SAP Puzzle Challenge Home Magic time boxes ★★★★★★★

Skip to end of metadata
Go to start of metadata

There are two magic time boxes. One box makes the time in it sqrt(2) times as fast as the time out of it. The other box makes the time in it sqrt(3) times as fast as the time out of it. You have two straight sticks, each is 1 meter long, and have two ants, they move at same speed. (Be aware, an ant seems faster when it is in a box.) The boxes are transparent, and you can make accurate comparison on distance which ants have moved between sticks even in different boxes. How to find 1/3 meter length? How to find 1/5 meter length?
----Designed by Hansen Chen in 2010( http://wiki.sdn.sap.com/wiki/display/SCNPC/Home )

There is no limitation on boxes’ size (you can enter a box if you want). You can mark the sticks. You are not allow to break them. You can do anything reasonable, for example: put a part of stick in one box, and put the rest of it in the other box.

• No labels

1 Comment

Hide/Show Comments
1. flyerbug
Posts: 73
Registered: 8/4/09
Forum Points: 0 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 12, 2010 10:03 PM     in response to: Hansen Chen Edit Reply
If I put one stick in a sqrt(3) box and one stick out side the box and have the ants run from the ends of the sticks, I could mark the outside stick where the outside ant is when the inside ant finished the stick. This mark is 1/sqrt(3) meter in length. I can then copy the 1/sqrt(3) mark to the inside stick. I restarts the ants from the begining and mark the outside stick where the outside ant is when the inside ant reaches the 1/sqrt(3) mark. This new mark should be 1/3 meter in length.
This is too simple and I start to doubt that I misread the problem. As Hansen always told us: this is a 9 stars problem, if you solve it too quickly, something is incorrect.
Hansen Chen
Posts: 350
Registered: 7/16/08
Forum Points: 46 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 12, 2010 10:18 PM     in response to: flyerbug Edit Reply
The 1/5 meter is the 9-star one. As it is very special, when I designed it yesterday, I didn't even believe it is possible at the begining. But believe me, there is a solution.
Anyone who firstly in the world solve it will have the honor to name the puzzle. For example, if flyerbug solve it firstly, the puzzle would be called Flyerbug's puzzle.
Chris C.
Posts: 58
Registered: 10/23/08
Forum Points: 48 Re: Magic time boxes ★★★★★★★★★
Posted: Apr 14, 2010 7:39 AM     in response to: Hansen Chen Edit Reply
Put one stick inside the sqrt(2) box and one stick outside. Both ants on both stick start at the end of a stick. Mark the point A1 in the outside stick when the ant in the inside stick reaches the end. Then switch sticks. Both ants on both stick start at the end of a stick. Mark the point B1 in the outside stick when the ant in the inside stick reaches A1. This B1 point is the half point.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ B1_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ = 15/30
Then we use the sqrt(3) box. Switch the sticks so that the B1 stick is inside. Both ants on both stick start at the end of a stick. Mark the point A2 in the outside stick when the ant in the inside stick reaches the end. Then switch sticks. The ant in the outside starts at B1. Mark the point B2 in the outside stick when the ant inside reaches A2. This is 1/2 + 1/3 point.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ B2 _ _ _ _ = 25/30
- - - - - - - - - Segment1- - - - - - - - - - - - - - - - - Segment2
The resulting segment2 is 1/5 of the resulting segment1. The distance from B2 to the end of segment2 can be covered by putting the sqrt(2) box inside the sqrt(3) box which will ultimately become a sqrt(6) box.
Put the B stick inside the sqrt(6) box. Both ants on both stick start at the end of a stick. Mark the point A3 in the outside stick when the ant in the inside stick reaches the end. Then switch sticks. The ant in the outside starts at the end of segment2. Take note of the time it takes outside T1 when the ant inside reaches A3. Get this time and divide it by 5.
