Shortest time in which P catches Q

  • Thread starter K Sengupta
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In summary, P is in pursuit of enemy Q who is hiding in one of 17 caves. Every night, Q moves to a cave adjacent to the one he was previously in. P can search two caves per day with no restrictions. The shortest time P can guarantee catching Q is 9 days, taking advantage of the fact that Q must switch between odd and even numbered caves every other day. However, this solution does not work if Q moves to the same cave multiple times in a row.
  • #1
K Sengupta
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P is hot in pursuit of his enemy Q, who is hiding in one of 17 caves.

The caves form a linear array, and every night Q moves from the cave he is into one of the caves on either side of it.

P can search two caves each day, with no restrictions on his choice.

For example, if P searches (1, 2), (2, 3), ..., (16, 17), then he is certain to catch Q, though it might take him 16 days.

What is the shortest time in which P can be guaranteed of catching Q?
 
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  • #2
Well, I can think of a way to get him in 14 days... not sure if it's optimal, though... Hmmm.

DaveE
 
  • #3
I had found a 14 days solution, too...
[2,4] [2,4] [4,6] [4,6] [6,8] [6,8] [8,10] [8,10] [10,12] [10,12] [12,14] [12,14] [14,16] [14,16]
But then, I realized it could be done in 11 days.
[2,4] [5,7] [8,10] [11,13] [14,16] [17,16] [15,13] [12,10] [9,7] [6,4] [3,1]
:smile:
 
  • #4
Ahhh yes, clever-- take advantage of the fact that if he's in an odd numbered cave on odd days, he has to be in an even numbered cave on even days, and visa versa.

DaveE
 
  • #5
I've found a 9 days solution I think...

[2,4] [2,4] [4,6] [6,8] [8,10] [10,12] [12,14] [14,16] [14,16]

Actually looking at it now, it's pretty much the 14 days solution but you don't have to double up on each pair of caves in consecutive days, just the second of the pair.
 
  • #6
Soca fo so said:
I've found a 9 days solution I think...

[2,4] [2,4] [4,6] [6,8] [8,10] [10,12] [12,14] [14,16] [14,16]

Actually looking at it now, it's pretty much the 14 days solution but you don't have to double up on each pair of caves in consecutive days, just the second of the pair.

It is not a solution...
Enemy Q moves:
[5] [6] [5] [4] [5] [4] [5] [4] [5]
 
  • #7
Ah yes, damn :smile:
 
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1. What is the concept of "Shortest time in which P catches Q"?

The concept of "Shortest time in which P catches Q" is a hypothetical scenario in which one object (P) is trying to catch or reach another object (Q) as quickly as possible. This can be applied to various situations, such as a race, a game, or a scientific experiment.

2. How is the shortest time in which P catches Q determined?

The shortest time in which P catches Q is determined by calculating the time it takes for P to reach Q, using various factors such as the distance between the two objects, the speed of P, and any obstacles or challenges that may be present.

3. Can the shortest time in which P catches Q be influenced by external factors?

Yes, the shortest time in which P catches Q can be influenced by external factors such as the environment, the physical capabilities of P and Q, and any external forces acting upon them. These factors can affect the speed and efficiency of P in catching Q.

4. Are there any strategies or techniques that can help P catch Q in the shortest time?

Yes, there are various strategies and techniques that can help P catch Q in the shortest time possible. These may include optimizing P's speed and agility, minimizing the distance between P and Q, and finding the most efficient route or path to reach Q.

5. How does the concept of "Shortest time in which P catches Q" apply to real-life situations?

The concept of "Shortest time in which P catches Q" can be applied to real-life situations, such as sports competitions, traffic navigation, and even animal hunting behaviors. It can also be used in scientific experiments to measure the speed and efficiency of certain processes or reactions.

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