I am just curious on how do I determine whether a simplified boolean expression is in a SOP form or POS form.
for example this question:
Question
the answer to this expression is : NOT B.D/ ⌝B.D
and this is in SOP form
Can anyone explain why?
I think this should be a 'philosophical' argument. ⌝B.D is the special case where the number of elements to be summed up becomes one.
You can think of ⌝B.D = ⌝B.D + ⌝B.B + ⌝D.D + 0.(anything) which makes it an SOP.
Terminology:
First the theory, you can further study on Wikipedia (DNF, CNF):
Sum of products = DNF (Disjunctive normal form) = disjunction (+) of conjunctions (·) ~ "the disjunction is not inside any bracket, but only
as the root operator(s)".
Product of sums = CNF (Conjunctive normal form) = conjuction of disjunctions ~ "the conjunction is not inside a bracket, but only
as the root operator(s)".
Full/Complete CNF/DNF = the terms (products/sums) contains all given variables in either direct or negated form; the terms are then maxterms/minterms.
Finding the right circles:
You can see, that the four circles in the Karnaugh map correspond to the four products in the original function in the same order (top to bottom, left to right).
Given function as a SOP:
The function is now in a form of sum of products, because you can literally see, that there are four products.
It is also in the form of sum of maxterms, because the four parts contain all variables in their direct or negated form.
f(a,b,c,d) = ¬a·¬b·¬c·d + ¬a·¬b·c·d + a·¬b·c·d + a·¬b·¬c·d
For example the first term: ¬a·¬b·¬c·d ~ if the variables a, b and c are logical zeros and only d is true, then the function's output is logical 1.
Minimized function as a SOP:
You can see, that the maxterms can be grouped and that creates the minimal sum of product: f(a,b,c,d) = ¬b·d, because all cells, where b is a logical 0 and d is a logical 1 are included.
The minimized function is indeed a SOP/DNF, because it certainly does contain only one product — the ¬b·d — and there is no + (disjunction) operator inside that product.
Minimized function as a POS:
The surprise might come when you realize, that circling and writing the function as a product of sums yields the same minimal form: f(a,b,c,d) = (¬b)·(d), because there are exactly two terms: ¬b (orange circle) and d (red circle).
Both are sums with only one operand. Because of that the minimized function is a product of sum.
Conclusion:
The minimized function f(a,b,c,d) = ¬b·d is both a SOP and a POS. You can check the correct solution by using wolframalpha.com.
Related
I am trying to find a (SoP)-expression using the embedded K-map. I have a box of size 4x4 which is a permitted use however I am having a hard time understanding how I could implement it.
To me the 4x4 box represents that the output is always 1 independet on any of the variables. Then I'd like to use the 2x4 box to the right and produce:
1 OR (Qc AND !Qd), but this does not produce the correct result.
I can see several alternative ways to produce the correct result. My questions are specifically:
Why can't I use the 4x4 box, or perhaps, how do I represent it correctly?
How do I know when I can represent parts of the output as a 4x4 box?
Perhaps Im missing something more fundamental.
Thx in advance.
The point of placing rectangles in a K-map is to eliminate variables from an expression. When the result of a rectangle is the same for the variable values X and X', then the variable X is not needed and can be removed. You do this by extending an existing rectangle by doubling the size and eliminating exactly one variable, where every other variable stays the same. For the common/normal K-map with four variables this works with every such rectangle because in a way the columns/rows are labelled/positioned. See the following example:
The rectangle has eliminated the variables A and B, one variable at a time when the size of the rectangle has been extended/doubled. This results in the function F(A,B,C,D) = C'D'. But check the following K-map of four variables:
Notice that the columns for the D variable has been changed (resulting in a different function overall). When you try to extend the red rectangle to catch the other two 1 values as well, you are eliminating two variables at the same time (B and D). As you cannot grow the rectangle anymore, you are left with two rectangles, resulting in the function F(A,B,C,D) = BC'D' + B'C'D (which can be simplified to C' * (BD' + B'D)).
The practice in placing rectangles in the K-map isn't just placing the biggest rectangle possible, but to eliminate variables in the right way. To answer your questions, you can always start with the smallest rectangle and extend/double its size to eliminate one variable. See the following example:
The green rectangle grows in these steps:
Start with A'BC'D'E
Eliminate the (only) variable A by growing "down", resulting in BC'D'E
Eliminate the (only) variable D by growing "right", resulting in BC'E.
But now, the rectangle cannot grow/double its size anymore because that would eliminate the variable E, but also somehow eliminate the variable C. You cannot eliminate the variable E, because you have 0 values to the left of the green rectangle and 1 values to the right of the green rectangle (all in the left half of the K-map, where you have the value C'). The only way to increase/grow the rectangle is to get the "don't care" values to eliminate the B variable (not shown here).
The overall function for this K-map would be F(A,B,C,D,E) = C'E + DE' + CD' (from three 2x4 rectangles).
I have this situation:
val a = listOf("wwfooww", "qqbarooo", "ttbazi")
val b = listOf("foo", "bar")
I want to determine if all items of b are contained in substrings of a, so the desired function should return true in the situation above. The best I can come up with is this:
return a.any { it.contains("foo") } && a.any { it.contains("bar") }
But it iterates over a twice. a.containsAll(b) doesn't work either because it compares on string equality and not substrings.
