How can I create a K-MAP by looking at this function.I dont know how to create one
Create the truth table, from that create the K-MAPs (Karnaugh maps), from that create the booleans, from that create the CKTs. One of the 6 outputs, Y5 (Y=f(x)), is completed below ...
Truth Table
------------
4s 2s 1s 32s 16s 8s 4s 2s 1s
X2 X1 X0 # | # Y5 Y4 Y3 Y2 Y1 Y1
0 0 0 0 | 0 x x x x x x [1]
0 0 1 1 | 0 0 0 0 0 0 0
0 1 0 2 | 2 0 0 0 0 1 0
0 1 1 3 | 6 0 0 0 1 1 0
1 0 0 4 | 12 0 0 1 1 0 0
1 0 1 5 | 20 0 1 0 1 0 0
1 1 0 6 | 30 0 1 1 1 1 0
1 1 1 7 | 42 1 0 1 0 1 0
[1] 0 input is not positive so is not allowed, so we "don't care" about these outputs
Y5 K-Map
--------
X2 \ X1 X0 00 01 11 10
\
0 x 0 0 0
1 0 0 1 0
Y5 Boolean (Product of Sums in this case, Sum of Products is another option)
----------
Y5 = X2 ^ X1 ^ X3 = (X2 ^ X1) ^ X3
Y5 Circuit
----------
______
X2 ---| |
| AND |---| ______
X1 ---|_____| |---| |
| AND |--- Y5
X2 -----------------|_____|
Etc...
Related
I have a matrix:
$cat ifile.txt
2 3 4 5 10 0 2 2 0 1 0 0 0 1
0 3 4 6 2 0 2 0 0 0 0 1 2 3
0 0 0 2 3 0 3 0 3 1 2 3 1 0
Here it has total 14 columns e.g. A1 B1 A2 B2 A3 B3 A4 B4 A5 B5 A6 B6 A7 B7. Each odd number columns correspond to A and even number columns correspond to B.
I would like to print all A in one column and all B in one column. So my desire file looks like:
$cat ofile.txt
2 3
0 3
0 0
4 5
4 6
0 2
10 0
2 0
3 0
2 0
0 0
0 3
....
It is possible for me to do manually in the following way, but I am looking for some more easy way to do it.
for c in 1 3 5 7 9 11 13;do
awk'{printf"%5s %5s",$c,$(c+1)} > A$c.txt
cat A1 A3 A5 A7 A9 A11 A13 > ofile.txt
$ cat tst.awk
{
for ( i=1; i<=NF; i++ ) {
a[NR,i] = $i
}
}
END {
for ( i=1; i<=NF; i+=2 ) {
for (j=1; j<=NR; j++ ) {
print a[j,i], a[j,i+1]
}
}
}
.
$ awk -f tst.awk file
2 3
0 3
0 0
4 5
4 6
0 2
10 0
2 0
3 0
2 2
2 0
3 0
0 1
0 0
3 1
0 0
0 1
2 3
0 1
2 3
1 0
If you want to generalize for more than 2 output columns:
$ cat tst.awk
BEGIN { n=(n ? n : 2) }
{
for (i=1; i<=NF; i++) {
a[NR,i] = $i
}
}
END {
for ( i=1; i<=NF; i+=n ) {
for (j=1; j<=NR; j++) {
for ( k=1; k<=n; k++ ) {
printf "%s%s", a[j,i+k-1], (k<n ? OFS : ORS)
}
}
}
}
.
$ awk -v n=2 -f tst.awk file
2 3
0 3
0 0
4 5
4 6
0 2
10 0
2 0
3 0
2 2
2 0
3 0
0 1
0 0
3 1
0 0
0 1
2 3
0 1
2 3
1 0
.
$ awk -v n=7 -f tst.awk file
2 3 4 5 10 0 2
0 3 4 6 2 0 2
0 0 0 2 3 0 3
2 0 1 0 0 0 1
0 0 0 0 1 2 3
0 3 1 2 3 1 0
We have a matrix:
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
I want to retain the data in these fields, but in the shape of a 2D-circle:
0 1 1 1 0
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
0 1 1 1 0
But this also scales up:
0 0 0 1 1 1 1 0 0 0
0 0 1 1 1 1 1 1 0 0
0 1 1 1 1 1 1 1 1 0
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 0
0 0 1 1 1 1 1 1 0 0
0 0 0 1 1 1 1 0 0 0
What would the greatest way to approach this?
cy and cx are center points
r is the radius
tiles is an empty grid
function MakeCircle(tiles, cx, cy, r):
for x in range(cx - r, cx + r):
for y in range(cy - r, cy + r):
if (distance(cx, cy, x, y) <= r):
tiles[x][y] = 1
return(tiles)
function distance(x1, y1, x2, y2):
return(sqrt((x1 - x2)**2) + (y1 - y2)**2))
This dynamically creates a circle of ones in a square-matrix regardless of size of matrix.
I have to generate variables entry_1, entry_2 and entry_3 which will adopt the value 1 if id_i for that particular month had entry=1.
Example.
id month entry entry_1 entry_2 entry_3
1 1 1 1 0 0
1 2 0 0 0 0
1 3 0 0 1 1
1 4 0 0 0 0
2 1 0 1 0 0
2 2 0 0 0 0
2 3 1 0 1 1
2 4 0 0 0 0
3 1 0 1 0 0
3 2 0 0 0 0
3 3 1 0 1 1
3 4 0 0 0 0
Would anyone be so kind to propose an idea of how to implement a loop in order to do this?
I am thinking of something like this:
forvalues i=1(1)3 {
gen entry`i'=0
replace entry`i'=1 if on that particular month id=`i' had entry=1
}
You could do something like this (although your data don't quite look right for the question you're asking):
forvalues i = 1/3 {
gen entry_`i' = id == `i' & entry == 1
}
This generates a dummy variable entry_i for each i in the forvalues loop where entry_i = 1 if id is i and entry is 1, and 0 otherwise.
