[math-fun] Integers shaped like right triangles
Hello Math-Fun, Write an integer somewhere on a grid, one digit per square: +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | 1 | 3 | 4 | 0 | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ The digits will now move vertically: - odd digit j will climb j squares - even digit k will sink k squares - zero stays were it is In our example, the digits of 1340 move like this: +---+---+---+---+ | | 3 | | | +---+---+---+---+ | | | | | +---+---+---+---+ | 1 | | | | +---+---+---+---+ | . | . | . | 0 | +---+---+---+---+ | | | | | +---+---+---+---+ | | | | | +---+---+---+---+ | | | | | +---+---+---+---+ | | | 4 | | +---+---+---+---+ Encircle now those new digits with an elastic band. The resulting shape is not a right triangle - which is a pity because we are precisely looking for integers moving into right triangles! The additional constraint being that ALL the new digits must be ON the elastic band. [you can see this more clearly there: http://www.cetteadressecomportecinquantesignes.com/RightTriangleNumbers.htm ] Here are a few good examples: Integer 130 (the right angle is on 1): +---+---+---+---+ | | 3 | | | +---+---+---+---+ | | | | | +---+---+---+---+ | 1 | | | | +---+---+---+---+ | . | . | 0 | | +---+---+---+---+ Integer 313333 (right angle on 1): +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | 3 | | 3 | 3 | 3 | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | 1 | | | | | +---+---+---+---+---+---+ | . | . | . | . | . | | +---+---+---+---+---+---+ Integer 2201022 (right angle on 1): +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ | | | | 1 | | | | +---+---+---+---+---+---+---+ | . | . | 0 | . | 0 | . | . | +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ | 2 | 2 | | | | 2 | 2 | +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ Here is a bad example: Integer 21000 (yes we have a triangle, yes there is a right angle --on 1--, but the band doesn't pass through ALL digits): +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ | | | 1 | | | | | +---+---+---+---+---+---+---+ | | . | . | 0 | 0 | 0 | | +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ | | 2 | | | | | | +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ Questions: a) Find the 40 first integers producing a right triangle b) Find the biggest integer producing a right triangle c) Find the smallest integer producing a rectangle. [Answers here in December 2010] Best, É.
What's the proper name for the following "4 dimensional donut"? Identifying opposite edges of a sheet of paper makes a torus, actually just the surface, either 0 or "1 cell" thick. Identifying 2 pairs of opposite sides of a 3-D rectangular solid (brick) makes a torus skin with some thickness, like the glaze of a donut. Now also identify the third pair of sides of the brick. This connects the inner surface of the glaze to the outer surface. That seems to require a fourth dimension, so I call this a "4 dimensional donut". But what is its correct name? I'm still working on domino-nets. The smallest "4-D donut" that might have solutions is 4x5x5, for donimos with 1 to 25 pips (no double dominos). This is probably too large for me to either find a solution or to prove there are none, unfortunately. Thanks, -- Mike Beeler
Just like a normal torus is a circle squared, and a cylinder is a circle multiplied by a line, I think of your torus as a circle raised to the 3rd power, with the 3-d surface being embedded in 6 dimensions (three orthogonal sets of two dimensions, each defining a plane that gives a circle as the cross-section). On Wed, Dec 1, 2010 at 13:16, Michael Beeler <mikebeeler@verizon.net> wrote:
What's the proper name for the following "4 dimensional donut"?
Identifying opposite edges of a sheet of paper makes a torus, actually just the surface, either 0 or "1 cell" thick. Identifying 2 pairs of opposite sides of a 3-D rectangular solid (brick) makes a torus skin with some thickness, like the glaze of a donut. Now also identify the third pair of sides of the brick. This connects the inner surface of the glaze to the outer surface. That seems to require a fourth dimension, so I call this a "4 dimensional donut". But what is its correct name?
I'm still working on domino-nets. The smallest "4-D donut" that might have solutions is 4x5x5, for donimos with 1 to 25 pips (no double dominos). This is probably too large for me to either find a solution or to prove there are none, unfortunately.
Thanks, -- Mike Beeler
-- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com
Six dimensions is overdoing it; as Mike pointed out, the object can be embedded in a four dimensional space. I am having a vague memory that this manifold (S1^3, as Robert points out) is surprisingly equivalent to some other 3-manifold, but I forget all details. Perhaps Bill Thurston knows this one? On Wed, Dec 1, 2010 at 1:33 PM, Robert Munafo <mrob27@gmail.com> wrote:
Just like a normal torus is a circle squared, and a cylinder is a circle multiplied by a line, I think of your torus as a circle raised to the 3rd power, with the 3-d surface being embedded in 6 dimensions (three orthogonal sets of two dimensions, each defining a plane that gives a circle as the cross-section).
