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Geek Culture / Countdown Number Round Challenge

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BatVink
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Posted: 28th Mar 2008 19:12 Edited at: 28th Mar 2008 19:13
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In the UK, we have a quiz program called Countdown. One of the rounds involves solves a number problem, and I wondered how you could go about creating a program that can find the solution. I have found other on the net, but it would still be interesting to come up with a coding algorithm that does it.

Here is how the challenge works.

You choose 6 random numbers, ranging between 1 and 100.
Another random number between 1 and 999 is generated. You must create a sum from the 6 numbers where the answer equals the generated answer. You can add, subtract, divide and multiply. Integer numbers only all the way.

For example:6 numbers: 3,3,4,8,10,50
Generated answer: 919

Solution

8 x 10 = 80
80 x 4 = 320
320 + 3 = 323
323 x 3 = 969
969 - 50 = 919

So, how do you create a program that can check all combinations and find a solution? You don't have to use all the numbers.
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Aertic
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Posted: 28th Mar 2008 19:14
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Sounds uhmmm, odd?
I'll chack through that again, also Ive watched countdown before, its good.

Just tell me if its over sized, its my first attempt at a sig.

[URL]www.Redarrow.forums1.net[/URL]
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BatVink
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Posted: 28th Mar 2008 19:17
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My first thought on creating a solution is this:

1. Work out all the different ways of arranging 6 individual numbers. e.g:

n1 + n2 + n3 + n4 + n5 + n6
n1 - n2 -n3...
n1 + n2 - n3 + n4...
n1 x n2 x n3...

This obviously needs optimising, it's just a rough outline.

2. repeat for 5 numbers. Before you can do this, you need to establish all the combinations of 5 numbers. This would be any 5 from the 6. But it would also be:

(n1 + n2) + the other 4
(n1 + n3) + the other 4
(n1 x n2) + the other 4
and so on...

3. Repeat for 4 numbers, 3 numbers, 2 numbers.

That would cover all options, but there is obviously a lot of padding to put into those statements.
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Agent Dink
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Posted: 28th Mar 2008 19:38
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Sounds like something I don't want to deal with

MISoft Studios - Silver-Dawn

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Aertic
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Posted: 28th Mar 2008 20:25
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Me either.
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BatVink
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Posted: 28th Mar 2008 21:24
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Rub a couple of brain cells together, create a spark. Coding puzzles like this are good for all sorts of situations. The answer, I'm sure, is in multiple recursive functions.
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Aertic
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Posted: 28th Mar 2008 21:27
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Yeah, but I may have DBP and never used it, not to mention installed it, yet to mention read them damn books I got, and not to metion I dont care about it, if I could do this in C# C++ the I will, but I dont know the varibles/actions/commands whatever there called?

But it is a cool idea.
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Venge
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Posted: 28th Mar 2008 22:40
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You might be able to do something like that involving factorials. Cycle through all possible combinations of numbers and operators...Man that would be a huge number.

Something along the lines of 6!*4, except you would multiply by 4 for every step to check for each operator...So it would actually be

(6*4)*(5*4)*(4*4)*(3*4)*(2*4)*(1) = 737,280 possible combinations. (I think)


Except you might not use every number, or might use a number twice...Agh.

There are over 100 billion galaxies in our observable universe, each containing up to a trillion stars.

Feeling small yet?
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flashing snall
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Posted: 29th Mar 2008 00:38
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oo sounds fun... i cant try right now, but i will once i finish my HW...
i dont think it would be easier, but what if DBRPro stared with the anser, anf tried to get to zero... i dont think it makes a difference, but just something to ponder...


This is my WIP, not even ready for a WIP thread yet though.http://smallgroupproductions.com/
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BatVink
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Posted: 29th Mar 2008 00:44 Edited at: 29th Mar 2008 00:46
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Quote: "or might use a number twice...Agh"


nope, not allowed, each number just once. Although the same number may appear twice.

My next thought is that you need to get over this scenario...

6 + 4 x 3

which could be

(6 + 4) x 3

or

6 + (4 x 3)

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Digital Awakening
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Posted: 29th Mar 2008 09:38
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Quick solution:

Add 6 numbers to an array
Randomize the order
Use a for/next loop
Randomly choose +,-,*,/
You can randomly check if / is possible like 50% of the times
You can weight * high when the factors are low
Check to make sure that the result is always 1-999
And at the end you have the random answer

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BatVink
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Posted: 29th Mar 2008 15:51
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I think you misunderstand. Given the 6 numbers and a final number, you must create a program that can work out if there is a possible answer, and what it is.
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Digital Awakening
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Posted: 29th Mar 2008 16:06
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Ah, that would be far trickier. But what to use it for? I'm no math wiz so this is not for me.

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Digital Awakening
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Posted: 29th Mar 2008 18:34
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You have 6 numbers and 4 operators. Best way would probably to write a code that goes through all the possibilities. You should split it up into two separate algorithms. One that makes a list of all the possible combinations of the 6 numbers (you could make it work for other amounts as well). And another one that runs all the combinations of operators.

I have a thought on the number sorter but it's hard to explain in text so try to follow my thoughts with this:

Flip 1st pair:
1234 > 2134

Flip 2nd pair, then 1st pair:
1324 > 3124 : 2314 > 3214

And keep going with the 3rd pair. If my thoughts are correctly this method should work on any amount of numbers, you just have to figure out a nice way of storing them up into a list.

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BatVink
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Posted: 29th Mar 2008 18:36
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Quote: "But what to use it for?"


http://www.channel4.com/entertainment/tv/microsites/C/countdown/index.html

A legend in British television history. Just ask Ric, he has a website dedicated to one of the panel.
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Digital Awakening
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Posted: 29th Mar 2008 18:40
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For the operators, start by making a an array equally long as the amount of numbers -1 and just count:

111, 211, 311, 411, 121, 221...

The numbers would represent the operators. Run both lists together in a 3rd algorithm.

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Digital Awakening
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Posted: 29th Mar 2008 18:43 Edited at: 29th Mar 2008 18:44
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BatVink:
Yes, but if you plan to make problems for the show then using my first solution should be fine. If I understand correctly the participants must use 6 numbers to get to the final one. Can't these just be premade?


Anyway, anyone think my method would work?

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BatVink
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Posted: 29th Mar 2008 18:44 Edited at: 29th Mar 2008 18:46
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That's a good way to start DA. It gets trickier as you go, though. For example, you can put operations in brackets. You can also use just 5 numbers, or 4...

Quote: "but if you plan to make problems for the show then using my first solution should be fine"


No, that's not the idea. Sometimes they get a puzzle that they cannot solve. The example I gave earlier was one of them. But with the right program, you can work out whether they should have been able to solve it, or if it was impossible.
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Digital Awakening
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Posted: 29th Mar 2008 18:55 Edited at: 29th Mar 2008 18:56
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6 + 4 x 3
(6 + 4) x 3
6 + (4 x 3)

This is not a problem as you run through all combinations you would get:

6+4 = 10x3 = 30 : (6 + 4) x 3
4x3 = 12+6 = 18 : 6 + (4 x 3)

Thus, running through all combinations of numbers and operators solves this problem.

As for only using a limited amount of the numbers simply take the list of number combination and run each combination through a process where you use 2 up to the final amount of numbers. You can premake lists of various amounts of operators to save processing time.

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