1. Standard memberXanthosNZ
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    08 Apr '06 00:41
    Originally posted by CrazyLilTing
    with all due respect:
    can you probe that they are true and why?
    I don't want to waste the rest of my life checking them using "general examples".
    Sincerely, I will appreciate it. May be induction is the key word.
    Honestly, I don't know.
    I must to confess that my skill at math problems is way too poor.

    Regards

    - Julia

    ps: I'm sure that common sense isn't the answer
    Common sense is in fact the answer. Say you have your square a,b,c,d you know set the values to the zeros or same numbers required. Then you find the next square and see what it contains, as you are working in general cases you have a proof.

    So say we are looking at two adjacent zeros. We have 0,b,c,0 (from top left clockwise, so the zeros are the two left numbers). Finding the next square we find 0,b,abs(b-c),c . So if b and c are different we have a new square with one zero (which is a different case) and if they are the same we have a new square with non-adjacent zeros (as abs(b-c) would now be zero also).

    That's all there is to it.
  2. Standard memberpsychopath42
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    08 Apr '06 04:504 edits
    I think Xanthos has a legit proof only for the case of integers, but not for the general case of real numbers.


    Summarizing the main points of Xanthos's solution:

    0) WOLOG we may assume that all the numbers are non-negative - the first iteration will yield 4 non-negative numbers because of the absolute value property.

    1) Let the maximum value of any set of 4 non-negative numbers be 'a'. Then, on the next iteration, the maximum value will be at most 'a'. There will also exist a future iteration where all numbers are nonzero.

    2) Furthermore, if none of the 4 numbers are zero (i.e. all numbers are positive), then, on the next iteration, the maximum value will be strictly less than 'a'.

    (For a proof, see Xanthos' post above 😀)


    Now, under the condition of restricting the original numbers to integers, the above statements certainly show that we will reach the (0,0,0,0) quartuplet in a finite number of iterations.

    However, under the condition of generalizing to real numbers, the above statements don't prove that (0,0,0,0) is ultimately achieved. It is possible to have a decreasing sequence of positive reals that do not approach 0 - e.g. take the sequence (1 + 1/k) for k=1,2,...

    On a side note, convergence (of the maximum element in the set) is guaranteed because we have a bounded decreasing sequence.

    I have nothing else to contribute at this point.. 🙄


    edits: minor editing for clarity.
  3. Joined
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    08 Apr '06 10:40
    Originally posted by psychopath42
    I think Xanthos has a legit proof only for the case of integers, but not for the general case of real numbers.
    Though if the proof holds for integers it also holds for rationals. If the four starting numbers are p1/q1, p2/q2, p3/q3 and p4/q4 then we can multiply top and bottom by q1.q2.q3.q4 giving:

    p1.q2.q3.q4/q1.q2.q3.q4, p2.q1.q3.q4/q1.q2.q3.q4 etc

    And Xanthos' proof shows that the numerators will necessarilly be reduced to zero.

    We obviously can say the same thing for the reals - pity they're such a large set of numbers.
  4. Standard memberpsychopath42
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    08 Apr '06 19:01
    Originally posted by howardbradley
    We obviously can say the same thing for the reals - pity they're such a large set of numbers.
    rationals i can agree with.

    but what about irrationals? transcendentals? what would the common denominator be?
  5. Standard memberXanthosNZ
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    09 Apr '06 00:21
    Originally posted by psychopath42
    rationals i can agree with.

    but what about irrationals? transcendentals? what would the common denominator be?
    As long as we are not dealing with infintesimals (1/infinity and the like) then we have no problem. We don't need a common factor (or multiple) for my proof to work. Why would we? My proof needs nothing but the few simple rules that I laid out to be true. Which ones aren't true in the irrational case?
  6. Standard memberpsychopath42
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    09 Apr '06 18:441 edit
    Originally posted by XanthosNZ
    As long as we are not dealing with infintesimals (1/infinity and the like) then we have no problem. We don't need a common factor (or multiple) for my proof to work. Why would we? My proof needs nothing but the few simple rules that I laid out to be true. Which ones aren't true in the irrational case?
    I think you nailed it on the dot. If we are not dealing with infinitesimals, then your proof works. There will be a finite number of values between '0' and any maximum 'a' at any point during the iterative process.

    But, how are we guaranteed that we are not dealing with infinitesimals? I don't think the answer to this question is as simple as it appears.

    Let me give an example. Let our initial set of 4 values be the numbers 'e','x','y','z' where x, y, and z are all positive integers, and e is the base of the natural logarithm. Then, on every iteration, all values for each iteration have the form be+c, where b and c are integers. Your proof states that as long as we don't deal with infinitesimals, the maximum value for each iteration will be non-increasing, and will eventually become 0.

    However, there is a decreasing subsequence of {be+c} that approaches 0 but never reaches 0. (This is due to e being irrational.) In fact, the only way {be+c} becomes 0 is if b=c=0. No part of your proof indicates that this must be the case.

    Apologies for playing the devil's advocate here! It is an intriguing problem, and if the above knick can be ironed out, then I believe your solution solves the problem completely.

    (Perhaps I am missing something, in which case, I will withdraw my statements!)
  7. Standard memberXanthosNZ
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    09 Apr '06 20:23
    Originally posted by psychopath42
    I think you nailed it on the dot. If we are not dealing with infinitesimals, then your proof works. There will be a finite number of values between '0' and any maximum 'a' at any point during the iterative process.

