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Water Logic

LOOLOOLO
WATER LOGIC

Origin of the Game of Pipes
The Game of Pipes originated within a remote island culture of the South Pacific. Natives to that culture call themselves Looloolo, and their island home Looloololo. In the Looloolo language, a looloo is a pipe and lolo means land. As you know, emigrants from Looloololo have established a chain of unique pizza parlors in their newly adopted countries. As you also know, the Looloolos are an exceptionally logical race of people, but you may not know that their craftsmen are also outstanding water mechanics. We would call them hydraulic engineers. They have managed over the past millenia to capture energy from their rivers and tides, and create intricate machines that control the waters falling from their mountains or flowing onto their beaches.

Over the centuries, Looloolos have developed water systems equivalent to our AND OR NOT logic. Firing up their solar powered pizza ovens, they bake ceramic pipes containing trap doors and bridges. If water flows thru pipe A or B, the water pressure is heavy enough to lower a bridge, and the water flows on; otherwise the water stops flowing, and not a drop flows thru to the other end.

In the AND structure, both A and B pipes must contribute water, or both bridges will not fall allowing the water to pour out on the right side.

AND00

In the OR structure, if either A or B pipes contribute water, a bridge will fall and water will pour out the right side.
OR00

In the NOT structure, if pipe A has no water, an external continuous stream pours thru the right side. Otherwise, if pipe A contains water, a trap door falls, flushing the water, while another door closes off the external stream, disabling the continuous flow out the right side.
NOT0

With these three structures, the Looloolos built extremely intricate water mazes covering their entire island. Eventually, they developed dynamic systems that they could modify continuously with buttons and switches. These buttons and switches were equivalent to our AND OR NOT logic. Over the years, they managed to shrink each of their three AND OR NOT switches into prefabricated modular blocks. We would call this technique micro-hydraulics. Systems built with Looloolo water logic may eventually replace the oil driven vehicles we drive today.

Square

The AND block is represented as a SQUARE.

Circle

The OR block is represented as a CIRCLE.

The NOT block is represented as a TRIANGLE.

Here is a Looloolo design equivalent to implication: A �> B

This diagram is equivalent to implication because it has the same truth table as implication.

 
A B | A�>B              A   B   | A�>B
�����������             ���������������
0 0 |  1                OFF OFF |  ON
0 1 |  1                OFF OFF |  ON
1 0 |  0                ON  OFF |  OFF
1 1 |  1                ON  ON  |  ON
�����������             ���������������

When water sources A and B are both off, water flows out the system thru the output pipe.
When A is off and B is on, water also flows out the output pipe.
When both A and B are on, water flows out the output pipe.
However, when A is on and B is off, no water flows out the output pipe.

ASSIGNMENT
Today's exercise reverses our last exercise. Today we start with the truth table, then draw the diagram. Return to the two expressions you selected in last week's Conditional assignment. You called the first expression P; and the second, Q. Remember? You then completed a truth table for P�>Q. Find that truth table and draw a diagram equivalent to it. You will discover there are many ways to diagram any truth table. Use as few blocks as possible. It may simplify your diagram to substitute equivalent expressions. If you have been observant, you should know by now, and from the set matcher, that some expressions are equivalent to others. That is to say, if two expressions are equivalent, they each output the same truth table. Example: You may have noticed already that P�>Q is the same as �P+Q. Why? They have the same truth table.

 
P Q | P�>Q              P Q | -P + Q
�����������             �������������
0 0 |  1                0 0 | 1  1 0
0 1 |  1                0 1 | 1  1 1
1 0 |  0                1 0 | 0  0 0
1 1 |  1                1 1 | 0  1 1
�����������             �������������
                                 ^

Here are the Steps to Convert an Implication into a Diagram:

Draw the diagram for P.
Invert P with a NOT block.
Draw the diagram for Q.
OR the output from NOT P with the output from Q.

Design your system with pencil and paper before you transfer it to the applet.
Post your integrated diagram and truth table in a comment.
As you did in the previous pipe assignment, tag your input and output pipes.

If you do it right, you earn 7 wonderful points.

Diagram for P.
Inversion of P with a NOT block.
Diagram for Q.
ORing the output from NOT P with the output from Q.
Tagging your input and output pipes.
Posting your P�>Q applet diagram in a comment.
Copying your P�>Q truth table to the same comment.
Show us the A's and B's that P and Q equal.

