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Topic: Machinarium (Read 2217 times) |
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alien2
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Machinarium
« on: Sep 15th, 2012, 12:57pm » |
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The riddle takes place in the eponymous city populated by robots. Robots called Josef and Alsatia have a date. They agree to meet by a cylindrical construction, around which a lap is 50 m long. They’re currently standing near the two opposite points of construction’s diameter. He is carrying a rose and she is wearing a dressy hat. But since they’re nervous, they’re experiencing defective functioning, which will be corrected instantly when he hands her the rose. They are aware of each other's status because they're equipped with unaffected and sophisticated sensors.
Their malfunction is as follows. It takes 5 s for him to travel 5 m. He’ll move around the construction 5 s clockwise, then 10 s counterclockwise, then 15 s clockwise and so on. Alsatia can only stand in place and wait for Josef.
From the moment he begins to walk, how many seconds will it take for the robots to function within established parameters?
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playful
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Re: Machinarium
« Reply #1 on: Sep 16th, 2012, 3:26pm » |
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Let see...
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With each direction change, he gains 5 meters, so to travel the 25 meters will mean 5 direction changes (counting the first movement as a change).
The time for direction change i is 5i, so the time for 5 changes is
5(1+2+...+5), i.e. 5(5x6/2), i.e. 75.
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How does that sound?
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towr
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Re: Machinarium
« Reply #2 on: Sep 16th, 2012, 10:31pm » |
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It doesn't sound entirely correct. After 5+10 seconds he's as far away as after 5 seconds, because he first undoes the gain of the first 5 seconds.
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| « Last Edit: Sep 16th, 2012, 10:32pm by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Grimbal
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Re: Machinarium
« Reply #3 on: Sep 17th, 2012, 8:18am » |
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It is easy to run the simulation by hand
time / total time / position
5 / 5 / 5
10 / 15 / -5
15 / 30 / 10
20 / 50 / -10
25 / 75 / 15
30 / 105 / -15
35 / 140 / 20
40 / 180 / -20
45 / 225 / 25
and here you are in 225 seconds.
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| « Last Edit: Sep 17th, 2012, 8:19am by Grimbal » |
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playful
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Re: Machinarium
« Reply #4 on: Sep 17th, 2012, 2:41pm » |
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Quote:| It doesn't sound entirely correct. |
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Nice. I love it when I overlook something "obvious". That's a nice ingredient in a puzzle. Thank you for the entertaining puzzle, alien2, and thank you for your gentle nudges on this puzzle that must be trivial for you, towr and Grimbal.
Quote:| It is easy to run the simulation by hand |
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Yes. So I'll try a more general case again, let's see if I'm more awake this morning.
D is the distance (or half perimeter). s is the meters per second as well as the increment (I guess these can be split into separate variables in another installment.) So for D and s, how long before the flowers are delivered?
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We gain s every second iteration (on 1, 3, 5...)
and we need to gain s a total of D/s times.
So we need (D/s) x 2 -1 iterations.
The distance traveled for iteration i is 5i. So the total distance is
5(1 + 2 + ... + (2D/s -1) ) =
5 ( ( [2D/s -1] x [2D/s] ) / 2) =
(10D^2 - 5Ds) / s^2
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I'm sure some of you will come up with variations to make the question considerably more complex.
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alien2
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Re: Machinarium
« Reply #5 on: Sep 20th, 2012, 6:18am » |
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Thank you for your thanks, till the next riddle.
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