Starting at B2 going to segment one, run T1 for the outside ant. The distance B3 it will cover is the 4/5th point of the stick. The resulting segment3 is the 1/5th point.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ B3 _ _ _ _ _ = 24/30
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -Segment3
Edited by: Chris C. on Apr 14, 2010 7:43 AM
flyerbug
Posts: 73
Registered: 8/4/09
Forum Points: 0 Re: Magic time boxes ★★★★★★★★★
Posted: Apr 14, 2010 5:15 PM     in response to: Chris C. Edit Reply
Chris,
Why bother with your lengthy procedure? You can simply do:
Take note of the time it takes an T1 when a ant runs a stick from one end to the other. Get this time and divide it by 5 (and still call it T1). Start the ant again and mark the stick when it run T1 time.
You see, the problem is that YOU DON'T HAVE A STOP WATCH to get the time.
flyerbug
Posts: 73
Registered: 8/4/09
Forum Points: 0 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 14, 2010 5:41 PM     in response to: Hansen Chen Edit Reply
According to my previous posting, we can mark a stick at 1/3 meters. The same thing can be done with sqrt(2) box to mark a stick at 1/2 meters. Further more, we can use 1/3 meter stick to mark at 1/3 * 1/2 = 1/6 meter point.
l ..............................................................l
l ...........................BOX............................l
-----------C----------------------B-------------------
.............l .....................0
.............l ................0
.............l ..........0
.............l ....0
............A
........0...l
...0........l
.............l
.............l
.............l
.............l
.............l
We can then put the sticks against the wall of one time box as in the figure above. Stick 1 is placed vertically against the wall with AC = 1/6 meters. Stick 2 is then placed with AB = 5/6 meters. Now we move stick 1 to the left (still keeping it vertically against the wall) until it touches the end of the stick 2:
l ..............................................................l
l ...........................BOX............................l
---Y-------------------------------B-------------------
....l ..............................0
....l .........................0
....l ...................0
....l .............0
....l.........0
....l....0
...X
....l
....l
....l
....l
....l
Now XY = 1/5 meters.
Hansen Chen
Posts: 350
Registered: 7/16/08
Forum Points: 46 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 14, 2010 8:43 PM     in response to: flyerbug Edit Reply
Where is the wall?
S. Huang
Posts: 97
Registered: 7/21/09
Forum Points: 0 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 14, 2010 9:01 PM     in response to: Hansen Chen Edit Reply
Hensan,
Is it allowed to draw straight lines on the floor with the stick? If yes, this puzzle can be solved by simply constructing two similar triangles with the help of one stick and two ants, no need for the time space boxes or anything else.
Edited by: S. Huang on Apr 14, 2010 11:11 PM
flyerbug
Posts: 73
Registered: 8/4/09
Forum Points: 0 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 14, 2010 10:02 PM     in response to: Hansen Chen Edit Reply
The wall is the outside wall of a time space box. It is assumed to be a flat surface.
Hansen Chen
Posts: 350
Registered: 7/16/08
Forum Points: 46 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 14, 2010 10:33 PM     in response to: flyerbug Edit Reply
Good puzzle deserves a good solution. If your solution doesn't look beautiful, try in different path.
As a hint, solution to a 9 star puzzle usually is not necessay to be complex. This is why I prefer 9 star puzzle than 10 star one.
flyerbug
Posts: 73
Registered: 8/4/09
Forum Points: 0 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 15, 2010 12:08 AM     in response to: Hansen Chen Edit Reply
It looks like Hansen doesn't like to use of the wall. How about a recursive solution:
1. Pick a spot near the center of a stick, let's say it is X from one end and X > 1/2.
2. Using the method for 1/3 meter solution, we can mark X/3 position.
3. Similarly, using the sqrt(2) box, we can mark (1-X)/2 position.
4. If X/3 position > (1-X)/2 position, we move X lower and repeat steps 1-3
5. If X/3 position < (1-X)/2 position, we move X higher and repeat steps 1-3
In the end, we will get a X position that makes
X/3 = (1-X)/2
This means X = 3/5
The (1-X) mark is then 2/5
The length between (1-X) and X marks is 1/5
Hansen Chen
Posts: 350
Registered: 7/16/08
Forum Points: 46 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 15, 2010 3:38 AM     in response to: flyerbug Edit Reply
It is not convining.
flyerbug
Posts: 73
Registered: 8/4/09
Forum Points: 0 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 15, 2010 4:26 AM     in response to: Hansen Chen Edit Reply
I am sure this solution will satify Hansen:
1. Put sqrt(2) box inside sqrt(3) box to create an effective sqrt(6) box
2. mark the sticks at 1/6 meters using the method for 1/3 meter solution.