    But, how are we guaranteed that we are not dealing with infinitesimals? I don't think the answer to this question is as simple as it a ...[text shortened]...
    (Perhaps I am missing something, in which case, I will withdraw my statements!)
    Actually if you take an example you can see what happens. Say we take e,1,3,7.

    The sequence goes:
    (e-1), 6, 4, (3-e)
    2, (7-e), 2, (1+e)
    (5-e), (5-e), (e-1), (e-1)
    And we have reached a known result (this will go to 0, x, 0, x and then to x, x, x, x and then 0, 0, 0, 0).

    We can see that I'm not actually dealing with the numbers becoming zero but with the differences between the numbers becoming zero. This is possible with non-zero portions of b and c (in your (be+c) notation).
  8. Argentina
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    09 Apr '06 21:141 edit
    [...]
    5. CONCLUSIONS, APPLICATIONS, AND EXTENSIONS. We now return to our original question: Does every diffy box converge to the zero box in a finite number of generations, and, if so, how many generations will it take? We can now reinterpret the results of our analysis on equivalence classes in terms of boxes as four-tuples.
    The answer to our question as phrased is no, for the diffy boxes belonging to the class containing the box [0 1 q(q − 1) q] associated with the fixed point of the map f require an infinite number of iterations to reach zero: any diffy box in this class has entries (vertex numbers) that become smaller and smaller but never actually reach zero. However, the counterexample class is unique, and all other diffy boxes converge to zero in finitely many generations (the number depends on how close their canonical forms are to the fixed point). In fact, for any specified longevity, there are classes of diffy boxes that take precisely that long to converge to zero. Nonmonotone boxes converge especially quickly (in no more than six generations), while nonotone boxes have longevity at least five. To determine the longevity of a monotone diffy box, it is simplest to put the box in canonical form and compare its coordinates (x, y) with a graph of the regions of various longevities identified in the previous section.We might also make the observation that (as seen by the regions into which S is subdivided in Figure 3) the use of “complicated” numbers such as radicals or transcendentals does not really prolong convergence much, since within a couple of generations the differences have propagated through the four vertices and get subtracted out.
    [...]

    This, in my humble opinion. should close the discusion. But, I have understand a 5% of what they said in their article. I think bear was referring to this.

    I'll post the URL of the complete article in my nest post.
    If someone can explain it to me I'll be very grateful.

    - J

    Ps: this is the url to the article:
    http://www.trinity.edu/vadim/difboxfinal.pdf
  9. Standard memberpsychopath42
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    09 Apr '06 22:27
    Originally posted by CrazyLilTing

    This, in my humble opinion. should close the discusion. But, I have understand a 5% of what they said in their article. I think bear was referring to this.

    I'll post the URL of the complete article in my nest post.
    If someone can explain it to me I'll be very grateful.

    - J

    Ps: this is the url to the article:
    http://www.trinity.edu/vadim/difboxfinal.pdf
    Neat article! Thanks for putting us out of our misery 🙄
  10. Argentina
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    09 Apr '06 23:161 edit
    Originally posted by psychopath42
    Neat article! Thanks for putting us out of our misery 🙄
    You are welcome. Sir.

    - Julia

    ps: as far as I seek in google, this is the most serious article on the theme. As my basic math knowledge lets me to understand, I'm reading it line after line trying to make sense of its content.

    I'm hard. But I'm sure I'll understand it completely in a few days!

    My regards

    (To you too, Xanthos, and the little bear. He is sick and I wish he will be get better soon)

    As a non subs my rec to the person who has originated this thread!
    (sorry... I can't remember his handle just now! 😳 )
  11. Joined
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    09 Apr '06 23:232 edits
    Thanks to all for their contributions. I must say that when I started the thread I had expected someone to say something like "Oh yeah, we did those in school. They take at most [small number] iterations because [simple explanation] QED". I didn't think it would generate quite so much math. And I'm certainly surprised by the result.

    I'm going to see how much of that URL I can understand.

    Thanks again,

    Howard.
  12. Argentina
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    09 Apr '06 23:342 edits
    Originally posted by howardbradley
    Thanks to all for their contributions. I must say that when I started the thread I had expected someone to say something like "Oh yeah, we did those in school. They take at most [small number] iterations because [simple explanation] QED". I didn't think it would generate quite so much math. And I'm certain surprised by the result.

    I'm going to see how much of that URL I can understand.

    Thanks again,

    Howard.
    You are welcome, Howard.

    Please (this is out off topic) pray in your own way for the health of the bear.

    My regards

    - Julia

    ps: do you have more problems like this? It was fascinating to put a humble effort to solve this. I feel it was painful to find the solution on the net, but it was previsible.


    Edits: my painful command of english language 😳
    (plus my typos 🙂 )
  13. Standard memberXanthosNZ
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    10 Apr '06 03:12
    I can't understand what the link is saying. It seems to say that 0, 1, q(q-1), q does not converge to 0, 0, 0, 0 in a finite number of steps. There is also something about the fixed point f.

    So there they seem to say there is a counter example, I would be interested in finding such a counterexample.
  14. Joined
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    10 Apr '06 11:161 edit
    For anybody who's interested here are those two numbers with a large number of decimal places:

    (1+e(l(19+3*sqrt(33))/3)+e(l(19-3*sqrt(33))/3))/3
    1.839286755214161132551852564653286600424178746097592246778758639404203222081966425738435418

    (1+e(l(19+3*sqrt(33))/3)+e(l(19-3*sqrt(33))/3))/3*((1+e(l(19+3*sqrt(33))/3)+e(l(19-3*sqrt(33))/3))/3-1)
    1.543689012692076361570855971801747986525203297650983935240804037831168673927973866485157910
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