
wife of the UH doesn't know that anything is going on and so (of
course) doesn't do anything the first night. The next day she gets
the notice, figures it all out, and her husband will be history come
that midnight. BUT... *every* other wife thought that there should
have been a shooting the first night, and since there wasn't there
must have been an additional UH, and it can only have been _her_
husband. So on the second night **ALL** of the husbands are shot.
Things are much more complicated if the mix of who gets the notice
when is less simple than the one I mentioned above, but it is always
wrong and/or tragic.
NOTE: if the wives *know* that the country courier service (or however
these things get delivered) is flaky, then they can avoid the
massacre, but unless the wives exchange notes no one will ever be shot
(since there is always a chance that rather than _your_ husband being
an UH, you could reason that it might be that the wife of one of the
UH's that you know about just hasn't gotten her copy of the scroll
yet). I guess you could call this case "unjust", too, since the UH's
evade punishment, despite the perfect logic of the wives.
==> logic/locks.and.boxes.p <==
You want to send a valuable object to a friend. You have a box which
is more than large enough to contain the object. You have several
locks with keys. The box has a locking ring which is more than large enough
to have a lock attached. But your friend does not have the key to any
lock that you have. How do you do it?
==> logic/locks.and.boxes.s <==
Attach a lock to the ring. Send it to her. She attaches her own lock
and sends it back. You remove your lock and send it back to her. She
removes her lock.
==> logic/mixing.p <==
Start with a half cup of tea and a half cup of coffee. Take one tablespoon
of the tea and mix it in with the coffee. Take one tablespoon of this mixture
and mix it back in with the tea. Which of the two cups contains more of its
original contents?
==> logic/mixing.s <==
Mixing Liquids
The two cups end up with the same volume of liquid they started with. The same
amount of tea was moved to the coffee cup as coffee to the teacup. Therefore
each cup contains the same amount of its original contents.
==> logic/number.p <==
Mr. S. and Mr. P. are both perfect logicians, being able to correctly deduce
any truth from any set of axioms. Two integers (not necessarily unique) are
somehow chosen such that each is within some specified range. Mr. S.
is given the sum of these two integers; Mr. P. is given the product of these
two integers. After receiving these numbers, the two logicians do not
have any communication at all except the following dialogue:
<<1>> Mr. P.: I do not know the two numbers.
<<2>> Mr. S.: I knew that you didn't know the two numbers.
<<3>> Mr. P.: Now I know the two numbers.
<<4>> Mr. S.: Now I know the two numbers.
Given that the above statements are absolutely truthful, what are the two
numbers?
==> logic/number.s <==
The answer depends upon the ranges from which the numbers are chosen.
The unique solution for the ranges [2,62] through [2,500+] is:
SUM PRODUCT X Y
17 52 4 13
The unique solution for the ranges [3,94] through [3,500+] is:
SUM PRODUCT X Y
29 208 13 16
There are no unique solutions for the ranges starting with 1,
and there are no solutions for ranges starting with numbers above 3.
A program to compute the possible pairs is included below.
#include <stdio.h>
/*
BEGINNING OF PROBLEM STATEMENT:
Mr. S. and Mr. P. are both perfect logicians, being able to correctly deduce
any truth from any set of axioms. Two integers (not necessarily unique) are
somehow chosen such that each is within some specified range. Mr. S.
is given the sum of these two integers; Mr. P. is given the product of these
two integers. After receiving these numbers, the two logicians do not
have any communication at all except the following dialogue:
<<1>> Mr. P.: I do not know the two numbers.
<<2>> Mr. S.: I knew that you didn't know the two numbers.
<<3>> Mr. P.: Now I know the two numbers.
<<4>> Mr. S.: Now I know the two numbers.
Given that the above statements are absolutely truthful, what are the two
numbers?
END OF PROBLEM STATEMENT
*/
#define SMALLEST_MIN 1
#define LARGEST_MIN 10
#define SMALLEST_MAX 50
#define LARGEST_MAX 500
long P[(LARGEST_MAX + 1) * (LARGEST_MAX + 1)]; /* products */
long S[(LARGEST_MAX + 1) + (LARGEST_MAX + 1)]; /* sums */
find(long min, long max)
{
long i, j;
/*
* count factorizations in P[]
* all P[n] > 1 satisfy <<1>>.
*/
for(i = 0; i <= max * max; ++i)
P[i] = 0;
for(i = min; i <= max; ++i)
for(j = i; j <= max; ++j)
++P[i * j];
/*
* decompose possible SUMs and check factorizations
* all S[n] == min - 1 satisfy <<2>>.
