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Old 2014-08-27, 05:37   #1
MattcAnderson
 
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"Matthew Anderson"
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Question Comedy Central

Hi Math people,

I have problem 1.4.8 from "The Heart of Mathematics" by Burger and Starbird.

It reads -
Comedy Central. A good comedian armed with strong material can really kill an audience. Bad comedians, on the other hand, are caught and brought to Comedy Central Correctional (CCC), where they do their timing. But moving C- and D-list comedians to lockdown is not as easy at it may appear. One day, in the outback of Los Angeles, three comedy correctional officers were escorting three bad comedians, who had just bombed the Laff Stop, to CCC. Suddenly, along their march to CCC, they came upon the banks of the mighty Los Angeles River and needed to cross its turbulent and deep waters. None of the six could swim, but all could row. Fortunately, on the river's shore was a small rowboat available for use.
Since the boat was small and the officers and comedians were all on the portly side, it was clear that only two persons could cross at one time. The problem was: While each comedian was not particularly great, if ever there were a moment when the comedians outnumbered the officers, then the comedians - together - were funny enough to kill their audience (i.e. the officers). Given this reality, the officers decided that being prudent was better than being dead, so they agreed that at no time would they allow any group of officers to be outnumbered by comedians during the crossing. For their part, the comedians did not fear being outnumbered by the officers because they realized that an excess of officders would result only in more discussion among the officers, thus relieving the comedians of the burden of being "on".
How do the officers and comedians all cross the river using only the one boat yet at no time letting the comedians outnumber the officers on either side of the river?

Thanks for reading this far. I found a related Wikipedia article -
http://en.wikipedia.org/wiki/Mission...nibals_problem
and reference 5 at the bottom seems to enumerate the steps.

It seems the boat cannot be sent across empty.

Regards,
Matt
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Old 2014-08-28, 00:14   #2
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Quote:
Originally Posted by MattcAnderson View Post
Hi Math people,

I have problem 1.4.8 from "The Heart of Mathematics" by Burger and Starbird.

It reads -
Comedy Central. A good comedian armed with strong material can really kill an audience. Bad comedians, on the other hand, are caught and brought to Comedy Central Correctional (CCC), where they do their timing. But moving C- and D-list comedians to lockdown is not as easy at it may appear. One day, in the outback of Los Angeles, three comedy correctional officers were escorting three bad comedians, who had just bombed the Laff Stop, to CCC. Suddenly, along their march to CCC, they came upon the banks of the mighty Los Angeles River and needed to cross its turbulent and deep waters. None of the six could swim, but all could row. Fortunately, on the river's shore was a small rowboat available for use.
Since the boat was small and the officers and comedians were all on the portly side, it was clear that only two persons could cross at one time. The problem was: While each comedian was not particularly great, if ever there were a moment when the comedians outnumbered the officers, then the comedians - together - were funny enough to kill their audience (i.e. the officers). Given this reality, the officers decided that being prudent was better than being dead, so they agreed that at no time would they allow any group of officers to be outnumbered by comedians during the crossing. For their part, the comedians did not fear being outnumbered by the officers because they realized that an excess of officders would result only in more discussion among the officers, thus relieving the comedians of the burden of being "on".
How do the officers and comedians all cross the river using only the one boat yet at no time letting the comedians outnumber the officers on either side of the river?

Thanks for reading this far. I found a related Wikipedia article -
http://en.wikipedia.org/wiki/Mission...nibals_problem
and reference 5 at the bottom seems to enumerate the steps.

It seems the boat cannot be sent across empty.

Regards,
Matt
Is this a puzzle or homework ? or were your previous puzzles homework and this is just a puzzle ?
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Old 2014-08-28, 00:43   #3
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Hi all math people,

I am auditing Math 130 at Willamette University.

This is a homework problem.

The other problems in the puzzles forum were just for fun.

According to "The Mathematical Gazette" June 1989 article by Pressman and Singmaster, this puzzle can be solved in 11 crossings.

Regards,
Matt
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Old 2014-08-28, 00:55   #4
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Quote:
Originally Posted by MattcAnderson View Post
Hi all math people,

I am auditing Math 130 at Willamette University.

This is a homework problem.

The other problems in the puzzles forum were just for fun.

According to "The Mathematical Gazette" June 1989 article by Pressman and Singmaster, this puzzle can be solved in 11 crossings.

Regards,
Matt
hint: hint deleted as it was too direct

Last fiddled with by science_man_88 on 2014-08-28 at 00:57
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Old 2014-08-28, 01:02   #5
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One strategy is to look at a state diagram with 10 legal states.

Regards,
Matt
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Old 2014-08-28, 01:03   #6
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Quote:
Originally Posted by MattcAnderson View Post
One strategy is to look at a state diagram with 10 legal states.

Regards,
Matt
I have the answer I guess I'm not seeing the point of the thread for some reason.
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Old 2014-09-08, 14:04   #7
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Wink

Here is an answer to the problem.

1) Send boat across with two comedians.
2) Boat returns with one comedian.
3) Boat crosses with two comedians.
4) Boat returns with one comedian.
5) Boat crosses with two officers.
6) Boat returns with one officer and one comedian.
7) Boat crosses with two officers.
8) Boat returns with one comedian.
9) Boat crosses with two comedians.
10) Boat returns with one comedian.
11) Boat corsses with two comedians.


With eleven crossings, I'm pretty sure that is optimal.
Regards,
Matt
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Old 2014-09-08, 14:46   #8
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Quote:
Originally Posted by MattcAnderson View Post
Here is an answer to the problem.

1) Send boat across with two comedians.
2) Boat returns with one comedian.
3) Boat crosses with two comedians.
4) Boat returns with one comedian.
5) Boat crosses with two officers.
6) Boat returns with one officer and one comedian.
7) Boat crosses with two officers.
8) Boat returns with one comedian.
9) Boat crosses with two comedians.
10) Boat returns with one comedian.
11) Boat corsses with two comedians.


With eleven crossings, I'm pretty sure that is optimal.
Regards,
Matt

what about:

1) Send boat across with an officer and comedian
2) Boat returns with one officer.
3) Boat crosses with an officer and comedian
4) Boat returns with an officer and a comedian.
5) Boat crosses with two officers.
6) Boat returns with one officer
7) Boat crosses with one officer and one comedian
8) Boat returns with one officer
9) Boat crosses with one officer and one comedian


I'm guessing number 3 ruins this possibility ?
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Old 2014-09-08, 18:25   #9
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Yes, after step 3 an officer will be on the far side of the river with two comedians and he will be overpowered. Then he will laugh his head off and that would be bad.

Thanks for contributing.

Regards,
Matt
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