Finite - 16 Tables

    ——————————————————————————————————————————————————————————————————————
    A B ||CON|NOR|NCV| —A|NEG| —B|XOR|NAN|AND|EQV| B |IMP| A |CNV| OR|TAU|
    ——————————————————————————————————————————————————————————————————————
1) 0 0 || 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 |
2) 0 1 || 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |
3) 1 0 || 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
4) 1 1 || 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
    ——————————————————————————————————————————————————————————————————————

We previously made truth tables for OR +, EXCLUSIVE OR Å, AND &, IF ®

            +                 Å                  &                  ®
    A B | OR          A B | XOR          A B | AND          A B | IMP
    _________          _________          _________          _________

0) 0 0 | 0           0 0 | 0           0 0 | 0           0 0 | 1
1) 0 1 | 1           0 1 | 1           0 1 | 0           0 1 | 1
2) 1 0 | 1           1 0 | 1           1 0 | 0           1 0 | 0
3) 1 1 | 1           1 1 | 0           1 1 | 1           1 1 | 1
    _________          _________          _________          _________

What other truth tables can we make with two simple statements A and B?
Here are twelve more AB truth tables to go with our previous four.
The simple statements A and B on the left are called the INPUTS.
The compound statement on the right is called the OUTPUT.

                             <=>
    A B | NOR          A B | EQV          A B | NAND         A B | NEG
    _________          _________          _________          _________

0) 0 0 | 1           0 0 | 1           0 0 | 1           0 0 | 0
1) 0 1 | 0           0 1 | 0           0 1 | 1           0 1 | 0
2) 1 0 | 0           1 0 | 0           1 0 | 1           1 0 | 1
3) 1 1 | 0           1 1 | 1           1 1 | 0           1 1 | 0
    _________          _________          _________          _________

NOR (NOT OR) is the negation (opposite) of OR.
—(A + B) = Nor is it true that Allman shaved himself or the Barber shaved him.

EQUIVALENCE (EQV) is the opposite of XOR.
Its special symbol is <=>, often abbreviated as IFF (if and only if).
A <=> B = Allman shaved himself if and only if the Barber shaved him.

NAND (NOT AND) is the opposite of AND.
—(A & B) = It is false that Allman shaved himself and the Barber shaved him also.

NEGATION (NEG) is the opposite of IF (i.e. IMPLICATION).
One of my students proposed a better term for it: NIF.
—(A —> B) = Allman shaving himself does not imply that the Barber shaved him.

    A B | CNV          A B | NCV          A B | TAU          A B | CON
    _________          _________          _________          _________

0) 0 0 | 1           0 0 | 0           0 0 | 1           0 0 | 0
1) 0 1 | 0           0 1 | 1           0 1 | 1           0 1 | 0
2) 1 0 | 1           1 0 | 0           1 0 | 1           1 0 | 0
3) 1 1 | 1           1 1 | 0           1 1 | 1           1 1 | 0
    _________          _________          _________          _________

CONVERSE (CNV) is the reverse of implication, not the negation.
B -> A = If the Barber shaved Allman, then Allman shaved himself.

CONVERSE NEGATION (NCV) is the negation of the converse.
-(B -> A) = The Barber shaving Allman does not imply that Allman shaved himself.

A TAUTOLOGY (TAU) is always true no matter what the inputs A and B.
(A + -A) + (B + -B) = Allman may or may not have shaved himself,
                      or the Barber may or may not have shaved him.

A CONTRADICTION (CON) is always false no matter what the inputs A and B.
(A & -A) & (B & -B) = Allman shaved himself and he did not shave himself;
                      the Barber shaved Allman and the Barber did not.

    A B | A           A B | —A           A B | B           A B | -B
    _________          _________          _________          _________

0) 0 0 | 0           0 0 | 1           0 0 | 0           0 0 | 1
1) 0 1 | 0           0 1 | 1           0 1 | 1           0 1 | 0
2) 1 0 | 1           1 0 | 0           1 0 | 0           1 0 | 1
3) 1 1 | 1           1 1 | 0           1 1 | 1           1 1 | 0
    _________          _________          _________          _________

The last 4 truth tables are simply A or B or the negation of A or B.
Each says that the output is independent of one of the inputs,
and totally depends on the other.
A = Allman shaving himself depends only on Allman, not on the Barber.
B = The Barber shaving Allman depends totally on the Barber,
    it doesn't matter what Allman does.

