The worlds first riddle!

Iacomus - I think you are clever enough; but it seems that this time I did nothing more than mention the von Neumann story; the Iacomus story comes from you.

Try : Jane's new phone number : the product of digits is 55125.

Football : 5.647 times per season.

I believe snow started 2 hours before noon (at 10 in the morning).

Correction to Snow puzzle
I just saw a mistake I made with the Snow puzzle: what I posted is 1/t instead of t.
It is really 1/2 of an hour before noon that it started to snow.
What you get out of the equations in the end is ratio Rs/H0 where H0 is height of snow at noon and Rs is the rate at which snow is falling.
Time it takes to get to H0 is H0/Rs , so i made a reverse answer.

Wine/water

You have a bottle of wine, containing 1 liter of that delicious drink. You also have an equal bottle with 1 liter of water. You pour a glass of wine from the wine bottle to the water bottle, and stirr the fluid to mix it completely. You now have a solution of wine in water. So you pour a glass of this mixture back to the wine bottle, and mix it well. Which bottle contains a purer mixture? Assume water was 100% pure and wine was 100% pure. Forget about wine already containing water - treat it as fundamentally different substance.


A moon M is orbiting a planet X in a distant system. Astronomers on X have determined that M is always showing the same side to observers on planet X. Astronomers have also determined that M appears at exactly the same spot in the sky every 5 days, as seen from the Royal Observatory in the capital city. What is the speed of rotation of M around it's axis?

A harder one:
You go to a party organised by Zings. Zings are funny looking logicians; every Zing has not more than three enemies. Prove that the party Zings can be divided into two groups in such a way that every Zing will have not more than one enemy to
contend with in the same group. (If A is the enemy of B, then B
is the enemy of A).

An ant?

I was hoping to post my answer for the Simpsons presents today but I left all my paper at home so, tomorrow then. Have been too busy arguing about a trolley to think it through to the finish. How was Lacomus' answer?

Can you provide the Chinese,otherwise I can help you to find out...

PMs sent to both Tryagain and Relative - not necessarily in that order.

I'd give up on that 'Jack and Jill' rubbish if I were you, Try. I got as far as 'up the hill to fetch a pail of water' and threw the book away. Whoever heard of water UP a hill? I wish these books would just GET IT RIGHT!!!!

Welcome Max. It is always good to see a new name in the Riddles section. If I failed to give the reasoning behind the China question. I do so now.

Intuitively, it may feel that the families are adopting a strategy favouring producing a son, but this is incorrect. The families expected number of sons is one, by the definition of the strategy. Nevertheless, the expected number of daughters does not drop. Consider:

There is a 1-in-2 chance that the son is the first-born, and so no daughters are born.
There is a 1-in-4 chance that the son is the second born, and so one daughter is born.
There is a 1-in-8 chance that the son is the third born, and so two daughters are born.

Therefore, the expected number of daughters is (.50 0 daughters) + (.25 1 daughter) + (.125 * 2 daughters)... = 0.5 daughters.



Hopeful answers to Relative's three questions.

Wine.
There is as much wine in the Water bottle as water in the Wine bottle.

Speed of rotation. (Don't laugh)
320 kph

Zings.

Three-Valued Logic
In three unnumbered pages from his unpublished notes written before 1910, Peirce developed what amounts to a semantics for three-valued logic. This is at least ten years before Emil Post's dissertation, which is usually cited as the origin of three-valued logic.

In his notes, Peirce experiments with three symbols representing truth values: V, L, and F. He associates V with "1" and "T", indicating truth. He associates F with "0" and "F", indicating falsehood. He associates L with "1/2" and "N", indicating perhaps an intermediate or unknown value.
Peirce defines a large number of unary and binary operators on these three truth values. The semantics for the operators is indicated by truth tables. Two examples are given here. First, the bar operator (indicated here by a minus sign) is defined as follows:
x V L F

-x F L V
Applied to truth the bar operator yields falsehood, applied to unknown it yields unknown and applied to falsehood it yields truth.
The Z operator is a binary operator which Peirce defines as follows:
V L F

V | V L F
L | L L F
F | F F F

Thus, the Z operator applied to a falsehood and anything else yields a falsehood. The Z operator applied to an unknown and anything but a falsehood yields an unknown. And the Z operator applied to a truth and some other value yields the other value.

The bar operator and the Z operator provide the essentials of a truth-functionally complete strong Kleene semantics for three-valued logic. In addition to these two strong Kleene operators, Peirce defines several other forms of negation, conjunction, and disjunction. The notes also provide some basic properties of some of the operators, such as being symmetric and being associative.

