• #### a fox in 5 (adjacent) holes

From henhanna@gmail.com@21:1/5 to All on Mon May 23 10:00:13 2022
(no point in anyone posting the Answers, because they are all over the Net)

Consider five holes in a line. One of them is occupied by a fox.

Each night, the fox moves to a neighboring hole, either to the left or to the right.

Each morning, you get to inspect a hole of your choice.

What strategy would ensure that the fox is eventually caught ?

______________________

a fox in 3 (adjacent) holes
a fox in 4 (adjacent) holes
a fox in 5 (adjacent) holes

if you have a proof that's simpler (or more elegant)
than the one(s) found here
https://gurmeet.net/puzzles/fox-in-a-hole/
pls let me know! HH

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• From leflynn@21:1/5 to henh...@gmail.com on Tue May 24 05:29:13 2022
On Monday, May 23, 2022 at 1:00:14 PM UTC-4, henh...@gmail.com wrote:
(no point in anyone posting the Answers, because they are all over the Net)

Consider five holes in a line. One of them is occupied by a fox.

Each night, the fox moves to a neighboring hole, either to the left or to the right.

Each morning, you get to inspect a hole of your choice.

What strategy would ensure that the fox is eventually caught ?

______________________

a fox in 3 (adjacent) holes
a fox in 4 (adjacent) holes
a fox in 5 (adjacent) holes

if you have a proof that's simpler (or more elegant)
than the one(s) found here
https://gurmeet.net/puzzles/fox-in-a-hole/
pls let me know! HH

The one-paragraph explanation of the solution looks pretty compact to me. The even / odd parity is well explained.
It may help to reformulate the problem to one where there is an unlimited supply of foxes and they always move to populate the (one or two) adjacent holes from one turn to another, rather than having a probabilistic element.
As given in in the discussion, the strategy for five holes is easily adapted to N holes.

The solution posited in the replies by "Prince" of 4, 4, 2, 2, 3, 4 is in error.
After you check hole #4 twice, the foxes are in 1, 2 or 3 but they then move to 1, 2, 3 or 4 so checking hole #2 on your third and fourth turns does not catch all the foxes that were in 1 or 2 at the end of your second turn.

L. Flynn

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• From henhanna@gmail.com@21:1/5 to leflynn on Thu May 26 17:39:12 2022
On Tuesday, May 24, 2022 at 5:29:14 AM UTC-7, leflynn wrote:
On Monday, May 23, 2022 at 1:00:14 PM UTC-4, henh...@gmail.com wrote:
(no point in anyone posting the Answers, because they are all over the Net)

Consider five holes in a line. One of them is occupied by a fox.

Each night, the fox moves to a neighboring hole, either to the left or to the right.

Each morning, you get to inspect a hole of your choice.

What strategy would ensure that the fox is eventually caught ?

______________________

a fox in 3 (adjacent) holes
a fox in 4 (adjacent) holes
a fox in 5 (adjacent) holes

if you have a proof that's simpler (or more elegant)
than the one(s) found here
https://gurmeet.net/puzzles/fox-in-a-hole/
pls let me know! HH

The one-paragraph explanation of the solution looks pretty compact to me. The even / odd parity is well explained.
It may help to reformulate the problem to one where there is an unlimited supply of foxes and they always move to populate the (one or two) adjacent holes from one turn to another, rather than having a probabilistic element.
As given in in the discussion, the strategy for five holes is easily adapted to N holes.

The solution posited in the replies by "Prince" of 4, 4, 2, 2, 3, 4 is in error.
After you check hole #4 twice, the foxes are in 1, 2 or 3 but they then move to 1, 2, 3 or 4 so checking hole #2 on your third and fourth turns does not catch all the foxes that were in 1 or 2 at the end of your second turn.

L. Flynn

you're right... the 1st proof (explanation) is very good.

it can be used for 7-holes, 9-holes, ... cases.

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• From leflynn@21:1/5 to henh...@gmail.com on Mon May 30 14:39:27 2022
On Thursday, May 26, 2022 at 8:39:14 PM UTC-4, henh...@gmail.com wrote:
On Tuesday, May 24, 2022 at 5:29:14 AM UTC-7, leflynn wrote:
On Monday, May 23, 2022 at 1:00:14 PM UTC-4, henh...@gmail.com wrote:
Consider five holes in a line. One of them is occupied by a fox.
Each night, the fox moves to a neighboring hole, either to the left or to the right.
Each morning, you get to inspect a hole of your choice.
What strategy would ensure that the fox is eventually caught ?
a fox in 3 (adjacent) holes
a fox in 4 (adjacent) holes
a fox in 5 (adjacent) holes

you're right... the 1st proof (explanation) is very good.
it can be used for 7-holes, 9-holes, ... cases.
Citing https://gurmeet.net/puzzles/fox-in-a-hole/
For even numbers of holes, the up and down strategy will work.
For example, for 6 holes the sequence 2,3,4,5,5,4,3,2 will work.
The first four capture all the foxes who started in even holes,
and the last four those who started in odd holes.
L. Flynn

