This seems to be the best place to post this thread.
Forums try these two easy tests of logical thinking.
Their are twelve identical featureless balls of equal volume and size. One differs minutely in mass or weight if you like.
The task is to establish in three weighing steps using a very sensitive balance scale, like the scale of Libra. Which ball is different and if it is heaver or lighter. You can use any combination , three and three, four against four, anything you like but you must solve the problem in three weighs. You cant use a bathroom scale this would be useless.
O O O O O O O O O O O O
Problem two
A person is rowing his bout upstream in a river flowing at three miles an hour at seven miles per hour relative, to the bank of the river. His hat falls off and only after 45 minutes does he notice this. He immediately turns around and rows at the same speed to get his beloved hat. (disregard the time taken for turning around for the purpose of this test).
How long does it take for him to catch up to his hat and retrieve it??
Last problem
A man sentenced to death is given a choice. He is put in a room with two PC computers, one is programed to only lie, and the other programed to only tell the truth. There are two exit doors , one leading to the death chamber and the other to freedom. He is only allowed to key in "one question" to only "one of the computers" and by this one question, he must establish the door to freedom or face execution.
Give it a go
alan
Problem two:
Also 45 minutes, assuming the hat is in the water, not his boat- you can view it as that he has a speed of 10 MPH with a fixed point in the water, which is seen as 7 MPH onshore, as the water is moving relative to the shore 3 MPH. So he has moved away from the hat at 10 MPH relative to its location for 45 minutes, which means he is 7.5 miles ( of water) away. As when he reverses his direction of travel, he is still moving 10 MPH with respect to the water (he is now moving 13 MPH with respect to the shore, but the hat is also traveling 3 MPH away, so with respect to the water, and the hat, the speed is still 10 MPH). Since it took 45 minutes to cover the 7.5 miles at 10 MPH, it should take the same amount of time to uncover that same distance at same speed.
So also 45 minutes- but I might be wrong if the solution is meant to be more abstract than this, or the solution is something cute or clever, like- "The hat fell in his boat, stupid", lol.
_______________________________
I will let others try the last two before I offer solutions, lol.
Stilwater
Correct simple logic
alan
Quote from: Alan McDougall on May 29, 2008, 04:15:51
Last problem
A man sentenced to death is given a choice. He is put in a room with two PC computers, one is programed to only lie, and the other programed to only tell the truth. There are two exit doors , one leading to the death chamber and the other to freedom. He is only allowed to key in "one question" to only "one of the computers" and by this one question, he must establish the door to freedom or face execution.
A question for one computer: "Would the other computer agree that the right door lead to freedom?" If the answer is "no", right door lead to freedom, other wise left door lead to freedom.
That's clever- have you seen a problem of this general type before?
I think the solution to this one was much more novel than problem 2 :wink:
Galaxystorm,
Correct good now try and solve the real problen namely the twelve balls
alan
12 balls:
Weigh 1: Weigh 6 against 6. One side will be heavier.
Weigh 2: Divide the heavier side into 3 and 3 and weigh. One side will again be heavier.
Weigh 3: Take the 3 heavier balls, remove 1, and weigh the remaining two. If they balance, the one you removed was heavier. If they don't balance, you still have your answer.
Quote from: Stillwater on May 29, 2008, 20:54:28
That's clever- have you seen a problem of this general type before?
Yes i had but I forgot the solution. I actually thought this up by myself now (you have a statement processed through two "units", and one of them negates it- it will always arrive negated)
no_leaf_clover: I also got to the 3 balls but didn't know how to find the heavier one... that was clever and simple!
No-leaf-clover
Quote12 balls:
Weigh 1: Weigh 6 against 6. One side will be heavier.
Wrong wrong wrong on your first weigh at that. You made the assumption that the ball was heavier. I never said that. It could be either heavier or lighter. The problem is more difficult than the others but can be solved be simple logic
keep tying
alan
Uhuuuu I got it i got it!! hehehe
Check it out:
1st weigh 6 balls - 3 on the A side of the balance x 3 on the B side of the balance.
case 1: If there's equilibrium, the different ball is not there, then paint these 6 balls black to help you.
case 2: If there's no equilibrium, you know that the different ball is on either side (A or B), then leave these balls white and paint black the balls you didn't weigh.
Now you have 6 black balls and 6 white balls, and know that the different ball is among the white ones.
