------ pls wait 3+ days (Longer if you find it easy or trivial) before posting answers or hints.
A flashlight takes 2 (working) batteries.
There are 8 batteries (on the Table), but some of them may be Dead (and Non-Conducting).
To guarantee that the flashlight gets turned on, what is the minimum number of battery pairs you need to test ?
(0) ... when you have Zero additional info2
(1) ... when exactly 1 of the batteries is Dead.
(2) ... when exactly 2 of the batteries are Dead.
(3) ... when exactly 3 of the batteries are Dead.
(4) ... when exactly 4 of the batteries are Dead.
(5) ... when exactly 5 of the batteries are Dead.
(6) ... when exactly 6 of the batteries are Dead.
(7) ... when exactly 7 of the batteries are Dead.0. It ain't gonna work.
On Tue, 30 Aug 2022 12:37:42 -0700 (PDT)
"henh...@gmail.com" <henhanna@gmail.com> wrote:
------ pls wait 3+ days (Longer if you find it easy or trivial) before posting answers or hints.
Feh.
A flashlight takes 2 (working) batteries.
There are 8 batteries (on the Table), but some of them may be Dead (and Non-Conducting).
To guarantee that the flashlight gets turned on, what is the minimum number of battery pairs you need to test ?
On Tue, 30 Aug 2022 12:37:42 -0700 (PDT)
"henh...@gmail.com" <henhanna@gmail.com> wrote:
------ pls wait 3+ days (Longer if you find it easy or trivial) before posting answers or hints.
Feh.
2
A flashlight takes 2 (working) batteries.
There are 8 batteries (on the Table), but some of them may be Dead (and Non-Conducting).
To guarantee that the flashlight gets turned on, what is the minimum number of battery pairs you need to test ?
(0) ... when you have Zero additional info
(1) ... when exactly 1 of the batteries is Dead.
(2) ... when exactly 2 of the batteries are Dead.
(3) ... when exactly 3 of the batteries are Dead.
(4) ... when exactly 4 of the batteries are Dead.
(5) ... when exactly 5 of the batteries are Dead.
(6) ... when exactly 6 of the batteries are Dead.
You'll probably need some stats nCx type combinatorics to get these.
(7) ... when exactly 7 of the batteries are Dead.0. It ain't gonna work.
On 8/31/2022 2:36 AM, Kerr-Mudd, John wrote:
On Tue, 30 Aug 2022 12:37:42 -0700 (PDT)
"henh...@gmail.com" <henh...@gmail.com> wrote:
------ pls wait 3+ days (Longer if you find it easy or trivial) before posting answers or hints.
Feh.
2
A flashlight takes 2 (working) batteries.
There are 8 batteries (on the Table), but some of them may be Dead (and Non-Conducting).
To guarantee that the flashlight gets turned on, what is the minimum number of battery pairs you need to test ?
(0) ... when you have Zero additional info
(1) ... when exactly 1 of the batteries is Dead.
(2) ... when exactly 2 of the batteries are Dead.
(3) ... when exactly 3 of the batteries are Dead.
(4) ... when exactly 4 of the batteries are Dead.
(5) ... when exactly 5 of the batteries are Dead.
(6) ... when exactly 6 of the batteries are Dead.
You'll probably need some stats nCx type combinatorics to get these.Some of these are simpler. Label the batteries A through H, then test
AB CD EF GH in that order:
(2) 3. Worst-case scenario is that the dead batteries are (one of AB)
and (one of CD).
Similarly, (3) 4. Worst-case scenario is (one of AB), (one of CD), and
(one of EF).
(4) and (5) are the most complex of the bunch.
* There are C(8, 2) = 28 unordered pairs of batteries.
* For (4), if we test AB CD EF GH and they all fail, then the dead
batteries are (one of AB), (one of CD), (one of EF), and (one of
GH). So then we can test AC AD BC BD in that order, and the
worst-case scenario is that A and C are dead. So we can succeed
with at most 8 tests; there may be a more efficient approach,
though I don't know what it would be.
