I know that I can say convert a 2-input AND gate into a NOR gate by simply inverting the two inputs because of DeMorgan's Theorem.
But how would you do the equivalent on a 3-input AND gate?
Say...
____
A___| \
B___| )___
C___|____ /
I'm trying to understand this because my homework asks me to take a circuit and convert it using NOR synthesis to only use nor gates, and I know how to do it with 2 input gates, but the gate with 3 inputs is throwing me for a spin.
DeMorgan's theorem for 2-input AND would produce:
AB
(AB)''
(A' + B')'
So, yes, the inputs are inverted and fed into a NOR gate.
DeMorgan's theorem for 3-input AND would similarly produce:
ABC
(ABC)''
(A' + B' + C')'
Which is, again, inputs inverted and fed into a (3-input) NOR gate:
___
A--O\ \
B--O ) )O---
C--O/___ /
#SailorChibi has truth tables that show equivalence.
If i haven't made any mistakes it is pretty much the same, invert all 3 of the inputs and you get a NOR
Table:
AND with inverted in is exact the same as
1 1 1 = 1
1 1 1 = 0
1 0 1 = 0
0 1 0 = 0
0 1 1 = 0
0 1 0 = 0
0 0 1 = 0
0 0 0 = 0
NOR with original input
0 0 0 = 1
0 0 1 = 0
0 1 0 = 0
1 0 1 = 0
1 0 0 = 0
1 0 1 = 0
1 1 0 = 0
1 1 1 = 0
Related
I know how to find the number of triangles in an adjacency matrix.
tri = trace(A^3) / 6
But i require to find the nodes so that i can finally find the value of the edges from adjacency matrix since it's a sign graph. Is there already existing function which does that?
Taking the power of the adjacency matrix loses information about the intermediate nodes. Instead of a 2-dimensional matrix, we need 3 dimensions.
Given a graph:
and its adjacency matrix:
A =
0 0 0 0 1 1 0 1 0 0
0 0 0 1 0 1 0 0 0 0
0 0 0 1 0 0 0 1 0 1
0 1 1 0 1 0 1 0 0 0
1 0 0 1 0 0 1 0 0 0
1 1 0 0 0 0 0 1 1 0
0 0 0 1 1 0 0 0 1 0
1 0 1 0 0 1 0 0 0 0
0 0 0 0 0 1 1 0 0 0
0 0 1 0 0 0 0 0 0 0
Compute the 3d matrix T such that T(i,j,k) == 1 iff there is a path in the graph i=>j=>k=>i.
T = and(A, permute(A, [3 1 2]))
This is the equivalent of squaring the adjacency matrix, but keeping the path information. and is used here instead of multiplication in case A is a weighted adjacency matrix. If you sum along the 2nd dimension, you'll get A^2:
>> isequal(squeeze(sum(T,2)), A^2)
ans = 1
Now that we've got the paths of length 2, we just need to filter so we keep only the paths that return to their starting points.
T = and(T, permute(A.', [1 3 2])); % Transpose A in case graph is directed
Now, if T(i,j,k) == 1, then there is a triangle starting at node i, through nodes j and k and returning to node i. If you want to find all such paths:
[M,N,P] = ind2sub(size(T), find(T));
P = [M,N,P];
P will be a list of all triangular paths:
P =
8 6 1
6 8 1
7 5 4
5 7 4
7 4 5
4 7 5
8 1 6
1 8 6
5 4 7
4 5 7
6 1 8
1 6 8
In this case we get 12 paths. All paths in an undirected graph have 6 duplicates: one starting at each triangle point, times 2 directions. This gives the same results as trace:
>> trace(A^3)
ans = 12
If you want to remove the duplicates, the simplest way for triangles is to simply sort the vertex ordering and then take the unique rows of the list. This works for triangles only because all permutations of the nodes in the cycle are present. For longer cycles, this will not work.
P = unique(sort(P, 2), 'rows');
P =
1 6 8
4 5 7
Here is a solution using matrix multiplication:
C = (A * A.') & A;
[x, y] = find(tril(C));
n = numel(x);
D = sparse([x; y], [1:n 1:n].', 1, size(A,1), n);
[X, ~, V] = find(C * D);
tri = [x y X(V == 2)]
tri = unique(sort(tri, 2), 'rows');
First we need to know what are triangle nodes. Two nodes are triangle nodes if they have a common neighbor and both of them are neighbor of each other.
We take the definition to compute an adjacency matrix C that only contains triangle nodes and all other node are removed.
The expression A * A.' selects nodes that have common neighbors and the & A operator says that those nodes that have common neighbors should by neighbor of each other.