--- SoupGate-Win32 v1.05
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• From henhanna@gmail.com@21:1/5 to leflynn on Wed Jun 1 06:42:00 2022
On Monday, May 30, 2022 at 2:39:28 PM UTC-7, leflynn wrote:
On Thursday, May 26, 2022 at 8:39:14 PM UTC-4, henh...@gmail.com wrote:
On Tuesday, May 24, 2022 at 5:29:14 AM UTC-7, leflynn wrote:
On Monday, May 23, 2022 at 1:00:14 PM UTC-4, henh...@gmail.com wrote:
Consider five holes in a line. One of them is occupied by a fox.

Each night, the fox moves to a neighboring hole, either to the left or to the right.

Each morning, you get to inspect a hole of your choice.
What strategy would ensure that the fox is eventually caught ?
a fox in 3 (adjacent) holes
a fox in 4 (adjacent) holes
a fox in 5 (adjacent) holes

you're right... the 1st proof (explanation) is very good.
it can be used for 7-holes, 9-holes, ... cases.

Citing https://gurmeet.net/puzzles/fox-in-a-hole/
For even numbers of holes, the up and down strategy will work.
For example, for 6 holes the sequence 2,3,4,5,5,4,3,2 will work.
The first four capture all the foxes who started in even holes,
and the last four those who started in odd holes.
L. Flynn

so i thought of 5 variants...
(if you have a definitive answer, pls wait a few days before posting it... thanks)

A. Each night, the fox doesn't move

B. Each night, the fox moves to the right, if possible. (adjacent hole)

C. Each night, the fox can move to the right, or stay put. (adjacent hole)

D. Each night, the fox can move to a neighboring hole, either to the left or to the right. (or stay put)

E. Each night, the fox moves to a (distance-2) neighboring hole, either to the left or to the right. (if possible)

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• From Anton Shepelev@21:1/5 to All on Wed Jun 1 17:47:17 2022
HH:

A. Each night, the fox doesn't move

This is explicitly precluded:

Each night, the fox moves to a neighboring hole,
either to the left or to the right.

--
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• From leflynn@21:1/5 to henh...@gmail.com on Thu Jun 2 07:54:10 2022
On Wednesday, June 1, 2022 at 9:42:02 AM UTC-4, henh...@gmail.com wrote:
On Monday, May 30, 2022 at 2:39:28 PM UTC-7, leflynn wrote:
On Thursday, May 26, 2022 at 8:39:14 PM UTC-4, henh...@gmail.com wrote:
On Tuesday, May 24, 2022 at 5:29:14 AM UTC-7, leflynn wrote:
On Monday, May 23, 2022 at 1:00:14 PM UTC-4, henh...@gmail.com wrote:
Consider five holes in a line. One of them is occupied by a fox.

Each night, the fox moves to a neighboring hole, either to the left or to the right.

Each morning, you get to inspect a hole of your choice.
What strategy would ensure that the fox is eventually caught ?
a fox in 3 (adjacent) holes
a fox in 4 (adjacent) holes
a fox in 5 (adjacent) holes

you're right... the 1st proof (explanation) is very good.
it can be used for 7-holes, 9-holes, ... cases.

Citing https://gurmeet.net/puzzles/fox-in-a-hole/
For even numbers of holes, the up and down strategy will work.
For example, for 6 holes the sequence 2,3,4,5,5,4,3,2 will work.
The first four capture all the foxes who started in even holes,
and the last four those who started in odd holes.
L. Flynn
so i thought of 5 variants...
(if you have a definitive answer, pls wait a few days before posting it... thanks)

A. Each night, the fox doesn't move

B. Each night, the fox moves to the right, if possible. (adjacent hole)

C. Each night, the fox can move to the right, or stay put. (adjacent hole)

D. Each night, the fox can move to a neighboring hole, either to the left or to the right. (or stay put)

For E.
Do you mean a fox in 3 can move to either 1 or 5 or can it move to one of 1,2,4 or 5?
E. Each night, the fox moves to a (distance-2) neighboring hole, either to the left or to the right. (if possible)

