Combinational Circuit design question.
A
____
| |
F | | B
| |
____
| G |
E | | C
| |
____
D
Suppose this is a LED display. It would take input of 4 bit
(0000)-(1111) and display the Hex of it. For example
if (1100) come in it would display C by turning on AFED and turning off BCG.
If (1010) comes in it would display A by turning on ABCEFG
and turn off D.
These display will all be Capital letters so there is no visual
difference between 0 and D and 8 and B.
Develop a truth table and an optimized expression using Karnaugh Maps.
I'm not exactly sure how to begin. For the truth table would I be using (w,x,y,z) as input variable or just the ABCDEFG variable since it's the one turning on and off?
input (1010)-->A--> ABCEFG~D (~ stand for NOT)
input (1011)-->B--> ABCDEFG
input (1100)-->C--> ADEF~B~C~G
So would I do for all hex 0-F then that would give me the min. term canonical then use Karnaugh Map to optimize it? Any help would be grateful!
1) Map your lights to bits:
ABCDEFG, so truth table will be:
ABCDEFG
input (1010)-->A-->1110110
and so on.
You will have big table (with 16 rows).
2) Then follow sample on wikipedia for every output light.
You need to do 7 of these: Each for one segment in the 7-segment display.
This figure is for illustration only. It doesn't necessarily map to any segment in your problem.
cd=00 01 11 10 <-- where abcd = 0000 for 0 : put '1' if the light is on
ab= 00 1 1 1 1 = 0001 for 1 : put '0' if it's off for
ab= 01 1 1 1 0 = 0010 for 2 ... the given segment
ab= 11 0 1 1 1
ab= 10 1 1 1 0 = 1111 for f
^^^^ = d=1 region
^^^^ = c==1 region
The two middle rows represent "b==1" region and the two last rows are a==1 region.
From that map find maximum size rectangles (that are of size [1,2 or 4] x [1, 2 or 4]); that can be overlapping. The middle 2x4 region is coded as 'd'. The top row is '~a~b'. The top left 2x2 square is '~a~c'. A bottom left square that wraps from row 4 to row 1 is '~b~c'. Finally the small 2x1 region that covers position x=4, y=3 is 'abc'.
This function would thus be 'd + ~a~b + ~a~c + ~b~c + abc'. If there are no redundant squares (that are completely covered by other squares), then this formula should be optimal canonical form. (not counting XOR operation). Repeat for 7 times for the real data!
Any selection/permutation of the variables should give the same logical circuit, whether you use abcd or dcba or acbd etc.
Related
Given 15 random hexadecimal numbers (60 bits) where there is always at least 1 duplicate in every 20 bit run (5 hexdecimals).
What is the optimal way to compress the bytes?
Here are some examples:
01230 45647 789AA
D8D9F 8AAAF 21052
20D22 8CC56 AA53A
AECAB 3BB95 E1E6D
9993F C9F29 B3130
Initially I've been trying to use Huffman encoding on just 20 bits because huffman coding can go from 20 bits down to ~10 bits but storing the table takes more than 9 bits.
Here is the breakdown showing 20 bits -> 10 bits for 01230
Character Frequency Assignment Space Savings
0 2 0 2×4 - 2×1 = 6 bits
2 1 10 1×4 - 1×2 = 2 bits
1 1 110 1×4 - 1×3 = 1 bits
3 1 111 1×4 - 1×3 = 1 bits
I then tried to do huffman encoding on all 300 bits (five 60bit runs) and here is the mapping given the above example:
Character Frequency Assignment Space Savings
---------------------------------------------------------
a 10 101 10×4 - 10×3 = 10 bits
9 8 000 8×4 - 8×3 = 8 bits
2 7 1111 7×4 - 7×4 = 0 bits
3 6 1101 6×4 - 6×4 = 0 bits
0 5 1100 5×4 - 5×4 = 0 bits
5 5 1001 5×4 - 5×4 = 0 bits
1 4 0010 4×4 - 4×4 = 0 bits
8 4 0111 4×4 - 4×4 = 0 bits
d 4 0101 4×4 - 4×4 = 0 bits
f 4 0110 4×4 - 4×4 = 0 bits
c 4 1000 4×4 - 4×4 = 0 bits
b 4 0011 4×4 - 4×4 = 0 bits
6 3 11100 3×4 - 3×5 = -3 bits
e 3 11101 3×4 - 3×5 = -3 bits
4 2 01000 2×4 - 2×5 = -2 bits
7 2 01001 2×4 - 2×5 = -2 bits
This yields a savings of 8 bits overall, but 8 bits isn't enough to store the huffman table. It seems because of the randomness of the data that the more bits you try to encode with huffman the less effective it works. Huffman encoding seemed to work best with 20 bits (50% reduction) but storing the table in 9 or less bits isnt possible AFAIK.