On 8/31/2022 2:36 AM, Kerr-Mudd, John wrote:
On Tue, 30 Aug 2022 12:37:42 -0700 (PDT)
"henh...@gmail.com" <henh...@gmail.com> wrote:
------ pls wait 3+ days (Longer if you find it easy or trivial) before posting answers or hints.
Feh.
2
A flashlight takes 2 (working) batteries.
There are 8 batteries (on the Table), but some of them may be Dead (and Non-Conducting).
To guarantee that the flashlight gets turned on, what is the minimum number of battery pairs you need to test ?
(0) ... when you have Zero additional info
(1) ... when exactly 1 of the batteries is Dead.
(2) ... when exactly 2 of the batteries are Dead.
(3) ... when exactly 3 of the batteries are Dead.
(4) ... when exactly 4 of the batteries are Dead.
(5) ... when exactly 5 of the batteries are Dead.
(6) ... when exactly 6 of the batteries are Dead.
You'll probably need some stats nCx type combinatorics to get these.Some of these are simpler. Label the batteries A through H, then test
AB CD EF GH in that order:
(2) 3. Worst-case scenario is that the dead batteries are (one of AB)
and (one of CD).
Similarly, (3) 4. Worst-case scenario is (one of AB), (one of CD), and
(one of EF).
(4) and (5) are the most complex of the bunch.
* There are C(8, 2) = 28 unordered pairs of batteries.
* For (4), if we test AB CD EF GH and they all fail, then the dead
batteries are (one of AB), (one of CD), (one of EF), and (one of
GH). So then we can test AC AD BC BD in that order, and the
worst-case scenario is that A and C are dead. So we can succeed
with at most 8 tests; there may be a more efficient approach,
though I don't know what it would be.
* For (5), if we test the C(7, 2) = 21 unordered pairs that *don't*
contain H, then we can succeed with at most 21 tests. (Even if H
works, so do two others.) Again, there may be a more efficient
approach, though I don't know what it would be.
On Monday, September 5, 2022 at 5:48:33 PM UTC-7, Edward Murphy wrote:
On 8/31/2022 2:36 AM, Kerr-Mudd, John wrote:
On Tue, 30 Aug 2022 12:37:42 -0700 (PDT)Some of these are simpler. Label the batteries A through H, then test
"henh...@gmail.com" <henh...@gmail.com> wrote:
------ pls wait 3+ days (Longer if you find it easy or trivial) before posting answers or hints.
Feh.
2
A flashlight takes 2 (working) batteries.
There are 8 batteries (on the Table), but some of them may be Dead (and Non-Conducting).
To guarantee that the flashlight gets turned on, what is the minimum number of battery pairs you need to test ?
(0) ... when you have Zero additional info
(1) ... when exactly 1 of the batteries is Dead.
(2) ... when exactly 2 of the batteries are Dead.
(3) ... when exactly 3 of the batteries are Dead.
(4) ... when exactly 4 of the batteries are Dead.
(5) ... when exactly 5 of the batteries are Dead.
(6) ... when exactly 6 of the batteries are Dead.
You'll probably need some stats nCx type combinatorics to get these.
AB CD EF GH in that order:
(2) 3. Worst-case scenario is that the dead batteries are (one of AB)
and (one of CD).
Similarly, (3) 4. Worst-case scenario is (one of AB), (one of CD), and
(one of EF).
(4) and (5) are the most complex of the bunch.
* There are C(8, 2) = 28 unordered pairs of batteries.
* For (4), if we test AB CD EF GH and they all fail, then the dead
batteries are (one of AB), (one of CD), (one of EF), and (one of
GH). So then we can test AC AD BC BD in that order, and the
worst-case scenario is that A and C are dead. So we can succeed
with at most 8 tests; there may be a more efficient approach,
though I don't know what it would be.
* For (5), if we test the C(7, 2) = 21 unordered pairs that *don't*
contain H, then we can succeed with at most 21 tests. (Even if H
works, so do two others.) Again, there may be a more efficient
approach, though I don't know what it would be.
(5 dead) ... when exactly 5 of the batteries are Dead. (and 3 are Alive)
method 5A requires 12 moves in the worst case.
method 5B requires 13 moves in the worst case. ---- could this one be BETTER in some way ?
in these problems, we are trying to reduce the # of moves in the worst case.