QUESTIONS:

1. How many rows are there in a truth table with 3 ABC inputs?
2. How many rows are there in a truth table with 4 ABCD inputs?
3. How many rows are there in a truth table with any number of inputs?
Let n represent any number.
4. Do these questions relate to any previous problems? Why?

Send me your answers to questions 1 thru 4 privately.
They are due in two weeks.
Discuss 5, 6, and 7 here on the wiki.
They are due by the end of the course,
and will be scored at the end of the course.

5. How many truth tables are possible with 3 ABC inputs?
The first to post the correct answer earns 2 points.
Explaining the answer earns 1 point.
Offering an alternative explanation earns 1 point.
Any enlightened comment earns 1 point.

6. How many truth tables are possible with 4 ABCD inputs?
The first to post the correct answer earns 2 points.
Explaining the answer earns 1 point.
Offering an alternative explanation earns 1 point.
Any enlightened comment earns 1 point.

7. How many truth tables are possible with any number of inputs?
Can you write a formula using n for the number of inputs?
The first to post the correct formula earns 2 points.
Explaining the formula earns 1 point.
Offering an alternative explanation earns 1 point.
Any enlightened comment earns 1 point.

I will not tell you if you are right or wrong about questions 5, 6, 7 until the end of the course.
Remember: the formula you derive must also work for truth tables with only 1 or 2 inputs.
You are on your own. Make it a team effort.

ABBREVIATIONS

CON = CONTRADICTION (ALWAYS FALSE)
NOR = NOR (NOT OR)
NCV = NEGATION OF THE CONVERSE
~A = NOT A
NEG = NEGATION OF IMPLICATION
~B = NOT B
XOR = EXCLUSIVE OR
NAN = NAND (NOT AND)
AND = AND &
EQV = EQUIVALENCE
B    = B
IMP = IMPLICATION
A    = A
CNV = CONVERSE
OR = OR +
TAU = TAUTOLOGY (ALWAYS TRUE)

Comments:

5. How many truth tables are possible with 3 ABC inputs? 256
6. How many truth tables are possible with 4 ABCD inputs? 65,536
7. How many truth tables are possible with any number of inputs? the formula is 2^2^n

So I found a website (http://www.qhull.org/ttcnf/) that goes through it all about 1/3 way down the page.

My previous logic that included the 12+2n logic (12 fixed tables and then 2 (+/-) tables that depend on how many inputs (n) are included still makes sense to me, but I'm okay with saying that it's wrong.

I'd really like to see all 256 truth tables for ABC inputs because I can't visualize it, although it makes sense why there would be that many.

DirtyBird -

Read my comment to PAC
and others' comments as well.

5. How many truth tables are possible with 3 ABC inputs?
18

6. How many truth tables are possible with 4 ABCD inputs?
20

7. How many truth tables are possible with any number of inputs?
TT=12+2n

Everytime you add more to the problem there will be an extra 2 tables to add to it.

Pac -
You left out the table that looks like this:

1
0
0
0
1
0
0
0

Plus a few more tables as well...

I've used the Truth Table Constructor applet to build the 18 truth tables that would go along with ABC inputs.

    A B C ||CON|NOR|NCV| —A|NEG| —B|XOR|NAN|AND|EQV| B |IMP| A |CNV| OR|TAU| -C| C |
    —————————————————————————————————————————————————————————————————————-----------
1) 0 0 0 || 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 |
2) 0 0 1 || 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 |
3)  0 1 0 || 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 |
4)  0 1 1 || 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |

5)  1 0 0 || 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 |
6)  1 0 1 || 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 |
7)  1 1 0 || 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 |
8)  1 1 1 || 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 |

The 3 questions in this problem do not ask you to name each of the extra tables.
Counting is sufficient.
However, you are not restricted in your discussions to any specific ideas
as long as they refer back to the questions.
You may just come up with something no one ever thought of before.

Ok this is a group activity so we will do it together. Some of you posted 256 for three inputs. That is correct but now it's time to explain it.

Each table will have a unique output combination. This output combination is made up of the eight possible true/false answers. There are 256 unique combinations that these answers can form. This is because true/false = 2 choices eight times = 2^8 = 256 Ok now that you've got that; you might be saying something like " yeah but what are the formulas for the 256 tables?". The formula for any table is whatever will make the output combination. If you wrote two different table formulas Example: ~(A&B&C) or the table (~A) + (~B) + (~C). Both of these tables have the same output and therefore are the same table. any equation you write will have one of the 256 possible table outputs. obviously if the equation you write has the same output as a different equation, they are both the same equation once reduced to the simplest common equation. I hope this is making sense to yall. If it is, you should be seeing how that any infinite number of equations you may write, when reduced to their simplest form, will only be a maximum of 256 tables.

some of the rest of you post about four inputs or five or ten. Its all right here in front of you.