--- SoupGate-Win32 v1.05
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• From henhanna@gmail.com@21:1/5 to leflynn on Thu Jun 2 08:20:30 2022
On Thursday, June 2, 2022 at 7:54:12 AM UTC-7, leflynn wrote:
On Wednesday, June 1, 2022 at 9:42:02 AM UTC-4, henh...@gmail.com wrote:
On Monday, May 30, 2022 at 2:39:28 PM UTC-7, leflynn wrote:
On Thursday, May 26, 2022 at 8:39:14 PM UTC-4, henh...@gmail.com wrote:
On Tuesday, May 24, 2022 at 5:29:14 AM UTC-7, leflynn wrote:
On Monday, May 23, 2022 at 1:00:14 PM UTC-4, henh...@gmail.com wrote:
Consider five holes in a line. One of them is occupied by a fox.

Each night, the fox moves to a neighboring hole, either to the left or to the right.

Each morning, you get to inspect a hole of your choice.
What strategy would ensure that the fox is eventually caught ?
a fox in 3 (adjacent) holes
a fox in 4 (adjacent) holes
a fox in 5 (adjacent) holes

you're right... the 1st proof (explanation) is very good.
it can be used for 7-holes, 9-holes, ... cases.

Citing https://gurmeet.net/puzzles/fox-in-a-hole/
For even numbers of holes, the up and down strategy will work.
For example, for 6 holes the sequence 2,3,4,5,5,4,3,2 will work.
The first four capture all the foxes who started in even holes,
and the last four those who started in odd holes.
L. Flynn
so i thought of 5 variants...
(if you have a definitive answer, pls wait a few days before posting it... thanks)

A. Each night, the fox doesn't move

B. Each night, the fox moves to the right, if possible. (adjacent hole)

C. Each night, the fox can move to the right, or stay put. (adjacent hole)

D. Each night, the fox can move to a neighboring hole, either to the left or to the right. (or stay put)

For E.
Do you mean a fox in 3 can move to either 1 or 5

yes... that's waht i mean.

E. Each night, the fox moves to a (distance-2) neighboring hole, either to the left or to the right. (if possible)

--- SoupGate-Win32 v1.05
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• From Richard Tobin@21:1/5 to leflynn@hotmail.com on Thu Jun 2 19:23:59 2022
leflynn <leflynn@hotmail.com> wrote:

For example, for 6 holes the sequence 2,3,4,5,5,4,3,2 will work.

With an even number of holes, you can imagine that after the first
pass you approach the row of holes from the other side so that the odd
holes become even and vice versa.

-- Richard

--- SoupGate-Win32 v1.05
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• From leflynn@21:1/5 to henh...@gmail.com on Sat Jun 4 10:44:59 2022
On Wednesday, June 1, 2022 at 9:42:02 AM UTC-4, henh...@gmail.com wrote:
On Monday, May 30, 2022 at 2:39:28 PM UTC-7, leflynn wrote:
On Thursday, May 26, 2022 at 8:39:14 PM UTC-4, henh...@gmail.com wrote:
On Tuesday, May 24, 2022 at 5:29:14 AM UTC-7, leflynn wrote:
On Monday, May 23, 2022 at 1:00:14 PM UTC-4, henh...@gmail.com wrote:
Consider five holes in a line. One of them is occupied by a fox.

Each night, the fox moves to a neighboring hole, either to the left or to the right.

Each morning, you get to inspect a hole of your choice.
What strategy would ensure that the fox is eventually caught ?
a fox in 3 (adjacent) holes
a fox in 4 (adjacent) holes
a fox in 5 (adjacent) holes

you're right... the 1st proof (explanation) is very good.
it can be used for 7-holes, 9-holes, ... cases.

Citing https://gurmeet.net/puzzles/fox-in-a-hole/
For even numbers of holes, the up and down strategy will work.
For example, for 6 holes the sequence 2,3,4,5,5,4,3,2 will work.
The first four capture all the foxes who started in even holes,
and the last four those who started in odd holes.
L. Flynn
so i thought of 5 variants...
(if you have a definitive answer, pls wait a few days before posting it... thanks)

A. Each night, the fox doesn't move

B. Each night, the fox moves to the right, if possible. (adjacent hole)

C. Each night, the fox can move to the right, or stay put. (adjacent hole)

D. Each night, the fox can move to a neighboring hole, either to the left or to the right. (or stay put)

E. Each night, the fox moves to a (distance-2) neighboring hole, either to the left or to the right. (if possible)
A. Each night, the fox doesn't move
Optimal search, 12345.

B. Each night, the fox moves to the right, if possible. (adjacent hole)
Optimal search 135 (or 225) (Does “if possible” mean “if a den is unoccupied” or just that “they move until they reach 5”?)

