Dominion Strategy Forum
Miscellaneous => Forum Games => Non-Mafia Game Threads => Topic started by: sudgy on November 29, 2013, 01:18:02 am
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Inspired by this (http://forum.dominionstrategy.com/index.php?topic=5727.msg319795#msg319795) post, here is a game. I write a sequence of numbers, and others have to try to guess what the sequence is. Guesses are made by adding more to the sequence (in spoilers). The winner will make the new sequence.
1, 0, 1, 0, 1, 2, 0, 1, 2, 4, 0, 1, 2, 4, 8, 16, 0, 1, 2, 4, 8, 16, 32, 64, 128, 0, 1, 2, 4,...
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8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 0, 1, ...
If that is correct, than the rule is
0,2^0,2^1,...,2^n-1, where n are Fibonacci numbers
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I agree that's the rule, but I believe your sequence should actually continue:
8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 0, 1, ...
Also I'm not really sure this is quite the same as Zendo. There's no information prodding, which is kind of a big part of Zendo
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Well, I didn't know how to play Zendo. I just got the idea. Maybe I should change the title.
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And, Tables got it right (sorry ghostofmars, you made a mistake!).
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Sure. Well, let's see what can be done that's interesting... hmm...
1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ...
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Ooh ooh I know this one it's the look and say sequence
...31131211131221, 13211311123113112211,...
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Yep, that's right. A really simple one once you know how it's constructed, but pretty tough to pick out if you don't.
You construct each term by reading the previous term from left to right, counting how many times a given number appears in a row. So e.g. 1211 is "one one, one two, two one(s)" and so the next term is 111221
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Ok let's go a bit more classic:
I have a rule
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...
I'm, uh, glad to hear it?
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Lol fine. Give me a minute and I'll post a damn sequence
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Lol fine. Give me a minute and I'll post a damn sequence
It's meant to be a numerical sequence...
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6,6,6 :P
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...6,6,6,6,6,...?
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...12, 30, 66, 126, 216. Sequence is x3-6x2+11x
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5, 13, 17, 25, 29, 37, 41, 53, 61, 65...
Too easy?
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5, 13, 17, 25, 29, 37, 41, 53, 61, 65...
Too easy?
Yep.
73, 77, 85, 89, 97, 101, 109, 113,... add 4, add 8, add 4, add 8...
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There is a number missing?
..., 41, 49, 53, ...
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0, 1, 2, 2, 3, 5, 7, 10, 15, 22, 32, 47,...
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Incorrect. 49 should not be included.
Not that easy!
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69, 101, 148, ...
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a_n = a_n-1 + a_n-3
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5, 13, 17, 25, 29, 37, 41, 53, 61, 65, 73, 85, 89, 97, 101, 109, 113, 125, 137, 145...
Too easy?
Added a few more terms
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69, 101, 148, ...
You're right, but Walrus' isn't done yet...
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5, 13, 17, 25, 29, 37, 41, 53, 61, 65, 73, 85, 89, 97, 101, 109, 113, 125, 137, 145...
Too easy?
Could you check the numbers again. I thought of two very similar rows
your: 5, 13, 17, 25, 29, 37, 41, 53, 61, 65, 73, 85, 89, 97, 101, 109, 113, 125, 137, 145, ...
1st: 5, 13, 17, 25, 29, 37, 49, 53, 65, 73, 77, 85, 97, 109, 113, 125, 133, 137, 149, ...
2nd: 5, 13, 17, 25, 37, 41, 53, 61, 65, 73, 85, 97, 101, 113, 121, 125, 137, 145, ...
Perhaps the idea helps someone else
1st: 2*prime number - 9 (starting from 7)
2nd: 2*prime number - 21 (starting from 13)
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Just double checked, I think that is correct.
ghostofmars, it doesn't have anything directly to do with prime numbers that I know of.
Here are a few clues if you want 'em:
5 is the lowest value in the sequence, you can't extrapolate it backwards any further
Arguably 65, 85, and 145 could be listed twice
Probably only the first three or four entries will be familiar to most
Just adding fours and eights (and now twelves) in convoluted ways won't do it...down the road there are larger differences between some successive numbers
More clues to follow if this stagnates.
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Are the differences always multiples of 4?
Edit: Hey asking things like this makes it more like Numerical Zendo!
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I believe so. At least through the first 100 terms. I wonder if I could prove it to be generally true.
Hint: This sequence is most directly inspired by geometry.
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Aha!