3. Place one stick (stick A) along the side of the sqrt(6) box
4. Put one ant inside the box and one ant outside the box and starts them at one end of the stick
5. Use the other stick (stick B) to measure the distance between the ants, ie: keep one end of stick B with the slow ant
6. Mark the position of the slow ant on stick A when the distance between the ants = 1/6 meters. This mark is
1/6 / (sqrt(6)-1) meters.
7. Copy the 1/6 /(sqrt(6)-1) mark to stick B 6 times to create 1/( sqrt(6) -1 ) meter mark.
8. Put stick B along the side of the sqrt(6) box
9. Put one ant outside the box at the end of the stick and put another ant inside the box at the 1/(sqrt(6)-1) mark. Starts the ants to run towards each other.
10. Mark the position where the ants meet. This position is:
1/(sqrt(6)-1) / (sqrt(6) +1 ) = 1/5 meters
Hansen Chen
Posts: 350
Registered: 7/16/08
Forum Points: 46 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 15, 2010 5:44 PM     in response to: flyerbug Edit Reply
flyerbug,
How do you want to name the puzzle? Your nick or your formal name?
By the way, I am not quite following your solution. I assuming it is correct. But for my solution, the ants only run two times. Would you please optimize it?
Thanks
Hansen
flyerbug
Posts: 73
Registered: 8/4/09
Forum Points: 0 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 15, 2010 7:42 PM     in response to: Hansen Chen Edit Reply
Sure. The key is to get 1/ (sqrt(6) - 1) and 1/ (sqrt(6) + 1). Let me try to do it in 2 runs:
1. Put stick A inside sqrt(6) box and stick B outside the box. Run the ants twards each other from the ends. When the ants meet, mark stick B. This is 1/ (sqrt(6) + 1) meter mark.
2. Now put stick B inside the box and stick A outside the box. Restart the ants from the ends of sticks, except now they are run in the same direction for a race. Because the inside ant run faster, we will keep moving the outside stick (stick A) forward to keep the ants neck-to-neck in the race. When the end of stick A reaches the 1/ (sqrt(6) + 1) meter mark on stick B, we mark where the ant is on stick A. This mark is: 1/ (sqrt(6) + 1) / (sqrt(6) - 1) = 1/5 meters.
This is an interesting puzzle. I am hornored to get the naming right. I prefer to call it flyerbug's time box. Also, I would like to suggest a couple of changes to the puzzle:
1. Change the ants to flyerbugs.
2. Add limitation that flyerbugs can only run on a stick. (to avoid attempts to run flyerbugs on the floor as Huang suggested)
Hansen Chen
Posts: 350
Registered: 7/16/08
Forum Points: 46 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 15, 2010 8:46 PM     in response to: flyerbug Edit Reply
Hi flyerbug Congratulation!
Now we call the puzzle as flyerbug's time box. Great job! I don't think in any other puzzle forums such as IBM, Microsoft, Mensa people are able to solve it. Your honor base on your super intelligence, you deserve it.
Thanks
Hansen
Hansen Chen
Posts: 350
Registered: 7/16/08
Forum Points: 46 Re: Super time space boxes ★★★★★★★★★
Posted: Apr 15, 2010 8:46 PM     in response to: Hansen Chen Edit Reply
By the way I am sharing my solution, a little bit more straightforward:
1. Put a box inside the other and this makes the sqrt(6) box.
2. Put stick A inside sqrt(6) box and stick B outside the box, side by side. Run the ants twards each other from the ends. When the ants meet, mark sticks. This is 1/ (sqrt(6) + 1) meter mark.