Do we have interesting (new) genre of problems, if we instead
tried to reduce the # of moves in the average case ?
what is the minimum number of battery pairs you need to test ?------ pls wait 3+ days (Longer if you find it easy or trivial) before posting answers or hints.
A flashlight takes 2 (working) batteries.
There are 8 batteries (on the Table), but some of them may be Dead (and Non-Conducting).
To guarantee that the flashlight gets turned on,
On 9/5/2022 10:15 PM, henh...@gmail.com wrote:
On Monday, September 5, 2022 at 5:48:33 PM UTC-7, Edward Murphy wrote:
On 8/31/2022 2:36 AM, Kerr-Mudd, John wrote:
On Tue, 30 Aug 2022 12:37:42 -0700 (PDT)Some of these are simpler. Label the batteries A through H, then test
"henh...@gmail.com" <henh...@gmail.com> wrote:
------ pls wait 3+ days (Longer if you find it easy or trivial) before posting answers or hints.
Feh.
2
A flashlight takes 2 (working) batteries.
There are 8 batteries (on the Table), but some of them may be Dead (and Non-Conducting).
To guarantee that the flashlight gets turned on, what is the minimum number of battery pairs you need to test ?
(0) ... when you have Zero additional info
(1) ... when exactly 1 of the batteries is Dead.
(2) ... when exactly 2 of the batteries are Dead.
(3) ... when exactly 3 of the batteries are Dead.
(4) ... when exactly 4 of the batteries are Dead.
(5) ... when exactly 5 of the batteries are Dead.
(6) ... when exactly 6 of the batteries are Dead.
You'll probably need some stats nCx type combinatorics to get these.
AB CD EF GH in that order:
(2) 3. Worst-case scenario is that the dead batteries are (one of AB)
and (one of CD).
Similarly, (3) 4. Worst-case scenario is (one of AB), (one of CD), and
(one of EF).
(4) and (5) are the most complex of the bunch.
* There are C(8, 2) = 28 unordered pairs of batteries.
* For (4), if we test AB CD EF GH and they all fail, then the dead
batteries are (one of AB), (one of CD), (one of EF), and (one of
GH). So then we can test AC AD BC BD in that order, and the
worst-case scenario is that A and C are dead. So we can succeed
with at most 8 tests; there may be a more efficient approach,
though I don't know what it would be.
* For (5), if we test the C(7, 2) = 21 unordered pairs that *don't*
contain H, then we can succeed with at most 21 tests. (Even if H
works, so do two others.) Again, there may be a more efficient
approach, though I don't know what it would be.
(5 dead) ... when exactly 5 of the batteries are Dead. (and 3 are Alive)
method 5A requires 12 moves in the worst case.
method 5B requires 13 moves in the worst case. ---- could this one be BETTER in some way ?I don't understand what specific methods you're labeling as 5A and 5B
here. (If they're different methods to approach the same scenario, then clearly 5B could "be better": you could replace it with 5A.)
Taking the same sort of constructive approach that I labeled (4) above
(4 dead / 4 alive), and applying it to the case of 5 dead / 3 alive:
* Test AB CD EF GH. Worst case is that they all fail, meaning that the
5 dead batteries include one from AB, one from CD, one from EF, and
one from GH.
* Test AC AD BC BD. Worst case is that they all fail, meaning that the
5 dead batteries include either (both A and B) or (both C and D).
* Test EG EH FG FH. Worst case is that the first three fail, meaning
that E and G are dead, in which case FH succeeds (test #12).
I'm guessing that this is what you meant by 5A. 5B remains unclear.
Fox eats goose and goose eats corn if left alone. Farmer can row the boat himself and doesn't …
A farmer is travelling with a fox, a sheep and a small sack of hay. He comes to a river with a small boat in it. The boat can only support the farmer and one other animal/item. If the farmer leaves …
Sysop: | Keyop |
---|---|
Location: | Huddersfield, West Yorkshire, UK |
Users: | 325 |
Nodes: | 16 (2 / 14) |
Uptime: | 63:46:12 |
Calls: | 7,123 |
Calls today: | 1 |
Files: | 12,524 |
Messages: | 5,521,179 |