4 ms·

Solving a “Layton Puzzle” with Prolog

110 points by Tomte 2y ago

curtisszmania 2y ago
[dead]
taeric 2y ago
Very fun post! If folks haven't seen it, Knuth's latest release looks at constraint satisfaction. Really fun discussion on the relations between different ways to model this sort of problem. Happy to add this book to my list! I would also challenge the author to consider puzzles a bit more. Sprinkling fun into the topics goes a long way. Specifically, do both! Talk about how you can reframe a "real world" problem in terms of some puzzles we may have played as kids. Invite us to play more as adults.
trainyperson 2y ago
Doing the puzzle with “pencil and paper” logic is actually quite approachable and fun! I recommend it. Hint: you don’t need to run constraint satisfaction! There are some insightful shortcuts to be made
NooneAtAll3 2y ago
if anyone wants a "human" solution 1) there are 4 questions that everyone answered the same way (let's say this earned everyone A points) 2) in the rest of the questions, 7-point and 5-point students gave opposite answers that means that 7-A (rest of 7-point student's correct answers) equals to 5-(4-A) (rest of 5-point student's incorrect answers) 7-A = 1+A => A=3 That means 3-point student scored 3 points together with everyone else, and answered rest of the questions wrong From this it's trivial to find the hidden score and why there are 4 possible answer keys
- thaumasiotes 2y ago
  It seems to have been designed to be easy to solve by hand. My approach: 1. Dan and Lisa agree on 8 out of 10 answers. 2. Dan, by changing two of Lisa's answers, has gained 20 points. Dan is right about both of those answers (6 and 9). 3. Mary is wrong about answers 6 and 9. 4. Mary has a total of three mistakes. 5. There are 8 possible answer keys, because Mary only has one mistake left to distribute among the other 8 answers. 6. On four of those possible answer keys, Lisa scores 50 points. Those are invalid. 7. Mary's mistake must be in answer to question 1, 3, 7, or 8. 8. On all four of those questions, Colin agrees with Mary. 9. Colin scores 30 points collectively on questions 1, 3, 7, and 8, plus 10 points each on questions 2, 5, and 9. 60 points.
  - moomin 2y ago
    It definitely has been designed to be solved by hand or possibly even in your head. Nintendo aren't in the habit of handing out Advent of Code questions in their games. Honestly, the sophistication of the puzzle design is extremely admirable.
deleted 2y ago
[deleted]
crtasm 2y ago
A puzzle from a game in this series https://en.wikipedia.org/wiki/Professor_Layton https://en.wikipedia.org/wiki/Professor_Layton
- mgnn 2y ago
  What nostalgia!I haven't played any since one of the early 3DS ones. And there's an incoming one for the switch too!
  - godot 2y ago
    One of my favorite series from when I was younger. I only ever played the DS ones, not the 3DS ones. That was the 2008-2011 timeframe -- 14-17 years ago!
- saagarjha 2y ago
  It’s this one from Professor Layton and the Diabolical Box: https://layton.fandom.com/wiki/Puzzle:Colin%27s_Score https://layton.fandom.com/wiki/Puzzle:Colin%27s_Score
jfengel 2y ago
The puzzle is easy enough to solve by brute force, but it seems like rather a lot to solve by hand for a very junior-targeted game. Heuristics could reduce trial-and-error to a reasonable search space, but is there a purely deductive way to solve this by hand?
- Jtsummers 2y ago
  With the 4 common answers (all 4 people answered them the same) we know that no more than 3 of them can be correct (because Lisa scored 3 points). We also know at least one must be correct (otherwise Mary can't score 7). Examine what Mary and Dan have in common besides those 4... Nothing. They disagreed on each of the remaining 6 answers. We can break this down into three cases, either 1, 2, or 3 of the common answers are correct: 1 - Impossible. Mary gets 6 of the remaining correct, Dan gets 4 of the 6 remaining correct, but they have none in common so this doesn't work. 2 - Impossible. Mary gets 5 of the remaining correct, Dan gets 3 of the remaining correct, but again their answers disagree so they cannot share a common answer out of the 6 remaining. 