Tuesday, 27 October 2015

Einstein’s Five House Riddle - Solution in 'e'

Over the weekend, I came across this riddle and thought of solving it using ‘e’. Later on I challenged our team members over the WhatsApp and they took up the challenge. They came up with their version of solution in SV and SVA. I thought of sharing this solution to show how constraints solver engine can solve these kind of riddles. During the course of solution, I observed specifics about constraint solving technology of Specman. But before that, let’s look at the riddle. It is called with different names like Zebra Riddle or Einstein’s Riddle. This riddle has different version of stories and none of it has any solid proof to prove its origin. I’ve taken this riddle from one of the student’s home page from Stanford University website. Here is his version of the riddle which we tried to solve.

Let us assume that there are five houses of different colors next to each other on the same road. In each house lives a man of a different nationality. Every man has his different favorite drink, his different favorite brand of cigarettes, and keeps different pets of a particular kind.
  1. The Englishman lives in the red house.
  2. The Swede keeps dogs.
  3. The Dane drinks tea.
  4. The green house is just to the left of the white one.
  5. The owner of the green house drinks coffee.
  6. The Pall Mall smoker keeps birds.
  7. The owner of the yellow house smokes Dunhills.
  8. The man in the center house drinks milk.
  9. The Norwegian lives in the first house.
  10. The Blend smoker has a neighbor who keeps cats.
  11. The man who smokes Blue Masters drinks bier.
  12. The man who keeps horses lives next to the Dunhill smoker.
  13. The German smokes Prince.
  14. The Norwegian lives next to the blue house.
  15. The Blend smoker has a neighbor who drinks water.
The question to be answered is: Who keeps fish?
Now to solve this riddle, I started thinking these clues as constraints in ‘e’. Let’s go step by step how I approached this problem.

Here, we have 5 men who has different characteristics like they live in specific house number which has specific color. Each one has unique nationality, favorite drink, favorite brand of cigarettes and pet. So, based on these attributes, we can create a struct which can have these 5 attributes. These five attributes can be enumerated and ‘man’ struct can be defined as shown below.

type nationality_t : [english, swede, dane, norweigian, german];
type house_color_t : [red, green, yellow, blue, white];
type cigarette_t : [dunhill, blend, blue_masters, prince, pall_mall];
type pet_t : [dog, bird, cat, horse, fish];
type drink_t : [tea, coffee, milk, bier, water];

struct man {
   nationality : nationality_t;
   house_color : house_color_t;
   cigarette   : cigarette_t;
   pet         : pet_t;
   drink       : drink_t;
};

Here, I’ve not captured the house number as an attribute since I thought I can create list of struct which automatically will fix the house number of each object.
Now, I’ve categorized the clues in 2 category. One category which does not have house number order dependency and the other category which has house number dependency. Like, clue number 1 “The Englishman lives in red house” does not have any reference to the number of house, whereas clue no. 4 “The gree house is just to the left of the white one” has reference to the house number order. Since I have not captured the house number as an attribute, I’ve to split the clues into two different categories of constraints. One without the reference of the house number can be part of the original struct definition. One with the house number dependency can go to the place where list of struct is created so that we have notion of the house orders using index number of the list.

Let’s first capture the clues without house number ordering as global constraints as part of struct declaration. Here is what I’ve captured.

extend man {
      //Englishman lives in the red house
      keep nationality == english      => house_color == red;

      //Swede keeps dog
      keep nationality == swede        => pet         == dog;

      //dane drinks tea
      keep nationality == dane         => drink       == tea;

      //owner of the green house drinks coffee
      keep house_color == green        => drink       == coffee;

      //Pall Mall smoker keeps bird
      keep cigarette   == pall_mall    => pet         == bird;

      //owner of the yellow house smokes dunhill
      keep house_color == yellow       => cigarette   == dunhill;

      //Man who smokes Blue Masters drinks bier
      keep cigarette   == blue_masters => drink       == bier;

      //German smokes prince
      keep nationality == german       => cigarette   == prince;

};

Now let’s move on to declaring this struct as 5 objects as shown below.

extend sys
{
   men : list of man;
     keep men.size() == 5;
};

Now over here, we can capture the clues with house number ordering as different constraints involving different objects. Here, we’ll go by clues.

