## Introduction

One technique to solve difficult problems using a computational system,
is to apply brute force search. This means to exhaustively search through
all possible combinations until a solution is found. In this article I
will present an implementation of a brute force search algorithm, that can
be applied to a variety of problems. The idea for this article was taken
by an
article in CodeProject, which presented a C++ program that solves
Einstein's Riddle. The functionality of this article was extended, in
order to exploit the benefits of Object Oriented Programming and to
produce a generic solver class, that can be used as base class from many
problem-specific solver classes. Many examples of problem-solver classes
will be presented, including one that solves Einstein's Riddle.

*Note: *The language of the implementation is C++. Although some
advanced features of the language are used (template classes, STL), the
technical details in this article are kept in a minimum, so that even the
less-experienced programmer can follow the logic of the brute force
algorithm and have fun with the example puzzles.

## Eight Queens Riddle

I will introduce the idea of the brute force search algorithm starting
from the Eight Queens Riddle. This is the well-known problem of placing 8
non-attacking queens on a chessboard. Source code for solving this puzzle
can be found in many software books. In the download section you will find,
a Java applet (java_applets_src.zip) that implements the riddle.

A computing system can help us solve the riddle quickly. Moreover it
can help us find all possible solutions. To do so we must instruct our
program to search all possible placements of 8 queens on the chessboard.
There is a total of 64!63!62!61!60!59!58!57! different
placements. It would be highly inefficient to try to search all these
different combinations. It is much better to solve the problem
inductively. That is to first solve the problem of placing one queen at
a 1x8 chessboard, then the problem of placing two queens at a 2x8
chessboard and continue until the full solution is found. In this way
large groups of apparently erroneous combinations are caught early in
the process and need not be tested. If there is a pair of attacking
queens at the first n columns of the chessboard, then no matter where
the rest 8-n queens are placed, the combination can not constitute a
solution.

The inductive searching algorithm works as follows. We start by placing
the first queen at a1. The second queen is placed at b3 (as b1 and b2 are
attacked by the first queen). Then we look for an unattacked square at
the third column. We continue in the same manner, until we reach a column,
which has no unattacked squares. In this case we return to the previous
column and we try to move the queen placed there to another unattacked
square. We only examine squares that are in a higher row than the previous
one. If this is impossible too, we move another column back and we try to
do the same. If a new unattacked square at a previous column is found, we
are ready to move forwards again. Everytime we are able to place a queen
at eighth row, a new solution is found. We increment a variable, store
the positions of the queens in a table and we return to column 7 to find
a new unattacked square. The searching is complete when we need to return
from column 2 to column 1 and the queen is placed at a8. A visualisation
of the algorithm can be presented by clicking "Solve" button at the
previous applet.

## Generalisation

The concept of the inductive searching algorithm can be used in more
general situations. To be able to re-use the functionality of the
algorithm we'll define an "abstract" riddle. This "abstract" riddle is
represented by a table of n cells, which must be filled by elements coming
from an ordered list (see Figure 1).

**Figure 1**: General Riddle
To solve the riddle inductively we first fill cell 1, then cell 2 and so
on until all cells are filled. The flowchart of the algorithm is presented
in the following figure:

**Figure 2**: Flowchart
For this technique to work we must define two functions:

*check_placement(m) :* which checks if the sub-riddle of first
m cells is solved
*get_next_element(e) :* which given an element e, returns the
element immediately after it in the ordered list.

The C++ class `riddle_solver<move_t>`

implements the
"abstract" riddle.

template <class move_t>
class riddle_solver
{
public:
riddle_solver<move_t>();
~riddle_solver<move_t>();
bool solve();
void print_all();
void set_solutions_count(int n);
void set_size(int s);
void set_null_element(move_t mn);
private:
bool make_next_move();
bool go_back();
protected:
virtual int check_placement();
virtual move_t get_next_element(int pos);
virtual void register_move(int pos, move_t move);
virtual void unregister_last_move();
virtual void print_solution(vector<move_t> sol) {;};
protected:
vector<move_t> moves_array;
int size;
int moves_count;
private:
move_t move_null;
int sol_number_max;
vector<vector<move_t> > sol_v;
};