On Wed, Dec 1, 2010 at 13:16, Michael Beeler <mikebeeler@verizon.net> wrote:
What's the proper name for the following "4 dimensional donut"?
Identifying opposite edges of a sheet of paper makes a torus, actually just the surface, either 0 or "1 cell" thick. Identifying 2 pairs of opposite sides of a 3-D rectangular solid (brick) makes a torus skin with some thickness, like the glaze of a donut. Now also identify the third pair of sides of the brick. This connects the inner surface of the glaze to the outer surface. That seems to require a fourth dimension, so I call this a "4 dimensional donut". But what is its correct name?
I'm still working on domino-nets. The smallest "4-D donut" that might have solutions is 4x5x5, for donimos with 1 to 25 pips (no double dominos). This is probably too large for me to either find a solution or to prove there are none, unfortunately.
Thanks, -- Mike Beeler
-- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
The standard name for an n-cube with opposite sides identified via translation = (S^1)^n is the n-torus, T^n. It's also equivalent to the quotient of the additive group of vectors in R^n by the integer lattice subgroup, T^n = R^n/Z^n. In general, the n-torus can be embedded in R^{n+1}. For the circle, this is immediate. Everything else can be done by induction: if the n-torus is smoothly embedded in R^{n+1}, then the boundary of an epsilon neighborhood in R^{n+2} gives a smooth embedding of T^{n+1} in R^{n+2}. Bill On Dec 1, 2010, at 1:54 PM, Allan Wechsler wrote:
Six dimensions is overdoing it; as Mike pointed out, the object can be embedded in a four dimensional space. I am having a vague memory that this manifold (S1^3, as Robert points out) is surprisingly equivalent to some other 3-manifold, but I forget all details. Perhaps Bill Thurston knows this one?
On Wed, Dec 1, 2010 at 1:33 PM, Robert Munafo <mrob27@gmail.com> wrote:
Just like a normal torus is a circle squared, and a cylinder is a circle multiplied by a line, I think of your torus as a circle raised to the 3rd power, with the 3-d surface being embedded in 6 dimensions (three orthogonal sets of two dimensions, each defining a plane that gives a circle as the cross-section).
On Wed, Dec 1, 2010 at 13:16, Michael Beeler <mikebeeler@verizon.net> wrote:
What's the proper name for the following "4 dimensional donut"?
Identifying opposite edges of a sheet of paper makes a torus, actually just the surface, either 0 or "1 cell" thick. Identifying 2 pairs of opposite sides of a 3-D rectangular solid (brick) makes a torus skin with some thickness, like the glaze of a donut. Now also identify the third pair of sides of the brick. This connects the inner surface of the glaze to the outer surface. That seems to require a fourth dimension, so I call this a "4 dimensional donut". But what is its correct name?
I'm still working on domino-nets. The smallest "4-D donut" that might have solutions is 4x5x5, for donimos with 1 to 25 pips (no double dominos). This is probably too large for me to either find a solution or to prove there are none, unfortunately.
Thanks, -- Mike Beeler
-- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Hey Bill, that makes a lot of sense. If I read you right, you're saying that just as we can expand a circle into a set of epsilon-sized circles which together make up a 2-torus (the surface of a donut whose "donut hole" is almost as big as the original circle) we can do the same thing in 4 dimensions to get a 3-torus surface, and so on. I still like to embed the 2-torus in 4 dimensions, because then nothing gets stretched. Probably just the way my mind works. Bill Thurston wrote:
The standard name for an n-cube with opposite sides identified via translation = (S^1)^n is the n-torus, T^n. It's also equivalent to the quotient of the additive group of vectors in R^n by the integer lattice subgroup, T^n = R^n/Z^n.
In general, the n-torus can be embedded in R^{n+1}. For the circle, this is immediate. Everything else can be done by induction: if the n-torus is smoothly embedded in R^{n+1}, then the boundary of an epsilon neighborhood in R^{n+2} gives a smooth embedding of T^{n+1} in R^{n+2}.
-- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com
participants (5)
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Allan Wechsler -
Bill Thurston -
Eric Angelini -
Michael Beeler -
Robert Munafo