My own equation is a bit longer, but the following example shows perfectly where I struggle at the moment.
So far I have been using the let() and letsimp() function
to substitute longer terms in an equation,
but in this example they have no effect:
(%i1) eq: ((2*u+a^2+d) * y+x)/2*a = x;
2
a ((2 u + d + a ) y + x)
(%o1) ------------------------ = x
2
(%i2) let(2*u+a^2+d, %beta);
2
(%o2) 2 u + d + a --> %beta
(%i3) letsimp(eq);
2
a ((2 u + d + a ) y + x)
(%o3) ------------------------ = x
2
What is the preferred way to replace 2*u+a^2+d with %beta in this sample equation?
And why has letsimp() no effect?
Thank you very much!
letsimp applies only to "*" expressions. You could try subst.
Related
I am working on simplifying the expression f = x'yz + xy'z + xyz' + xyz. Actually, it may not be this expression. The question is: simplify the boolean expression for a voting system, the system being: three people vote on multiple candidates, and two or more people should agree(true) on the candidate in order to pass. So I think the answer would be xy + yz + xz, but I can't figure out the process between. Can anyone explain?
From the idempotent/identity law, we have x + x = x, and so xyz + xyz = xyz. Applying this principle, we can rewrite your expression as:
f = x'yz + xy'z + xyz' + xyz
=> f = x'yz + xy'z + xyz' + xyz + xyz + xyz --OR with xyz twice without affecting the value
=> f = x'yz + xyz + xy'z + xyz + xyz' + xyz --Rearrange
=> f = yz (x + x') + xz (y + y') + xy(z' + z) --Group
=> f = yz + xz + xy --Since x+x' = 1
That said, as the diagram clearly shows, you can simply take AND together each pair of inputs, and OR them together to get the same result. By doing this, you ensure that:
If any 2 of the 3 inputs are true, your overall result is true
When all 3 are true, the result is still true
The advantage of expressing it in this way is that you can just focus on each pair of inputs at one time, without worrying about the impact of the third one.
A simple way without involved logical reasoning
Write a truth table. For three inputs, there are 2^3 = 8 rows.
Four rows correspond to the given terms in your sum-of-products expression.
Enter the eight values of your expression into a Karnaugh map:
Group adjacent 1-terms to blocks as shown.
A pair of cells can be merged into a bigger block, if they just differ in one input. This way, the blocks double their cell-count and reduce their input-count by one in every merge step.
Each of the resulting blocks corresponds to one implicant term in the minimized expression.
Drawing the map and finding the blocks can be done automatically using a nice online tool of Marburg University.
I'm trying to understand 1-dimensional convolution. Can someone explain--potentially step by step--how convolve(A,B) works for vectors A=[a,b,c] and B=[d,e,f]. Thanks!
I'm not sure what exacly you wan to do with this convolution, but here you can find amazing pdf about this. In first chapter there's example for 1-dimensional data. Later there are some examples and explanations form image processing.
Hope it will help you!
For the linear convolution result [v,w,x,y,z],
first reverse [d,e,f] to [f,e,d] , then input the first element of [a,b,c] :
v = f * 0 + e * 0 + d * a
for the next output, shift 0,0,a,b,c to the left by 1 and use the same f,e,d
w = f * 0 + e * a + d * b
for the next output, shift 0,a,b,c to the left by 1 and use the same f,e,d
x = f * a + e * b + d * c
for the next output, shift a,b,c to the left by 1 and use the same f,e,d
y = f * b + e * c + d * 0
for the next output, shift a,b,c to the left by 1 and use the same f,e,d
z = f * c + e * 0 + d * 0
For circular convolution replace the zeros with inputs that are wrapped around elements inside [a,b,c].
As we all know usually negative numbers in memory represents as two's complement numbers like that
from x to ~x + 1
and to get back we don't do the obvious thing like
~([~x + 1] - 1)
but instead we do
~[~x + 1] + 1
can someone explain why does it always work? I think I can proof it with 1-bit, 2-bit, 3-bit numbers and then use Mathematical induction but it doesn't help me understand how exactly that works.
Thanks!
That's the same thing anyway. That is, ~x + 1 == ~(x - 1). But let's put that aside for now.
f(x) = ~x + 1 is its own inverse. Proof:
~(~x + 1) + 1 =
(definition of subtraction: a - b = ~(~a + b))
x - 1 + 1 =
(you know this step)
x
Also, ~x + 1 == ~(x - 1). Why? Well,
~(x - 1) =
(definition of subtraction: a - b = ~(~a + b))
~(~(~x + 1)) =
(remove double negation)
~x + 1
And that (slightly unusual) definition of subtraction, a - b = ~(~a + b)?
~(~a + b) =
(use definition of two's complement, ~x = -x - 1)
-(~a + b) - 1 =
(move the 1)
-(~a + b + 1) =
(use definition of two's complement, ~x = -x - 1)
-(-a + b) =
(you know this step)
a - b
This is because if you increment ~x (assuming no overflow). Then converting it to back to x, you've incremented relative to ~x, but decremented relative to x. Same thing applies vice versa. Assuming your variable x has a specific value, every time you increment it, relative to ~x you'll notice it decrements.