Calculus of Relations
Building on ideas of De Morgan, Peirce fruitfully applied the concepts of Boolean algebra to relations. Boolean algebra is concerned with operations on general or class terms. Peirce applied the same idea to what he called "relatives" or "relative terms." While his ideas evolved continually over time on this subject, fairly definitive presentations are found in Peirce (1870) and Peirce (1883). The calculus of relatives is developed further in Tarski (1941). A history of work on the subject is Maddux (1990).

Given relative terms such as "friend of" and "enemy of" (more briefly "f" and "e"), Peirce studied various operations on these terms such as the following:
(union) friend of or enemy of
A pair <a, b> stands in this relation if and only if if stands in one or both of the relations. In symbols "f + e".
(intersection) friend of and enemy of
A pair <a, b> stands in this relation if and only if if stands in both of the relations. In symbols "f . e".
(relative product) friend of an enemy of
A pair <a, b> stands in this relation if and only if there is a c such a is a friend of c and c is an enemy of b. In symbols "f ; e".
(relative sum) friend of every enemy of
A pair <a, b> stands in this relation if and only if a is the friend of every object c that is the enemy of b. In symbols "f , e" (Peirce uses a dagger rather than a comma)
(complement) is not a friend of
A pair <a, b> stands in this relation if and only if <a, b> does not stand in the friend-of relation. In symbols "-f" (Peirce places a bar over the relative term).
(converse) is one to whom the other is friend
A pair <a, b> stands in this relation if and only if b is a friend of a. In symbols "~f" (Peirce places an upwards facing semi-circle over the relative term).
Peirce presented numerous theorems involving his operations on relative terms. Examples of the numerous such laws identified by Peirce are:
~(r + s) = ~r + ~s
-(r ; s) = -r , -s
(r . s) , t = (r , s) . (r , t)

Peirce's calculus of relations has been criticized for remaining unnecessarily tied to previous work on Boolean algebra and the equational paradigm in mathematics. It has been frequently claimed that real progress in logic was only realized in the work of Frege and later work of Peirce in which the equational paradigm was dropped and the powerful expressive ability of quantification theory was realized.

Nevertheless, Peirce's calculus of relations has remained a topic of interest to this day as an alternative, algebraic approach to the logic of relations. It has been studied by Lowenheim, Tarski and others. Lowenheim's famous theorem was originally a result about the calculus of relations rather than quantification theory, as it is usually presented today.

Therefore, you can see, I have no idea.

Logicians' jokes

A logician saves the life of a space alien and is rewarded with an offer to
answer any question. After a thought he asks: What is the best question to ask and the correct answer to it? After a brief panic the alien consults her computer and says:

" The best question to ask is the one you just did and the correct answer to it is the one I gave."

In theory, there is no difference between theory and practice.

But, in practice, there is.


Complete the following equations with +, -, *, /. Each number must appear in your equations (and only once).

1 2 3 4 = 28
e.g. [(2*3)+1]*4 = 28

2 3 4 5 = 28

3 4 5 6 = 28

4 5 6 7 = 28


11 scientists want to lock up important documents in a box. They want to attach padlocks to this box so that (any) 6 or more scientists can open the box, but 1 to 5 scientists should not be able to open it.

How many locks must be installed and how many keys must be distributed among the scientists

Try

Such a PM was definitely sent. I will check on it and, in the meantime, I will send a PM regarding the equations.

Worm puzzle - the answer
Well, here it is - an attempt to clarify the worm puzzle posted a while back.

I tried to find an easy explanation, alas I could not.

The worm will get to the end of the rubber band, although it seems impossible! The reasoning can go like this (a little cheat here - the worm moves first, then the rubber stretches):
After one second, the worm will move 1 centimeter, then the rubber band will expand one meter, carrying the worm with it. The 99 centimeters between the worm and the far side of the rubber band will expand 99% of that meter, while the 1% of the expanded meter will be already behind the worm.
Percentage of already traversed rubber band is not affected by the rubber stretching, but is affected by the worm moving, making it less and less. Also because of constant stretching velocity the space between worm and the far end increases by less every second - the relative speed between far end and the worm decreases.
Eventually it should become close to zero, and the worm will catch up.


For those not quite satisfyed with this (like myself), here are the equations. I could not make them simpler - in fact I struggled with them for quite some time and I'm happy that I finally got them at all

t - time
L(t) - length of rubber band, depends on time
l0=L(0) - initial length of band = 1 meter
v0 - velocity of far side of rubber band = 1 m/s
X(t) - position of worm on rubber band, depends on time
v1 - relative local velocity of worm relating to rubber band (1/100 m/s)
W(t) = X'(t) - velocity of worm

L(0) = l0 = 1 m
v0 = 1 m/s
X(0) = 0 m
v1 = 0.01 m/s

We can write the following equation:

W(t) = X'(t) = v1 (worm relative to band locally) + vs(velocity of stretching of rubber band at position of worm)

vs can be expressed as v0*X(t)/L(t) since the velocity of stretching is proportional to offset from fixed end of the band, and is v0 at the free (moving) end.