That table would be ~(B&C). Your table also implies compound formulas. Based on that, I have a theory on how many tables there could be.

Thinking . . .

BAM!

I have just solved this . . . Eureka has struck!!! . . . I will post it as soon as I get the math done. ( I have to go to work now.)

PS. somebody better hurry up and post it before tomorow morning or I am going to snatch up those points!

Thanks, Superduke for making us some tables.
What name would you give this one, if you had to give it a name?

A B C output
0 0 0 1
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 0
1 1 1 0

As you can see I haven't filled in any of the 4 kinds of implication tables. A implication table would be " if A then B" with 2 inputs. Does that mean it would be "If A then B, If B then C"? Or does it mean "IF A then B and C"? Or does it mean "If A then B or C"? Or does it mean "IF A and B then C"?

As you can see I think this is an invalid type of table for this exercise. Anyone have a different theory?

Most of you agree that there are 18 truth tables with 3 inputs,
but no one has actually shown me the 18 truth tables.
All I want to see are the 18 output columns. What do they look like?
How many rows will there be in each truth table?
In each of the 16 tables, there are 4 rows.
Will your 18 truth tables each have 4 rows also?
Enquiring minds want to know...

Ok here's the deal. I am going to fill in the ones that can be made to work. If one of you see's one that you think will work that is not on here then post that table with your proof and I will update this table. If you see one that we put on here that is incorrect post the proof as to why it is wrong and I will remove it. We should be able to resolve this if we work together using a community table. Let's all pitch in here please.

A	B	C	and	or	Xor	imp	Nor	eqv	nand	neg	cnv	ncv	tau	con	a	~a	b	~b	c	~c
0	0	0	0	0	0		1	1	1				1	0	0	1	0	1	0	1
0	0	1	0	1	1		0	0	1				1	0	0	1	0	1	1	0
0	1	0	0	1	1		0	0	1				1	0	0	1	1	0	0	1
0	1	1	0	1	0		0	0	1				1	0	0	1	1	0	1	0
1	0	0	0	1	1		0	0	1				1	0	1	0	0	1	0	1
1	0	1	0	1	0		0	0	1				1	0	1	0	0	1	1	0
1	1	0	0	1	0		0	0	1				1	0	1	0	1	0	0	1
1	1	1	1	1	0		0	1	0				1	0	1	0	1	0	1	0

5. How many Truth Tables are possible with 3 ABC inputs? 18

6. How many Truth Tables are possible with 4 ABCD inputs? 20

7. How many Truth Tables are possible with 'n' inputs? TT=12+2n

For every element that is added to the 12 original truth tables, you have to have a positive and negitive side of the element included with it.

ex: if i were to add C to the 12 original truth tables i must have a +C and a -C to prove the truth table correct. If I added C & D i would have to have a +C & -C and also a +D & -D.

the formula to figure out how many truth tables you would need if you had ' n ' elements, you would have to use TT= 12+2n

When I suggested you list your tables, I meant list the 1 and 0 patterns in each table.
The names of the tables dont mean anything unless you define them.

Question: what would an IMP table look like with 3 variables?
We know what it means using 2 variables?
But does it have any meaning with 3?

THIS PAGE WILL NOT CLOSE TILL DECEMBER 15TH

KEEP DISCUSSION GOING IF YOU WANT 3 POINTS

WHOEVER PRESENTS AND EXPLAINS THE CORRECT ANSWERS
EARNS BONUS POINTS!

To list less than 20, I said 18 for 3 inputs and came up with OR, XOR, AND, IMP, NOR, EQV, NAND, NEG, CNV, NCV, TAU, CON, A, -A, B, -B, C, -C for a total of 18.

I came up with 13 tables for 3 inputs and 15 tables for 4 inputs.

Golfgirl,

how did you come up with these answers and this equation? Where did this 7 come from? And are you getting the number of inputs confused with the number of truth tables? I cannot see how there would be 256 truth tables with 3 inputs? Or are you trying to say that there would be a final count of 256 inputs numbers throughout the tables? I am really confused because these numbers are way to large I think. Aren't we supposed to be figuring out the number of truth tables and not the inputs?

