1. Standard memberSwissGambit
    Caninus Interruptus
    2014.05.01
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    20 Aug '07 21:581 edit
    Start with four numbers:
    a b c d

    Fill in subsequent rows by taking the absolute value of the number above minus the one above and to the right:
    |a-b| |b-c| |c-d| |d-a| (d has nothing to the right, so we 'wrap around' and use a)

    The challenge is: Find numbers a, b, c, and d that yield 20 rows where at least one number on each row is not zero. (not counting the initial row)

    Example:

    If we start with 1, 2, 3, 4, we get:
    1 1 1 3
    0 0 2 2
    0 2 0 2
    2 2 2 2
    0 0 0 0

    This counts as 4 iterations.
  2. Standard memberagryson
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    20 Aug '07 22:12
    Originally posted by SwissGambit
    Start with four numbers:
    a b c d

    Fill in subsequent rows by taking the absolute value of the number above minus the one above and to the right:
    |a-b| |b-c| |c-d| |d-a| (d has nothing to the right, so we 'wrap around' and use a)

    The challenge is: [b]Find numbers a, b, c, and d that yield 20 rows where at least one number on each row is not ...[text shortened]... 2, 3, 4, we get:
    1 1 1 3
    0 0 2 2
    0 2 0 2
    2 2 2 2
    0 0 0 0

    This counts as 4 iterations.
    100, 3, 2, 1

    ?
    did I understqnd the question right?
  3. Standard memberSwissGambit
    Caninus Interruptus
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    20 Aug '07 22:271 edit
    Originally posted by agryson
    100, 3, 2, 1

    ?
    did I understqnd the question right?
    Using those numbers, i get

    100 3 2 1
    97 1 1 99
    96 0 98 2
    96 98 96 94
    2 2 2 2
    0 0 0 0

    ...which is only 4 iterations.

    Order of operations is to subtract first, and then take the absolute value of that result.
  4. Standard memberagryson
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    21 Aug '07 00:05
    Originally posted by SwissGambit
    Using those numbers, i get

    100 3 2 1
    97 1 1 99
    96 0 98 2
    96 98 96 94
    2 2 2 2
    0 0 0 0

    ...which is only 4 iterations.

    Order of operations is to subtract first, and then take the absolute value of that result.
    hmm... ok, I'll work on it tomorrow,
  5. Subscriberinvigorate
    Only 1 F in Uckfield
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    21 Aug '07 13:59
    Originally posted by SwissGambit
    [b]Start with four numbers:
    a b c d
    Spot the mistake?
  6. Standard memberSwissGambit
    Caninus Interruptus
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    21 Aug '07 19:13
    Originally posted by invigorate
    Spot the mistake?
    The mistake is tackling a problem like this without a basic knowledge of algebra.
  7. Joined
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    21 Aug '07 20:55
    Originally posted by invigorate
    Spot the mistake?
    I don't. What is the mistake?

    I like this problem! Really!
    Now I have something to think about during boring meetings.
  8. B is for bye bye
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    21 Aug '07 23:41
    The mistake is a b c and d are letters not numbers!!!

    HA HA HA!
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    22 Aug '07 04:16
    Originally posted by Gastel
    The mistake is a b c and d are letters not numbers!!!

    HA HA HA!
    Who says they are letters?
    He says from the beginning: "Start with four numbers: a b c d"
    If he says the symbols are numbers they are, believe me. They perhaps look like letters but they are in fact numbers.

    Can you solve the equation 2x=4 not realizing that the x stands for a number?
  10. Joined
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    22 Aug '07 09:14
    I think he was trying to be funny.

    Trying.
  11. Joined
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    22 Aug '07 09:37
    Originally posted by mtthw
    I think he was trying to be funny.

    Trying.
    Who? invigorate or Gastel or both? Exactly the same kind of humour? Strange...
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    22 Aug '07 11:11
    a=pi
    b=e
    c=sqrt(2)
    d=sqrt(3)
  13. Joined
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    22 Aug '07 15:13
    Originally posted by David113
    a=pi
    b=e
    c=sqrt(2)
    d=sqrt(3)
    Doesn't work! Hits zero in four iterations 🙂. Try it out.
  14. Joined
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    22 Aug '07 16:37
    I spent some time trying to come up with a pattern that worked. No luck so far.

    So went for the brute force approach 🙂. Best so far is [4910, 4163, 2789, 262], which gives the 20 iterations you wanted.

    I'm sure there's a better way.
  15. Standard memberSwissGambit
    Caninus Interruptus
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    22 Aug '07 16:521 edit
    Originally posted by David113
    a=pi
    b=e
    c=sqrt(2)
    d=sqrt(3)
    Oops...the stipulation should allow only whole numbers.
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