*/
for(i = min + min; i <= max + max; ++i) {
for(j = i / 2; j >= min; --j)
if(P[j * (i - j)] < 2)
break;
S[i] = j;
}
/*
* decompose SUMs which satisfy <<2>> and see which products
* they produce. All (P[n] / 1000 == 1) satisfy <<3>>.
*/
for(i = min + min; i <= max + max; ++i)
if(S[i] == min - 1)
for(j = i / 2; j >= min; --j)
if(P[j * (i - j)] > 1)
P[j * (i - j)] += 1000;
/*
* decompose SUMs which satisfy <<2>> again and see which products
* satisfy <<3>>. Any (S[n] == 999 + min) satisfies <<4>>
*/
for(i = min + min; i <= max + max; ++i)
if(S[i] == min - 1)
for(j = i / 2; j >= min; --j)
if(P[j * (i - j)] / 1000 == 1)
S[i] += 1000;
/*
* find the answer(s) and print them
*/
printf("[%d,%d]\n",min,max);
for(i = min + min; i <= max + max; ++i)
if(S[i] == 999 + min)
for(j = i / 2; j >= min; --j)
if(P[j * (i - j)] / 1000 == 1)
printf("{ %d %d }: S = %d, P = %d\n",
i - j, j, i, (i - j) * j);
}
main()
{
long min, max;
for (min = SMALLEST_MIN; min <= LARGEST_MIN; min ++)
for (max = SMALLEST_MAX; max <= LARGEST_MAX; max++)
find(min,max);
}
-------------------------------------------------------------------------
= Jeff Kenton (617) 894-4508 =
= jkenton@world.std.com =
-------------------------------------------------------------------------
==> logic/riddle.p <==
Who makes it, has no need of it. Who buys it, has no use for it. Who
uses it can neither see nor feel it.
Tell me what a dozen rubber trees with thirty boughs on each might be?
As I went over London Bridge
I met my sister Jenny
I broke her neck and drank her blood
And left her standing empty
It is said among my people that some things are improved by death.
Tell me, what stinks while living, but in death, smells good?
All right. Riddle me this: what goes through the door without
pinching itself? What sits on the stove without burning itself? What
sits on the table and is not ashamed?
What work is it that the faster you work, the longer it is before
you're done, and the slower you work, the sooner you're finished?
Whilst I was engaged in sitting I spied the dead carrying the living.
I know a word of letters three. Add two, and fewer there will be.
I give you a group of three. One is sitting down, and will never get
up. The second eats as much as is given to him, yet is always hungry.
The third goes away and never returns.
Whoever makes it, tells it not. Whoever takes it, knows it not. And
whoever knows it wants it not.
Two words, my answer is only two words.
To keep me, you must give me.
Sir, I bear a rhyme excelling
In mystic force and magic spelling
Celestial sprites elucidate
All my own striving can't relate
There is not wind enough to twirl
That one red leaf, nearest of its clan,
Which dances as often as dance it can.
Half-way up the hill, I see thee at last
Lying beneath me with thy sounds and sights --
A city in the twilight, dim and vast,
With smoking roofs, soft bells, and gleaming lights.
I am, in truth, a yellow fork
From tables in the sky
By inadvertent fingers dropped
The awful cutlery.
Of mansions never quite disclosed
And never quite concealed
The apparatus of the dark
To ignorance revealed.
Many-maned scud-thumper,
Maker of worn wood,
Shrub-ruster,
Sky-mocker,
Rave!
Make me thy lyre, even as the forests are.
What if my leaves fell like its own --
The tumult of thy mighty harmonies
Will take from both a deep autumnal tone.
This darksome burn, horseback brown,
His rollock highroad roaring down,
In coop and in comb the fleece of his foam
Flutes and low to the body falls home.
I've measured it from side to side,
'Tis three feet long and two feet wide.
It is of compass small, and bare
To thirsty suns and parching air.
My love, when I gaze on thy beautiful face,
Careering along, yet always in place --
The thought has often come into my mind
If I ever shall see thy glorious behind.
Then all thy feculent majesty recalls
The nauseous mustiness of forsaken bowers,
The leprous nudity of deserted halls --
The positive nastiness of sullied flowers.
And I mark the colours, yellow and black,
That fresco thy lithe, dictatorial thighs.
When young, I am sweet in the sun.
When middle-aged, I make you gay.
When old, I am valued more than ever.
I am always hungry,
I must always be fed,
The finger I lick
Will soon turn red.
All about, but cannot be seen,
Can be captured, cannot be held,
No throat, but can be heard.
I am only useful
When I am full,
Yet I am always
Full of holes.