3. Mary needs 4 of the remaining 6 and Dan needs 2, this is feasible. At this point you can ignore the 4 common ones other than to say Colin has at least a score of 3. Since we now know that everything outside those common ones are incorrect for Lisa, cross everything off that Lisa and Colin share of the 6 non-common answers. There are only 3 that disagree with Lisa so they must be correct (because everything Lisa answered here is wrong). This gives Colin a final score of 6, three in common with everyone and three from disagreeing with Lisa. The reason there are 4 keys in Hillel's solution is actually kind of funny, they don't matter. Those 4 keys only disagree on which of the 4 common answers are incorrect. They agree everywhere else. EDIT: Presentation and some typos
  - flanbiscuit 2y ago
    I'm more of a visual person when it comes to puzzles like these so here's my take: 1 2 3 4 5 6 7 8 9 10 Mary B B A B A B B A B B 7 Dan B A A A B A B A A A 5 Lisa B A A A B B B A B A 3 Colin B B A A A B B A A A ? First I remove the answers that everyone agreed on. We know that one of these 4 answers is wrong because of Lisa. 1 2 3 4 5 6 7 8 9 10 Mary _ B _ B A B _ _ B B 7 Dan _ A _ A B A _ _ A A 5 Lisa _ A _ A B B _ _ B A 3 Colin _ B _ A A B _ _ A A ? Next we can assume anything matching Lisa's remaining answers are wrong, so I will put an x for those. These we 100% know are wrong answers. 1 2 3 4 5 6 7 8 9 10 Mary _ B _ B A x _ _ x B 7 Dan _ x _ x x A _ _ A x 5 Lisa _ x _ x x x _ _ x x 3 Colin _ B _ x A x _ _ A x ? If you count both the remaining answers and the blanks, you'll see they equal to their score + 1. That 1 being that unknown of the 4 original blank columns where everyone agreed. So the answer for Colin would by 6 because he has 7 remaining answers.
AceJohnny2 2y ago
Related: https://news.ycombinator.com/item?id=35623625 https://news.ycombinator.com/item?id=35623625 (2023) "Why Did Prolog Lose Steam (2010)" Anecdotally, some years ago the Zebra Puzzle [1] made the rounds in my team. Two people solved it: myself, a young intern, who mapped out the constraints as a physical puzzle that I was able to solve visually, and a more seasoned colleague who used Prolog. [1] https://en.wikipedia.org/wiki/Zebra_Puzzle https://en.wikipedia.org/wiki/Zebra_Puzzle
- tannhaeuser 2y ago
  FYI, a very compact Prolog version of the Zebra/Einstein puzzle can be loaded into the editor for execution using "Load example" on [1]. [1]: https://quantumprolog.sgml.net/browser-demo/browser-demo.html https://quantumprolog.sgml.net/browser-demo/browser-demo.htm...
superlopuh 2y ago
I once used Sentient[0] to solve a similar puzzle in a different game that I think made a puzzle that required a constraint solver as a joke on the player. I'm a bit saddened to see that the repo hasn't been updated in 6 years, and a node update broke the binary installed on my computer, I find it a much more ergonomic environment than Prolog, hopefully someone else will pick up the mantle. [0]: https://sentient-lang.org/ https://sentient-lang.org/
tannhaeuser 2y ago
Don't understand why the article uses SWI's constraint libs when the problem is easily solved using pure ISO Prolog with code readily linked from the article even ie Pablo's solution at [2]. Just paste that code into [1] and append the line (or place it at the start, doesn't matter) :- initialization(colin_score(X)). to tell Prolog which top-level goal to solve and hit Execute to see the solution in your browser. [1]: https://quantumprolog.sgml.net/browser-demo/browser-demo.html https://quantumprolog.sgml.net/browser-demo/browser-demo.htm... [2]: https://morepablo.com/2010/09/some-professor-layton-prolog.html https://morepablo.com/2010/09/some-professor-layton-prolog.h...
- Avshalom 2y ago
  It's hardly SWI's lib, clpFD is available almost identically in Sicstus, Ciao, GNU, B† and others. As to why? I dunno I guess for a lot of us constraint programming just comes to hand very naturally. It often feels more Prolog-y to us than non-constraint approaches. †B-Prolog of course morphed into Picat which is also mentioned and Picat also leans heavy on clp so if the author is more familiar with Picat it would be even more natural to use clpFD.
- jodrellblank 2y ago
