Skip to main content

programming - How to improve this code for solving the "Mr.S and Mr.P" puzzle?




Mr.S and Mr. P puzzle — "Formalization of two Puzzles Involving Knowledge", McCarthy, John (1987)


We pick two numbers $a$ and $b$, such that $a\geq b$ and both numbers are within the range $(2,99)$. We give Mr.P the product $a b$ and give Mr.S the sum $a+b$. Then following dialog takes place:


Mr.P: I don't know the numbers
Mr.S: I knew you didn't know. I don'tknow either.
Mr.P: Now I know the numbers
Mr.S: Now I know them too


Can we find the numbers $a$ and $b$?



I tried to do this, but it is very slow. I am sure there must be a simpler way.


Clear[pool, f1, f2, f3];

pool = Join @@ Table[{i, j}, {i, 2, 99}, {j, 2, i}];

f1[x_] := Length@Select[pool, Times @@ # == x &] != 1
f2[x_] := Length@Select[pool, Plus @@ # == x &] != 1
f3[x_] := And @@ (f1 /@ (Times @@@ Select[pool, Plus @@ # == x &]))
f4[x_] := Length@Select[Select[pool, Times @@ # == x &], f3[#[[1]] + #[[2]]] &] == 1
f5[x_] := Length@Select[Select[pool, Plus @@ # == x &], f4[#[[1]]*#[[2]]] &] == 1

Select[pool, 
  f1[#[[1]]*#[[2]]] && f2[#[[1]] + #[[2]]] && f3[#[[1]] + #[[2]]] && 

    f4[#[[1]]*#[[2]]] && f5[#[[1]] + #[[2]]] &] // Timing

Answer



I tried to understand the other two solutions, but honestly, I couldn't. So I tried to write a version that is easier to understand.


EDIT: I've refactored the code a little, primarily pulling out the "knowledge operators" personKnowsSolution and personKnowsProperty. I'm not really following McCarthy's axiomatization of knowledge, this is just my ad-hoc way of expressing "knowledge" in Mathematica.


Clear[personKnowsSolution, personKnowsProperty, mrP, mrS]
personKnowsSolution[informationFilter_, possibilities_] := 
 Join @@ Select[GatherBy[possibilities, informationFilter], Length[#] == 1 &]
personKnowsProperty[informationFilter_, possibilities_, property_] := 
 Select[possibilities, property[informationFilter[#]] &]
mrP[{a_, b_}] := a*b

mrS[{a_, b_}] := a + b
(
  allPossibilities = Join @@ Table[{i, j}, {i, 2, 99}, {j, 2, i}];

  (* Mr. P doesn't know the solution *)
  mrPWouldKnowSolution = personKnowsSolution[mrP, allPossibilities];
  mrPDoesntKnowSolution = Complement[allPossibilities, mrPWouldKnowSolution];

  (* Mr. S doesn't know the solution *) 
  mrSWouldKnowSolution = personKnowsSolution[mrS, allPossibilities];

  mrSDoesntKnowSolution = Complement[allPossibilities, mrSWouldKnowSolution];

  (* Mr. S knows Mr. P doesn't know the solution *)
  sumsWhereMrPWouldKnowTheSolution = Union[mrS /@ mrPWouldKnowSolution];
  mrSKnowsMrPDoesntKnow = 
   personKnowsProperty[mrS, mrSDoesntKnowSolution, 
    Not[MemberQ[sumsWhereMrPWouldKnowTheSolution, #]] &];

  (* Given that, Mr. P knows the solution *)
  mrPKnowsTheSolution = personKnowsSolution[mrP, mrSKnowsMrPDoesntKnow];


  (* Given that, Mr. S knows the solution *)
  mrSKnowsTheSolution = personKnowsSolution[mrS, mrPKnowsTheSolution]
) // Timing

Output: {0.063, {{13, 4}}}


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