#5. How many Truth Tables are possible with 3 ABC inputs? 18

#6. How many Truth Tables are possible with 4 ABCD inputs? 20

#7. How many Truth Tables are possible with 'n' inputs? 2n + 12

When you add a new input then you are only adding a new truth table with the new input and the negation of the input. For example if you have ABC inputs then the only new truth tables that you would add would be c and -c for the positive input and the negation of that input. Also the same with the ABCD inputs you would start off with 16 truth tables with 2 inputs then add c and d and -c and -d which would add 4 new truth tables to the existing 16 and make 20 truth tables. You can use the 2n + 12 to figure out the equation because when you have 2 inputs you end up with 16 truth tables. So 2(2) +12= 16 truth tables, the same with 3 inputs. The equation would go 2(3) + 12= 18 truth tables. Which proves my answers for #'s 5 and 6. you would get the number 12 because if you had no inputs then there would still be 12 truth tables, because with 2 inputs you have 16 truth tables which include A,B,-A and -B. So if you had no inputs then you would not have those 4 truth tables so you would still have 12 truth tables. The n comes from the number of inputs you have the the 2 before the n comes from the 2 different truth table the positive input and the negation of that positive input. Like A and -A.

5. How many truth tables are possible with 3 ABC inputs?

** 256 inputs because 2^(n^2)

6. How many truth tables are possible with 4 ABCD inputs.

** 65,556 because 2^(n^2)

7. How many truth tables are possible with any number of inputs? 7+(n*2)

i have read everyones statements and studdied on this for days and the only thing i can come up with is for # 5 there are 256 inputs and #2 ther are 4096 and #3 the formula is 2xnx2 but i am probably wrong.

5. How many truth tables are possible with 3 ABC inputs?

With 3 inputs there are 4 rows and 256 inputs.

6. How many truth tables are possible with 4 ABCD inputs?

With 4 inputs there are 16 rows and 65,536 inputs.

7. How many truth tables are possible with any number of inputs?

With n inputs, we have 2 x N rows and 2xNx2 inputs.

5. How many truth tables are possiblewith 3 inputs? 13 ( CON, NOR, ~A, ~B, XOR, NAN, AND, B, A, OR, TAU, ~C, and C.

6. 15 All of those from above plus ~D and D.

7. n= 7 + (N*2)

All i can figure out is the same thing that Boki posted 1st. After racking my brain. This stuff is still very much greek to me. but anyway.

5. = 256 because 2^(n^2)

6. = 65536 because 2^(n^2)

7. = any depending on what you make N equal use the same formula 2^(n^2)

This stuff is some what over my head. I hope these are right.

n² is a number of rows. So, if n=3 (3 inputs) we will have 2^3=8 rows. Multiplying rows by columns we have all possible inputs which are 24. Each row has at least 1 output (if we use simply AND, OR, or NOT), so we will add 8 outputs to 24 inputs which makes 32 all together. If we have 4 inputs, we will have 16 rows and all possible combinations will be 64. Adding to this 16 outputs, we will have 80 all together.

I've been staring at these postings for about an hour straight, and I can't seem to come up with anything. I don't quite understand this. I'll keep working on it though.

If your count is less than 20, list your tables to prove your conclusions.

So SuperDuke, in response to your comment about combining inputs in order to get to an output which makes the particular table (if and only if I think was your example), wouldn't you just move from left to right - first assess A and B and then B and C? That's how I'd do it, but I'm not willing to stake my life on it being right.

I figured up the answers on my own to 5, 6, and 7 but I'm not the first to post what I think the answers are... Its been difficult for me to stop overthinking things so far this course; hopefully I'm right.

With 3 inputs (ABC) - there are 12 "fixed" truth tables, then there are the ones that are dependent upon how many inputs you have (A, -A, B, -B, etc.). So, with 3 inputs, you'd have 18 truth tables.

With 4 inputs (ABCD), the same as three holds true. You have your 12 "fixed" tables and then you have to add A, -A, B, -B, C, -C, D, -D. That gives you 20 because you just took what you had from above (18 tables from A, B, C inputs, and added D (2 extra tables for D and -D).

With n inputs (ABCD....MNOP....whatever), it gets pseudo tricky :-) You have 12 fixed plus 2 for each input. That's 12+2n. So, with all of the letters of the alphabet, you've got 26 inputs. That's 12+(2*26) or 64 tables. This formula even holds true with one input (A) - you've got 12, then A and -A, which is 14. The formula 12+(2*1) also equals 14!