2n: Now you can, for example, replace the balls on the A side of the balance with 3 black balls.
case 1: If there's equilibrium, you know that the different ball is among the 3 whites you had removed.
case 2: If there's no equilibrium, you know it's among the 3 whites on the B side of the balance.
Doing this you're left with 3 white balls, and already knows if the different one is lighter or heavier.
3rd: From the 3 whites left, pick one up and set it apart of the others. Now weigh one white with one black on the A side of the balance, and one white and one black on the B side of the balance.
case 1: If there's equilibrium, the different ball is the one you had set apart.
case 2: If there's no equilibrium, you know which one of the whites is the different one, because you knew whether it was heavier or lighter from the prior weighing.
:-D
Oops, no, I think I made a little mistake, I should have made another group from the 2nd weighing.
Andrew the singer,
The answer is incorrect you cant solve the problem by weighing three balls against three. How are you going to establish wether the ball is heavier or lighter with six still unknown balls. It cant be done. You are on the correct track and assumptions are the way to solve the problem
I will have another look at your solution in the morning as I am keying in this response late at night
regards
Alan
Andrew,
I came up with the same thing you did, except I used numbers to distinguish them instead of colors. I thought I finally had it at first, too. >.<
This is a bastard of a logic problem, lol. I buy books of these things, though. I love them. They all relate to "real world" logic somehow and reveal some kind of truth.
Hehe I followed your initiative of removing one ball in the end, I think that's part of the solution. There may be something very tricky about the 2nd weighing.
Let's see, I don't know if this is the right answer but it might work.
After making groups of black and white, the 2nd step is to weigh the 6 black x 6 white, then you'll know if the different ball is heavier or lighter.
You're left with 6 white balls still, then for the 3rd step weigh 3 white x 3 white. Now you're with 3 white balls and can't weigh anymore. Here's the cheat lol ...
take a bottle a fill it up with oil, drop the first ball in it and see how long it takes to hit the bottom. Do the same with the other 2 balls. The one that hits first or last is your ball! :wink:
But that's probably not the right answer :roll:
Andrewthesinger,
QuoteAfter making groups of black and white, the 2nd step is to weigh the 6 black x 6 white, then you'll know if the different ball is heavier or lighter
Andrew you receded ,further from the solution by the above, by weighing 6 balls against 6 balls you achieve absolutely nothing except waisting you first weigh. You do not know if the ball is "heavier" or "lighter" it could be "either" maybe amongst the 6 balls residing in the lower part of the scale as a "heavier ball" or it might be one of the 6 balls on that is "lighter ball" in the high part of the scale.
Your first attempt was much closer , but it relied on luck to know which was the correct ball. You could identify the odd ball by this route but not know it is is heavier or lighter
There is only one odd ball.
Luck must play no part in the solution it must by foolproof
Regards
Alan
No Alan, first I had established in which group the odd ball was. I knew it was among one of the whites. So when I put white against black there will not be equilibrium, if the black side is heavier then I know the odd ball is lighter. If the white side is heavier then I know the odd ball is heavier, it works. But I reckon the bottle and oil in the end might not be what the problem wanted
Hi Andrew,
QuoteNo Alan, first I had established in which group the odd ball was. I knew it was among one of the whites. So when I put white against black there will not be equilibrium, if the black side is heavier then I know the odd ball is lighter. If the white side is heavier then I know the odd ball is heavier
,
Andrew then you must have weighed all the balls one by one untill you found the odd ball by some other method, before you began the test of logic. This is not how it must be solved, you have to do this test of logic completely blind, "on only this one scale". no forsight, not knowing where, which, and wether the odd ball is lighter or heavier. The answer could be either. "And luck must play no part in the solution"
This is a complex test of logic
Regards
Alan
OK, guys,
Here is the solution, I have given this test of logic to many people over the past 30 years and only two have solved it.
Before I explain some conventions in setting out the solution to the Twelve Balls problem, here again is a statement of the task.
There are twelve balls, all the same size, shape and color. All weigh the same, except that one ball is slighterly different in weight, but not noticeably so in the hand. Moreover, the odd ball might be lighter or heavier than the others.
Your challenge is to discover the odd ball and whether it is lighter or heavier. You must use a beam balance only, and you are restricted to three weighing operations.
Note!! ANDREW Starting with 6 against six 5 against 5 or 3 against 3. would not lead you to the perfect solution
The only way to solve this rather complex problems is by starting by weighing 4 balls against 4 as you will see by the solution below
At every weighing one of three things theoretically can happen: the pans can balance, the left pan can go down or the left pan can go up.