*Note: *`riddle_solver<move_t>`

is a template
class. `move_t`

type defines the type of the elements that
fill the cells of the table. In all the examples of this article ```
riddle_solver
```

will be instantiated with int as `move_t`

.
However the class can be used with any other native type (such as float or double)
or user-defined class that overloads = operator.

Using `riddle_solver<int>`

as base class, the eight
queens riddle can be solved with a class as simple as that:

q8_riddle_solver::q8_riddle_solver()
{
set_null_element(-1);
set_size(8);
}
q8_riddle_solver::~q8_riddle_solver() {}
int
q8_riddle_solver::get_next_element(int pos)
{
if (moves_array[pos] == 7)
return(-1);
else
return(moves_array[pos]+1);
return(-1);
}
int
q8_riddle_solver::check_placement()
{
int i, j;
for(i=0;i<moves_count;i++)
{
for(j=i+1;j<moves_count;j++)
{
if (moves_array[i] == moves_array[j])
return(0);
if (i + moves_array[i] == j + moves_array[j])
return(0);
if (i - moves_array[i] == j - moves_array[j])
return(0);
}
}
return(moves_count);
}
void
q8_riddle_solver::print_solution(vector<int> sol)
{
char str[4];
str[2] = ' ';
str[3] = '\0';
for(char i = 0; i < 8; i ++)
{
str[0] = 'a' + i;
str[1] = '1' + sol[i];
cout << str;
}
cout << endl;
};

In the following paragraphs `riddle_solver`

class will be used
for solving more puzzles.

## Einstein's Riddle

This is the riddle that originally gave me the motivation for this article. This
is how it goes:

- There are 5 houses in five different colors.
- In each house lives a person with a different nationality.
- These five owners drink a certain type of beverage, smoke a certain
brand of cigar and keep a certain pet.
- No owners have the same pet, smoke the same brand of cigar or drink
the same beverage.

For this bizarre neighbourhood we know the following:

- the Brit lives in the red house.
- the Swede keeps dogs as pets.
- the Dane drinks tea.
- the green house is on the left of the white house.
- the green house's owner drinks coffee.
- the person who smokes Pall Mall rears birds.
- the owner of the yellow house smokes Dunhill.
- the man living in the center house drinks milk.
- the Norwegian lives in the first house.
- the man who smokes blends lives next to the one who keeps cats.
- the man who keeps horses lives next to the man who smokes Dunhill.
- the owner who smokes BlueMaster drinks beer.
- the German smokes Prince.
- the Norwegian lives next to the blue house.
- the man who smokes blends has a neighbour who drinks water.

The question is: who owns the fish? To answer this question one must find
the nationality, colour, pet, beverage and smoke that correspond to
all the houses.

**Figure 3**: Einstein's riddle neighbourhood
To adapt the riddle to the form of our abstract riddle, we define a table
of size 25. First 5 cells contain the nationality of the owners, the next five
the colour of the houses, then the pets the beverages and the smokes. This
means that while we construct a solution, we will first find the nationalities
of all owners, then move to the colours of the houses and so on. The elements
list consists of numbers 1 to 5, with the obvious ordering. We have 5 property
types (nationality, colour, pet, beverage and smoke) with 5 property values each.
We map each property value to a number from 1 to 5. The cells of the table are
filled with numbers from 1 to 5, however there is the restriction that two houses
can't contain the same value for any given property. The details of the implementation
appear in files eisteins_riddle.cpp and einsteins_riddle.hpp.

## Magic Squares

A magic square of order n is a nxn table, containing all numbers from 1
to n² in such a way, that the sum of all rows, columns and primary
diagonals is the same. Algorithms for constructing a magic square of an
arbitrary order exist [1]. There is also an
article in
CodeProject that will do the job. The problem of enumerating
all magic squares of a specific order sounds more challenging
([2]). We will try to use `riddle_solver`

class to enumerate the magic squares of order 3, 4 and 5.

In the concept of the abstract riddle presented above, we are going to
construct the magic square cell by cell, visiting each one of them with
the order presented below.

**Figure 4**: Visiting order for magic squares
The following listing presents the implementation of `check_placement`

routine. Whenever a cell is filled with a number, the routine
checks if a row, a column or a primary diagonal is completed. If this has
happened, it calculates the sum and tests if it is the proper one. The sum
of all rows, columns and primary diagonals of a magic square nxn should be
n[n²+1]/2.

int
magic_square_solver::check_placement()
{
int r, c;
if (moves_count == 0)
return(0);
r = (moves_count-1) / square_size;