In the worst-case for a 60 bit string there are still at least 3 duplicates, the average case there are more than 3 duplicates (my assumption). As a result of at least 3 duplicates the most symbols you can have in a run of 60 bits is just 12.
Because of the duplicates plus the less than 16 symbols, I can't help but feel like there is some type of compression that can be used
If I simply count the number of 20-bit values with at least two hexadecimal digits equal, there are 524,416 of them. A smidge more than 219. So the most you could possibly save is a little less than one bit out of the 20.
Hardly seems worth it.
If I split your question in two parts:
How do I compress (perfect) random data: You can't. Every bit is some new entropy which can't be "guessed" by a compression algorithm.
How to compress "one duplicate in five characters": There are exactly 10 options where the duplicate can be (see table below). This is basically the entropy. Just store which option it is (maybe grouped for the whole line).
These are the options:
AAbcd = 1 AbAcd = 2 AbcAd = 3 AbcdA = 4 (<-- cases where first character is duplicated somewhere)
aBBcd = 5 aBcBd = 6 aBcdB = 7 (<-- cases where second character is duplicated somewhere)
abCCd = 8 abCdC = 9 (<-- cases where third character is duplicated somewhere)
abcDD = 0 (<-- cases where last characters are duplicated)
So for your first example:
01230 45647 789AA
The first one (01230) is option 4, the second 3 and the third option 0.
You can compress this by multiplying each consecutive by 10: (4*10 + 3)*10 + 0 = 430
And uncompress it by using divide and modulo: 430%10=0, (430/10)%10=3, (430/10/10)%10=4. So you could store your number like that:
1AE 0123 4567 789A
^^^ this is 430 in hex and requires only 10 bit
The maximum number for the three options combined is 1000, so 10 bit are enough.
Compared to storing these 3 characters normally you save 2 bit. As someone else already commented - this is probably not worth it. For the whole line it's even less: 2 bit / 60 bit = 3.3% saved.
If you want to get rid of the duplicates first, do this, then look at the links at the bottom of the page. If you don't want to get rid of the duplicates, then still look at the links at the bottom of the page:
Array.prototype.contains = function(v) {
for (var i = 0; i < this.length; i++) {
if (this[i] === v) return true;
}
return false;
};
Array.prototype.unique = function() {
var arr = [];
for (var i = 0; i < this.length; i++) {
if (!arr.contains(this[i])) {
arr.push(this[i]);
}
}
return arr;
}
var duplicates = [1, 3, 4, 2, 1, 2, 3, 8];
var uniques = duplicates.unique(); // result = [1,3,4,2,8]
console.log(uniques);
Then you would have shortened your code that you have to deal with. Then you might want to check out Smaz
Smaz is a simple compression library suitable for compressing strings.
If that doesn't work, then you could take a look at this:
http://ed-von-schleck.github.io/shoco/
Shoco is a C library to compress and decompress short strings. It is very fast and easy to use. The default compression model is optimized for english words, but you can generate your own compression model based on your specific input data.
Let me know if it works!
While going through the OpenCV source code, I noticed that for iterations more than one it just creates a kernel of bigger size and do a single iteration.
So my question is if we take SQUARE structuring element of 3x3 size and dilate/erode it in three iterations, will it be same as dilating/eroding it with a 9x9 kernel once.
if( iterations > 1 && countNonZero(kernel) == kernel.rows*kernel.cols )
{
anchor = Point(anchor.x*iterations, anchor.y*iterations);
kernel = getStructuringElement(MORPH_RECT,
Size(ksize.width + (iterations-1)*(ksize.width-1),
ksize.height + (iterations-1)*(ksize.height-1)),
anchor);
iterations = 1;
}
Refering to Jordi's Answer:
[Quoted] ... Note, however, that this does not hold for all structuring elements...
In fact, it holds, in the following way (not in Jordi's example):
First step, calculate the 5x5 kernel by dilation twice in 3x3 kernel on a single center point 5x5 source image:
00000 00000 00100
00000 010 00100 010 01110
00100 + 111 -> 01110 + 111 -> 11111 ===> this is the equivalent 5x5 kernel for 2x 3x3 dilation
00000 010 00100 010 01110
00000 00000 00100
Then applying twice of 3x3 original dilation kernel is equivalent to applying this 5x5 dilation kernel on a bigger image. For example:
0000000000 0000000000 00100
0000000000 010 010 0000000000 01110
0011100000 + 111 + 111 === 0011100000 + 11111
0000001000 010 010 0000001000 01110
0000000000 0000000000 00100
0000000000 0000000000
This does not directly answer your question though. However, I can not just use 'comment' as it is very hard (if not impossible) to format all these equations/explanations.