149, 157, 169, 173, 181
Explanation: The hypotenuse in Pythagorean Triples. Some numbers are arguably repeated because they can be formed from two different triangles, e.g. 162 + 632 = 332 + 562 = 652
Edit: I included a link but links are visible through spoiler tags. You can google Pythagorean Triples if you are curious though.
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If the above is correct (which I'm reasonably certain it is) then Ghostofmars can go next, given I kind of sniped his first answer.
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Nice! Yep that's it.
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Let's try this
1, 6, 9, 10, 19, 26, 31, 44, 65, 74, 91, ...
It's perhaps a bit harder, so asking questions is encouraged :)
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Hmmm... are you sure that 65 isn't a 55?
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Observation: The difference between successive terms is odd
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Hmmm... are you sure that 65 isn't a 55?
I am also wondering this.
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Hmmm... are you sure that 65 isn't a 55?
Yes, 65 is correct.
Observation: The difference between successive terms is odd
That is true for the whole series.
I've extended the series a bit:
1, 6, 9, 10, 19, 26, 31, 44, 65, 74, 91, 116, 139, 160, 189, 216, ...
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A few more numbers
1, 6, 9, 10, 19, 26, 31, 44, 65, 74, 91, 116, 139, 160, 189, 216, 241, 274, 305, 334, 371, 406, 449, 490, 539, 576, ...
and the first tip
- can you find periodic features?
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First differences: [5,3,1,9,7,5,13,21,9,17,25,23,21,29,27,25,33,31,29,37,35,43,41,49,37]
Second differences: [-2,-2,8,-2,-2,8,8,-12,8,8,-2,-2,8,-2,-2,8,-2,-2,8,-2,8,-2,8,-12]
Third differences: [0,10,-10,0,10,0,-20,20,0,-10,0,10,-10,0,10,-10,0,10,-10,10,-10,10,-20]
Nothing that lets me guess a term yet.
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1, 6, 9, 10, 19, 26, 31, 44, 65, 74, 91, 116, 139, 160, 189, 216, 241, 274, 305, 334, 371, 406, 449, 490, 539, 576, 621, 674, 725, 774, 821, 886, 949, 1000, ...
2nd tip: Plot the numbers or the differences.
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(http://i.imgur.com/zJPDvDQ.png)
I can't figure anything out from it. The one on the left is the original numbers and the one on the right is the differences.
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Judging by the (psuedo)derivative the function appears to be roughly quadratic. Perhaps see what the best approximation to a quadratic equation is and see if that gives any rough hints?
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Did it myself. Excel's best approximation is (roughly) 0.91x2 - 1.4x + 1.8. Comparing that to the data... I don't see anything of relevance.
Can I get a quick check: Are these numbers correct: ...774, 821, 886, ... ?
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Okay well I had a theory but it isn't panning out, so I'll just throw my observation/question out to the masses:
The second derivative is always of the form 10n - 2 for integer n - i.e. -12, -2, 8, 18 etc. Is this always the case?
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...1069, 1136, 1201, 1274?
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Judging by the (psuedo)derivative the function appears to be roughly quadratic. Perhaps see what the best approximation to a quadratic equation is and see if that gives any rough hints?
Yes, it's quadratic +?. And the +? is the difficult part.
The second derivative is always of the form 10n - 2 for integer n - i.e. -12, -2, 8, 18 etc. Is this always the case?
Yes.
Can I get a quick check: Are these numbers correct: ...774, 821, 886, ... ?
I generate the numbers automatically. I can share the sheet (https://docs.google.com/spreadsheet/ccc?key=0Aq-uBisGva7kdEM2QV94bjBTU0ZNcHM5R1NFU1VIN0E&usp=sharing). I hope, I set the privacy settings correctly ;)
...1069, 1136, 1201, 1274?
No, 1069, 1126, 1181, 1254
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Well, as I am prone to do, I cheated and found the answer in the google doc. The secret is to make a copy.
Anyway, it was pretty complicated.
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Well, as I am prone to do, I cheated and found the answer in the google doc. The secret is to make a copy.
Anyway, it was pretty complicated.
Actually, the generation in the spreadsheet, is more complicated than the generation by hand.
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It's very easy to cheat with google docs based information sharing. The general unwritten rule in PBF games is don't :).
Hmm...
I've added all the cells up to 100. You initially gave 11 numbers. Do you think that's enough to spot the 'complex' bit that's added completely?
Looks like the Quadratic without the extra bit, adding in those new terms, is most closely approximated by about 0.9x2 - 1.2x + 0.85. Is this correct?
Further looking, the period of the extra term seems to be 50. Also correct?
I feel like I've deduced (almost) everything I need now and just need to piece it together correctly to solve this now...
