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Procedure sequence puzzle

  1. Jun 16, 2005 #1
    I had this idea this morning and was going to make another "figure-out-the-sequence" out of it, until it occurred to me that it was slightly more interesting than that.

    Consider the following procedure: start a line with 0 2, and add on additional numbers as you please so long as the following condition is met for each additional number: label the preceding two numbers from the one you are adding on a and b (in order) and the new number must be a - b, a + b, a / b, or a * b.

    Here are lines made by that procedure that end in the numbers 1-4 respectively:
    0 2 2 1
    (0 + 2 = 2, 2 / 2 = 1)
    0 2
    (no additional numbers added)
    0 2 2 1 3
    (0 + 2 = 3, 2 / 2 = 1, 2 + 1 = 3)
    0 2 2 4
    (0 + 2 = 2, 2 * 2 = 4)

    I ended it here because 5 is the first one that's a little bit harder.


    Puzzlers:
    1. Find lines ending in numbers as high as you can (say, shoot for 20; I just did 11).

    2. Given that the last number so far on a line is x, how can you append numbers so that the number -x is the last on the line?

    3. Given that the last two numbers in a line are arbitrary x y, can you append numbers until the line ends in y x? (note: if the last two numbers are x y, you can append x - y but not y - x)

    4. Given that the last number on a line is arbitrary x, can you append numbers until the line ends in x 1 or in 1 x?

    5. Can a line be constructed that ends in any positive integer whatsoever?


    I do not know the answers to 3 and 4 at present. If, in thinking about this, you come up with other puzzlers about this type of sequence, feel free to add them to the thread.
     
    Last edited: Jun 16, 2005
  2. jcsd
  3. Jun 17, 2005 #2
    Here's some more idea of what I mean, in the form of lines ending in numbers from 1 to 9. Complete these up to 30:

    0 2 2 1
    0 2
    0 2 2 1 3
    0 2 2 4
    0 2 2 1 3 -2 5 (slightly tricky)
    0 2 2 4 6
    0 2 2 1 3 4 7
    0 2 2 4 8
    0 2 2 1 3 3 9


    ---------------------
    To be clear, this is how you can make 5:
    0 2
    (start of line) Now the next number can be 2 (= 0 + 2), -2 (=0 - 2), 0 (=0 * 2), or 0 (= 0 / 2). In this case I chose 2.
    0 2 2
    Now the next number can be 4 (= 2 + 2), 0 (= 2 - 2), 4 (= 2 * 2), or 1 (= 2 / 2). In this case I chose 1.
    0 2 2 1
    Now the next number can be 3 (= 2 + 1), 1 (= 2 - 1), 2 (= 2 * 1), or 2 (= 2 / 1). In this case I chose 3.
    0 2 2 1 3
    Now the next number can be 4 (= 1 + 3), -2 (= 1 - 3), 3 (= 1 * 3), or 1/3 (= 1 / 3). In this case I chose -2.
    0 2 2 1 3 -2
    Now the next number can be 1 (= 3 + (-2)), 5 (= 3 - (-2)), -6 (= 3 * (-2)), or -1.5 (= 3 / (-2)). In this case I chose 5.
    0 2 2 1 3 -2 5
     
    Last edited: Jun 17, 2005
  4. Jun 17, 2005 #3
    Thanks for the clarification bicycletree. I've gotten pretty high numbers, obviously, there isn't a limit to the series. Those number with a lot of factors are particularly easy. I can't seem to get any of the higher primes(like 17, 31, 53, etc.). I can also get quite a large number of fractions. I'm still working on it though. :frown:
     
  5. Jun 19, 2005 #4
    Well, any integer can be constructed--in fact I figured out a general method that will do so, although the line it makes for a given number n is somewhat longer than n is large. Doing 17 (without that method) was a little tricky:
    0 2 2 1 3 4 7 11 -4 7 3 10 -7 17

    You can get 31 from 17:
    0 2 2 1 3 4 7 11 -4 7 3 10 -7 17 -24 -7 -31 22 -7 31

    And you can get 53 from 31:
    0 2 2 1 3 4 7 11 -4 7 3 10 -7 17 -24 -7 -31 22 -7 31 22 53
     
    Last edited: Jun 19, 2005
  6. Jun 19, 2005 #5
    I have 4 procedures for negating a number. Here's one of them, the second one I discovered:

    a b (a+b) (a+2b) -b

    What are the other 3?
     