EXAMPLE:

Start with your truth table from the Conditionals exercise.
Here is the truth table I used in the previous Pipe Game exercise:


   A B | OUT = �(A+B)+(A&B)
   �������������������������
   0 0 |  1
   0 1 |  0
   1 0 |  0
   1 1 |  1
   �������������������������

As noted in the Pipe Game exercise, �(A+B)+(A&B) is equal to (A+B)�>(A&B)
because they have the same truth table.

Let the Premise P = A+B
Let the Conclusion Q = A&B
Implement (A+B) �> (A&B)

Draw the diagram for P
Invert P with a NOT block
Draw the diagram for Q
OR the NOT P diagram with the Q diagram

This last diagram is the one to post in your comment.
Make sure this diagram shows your A and B logic, or you will do it again.
Translated into words, �(A+B)+(A&B) says Neither A nor B, or A and B.
The truth table for �(A+B)+(A&B) says Either 0 and 0, or 1 and 1.


   P=�(A+B) Q=A&B | OUT = P->Q = �(A+B)+(A&B)
   ���������������������������������������������
   0        0     |  1
   0        1     |  0
   1        0     |  0
   1        1     |  1
   ���������������������������������������������

When you copy your table from the Conditionals page,
delete all but the the first 3 columns: P, Q, and P->Q
Make sure to show what P and Q equal in terms of A and B.

Click here for a Water Logic Demo.

Comments:

From wHolt - 12/15/06 5:12 PM

OK except that rightmost AND should be an OR to equal your conditional.
Review the pattern, and what this exercise is supposed to do.

From BassLady - 12/15/06 2:45 PM

Water Logic #100

A B - (A & -B) & (A + B)
0 0 1 0 0 0
0 1 1 0 1 1
1 0 0 1 0 1
1 1 1 0 1 1

From wHolt - 12/15/06 10:15 AM

BassLady - almost
except you ignored this:
~(A&B) should have a NOT on the output from A&B.
why?
~(A&B) does not equal ~A & ~B.
You did ~A & ~B.
Fix.

From BassLady - 12/14/06 8:50 AM

Water Logic

A B - (A & B) & (A + B)
0 0 1 0 0 0
0 1 1 0 1 1
1 0 1 0 1 1
1 1 0 1 0 1

From wHolt - 12/12/06 12:05 AM

BassLady -
Use AND blocks for &

~(A&B) should have a NOT on the output from A&B

From BassLady - 12/11/06 9:33 AM

A	B	(A & -B)	&	(A & B)
0	0	0	0	0
0	1	0	0	0
1	0	0	0	0
1	1	1	1	1

This is a conditional table.

From wHolt - 11/11/06 12:09 PM

BassLady - i dont see pipe B
or an AND block for the &
also use Brian's applet for your table.
it makes everything easier.

From BassLady - 11/10/06 9:57 PM

Water Logic

A B -(A & B ) + (-A & -B)
0 0 1 0 1 1 1 1 1
0 1 1 0 0 1 1 0 0
1 0 0 1 1 0 0 0 1
1 1 1 0 0 0 0 0 0

From Poovey - 11/5/06 7:41 PM

trying this again....... p->q=(a&-b)->(-a&-b)

From wHolt - 10/24/06 11:03 AM

A simple edit will do.
You dont want to confuse anyone else with an incorrect example.
Wouldnt be prudent.

From DirtyBird - 10/23/06 7:54 PM

Do you want me to redo it?

From wHolt - 10/23/06 1:20 PM

DirtyBird
-P+Q = -(-A&-B) + (-A&B)
But I wont tell anyone if you dont.

SkoolGirl - OK except for some typos:
-P+Q = -(-A+-B) + -(A+B)

From SkoolGirl - 10/23/06 1:42 AM

Water Logic

P Q P -> Q -P->Q
1 1      1        1
1 0      0     0
1 0      0     0
0 0      1     1

P->Q (-A+-B)->-(A+B)

-p->Q -(-A+-B) + -(A+B)

From DirtyBird - 10/22/06 12:11 AM

P=>Q= (-A&-B) => (-A&B)

-P+Q= -(-A&-B) + (-A&B)

P Q P=>Q -P+Q
1 0    0    0
0 1    1    1
0 0    1    1
0 0    1    1

From wHolt - 10/21/06 10:46 AM

SkoolGirl- sorry I dont see the change.
This expression has typos
-p->Q -(-A+-B)->+ -(A+B)
should be
-P+Q = -(-A+-B) + -(A+B)
Hope you see the difference...

From SkoolGirl - 10/20/06 7:50 PM

Yes you showed me already......I changed it!