C. Each night, the fox can move to the right, or stay put. (adjacent hole) Optimal search 12345 (not in terms of quickest average time to catch a single random fox. Need probability of stay/move.)

D. Each night, the fox can move to a neighboring hole, either to the left or to the right. (or stay put)
No single hole checking strategy can guarantee getting the fox.

E. Each night, the fox moves to a (distance-2) neighboring hole, either to the left or to the right. (if possible)
Do you mean a fox in 3 can move to either 1 or 5 or can it move to one of 1,2,4 or 5?
If you mean the steps have to be 2 holes, then we just have even and odd fox populations and they never mix. We can solve the sub-problems sequentially once subset at a time just like the original problem.
If you mean the steps can be 1 or 2 holes, then we are out of luck.

What happens to these variants is the holes are arranged in a circle?

--- SoupGate-Win32 v1.05
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• From Edward Murphy@21:1/5 to leflynn on Sun Jun 5 12:35:42 2022
On 6/4/2022 10:44 AM, leflynn wrote:
On Wednesday, June 1, 2022 at 9:42:02 AM UTC-4, henh...@gmail.com wrote:
On Monday, May 30, 2022 at 2:39:28 PM UTC-7, leflynn wrote:
On Thursday, May 26, 2022 at 8:39:14 PM UTC-4, henh...@gmail.com wrote: >>>> On Tuesday, May 24, 2022 at 5:29:14 AM UTC-7, leflynn wrote:
On Monday, May 23, 2022 at 1:00:14 PM UTC-4, henh...@gmail.com wrote: >>>>>> Consider five holes in a line. One of them is occupied by a fox.

Each night, the fox moves to a neighboring hole, either to the left or to the right.

Each morning, you get to inspect a hole of your choice.
What strategy would ensure that the fox is eventually caught ?
a fox in 3 (adjacent) holes
a fox in 4 (adjacent) holes
a fox in 5 (adjacent) holes

you're right... the 1st proof (explanation) is very good.
it can be used for 7-holes, 9-holes, ... cases.

Citing https://gurmeet.net/puzzles/fox-in-a-hole/
For even numbers of holes, the up and down strategy will work.
For example, for 6 holes the sequence 2,3,4,5,5,4,3,2 will work.
The first four capture all the foxes who started in even holes,
and the last four those who started in odd holes.
L. Flynn
so i thought of 5 variants...
(if you have a definitive answer, pls wait a few days before posting it... thanks)

A. Each night, the fox doesn't move

B. Each night, the fox moves to the right, if possible. (adjacent hole)

C. Each night, the fox can move to the right, or stay put. (adjacent hole) >>
D. Each night, the fox can move to a neighboring hole, either to the left or to the right. (or stay put)

E. Each night, the fox moves to a (distance-2) neighboring hole, either to the left or to the right. (if possible)
A. Each night, the fox doesn't move
Optimal search, 12345.

B. Each night, the fox moves to the right, if possible. (adjacent hole) Optimal search 135 (or 225) (Does “if possible” mean “if a den is unoccupied” or just that “they move until they reach 5”?)

C. Each night, the fox can move to the right, or stay put. (adjacent hole) Optimal search 12345 (not in terms of quickest average time to catch a single random fox. Need probability of stay/move.)

D. Each night, the fox can move to a neighboring hole, either to the left or to the right. (or stay put)
No single hole checking strategy can guarantee getting the fox.

E. Each night, the fox moves to a (distance-2) neighboring hole, either to the left or to the right. (if possible)
Do you mean a fox in 3 can move to either 1 or 5 or can it move to one of 1,2,4 or 5?
If you mean the steps have to be 2 holes, then we just have even and odd fox populations and they never mix. We can solve the sub-problems sequentially once subset at a time just like the original problem.
If you mean the steps can be 1 or 2 holes, then we are out of luck.

What happens to these variants is the holes are arranged in a circle?

Translating left/right to clockwise/counterclockwise (in some order):

Original. No guarantee. Say you check 1 and the fox isn't there, so on
that day it was in one of 2345, but on the next day it could be in any
of them (after moving in one direction to one of 1234 respectively, or
in the other direction to one of 3451 respectively).

A. Optimal search 12345.

B. Optimal search 11111.

C. No guarantee. Say you check 1 and the fox isn't there, so on that
day it was in one of 2345, but on the next day it could be in any
of them (after staying put in one of 2345 respectively, or moving
from 5 to 1).

D. No guarantee.

E. This is equivalent to the original (after re-labeling the holes
appropriately), so again, no guarantee.

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