Hotrod - good try at listing the inputs. now how many outputs are possible from those inputs?
p.s. row 3 and row 8 are the same. (I counted from 0)

Okay guys, I've come up with something totally different. In the question he asked for ROWS not COLUMNS which to me means that with 3 inputs you would get 9 rows, with 4 inputs you would get 16 rows and with n number of inputs you would get n² number of rows. See graphs below.

A B C A B C D
1 1 1 1 1 1 1
1 1 0 1 1 1 0
1 0 0 1 1 0 0
1 0 1 1 0 0 0
0 1 0 1 1 0 1
0 1 1 1 0 0 1
0 0 1 1 0 1 0
1 0 1 1 0 1 1
0 0 0 0 1 1 1
0 1 0 1
0 1 1 0
0 1 0 0
0 0 1 0
0 0 1 1
0 0 0 1
0 0 0 0

5. How many truth tables are possible with 3 ABC inputs?
13
A, B, C, -A, -B, -C, tau, nor, nand, and, xor, or

3 inputs limit the amount of statements being used; add the number of statements and the number of inputs (both positive and negative) and you will have the total number of tables. Conditionals are limited to two inputs and should not be counted.

6. How many truth tables are possible with 4 ABCD inputs?
15

the same rules apply as in number five. Take the number of statements, take out the conditionals that will only work with 2 inputs and add the number of inputs (both positive and negative).

7. How many truth tables are possible with any number of inputs?

This is the same rule as 5 and 6.
Take the number of statements minus the conditionals; add number of inputs both negative and positive.
You start with the number of statements left which = 7
N = inputs (we multiply by 2 to give both negative and positive values).

7 + (n*2) = number of tables

Sorry Superduke and everyone else. I still do not have a grasp on this stuff and what you have said thus far is pretty much Greek to me. Maybe I will get my brain wrapped around this stuff then I can help. I spend hours trying to figure this out than doing the actual homework.

Hey Guys,

On number 5, Is the relation of c to a the same as the relationship of b to a.
ie a&b&c, a&b or c

Thanks, Superduke, for encouraging everyone to join the dance.

Here is a question:
if the 16 tables include an IF table, and also an IF AND ONLY IF table,
should you count IF tables with 3 inputs in your count?

Notice that each of the 16 tables is unique
and represents one possible set of outcomes for 2 inputs.
Also, together the output from the 16 tables represent ALL possible outputs from 2 inputs.
When making your set of tables with 3 inputs,
make sure you have included all possible outputs from 3 inputs.

Oh here's another quick point. If you have 1 input you would have 14 tables not sixteen. B and -B would not exist. 2^(2^1) = 4 INVALID

16^n-1 for 1 input would be 16^1 -1 = 15=INVALID or if you meant 16^(1-1) it would = 16^0 which would be 0=invalid

why do i feel like I am talking to myself here? comon people tell me what you guys are thinking. I am think some of you are looking at these questions and just have not posted yet. He said "group" here so why don't you post what you have come up with so far. Or what direction you are taking in trying to reach you solution. The home page clearly states you will not be shot!

PLEASE

Disproving here.

A couple of you say 265 for two actions between 3 inputs. That would be incorrect. you base that on 16 actions with 16 additional actions to have with each of those. thats right 16x16= 256.

Its not 16, its 18 now..... look at your last 4 tables above, now add c and add -c.

its not 4096 for 4 inputs either because you would have 20 tables not sixteen. 16x16x16=4096

its not 65636 either "16 x 16 x 16 x 16 = 65536" invalid your formula dosent account for the extra tables either.

When you guys plug in your formulas they give you your numbers. The problem is the math that let you arrive at those numbers to start with was wrong.

Also your formulas exclude the notion of using one action. Why are we stuck on "A {action} B {action} C" wouldn't a "and" or a "or" apply to all three or four or a gillion by itself. If thats the case your formulas are invalid on those reasons also.

You guys are probably getting tired of me by now but I am trying to demonstrate how that you cannot apply these formulas because you are missing rules. The information is not here, we cannot simply make assumptions about what is and what is not valid. If you can disprove it it is invalid.