  7. Jun 20, 2005 #6
    I did some more thinking about this last night and discovered that yes, it is possible to create a line ending in x 1 or 1 x from a line ending in x.

    So the only question of the original five that I don't know the answer to is whether you can create a line ending in y x from a line ending in x y. My feeling at the moment is no; in fact, I know that you can't do it in a straightforward manner using addition and subtraction only. Looking at each new appended number algebraically as (nx+my), you can't get y x explicitly. Puzzle: how do you prove that?

    Also, here is an additional question: can you create any rational number whatsoever? Or, more broadly, can you create a line ending in any two integers whatsoever?
     
  8. Jun 20, 2005 #7

    AKG

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    Getting 20:
    0 2
    0 2 2 +
    0 2 2 1 /
    0 2 2 1 3 +
    0 2 2 1 3 4 +
    0 2 2 1 3 4 12 x
    0 2 2 1 3 4 12 16 +
    0 2 2 1 3 4 12 16 -4 -
    0 2 2 1 3 4 12 16 -4 20 -
     
  9. Jun 20, 2005 #8

    (+, +, -) a b a+b a+2b -b
    (-, +, -) a b a-b a -b
    (*, +, -) a b ab b+ab -b
    (/, +, -) a b a/b b+a/b -b

    (*, /, *, +, +, -) a b ab 1/a b (1/a)+b (1/a)+2b -b
    (*, /, *, -, +, -) a b ab 1/a b (1/a)-b 1/a -b
    (*, /, *, *, +, -) a b ab 1/a b b/a b+b/a -b
    (*, /, *, /, +, -) a b ab 1/a b 1/(ab) b+1/(ab) -b

    (/, /, *, *, +, +, -) a b a/b bb/a b b+bb/a 2b+bb/a -b
    (/, /, *, *, -, +, -) a b a/b bb/a b b-b/a 2b-b/a -b
    (/, /, *, *, *, +, -) a b a/b bb/a b bbb/a b+bbb/a -b
    (/, /, *, *, /, +, -) a b a/b bb/a b b/a b+b/a -b


    (*, *, /, *, *, -, +, -, /) a b ab abb 1/b ab a ab-a ab -a -b

     
  10. Jun 20, 2005 #9
    Yep, Gerben--the first four of those are what I had in mind. The next eight are just variations on the first four (you just did some operations to get b again, and then applied one of the first four) but the last one is interesting.

    Here's my sequence for all the integers:
    + + - - - + + + - ...
    a b a+b a+2b -b a+3b -a-4b -b -a-5b -a-6b b ...
    When a = 0 and b = 1, which you can get from 0 2 2 1 1 0 1, this creates every integer or the negation of that integer. Although it does take a long time for larger integers. Anyone know a shorter way (besides brute force)?
     
  11. Jun 22, 2005 #10
    I have figured out the remaining puzzles. So here is the remaining list of questions, which I have the answer to:

    1. Given that the last two numbers in a line are arbitrary x y, can you append numbers until the line ends in y x? (note: if the last two numbers are x y, you can append x - y but not y - x) (yes)

    2. Given that the last number on a line is arbitrary x, can you append numbers until the line ends in x 1 or in 1 x? (yes)

    3. Can a line be constructed that ends in any positive integer whatsoever? (yes--I have answered this one though a quicker answer may exist, and it's necessary in order to solve the other questions)

    4. Can any rational number be constructed? (yes)

    5. Can a line be constructed that ends in any two integers whatsoever? (yes)

    6. Given that a line ends in x y, does there exist a single sequence of operations involving only addition and subtraction that can always be applied to make the line end in y x? (no)

    And here is a puzzle I have not solved:
    7. Given that a line ends in x y, does there exist a single sequence of operations that can always be applied to make the line end in y x?
     
  12. Jun 22, 2005 #11

    Alkatran

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    Trivial answer:

    0 2 2 1 2 2 1 2 2 1 2 2 1 2 2 1

    0+2
    2/2
    2/1
    1*2
    2/2
    2/1
    1*2
    ...repeat
     
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