In fact, a proof for binary image (image with only value 0 or 1 in each pixel) for the larger combined kernel for dilation is easy:
Let's define the binary operator + to be the dilation operator, where the 1st operand is the kernel, and the second operand is the image to be dilated.. So, if we want to do dilation on image I with kernel K, we write dilated-image = K + I
Let's define binary operator U to be the union operator, or, in other word, the binary 'OR' operator for each pixel, where the two operand of U must be binary images in the same dimension. For example: A U B means doing -OR- on each corresponding pixel of A and B:
A= 0 0 1 B= 0 1 1
1 0 1 1 1 1
1 1 0 0 1 0
Then
A U B = 0 1 1
1 1 1
1 1 0
We also define U A(i), i=1, ..., n to be A(1) U A(2) U ... U A(n).
Let's define K^n to be the dilation-styled larger kernel by applying n times of kernel K on a single center point image.
Note that any image I, we can decompose it into union of single point images. For example,
0 1 0 0 1 0 0 0 0 0 0 0
I = 0 0 0 === 0 0 0 U 0 0 0 U 0 0 0
1 0 1 0 0 0 1 0 0 0 0 1
Now it's time to prove it:
For any image I, we define D(i), i = 1, ..., n to be the single point decomposition of I,
and thus I = U D(i), i = 1, ..., n
By definition of the binary dilation, K + I == K + (U D(i)) == U (K+D(i)).
(Remember that dilation is to mask kernel K on each pixel of I, and mark all corresponding 1's).
Now, let's see what is K + (K + I):
K + (K + I) == K + U (K + D(i))
== U(K + (K + D(i))) (Note: this is tricky. see Theorem 1 below)
== U (K^2 + D(i)) (by definition of K^2)
== K^2 + U D(i) (by definition of the dilation)
== K^2 + I (since I = U D(i))
Now, we already know K + (K + I) == K^2 + I, and it's easy to apply mathematical induction to prove that K + K + K .. + K + I = K^n + I (Note: please apply right association, as I have drop the parenthesis).
Theorem 1: Proof of the deduction from K + U (K + D(i)) to U(K + (K+D(i)))
It's suffice to just prove that for any two binary images A and B in a same dimension,
K + (A U B) = (K+A) U (K+B)
It's quite easy to see that, if we decompose image A and B, and apply kernel K on the decomposed images, those common points (i.e. the intersection points of A and B, or the common 1's point of A and B), will contribute the same resulting points after applying kernel K. And by the definition of dilation, we need to union all points contributed by each decomposed image of A and B. Thus Theorem 1 holds.
=== UPDATE ===
Regarding to kid.abr's comment "27 operations compared to 7x7 kernel with 49 operations":
Generally speaking, it is not 27 operations. It depends. For example, a source image of 100x100 pixels,
with 20 singular point (1's) sparsely distributed. Applying a 3x3 solid kernel (i.e. All 1's) 3 times on it
requires the following steps for each of the 20 singular point:
Loop 1: 9 operations, and generate 9 points.
Loop 2: For each of the 9 points generated, it needs 9 operations => 9 x 9 = 81 steps. And it generates 25 points
Loop 3: For each of the 25 points generated, it needs 9 operations => 25 x 9 = 225 steps.
Total: 9 + 81 + 225 = 315 steps.
Please note that when we visit a pixel with 0 value in the source image, we don't need to apply the kernel
on that point, right?
So, the same case applying the larger kernel, it requires 7x7 = 49 steps.
Yet, if the source image has a large solid area of 1's, the 3-step method wins.
Short answer: with a square structuring element, yes.
Long answer: you need to consider what the erosion/dilation operations do. Dilation, for instance, moves the kernel over the image and sets its centre to 1 whenever any of its grid positions are 1 (I'm assuming binary images, it works the same for greyscale). Increasing the distance between the centre of the structuring element and its edges is then the same as increasing the size of the kernel.
Note, however, that this does not hold for all structuring elements. Suppose you take a structuring element that is just a stretched plus, obviously dilating twice with size 3 is not the same as dilating once with size 5:
00000 00000 00100
00000 010 00100 010 01110
00100 + 111 -> 01110 + 111 -> 11111
00000 010 00100 010 01110
00000 00000 00100
00000 00100 00100
00000 00100 00100
00100 + 11111 -> 11111
00000 00100 00100
00000 00100 00100
Of course, this does work if we define the scaled version of plus as a square without its corners (as it usually would be). I think that in general this shortcut works when the kernel of size k+1 is the dilated version of the kernel of size k, but I have no proof for this.