c = (moves_count-1) % square_size;
if (c == square_size-1)
{
if (!test_row(r))
return(0);
}
if (r == square_size - 1)
{
if (c == 0)
{
if (!test_reverse_diagonal())
return(0);
}
if (c == square_size - 1)
{
if (!test_diagonal())
return(0);
}
if (!test_column(c))
return(0);
}
return(moves_count);
}

Although this code works well for 3x3 squares, you will notice that it
takes way too long to be of any use for 4x4 squares. To make it faster
we have to enhance `check_placement`

routine in order to detect
erroneous combinations earlier. We achieve this by modifying `check_placement`

,
so that it makes tests for every new cell filled, not just for the cells that
complete a row, a column or a primary diagonal. For every newly filled
cell we calculate the sums of the partially filled row and column it
belongs to. If it is a part of primary diagonal, we also calculate the
sum of the diagonal so far. Then we check, if any of these partial sums
is already bigger from the desired sum or if it is less than the minimum it
should be. If this is the case, we return one cell back. The minimum sum
for a partially filled row/column/diagonal is calculated as follows.
If n-m cells are filled, the partial sum should be equal or greater of
n[n²+1]/2- n² - (n²-1) - ... - (n²-m+1).

int
magic_square_solver::check_placement()
{
int r, c;
if (moves_count == 0)
return(0);
r = (moves_count-1) / square_size;
c = (moves_count-1) % square_size;
if (c == square_size-1)
{
if (!test_row(r))
return(0);
}
if (r == square_size - 1)
{
if (c == 0)
{
if (!test_reverse_diagonal())
return(0);
}
if (c == square_size - 1)
{
if (!test_diagonal())
return(0);
}
if (!test_column(c))
return(0);
}
*else
{
if (c != square_size-1 && !test_partial_row(r, c))
return(0);
if (!test_partial_column(c, r))
return(0);
if (c == r && !test_partial_diagonal(r))
return(0);
if ((c == square_size - r - 1) && !test_partial_reverse_diagonal(r))
return(0);
}*
return(moves_count);
}

With the enhanced `check_placement`

routine you will be able to enumerate
the 4x4 squares in a couple of minutes. Compare the result you will find with that
presented in [2]. Note that if we have a solution,
we can easily find 7 more by means of reflections and rotations. If these
variations of the same solution aren't counted, we must divide our result
with 8.

When we move to 5x5 squares, we notice that even the enhanced version is
inefficient. Actually there are 275305224*8 magic squares of 5-order. If
the algorithm were able to find a solution every 1ms, it would still
require 25 and half days to enumerate all the solutions! No need to get
discouraged though. With a simple trick we can boost the speed of the
algorithm. All we have to do is to change the order we visit the cells:

**Figure 5**: Visiting order for 5x5 magic squares
The `magic_square_5`

class solves the problem of 5x5 magic squares
following the visiting order of the above figure. The class performs only
basic tests, when a row, a cell or a primary diagonal is completed. In my
machine it was able to find about one or two solutions every second. It is
still very slow but I believe that with a few optimisations it could be
tweaked to enumerate all magic squares of order 5 in a decent period of
time.

From this example one can conclude that is very important to catch invalid
combinations as early as possible, even if that means that `check_placement`

routine must be made more complicated and thus more time consuming. The benefit
from limiting the number of combinations that must be tested could be of
much greater importance than the time spent in `check_placement`

.
The same effect (limiting the number of combinations) can be achieved by
changing the order of filling the table.

## Peg Solitaire

This is a classic board game. To win the game you must remove all the pegs
off the board, except of the last one. A peg can move to a free position, two
positions at the left, right, up or down, jumping over another peg. The
peg that was jumped over is removed. You can play the game with a Java applet
you will find in the download section.

To solve this riddle using `riddle_solver`

class, the riddle's
table and the elements list must be defined. The riddle is solved when 31
moves are made, therefore the riddle's table should be of size 31 and contain
all the moves required to win the game. We must also find a way to represent
a move. This is done based on the position of the peg that is moving and
the direction of the move (see Figure 4).

**Figure 6**: Representation of moves
The moves are ordered based on their numeric values, that is a move with a
higher numerical value is tried after a move with a lower numerical value.
The details of the implementation can be found in solo.cpp and solo.hpp files.

## Links

## History

- 3 June 2004. Initial release