3. Now put the ant on A to the other end, and leave the other ant at the mark. Make them run again in the same direction. When they meet again. The distance that the ant run at the second time is: 1/ (sqrt(6) + 1) / (sqrt(6) - 1) = 1/5 meters.
S. Huang
Posts: 97
Registered: 7/21/09
Forum Points: 0 Re: Magic time boxes ★★★★★★★★★
Posted: Apr 16, 2010 5:54 AM     in response to: Hansen Chen Edit Reply
I have a variant for this puzzle.
You have a straight stick which is 2 meter long. You can mark on the stick, and you can draw straight line on the floor with a pencil and the stick.
How to find 1/N meter length?
Hansen Chen
Posts: 350
Registered: 7/16/08
Forum Points: 46 Variation
Posted: Apr 16, 2010 10:24 PM     in response to: Hansen Chen Edit Reply
The variation is based on the same idea, so the solution might be similar, just for fun.
There are two flyerbug space boxes. One box makes the length in it 1/sqrt(2) times as long as the length out of it. The other box makes the length in it 1/sqrt(3) times as long as the length out of it. You have two straight sticks, each is 1 meter long, and have two flyerbugs, they move at same speed. The boxes are transparent, and you can make accurate comparison on distance which ants have moved between sticks even in different boxes. How to find 1/3 meter length? How to find 1/5 meter length?
flyerbug
Posts: 73
Registered: 8/4/09
Forum Points: 0 Re: Variation
Posted: Apr 19, 2010 5:49 PM     in response to: Hansen Chen Edit Reply
To get 1/3 meters, no flyerbugs are needed:
Put a stick inside sqrt(3) box, it will look shorter. Compare it to the stick outside the box, we get a mark of 1/sqrt(3) on the outside stick. Exchange the sticks, we can get a mark for 1/sqrt(3) * 1/sqrt(3) = 1/3 meters.
To get 1/5 meters:
We can put sqrt(2) box inside sqrt(3) box to create a sqrt(6) box.
We put a stick partially in the box, and starts flyerbugs from two ends to run toward each other. We keep adjusting the stick so that the flyerbugs never cross the edge of the box. When they meet, they will be right at the edge of the box. We mark that position on the stick. This is 1/(sqrt(6) + 1).
We now put a stick partially in the box and lineup the box edge with the 1/(sqrt(6)+1) mark so that ( 1 - 1/(sqrt(6) + 1) ) meters is at the outside of the box. We put a flyerbug at the stick end inside the box, a flyerbug right on the box edge but outside the box. We run the flyerbugs in the same direction towards outside of the box. During the run, we keep pushing the stick into the box so that the flyerbug at the box edge remains at the box edge. When the two flyerbugs meet, we mark the stick. The distance from the 1/(sqrt(6)+1) mark to this new mark is 1/(sqrt(6)+1) / (sqrt(6)-1) = 1/5 meters.
Hansen Chen
Posts: 350
Registered: 7/16/08
Forum Points: 46 Why is it 9 star?
Posted: Apr 20, 2010 12:40 AM     in response to: flyerbug Edit Reply
The solution to the puzzle is simple and straightforward, someone might doubt why it is a 9 star puzzle. The fact is that it is easy to understand the solution, but it is not easy to create the solution. Most of my puzzles would inspire creation, since you have to create new methods which cannot be found in any book.
flyerbug
Posts: 73
Registered: 8/4/09
Forum Points: 0 Re: Variation
Posted: Apr 20, 2010 3:42 PM     in response to: Hansen Chen Edit Reply
Both the original puzzle and the variation can be changed to have only one stick.
To be complete, how about another variation:
There are two flyerbug time-space boxes. One box makes the length in it 1/sqrt(2) times as long as the length out of it AND makes the time sqrt(2) faster. The other box makes the length in it 1/sqrt(3) times as long as the length out of it AND makes the time sqrt(3) faster. You have two straight sticks, each is 1 meter long, and have two flyerbugs, they move at same speed. The boxes are transparent, and you can make accurate comparison on distance which ants have moved between sticks even in different boxes. How to find 1/3 meter length? How to find 1/5 meter length?