  It can be more briefly solved without constraints than the linked code, brute-forcing through the possible correct-answer permutations: member_of(List, X) :- % helper swaps args for use with maplist. member(X, List). score(Given, Correct, Adder, NewScore) :- (Given = Correct % Prolog if/else -> NewScore is Adder + 10 ; NewScore = Adder). puzzle(ColinScore) :- length(CorrectAnswers, 10), % ten unknown correct answers, maplist(member_of([a,b]), CorrectAnswers), % each 'a' or 'b' % compare player answers to correct answers, add their scores. foldl(score, [b,b,a,b,a,b,b,a,b,b], CorrectAnswers, 0, 70), % Mary foldl(score, [b,a,a,a,b,a,b,a,a,a], CorrectAnswers, 0, 50), % Dan foldl(score, [b,a,a,a,b,b,b,a,b,a], CorrectAnswers, 0, 30), % Lisa foldl(score, [b,b,a,a,a,b,b,a,a,a], CorrectAnswers, 0, ColinScore). Then: ?- puzzle(ColinScore). ColinScore = 60 It's fine for a toy puzzle and it uses fewer resources than invoking the constraint solver, but it's a poor way to approach a puzzle or real world task with a larger search space.
a_e_k 2y ago
It was fun to try to golf this a bit with Z3Py. I'm not an expert at writing for Z3, so it might be possible to do better, but here is a solution in 7 lines: import z3 res = [("bbababbabb", 7), ("baaababaaa", 5), ("baaabbbaba", 3), ("bbaaabbaaa", z3.Int("x"))] key = [z3.Int(f"k{i}") for i in range(10)] z = z3.Solver() z.add([z3.Sum([ord(a) == k for a, k in zip(r[0], key)]) == r[1] for r in res]) z.check() print(z.model()[res[-1][1]])
- lheck 2y ago
  almost-one-liner (though it will give you the full model): import z3; z3.solve([z3.Sum([ord(a) == z3.Int(f"k{i}") for i, a in enumerate(r[0])]) == r[1] for r in [("bbababbabb", 7), ("baaababaaa", 5), ("baaabbbaba", 3), ("bbaaabbaaa", z3.Int("x"))]])
  - happyhacks 2y ago
    my take - more writeup style for people new to z3 https://happyhacks.bearblog.dev/solving-a-layton-puzzle-with-z3/ https://happyhacks.bearblog.dev/solving-a-layton-puzzle-with...
jodrellblank 2y ago
I got to writing this Prolog/constraint version which you can try on SWISH by clicking https://swish.swi-prolog.org/p/FYMhJEmj.pl https://swish.swi-prolog.org/p/FYMhJEmj.pl then clicking the blue 'Run' button in the lower right corner. :-use_module(library(clpfd)). % score/3 has mode ++- and maps % a given answer and a correct answer to a score. score(Given, Correct, Score) :- zcompare(Order, Given, Correct), score_(Order, Score). score_(=, 10). % 10 points if given matches correct, score_(<, 0). % else 0 points score_(>, 0). puzzle(ColinTotal, ColinScores, CorrectAnswers) :- A = 0, % false B = 1, % true length(CorrectAnswers, 10), % ten unknown correct answers, CorrectAnswers ins 0..1, % all either true or false. % compare player answers with correct answers, add their scores. maplist(score, [B,B,A,B,A,B,B,A,B,B], CorrectAnswers, Mary), sum(Mary, #=, 70), maplist(score, [B,A,A,A,B,A,B,A,A,A], CorrectAnswers, Dan), sum(Dan, #=, 50), maplist(score, [B,A,A,A,B,B,B,A,B,A], CorrectAnswers, Lisa), sum(Lisa, #=, 30), maplist(score, [B,B,A,A,A,B,B,A,A,A], CorrectAnswers, ColinScores), sum(ColinScores, #=, ColinTotal). Then this query: ?- puzzle(ColinTotal, ColinScores, CorrectAnswers). Solves without a constraint solver search and gives four possible solutions, each giving the same score for Colin which I've aligned by hand for this comment: ColinTotal = 60, ColinScores = [10, 10, 10, 0, 10, 0, 10, 0, 10, 0], CorrectAnswers = [ 1, 1, 0, 1, 0, 0, 1, 1, 0, 1] ColinTotal = 60, ColinScores = [10, 10, 10, 0, 10, 0, 0, 10, 10, 0], CorrectAnswers = [ 1, 1, 0, 1, 0, 0, 0, 0, 0, 1] ColinTotal = 60, ColinScores = [10, 10, 0, 0, 10, 0, 10, 10, 10, 0], CorrectAnswers = [ 1, 1, 1, 1, 0, 0, 1, 0, 0, 1] ColinTotal = 60, ColinScores = [ 0, 10, 10, 0, 10, 0, 10, 10, 10, 0], CorrectAnswers = [ 0, 1, 0, 1, 0, 0, 1, 0, 0, 1] Questions 2,4,5,6,9,10 have a fixed correct answer and 1,3,7,8 can vary and the puzzle is still coherent and solvable.
Avshalom 2y ago
well as long as we're swaping our own answers here is mine: %%% This works in swi as is %%% in ciao we need to include some modules %:- use_module(library(numlists)). %:- use_module(library(hiordlib)). %:- use_package(clpfd). main :- M = [b,b,a,b,a,b,b,a,b,b], D = [b,a,a,a,b,a,b,a,a,a], L = [b,a,a,a,b,b,b,a,b,a], C = [b,b,a,a,a,b,b,a,a,a], score(K,M,7), score(K,D,5), score(K,L,3), score(K,C,S), write(S). score(K,T,S) :- maplist(eq_OZ,K,T,SL), sum_list(SL,S). eq_OZ(X,X,1). eq_OZ(X,Y,0) :- dif(X,Y). %%% ciao doesn't have dif/2 so we use this %eq_OZ(X,Y,0) :- X #\= Y.

