??? 02/25/07 03:37 Modified: 02/25/07 05:27 Read: times |
#133651 - One solution... Responding to: ???'s previous message |
Ok, I wrote:
Kai said:
Means, if we want to decide the status of one signal, it should be enough to ask, what the two others do (without differentiating between them!) in correlation to what all the three do (again not differentiating between them!). But if we cannot differentiate between the individual signals, the only that we have, finally, is the number of "1" ("Ones") in the set of input signals A,B,C. So the truth table, which looks like A B C !A ------------- 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 (if we focus on only !A, for the sake of clarity), tells us: !A = zero"1" + one"1" & (B+C) + two"1" & (B&C) In words: !A is to be set to "1", if no "1" are in the set of input signals A,B,C, or if one "1" is there and at the same time either B or C is "1" or if two "1" are there and at the same time B and C together are "1". So, we should create help signals "S0" and "S1", which binarily count the number of "1" in the set of input signals A,B,C, as follows: A B C !A S1 S0 --------------------- 0 0 0 1 0 0 0 0 1 1 0 1 0 1 0 1 0 1 0 1 1 1 1 0 1 0 0 0 0 1 1 0 1 0 1 0 1 1 0 0 1 0 1 1 1 0 1 1 We see immediately that S1 = A&B + A&C + B&C and, as "S0" is "nearly" !S1: S0 = !S1 & (A+B+C) + A&B&C This means, that !A becomes: !A = !S0 & !S1 + S0 & !S1 & (B+C) + !S0 & S1 & B & C With this scheme only two inverters are needed, to invert the help signals "S0" and "S1". The whole solution becomes then: !A = !S0 & !S1 + S0 & !S1 & (B+C) + !S0 & S1 & B & C !B = !S0 & !S1 + S0 & !S1 & (A+C) + !S0 & S1 & A & C !C = !S0 & !S1 + S0 & !S1 & (A+B) + !S0 & S1 & A & B This would be the circuit for !A: The upper ten gates generate the help signals S0, !S0, S1 and !S1, while the lower five gates generate !A. Again five gates had to be added for the generation of !B and again five gates for !C. How to save one inverter... Kai |
Topic | Author | Date |
weekend brainteaser | 01/01/70 00:00 | |
here goes my weekend. | 01/01/70 00:00 | |
Dont worry once you see the answer its obvious. | 01/01/70 00:00 | |
Logically thinking... | 01/01/70 00:00 | |
you are quite close | 01/01/70 00:00 | |
One solution... | 01/01/70 00:00 | |
yey we have a winner | 01/01/70 00:00 | |
Very Nice | 01/01/70 00:00 | |
Thanks | 01/01/70 00:00 | |
An Alternate Solution | 01/01/70 00:00 | |
Yes... | 01/01/70 00:00 | |
Ive run out ;-( | 01/01/70 00:00 | |
Congratulations, Kai! | 01/01/70 00:00 | |
Thanks | 01/01/70 00:00 | |
Now it is clearer... | 01/01/70 00:00 |