L(t) = v0*t+l0

We have :

X'(t) = X(t)*v0/(v0*t + l0) + v1

after some rearranging

(eq. 1) X'(t) = X(t)/(t + l0/v0) + v1

Then we solve this differential equation or we simply see that derivative of

y(x) = A*x*ln(x)
is
y'(x) = A*ln(x) + A= y(x)/x + A

When we use this insight on eq. 1 we obtain equation of motion for the worm:

(eq. 2) X(t) = v1*(t + l0/v0)*ln(t + l0/v0)


Now when the worm has reached the far end of the band, the following will hold:

X(t) = L(t)

or

v1*(t + l0/v0)*ln(t + l0/v0) = v0*t + l0 = v0 (t + l0/v0)

If we substitute z = (t + l0/v0)

v1*z*ln(z) = v0 * z

If z<>0, then

ln(z) = v0/v1 => z= e^(v0/v1) = e^100

t = z - l0/v0 = e^100 - 1

So the worm will reach the far end in e^100 - 1 seconds, which is a fascinating number - I think it is much more than the age of the universe!

This problem maybe relates to the movement in our expanding universe, where the far end of the rubber band is a distant galaxy and the worm is the spaceship.

Wine : as Iacomus & Try have pointed out, there is the same amount of water in wine as there is wine in water. This can be seen best if we think of boys and girls instead:
You have a group of ten boys and a group of ten girls. Four boys mix with girls and then four children are returned back to the boy's group. Are there more boys among girls or girls among boys? Of course, the same number. As many boys as are missing in the boys' group were replaced by girls.

The same trick is sometimes applied to a deck of cards to do some 'magic'.

Moon : the main question is "What is the speed of rotation of M around it's axis?". A revolutions/day result is expected. I will wait some more for the correct answer. Nice Try, but Tryagain .

I like the 'Calculus of Relatives' .
But the Zings are still waiting to be grouped.

Is this stretching creditability too far?

Original question.

"At one end of the band is a snail, travelling at a speed of 1cm/second (a supersnail) trying to get to the other side. But, alas, the rubber band is stretching, it's far end moving at a horrible velocity of 1 meter/second. Now this band is a superband also, and is infinitely stretchable."

Note: No mention of who goes first, or, more importantly. "Percentage of already traversed rubber band is not affected by the rubber stretching "

Would this mean that the band Never stretches 1 meter, or, it's first move would be .99? However, even with these restrictions;

The worm is one meter behind at the start. After the first second, the worm is 1.89 behind. The band having expanded .99 + its start length, and pulling the snail along another .9cm. The 2nd second the snail is 2.89.

Nevertheless, would the band now continue to expand at a full meter. Therefore the worm will never catch-up.

On the other hand, if, the band can only increase a diminishing percentage of 1 meter, and the worm gains 1cm+an increasing percentage of the bands increase; he/she will be there before I finish digging this field.

Well, one out of three is good for me :wink:
Damn, those Zings.

Rubber band:

The band stretches 1 meter every second. The snail crawling along the band is crawling at a speed 1 cm/second locally, that is relative to the point on the rubber band that he currently is at.
That means that the stretching of the rubber band carries the snail with it : imagine the moment when the snail is just 1 cm or so before the end. The rubber band is still stretching at 1m/second, but the 1cm space between the snail and the rubber band end will only stretch for a small fraction. The stretch appears at all points on the band so small sections of it will stretch small amounts. Now how can the snail get to this great position only 1 cm to the end? By being about 2 cm to the end at a previous moment.. so imagine a backwards process, the rubber band contracting this time and the snail getting further away every second.



There is no mention of who goes first because it happens simultaneously. The reasoning is not correct, it is only an illustration. If you imagine, however, that the steps performed (snail-band-snail...) get smaller and smaller, you have something called 'a limit process' which is an essential tool for analysing dynamical systems.

Try

I will send you a PM concerning those locks. I hesitate to call it a solution because I find it intuitively unacceptable. (It just 'smells' wrong, but the numbers seem to work)

Something a little easier.

Abel, Mabel, and Calib went bird watching. Each of them saw one bird that none of the others did. Each pair saw one bird that the third did not. In addition, one bird was seen by all three.
Of the birds, Abel saw, two were yellow. Of the birds, Mabel saw, three were yellow. Of the birds, Calib saw, four were yellow.