110 points by Tomte 2y ago

[dead]

Very fun post! If folks haven't seen it, Knuth's latest release looks at constraint satisfaction. Really fun discussion on the relations between different ways to model this sort of problem. Happy to add this book to my list! I would also challenge the author to consider puzzles a bit more. Sprinkling fun into the topics goes a long way. Specifically, do both! Talk about how you can reframe a "real world" problem in terms of some puzzles we may have played as kids. Invite us to play more as adults.

trainyperson 2y ago

Doing the puzzle with “pencil and paper” logic is actually quite approachable and fun! I recommend it. Hint: you don’t need to run constraint satisfaction! There are some insightful shortcuts to be made

NooneAtAll3 2y ago

if anyone wants a "human" solution 1) there are 4 questions that everyone answered the same way (let's say this earned everyone A points) 2) in the rest of the questions, 7-point and 5-point students gave opposite answers that means that 7-A (rest of 7-point student's correct answers) equals to 5-(4-A) (rest of 5-point student's incorrect answers) 7-A = 1+A => A=3 That means 3-point student scored 3 points together with everyone else, and answered rest of the questions wrong From this it's trivial to find the hidden score and why there are 4 possible answer keys

thaumasiotes 2y ago
It seems to have been designed to be easy to solve by hand. My approach: 1. Dan and Lisa agree on 8 out of 10 answers. 2. Dan, by changing two of Lisa's answers, has gained 20 points. Dan is right about both of those answers (6 and 9). 3. Mary is wrong about answers 6 and 9. 4. Mary has a total of three mistakes. 5. There are 8 possible answer keys, because Mary only has one mistake left to distribute among the other 8 answers. 6. On four of those possible answer keys, Lisa scores 50 points. Those are invalid. 7. Mary's mistake must be in answer to question 1, 3, 7, or 8. 8. On all four of those questions, Colin agrees with Mary. 9. Colin scores 30 points collectively on questions 1, 3, 7, and 8, plus 10 points each on questions 2, 5, and 9. 60 points.
- moomin 2y ago
  It definitely has been designed to be solved by hand or possibly even in your head. Nintendo aren't in the habit of handing out Advent of Code questions in their games. Honestly, the sophistication of the puzzle design is extremely admirable.

