This question is similar to:    and    but need to do this programmatically with JTS.  This question is not for overlapping polygons or touching polygons but for adjacent polygons that is the difference when they don't touch but next to each other. the answer to this question  , worked for intersecting..but not for adjacent.  I am trying to  implement this using JTS/GeoTools.. i found the .dissolve(..) thinking that would do the trick but it is not working.  The original GeoJson( //loadAdjacentGeoJson() )     Updated Solution:   for (GeoJsonPolygon geoJsonPolygon : geoJsonPolygonList) {                 Coordinate[] coordinates = toCoordinateArray(geoJsonPolygon);                 SimpleFeatureBuilder simpleFeatureBuilder = new SimpleFeatureBuilder(simpleFeatureType);                 simpleFeatureBuilder.add(inflate(geometryFactory.createPolygon(coordinates)));  //add inflated polygons                  SimpleFeature simpleFeature = simpleFeatureBuilder.buildFeature(Integer.toString(index));                 simpleFeature.getUserData().put("zipcode", postal.getZipCode());                 index = index + 1;                 geometryList.add((Geometry)simpleFeature.getDefaultGeometry());             }          Object obj = geometryFactory.buildGeometry(geometryList);          GeometryCollection geometryCollection = (GeometryCollection) obj;          Geometry union = geometryCollection.union();  //perform union          LOGGER.info(new GeoJsonWriter().write(geometryCollection)); //shows inflated polygons with overlap          LOGGER.info(new GeoJsonWriter().write( deflate( union ))); //deflate back.   inflated polygons so union will work.:    inlate/deflate methods:  private Geometry deflate(Geometry geom) {         BufferParameters bufferParameters = new BufferParameters();         bufferParameters.setEndCapStyle(BufferParameters.CAP_ROUND);         bufferParameters.setJoinStyle(BufferParameters.JOIN_MITRE);         Geometry buffered = BufferOp.bufferOp(geom,-.0001, bufferParameters);         buffered.setUserData(geom.getUserData());         return buffered;     }      private Geometry inflate(Geometry geom) {         BufferParameters bufferParameters = new BufferParameters();         bufferParameters.setEndCapStyle(BufferParameters.CAP_ROUND);         bufferParameters.setJoinStyle(BufferParameters.JOIN_MITRE);         Geometry buffered = BufferOp.bufferOp(geom,.0001, bufferParameters);         buffered.setUserData(geom.getUserData());         return buffered;     }   results:
Is there a way to remove an overlap from multiple polygons to make one final polygon?..programmatically, I am using the GeoTools or/and OpenGIS/JTS.. any ideas to put me on the right path...  example initial:  final:
WPF TextBox and Scroll behavior I have a problem. I need to host grid with controls in ScrollViewer to prevent textbox from being either truncated or collapsed to zero-with at the UI. Also I want the with of textbox to be expanded when user change windows width. I'm setting Window's content to following code  &lt;DockPanel&gt;     &lt;TreeView DockPanel.Dock="Left" Width="150"/&gt;     &lt;ScrollViewer VerticalScrollBarVisibility="Auto" HorizontalScrollBarVisibility="Auto"&gt;         &lt;Grid&gt;             &lt;Grid.ColumnDefinitions&gt;                 &lt;ColumnDefinition Width="Auto"/&gt;                 &lt;ColumnDefinition /&gt;             &lt;/Grid.ColumnDefinitions&gt;             &lt;Grid.RowDefinitions&gt;                 &lt;RowDefinition Height="Auto"/&gt;             &lt;/Grid.RowDefinitions&gt;             &lt;TextBlock Text="Name"                         Margin="5"                         VerticalAlignment="Center"/&gt;             &lt;TextBox Grid.Column="1"                      Text="Some Name"                      Margin="5"                      VerticalAlignment="Center"                      MinWidth="200"/&gt;         &lt;/Grid&gt;     &lt;/ScrollViewer&gt; &lt;/DockPanel&gt;   All work fine, but when user types very long text in TextBox it is being expanded and horizontal scroll appears. Is there any easy way to limit TextBox maximum width and allow it to be expanded only when user changes window size. I came across a user with a notification that states:     This account is temporarily suspended network-wide. The suspension period ends on Dec 2 '18 at 2:35.   Screencap for posterity:    Unless this user is  (in which case the full year should be shown, or at least a tooltip), it seems like they should have been unbanned 4 months ago. I can join the server and sometimes the map would load for the server but i would just fall through the map and it would dissconect me and say timed out, any solutions? What is the yield keyword used for in C#? What is the name of an 80s TV movie where a man acquires endurance powers and quick self-healing ability after a crash with a vehicle containing radioactive material? Why is polynomial regression considered a special case of multiple linear regression? If polynomial regression models nonlinear relationships, how can it be considered a special case of multiple linear regression?  Wikipedia notes that "Although polynomial regression fits a nonlinear model to the data, as a statistical estimation problem it is linear, in the sense that the regression function $\mathbb{E}(y | x)$ is linear in the unknown parameters that are estimated from the data."  How is polynomial regression linear in the unknown parameters if the parameters are coefficients for terms with order $\ge$ 2? Was the singularity at Big Bang perfectly uniform and if so, why did the universe lose its uniformity? $\forall a,b,c \in \mathbb{N}\!:\ a\mid b \Rightarrow ac\mid bc. $ Divisibility Proof Could someone verify this proof?  