Let’s look at the first constraint which has dependency on the house number ordering. This clue is clue number 4 “The green house is just to the left of the white one”. This can be captured as shown below. Here I’ve assumed house numbers starts from 0 to 4 runs from left to right. Due to this assumption, house number 0 does not have any house on left side of it.

extend sys {
   //The green house is just to the left of the white one.
   keep men[1].house_color == white => men[0].house_color == green;
   keep men[2].house_color == white => men[1].house_color == green;
   keep men[3].house_color == white => men[2].house_color == green;
   keep men[4].house_color == white => men[3].house_color == green;
};

Next clue which has dependency on the house number ordering is clue number 8 “The man in the center house drinks milk”. This can be captured easily as we know the house number 2 is in center.

extend sys {
   //The man in the center house drinks milk.
   keep men[2].drink       == milk;
};

Next such clue is clue number 9 “The Norwegian lives in the first house”. Again it is simple to capture.

extend sys {
   //Norweigian lives in the first house
   keep men[0].nationality == norweigian;
};

Next such clue is clue number 10 “The Blend smoker has a neighbor who keeps cats”. This is how I captured it. It is obvious that man staying in house 0 or 4 have only one neighbor and rest men have 2 neighbour.

extend sys {
   //The Blend smoker has a neighbor who keeps cats.
   keep men[0].pet         == cat   => men[1].cigarette == blend;
   keep men[1].pet         == cat => men[0].cigarette == blend or men[2].cigarette == blend;
   keep men[2].pet         == cat => men[1].cigarette == blend or men[3].cigarette == blend;
   keep men[3].pet         == cat => men[2].cigarette == blend or men[4].cigarette == blend;
   keep men[4].pet         == cat => men[3].cigarette == blend;
};

Next clue is clue number 12 “The man who keeps horses lives next to the Dunhill smoker”. Again it is captured just like the above constraints for clue number 10.

extend sys {
   //The man who keeps horses lives next to the Dunhill smoker.
   keep men[0].pet         == horse => men[1].cigarette == dunhill;
   keep men[1].pet         == horse => men[0].cigarette == dunhill or men[2].cigarette == dunhill;
   keep men[2].pet         == horse => men[1].cigarette == dunhill or men[3].cigarette == dunhill;
   keep men[3].pet         == horse => men[2].cigarette == dunhill or men[4].cigarette == dunhill;
   keep men[4].pet         == horse => men[3].cigarette == dunhill;
};

Next clue in this category is clue number 14 “The Norwegian lives next to the blue house”.  Since it is already declared in clue number 9 “The Norwegian lives in the first house”, it is obvious that the blue house is house number 1. The Here is how I’ve captured it.

extend sys {
   //The Norwegian lives next to the blue house.
   keep men[1].house_color == blue;
};

If we do not want to capture it by interpreting it by ourselves and want tool to interpret for us, we can capture it just like we captured clue number 10 and 12. Here is how clue number 14 can be captured in alternate way.

extend sys {
   //The Norwegian lives next to the blue house.
   keep men[0].nationality == norweigian => men[1].house_color == blue;
   keep men[1].nationality == norweigian => men[0].house_color == blue or men[2].house_color == blue;
   keep men[2].nationality == norweigian => men[1].house_color == blue or men[3].house_color == blue;
   keep men[3].nationality == norweigian => men[2].house_color == blue or men[4].house_color == blue;
   keep men[4].nationality == norweigian => men[2].house_color == blue or men[4].house_color == blue;
};

Last clue that is clue number 15 “The Blend smoker has a neighbor who drinks water” is again captured the way we have captured clue number 10 and 12.

extend sys {
   //The Blend smoker has a neighbor who drinks water.
   keep men[0].cigarette   == blend => men[1].drink == water;
   keep men[1].cigarette   == blend => men[0].drink == water or men[2].drink == water;
   keep men[2].cigarette   == blend => men[1].drink == water or men[3].drink == water;
   keep men[3].cigarette   == blend => men[2].drink == water or men[4].drink == water;
   keep men[4].cigarette   == blend => men[3].drink == water;
};

Now we have captured all the clues as constraints. But when we load this file into Specman and try to solve it, we get constraint error. The reason is that we have not capture an important piece of information into this solution, that is each man has unique values of each 5 attributes. Now capturing this and informing Specman to generate uniq values for all 5 objects of man struct is bit tricky in ‘e’. In system Verilog, randc operator helps us capture it very easily, but in ‘e’ we have to do it with lot of codes. Here is how I’ve done it.

