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Thread: 187. Space objects evenly by adding “1”

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    Supporting Member Make Things's Avatar
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    187. Space objects evenly by adding “1”


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    Supporting Member mklotz's Avatar
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    The " + 1 " appears automatically if you think about how many spaces you need to create.

    If you arrange N screws in a row you automatically create N - 1 spaces. Two screws create one space between them, three screws create two spaces, four create three, and so on. However you need to add the two spaces at the ends, one between the leftmost screw and the leftmost edge of the board, one between the rightmost screw and the rightmost edge of the board. Thus, you need

    N - 1 + 2 = N + 1

    spaces and voilà, there's your " + 1 ", and dividing N + 1 into the length of the board will give you the size of the required space(s).

    Note that this formulation works for the case of only one screw, yielding a space equal to half the width of the board, i.e. put the screw in the middle of the board.

    It also works for zero screws, yielding a spacing equal to the board width.

    While the latter two cases aren't of practical interest, they're mathematically satisfying showing the formulation would work for all possible values of N.

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    Quote Originally Posted by mklotz View Post
    The " + 1 " appears automatically if you think about how many spaces you need to create.

    If you arrange N screws in a row you automatically create N - 1 spaces. Two screws create one space between them, three screws create two spaces, four create three, and so on. However you need to add the two spaces at the ends, one between the leftmost screw and the leftmost edge of the board, one between the rightmost screw and the rightmost edge of the board. Thus, you need

    N - 1 + 2 = N + 1

    spaces and voilà, there's your " + 1 ", and dividing N + 1 into the length of the board will give you the size of the required space(s).

    Note that this formulation works for the case of only one screw, yielding a space equal to half the width of the board, i.e. put the screw in the middle of the board.

    It also works for zero screws, yielding a spacing equal to the board width.

    While the latter two cases aren't of practical interest, they're mathematically satisfying showing the formulation would work for all possible values of N.
    It's called a "fence post problem" in some circles. I've got parts of a book case here that I made many years ago before I learned this. Plan was to make folding book cases for standard paper-back books, of which I had thousands. 6' tall by 2' wide, made from 1x5 lumber, IIRC. Wound up with 6 perfect shelves, and one very short shelf... They were to be backed with 1/4" plywood, and hinged so they closed to form a box around the books. I was gonna do a whole bunch of them! That idea crashed and burned on my very limited math skills. (Hah!) Haven't gotten around to it since them, either. That was in preparation for my 36th move, I think. I'm around 70 or 80 moves, now. And the book herd has been thinned a few times, by floods, children, and natural wear and tear on cheap paperbacks. Though I've been in this place for 27 years middle of next month. Before that, longest I'd ever lived in one place was 5 years, and that was 3 different residences.

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    Supporting Member mklotz's Avatar
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    A large part of mathematical expertise is being able to see the problem at hand from different perspectives. Gauss, one of the three greatest mathematicians of the modern era, provided an excellent example of that skill.

    The story is a tale from when Gauss was still at primary school. One day Gauss' teacher asked his class to add all the numbers from 1 to 100, assuming that this task would occupy them for quite a while. He was shocked when young Gauss, after a few seconds thought, wrote down the answer 5050. The teacher couldn't understand how his pupil had calculated the sum so quickly in his head, but the eight year old Gauss pointed out that the problem was actually quite simple.

    He had added the numbers in pairs - the first and the last, the second and the second to last and so on, observing that 1+100=101, 2+99=101, 3+98=101, ...so the total would be 50 lots of 101, which is 5050.

    Later, Gauss generalized this approach with the formula we still use today for the sum of the numbers from 1 to N, N*(N + 1) / 2.
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    Supporting Member mklotz's Avatar
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    Quote Originally Posted by mklotz View Post
    A large part of mathematical expertise is being able to see the problem at hand from different perspectives. Gauss, one of the three greatest mathematicians of the modern era, provided an excellent example of that skill.

    The story is a tale from when Gauss was still at primary school. One day Gauss' teacher asked his class to add all the numbers from 1 to 100, assuming that this task would occupy them for quite a while. He was shocked when young Gauss, after a few seconds thought, wrote down the answer 5050. The teacher couldn't understand how his pupil had calculated the sum so quickly in his head, but the eight year old Gauss pointed out that the problem was actually quite simple.

