The worlds first riddle!

Frankapisa

My attempt at the logic puzzle:

It seems to me that if he said "You will refuse me both the picture and a kiss".
If she then refuses him both he has made a true statement and she must give him the picture.
But if she gives him the picture it becomes false and she cannot give him the picture. A paradox, but this paradox can be broken.
If she gives him the kiss then the statement was false and she does not have to give him the picture.

Try

I offer this for your consideration. Your solution to the diamond problem might be questionable in physics even though it works in geometry, and for the following reason:

The dimensions you gave give a volume of 3.5 cu ft per board.
Wood averages at 30 lbs per cu ft, giving these boards a weight that can safely be assumed to be 100lbs each or more.

In your solution the diamondeer is standing on a 16 in board over acid. He must bridge the space between where he is standing and the island. The bpoard weighs around 100 lbs and he has to hold it somewhere in the region of one end else he will not reach the island with it.

Assume he has one hand two foot from the end and the other hand at the end (This assumes that the distance he has to span is no more then 14 ft)

The leverage of the extended part of the board will result in the diamondeer having to apply over 300 lbs of force with each hand, and this is just to support the board. The turning force that has to be resisted would be a lot more than this and he has a base to work from of no more than sixteen inches. The ratio of 16ins to 16 ft is clearly 12:1. The board weighs at least 100 lbs, so he has to resist a torque of approaching - or even exceeding - 1200 lbs, or around half a ton. And this is over a bath of acid.

I have a better idea than collecting the diamond. I will give everyone who asks a copy of your proof, point out the diamond to them, then sit back and wait to collect the money I insured them for.

Excellent. You got it.

Suppose the guy not only wanted the kiss -- but actually wanted the picture as well.

Can you come up with a statement that would get him both?

(rope arounr the world)
+ 2PI*1yard approx 6 yards.

Here's another: one guy decided not to make it elevated 1 yard all around, just support it with a stick 1 meter high. What now?

Relative

And the roller prob: 4 cubits.

4 quarters ?

And I don't understand that self-referencing prose..

Try
Pyramid: The block moves 4 cubits.
Quarters: There can be four quarters so that each touches every one of the others.
Paragraph: Letter frequency.

Diamond: The point is not whether my solution would or not work; the point is that the solution - AS GIVEN - could not. It is nothing to do with my ego but a continuation of the puzzle. You say we have to work with what we have and in puzzles that usually means 'within the information given'. In this case these dimensions of the boards were given as part of the puzzle!
You say 'what if it were a scaffold board at less weight?' I am willing to bet you at 10:1 in your favor that even with an ordinary scaffold board of say, 10 ft length, you could not lift it clear of the ground by holding it only at one end. And that is within the information given in the puzzle.

Frank

I'm still thinking about your alternative. There would still need to be a paradox that could be 'defused' and it seems that the one used in the first form of the problem has to be it. This is far more difficult - for me at least - than the original form.

Frank

The kiss and the photograph.

The best I can figure is something like this:

His statement is that she will either refuse him the photograph OR give him the kiss OR both.

As he does not have the photograph, and as no kiss has been given, his statement is true and she must give him the photograph. But his statement is now false and he has to relinquish the photograph, whereupon it is true again. The paradox.

The only other option open to her is to kiss him. If she does this the statement becomes true 'for all time' and he gets the photograph too. So either she is caught in the paradox, or he gets both the kiss and the photograph

If there are any more of these I suggest we start using some kind of algebraic notation.

Not really sure -- but the problem bears a resemblance to a little test golfers sometimes quiz each other about.

How far does a golf ball (American sized) go in a full turn on the green.

The distance is huge.

A golf ball has a diameter of about an inch and a half -- but if put on the ground and rolled one full turn -- it goes about 6 inches.

*****



I can get five more to touch.


****



Just about instantly, Tryagain.




(My response up above continues the unusual.)




One extra yard will do the trick.

Try putting that into a statement. It truly does not work -- in part because the "kiss" is not introduced until the statement is made.

But we can look at it if you actually make a statement that the student can make to the young lady.

Remember, the student has merely gotten the young lady to promise that if his statement is true, she will give him her photograph -- and if the statement is false, she promises not to give him her photo.

So what statement can he make to her that will get him both a kiss (as you discovered earlier) and the photo?

By the way -- these are Smullyan riddles -- and ALL Smullyan riddles can be solved by algebra.

But I do not know an "x" from a "y" -- so I have to use reasoning.

I did manage to reason these riddles out -- but he has some that algebra absolutely has to be used.

The diamond problem again. (I'm getting more 'brain-teasing' out of deconstructing the puzzle that I did from the puzzle itself)

If three of the boards are floated in the acid, parallel and equidistant, so that the outer edges are 16 ft apart, and if the three other boards are laid on these at right angles to the other boards, parallel and equidistant; there will be a square raft of 16 ft sides.

The diagonal of this raft will be in the close order of 22.5 ft, which is greater than the distance to the island. With the raft correctly positioned, he can walk to the island and collect the diamond. The 21cu ft of timber will provide plenty of buoyancy.

None of this solution goes outside of the information given in the problem and none of it is beyond human musculature. No animals were injured in the making of this solution.