Short answer for a general kernel: Yes for dilation/erosion, but not necessarily with an equivalent kernel.
From wikipedia:
Dilation: (A⊕B)⊕C = A⊕(B⊕C)
Erosion: (A⊖B)⊖C = A⊖(B⊕C)
Where ⊕ denotes the morphological dilation, and ⊖ denotes the morphological erosion.
Basically, performing erosion/dilation on image A with kernel B and then kernel C is equivalent to performing erosion/dilation on image A with the kernel obtained by dilating B with C. This can easily be expanded to an arbitrary number erosions/dilations.
I'm studying fundamental of data structure (Queue) , so far I understand the flow of Queue but I don't understand whenever queue is applying with Mod operator. There a several question which confusing my brain. How to answer this question (refer to picture)?
The best method for handling circular queues is to draw them out. Since circles don't post very well with ASCII art, I'll use a linear array.
+---+---+---+---+---+
| | | | | |
+---+---+---+---+---+
0 1 2 3 4
^
Rear
The REAR is at index 4.
Let's perform the operation step by step.
First: Add 1 to REAR. This makes REAR point beyond the array:
+---+---+---+---+---+
| | | | | |
+---+---+---+---+---+
0 1 2 3 4 5
^
Rear
Applying the modulo operation, %, this will give us the remainder of 5 / 5 which is zero:
+---+---+---+---+---+
| | | | | |
+---+---+---+---+---+
0 1 2 3 4
^
Rear
Thus the modulo operation wraps around the array, like a circle.
The next question is for you to solve. Remember draw the array or queue. You can use circles (think of a pie sliced or a pizza sliced).
Edit 1: Modulo details
The modulo operation will give a value in the range 0..N, when N is the divisor.
Given N == 4, here are some results for modulo:
Index result
0 0
1 1
2 2
3 3
4 0 --> The remainder of 4 / 4 == 0.
5 1
6 2
7 3
8 0 --> The remainder of 8 / 4 == 0.
Modulus returns the remainder of the two operands. For example, 4%2=0 since 4/2=2 with no remainder, while 4%3=1 since 4/3=1 with remainder 1. Since you can never have a remainder higher than the right operand, you have an effective "range" of answers for any modulus of 0 to (n-1). With that in mind, just plug in the numbers for the variables ((4+1)%5=? and (1+1)%4=?). Usually to find the remainder you would use long division, but one useful thing to remember is that any number divided by itself has a remainder of 0, and any number divided by a larger number will have a remainder equal to itself.
I'm looking for some regex/automata help. I'm limited to + or the Kleene Star. Parsing through a string representing a ternary number (like binary, just 3), I need to be able to know if the result is 1-less than a multiple of 4.
So, for example 120 = 0*1+2*3+1*9 = 9+6 = 15 = 16-1 = 4(n)-1.
Even a pointer to the pattern would be really helpful!
You can generate a series of values to do some observation with bc in bash:
for n in {1..40}; do v=$((4*n-1)); echo -en $v"\t"; echo "ibase=10;obase=3;$v" | bc ; done
3 10
7 21
11 102
15 120
19 201
23 212
27 1000
31 1011
...
Notice that each digit's value (in decimal) is either 1 more or 1 less than something divisible by 4, alternately. So the 1 (lsb) digit is one more than 0, the 3 (2nd) digit is one less than 4, the 9 (3rd) digit is 1 more than 8, the 27 (4th) digit is one less than 28, etc.
If you sum up all the even-placed digits and all the odd-placed digits, then add 1 to the odd-placed ones (if counting from 1), you should get equality.
In your example: odd: (0+1)+1, even: (2). So they are equal, and so the number is of the form 4n-1.
I am working on a project where I need to create a boundary around a group of rectangles.
Let's use this picture as an example of what I want to accomplish.
EDIT: Couldn't get the image tag to work properly, so here is the full link:
http://www.flickr.com/photos/21093416#N04/3029621742/
We have rectangles A and C who are linked by a special link rectangle B. You could think of this as two nodes in a graph (A,C) and the edge between them (B). That means the rectangles have pointers to each other in the following manner: A->B, A<-B->C, C->B
Each rectangle has four vertices stored in an array where index 0 is bottom left, and index 3 is bottom right.
I want to "traverse" this linked structure and calculate the vertices making up the boundary (red line) around it. I already have some small ideas around how to accomplish this, but want to know if some of you more mathematically inclined have some neat tricks up your sleeves.