Now we can use [x, y] = find(tril(C)); to extract the first and the second points of each triangle as x and y respectively.
For the third node we need to find a node that has x and y as its neighbors. As before we can use the multiplication of boolean matrix trick to speed up the computation.
Finally the result tri has duplicates that should be remove using unique and sort.
I'm currently trying to convert 44/7 to half-precision floating point format.
I'm not sure if I've done it correctly so far, so I'd really appreciate it if someone could have a look at it.
44/7 = 6,285714285714...
6 in dual -> 110;
0.285714 * 2 = 0,571428 -> 0
0.571428 * 2 = 1.142856 -> 1
0.142856 * 2 = 0.285714 -> 0
... -> 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1...
-> 110, 01001001001001
-> 1,1001001001001001 -> exponent: 2;
Bias + Exponent : 2+15 = 17 => 1 0 0 0 1
All stitched together: 0 1 0 0 0 1 1 0 0 1 0 0 1 0 0 1
I've never converted decimal to 16bit IEEE754, is this the correct way of converting it?
Thanks a lot!
Correct. As you might expect, it is quantized to 6.28515625.
0100011001001001(base 2) = 4649(base 16)
6.2857142857139996
= H(4649)
= F(40C92492)
= D(40192492 49249107)
= A(0X1.92492492491070P+2)
6.28515625
= H(4649)
= F(40C92000)
= D(40192400 00000000)
= A(0X1.92400000000000P+2)
Other data points:
+0. 0000
-0. 8000
-1. BC00
+1. 3C00
+2. 4000
+4. 4400
+8. 4800
+16. 4C00
+32768. 7800
+Max 7BFF 65504
+.5f 3800
+.25f 3400
+.125f 3000
+.0625f 2C00
+MinNorm 0400 +6.103515625e-05
-MinNorm 8400 -6.103515625e-05
+MinDenorm 0001 +5.9604644775390625e-08
-MinDenorm 8001 -5.9604644775390625e-08
+Infinity 7C00
-Infinity FC00
+NaN(0) 7E00
-NaN(0) FE00
I have a fairly simple question that i cannot find an example of online. I understand that this can simplify to A^B but I have not yet covered that section. What is the correct value of the boolean expression (A'B + AB')?
Lets look at the truth table
A B A'B AB' A'B + AB'
-----------------------------
0 0 0 0 0
0 1 1 0 1
1 0 0 1 1
1 1 0 0 0
This simply computes the xor of A and B.
and hence this is our answer.
The definition of the symbol XOR (^) is a^b = a'b + ab', i.e. one or the other but not both must be true for the expression to be true. Therefore there are no intermediate steps to convert between the two expressions. This is because a'b and ab' are prime implicants of the boolean function.
Another (not necessarily more simplified) way to define XOR is (A+B).(A'+B')
A B A+B A' B' A'+B' (A+B).(A'+B')
----------------------------------------
0 0 0 1 1 1 0
0 1 1 1 0 1 1
1 0 1 0 1 1 1
1 1 1 0 0 0 0
I have a problem grasping the fact that A + AB = A and A(A + B) = A
Can anyone tell me how?
To understand this, look at the truth tables (1=true, 0=false):
"A" "B" "AB" "A+B" "A + AB" "A(A+B)"
0 0 0 0 0 0
0 1 0 1 0 0
1 0 0 1 1 1
1 1 1 1 1 1
is it possible to simply this Boolean function
(!A*!B*!C) + (!A*!B*C*!D) + (A*!B*!C*D) + (A*!B*C*!D) + (A*B*!C*!D)
I came up with this:
(!B*(!A*(!C+!D))+A*(C XOR D)) + (A*B*!C*!D)
Messy to look at, but there are fewer terms.
Look at the truth table:
A B C D X
0 0 0 0 1
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0
0 1 0 0 0
0 1 0 1 0
0 1 1 0 0
0 1 1 1 0
1 0 0 0 0
1 0 0 1 1
1 0 1 0 1
1 0 1 1 0
1 1 0 0 1
1 1 0 1 0
1 1 1 0 0
1 1 1 1 0
It looks like you can take the three parts of the table where X = 1 and simplify this to the sum of three terms:
!A*!B*!(C*D) + A*!B*(C^D) + A*B*!C*!D
Note that I've use XOR (^) in the second term. If you can't use XOR then you'll need to expand the second term a little.
You can reduce the number of terms further by factoring out either !B or A for two of the terms, e.g.
!B*(!A*!(C*D) + A*(C^D)) + A*B*!C*!D
or:
!A*!B*!(C*D) + A*(!B*(C^D) + B*!C*!D)