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Jan 14th, 2010, 04:08 PM
#1
Bingo Minimum
I was watching a game show which fills in a 5x5 "bingo"-style grid, one element at a time. At the end of the game show, contestants try to get a bingo (that is, 5 filled-in elements in a row either horizontally, vertically, or diagonally). The grid starts with 12 squares filled in, and, based on the contestants' previous performance, they draw randomly to fill in the remaining squares.
Contestants who do reasonably well get around 8 draws. I was wondering if, at some point, these extra draws become unnecessary. The rules of the game cap the number of draws at 10, leaving at least 3 grid elements unfilled. To put it another way, what is the minimum number of elements on a 5x5 grid that need to be filled in to ensure that there will be a bingo?
I know that 20 isn't quite enough, from the following counterexample:
Code:
- x x x x
x - x x x
x x - x x
x x x - x
x x x x -
x = filled-in
- = open
[I posted this without trying to analyze it myself since I was thinking somebody might be interested in it as a puzzle.]
The time you enjoy wasting is not wasted time.
Bertrand Russell
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Jan 15th, 2010, 12:46 AM
#2
Re: Bingo Minimum
Maybe I don't really get it, but it seems to me that the minimum of filled-in numbers to get a BINGO is 5. That's the case when all those five numbers are in single row/column/diagonal.
So a grid with 12 numbers filled-in could already be a solution, however such a grid wouldn't be handed out.
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Jan 15th, 2010, 10:32 AM
#3
Re: Bingo Minimum
Sorry if I wasn't clear. I meant to ask the opposite question--not the minimum number of filled in numbers to possibly have a bingo, but the minimum number of filled in numbers to guarantee a bingo.
The time you enjoy wasting is not wasted time.
Bertrand Russell
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Jan 16th, 2010, 06:44 AM
#4
Re: Bingo Minimum
Does your counter example not pretty much answer the question? It proves 20 is not quite enough but with 21 or 4 open it is impossible to have a blank in all 5 columns and all 5 rows.
A no-brainer that one, Jem.
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Jan 16th, 2010, 05:11 PM
#5
Re: Bingo Minimum
 Originally Posted by Milk
Does your counter example not pretty much answer the question? It proves 20 is not quite enough but with 21 or 4 open it is impossible to have a blank in all 5 columns and all 5 rows.
A no-brainer that one, Jem.
Yup, you're right. As I said, originally I hadn't analyzed it myself, past the 20 counterexample--in computer terms, I killed the process. [Though to be honest in writing in this thread I had accidentally figured it out; I figured someone might have found it interesting regardless, though, especially if they've seen the show.]
The time you enjoy wasting is not wasted time.
Bertrand Russell
<- Remember to rate posts you find helpful.
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