extend sys {
   //Let us assume that there are five houses of different colors next to each other on the same road. In each house lives a man of a different nationality. Every man has his favorite drink, his favorite brand of cigarettes, and keeps pets of a particular kind.
   //Unique nationality, house_color, cigarette, pet and drink
   keep men[0].nationality != men[1].nationality;
   keep men[0].nationality != men[2].nationality;
   keep men[0].nationality != men[3].nationality;
   keep men[0].nationality != men[4].nationality;
   keep men[0].house_color != men[1].house_color;
   keep men[0].house_color != men[2].house_color;
   keep men[0].house_color != men[3].house_color;
   keep men[0].house_color != men[4].house_color;
   keep men[0].cigarette   != men[1].cigarette;
   keep men[0].cigarette   != men[2].cigarette;
   keep men[0].cigarette   != men[3].cigarette;
   keep men[0].cigarette   != men[4].cigarette;
   keep men[0].pet         != men[1].pet;
   keep men[0].pet         != men[2].pet;
   keep men[0].pet         != men[3].pet;
   keep men[0].pet         != men[4].pet;
   keep men[0].drink       != men[1].drink;
   keep men[0].drink       != men[2].drink;
   keep men[0].drink       != men[3].drink;
   keep men[0].drink       != men[4].drink;

   keep men[1].nationality != men[2].nationality;
   keep men[1].nationality != men[3].nationality;
   keep men[1].nationality != men[4].nationality;
   keep men[1].house_color != men[2].house_color;
   keep men[1].house_color != men[3].house_color;
   keep men[1].house_color != men[4].house_color;
   keep men[1].cigarette   != men[2].cigarette;
   keep men[1].cigarette   != men[3].cigarette;
   keep men[1].cigarette   != men[4].cigarette;
   keep men[1].pet         != men[2].pet;
   keep men[1].pet         != men[3].pet;
   keep men[1].pet         != men[4].pet;
   keep men[1].drink       != men[2].drink;
   keep men[1].drink       != men[3].drink;
   keep men[1].drink       != men[4].drink;

   keep men[2].nationality != men[3].nationality;
   keep men[2].nationality != men[4].nationality;
   keep men[2].house_color != men[3].house_color;
   keep men[2].house_color != men[4].house_color;
   keep men[2].cigarette   != men[3].cigarette;
   keep men[2].cigarette   != men[4].cigarette;
   keep men[2].pet         != men[3].pet;
   keep men[2].pet         != men[4].pet;
   keep men[2].drink       != men[3].drink;
   keep men[2].drink       != men[4].drink;

   keep men[3].nationality != men[4].nationality;
   keep men[3].house_color != men[4].house_color;
   keep men[3].cigarette   != men[4].cigarette;
   keep men[3].pet         != men[4].pet;
   keep men[3].drink       != men[4].drink;
};

Now we are ready to solve it. Let’s add some debug and solution print statements.

extend {
   post_generate() is also
   {
      for i from 0 to 4 {
         outf ("\n");
         outf ("men[%d].nationality = %s\n", i, men[i].nationality);
         outf ("men[%d].house_color = %s\n", i, men[i].house_color);
         outf ("men[%d].cigarette   = %s\n", i, men[i].cigarette  );
         outf ("men[%d].pet         = %s\n", i, men[i].pet        );
         outf ("men[%d].drink       = %s\n", i, men[i].drink      );
      };

      outf ("===================================\n");
      for i from 0 to 4 {
         if (men[i].pet == fish)
         {
            outf ("\nAnswer to the quesiton 'Who keeps fish?' is %s who lives in house no %d with house color %s and smokes %s and drinks %s has %s as pet...\n", men[i].nationality, i, men[i].house_color, men[i].cigarette, men[i].drink, men[i].pet);
            outf ("\n");
            outf ("men[%d].nationality = %s\n", i, men[i].nationality);
            outf ("men[%d].house_color = %s\n", i, men[i].house_color);
            outf ("men[%d].cigarette   = %s\n", i, men[i].cigarette  );
            outf ("men[%d].pet         = %s\n", i, men[i].pet        );
            outf ("men[%d].drink       = %s\n", i, men[i].drink      );

         };
      };
   }; //post_generate() is also
}; //extend sys

After loading it into Specman and solving it, we get following output.

Loading einsteins_puzle.e ...

Doing setup ...
Generating the test with IntelliGen using seed 1966611192...

men[0].nationality = norweigian
men[0].house_color = yellow
men[0].cigarette   = dunhill
men[0].pet         = cat
men[0].drink       = water

men[1].nationality = dane
men[1].house_color = blue
men[1].cigarette   = blend
men[1].pet         = horse
men[1].drink       = tea

men[2].nationality = english
men[2].house_color = red
men[2].cigarette   = pall_mall
men[2].pet         = bird
men[2].drink       = milk

men[3].nationality = german
men[3].house_color = green
men[3].cigarette   = prince
men[3].pet         = fish
men[3].drink       = coffee

men[4].nationality = swede
men[4].house_color = white
men[4].cigarette   = blue_masters
men[4].pet         = dog
men[4].drink       = bier
===================================

Answer to the quesiton 'Who keeps fish?' is german who lives in house no 3 with house color green and smokes prince and drinks coffee has fish as pet...

men[3].nationality = german
men[3].house_color = green
men[3].cigarette   = prince
men[3].pet         = fish
men[3].drink       = coffee

Starting the test ...
Running the test ...
No actual running requested.
Checking the test ...
Checking is complete - 0 DUT errors, 0 DUT warnings.

