Design an efficient algorithm to sort 5 distinct keys in fewer than 8 comparisons

Question

Design an efficient algorithm to sort 5 distinct - very large - keys less than 8 comparisons in the worst case. You can\'t use radix sort.

Answer 1

Here is C++ implementation which sorts 5 elements in <= 7 comparisons. Was able to find 8 cases which can be sorted in 6 comparisons. That makes sense if we imagine full binary tree with 120 leaf nodes, there will be 112 nodes at level 7 and 8 leaf nodes at level 6. Here is the full code that is tested to work for all possible permutations.

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

ostream& operator << ( ostream& os, vector v )
{
    cout << "[ ";
    for ( auto x: v ) cout << x << ' ';
    cout << "]";
    return os;
}

class Comp {
    int count;
public:
    Comp(): count{0}{}
    bool isLess( vector v, int i, int j ) {
        count++;
        //cout << v << "Comparison#" << count << ": " << i << ", " << j << endl;
        return v[ i ] < v[ j ];
    }
    int getCount() { return count; }
};


int mySort( vector &v )
{
    Comp c;
    if ( c.isLess( v, 1, 0 ) ) {
        swap( v[ 0 ], v[ 1 ] );
    }
    if ( c.isLess( v, 3, 2 ) ) {
        swap( v[ 2 ], v[ 3 ] );
    }
    // By now (0, 1) (2, 3) (4)
    if ( c.isLess( v, 0, 2 ) ) {
        // ( 0, 2, 3 ) (1)
        swap( v[ 1 ], v[ 2 ] );
        swap( v[ 2 ], v[ 3 ] );
    } else {
        // ( 2, 0, 1 ) ( 3 )
        swap( v[ 1 ], v[ 2 ] );
        swap( v[ 0 ], v[ 1 ] );
    }
    // By now sorted order ( 0, 1, 2 ) ( 3 ) ( 4 ) and know that 3 > 0
    if ( c.isLess( v, 4, 1 ) ) {
        if ( c.isLess( v, 4, 0 ) ) {
            // ( 4, 0, 1, 2 ) ( 3 ) ( ... )
            v.insert( v.begin(), v[4] );
            // By now ( 0, 1, 2, 3 ) ( 4 ) ( ... ) and know that 4 > 1
            if ( c.isLess( v, 4, 2 ) ) {
                // ( 0, 1, 4, 2, 3 ) ( ... )
                v.insert( v.begin() + 2, v[4] );
            } else {
                if ( c.isLess( v, 4, 3 ) ) {
                    // ( 0, 1, 2, 4, 3 ) ( ... )
                    v.insert( v.begin() + 3, v[4] );
                } else {
                    // ( 0, 1, 2, 3, 4 ) ( ... )
                    v.insert( v.begin() + 4, v[4] );
                }
            }
            // ( 1 2 3 4 5 ) and trim the rest
            v.erase( v.begin()+5, v.end() );
            return c.getCount(); /////////// <--- Special case we could been done in 6 comparisons
        } else {
            // ( 0, 4, 1, 2 ) ( 3 ) ( ... ) 
            v.insert( v.begin() + 1, v[4] );
        }
    } else {
        if ( c.isLess( v, 4, 2 ) ) {
            // ( 0, 1, 4, 2 ) ( 3 ) ( ... )
            v.insert( v.begin() + 2, v[4] );
        } else {
            // ( 0, 1, 2, 4 ) ( 3 ) ( ... )
            v.insert( v.begin() + 3, v[4] );
        }
    }
    // By now ( 0, 1, 2, 3 ) ( 4 )( ... ): with 4 > 0
    if ( c.isLess( v, 4, 2 ) ) {
        if ( c.isLess( v, 4, 1 ) ) {
            // ( 0, 4, 1, 2, 3 )( ... )
            v.insert( v.begin() + 1, v[4] );
        } else {
            // ( 0, 1, 4, 2, 3 )( ... )
            v.insert( v.begin() + 2, v[4] );
        }
    } else {
        if ( c.isLess( v, 4, 3 ) ) {
            // ( 0, 1, 2, 4, 3 )( ... )
            v.insert( v.begin() + 3, v[4] );
        } else {
            // ( 0, 1, 2, 3, 4 )( ... )
            v.insert( v.begin() + 4, v[4] );
        }
    }
    v.erase( v.begin()+5, v.end() );
    return c.getCount();
}

#define TEST_ALL
//#define TEST_ONE

int main()
{
#ifdef TEST_ALL
    vector v1(5);
    iota( v1.begin(), v1.end(), 1 );
    do {
        vector v2 = v1, v3 = v1;
        int count = mySort( v2 );
        if ( count == 6 )
            cout << v3 << " => " << v2 << " #" << count << endl;
    } while( next_permutation( v1.begin(), v1.end() ) );
#endif

#ifdef TEST_ONE
    vector v{ 1, 2, 3, 1, 2};
    mySort( v );
    cout << v << endl;
#endif
}