It will be necessary to refer to a given ball as definitely normal (N), potentially "heavier" (H) or potentially "lighter" (L). Often our identification of a ball in this way will be as part of a group (= "This group contains a heavier/lighter ball"), and will depend on what we learn from a previous weighing. At the start, all balls have a status of unknown (U).
To show at each weighing what is being placed in each pan, represent the situation as per the following examples:
.
First Weighing UUUU ——— UUUU
Pans balance All these U's are now known to be N's; the odd ball is one of the remaining unweighed four (call them UUUU from now on).
Proceed to Second Weighing: Case 1
Left pan down One of the four balls in the left pan might be heavier (call them HHHH from now on) or one of the four balls in the right pan might be lighter (call them LLLL from now on).
Proceed to Second Weighing — Case 2
Left pan up One of the four balls in the left pan might be lighter (call them LLLL from now on) or one of the four balls in the right pan is heavier (call them HHHH from now on).
Proceed to Second Weighing — Case 2
Second Weighing
Case 1 UUU ——— NNN
Pans balance All these U's are now known to be N's; the odd ball is the remaining unweighed U, but we don't yet know if it's heavier or lighter than normal.
Proceed to Third Weighing — Case 1
Left pan down One of these U's is heavier than normal, but we don't yet know which one (call them HHH from now on).
Proceed to Third Weighing — Case 2
Left pan up One of these U's is lighter than normal, but we don't yet know which one (call them LLL from now on).
Proceed to Third Weighing — Case 3
Case 2 HHL ——— HLN
Pans balance All these H's and L's are now known to be N's; the odd ball is one of the remaining unweighed H or two L's.
Proceed to Third Weighing — Case 4
Left pan down The odd ball is one of the left two H's or the right L.
Proceed to Third Weighing — Case 5
Left pan up The odd ball is either the right H or the left L.
Proceed to Third Weighing — Case 6
Third Weighing
Case 1 U ——— N
Pans balance Not possible
Left pan down The odd ball is this U, and it's heavier
Left pan up The odd ball is this U, and it's lighter
Case 2 H ——— H
Pans balance The odd ball is the remaining unweighed H (heavier)
Left pan down The odd ball is the left H (heavier)
Left pan up The odd ball is the right H (heavier)
Case 3 L ——— L
Pans balance The odd ball is the remaining unweighed L (lighter)
Left pan down The odd ball is the right L (lighter)
Left pan up The odd ball is the left L (lighter)
Case 4 L ——— L
Pans balance The odd ball is the remaining unweighed H (heavier)
Left pan down The odd ball is the right L (lighter)
Left pan up The odd ball is the left L (lighter)
Case 5 H ——— H
Pans balance The odd ball is the remaining unweighed L (lighter)
Left pan down The odd ball is the left H (heavier)
Left pan up The odd ball is the right H (heavier)
Case 6 H ——— N
Pans balance The odd ball is the remaining unweighed L (lighter)
Left pan down The odd ball is this H (heavier)
Left pan up Not possible .
Painting the balls would be a good idea but some balls would have to be given two colors
alan
Yep, that's pretty complex. I thought it would be better if I kept it simple by having only 2 groups, either 'there' or 'not there', but then it was impossible to find out in 3 steps. Since I liked that line of thought I didn't see the other possibilities. That was nice, thanks Alan.
I agree that starting out with four balls each is the way to go. I would do it this way:
1st weighing- 4 balls each. This will reveal which group the ball in question is part of.
2nd weighing- 2 balls each, using balls only from the group with the different ball in it.
3rd weighing- 1 ball each, from the side with the ball that is different.
Note: for each weighing, eliminate and keep separate the balls from the side that does not have the different ball in it. This way, you don't have to paint them.
And for the last problem, with the man sentenced to death, he could ask the computers this:
"If you were the other computer, which door would you say is the right one?
Both computers would name the false door.
Ambient Sound
You said
QuoteI agree that starting out with four balls each is the way to go. I would do it this way:
1st weighing- 4 balls each. This will reveal which group the ball in question is part of.
2nd weighing- 2 balls each, using balls only from the group with the different ball in it.
3rd weighing- 1 ball each, from the side with the ball that is different.
With respect this would not solve the problem, remember you do not know if the odd ball is heavier or lighter. You cannot use luck the problem can only be soled by a process of deduction and elimination. Have a rethink.
You are correct on the other problem
Alan