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I stopped playing massive game so I'll keep u guys entertained with these puzzles. This is round 1. The first to have the right answer with correct methods and intuitions(and put into a spoiler here) will be in honorary mention in the next round. I'll give more clarifications as I see people making different assumptions than what is intended(there are some math conventions not being explicit but expected).
Should be fun, shall we go?
You have a stick that is 100cm long, and you can place ants on it. The ants can be placed anywhere on the stick, and facing anyway.
Important facts:
-Ants takes no length, you can think of them as dots.
-Ants crawls at 1cm/s.
-When 2 ants run into each other, they immediately turn back on each other and crawls away from each other.
-Ants will fall off at the edge.
Your Experiment:
Place 100 ants on the stick anyway you want, i.e. any direction/position. And start stopwatch. When the clock begins to tick, the ants will begin to crawl, collide, and fall off the edge. When the last ant falls off the stick, stop stopwatch.
Your Job:
Maximize the time recorded on your watch with a particular configuration of ants.
Good Luck!
What I hope:
People posting and discussing the problem. I hate writing blogs with no replies.
Clarifications:
-Stick is 1-dimentional
-Ants CAN be placed at the same spot in a sense that since ants are points, you can have the distances between two points as small as possible, while maintaining that one point is on the left/right of another.
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question of clarification: can multiple ants occupy the same spot?
+ Show Spoiler +
Assuming ants take up 1cm each:
Start at the left of the stick and place ants starting at 0cm to 100cm on the stick, with no ants at 50cm. The ants from 0cm-49cm should be facing left, and the ants from 51cm to 100cm should be facing right. This way, there are no ant collisions and each second, two ants fall off the stick. THere should be no ants on the stick by just after 49 seconds
Assuming ants don't really take up space and you are really precise with placing ants:
Place an equal number of ants at both 0cm and 100cm in such a way that each ant is incrementally spaced between itself and another ant. Set them up facing their respective ends of the sticks. Almost right after you start, there should be no ants on the stick.
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On April 28 2009 02:28 phase wrote:question of clarification: can multiple ants occupy the same spot? + Show Spoiler +
Assuming ants take up 1cm each:
Start at the left of the stick and place ants starting at 0cm to 100cm on the stick, with no ants at 50cm. The ants from 0cm-49cm should be facing left, and the ants from 51cm to 100cm should be facing right. This way, there are no ant collisions and each second, two ants fall off the stick. THere should be no ants on the stick by just after 49 seconds
Assuming ants don't really take up space and you are really precise with placing ants:
Place an equal number of ants at both 0cm and 100cm in such a way that each ant is incrementally spaced between itself and another ant. Set them up facing their respective ends of the sticks. Almost right after you start, there should be no ants on the stick.
Your Job:
Maximize the time recorded on your watch with a particular configuration of ants.
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+ Show Spoiler +
Well by using the picture you gave us as an idea of the nature of the stick - i.e. round and presumably having edges only at the two cut ends...
Have all the ants facing perpendicular to the length of stick. They will all walk the circumference of the stick ad infinitum, and never fall off.
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On April 28 2009 02:33 LiAlH4 wrote:+ Show Spoiler +
Well by using the picture you gave us as an idea of the nature of the stick - i.e. round and presumably having edges only at the two cut ends...
Have all the ants facing perpendicular to the length of stick. They will all walk the circumference of the stick ad infinitum, and never fall off.
GG
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+ Show Spoiler +50 ants from 0cm-49cm, spaced 1cm apart, all facing 100 cm, and 50 ants from 51cm-100cm, spaced 1cm apart, all facing 0cm. When it starts the first pair (49cm and 51cm) will collide and turn around, and then collide with the ants directly behind them, sending a wave in either direction that will lead to a pair of ants falling off each end and the rest traveling back towards the center, ending with the middle pair being the last ants to fall off (after 50 revolutions of the situation above.)
Oddly enough, I believe this will only last 2 seconds longer than just putting a single ant at 0cm facing 100cm, but I could be wrong.
There might be more efficient ways to do it as far as positioning, but the above is the general idea.
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Calgary25991 Posts
Oh I thought the stick was two dimensional haha
Anyways I suck at these kind of things but I'll be reading the replies.
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100s
+ Show Spoiler +when two ants bump into eachother and reverse direction, you can view this as two ants passing through eachother. That is, you can just imagine placing ants and they walk mindlessly straight until they fall off. So, just place one ant at the far end of the stick facing the other direction, and place the other ants however the hell you want. all the ants will walk off in <=100s with at least one taking exactly 100s
hot damn i'm good
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On April 28 2009 02:52 Day[9] wrote:100s + Show Spoiler +when two ants bump into eachother and reverse direction, you can view this as two ants passing through eachother. That is, you can just imagine placing ants and they walk mindlessly straight until they fall off. So, just place one ant at the far end of the stick facing the other direction, and place the other ants however the hell you want. all the ants will walk off in <=100s with at least one taking exactly 100s hot damn i'm good
+ Show Spoiler +Damn, you are right, my solution assumed 102 ants and a 102 centimeter stick because I am retarded. 
