Hi, Bean. Straight homework help here. Cool.

Beginning digital logic classes are given three tools -- truth table, Karnaugh

mapping, and Demorgan, right? First thing I did was just draw a truth table.

Inspection shows there are only five cases where F is not true, so I decided to

do this inverted -- that is, solve for what's not true, rather than what's

true.

Let Q = (A! AND B! AND C!)

Let R = (A! AND B! AND D!)

Let S = (A! AND C! AND D!)

Let T = (B! AND C! AND D!)

It should be obvious looking at the table that F! = Q + R + S + T (with A! AND

B! AND C! AND D! being the redundant case which is true for all). Now, you can

Demorganize that, and come up with what you have below, with the last NAND

being used to invert again from F! to F. View in fixed font (like M$

Notepad)...

.---.

A!----| |

B!----| |o----------

C!----|& | |

| | |

'---' |

.---. |

A!----| | | .-----.

B!----| |o------- | | |

D!----|& | | '-- | | __

| | | | | |-----| |

'---' '----- | |o------| |& |o-

.---. | & | ------|__|

A!----| | .------| |

C!----| |o------- | |

D!----|& | .---| |

| | | '-----'

'---' |

.---. |

B!----| | |

C!----| |o----------

D!----|& |

| |

'---'

created by Andy´s ASCII-Circuit v1.22.310103 Beta

www.tech-chat.de
This is pretty similar to factoring quadratics back in high school -- your

ability improves with practice. Try solving problems in different ways, and

just do what works.

Good luck.

Chris