I would like to know how can I convert the following boolean function into a truth table and accordingly construct the k-map


thanks in advance :)

  • \$\begingroup\$ Any particular part? \$\endgroup\$ Apr 5, 2014 at 2:34
  • \$\begingroup\$ @IgnacioVazquez-Abrams just to construct the truth table and if there is a way to construct the k-map without the truth table \$\endgroup\$ Apr 5, 2014 at 2:38
  • \$\begingroup\$ Did you try putting 1s in the intersections? \$\endgroup\$ Apr 5, 2014 at 2:48
  • \$\begingroup\$ No there are still some missing pieces of the puzzle. I gonna got some sleep now. thanks for the support \$\endgroup\$ Apr 5, 2014 at 2:53

2 Answers 2


To construct the truth table, you need to manually assess each combination. A table works well, hence the name "truth table"! I assume you understand logical ANDs and ORs, to make sense of this answer.

First, you want to solve each ANDed group separately. Boolean algebra has the same order of precedence as standard algebra, with AND treated like multiplication, and OR treated like addition. Put these answers in a table. Don't worry, I'll attach a picture to demonstrate. Once you have all of these statements figured out, then you can OR them together. Follow the red lines on the following table:

Truth table

Now that the table is completed, you can build a map. One of the standard configurations is shown below. You have two bits defining the columns, and the other two bits defining the rows. Find the square that intersects the binary inputs (A, B, C, and D), and fill in the answer from your truth table. I've done two of them, in Purple and Orange:

enter image description here

I'll leave the rest for you! You didn't ask how to solve the K-Map. I assume you know how?

Take care!

(P.S. I've included a typo in the truth table. Can you find it?)

  • \$\begingroup\$ The typo is that 0111 should be 0 NOT 1 \$\endgroup\$ Apr 5, 2014 at 11:50
  • \$\begingroup\$ Can I ask you which software did you use to draw the truth table? \$\endgroup\$ Apr 5, 2014 at 11:52
  • \$\begingroup\$ Excellent :) I used good old Excel, and typed it all by hand. I normally would just write it on paper, but I wanted to post it online :) \$\endgroup\$
    – bitsmack
    Apr 5, 2014 at 13:54
  • \$\begingroup\$ Again, thank you mister for your time and effort. God bless you :) \$\endgroup\$ Apr 5, 2014 at 15:31

To draw truth table with four input A,B,C,D and output F

Output will be at logic 1 at

A'B'C'(000x) => 0000 & 0001

B'CD' (x010) => 0010 & 1010

A'BCD' (0110) => 0110

AB'C' (100x) => 1000 & 1001

For all other states output will be at logic 0

  • \$\begingroup\$ Perhaps you should explain why you set the output to 1. \$\endgroup\$ Apr 5, 2014 at 3:40
  • \$\begingroup\$ @user3490561 let me know if you need any clarification in any part of the answer. \$\endgroup\$ Apr 5, 2014 at 3:49

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