The reason I post this here is just that someone might have solved a similar problem before, and have some ideas I could use. I don't expect anyone to sit down and think this through long and hard. I'm going to work on a solution in parallell as I wait for answers.
Any input is greatly appreciated.
Using the example, where rectangles are perpendicular to each other and can therefore be presented by four values (two x coordinates and two y coordinates):
1 2 3 4 5 6
1 +---+---+
| |
2 + A +---+---+
| | B |
3 + + +---+---+
| | | | |
4 +---+---+---+---+ +
| |
5 + C +
| |
6 +---+---+
1) collect all the x coordinates (both left and right) into a list, then sort it and remove duplicates
1 3 4 5 6
2) collect all the y coordinates (both top and bottom) into a list, then sort it and remove duplicates
1 2 3 4 6
3) create a 2D array by number of gaps between the unique x coordinates * number of gaps between the unique y coordinates. It only needs to be one bit per cell, so in c++ a vector<bool> with likely give you a very memory-efficient version of this
4 * 4
4) paint all the rectangles into this grid
1 3 4 5 6
1 +---+
| 1 | 0 0 0
2 +---+---+---+
| 1 | 1 | 1 | 0
3 +---+---+---+---+
| 1 | 1 | 1 | 1 |
4 +---+---+---+---+
0 0 | 1 | 1 |
6 +---+---+
5) for each cell in the grid, for each edge, if the cell beside it in that cardinal direction is not painted, draw the boundary line for that edge
In the question, the rectangles are described as being four vectors where each represents a corner. If each rectangle can be at arbitrary and different rotation from others, then the approach I've outlined above won't work. The problem of finding the path around a complex polygon is regularly solved by vector graphics rasterizers, and a good approach to solving the problem is using a library such as Cairo to do the work for you!
The generalized solution to this problem is to implement boolean operations in terms of a scanline. You can find a brief discussion here to get you started. From the text:
"The basis of the boolean algorithms is scanlines. For the basic principles the book: Computational Geometry an Introduction by Franco P. Preparata and Michael Ian Shamos is very good."
I own this book, though it's at the office now, so I can't look up the page numbers you should read, though chapter 8, on the geometry of rectangles is probably the best starting point.
Calculate the sum of the boundaries of all 3 rectangles seperately
calculate the overlapping rectangle of A and B, and subtract it from the sum
Do the same for the overlapping rectangle of B and C
(to get the overlapping rectangle from A and B take the middle 2 X positions, together with the middle 2 Y positions)
Example (x1,y1) - (x2,y2):
Rectangle A: (1,1) - (3,4)
Rectangle B: (3,2) - (5,4)
Rectangle C: (4,3) - (6,6)
Calculation:
10 + 8 + 10 = 28
X coords ordered = 1,3,3,5 middle two are 3 and 3
Y coords ordered = 1,2,4,4 middle two are 2 and 4
so: (3,2) - (3,4) : boundery = 4
X coords ordered = 3,4,5,6 middle two are 4 and 5
Y coords ordered = 2,3,4,6 middle two are 3 and 4
so: (4,3) - (5,4) : boundery = 4
28 - 4 - 4 = 20
This is my example visualized:
1 2 3 4 5 6
1 +---+---+
| |
2 + A +---+---+
| | B |
3 + + +---+---+
| | | | |
4 +---+---+---+---+ +
| |
5 + C +
| |
6 +---+---+
A simple trick should be:
Create a region from the first rectangle
Add the other rectangles to the region
Get the boundary of the region (somehow? :P)
After some thinking I might end up doing something like this:
Pseudo code:
LinkRectsConnectedTo(Rectangle rectangle,Edge startEdge) // Edge can be West,North,East,South
for each edge in rectangle starting with the edge facing last rectangle
add vertices in the edge to the final boundary polygon
if edge is connected to another rectangle
if edge not equals startEdge
recursively call LinkRectsConnectedTo(rectangle,startEdge)
Obvisouly this pseudo code would have to be refined a bit and might not cover all cases, but I think I might have solved my own problem.
I haven't thought this out completely, but I wonder if you couldn't do something like:
Make a list of all the edges.
Get all the edges where P1.X = P2.X
In that list, get the pairs where X are equal
For each pair, replace with one or two edges for the parts where they DON'T overlap
Do something clever to get the edges in the right order
Will your rectangles always be horizontally aligned, if not you'd need to do the same thing but for Y too?
And are they always guaranteed to be touching? If not the algorithm wouldn't be broken, but the 'right order' wouldn't be definable.