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

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...