How many yellow birds were seen in all
How many non-yellow birds were seen in all


A gambler bet on a horse race, but the bookie would not tell him the results of the race. The bookie gave clues as to how the five horses finished -- which may have included some ties -- and wouldn't pay the gambler off unless the gambler could determine how the five horses finished based on the following clues:

Penuche Fudge finished before Near Miss and after Whispered Promises.
Whispered Promises tied with Penuche Fudge if and only if Happy Go Lucky did not tie with Skipper's Gal.

Penuche Fudge finished as many places after Skipper's Gal as Skipper's Gal finished after Whispered Promises if and only if Whispered Promises finished before Near Miss.

The gambler thought for a moment, then answered correctly.
How did the five horses finish the race

Double damn those Zings. I still do not have the answer.

Try

PM sent for Birds and horses.

As for Zings - I no longer care what Zings do at parties, but I am losing sleep over HOW the solution is arrived at. To blazes with the result, I want the method!

Ok, Zings will explain it all
tomorrow!

They will give you the following solution:

"We discovered early on that any party with just three Zings can be split into two groups.
From then on, we invite one more Zing to each subsequent party.
We divide the previous members as we did the previous time, and then the new Zing chooses his
(at most) three enemies. In the process, some of his new enemies might be forced to give up
some of their old enemies - which is a good thing.

There are two options:
1. All his enemies are in the same group. In this case, we assign him to the other group. And everybody is happy!

2. One of his enemies (say Z1) is in group A, others in group B. Assign him to house A.
This situation needs more attention:

Lets mark the new Zing as Z0. We have:
2a. Z1 has no other enemies in A but Z0. The rules are satisfied.
2b. Z1 already has an enemy in house A. In this case, move Z1 to B.
Now Z1 has at most 1 enemy in B, since he already has 2 in A.
Let's call this zing Z2. Now the same situation as in step 2 happens.
Either Z2 has no other enemy in B and we are done, or he has one,
in this case we move him. There, his enemy Z3 might be waiting and so on.

There are two possible outcomes of this chain-swap:

2b1.The swap ends with some Zx having a single enemy (the Z(x-1)) in the group.

2b2.The swap encounters a Zing that was already encountered before. In this case we have
a circular path, starting with Zc1, then coutinuing Zc1-Zc2-..Zcl--Zc1.
How did we encounter the circle? Let's say via Zb. He was moved, say from A to B,
because in A he already had two enemies. Starting from Zc1, all members of the circle
switch sides. The circle is finished by moving Zclast. Now Zclast (after the move) is either
on the same side as Zc1, or opposite.
- If Zc1 and Zclast are on the opposite sides, we are done!
- If on the same side, we are finished too. Zclast has only one enemy in his group, and this is Zc1.
The other two enemies are Zc(one but last), which is on opposite side because of circular swap,
and another one that was in the same group before the move (the reason for it), so he is in the other group too.
Zc1 also has one enemy in this group only : Zclast. The other enemies are Zc2 (moved across) and Zcb
(also in the other group).
- This is just the same if Z0 is the chain-starter. Since he only had Z1 on the same side before the swaps,
he can only have one after the swaps.


We have established this procedure, and this allows us to split up Zings on just _every_ and _any_ party! Cheers! "


Note that I had to go to the Zing-party myself to find out this for myself. The mighty Google only revealed the problem, but Zings would not say how they do it! So this is my best effort at interpreting the great Zings, any comments/corrections are welcome!

The Moon problem:

I must admit I invented this puzzle myself - which is probably no great secret - and I regret it. It turned out that the solution is neither elegant nor simple, it is not very interesting and is further spoiled by not being a single solution at all. WHAT was I thinking???
Maybe we could create an 'Original Riddles' thread and see what happens..

Anyway, here is an attempt at a solution.-.

Astronomers say that they observe M at the same spot in the sky every 5 days. One day on a planet is of course measured in that planet's rotations around it's axis.
If the M rotates around the P around the same axis as planet's own rotation, then it must rotate in the same direction also - otherwise it would fly over a point on planet's surface at least once per day!
If we imagine a point on planet's surface where the Royal Observatory is, the line from center through this point hits moon at starting moment.
after that the moon lags behind - it's rotation being slower than planet's rotation - until after 5 days it is 'caught' by the spot! The planet rotated 5 times, and the moon rotated 4 times around the planet during that time.
The other option is that the moon is faster than the planet - again one lap in 5 days - making moon rotate 6 times around the planet in five days.

Rotation of M around P matches rotation of M around it's axis because it always shows the same side to M!

This means that M turns around it's axis 4 or 6 times in 5 days!


***ADDITIONS***
****Now in case that M's orbit is at an angle relative to P's axis, things are different. M rotates with frequency f so that in 5 days it makes 5*f rotations, where 5*f is a whole number, and f is not a whole number. So f is k/5 where k is not a multiple of 5.

So you see, I will not make any more puzzles.