So the answer is “German keeps the Fish”. Now try to solve this solution multiple times with random seed and you will get the same answer. That way we can be sure that the constraints we’ve captured are correct.

One last observation I’ve made is that this solution can be solved in IntelliGen Specman solver only. It does not work with PGen Specman solver engine. The reason is quiet obvious. PGen depends on the order of constraints declaration whereas IntelliGen is truly bi-directional constraints solver which is natural way of solving all the constraints at the same time.

Hope you like this riddle and the solution. I am sure this is not the optimal way of solving the problem, but it’s my lazy weekend stuff, so I wrote whatever came to my mind first without thinking much about optimizing it. If you have any suggestion to improve it, please do comment. If you have solutions in other forms (XLS, Perl, tcl, python, C, C++, VB, PHP or any other your favorite language) then please do post it in the comment.

1 comment:

Aurelian Ionel Munteanu said...

Hi, Sandeep.

Such challenges are always nice to follow.

Here is a simplified version of the riddle in e-language code. At least the uniqueness constraint is solved in just 5 lines of code.

There is also a SystemVerilog version of this riddle. Check out the first recommended article from here:
http://www.amiq.com/consulting/2014/08/08/recommended-articles-july-2014/


<'
type nationality_t : [englishman, swede, dane, norwegian, german];
type house_color_t : [red, green, yellow, blue, white];
type cigarette_t : [dunhill, blend, blue_masters, prince, pall_mall];
type pet_t : [dog, bird, cat, horse, fish];
type drink_t : [tea, coffee, milk, bier, water];

define NUM_OF_HOUSES 5;

struct riddle_s {
nationality[NUM_OF_HOUSES] : list of nationality_t;
house_color[NUM_OF_HOUSES] : list of house_color_t;
cigar[NUM_OF_HOUSES] : list of cigarette_t;
pet[NUM_OF_HOUSES] : list of pet_t;
drink[NUM_OF_HOUSES] : list of drink_t;

keep nationality.is_a_permutation(all_values(nationality_t));
keep house_color.is_a_permutation(all_values(house_color_t));
keep pet.is_a_permutation(all_values(pet_t));
keep cigar.is_a_permutation(all_values(cigarette_t));
keep drink.is_a_permutation(all_values(drink_t));

keep for each in nationality{
it == englishman => house_color[index] == red; //#1
it == swede => pet[index] == dog;//#2
it == dane => drink[index] == tea;//#3
it == norwegian => index == 0;//#9
it == german => cigar[index] == prince;//#13
index in [1..NUM_OF_HOUSES-2] and it == norwegian => house_color[index+1] == blue or house_color[index-1] == blue;//#14
index == 0 and it == norwegian => house_color[index+1] == blue;//#14
index == NUM_OF_HOUSES-1 and it == norwegian => house_color[index-1] == blue;//#14
};

keep for each in house_color {
it == green => index != NUM_OF_HOUSES-1;//#4
it == green => house_color[index+1]== white;//#4
it == green => drink[index] == coffee;//#5
it == yellow => cigar[index] == dunhill;//#7
};

keep for each in cigar {
it == pall_mall => pet[index] == bird;//#6
index in [1..NUM_OF_HOUSES-2] and it == blend => pet[index+1] == cat or pet[index-1] == cat;//#10
index == 0 and it == blend => pet[index+1] == cat;//#10
index == NUM_OF_HOUSES-1 and it == blend => pet[index-1] == cat;//#10
it == blue_masters => drink[index] == bier;//#11
index in [1..NUM_OF_HOUSES-2] and it == dunhill => pet[index+1] == horse or pet[index-1] == horse;//#12
index == 0 and it == dunhill => pet[index+1] == horse;//#12
index == NUM_OF_HOUSES-1 and it == dunhill => pet[index-1] == horse;//#12
index in [1..NUM_OF_HOUSES-2] and it == blend => drink[index+1] == water or drink[index-1] == water;//#15
index == 0 and it == blend => drink[index+1] == water;//#15
index == NUM_OF_HOUSES-1 and it == blend => drink[index-1] == water;//#15
};

keep for each in drink {
it == milk => index == NUM_OF_HOUSES/2;//#8
};
};

extend sys {
run() is also {
var solution : riddle_s;
var position : int;
gen solution;
position = solution.pet.first_index(it == fish);
messagef(LOW, "Solution is: %s lives in the %s house #%0d, keeps a %s, drinks %s and smokes %s\n",
solution.nationality[position], solution.house_color[position], position, solution.pet[position], solution.drink[position], solution.cigar[position]);
};
};