From wHolt - 10/19/06 1:20 PM

THE FOLLOWING WATER LOGICIANS EARNED 7 POINTS:

7Iron
Boki
Centerfield
Cheana
Draco
Fro
Harkar
Houdini
JooJoo
Kathi
LexiOwen
Melewen
Pac
Pringle
Slick
Sunshine
SuperDuke
Tiger
Zonino

From wHolt - 10/19/06 1:18 PM

SkoolGirl- did i show you this already?
P->Q = (-A+-B)->-(A+B) = -P+Q = -(-A+-B) + -(A+B)

Butterfly - try the pipe game first.

Soller - define your P and Q in terms of A and B
or we cannot tell what you are doing

Fro - thanks, we can now see what P and Q are.

From 7Iron - 10/18/06 9:19 PM

Do you mean I got this problem right by saying "welcome to the plumbers union?"

From Soller - 10/18/06 9:03 PM

P	Q	P->Q	-P+Q
1	0	1	0
1	1	1	0
1	1	1	0
1	1	1	0

From Fro - 10/18/06 7:18 PM

P=~(A+B)

Q=(~A+B)

P→Q=~(A+B)→ (-A+B)

P->Q= ~P+Q

P & Q don't apply because they are the same.

From Soller - 10/18/06 7:12 PM

P	Q	P->Q	-P+Q
1	0	1	0
1	1	1	0
1	1	1	0
1	1	1	0

From Butterfly - 10/18/06 6:02 PM

A	B			-A+ -B	=P		A		B	-(-A & -B)	=Q
0	0			1 1	1		0		0	0 0	0
0	1			1 0	1		0		1	0 1	0
1	0			0 1	1		1		0	1 0	0
1	1			0 0	0		1		1	1 1	1
P		Q	P -> Q			Q	P	Q -> P
1		0	1 0			0	1	0 0
1		0	1 0			0	1	0 0
1		0	1 0			0	1	0 0
0		1	0 0			1	0	1 0

From SkoolGirl - 10/18/06 12:32 PM

Water Logic

P Q P -> Q -P->Q
1 1      1        1
1 0      0     0
1 0      0     0
0 0      1     1

P->Q (-A+-B)->-(A+B)

-p->Q -(-A+-B)->+ -(A+B)

From JooJoo - 10/18/06 9:42 AM

Here is my remaining table.

Is this all else that I need to complete?

-A +B -> -(-A + -B)

From BassLady - 10/18/06 1:32 AM

Water Logic

Q P (Q -> P) (Q -> -P) (-Q + -P) (Q & P)
0 0 0 0 0 0
0 1 0 1 1 1
1 0 0 1 1 1
1 1 1 1 1 1

From wHolt - 10/18/06 12:14 AM

Most of you are catching on very well.
It takes a little practice to think visually with logic.
You have worked on a lot of problems now
that require symbols, tables, diagrams, and English.
Linking all these together is mental exercise you may not have had before.
Be patient. There may be some easier ones up ahead.

Cheana, Draco, Tiger, Zonino, Centerfield, LexiOwen - Good ones

GolfGirl- ANDs require two inputs.

Poovey -
~(A+B) does not equal ~A+~B
likewise for ~(A&B)
Use Brians applet. Fix

Bubba - start over. Reread the example in the lesson.
What comes out of an AND block is A&B.

SkoolGirl- diagram is good but you are confused on the not ~ label:
P->Q = (-A+-B)->-(A+B) = -P+Q = -(-A+-B) + -(A+B)

BassLady - review your pipe rules so that blocks do not have 2 outputs.
Then study the correct examples above.

JooJoo- put your table with your diagram. someone may need your good example.

From SkoolGirl - 10/17/06 10:02 PM

Water Logic

P Q P -> Q -P->Q
1 1      1        1
1 0      0     0
1 0      0     0
0 0      1     1

P->Q (-A+-B)->-(A+B)

-p->Q -(-A+-B)->-(A+B)

From Lexiowen - 10/17/06 9:43 PM

water

P->=-A+B->-(A&-B)

-P+Q=-(-A+B)+-(A&-B)

P Q P->Q -P+Q
1 1    1     1
1 1    1     1
0 0    1     1
1 1    1     1

From CenterField - 10/17/06 8:28 PM

From Bubba - 10/17/06 7:47 PM

A B A+-B A B -A&B
0 0   0 1   1 0 0 1   O 0
0 1 0    0     0 0 1 1   1 1
1 0 1    1 1 1 0 0   0 0
1 1    1 0 1 1 1 0   1 0
P Q P -> Q T Q P Q -> P
1 0      1 0 0 1
0 1      1 0 1 0
1 0      1 0 0 1
1 0      1 0 0 1

