Re: Algorithm for cancelling common factors between two lists of multiplicands

in reply to Algorithm for cancelling common factors between two lists of multiplicands

The same exampled worked through:

@a = ( 10, 20, 33, 45, 60 );

@b = ( 2, 5, 10, 12, 16, 23, 45 );

Eliminate common elements:

@a = ( 20, 33, 60 );

@b = ( 2, 5, 12, 16, 23 );

Cancel 2 with 20 to give 1 and 10 (discard the 1):

@a = ( <strike>20</strike><ins>10</ins>, 33, 60 );

@b = ( <strike>2</strike><ins>1</ins>, 5, 12, 16, 23 );

Cancel 5 with 10 to give 1 and 2:

@a = ( <strike>10</strike><ins>2</ins>, 33, 60 );

@b = ( <strike>5</strike><ins>1</ins>, 12, 16, 23 );

Cancel 2 with 12 to give 1 and 6 (discard the 1 )

@a = ( 2, 33, 60 );

@b = ( 12, 16, 23 );

Cancel 6 with 60 to give 1 and 10 (discard the 1 )

@a = ( 33, 60 );

@b = ( 6, 16, 23 );

Cancel 16 with 10 to give 8 and 5

@a = ( 33, 10 );

@b = ( 16, 23 );

Result: (Other reductions are possible!)

@a = ( 33, 5 );

@b = ( 8, 23 );

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