Frank

I don't know that I agree with you. 'OR' has always been an operator in logic and this is merely an 'inclusive or' statement. It could be written algebraically as (with 's' being 'statement', 'p' being 'photograph', 'k' being 'kiss', and '-' being 'not')

"s = k OR -p "

But you did say 'no algebra' so, in words:

The statement, m'dear, is "Either you will kiss me or you will refuse me the photograph"

That seems like a straightforward enough statement to me.

The diamond problem again. (I'm getting more 'brain-teasing' out of deconstructing the puzzle that I did from the puzzle itself)

"If three of the boards are floated in the acid."

Unfortunately, the lake is not water but highly concentrated acid , the result of ignoring paranoid environmentalists.

This is of cause part of the problem with words. Whilst, I agree with you it would be impossible to hold it out over the acid. . Adrian's excellent answer would still apply if the last plank was held vertically, and allowed to fall to the island with only say a foot to stop its bounce.
However, as I seems slightly unlikely that we will ever find ourselves in a similar situation perhaps we will never know for sure. :wink:

The pyramids.
The block moves about 4 cubits, or a little over two yards. If you were thrown off by this deceptively simple problem, you probably forgot that while a point on the circumference of the roller moves two cubits relative to a point on the ground per rotation, that point on the roller also moves two cubits relative to a point on the block.

Five (5) is the maximum
You can arrange up to five quarters in a way that each quarter touches every other quarter. Starting with one horizontal quarter as the base, arrange two additional quarters to form a small "tent" over the first, resting lightly on top of it. The tricky part is adding two more. These two must touch each other at the tip, with their bases just barely filling the gap left by the others.


Relative wrote "
"And I don't understand that self-referencing prose.."
Nor do I. Too what do refer? Are you by chance, a relative of Adrian?

Rope around the world.
The circumference of a circle is 2 x pi x r, where r is the radius of the circle. If you want a rope that is one yard above the ground, this radius is larger by one yard. Let R be this new radius. So R = r + 1.
Let x be the amount of extra rope required by the eccentric's rival. So:
x = (2 x pi x (r + 1)) - (2 x pi x r) x = (2 x pi x r) + (2 x pi) - (2 x pi x r) x = 2 x pi
So x is about 6.2832 yards.

Note that this answer does not depend on the radius of the circle. If the eccentric and his rival were attempting to tie up a baseball rather than the earth, the amount of additional required rope would be the same amount.

Relative quote "Here's another: one guy decided not to make it elevated 1 yard all around, just support it with a stick 1 meter high. What now?"
Very little difference.

I leave you with;
You've angered the spirit of Halloween by failing to revere the Great Pumpkin, and now a curse has befallen you. On the walkway to your house is a Ward of Seven Jack O' Lanterns arranged in a circle. If midnight comes and any of the seven are still lit, a dark reaper and seven dark horses with seven dark riders shall visit thy abode. They shall surround thy domicile and circle it seven times seven times. And the seven riders having circled thy dwelling seven times seven times, they shall proceed with the throwing of the eggs and the cream of shaving. And come morn there will be a great mess to be reckoned with. Verily.

Therefore, you had better get those lanterns out.
You quickly discover something odd about these lanterns. When you blow out the first one, the lanterns on either side extinguish as well! But there's more- if you blow out a lantern adjacent to one that is extinguished, the extinguished one(s) will relight. It seems that blowing on any lantern will change the state of three- the one you blew on, and its two neighbours. Finally, you can blow on an extinguished lantern and it will relight, and its two neighbours will extinguish/ignite.

After a frustrating exercise of playing "Whack-A-Mole" with the lanterns, you find yourself back where you started, with all seven lit. Being the excellent puzzler you are, you sit down and examine the puzzle logically. However, be quick, hoof steps approach.

Lanterns:
I can do it in seven blows. Does anyone have a solution that takes fewer?

Try.

You are of course right about the acid, which just goes to show that disregarding some of the given information makes the solution doubtful regardless of who does it. Don't you agree? :wink:

That coin problem - the one where I could see it had to be five but only after everyone else had told me so - here is a variation:

If instead of coins it were cigarettes (or, for non-smokers, identical cylinders with the diameter only a small fraction of the length) how many of these can be arranged so that every one is in contact with every other?

a) I doubt it. :wink:
b) I sure do.
c) I can make six.

This one may well start an argument.
Can you put the ten coins into three glasses in such a way that each glass contains an odd number of coins

Tryagain, my mod of the riddle with the rope is not the same as the original. While the original requests the rope to be circular, floating 1 yard above the ground, my version requires just a 'tent' supported by a stick 1 meter high, so that a man may go under it in one place only. The rest of the rope fits the Earth perfectly.

And I did not think five coins could be arranged .. that one is cute.

Relative

Yes, that is straightforward.

But that is not what you said earlier -- and what you said earlier was not so straightforward.

I gotta think this one out.

It sounds right -- but I am not working with an answer sheet -- and I have got to test it. Be back later.

Try

The ten coins in three glasses.
If these are stackable glasses and a coin is in a glass that is inside another glass, does this coin count as being inside two glasses?

Frank.

Well, I thought I said the same things both times, but if you say 'not', that is goods enough for me.