Given the following definition of divisibility $(\forall{a,b \in \mathbb{N}})[a \mid b \equiv (\exists{c \in \mathbb{N}})(ac=b)]$.  Statement: $(\forall{a,b,c \in \mathbb{N}})[(a\mid b) \rightarrow (ac\mid bc)] $ .  Proof:  Let $a,b \in \mathbb{N}$ be arbitrary.  Assuming that $a\mid b$, we can conclude that $\exists{d \in \mathbb{N}}$ such that $ad = b$.  By Algebra we can deduce the following sequence of statements.  If $(ad)=b$, then $(ad)c=bc$.  Therefore $(ac)d=bc$.  Hence by the definition of divisibility we know that $ac\mid bc$. MacBook [Pro] - how many displays, what resolution/frequency? There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$? There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$, what I tried is if possible then define $g(z)=f(z)-{1\over 2^{1\over z}}$ Then $g({1\over n})=0$ and so $g(z)$ is entire and its $0$ set has limit point in it and so $f(z)={1\over 2^{1\over z}}$ which is not analytic at $0$? Please help!  Edit: OOps! the way I defined $g(z)$ that is not entire! could any one give me hint? Is it fair to buy a return ticket I know I will never use just because it's cheaper? Books about matroids Could you recommend any approachable books/papers/texts about matroids (maybe a chapter from somewhere)? The ideal reference would contain multiple examples, present some intuitions and keep formalism to a necessary minimum.  I would appreciate any hints or appropriate sources. What causes the 'Connection Refused' message? The identification of an electron as a particle and the positron as an antiparticle is a matter of convention. We see lots of electrons around us so they become the normal particle and the rare and unusual positrons become the .  My question is, when you have made the choice of the electron and positron as particle and anti-particle does this automatically identify every other particle (every other fermion?) as normal or anti?  For example the proton is a particle, or rather the quarks inside are. By considering the interactions of an electron with a quark inside a proton can we find something, e.g. a conserved quantity, that naturally identifies that quark as a particle rather than an antiparticle? Or do we also just have to extend our convention so say that a proton is a particle rather than an antiparticle? To complete the family I guess the same question would apply to the neutrinos. Does the Planck scale imply that spacetime is discrete? On a quantum scale the smallest unit is the , which is a discrete measure.  There several question that come to mind:   Does that mean that particles can only live in a discrete grid-like structure,  i.e. have to "magically" jump from one pocket to the next? But where are they in between? Does that even give rise to the old paradox that movement as such is impossible (e.g. )? Does the same hold true  for time (i.e. that it is discrete) - with all the ensuing paradoxes? Mathematically does it mean that you have to use difference equations instead of differential equations? (And sums instead of integrals?) From the point of view of the space metric do you have to use a discrete metric (e.g. the ) instead of good old Pythagoras?   Thank you for giving me some answers and/or references where I can turn to.  Update: I just saw this call for papers - it seems to be quite a topic after all: Is Reality Digital or Analog? FQXi Essay Contest, 2011. , , . One can find some pretty amazing papers over there. Which is stronger, a rope without knots or a rope with knots? Suppose I have two identical ropes, one of which I manually tie some knots in. If I use them to hang clothes, which one is more durable?      Personally I think that rope with knots will be more durable, but I can't come up with a satisfactory reason. How do I make Ubuntu not require a password to supend? After the therapy, eight children (43%) became able to crawl/move on their back.   Or should I use "on their backs"?   Singular because each child only has one back, or plural because we're dealing with eight backs? You "show" someone a picture. You "---" someone a song? In , we have a verb meaning &quot;to show&quot; corresponding to &quot;to see/to look&quot;, and we have a different verb corresponding to &quot;to hear/to listen&quot;:  inti tara stampa (you look at a picture.) ---- jiena nurilek stampa (I show you a picture) inti tisma' diska (you listen to a song.) ----- jiena nsemmagħlek diska (I 'show' you a song)  In English, do you &quot;show someone a song&quot; (sounds weird to me) or is there a verb that corresponds more directly to the Maltese insemmgħek? Evaluate   $$\lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n}.$$   $\underline{\bf{My\;\;Try}}::$ Let $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n} = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \left(\frac{(n+1)\cdot (n+2)\cdot (n+3)\cdots (n+n)}{(1)\cdot (2)\cdot (3)\cdots (n)}\right)$  $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \left\{\ln \left(\frac{n+1}{1}\right)+\ln \left(\frac{n+2}{2}\right)+\ln \left(\frac{n+3}{3}\right)+\cdots+\ln \left(\frac{n+n}{n}\right)\right\}$  $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{n+r}{r}\right) = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{1+\frac{r}{n}}{\frac{r}{n}}\right)$  Now Using Reinman Sum  $\displaystyle y = \frac{1}{2}\int_{0}^{1}\ln \left(\frac{x+1}{x}\right)dx = \frac{1}{2}\int_{0}^{1}\ln (x+1)dx-\frac{1}{2}\int_{0}^{1}\ln (x)dx =  \ln (2)$  My Question is , Is there is any method other then that like Striling Approximation OR Stolz–Cesàro theorem OR Ratio Test  If yes then please explain here  Thanks