Harvard Chart Method

Harvard Chart Method of Logical Equation Reduction was developed to address the need to automate the process of logical equation reduction in the early days of computer hardware and switching circuit development. Large scale production of computer circuitry entailed many more variables than could be reasonably handled by manual methods using Boolean logic, Venn diagrams, etc. Automated methods of logical equation reduction were necessary to minimize logic circuits and thereby reduce the number of logic gates which used vacuum tubes as switches due to their relatively high cost and excessive thermal emission.

The Harvard Chart Method of logical equation reduction is capable of reducing binary logical equations having five or more variables to minimum form. As the number of variables increase, the number of operations required to minimize a logical equation increases exponentially, placing practical limits on performing logical equation reduction by hand. The Harvard Chart Method was developed to solve this problem by using a computerized method to automate the process of logical reduction. The method represents one of the first instances where computers were able to assist in the design of the circuits of which they are made. Application of this method is limited only by the logical speed and size of the computer (or computer network) on which it is run. A method based on the Harvard Chart Method has since been developed to reduce multi-valued logical equations to minimum form.

Logical Equation Reduction, Simplification or Minimization
 (note this chart uses upper case to represent the logic state of "TRUE" and lower case to represent the logic state of "FALSE" 

Lets simplify the following equation:

f = ABc + ABC + aBC + aBc + AbC

The Harvard Chart (style 2)

 * {| class="wikitable" style="text-align:center"

! 1 || 2 || 3 || 4 || 5 || 6 || 7 || ! A  ||  B  ||  C  ||  AB  ||  AC  ||  BC  ||  ABC ||
 * +The Harvard Chart
 * a || b || c || ab || ac || bc || abc || Row1
 * a || b || C || ab || aC || bC || abC || Row2
 * a || B || c || aB || ac || Bc || aBc || Row3
 * a || B || C || aB || aC || BC || aBC || Row4
 * A || b || c || Ab || Ac || bc || Abc || Row5
 * A || b || C || Ab || AC || bC || AbC || Row6
 * A || B || c || AB || Ac || Bc || ABc || Row7
 * A || B || C || AB || AC || BC || ABC || Row8
 * }
 * A || b || c || Ab || Ac || bc || Abc || Row5
 * A || b || C || Ab || AC || bC || AbC || Row6
 * A || B || c || AB || Ac || Bc || ABc || Row7
 * A || B || C || AB || AC || BC || ABC || Row8
 * }
 * A || B || C || AB || AC || BC || ABC || Row8
 * }
 * }

The Reduction Steps

 * 1) Draw a line through all rows whose terms are not contained in the expression being simplified (rows 1,2 and 5).
 * 2) Starting with the left column (column 1) cross out all terms which have been lined out in step 1. (a is lined out in rows 1 and 2, and A in row 5; thus, all terms are lined out in the left column of this example.)
 * 3) In column 2 only b (lower case) is eliminated. Circle all B's for easy identification as a part of the final answer.
 * 4) Going to the right, cross out all terms containing B in all the rows with a circled B. For example, in row 4 the terms AB, BC, and aBC are lined out.
 * 5) Continue with lines 3 and 4.
 * 6) In column 5 the term AC is not crossed out, so it must be circled.
 * 7) Going to the right in rows containing AC, cross out all other terms containing AC.
 * 8) All terms in columns 6 and 7 (column BC and ABC) are now crossed out and the process ends. Only B and AC are left.


 * Therefore the answer is: f = B + AC

External Resources

 * Equation Reduction using the Harvard Chart Method