From a programmer's point of view, this is what you'd essentially witness.
Let short int x = 1 (0x0001)
then ~x = 65534 (0xFFFE)
~x + 1 = 65534 + 1 (0xFFFF)
~(~x+1) = 0 (0x0000)
~(~x+1) + 1 = 0 + 1 (0x0001)
I'm trying to solve an equation with 5 unknowns in Mathcad 14. My equations look like this:
Given
0 = e
1 = d
0 = c
-1 = 81a + 27b + 9c + 3d + e
0 = 108a + 27b + 6c + d
Find(a,b,c,d,e)
Find(a,b,c,d,e) is marked as red and says "pattern match exception". What is the problem?
In mathcad you need to do something similar to:
c:=0
d:=1
e:=0
a:=1
b:=1
Given
81*a + 27*b + 9*c + 3*d + e = -1
108*a + 27*b + 6*c + d = 0
Find(a,b,c,d,e) = (0,0,0,0,-1)
Now, what I have done here is to define the variables BEFORE the Solve Block (Given...Find), you have to give initial values which you think are close to the solution you require in order for the iteration to succeed.
Tips: To get the equals sign in the Solve Block, use ctrl and '='. If your looking to solve for 5 unknowns then you need 5 equations, the original post looked like you knew 3 of the variables and were looking for a and b, in this case you would do the following:
c:=0
d:=1
e:=0
a:=1
b:=1
Given
81*a + 27*b + 9*c + 3*d + e = -1
108*a + 27*b + 6*c + d = 0
Find(a,b) = (0.111,-0.481)
This has held c, d and e to their original values and iterated to solve for a and b only.
Hope this helps.
I'm trying to get x and y coordinates for points along a line (segment) at even intervals. In my test case, it's every 16 pixels, but the idea is to do it programmatically in ActionScript-3.
I know how to get slope between two points, the y intercept of a line, and a2 + b2 = c2, I just can't recall / figure out how to use slope or angle to get a and b (x and y) given c.
Does anyone know a mathematical formula to figure out a and b given c, y-intercept and slope (or angle)? (AS3 is also fine.)
You have a triangle:
|\ a^2 + b^2 = c^2 = 16^2 = 256
| \
| \ c a = sqrt(256 - b^2)
a | \ b = sqrt(256 - a^2)
| \
|__________\
b
You also know (m is slope):
a/b = m
a = m*b
From your original triangle:
m*b = a = sqrt(256 - b^2)
m^2 * b^2 = 256 - b^2
Also, since m = c, you can say:
m^2 * b^2 = m^2 - b^2
(m^2 + 1) * b^2 = m^2
Therefore:
b = m / sqrt(m^2 + 1)
I'm lazy so you can find a yourself: a = sqrt(m^2 - b^2)
Let s be the slop.
we have: 1) s^2 = a^2/b^2 ==> a^2 = s^2 * b^2
and: 2) a^2 + b^2 = c^2 = 16*16
substitute a^2 in 2) with 1):
b = 16/sqrt(s^2+1)
and
a = sqrt((s^2 * 256)/(s^2 + 1)) = 16*abs(s)/sqrt(s^2+1)
In above, I assume you want to get the length of a and b. In reality, your s is a signed value, so a could be negative. Therefore, the incremental value of a will really be:
a = 16s/sqrt(s^2+1)
The Point class built in to Flash has a wonderful set of methods for doing exactly what you want. Define the line using two points and you can use the "interpolate" method to get points further down the line automatically, without any of the trigonometry.
http://help.adobe.com/en_US/FlashPlatform/reference/actionscript/3/flash/geom/Point.html#interpolate()
The Slope is dy/dx. Or in your terms A/B.
Therefore you can step along the line by adding A to the Y coordinate, and B to the X coordinate. You can Scale A and B to make the steps bigger or smaller.
To Calculate the slope and get A and B.
Take two points on the line (X1,Y1) , (X2,Y2)
A= (Y2-Y1)
B= (X2-X1)
If you calculate this with the two points you want to iterate between simply divide A and B by the number of steps you want to take
STEPS=10
yStep= A/STEPS
xStep= B/STEPS
for (i=0;i<STEPS;i++)
{
xCur=x1+xStep*i;
yCur=y1+yStep*i;
}
Given the equation for a line as y=slope*x+intercept, you can simply plug in the x-values and read back the y's.
Your problem is computing the step-size along the x-axis (how big a change in x results from a 16-pixel move along the line, which is b in your included plot). Given that you know a^2 + b^2 = 16 (by definition) and slope = a/b, you can compute this:
slope = a/b => a = b * slope [multiply both sides by b]
a^2 + b^2 = 16 => (b * slope)^2 + b^2 = 16 [by substitution from the previous step]
I'll leave it to you to solve for b. After you have b you can compute (x,y) values by:
for x = 0; x += b
y = slope * x + intercept
echo (x,y)
loop