If you break me
I do not stop working,
If you touch me
I may be snared,
If you lose me
Nothing will matter.
If a man carried my burden
He would break his back.
I am not rich,
But leave silver in my track.
Until I am measured
I am not known,
Yet how you miss me
When I have flown.
I drive men mad
For love of me,
Easily beaten,
Never free.
When set loose
I fly away,
Never so cursed
As when I go astray.
I go around in circles
But always straight ahead,
Never complain
No matter where I am led.
Lighter than what
I am made of,
More of me is hidden
Than is seen.
I turn around once,
What is out will not get in.
I turn around again,
What is in will not get out.
Each morning I appear
To lie at your feet,
All day I will follow
No matter how fast you run,
Yet I nearly perish
In the midday sun.
Weight in my belly,
Trees on my back,
Nails in my ribs,
Feet I do lack.
Bright as diamonds,
Loud as thunder,
Never still,
A thing of wonder.
My life can be measured in hours,
I serve by being devoured.
Thin, I am quick
Fat, I am slow
Wind is my foe.
To unravel me
You need a simple key,
No key that was made
By locksmith's hand,
But a key that only I
Will understand.
I am seen in the water
If seen in the sky,
I am in the rainbow,
A jay's feather,
And lapis lazuli.
Glittering points
That downward thrust,
Sparkling spears
That never rust.
You heard me before,
Yet you hear me again,
Then I die,
'Till you call me again.
Three lives have I.
Gentle enough to soothe the skin,
Light enough to caress the sky,
Hard enough to crack rocks.
You can see nothing else
When you look in my face,
I will look you in the eye
And I will never lie.
Lovely and round,
I shine with pale light,
grown in the darkness,
A lady's delight.
At the sound of me, men may dream
Or stamp their feet
At the sound of me, women may laugh
Or sometimes weep
When I am filled
I can point the way,
When I am empty
Nothing moves me,
I have two skins
One without and one within.
My tines be long,
My tines be short
My tines end ere
My first report.
What am I?
==> logic/riddle.s <==
Who makes it, has no need of it. Who buys it, has no use for it. Who
uses it can neither see nor feel it.
coffin
Tell me what a dozen rubber trees with thirty boughs on each might be?
months of the year
As I went over London Bridge
I met my sister Jenny
I broke her neck and drank her blood
And left her standing empty
gin
It is said among my people that some things are improved by death.
Tell me, what stinks while living, but in death, smells good?
pig
All right. Riddle me this: what goes through the door without
pinching itself? What sits on the stove without burning itself? What
sits on the table and is not ashamed?
the sun
What work is it that the faster you work, the longer it is before
you're done, and the slower you work, the sooner you're finished?
roasting meat on a spit
Whilst I was engaged in sitting I spied the dead carrying the living.
a ship
I know a word of letters three. Add two, and fewer there will be.
'few'
I give you a group of three. One is sitting down, and will never get
up. The second eats as much as is given to him, yet is always hungry.
The third goes away and never returns.
stove, fire, and smoke
Whoever makes it, tells it not. Whoever takes it, knows it not. And
whoever knows it wants it not.
counterfeit money
Two words, my answer is only two words.
To keep me, you must give me.
your word
Sir, I bear a rhyme excelling
In mystic force and magic spelling
Celestial sprites elucidate
All my own striving can't relate
???
There is not wind enough to twirl
That one red leaf, nearest of its clan,
Which dances as often as dance it can.
the sun, Samuel Taylor Coleridge
Half-way up the hill, I see thee at last
Lying beneath me with thy sounds and sights --
A city in the twilight, dim and vast,
With smoking roofs, soft bells, and gleaming lights.
the past, Longfellow
I am, in truth, a yellow fork
From tables in the sky
By inadvertent fingers dropped
The awful cutlery.
Of mansions never quite disclosed
And never quite concealed
The apparatus of the dark
To ignorance revealed.
lightning, Emily Dickinson
Many-maned scud-thumper,
Maker of worn wood,
Shrub-ruster,
Sky-mocker,
Rave!
Portly pusher,
Wind-slave.
the ocean, John Updike
Make me thy lyre, even as the forests are.
What if my leaves fell like its own --
The tumult of thy mighty harmonies
Will take from both a deep autumnal tone.
the west wind, Percy Bysshe Shelley
This darksome burn, horseback brown,
His rollock highroad roaring down,
In coop and in comb the fleece of his foam
Flutes and low to the body falls home.
river, Gerard Manley Hopkins
I've measured it from side to side,
'Tis three feet long and two feet wide.
