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| Christmas Investigations Part Six |
How do you decide whether to release one of your locked-in teammates?
Louise is the team captain of a team being led around The Crystal Maze by Richard O'Brien. Things were going quite well at first but in the last five games the team have suffered five lock-ins. Louise is the only person left and there is just time for her to buy out any teammates she wants to take to the Crystal Dome to help collect credits.
She does a quick bit of maths and realises that the best thing she can do is buy out all of her teammates since buying out four, or five has the same effect and if she buys out all five everyone gets the experience of being in the Crystal Dome.
Given all of the above information, how many games did the team win before their disastrous run of lock ins?
So, how exactly do we go about solving this? Well, unlike the first two investigations, which I wrote especially for the festive period, this is a puzzle I've been pondering for a while.
The method behind the answer is simple:
Given x and y variable and K constant where x+y=K
As x tends towards y
xy tends towards a maximum.
xy tends towards a maximum.
Ok, so it's simple to write, but perhaps I need to start at the beginning.
We have two variables which will contribute to the overall success of a team in the Crystals Dome. These are the number of players available to collect tokens, and the number of crystals they have.
If we have a fence to paint and we know it takes a painter 8 hours to paint it, then we can say that the fence takes 8 man-hours to paint. We can fulfil this in a number of ways, two people for four hours, eight people for an hour each and so on. Each successful painting of the fence requires the number of painters, multiplied by the hours they paint for, to equal the required number of man-hours.
This is analogous to our situation in the Crystal Dome. We have a number of collectors, which we can call 'x' and a number of crystals which we can call 'y'. If we multiply x by y, and then by 5, we get the number of 'contestant seconds' available for a given situation.
Still with me, good, because now we need to take a look at some rectangles.
The area of a rectangle is it's length, multiplied by it's width. while the perimeter is twice the length plus twice the width. Or, to put it simply.
Area = L x W
Perimeter - 2L + 2W
Now, suppose we make the perimeter constant, but try different values of L and W.
2L + 2W = 24
dividing by two to make things simple:
L + W = 12
Using just whole number values for L and W, here are all the different rectangles we can draw.
As you can see, as L and W get closer together, the size of the rectangle increases.
Now, compare this to our problem of fence painting. Suppose that our funds for paying the painters, are limited. We can only pay for a number of hours and a number of painters that together total 12. Suddenly our fence problem is analogous to our rectangles. We can get much more work done by paying six painters to work for six hours each, than we can for paying one painter to work for eleven hours.
To put it another way:
Given x and y variable; and K constant where x+y=K
As x tends towards y
xy tends towards a maximum.
xy tends towards a maximum.
So by now you should have realised that when it comes to crystals and prisoners, the best situation to be in is one where the number of crystals is about the same as the number of contestants collecting.
Our question to you took this a stage further though, we told you that Louise had a number of crystals and five locked in teammates. She decided it was better to buy everyone out, and she was right, so how many crystals did she have.
Oddly what you need to solve this is a multiplication table.
Here's the one I used to originally solve this problem, let's look at an example to see how it could help to decide whether to buy out a prisoner.
For instance, if you have eight crystals and only five contestants you are better releasing a contestant.
The cell for seven crystals against six contestants is 42, vs 40 for eight crystals and five contestants. It's a small but significant gain.
The next step is to work out which moves are gains and which moves are losses in this chart. This of course could be done on paper, ir in your head, but being the kind of puzzler that I am, I composed another chart.
So this master document is showing something slightly different, each cell is showing the number of contestant-seconds you would gain, or lose if, from that situation, you exchanged a crystal for a prisoner.
So, we know from the original problem that Louise had five prisoners, or, to fit in with this chart, just one contestant. We also know that buying back four, or five teammates will have the same mathematical effect.
On our chart then this puts us at the intersection of six crystals and five contestants; which is the only situation on the chart where there is just one more contestant to buy back, and doing so would have no gain or loss.
Now all we need to do is travel diagonally down the diagram through each of the previous buy-back situations. Her second to last decision was with seven crystals and four contestants, with a gain of 10 person-seconds. Before that she had eight crystals and three teammates and the buy back she performed gained 20 person-seconds.
The move before that was from nine crystals and two teammates with a gain of 30 person-seconds when she bought out a third.
Finally the position she started in, just herself with ten crystals, she gained 40 person-seconds from that first move.
So we have an answer, 10 crystals, and a not insignificant amount of mathematics too.
That's it for this series of Christmas Investigation explanations, until next time, keep puzzling.
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