Skip to main content

programming - Efficient code for the Ten True Sentences puzzle


I am trying to solve Ten True Sentences Puzzle.



Take a look at the following sentences:




  • The number of times the digit 0 appears in this puzzle is _.

  • The number of times the digit 1 appears in this puzzle is _.

  • The number of times the digit 2 appears in this puzzle is _.

  • The number of times the digit 3 appears in this puzzle is _.

  • The number of times the digit 4 appears in this puzzle is _.

  • The number of times the digit 5 appears in this puzzle is _.

  • The number of times the digit 6 appears in this puzzle is _.

  • The number of times the digit 7 appears in this puzzle is _.

  • The number of times the digit 8 appears in this puzzle is _.

  • The number of times the digit 9 appears in this puzzle is _.



Fill these sentences with digits such that all the sentences holds true.



I wrote the following code, but it didn't work.


Do[
  If[Tally[Range[0, 9] ~Join~ x][[All, 2]] == x, Print@x],
  {x, Tuples[Range[0, 9], 10]}] // AbsoluteTiming

How can I imporve my code?


Updated



Compile[{},
   NestWhile[Join[{1, 7, 3, 2}, RandomInteger[10, 6]] &, Range[0,9], 
      Tally[Range[0, 9] ~Join~ #][[All, 2]] != # &], 
   CompilationTarget -> "C", RuntimeOptions -> "Speed"
   ][] // AbsoluteTiming

Answer



Since the finished puzzle will contain exactly 20 numbers, we can assert that the solution for x will satisfy Total[x] == 20. We can therefore consider the IntegerPartitions of 20, which are of length 10, and limited to the numbers 1-9:


parts = IntegerPartitions[20, {10}, Range[1, 9]];

The solution for x must be a permutation of one of these, so the complete candidate set for x is:



can = Flatten[Permutations /@ parts, 1];

There are 92278 of these, sufficiently few to use the OP's code straight away. However, we can also note that the solution must satisfy (x-1).Range[0,9] == 20, so we can further reduce the candidate set:


can = can[[Flatten@Position[(can - 1).Range[0, 9], 20]]];

This leaves just 391 possibilities for x, making the OP's code very fast:


Do[If[Tally[Range[0, 9]~Join~x][[All, 2]] == x, Print@x], {x, can}]
(*  {1,7,3,2,1,1,1,2,1,1}  *)

Comments

Post a Comment

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...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

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