Boolean Expression Represented as a Truth Table Example 2

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  • Опубліковано 13 гру 2024

КОМЕНТАРІ • 9

  • @fuzwuzzit
    @fuzwuzzit 2 роки тому +1

    Omg wish my lecturer could explain this as clearly and as interesting as this.

  • @asifirfan5232
    @asifirfan5232 4 роки тому +2

    This topic is really interesting and fun as well

    • @MathsAndStats
      @MathsAndStats  4 роки тому

      Hi Asif, I am glad you liked it. Please share. Regards. Jonathan.

  • @jaysondemarco
    @jaysondemarco Рік тому

    My only question is, how do you know which term to put next? its like so randomized i dont get that part, everything else makes since although

  • @sanjayapiyumal1110
    @sanjayapiyumal1110 6 років тому +1

    Well explained. thumbs up

  • @ahmedsliem5936
    @ahmedsliem5936 4 роки тому +1

    thanks

  • @wesaultman8220
    @wesaultman8220 4 роки тому +1

    i cannot find one with 4 terms

    • @wesaultman8220
      @wesaultman8220 4 роки тому +1

      like a b c d

    • @MathsAndStats
      @MathsAndStats  4 роки тому +9

      Hi Wes. The truth table would contain four parameters, lets call them P, Q, R, and S. for example, lets say the boolean function is:
      F = (P v Q) ^ (R ^ S)
      The truth table would look like the following:
      P Q R S | P v Q | R ^ S | (P v Q) ^ (R ^ S)
      -------------------------------------------------------------
      0 0 0 0 | 0 | 0 | 0
      0 0 0 1 | 0 | 0 | 0
      0 0 1 0 | 0 | 0 | 0
      0 0 1 1 | 0 | 1 |
      0 1 0 0 | 1 | 0 | 0
      0 1 0 1 | 1 | 0 | 0
      0 1 1 0 | 1 | 0 | 0
      0 1 1 1 | 1 | 1 | 1
      1 0 0 0 | 1 | 0 | 0
      1 0 0 1 | 1 | 0 | 0
      1 0 1 0 | 1 | 0 | 0
      1 0 1 1 | 1 | 1 | 1
      1 1 0 0 | 1 | 0 | 0
      1 1 0 1 | 1 | 0 | 0
      1 1 1 0 | 1 | 0 | 0
      1 1 1 1 | 1 | 1 | 1
      I hope this helps. Please share the channel and videos with your class friends.
      Kindest regards.
      Jonathan.