Skip to main content

FindInstance (and Solve, ...) abysmally slow on a fully determined system of linear equations and inequalities: why?


I took a quick look for FindInstance and Solve-related questions but came up empty for the aspect that I am curious about. So here is my newbie question:


FindInstance[{

  s==9,r==8,d==7,n==6,e==5,y==2,m==1,o==0
},{s,e,n,d,m,o,r,y},Integers]

runs as quickly as one would expect. But


FindInstance[{
  s!=e!=n!=d!=m!=o!=r!=y,
  s==9,r==8,d==7,n==6,e==5,y==2,m==1,o==0
},{s,e,n,d,m,o,r,y},Integers]

takes ages -- long enough that I ran out of patience before MMA was done.



And now I am curious: Why would Mathematica suddenly display less maths ability than a third-grader?



bill_s suggested something much faster, but it is not equivalent to the original problem. The original inequality required all variable values to be different from all other variable values, such that each value can only be used for one variable. The FindInstance statement in Bill's post only compares x[i] to the ith value, but not to the other n-1 values.


In other news,


FindInstance[{
  (*s!=e,s!=n,s!=d,s!=m,s!=o,s!=r,s!=y,*)
  (*e!=n,e!=d,e!=m,e!=o,e!=r,e!=y,*)
  n!=d,n!=m,n!=o,n!=r,n!=y,
  d!=m,d!=o,d!=r,d!=y,
  m!=o,m!=r,m!=y,

  o!=r,o!=y,
  r!=y,
  s==9,r==8,d==7,n==6,e==5,y==2,m==1,o==0
  },{s,e,n,d,m,o,r,y},Integers]

~1.1 seconds (on my laptop).


FindInstance[{
  (*s!=e,s!=n,s!=d,s!=m,s!=o,s!=r,s!=y,*)
  e!=n,e!=d,e!=m,e!=o,e!=r,e!=y,
  n!=d,n!=m,n!=o,n!=r,n!=y,

  d!=m,d!=o,d!=r,d!=y,
  m!=o,m!=r,m!=y,
  o!=r,o!=y,
  r!=y,
  s==9,r==8,d==7,n==6,e==5,y==2,m==1,o==0
  },{s,e,n,d,m,o,r,y},Integers]

2nd comment uncommentd. ~20 s.


FindInstance[{
  s!=e,s!=n,s!=d,s!=m,s!=o,s!=r,s!=y,

  e!=n,e!=d,e!=m,e!=o,e!=r,e!=y,
  n!=d,n!=m,n!=o,n!=r,n!=y,
  d!=m,d!=o,d!=r,d!=y,
  m!=o,m!=r,m!=y,
  o!=r,o!=y,
  r!=y,
  s==9,r==8,d==7,n==6,e==5,y==2,m==1,o==0
  },{s,e,n,d,m,o,r,y},Integers]

Both comments uncommented. ~1 minute.



The original version (multiple inequalities in one condition):


Timing[FindInstance[{
    s!=e!=n!=d!=m!=o!=r!=y,
    s==9,r==8,d==7,n==6,e==5,y==2,m==1,o==0},
    {s,e,n,d,m,o,r,y},Integers]]

150 seconds!



Answer



No one has upgraded this post in a long time and as Danny promised all the problems the OP posted have been dealt with. In v 11.1 for instance, the last one now runs in a lot less than 150 seconds.


Timing[FindInstance[{

    s!=e!=n!=d!=m!=o!=r!=y,
    s==9,r==8,d==7,n==6,e==5,y==2,m==1,o==0},
    {s,e,n,d,m,o,r,y},Integers]]


{0.01, {{s -> 9, e -> 5, n -> 6, d -> 7, m -> 1, o -> 0, r -> 8, y -> 2}}}

Comments

Post a Comment

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
Post a Comment
Read more

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....
Post a Comment
Read more