    He had added the numbers in pairs - the first and the last, the second and the second to last and so on, observing that 1+100=101, 2+99=101, 3+98=101, ...so the total would be 50 lots of 101, which is 5050.

    Later, Gauss generalized this approach with the formula we still use today for the sum of the numbers from 1 to N, N*(N + 1) / 2.
    Another, related, example of Gauss' lateral thinking involves a formula for the sum of the numbers between n1 and n2 (where n1 > 1 and n2 > n1).

    Now you can derive this formula using the above expression for the sum from 1 to n, although on your first try you'll probably make a very common mistake.

    Gauss noted that the formula for a common average...

    A = S / N

    where:

    A = average
    S = sum of items
    N = number of items

    could be solved for the sum

    S = A * N

    and A and N are easily computed...

    A = (n1 + n2) / 2
    N = (n2 - n1 +1)

    so:

    S = (n1 + n2) * (n2 - n1 +1) / 2

    which is identical to what you would get using the expression for the sum from 1 to n
    ---
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    Supporting Member imohtep56's Avatar
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    Quote Originally Posted by WmRMeyers View Post
    It's called a "fence post problem" in some circles. I've got parts of a book case here that I made many years ago before I learned this. Plan was to make folding book cases for standard paper-back books, of which I had thousands. 6' tall by 2' wide, made from 1x5 lumber, IIRC. Wound up with 6 perfect shelves, and one very short shelf... They were to be backed with 1/4" plywood, and hinged so they closed to form a box around the books. I was gonna do a whole bunch of them! That idea crashed and burned on my very limited math skills. (Hah!) Haven't gotten around to it since them, either. That was in preparation for my 36th move, I think. I'm around 70 or 80 moves, now. And the book herd has been thinned a few times, by floods, children, and natural wear and tear on cheap paperbacks. Though I've been in this place for 27 years middle of next month. Before that, longest I'd ever lived in one place was 5 years, and that was 3 different residences.
    Just take that very small shelf and add a stained glass panel to the front and backlight it...

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    When I want to evenly space screws or nails in a constructed thing I mark for the first screw/nail at each end, then divide the distance between those marks instead of the entire span.

    Neil

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    Quote Originally Posted by sossol View Post
    When I want to evenly space screws or nails in a constructed thing I mark for the first screw/nail at each end, then divide the distance between those marks instead of the entire span.

    Neil
    And, if you're going to put N screws between the first two, and the distance between the first two is A, you calculate the spacing, S, as...

    S = A / (N + 1)

    because the N screws are going to create N + 1 spaces between the first two screws.
    ---
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    Experience is always far worse than pessimism

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    Quote Originally Posted by mklotz View Post
    And, if you're going to put N screws between the first two, and the distance between the first two is A, you calculate the spacing, S, as...

    S = A / (N + 1)

    because the N screws are going to create N + 1 spaces between the first two screws.
    I’m going to assume that formula works. I can read them, but they get jumbled on their way to the processing center of my brain.

    Neil

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    Quote Originally Posted by sossol View Post
    I’m going to assume that formula works. I can read them, but they get jumbled on their way to the processing center of my brain.

    Neil
    My point was that your technique (as in post # 7) leads to the same formula that was discussed in the first post. Your initial two screws merely redefine the effective width into which the screws must be uniformly spaced.

    As an example, let's say you start out with seven screws and a 10" wide board, then place two screws two inches from the edges of the board. Now you're faced with the problem of uniformly spacing five screws into a space that is 6" wide. That's no different than the original problem of spacing five screws into a 6" wide board and you'll still need the same formula to compute the proper screw spacing to do that.

    There's a good reason algebra is the first advanced math subject taught after arithmetic. Algebra and the notation it uses is the language of all higher mathematics; if you don't understand it, you're not going to understand what follows. Get yourself an elementary algebra text and start reading it. Even if you don't finish the book, you'll have learned a bit about the notation and the rules that govern it.

    Remember too that there are lots of people here, myself included, who can help you when you have questions or need clarification.
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    Experience is always far worse than pessimism

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