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Further observation: Whatever the extra term on top of that 50 period cycle, can also be simplified to a 25 period cycle that reverses at the end. The extra term from the 9th to the 34th is the same as the one from the 34th to 59th but in the opposite direction (so t(33)=t(35), t(32)=t(36) etc.)
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Further observation: Whatever the extra term on top of that 50 period cycle, can also be simplified to a 25 period cycle that reverses at the end. The extra term from the 9th to the 34th is the same as the one from the 34th to 59th but in the opposite direction (so t(33)=t(35), t(32)=t(36) etc.)
Compare this to the property of nx^2, where n is any integer number.
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It's very easy to cheat with google docs based information sharing. The general unwritten rule in PBF games is don't :).
Hmm...
I've added all the cells up to 100. You initially gave 11 numbers. Do you think that's enough to spot the 'complex' bit that's added completely?
Looks like the Quadratic without the extra bit, adding in those new terms, is most closely approximated by about 0.9x2 - 1.2x + 0.85. Is this correct?
Further looking, the period of the extra term seems to be 50. Also correct?
I feel like I've deduced (almost) everything I need now and just need to piece it together correctly to solve this now...
With the initial numbers, you can figure it out if you recognize the pattern. Perhaps some later elements are easier
n a_n
100 9061
1000 900601
This assumes starting that the first element is a_0.
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Hmm... that's not a property I'd ever heard of before, but I see it. I presume what you're referring to is looking at the last two digits of nx2? They appear to exhibit the required behavior - cyclic with period 50 and inverses after 25 terms. There's an offset (this reverses at 0 and 25, not 9 and 34) but that's probably a secondary concern.
Now can I derive the extra term from this data? Let's have a look...
PPE: Okay correcting for assuming term 1 is generated from x=0 gives an actually slightly nicer base formula of 0.9x2+0.6+0.75. No major changes elsewhere. Pondering that new information through now.
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Urgh. I tried using the above property, and it gives me graphs that look similar to the one generated by difference in terms from the original equation, but not quite the same. So I looked for patterns and noticed something important shared between them: The value every 10th term is the same along two sequences: 8-18-28... (=1.85) and 13-23-33... (=-0.65) for the differences and of course, 0-10-20... (=0) and
5-15-25... (=25) for the nx^2. So that's a clear similarity in the patterns.
But I can't make them match up. Trying any sort of linear mapping approach kind of works, but I get numbers out ranging over 152 possible values, including negatives - which if I'm taking the result
As I typed this up I noticed, while the 5-15-25... sequence always equals SOMETHING, it doesn't have to be 25. Let's see what I can do now...
Also if I'm barking up the wrong tree knowing that would be nice I suppose...
Edit: Nope, this still doesn't work. The only other results make things worse. Short version of the issue: I should be able to map into a range of 100 numbers. I can't. I can take abs, but then I also need to explain why and when negatives are used, which considering the nice 3rd derivatives, seems like a bad idea.
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There is no floating point number necessary. As usual, you shouldn't think to complicated. If you look at the last digit, you will notice that some numbers are missing. Why?
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I presume what you're referring to is looking at the last two digits of nx2? They appear to exhibit the required behavior - cyclic with period 50 and inverses after 25 terms. There's an offset (this reverses at 0 and 25, not 9 and 34) but that's probably a secondary concern.
Note that (50+x)^2=2,500+100x+x^2. Since only the x^2 term affects the last two digits (the other terms are multiples of 100), then (50+x)^2 will have the same final two digits as x^2; hence the cyclic behavior of period 50.
Similarly, (50-x)^2=2,500-100x+x^2. Again, only the x^2 term affects the last two digits, so (50-x)^2 will have the same final two digits as x^2; hence the reversal after 25 terms.
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I think, you have basically all you need to solve this:
a) the trend for large n is 0.9n^2 + 0.6n + 1
b) the row shows some periodic behavior similar to n^2
c) I gave you the hint that the last digit is important
One final tip:
Write down the n^2 series and compare to the one I gave you.
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Alright, necroing an age-old thread because I have an interesting series. I use it all the time to pass the time when there's lots of numbers around me.
The series only applies to natural numbers. I'll select certain parts to show you of the series:
1, 2, 3, ... (all numbers included up to) 94, 95, 96, 98, 100, 101, 102, ...
A random point later on: ... 28482, 28483, 28486, 28487, 28490, ...
The frequency of natural numbers in the series diminishes as the number gets larger, with the last number in the series being 13,876,543,210.
This isn't nearly enough information to go off of, so feel free to ask questions about other numbers.