[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 579
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You can't just knock out all the lights, you don't have the whole picture. You need a criterion by which we can judge whether we have beaten all the horses. There could be oooooo many of them.
But is the solution right or not?
And in general - there was no optimality requirement in the problem condition.
You can't just knock out all the lights, you don't have the whole picture. You need a criterion by which we can judge whether we have beaten all the horses. There could be a hell of a lot of them.
But is it the right decision or not?It seems to me that your solution is correct, but highly suboptimal. The number of steps will grow as the square of the number of cars.
If you want, I can give you my solution.
If the horses stand in a row. We come in with a theodolite. Turn on 3 adjacent torches and measure the angles between them. From the angles calculate the number.
Dima, the only measuring device you have is the switch on the horse and your brain. You have no slingshots, theodolites or multimeters :)
Horses don't have to be in a row, but they can be arranged.
sand: Если хотите я могу привести мое решение.
Yeah, I don't mind.
Idea: Make sure that only one bulb is lit, then by counting the bulbs from the only "lit" bulb to the next "lit" bulb we will actually go around in a circle and count all the bulbs
(1) Choose the first lit bulb we see and the direction to go around
(2) Carry out a loop by switching off all other N light bulbs. At first N = 1
(3) Once we reach the N+1 light bulb, do the following
---- (3a) If the bulb is not lit, light it, increase N by one and go back to (2)
---- (3b) If the light is on, it may be the very first light. To check this, go back two steps, i.e. N + N+1 bulbs backwards.
-------- (3bi) If there is no light, this is the end of the N-bulb.
-------- (3bii) If the light bulb is still on, go back to the first point (3b), increase N by one, and go back to point (2).
P.S. Update
Yeah, I definitely don't mind.
You are in a certain wagon. If the lights in the carriage are off, you turn them on and start counting the cars. You count using two "counters". The first one keeps adding up. If you meet a carriage with the light off, you add 1 to the second counter. If the lights are on you "reset" the second counter and turn the lights off. If it turns out that the first counter is twice as big as the second counter, you have bypassed the entire train and the second counter contains the number of cars. You will need to go around the train 2 times to count it.
The first counter is the total number of cars travelled and the second counter is the accumulated number of cars without lights travelling in succession. Is that right?
P.S. Example: the first counter is 4, the second is 2. Passed 4 carriages, the last two had no light, and the first two had light. There are 1000 cars in the circle (you do not know it, I do).
How do you know for sure that we have completed the round?
Если свет горит вы "сбрасываете" второй счетчик и выключаете свет.
Here is where it is not clear. Will it be the first wagon with lights off (after resetting the second counter) - or it does not count?
If the horses stand in a row. Coming in with a theodolite...
The first counter is the total number of cars travelled and the second counter is the accumulated number of cars without lights travelling in succession. Is that right?
P.S. Example: first counter is 4, second counter is 2. Passed 4 carriages, there was no light in the last two. There are 1000 carriages in the circle (you do not know this, I do).
How can you be sure that we have completed the rounds?
Yes, you seem to be right. However, that answer was credited more than a year ago ))
I will think about this task.