It is of compass small, and bare
To thirsty suns and parching air.
the grave of a child, Wordsworth
My love, when I gaze on thy beautiful face,
Careering along, yet always in place --
The thought has often come into my mind
If I ever shall see thy glorious behind.
the moon, Sir Edmund Gosse
Then all thy feculent majesty recalls
The nauseous mustiness of forsaken bowers,
The leprous nudity of deserted halls --
The positive nastiness of sullied flowers.
And I mark the colours, yellow and black,
That fresco thy lithe, dictatorial thighs.
spider, Francis Saltus Saltus
When young, I am sweet in the sun.
When middle-aged, I make you gay.
When old, I am valued more than ever.
wine
I am always hungry,
I must always be fed,
The finger I lick
Will soon turn red.
fire
All about, but cannot be seen,
Can be captured, cannot be held,
No throat, but can be heard.
wind
I am only useful
When I am full,
Yet I am always
Full of holes.
sieve (or sponge)
If you break me
I do not stop working,
If you touch me
I may be snared,
If you lose me
Nothing will matter.
heart
If a man carried my burden
He would break his back.
I am not rich,
But leave silver in my track.
snail
Until I am measured
I am not known,
Yet how you miss me
When I have flown.
time
I drive men mad
For love of me,
Easily beaten,
Never free.
gold
When set loose
I fly away,
Never so cursed
As when I go astray.
?
I go around in circles
But always straight ahead,
Never complain
No matter where I am led.
wagon wheel
Lighter than what
I am made of,
More of me is hidden
Than is seen.
iceberg
I turn around once,
What is out will not get in.
I turn around again,
What is in will not get out.
stopcock
Each morning I appear
To lie at your feet,
All day I will follow
No matter how fast you run,
Yet I nearly perish
In the midday sun.
shadow
Weight in my belly,
Trees on my back,
Nails in my ribs,
Feet I do lack.
ship
Bright as diamonds,
Loud as thunder,
Never still,
A thing of wonder.
waterfall? (fireworks?)
My life can be measured in hours,
I serve by being devoured.
Thin, I am quick
Fat, I am slow
Wind is my foe.
candle
To unravel me
You need a simple key,
No key that was made
By locksmith's hand,
But a key that only I
Will understand.
cipher
I am seen in the water
If seen in the sky,
I am in the rainbow,
A jay's feather,
And lapis lazuli.
blue
Glittering points
That downward thrust,
Sparkling spears
That never rust.
icicle
You heard me before,
Yet you hear me again,
Then I die,
'Till you call me again.
echo
Three lives have I.
Gentle enough to soothe the skin,
Light enough to caress the sky,
Hard enough to crack rocks.
water
You can see nothing else
When you look in my face,
I will look you in the eye
And I will never lie.
your reflection
Lovely and round,
I shine with pale light,
grown in the darkness,
A lady's delight.
pearl
At the sound of me, men may dream
Or stamp their feet
At the sound of me, women may laugh
Or sometimes weep
music
When I am filled
I can point the way,
When I am empty
Nothing moves me,
I have two skins
One without and one within.
sails?
My tines be long,
My tines be short
My tines end ere
My first report.
What am I?
lightning
==> logic/river.crossing.p <==
Three humans, one big monkey and two small monkeys are to cross a river:
a) Only humans and the big monkey can row the boat.
b) At all times, the number of human on either side of the
river must be GREATER OR EQUAL to the number of monkeys
on THAT side. ( Or else the humans will be eaten by the monkeys!)
==> logic/river.crossing.s <==
The three columns represent the left bank, the boat, and the right bank
respectively. The < or > indicates the direction of motion of the boat.
HHHMmm . .
HHHm Mm> .
HHHm <M m
HHH Mm> m
HHH <M mm
HM HH> mm
HM <Hm Hm
Hm HM> Hm
Hm <Hm HM
mm HH> HM
mm <M HHH
m Mm> HHH
m <M HHHm
. Mm> HHHm
. . HHHMmm
==> logic/ropes.p <==
Two fifty foot ropes are suspended from a forty foot ceiling, about
twenty feet apart. Armed with only a knife, how much of the rope can
you steal?
==> logic/ropes.s <==
Almost all of it. Tie the ropes together. Climb up one of them. Tie
a loop in it as close as possible to the ceiling. Cut it below the
loop. Run the rope through the loop and tie it to your waist. Climb
the other rope (this may involve some swinging action). Pull the rope
going through the loop tight and cut the other rope as close as
possible to the ceiling. You will swing down on the rope through the
loop. Lower yourself to the ground by letting out rope. Pull the
rope through the loop. You will have nearly all the rope.
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