From Zonino - 10/17/06 7:46 PM

P	Q	P -> Q	-P+Q
1	0	0	0
0	1	1	1
1	0	0	0
1	0	0	0

P = A + -B

Q = -A & B

P->Q = (A + -B) -> (-A & B)

-P+Q = -(A+-B) + (-A & B)

From Tiger - 10/17/06 6:03 PM

P Q P->Q -P+Q
0 1 1 1
1 1 1 1
0 0 1 1
0 1 1 1

From Poovey - 10/17/06 6:02 PM

P->Q= (A&-B)->(-A+-B)

From Draco - 10/17/06 5:32 PM

P=-A+B

Q=-A&-B

OUT=-P+Q

P Q P -> Q -P+Q
1 1      1 1
1 0      0 0
0 0      1 1
1 0      0 0

From Cheana - 10/17/06 5:28 PM

From GolfGirl - 10/17/06 3:16 PM

From wHolt - 10/17/06 2:52 PM

CatsEyes- your first NOT should input to the bottom OR.
Also , post your A and B tables. PQ table is not relevant.

Slick and 7Iron - welcome to the plumbers union.

From CatsEyes - 10/16/06 7:41 PM

P	Q	P -> Q	-P+Q
1	0	0	0
0	1	1	1
0	0	1	1
0	0	1	1

From 7Iron - 10/16/06 5:48 PM

P= (-A+B)
Q= (-A&B)
P->Q = (-A+B) -> (-A&B)
-P+Q= -(-A+B) + (-A&B)

	P	Q	P->Q	-P + Q
0)	0	0	0	0
1)	0	1	1	1
2)	1	0	1	1
3)	1	1	0	0

From Slick - 10/16/06 3:39 PM

P	Q	P -> Q	Q	P	-P+Q
1	0	0	0	1	1
0	0	1	0	0	1
1	1	1	1	1	1
1	0	0	0	1	0

P->Q (A+-B) -> (A&-B)

-P+Q -(A+-B)+(A&-B)

From wHolt - 10/16/06 1:34 PM

Fro -
P=~(A+B)
Q=(~A+B)
P=>Q = (A+B)=>(-A+B) (do this table)
your PQ table does not apply. Why?

7Iron -
(-A+B) -> (-A&B) (do this table)
your PQ table does not apply. Why?

JooJoo -
-A +B -> -(-A + -B) (do this table)

From Pringle - 10/16/06 1:01 PM

P : -A + -B

Q : -A & -B

P -> Q : ( -A + -B) -> (-A & -B)

-p + Q : -(-A + -B) + (-A & -B)

From Sunshine - 10/16/06 10:00 AM

From Fro - 10/15/06 8:26 PM

P=~(A+B)

Q=(~A+B)

P→Q=~(A+B)→ (-A+B)

P->Q= ~P+Q

From JooJoo - 10/15/06 2:11 PM

P	Q	P→Q	-P+Q
1	1	1	0
1	1	1	0
0	1	1	1
1	0	0	1

P-> Q= -A +B -> -(-A + -B)

-P +Q = -(-A +B) + -(-A + -B)

From wHolt - 10/15/06 1:02 PM

Phoenix- diagram good, but table wrong
Use Brian's truth constructor to tabulate ~(A&~B)+(~A&B)

From Phoenix - 10/15/06 7:15 AM

water

P-->Q = (A&~B)-->(~A&B)

~P+Q = ~(A&~B)+(~A&B)

PQ P-->Q ~P+Q
00 1 1
01 1 1
10 0 0
11 1 1

From SuperDuke - 10/14/06 9:25 PM

P	Q	P->Q	-P + Q
1	1	1	1
0	0	1	1
1	0	0	0
1	0	0	0

P=>Q = (A+~B)=>(~A&~B)

-P+Q = ~(A+~B)+(~A&~B)

Mr. Holt this appears correct to me.