deleted 2y ago

[deleted]

crtasm 2y ago

A puzzle from a game in this series https://en.wikipedia.org/wiki/Professor_Layton https://en.wikipedia.org/wiki/Professor_Layton

mgnn 2y ago
What nostalgia!I haven't played any since one of the early 3DS ones. And there's an incoming one for the switch too!
- godot 2y ago
  One of my favorite series from when I was younger. I only ever played the DS ones, not the 3DS ones. That was the 2008-2011 timeframe -- 14-17 years ago!
saagarjha 2y ago
It’s this one from Professor Layton and the Diabolical Box: https://layton.fandom.com/wiki/Puzzle:Colin%27s_Score https://layton.fandom.com/wiki/Puzzle:Colin%27s_Score

jfengel 2y ago

The puzzle is easy enough to solve by brute force, but it seems like rather a lot to solve by hand for a very junior-targeted game. Heuristics could reduce trial-and-error to a reasonable search space, but is there a purely deductive way to solve this by hand?

Jtsummers 2y ago
With the 4 common answers (all 4 people answered them the same) we know that no more than 3 of them can be correct (because Lisa scored 3 points). We also know at least one must be correct (otherwise Mary can't score 7). Examine what Mary and Dan have in common besides those 4... Nothing. They disagreed on each of the remaining 6 answers. We can break this down into three cases, either 1, 2, or 3 of the common answers are correct: 1 - Impossible. Mary gets 6 of the remaining correct, Dan gets 4 of the 6 remaining correct, but they have none in common so this doesn't work. 2 - Impossible. Mary gets 5 of the remaining correct, Dan gets 3 of the remaining correct, but again their answers disagree so they cannot share a common answer out of the 6 remaining. 3. Mary needs 4 of the remaining 6 and Dan needs 2, this is feasible. At this point you can ignore the 4 common ones other than to say Colin has at least a score of 3. Since we now know that everything outside those common ones are incorrect for Lisa, cross everything off that Lisa and Colin share of the 6 non-common answers. There are only 3 that disagree with Lisa so they must be correct (because everything Lisa answered here is wrong). This gives Colin a final score of 6, three in common with everyone and three from disagreeing with Lisa. The reason there are 4 keys in Hillel's solution is actually kind of funny, they don't matter. Those 4 keys only disagree on which of the 4 common answers are incorrect. They agree everywhere else. EDIT: Presentation and some typos
- flanbiscuit 2y ago
  I'm more of a visual person when it comes to puzzles like these so here's my take: 1 2 3 4 5 6 7 8 9 10 Mary B B A B A B B A B B 7 Dan B A A A B A B A A A 5 Lisa B A A A B B B A B A 3 Colin B B A A A B B A A A ? First I remove the answers that everyone agreed on. We know that one of these 4 answers is wrong because of Lisa. 1 2 3 4 5 6 7 8 9 10 Mary _ B _ B A B _ _ B B 7 Dan _ A _ A B A _ _ A A 5 Lisa _ A _ A B B _ _ B A 3 Colin _ B _ A A B _ _ A A ? Next we can assume anything matching Lisa's remaining answers are wrong, so I will put an x for those. These we 100% know are wrong answers. 1 2 3 4 5 6 7 8 9 10 Mary _ B _ B A x _ _ x B 7 Dan _ x _ x x A _ _ A x 5 Lisa _ x _ x x x _ _ x x 3 Colin _ B _ x A x _ _ A x ? If you count both the remaining answers and the blanks, you'll see they equal to their score + 1. That 1 being that unknown of the 4 original blank columns where everyone agreed. So the answer for Colin would by 6 because he has 7 remaining answers.

