Search problem:The m-tuple satisfying the above condition is called a solution. The term compatibility was first introduced by Carter in 1974 [9]. This version is restricted because the compatibility function is only defined on ordered pairs rather than on all k-tuples,
Given a collection of sets of candidatesand a boolean compatibility function
defined for all
and
, find an m-tuple
with
such that
is true for all
.
.
 and let 
 be the set of all
candidates for column i of the incidence matrix of a projective plane,
then 
 can be defined as

where 
 denotes the inner product of columns x and
y. A solution is then a complete incidence matrix.
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In a backtrack approach, we generate  k-tuples with  
. A
 k-tuples
 is a  partial solution at level k
if 
 is true for all 
. The basic idea of the
backtrack approach is to  extend a partial solution at level k to
one at level k+1, and if this extension is impossible, then to go back
to the partial solution at level k-1 and attempt to generate a different
partial solution at level k.
 is
taken as the root, and the partial solution 
 is
represented as the child of the partial solution 
.
Following the computer science terminology, we often called a partial
solution a  node. It is often true that the computing cost of
processing a node is independent of its level k. Under this assumption,
the total computing cost of a search is equal to the number of nodes in
the search tree times the cost of processing a node.
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choices for the first column, corresponding to the number of ways of
placing 11 ones in the 111 rows. By using the row permutations, we can
assume that all these ones are placed on the first 11 rows, reducing the
number of partial solutions 
 from 
 to 1.
Mathematicians love to use the phrase ``without loss of generality'' to
indicate a simplification by symmetry operations. So without loss of
generality, the second column has only one choice --- with a one in the
first row and the remaining 10 ones in rows 12 to 21. Now, row 1 has nine
remaining ones. By permuting columns, we can assume that these remaining
ones of row 1 are in columns 3 to 11. Next, by row permutation, the
remaining 10 ones of column 3 can be placed in rows 22 to 31. Continuing
in this manner, it is not difficult to show that there is only one choice
up to column 21. Beyond this point, symmetry operations are difficult to
visualize, because they often involve combinations of row and column
permutations.
 nodes, or about 
 nodes per second.
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This was the situation when we entered the picture. While we knew
, the search for weight 16 codewords was about three-quarter
done and the search for weight 12 codewords was presumed to be too
difficult. The person who suggested the problem to us was John Thompson.