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On April 28 2009 02:52 Day[9] wrote:100s + Show Spoiler +when two ants bump into eachother and reverse direction, you can view this as two ants passing through eachother. That is, you can just imagine placing ants and they walk mindlessly straight until they fall off. So, just place one ant at the far end of the stick facing the other direction, and place the other ants however the hell you want. all the ants will walk off in <=100s with at least one taking exactly 100s hot damn i'm good
win
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On April 28 2009 02:52 Day[9] wrote:100s + Show Spoiler +when two ants bump into each other and reverse direction, you can view this as two ants passing through each other. That is, you can just imagine placing ants and they walk mindlessly straight until they fall off. So, just place one ant at the far end of the stick facing the other direction, and place the other ants however the hell you want. all the ants will walk off in <=100s with at least one taking exactly 100s hot damn i'm good
brilliant
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On April 28 2009 02:52 Day[9] wrote:100s + Show Spoiler +when two ants bump into eachother and reverse direction, you can view this as two ants passing through eachother. That is, you can just imagine placing ants and they walk mindlessly straight until they fall off. So, just place one ant at the far end of the stick facing the other direction, and place the other ants however the hell you want. all the ants will walk off in <=100s with at least one taking exactly 100s hot damn i'm good
Win. Shame I haven't even thought of the problem this way
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Day[9] wins the day. I was too busy trying to find some flaw about collision detection since the ants were minuscule to even begin to think about what a 1-dimensional stick really entailed 
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Day[9]: Lateral thinking FTW.
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On April 28 2009 02:52 Day[9] wrote:100s + Show Spoiler +when two ants bump into eachother and reverse direction, you can view this as two ants passing through eachother. That is, you can just imagine placing ants and they walk mindlessly straight until they fall off. So, just place one ant at the far end of the stick facing the other direction, and place the other ants however the hell you want. all the ants will walk off in <=100s with at least one taking exactly 100s hot damn i'm good
this is a neat problem because we naturally think of the problem by simulating it from an individual ant's perspective, which can get really complex. it takes a zerg player's hive mind to think of the set of ants this way. 
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I think if you place 50 ants at left end, walking towards right and 50 ants at right end walking towards left, you maximize the time. You can think of it Day's way.
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Another way to think of Day[9]'s solution is
+ Show Spoiler +to have all the ants lined up from one end to the other and facing the same direction. It's essentially that one ant on the end walking across the entire length of the stick.
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It's a true pleasure to read and hear Day[9]'s mind at work, imo.
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+ Show Spoiler +I'm thinking that the time it takes between each ant falling off is around quadratic, but then I thought the integral of log x would be nice. So I'm guessing x(log(x)-1)-100=0, where x is seconds. So that would be around a hundred. I'd try for one better, but I'm busy ^^
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On April 28 2009 03:45 Polemarch wrote:Show nested quote +On April 28 2009 02:52 Day[9] wrote:100s + Show Spoiler +when two ants bump into eachother and reverse direction, you can view this as two ants passing through eachother. That is, you can just imagine placing ants and they walk mindlessly straight until they fall off. So, just place one ant at the far end of the stick facing the other direction, and place the other ants however the hell you want. all the ants will walk off in <=100s with at least one taking exactly 100s hot damn i'm good this is a neat problem because we naturally think of the problem by simulating it from an individual ant's perspective, which can get really complex. it takes a zerg player's hive mind to think of the set of ants this way.
or just a math major :p
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nice puzzle and nice solution.
gj evan and sean
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It's never fun to start a puzzle when there are + show spoiler + all over the place and comments of admiration and awe. I see these threads too late.
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On April 28 2009 06:34 stenole wrote:
It's never fun to start a puzzle when there are + show spoiler + all over the place and comments of admiration and awe. I see these threads too late.
why? you just don't open the spoilers and you can try yourself =D
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Btw,
(a^n)(a^m)=a^(n+m)=a^(m+n)=(a^m)(a^n).
I just had to read the section first :-P
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On April 28 2009 06:48 edahl wrote:
Btw,
(a^n)(a^m)=a^(n+m)=a^(m+n)=(a^m)(a^n).
I just had to read the section first :-P
dude wtf......thats so useless haha
3*5 = 5*3 has the same message^^
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On April 28 2009 06:50 MasterReY wrote:Show nested quote +On April 28 2009 06:48 edahl wrote:
Btw,
(a^n)(a^m)=a^(n+m)=a^(m+n)=(a^m)(a^n).
I just had to read the section first :-P dude wtf......thats so useless haha
No it's not. It's proof that all cyclic groups are abelian.
EDIT: http://en.wikipedia.org/wiki/Cyclic_group
Evan challenged me to prove it quickly, you see, but I didn't know how yet :-P
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On April 28 2009 06:48 edahl wrote:
Btw,
(a^n)(a^m)=a^(n+m)=a^(m+n)=(a^m)(a^n).
I just had to read the section first :-P
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yea ok but what has that to do with the riddle?
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A group btw is a set G with a binary operation * satisfying certain axioms, making the tuple <G, *>.
The axioms are
1. (a*b)*c=a*(b*c) (associative property)
2. there is an element e in G so that a*e=e*a=a
3. for every a in G, there is an inverse called a^(-1), so that a*a^(-1)=a^(-1)*a=e
(4. if a*b=b*a (commutative property) holds true for all a, b in G, the group is called abelian.)
A cyclic group is a group generated by one element a in <a>, so that a^n is in the group for every n, where n is a whole number. n means the same as with exponents: repeated operation, so a^3 = a*a*a, and a^(-2)=a^(-1)*a^(-1).
PS: It's got nothing to do with the riddle. Evan just asked me to prove it yesterday or something :-P
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oh well. i had that stuff in german, but i can't recall with the english version now
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pffttt thats so lame, who ever though about maximizing the time. I (EECS) always want the shortest time.
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day[9] is fucking awesome
I didn't read the thread or anthing in it
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EPT![[image loading]](http://www.iayork.com/Images/2008/3-16-08/Leaf-cutter_ant.png)