AceJohnny2 2y ago

Related: https://news.ycombinator.com/item?id=35623625 https://news.ycombinator.com/item?id=35623625 (2023) "Why Did Prolog Lose Steam (2010)" Anecdotally, some years ago the Zebra Puzzle [1] made the rounds in my team. Two people solved it: myself, a young intern, who mapped out the constraints as a physical puzzle that I was able to solve visually, and a more seasoned colleague who used Prolog. [1] https://en.wikipedia.org/wiki/Zebra_Puzzle https://en.wikipedia.org/wiki/Zebra_Puzzle

tannhaeuser 2y ago
FYI, a very compact Prolog version of the Zebra/Einstein puzzle can be loaded into the editor for execution using "Load example" on [1]. [1]: https://quantumprolog.sgml.net/browser-demo/browser-demo.html https://quantumprolog.sgml.net/browser-demo/browser-demo.htm...

superlopuh 2y ago

I once used Sentient[0] to solve a similar puzzle in a different game that I think made a puzzle that required a constraint solver as a joke on the player. I'm a bit saddened to see that the repo hasn't been updated in 6 years, and a node update broke the binary installed on my computer, I find it a much more ergonomic environment than Prolog, hopefully someone else will pick up the mantle. [0]: https://sentient-lang.org/ https://sentient-lang.org/

tannhaeuser 2y ago

Don't understand why the article uses SWI's constraint libs when the problem is easily solved using pure ISO Prolog with code readily linked from the article even ie Pablo's solution at [2]. Just paste that code into [1] and append the line (or place it at the start, doesn't matter) :- initialization(colin_score(X)). to tell Prolog which top-level goal to solve and hit Execute to see the solution in your browser. [1]: https://quantumprolog.sgml.net/browser-demo/browser-demo.html https://quantumprolog.sgml.net/browser-demo/browser-demo.htm... [2]: https://morepablo.com/2010/09/some-professor-layton-prolog.html https://morepablo.com/2010/09/some-professor-layton-prolog.h...

Avshalom 2y ago
It's hardly SWI's lib, clpFD is available almost identically in Sicstus, Ciao, GNU, B† and others. As to why? I dunno I guess for a lot of us constraint programming just comes to hand very naturally. It often feels more Prolog-y to us than non-constraint approaches. †B-Prolog of course morphed into Picat which is also mentioned and Picat also leans heavy on clp so if the author is more familiar with Picat it would be even more natural to use clpFD.
jodrellblank 2y ago
It can be more briefly solved without constraints than the linked code, brute-forcing through the possible correct-answer permutations: member_of(List, X) :- % helper swaps args for use with maplist. member(X, List). score(Given, Correct, Adder, NewScore) :- (Given = Correct % Prolog if/else -> NewScore is Adder + 10 ; NewScore = Adder). puzzle(ColinScore) :- length(CorrectAnswers, 10), % ten unknown correct answers, maplist(member_of([a,b]), CorrectAnswers), % each 'a' or 'b' % compare player answers to correct answers, add their scores. foldl(score, [b,b,a,b,a,b,b,a,b,b], CorrectAnswers, 0, 70), % Mary foldl(score, [b,a,a,a,b,a,b,a,a,a], CorrectAnswers, 0, 50), % Dan foldl(score, [b,a,a,a,b,b,b,a,b,a], CorrectAnswers, 0, 30), % Lisa foldl(score, [b,b,a,a,a,b,b,a,a,a], CorrectAnswers, 0, ColinScore). Then: ?- puzzle(ColinScore). ColinScore = 60 It's fine for a toy puzzle and it uses fewer resources than invoking the constraint solver, but it's a poor way to approach a puzzle or real world task with a larger search space.

a_e_k 2y ago

It was fun to try to golf this a bit with Z3Py. I'm not an expert at writing for Z3, so it might be possible to do better, but here is a solution in 7 lines: import z3 res = [("bbababbabb", 7), ("baaababaaa", 5), ("baaabbbaba", 3), ("bbaaabbaaa", z3.Int("x"))] key = [z3.Int(f"k{i}") for i in range(10)] z = z3.Solver() z.add([z3.Sum([ord(a) == k for a, k in zip(r[0], key)]) == r[1] for r in res]) z.check() print(z.model()[res[-1][1]])

lheck 2y ago
almost-one-liner (though it will give you the full model): import z3; z3.solve([z3.Sum([ord(a) == z3.Int(f"k{i}") for i, a in enumerate(r[0])]) == r[1] for r in [("bbababbabb", 7), ("baaababaaa", 5), ("baaabbbaba", 3), ("bbaaabbaaa", z3.Int("x"))]])
- happyhacks 2y ago
  my take - more writeup style for people new to z3 https://happyhacks.bearblog.dev/solving-a-layton-puzzle-with-z3/ https://happyhacks.bearblog.dev/solving-a-layton-puzzle-with...

