Einstein and puzzles

Replies

• 

  skipper

  For all you Prolog fans, here's a recursive-descent (with pullback); coded succinctly.
  
  

  next_to(X, Y, List) :- iright(X, Y, List).
  next_to(X, Y, List) :- iright(Y, X, List).
  
  iright(L, R, [L | [R | _]]).
  iright(L, R, [_ | Rest]) :- iright(L, R, Rest).
  
  einstein(Houses, Fish_Owner) :-
  =(Houses, [[house, norwegian, _, _, _, _], _, [house, _, _, _, milk, _], _, _]),
  member([house, brit, _, _, _, red], Houses),
  member([house, swede, dog, _, _, _], Houses),
  member([house, dane, _, _, tea, _], Houses),
  iright([house, _, _, _, _, green], [house, _, _, _, _, white], Houses),
  member([house, _, _, _, coffee, green], Houses),
  member([house, _, bird, pallmall, _, _], Houses),
  member([house, _, _, dunhill, _, yellow], Houses),
  next_to([house, _, _, dunhill, _, _], [house, _, horse, _, _, _], Houses),
  member([house, _, _, _, milk, _], Houses),
  next_to([house, _, _, marlboro, _, _], [house, _, cat, _, _, _], Houses),
  next_to([house, _, _, marlboro, _, _], [house, _, _, _, water, _], Houses),
  member([house, _, _, winfield, beer, _], Houses),
  member([house, german, _, rothmans, _, _], Houses),
  next_to([house, norwegian, _, _, _, _], [house, _, _, _, _, blue], Houses),
  member([house, Fish_Owner, fish, _, _, _], Houses).
  
  
  Here's a run through the Prolog REPL. I pretty-printed the output by hand so as to not run off the right side of the screen.
  
  1 ?- consult('einstein').
  % /tmp/einstein compiled 0.00 sec, 4,740 bytes
  
  Yes
  2 ?- einstein(Houses, Fish_Owner).
  
  Houses = [[house, norwegian, cat, dunhill, water, yellow],
  [house, dane, horse, marlboro, tea, blue],
  [house, brit, bird, pallmall, milk, red],
  [house, german, fish, rothmans, coffee|...],
  [house, swede, dog, winfield|...]]
  
  Fish_Owner = german
  
  Yes
  3 ?-
  The German owns the fish!
  --https://sandbox.rulemaker.net/ngps/119

  For puzzle fans, the way to construct a puzzle which is mechanical and is a faithful representation, is to build a colored icosahedron = 4x5 sides and use 4 colors. Or just wait for Erno Rubik to invent a new mechanical puzzle. I spotted it in a K-mart and recognized what it was straight-off (it must have looked a bit nutso, me getting all excited about a plastic toy). Anyhoo, the "Magic Number" sphere, which is like the Tetraminx only simpler, can solve Einstein's riddle since, the Englishman is the red piece you remove from the 'color-surface' so you can move the other pieces. E hides one of his 5 pieces (#1). It's pretty cool.
  
  The remaining exercise is I suppose, to use the constraint table above to write down the code map that will construct a sphere with 'tracks' for triangular colored sections, each 1/20 the area of the sphere; or a good approximation. Ensure the pieces can be slid along the tracks, and that all 4 colors are preserved. Number each piece separately (for separation of number-states and angles, over the sphere).
  
  Someone beat me to it. Darn.
  
  p.s. if you areinto the Floyd, Dark Side Of The Moon has an album cover (call this A) black, but with a triangular figure, T, which has a beam at left (input I) 'breaking' into the sort of thing, a spectrum of color, that a toy company has built into its mechanical solution. Notice, the area on the right is also 3-sided (output I'). Red is the 'first' or upper string of color, violet is the last. E is 'across the spectrum' of colors I(e), as apparent edge-to-edge transitions. This is the key, like the piece of rectified, red triangular plastic you remove from the puzzle made by humans H, to tune the instrument (I*).
  
  It looks a bit like a sail against a black surface, regular sides and equal angles, with a wind blowing through it that divides into parts because the sail is equally partitioned over sides and angles.
  It's a kind of 3-way foil or 'sailing register' that counts strings of color, E, depending on the shape and input E. You might be able to twang some strings to see what happens.