From Pac - 10/14/06 7:16 PM

P = (-A+B)

Q = -(A+B)

P->Q = (-A+B)->(-(A+B))

-P+Q = (-(-A+B))+(-(A+B))

P Q P->Q -P+Q
1 1 1 1
1 0 0 0
0 0 1 1
1 0 0 0

From wHolt - 10/14/06 2:56 PM

Slick - 0+1=1. You said 0+1=0.
P->Q = (A+-B) -> (A&-B)
-P+Q = -(A+-B)+(A&-B)

From Slick - 10/14/06 9:45 AM

P	Q	P -> Q	Q	P	-P+Q
1	0	0	0	1	0
0	0	1	0	0	1
1	1	1	1	1	1
1	0	0	0	1	0

P->Q (A+-B) + (A&-B)

-P+Q -(A+-B)+(A&-B)

water logic

From wHolt - 10/13/06 2:16 PM

Fro - nicely laid out pipe diagram.
However, ~A+~B does not equal ~(A+B)
You should have checked before posting.
And when you check, check (-A+B) => -(A+B),
not P=>Q

From Fro - 10/12/06 8:54 PM

P=(~A+B)

Q=~(A+B)

P→Q=(-A+B)→ -(A+B)

P->Q= ~P+Q

From wHolt - 10/12/06 6:28 PM

THE FOLLOWING WATER LOGICIANS EARNED 7 POINTS:

Boki
Harkar
Houdini
Kathi

From wHolt - 10/12/06 6:25 PM

Melewen - what are your P and Q? Copy table.
also that's P->Q , not ~P->Q

From Melewen - 10/12/06 11:23 AM

From Harkar - 10/11/06 9:02 PM

P Q P -> Q -P+Q
0 0     1 1
1 0     0 0
0 1     1 1
0 0     1 1

From Kathi - 10/11/06 12:28 PM

P --> Q = (-A + B) --> (A + -B)

-P+Q= -( -A+ B)+(A+-B)

P	Q	\|	P --> Q		-P + Q
1	1	\|	1	\|	1
1	0	\|	0		0
0	1	\|	1		1
1	1	\|	1		1

From wHolt - 10/11/06 11:39 AM

Thanks, Boki. You have proven that persistence pays off! 7 Points!

From Boki - 10/10/06 11:10 AM

P	Q	P -> Q	-P + Q
1	1	1	1
0	0	1	1
0	0	1	1
0	0	1	1

P= ~A&~B
Q= ~(AvB)
P -> Q= (~A&~B) -> ~(AvB)
P -> Q= ~P + Q

From wHolt - 10/10/06 10:25 AM

Melewen - you cannot have 2 pipes coming out of a block.
Make a pipe branch from the output pipe of the first NOT

Boki - start over. copy your problem from the conditional page.
AND blocks output &
OR blocks output + (v)

From Boki - 10/9/06 5:56 PM

P	Q	P -> Q	-P + Q
1	1	1	1
0	0	1	1
0	0	1	1
0	0	1	1

P=~A&~B

Q=~(AvB)

P -> Q= ~P + Q

From Melewen - 10/9/06 4:35 PM

P -> Q = (-A + B) -> (-A + -B)

P = (-A + B)
Q = (-A + -B)

P	Q	\|	P->Q	\|	-P+Q
1	1	\|	1	\|	1
1	1	\|	1	\|	1
0	1	\|	1	\|	1
1	0	\|	0	\|	0

From wHolt - 10/9/06 1:49 PM

Seems you have the right ideas, but find this new symbolism awkward.
For example ~(A+B) means to OR A with B, then negate it.
Don't negate before the ORing.
Keep at it...

Boki - that first OR block has two outputs. fix it.

Kathi - your first OR block has only one input. Fix.

SuperDuke - (~A+~B) does not equal (~A&~B). Fix Q.

From Boki - 10/9/06 11:48 AM

P=~A&~B

Q=~(AvB)

P->Q=(~A&~B) -> ~(AvB)

-P v Q= ~(~A&~B) v ~(AvB)

P -> Q= ~P + Q

P	Q	P -> Q	-P + Q
1	1	1	1
0	0	1	1
0	0	1	1
0	0	1	1

From Kathi - 10/9/06 9:57 AM

P --> Q = (-A + B) --> (A + -B)

-P+Q= -( -A+ B)+(A+-B)

P	Q	\|	P --> Q		-P + Q
1	1	\|	1	\|	1
1	0	\|	0		0
0	1	\|	1		1
1	1	\|	1		1

From wHolt - 10/8/06 12:05 PM

Boki - your symbols here are confusing: -P -> Q =(AvB) ->~(AvB)
Remember: P->Q = ~P+Q
Your negations are in the wrong place.
Your table is miscopied also.
Also , your diagram at the top shows ~A , ~B going into an OR block.
What comes out must be ~A+~B which does not equal A+B.
Rethink your table and diagram.
The only reason you are getting this attention is that no one else seems to want to try.

From Boki - 10/8/06 10:35 AM

P -> Q = (~A & ~B) -> ~(A v B)

P=~A&~B -P=~(~(AvB))=AvB

Q=~(AvB)

-P -> Q =(AvB) ->~(AvB)