jodrellblank 2y ago

I got to writing this Prolog/constraint version which you can try on SWISH by clicking https://swish.swi-prolog.org/p/FYMhJEmj.pl https://swish.swi-prolog.org/p/FYMhJEmj.pl then clicking the blue 'Run' button in the lower right corner. :-use_module(library(clpfd)). % score/3 has mode ++- and maps % a given answer and a correct answer to a score. score(Given, Correct, Score) :- zcompare(Order, Given, Correct), score_(Order, Score). score_(=, 10). % 10 points if given matches correct, score_(<, 0). % else 0 points score_(>, 0). puzzle(ColinTotal, ColinScores, CorrectAnswers) :- A = 0, % false B = 1, % true length(CorrectAnswers, 10), % ten unknown correct answers, CorrectAnswers ins 0..1, % all either true or false. % compare player answers with correct answers, add their scores. maplist(score, [B,B,A,B,A,B,B,A,B,B], CorrectAnswers, Mary), sum(Mary, #=, 70), maplist(score, [B,A,A,A,B,A,B,A,A,A], CorrectAnswers, Dan), sum(Dan, #=, 50), maplist(score, [B,A,A,A,B,B,B,A,B,A], CorrectAnswers, Lisa), sum(Lisa, #=, 30), maplist(score, [B,B,A,A,A,B,B,A,A,A], CorrectAnswers, ColinScores), sum(ColinScores, #=, ColinTotal). Then this query: ?- puzzle(ColinTotal, ColinScores, CorrectAnswers). Solves without a constraint solver search and gives four possible solutions, each giving the same score for Colin which I've aligned by hand for this comment: ColinTotal = 60, ColinScores = [10, 10, 10, 0, 10, 0, 10, 0, 10, 0], CorrectAnswers = [ 1, 1, 0, 1, 0, 0, 1, 1, 0, 1] ColinTotal = 60, ColinScores = [10, 10, 10, 0, 10, 0, 0, 10, 10, 0], CorrectAnswers = [ 1, 1, 0, 1, 0, 0, 0, 0, 0, 1] ColinTotal = 60, ColinScores = [10, 10, 0, 0, 10, 0, 10, 10, 10, 0], CorrectAnswers = [ 1, 1, 1, 1, 0, 0, 1, 0, 0, 1] ColinTotal = 60, ColinScores = [ 0, 10, 10, 0, 10, 0, 10, 10, 10, 0], CorrectAnswers = [ 0, 1, 0, 1, 0, 0, 1, 0, 0, 1] Questions 2,4,5,6,9,10 have a fixed correct answer and 1,3,7,8 can vary and the puzzle is still coherent and solvable.

Avshalom 2y ago

well as long as we're swaping our own answers here is mine: %%% This works in swi as is %%% in ciao we need to include some modules %:- use_module(library(numlists)). %:- use_module(library(hiordlib)). %:- use_package(clpfd). main :- M = [b,b,a,b,a,b,b,a,b,b], D = [b,a,a,a,b,a,b,a,a,a], L = [b,a,a,a,b,b,b,a,b,a], C = [b,b,a,a,a,b,b,a,a,a], score(K,M,7), score(K,D,5), score(K,L,3), score(K,C,S), write(S). score(K,T,S) :- maplist(eq_OZ,K,T,SL), sum_list(SL,S). eq_OZ(X,X,1). eq_OZ(X,Y,0) :- dif(X,Y). %%% ciao doesn't have dif/2 so we use this %eq_OZ(X,Y,0) :- X #\= Y.