The code can be simplified down to at most one loop.
clear
input id month entry entry_1 entry_2 entry_3
1 1 1 1 0 0
1 2 0 0 0 0
1 3 0 0 1 1
1 4 0 0 0 0
2 1 0 1 0 0
2 2 0 0 0 0
2 3 1 0 1 1
2 4 0 0 0 0
3 1 0 1 0 0
3 2 0 0 0 0
3 3 1 0 1 1
3 4 0 0 0 0
end
forval j = 1/4 {
egen entry`j' = total(entry & id == `j'), by(month)
}
list id month entry entry? , sepby(id)
+--------------------------------------------------------+
| id month entry entry1 entry2 entry3 entry4 |
|--------------------------------------------------------|
1. | 1 1 1 1 0 0 0 |
2. | 1 2 0 0 0 0 0 |
3. | 1 3 0 0 1 1 0 |
4. | 1 4 0 0 0 0 0 |
|--------------------------------------------------------|
5. | 2 1 0 1 0 0 0 |
6. | 2 2 0 0 0 0 0 |
7. | 2 3 1 0 1 1 0 |
8. | 2 4 0 0 0 0 0 |
|--------------------------------------------------------|
9. | 3 1 0 1 0 0 0 |
10. | 3 2 0 0 0 0 0 |
11. | 3 3 1 0 1 1 0 |
12. | 3 4 0 0 0 0 0 |
+--------------------------------------------------------+
I don't know if this question is considered to be related to stackoverflow (I'm sorry if it's not but I have searched and did not find an answer anywhere).
I have coded a full adder
Output:
Truth Table :
a1 a2 b1 b2 S1 S2 C
______________________________
0 0 0 0 0 0 0
0 0 0 1 0 1 0
0 0 1 0 1 0 0
0 0 1 1 1 1 0
0 1 0 0 0 1 0
0 1 0 1 0 0 1
0 1 1 0 1 1 0
0 1 1 1 1 0 1
1 0 0 0 1 0 0
1 0 0 1 1 1 0
1 0 1 0 0 1 0
1 0 1 1 0 0 1
1 1 0 0 1 1 0
1 1 0 1 1 0 1
1 1 1 0 0 0 1
1 1 1 1 0 1 1
If somebody has ever calculated this, can they tell me if my output is correct
a1 a2 b1 b2 S1 S2 C a b s c
______________________________
0 0 0 0 0 0 0 0 0 0 0 nothing plus nothing is nothing
0 0 0 1 0 1 0 0 2 2 0 nothing plus two is two
0 0 1 0 1 0 0 0 1 1 0 nothing plus one is one
0 0 1 1 1 1 0 0 3 3 0 nothing plus three is three
0 1 0 0 0 1 0 2 0 2 0 two plus nothing is two
0 1 0 1 0 0 1 2 2 0 1 two plus two is four (four not in 0-3)
0 1 1 0 1 1 0 2 1 3 0 two plus 1 is three
0 1 1 1 1 0 1 2 3 1 1 two plus three is five (one and four)
1 0 0 0 1 0 0 1 0 1 0 one plus nothing is one
1 0 0 1 1 1 0 1 2 3 0 one plus two is three
1 0 1 0 0 1 0 1 1 2 0 one plus one is two
1 0 1 1 0 0 1 1 3 0 1 one plus three is four
1 1 0 0 1 1 0 3 0 3 0 three plus nothing is three
1 1 0 1 1 0 1 3 2 1 1 three plus two is five (one and four)
1 1 1 0 0 0 1 3 1 0 1 three plus one is four
1 1 1 1 0 1 1 3 3 2 1 three plus three is 6 (two and four)
Looks right. Ordering your 16 rows a little differently would make them flow in a more logical order.
It's an adder! Just check if it's adding. Let's take this row:
a2 a1 b2 b1 C S2 S2
1 0 1 1 1 0 1
Here I have reordered the columns in an easier to read manner: higher order bits first.
The a input is 10 = 2 (base 10). The b input is 11 = 3 (base 10). The output is 101, which
is 5 (base 10). So this one is right: 2 + 3 == 5.
I'll let you check the other rows.
I have to implement the function below with XOR gates. I drew the Karnaugh map and wrote down the resulting minimized function. But now I'm stuck with AND and OR gates, what should I do in order to get XOR gates?
My solution looks as follows:
F = Sum(1,2,4,7,8,11,13,14);
F = A' B' C' D + A' B' C D' + A' B C' D' + A' B C D + A B' C' D' + A B' C D + A B C' D + A B C D';
F = XOR(C, D) & A' B' + XOR(C, D) & A B + XOR(A, B) & C' D' + XOR(A, B) & C D;
F = XOR(C, D) & XOR(A, B)' + XOR(A, B) & XOR(C, D) ';
F = XOR(XOR(A, B), XOR(C, D));
A B C D XOR(A, B) XOR(C, D) F
00 0 0 0 0 0 0 0
01 0 0 0 1 0 1 1
02 0 0 1 0 0 1 1
03 0 0 1 1 0 0 0
04 0 1 0 0 1 0 1
05 0 1 0 1 1 1 0
06 0 1 1 0 1 1 0
07 0 1 1 1 1 0 1
08 1 0 0 0 1 0 1
09 1 0 0 1 1 1 0
10 1 0 1 0 1 1 0
11 1 0 1 1 1 0 1
12 1 1 0 0 0 0 0
13 1 1 0 1 0 1 1
14 1 1 1 0 0 1 1
15 1 1 1 1 0 0 0
A handy tool for such questions is "Logic Friday 1"