Re: Locating text in Show
- To: mathgroup at smc.vnet.net
- Subject: [mg8326] Re: [mg8205] Locating text in Show
- From: hattons at CPKWEBSER5.ncr.disa.mil (Steven T. Hatton)
- Date: Sun, 24 Aug 1997 13:24:41 -0400
- Organization: Logicon supporting DISA
- Sender: owner-wri-mathgroup at wolfram.com
--------------68EC8A1FF4F750A615D4533D Sherman, Thank you for taking the time to respond. I have an earlier response from someone else on this same issue. I need to fish that out and respond to it as well. What I finally came up with worked just by combining the graphs together. I still don't understand what was going on. When I plotted just the text it came out wierd. When I combined it with the rest of the graphic it did just what you said. I got the text placed considently with the points. I did some manipulation to get what I wanted. This is what I ended up with: labels1=RotateRight[Thread[A^-Range[-7,0]],2]; labels2=RotateRight[Thread[-Range[-7,0]],2]; displace=1/5 Thread[{Cos[(\[Pi]/4) Range[0,7]],Sin[(\[Pi]/4) Range[0,7]]}]; << Geometry`Polytopes` dots=Vertices[Octagon]; group1=Graphics[ { {PointSize[.03], Point /@ dots},{Circle[{0,0},1]}, {Thread[ Text[ labels1,dots[[Thread[Range[8] ]]] + displace[[Thread[Range[8] ]]] ] ] } }, AspectRatio -> 1,DisplayFunction->Identity]; group2=Graphics[ { {PointSize[.03], Point /@ dots},{Circle[{0,0},1]}, {Thread[ Text[ labels2,dots[[Thread[Range[8] ]]] + displace[[Thread[Range[8] ]]] ] ] } }, AspectRatio -> 1, DisplayFunction->Identity]; Show[GraphicsArray[{group1,group2}]] I don't know what happens to attachements in news groups. I don't even know if this will get to the news group. I'm new at all of this. Nonetheless, I attached the .nb of what I was trying to do. One thing I wold like to have been able to do is thread accross three lists. I'm just learning so I'm not sure how to describe that any better. Thanks again, Steve Sherman.Reed wrote: > STEVE, SEE NOTE BELOW. > SURE, THE POINTS ARE THE LOCATION ON THE UNIT CIRCLE WHERE THE > VERTICES OF THE OCTAGON ARE LOCATED. THE POINTS BEGIN IN MID > FIRST QUAD AND ROTATE COUNTER CLOCKWISE AND ENDUP AT THE > BEGINNING OF THE FIRST QUAD. PLOT YOUR POINTS MANUALLY AND > YOU WILL SEE THEY DO JUST WHAT YOU WANT, ALMOST. > > THE VERTICES OF THE OCTAGON AND THE TEXT OCCUPY THE SAME SPACE > AND THE VERTICES OF THE OCTAGON OVERLAY THE TEXT FOR THE POINTS. > > SOLUTION IS TO EITHER SHRINK OR EXPAND THE CIRCLE THAT CONTAINS > THE TEXT DATA. EASY TO DO WITH TRIG. IF YOU HAVE A PROBLEM,SEND > ME A NOTE. > > Sherman C. Reed > sherman.reed at worldnet.att.net > --------------68EC8A1FF4F750A615D4533D (*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info at wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 31388, 760]*) (*NotebookOutlinePosition[ 32101, 785]*) (* CellTagsIndexPosition[ 32057, 781]*) (*WindowFrame->Normal*) Notebook[{ Cell[TextData[{ StyleBox["Theorem:", FontWeight->"Bold"], " If G is a cyclic group of order n, then G is isomorphic to ", StyleBox[Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\_n\)]]], ".\n\n", StyleBox["Proof:", FontWeight->"Bold"], " Let G be a cyclic group of order n generated by ", StyleBox["a", FontWeight->"Bold"], " and define a map \[Phi] : ", StyleBox[Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\_n\)]]], "\[Rule]G by \[Phi] : k \[Rule] ", StyleBox[Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["a", FontWeight->"Bold"], "k"], TraditionalForm]]]], " where 0 \[LessEqual] k < n. \nTo show that \[Phi] is onto we must show \ that \[Phi](k) = \[Phi](h) \[Implies] k = h. Assume to the contrary, that k \ \[NotEqual] h but ", StyleBox[Cell[BoxData[ \(TraditionalForm\`a\^k = \ \(a\^h.\)\)]]], " Without loss of generality we say that h > k. This gives ", StyleBox[Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(a\^k\), "TraditionalForm"], \(a\^\(-k\)\)}], TraditionalForm]]]], "= ", StyleBox[Cell[BoxData[ \(TraditionalForm\`a\^0\)]]], "= e = ", StyleBox[Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(a\^h\), "TraditionalForm"], \(a\^\(-k\)\)}], TraditionalForm]]]], "= ", StyleBox[Cell[BoxData[ \(TraditionalForm\`a\^\(h - k\)\)]]], " where 0 < h - k < n. Let m = h - k and find q and r such that\n\n\ti = \ mq + r \tfor 0 \[LessEqual] r < m\n\t\nby the division algorithm. Then \n\n\ \t", StyleBox[Cell[BoxData[ \(TraditionalForm\`a\^i\)]]], "= ", StyleBox[Cell[BoxData[ \(TraditionalForm\`a\^\(mq\ + \ r\)\)]]], " = ", StyleBox[Cell[BoxData[ \(TraditionalForm\`\(\((a\^m)\)\^q\) a\^r\)]]], "= ", StyleBox[Cell[BoxData[ \(TraditionalForm\`\(\(e\^q\) a\^r\ \)\)]]], "= ", StyleBox[Cell[BoxData[ \(TraditionalForm\`a\^r\)]]], "\n\t\nfor 0 \[LessEqual] r < m. This means that there are only m < n \ elements in the group contradicting the assertion that the group is of order \ n. We have shown that \[Phi] is 1\[Dash]1. \nTo show that the mapping is \ onto we need to show that \n\n\t\[ForAll] ", StyleBox[Cell[BoxData[ \(TraditionalForm\`a\^k\)]]], "\[Exists] k \[Element] ", StyleBox[Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\_n\)]]], "\[SuchThat] ", StyleBox[Cell[BoxData[ \(TraditionalForm\`\(\[Phi]\^\(-1\)\)(a\^k)\)]]], " = k. \n\nThis follows from that fact that the generation of the ", StyleBox[Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["a", FontWeight->"Bold"], \(k\ \)], TraditionalForm]]]], "produces exponents sequentially in unitary increments beginning with k=1. \ We thus see that there are n-1 ", StyleBox[Cell[BoxData[ \(TraditionalForm\`a\^k\)]]], "\[NotEqual] e with 1\[LessEqual]k<n, and one ", StyleBox[Cell[BoxData[ \(TraditionalForm\`a\^k\)]]], "= e, namely ", StyleBox[Cell[BoxData[ \(TraditionalForm\`\(a\^0.\)\)]]], " \nIt has been shown that the mapping is one-to-one and onto. In order \ to show \[Phi] is a homomorphisim we need to show that \[Phi](i ", StyleBox[Cell[BoxData[ \(TraditionalForm\`+\_n\)]]], " j) = ", StyleBox[Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["a", FontWeight->"Bold"], RowBox[{"(", RowBox[{"i", " ", FormBox[\(+\_n\), "TraditionalForm"], " ", "j"}], ")"}]], TraditionalForm]]]], "=", StyleBox[Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["a", FontWeight->"Bold"], \(i\ \)], TraditionalForm]]]], StyleBox[Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SuperscriptBox[ StyleBox["a", FontWeight->"Bold"], \(j\ \)], "=", " ", RowBox[{\(\[Phi](i\ )\), FormBox[\(\[Phi](\), "TraditionalForm"], " ", "j"}]}], ")"}], TraditionalForm]]]], ". Decomposing this equivalence: \[Phi](i ", StyleBox[Cell[BoxData[ \(TraditionalForm\`+\_n\)]]], " j) = ", StyleBox[Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["a", FontWeight->"Bold"], RowBox[{"(", RowBox[{"i", " ", FormBox[\(+\_n\), "TraditionalForm"], " ", "j"}], ")"}]], TraditionalForm]]]], "is by the definition of \[Phi]. ", StyleBox[Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["a", FontWeight->"Bold"], RowBox[{"(", RowBox[{"i", " ", FormBox[\(+\_n\), "TraditionalForm"], " ", "j"}], ")"}]], TraditionalForm]]]], "=", StyleBox[Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["a", FontWeight->"Bold"], \(i\ \)], TraditionalForm]]]], StyleBox[Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox[ StyleBox["a", FontWeight->"Bold"], \(j\ \)], FormBox["", "TraditionalForm"]}], TraditionalForm]]]], "follows from the laws of exponents discussed on page 64 of the text, and \ from the 1\[Dash]1 and onto mapping of property \[Phi]:", StyleBox[Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\_n\)]]], "\[Rule] G. ", StyleBox[Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["a", FontWeight->"Bold"], \(i\ \)], TraditionalForm]]]], StyleBox[Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SuperscriptBox[ StyleBox["a", FontWeight->"Bold"], \(j\ \)], "=", " ", RowBox[{\(\[Phi](i\ )\), FormBox[\(\[Phi](\), "TraditionalForm"], " ", "j"}]}], ")"}], TraditionalForm]]]], " follows from the definition of \[Phi].\n\nSee the graphical output \ below:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(labels1 = RotateRight[Thread[A^\(-Range[\(-7\), 0]\)], 2]; \n labels2 = RotateRight[Thread[\(-Range[\(-7\), 0]\)], 2]; \n\n displace = 1/5\ Thread[{Cos[\((\[Pi]/4)\)\ Range[0, 7]], Sin[\((\[Pi]/4)\)\ Range[0, 7]]}]; \n\n<< \ Geometry`Polytopes`\n \), \(dots = Vertices[Octagon]; \n\n group1 = Graphics[\n \t\t{\n\t\t\t{PointSize[.03], \ Point\ /@\ dots}, { Circle[{0, 0}, 1]}, \n \t\t\t\t{ Thread[\n\t\t\t\t\t Text[\n\t\t\t\t\t\tlabels1, dots[\([Thread[Range[8]\ ]]\)]\ + \t displace[\([Thread[Range[8]\ 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Hatton email;internet: hattons at cpkwebser5.ncr.disa.mil note: this is a test x-mozilla-cpt: ;0 x-mozilla-html: FALSE end: vcard --------------68EC8A1FF4F750A615D4533D--