Let me ask you all a question. Of the sixteen truth tables how many of those tables could be applied to 3 inputs? ( don't assume here and don't start combinding actions, just answer what is asked)

Now as you go thru the tables you will see some could answer all three inputs, but some can't. you can't apply a "if then" to abc because you must tie 2 of them together somehow to make that work. So an implication table would not be a valid table. ( remember now guys we are talking about only these tables... we can't just assume that we can add some other stuff in to make this work... assumptions lead to incorrect theories)

Now look at the last 4 tables A, -A, B, -B. Ok these tables are based on the inputs A and B. By adding more inputs we increase the table count. so for ABC we must now include C, -C. I did not assume this, I must add these tables because they are clearly Valid and required.

Now tell me how many tables for 3 inputs? how many for 4?

1. How many truth tables are possible with 3 ABC inputs? 256

2. How many truth tables are possible with 4 ABCD inputs? 4,096

3. How many truth tables are possible with any number of inputs? 16^n-1

Good - a dialog!

To test your ideas, or disprove another's, make a list with n=1, or n=3.
You already know the results for n=2.
See if you can get a pattern going.

Boki,

I see now you are combinding actions. Ex: A + (B & C). I am sorry friend but I cannot see where this puzzle indicates that multiple actions can be used. I have not seen Mr. Holt give an example of two actions combined on a table in this lesson. Also the variables don't work out in your formula. look at the last four tables he gives. if you are combinding them for this lesson you could have -a & b or you could have -a +b or you could have -a + B -c ->a. The point I am trying to get across here is you are saying two actions for three inputs----- ( where did you get that idea?) If I can use two I can use ten. If I use ten in one table and in all the different ways they can interact we just went into the MILLIONS.

YOU might say hog wash but let me give you an example: (-(a + b ) & a) -> ((c & a) -b) .

In order for 256 to be correct, examples like I just gave cannot be included because the interaction between them can be reaaranged into many many ways that goes far beyond 256.

But I believe there would be a pattern there also and a formula could be wrote. I will give you this formula tommorow. None the less I respectfully declare your solution outside the scope of the lesson.

PS- I am defending my position zealously because I feel I have a grasp of this and I am correct. I am not trying to be confrontational and I am not trolling. Please look mine over and if you see a flaw or inconsistency that would rule it out please post it.

Anyone else want to weigh in on this.

SuperDuke,

You need to use this formula 2^(2^n).

We have n=3, then 2^(2^n)=2^(2^3)=2^8=2*2*2*2*2*2*2*2=256.

Boki,

Could you show a comment on how you arrive at 256 for 3 inputs. I am assuming you are including compound statements ( combined functions ) to arrive at this number. If this is the case wouldn't the number of tables be a MUCH MUCH larger number?

Everybody,

My comprehension of this puzzle is "how many tables can you make from these sixteen" with 3 inputs instead of two. He said group effort: can I get some comments here?

5. How many truth tables are possible with 3 ABC inputs?

For n=3 (number of inputs) we will have 256 truth tables (or different functions).

6. How many truth tables are possible with 4 ABCD inputs?
For n=4 (the number of inputs) we will have 65536 truth tables (or different functions).

7. How many truth tables are possible with any number of inputs?
With any number of inputs (for the n inputs) we can find the number of truth tables (different functions, or different ways we can assign true and false) using this formula:

2^(2^n).

With n variables, we have 2^n rows and a total of 2^(2^n) different functions.

#5. 13 . . . Of the Sixteen tables only eleven of the statements can be applied to 3 inputs. Additionally you would have two new tables of "C" and "-C". The total of thirteen are: or , xor , and , nor , nand , tau , con , A , B , C , -A , -B , -C .

These statements work because they are not limiting or conditional in any way. Example: if you say "and" in a AB table it would be A & B. In a ABC table it would be A & B & C. Or could be A or B and just as easily in a ABC table it would be A or B or C.

You will notice I excluded the five statements which are conditional. The reason for this is because they are limited to 2 parts. If this then that. If that then not this. Not this if not that. you get the picture.

They cannot describe 3 items, it would take a compound statement to do this. Example: If A then (B and C). As you can see I combined two actions, the "if" and the "and" to make this statement.

#6. with four inputs we would have the same 13 tables plus we would have d , -d . for a total of 15 tables.

#7. Here's the neat part the exact same rules apply even if it was 100 inputs. Example: in a table with 6 inputs of ABCDEF: if it was a "nor" table it would be A NOR B NOR C NOR D NOR E NOR F.

You would have 7 tables plus the number of inputs X2. ( remember for each input you have 2 possible tables such as A , -A ) So if N=inputs the formula would be: 9 + N x 2 = # of tables