Not sure if there is any way of doing that without iterating over a the same amount as b.size. Because if you only want 1 iteration of a, you will have to check all the elements on b and now you are iterating over b a.size times plus, in this scenario, you also need to keep track of which item in b already had a match, and not check them again, which might be worse than just iterating over a, since you can only do that by either removing them from the list (or a copy, which you use instead of b), or by using another list to keep track of the matches, then compare that to the original b.
So I think that you are on the right track with your code there, but there are some issues. For example you don't have any reference to b, just hardcoded strings, and doing it like that for all elements in b will result in quite a big function if you have more than 2, or better yet, if you don't already know the values.
This code will do the same thing as the one you put above, but it will actually use elements from b, and not hardcoded strings that match b. (it will iterate over b b.size times, and partially over a b.size times)
return b.all { bItem ->
a.any { it.contains(bItem) }
}
Alex's answer is by far the simplest approach, and is almost certainly the best one in most circumstances.
However, it has complexity A*B (where A and B are the sizes of the two lists) — which means that it doesn't scale: if both lists get big, it'll get very slow.
So for completeness, here's a way that's more involved, and slower for the small cases, but has complexity proportional to A+B and so can cope efficiently with much larger lists.
The idea is to preprocess the a list, to generate a set of all the possible substrings, and then scan through the b list just checking for inclusion in that set. (The preprocessing step takes time proportional* to A. Converting the substrings into a set means that it can check whether a string is present in constant time, using its hash code; so the rest then takes time proportional to B.)
I think this is clearest using a helper function:
/**
* Generates a list of all possible substrings, including
* the string itself (but excluding the empty string).
*/
fun String.substrings()
= indices.flatMap { start ->
((start + 1)..length).map { end ->
substring(start, end)
}
}
For example, "1234".substrings() gives [1, 12, 123, 1234, 2, 23, 234, 3, 34, 4].
Then we can generate the set of all substrings of items from a, and check that every item of b is in it:
return a.flatMap{ it.substrings() }
.toSet()
.containsAll(b)
(* Actually, the complexity is also affected by the lengths of the strings in the a list. Alex's version is directly proportional to the average length, while the preprocessing part of the algorithm above is proportional to its square (as indicated by the map nested in the flatMap). That's not good, of course; but in practice while the lists are likely to get longer, the strings within them probably won't, so that's unlikely to be significant. Worth knowing about, though.
And there are probably other, still more complex algorithms, that scale even better…)
What is the differences in following functions available in org.gdal.ogr.Geometry.
Within() vs Contains()
Crosses() vs Intersect()
The within and contains functions are reciprocal functions. a contains b, if and only if, b is within a. The C++ source code, says the same:
bool
Geometry::within(const Geometry* g) const
{
return g->contains(this);
}
The crosses() and intersects() function have different implementation.
According to this documentation page, intersects return true if disjoint returns false, and the disjoint predicate means that the two geometries have no point in common. We can conclude that intersects means that the two geometries have at least one point in common.
In the other hand, the crosses predicate means that the geometries have some but not all interior points in common. Which is more restrictive than the intersects predicate.
Ortomala Lokni is correct about crosses and intersects, Here is a visual explanation of how within and contains work.
Say that you have two geometries: The point A and the polygon B like shown on the image
Then A is within B, while B contains A.
The inverse is not true. A does not contains B and B is not within A.
the following two statements are true:
A.within(B) -> B.contains(A)
A.contains(B) -> B.within(A)
I am new to Z notation,
Lets say I have a function f defined as X |--> Y ,
where X is string and Y is number.
How can I get highest Y value in this function? Does 'loop' exist in formal method so I can solve it using loop?
I know there is recursion in Z notation, but based on the material provided, I only found it apply in multiset or bag, can it apply in function?
Any extra reference application of 'loop' or recursion application will be appreciated. Sorry for my English.
You can just use the predefined function max that takes a set of integers as input and returns the maximum number. The input values here are the range (the set of all values) of the function:
max(ran(f))
Please note that the maximum is not defined for empty sets.
Regarding your question about recursion or loops: You can actually define a function recursively but I think your question aims more at a way to compute something. This is not easily expressed in Z and this is IMO a good thing because it is used for specifications and it is not a programming language. Even if there wouldn't be a max or ran function, you could still specify the number m you are looking for by:
\exists s:String # (s,m):f /\
\forall s2:String, i2:Z # (s2,i2):f ==> i2 <= m
("m is a value of f, belonging to an s and all other values i2 of f are smaller or equal")
After getting used to the style it is usually far better to understand than any programming language (except your are trying to describe an algorithm itself and not its expected outcome).#
Just for reference: An example of a recursive definition (let's call it rmax) for the maximum would consist of a base case:
\forall e:Z # rmax({e}) = e
and a recursive case:
\forall e:Z; S:\pow(Z) #
S \noteq {} \land
rmax({e} \cup S) = \IF e > rmax(S) \THEN e \ELSE rmax(S)
But note that this is still not a "computation rule" of rmax because e in the second rule can be an arbitrary element of S. In more complex scenarios it might even be not obvious that the defined relation is a function at all because depending on the chosen elements different results could be computed.
Given:
I have no idea what the accepted language is.
From looking at it you can get several end results:
1.) bb
2.) ab(a,b)
3.) bbab(a, b)
4.) bbaaa
How to write regular expression for a DFA
In any automata, the purpose of state is like memory element. A state stores some information in automate like ON-OFF fan switch.
A Deterministic-Finite-Automata(DFA) called finite automata because finite amount of memory present in the form of states. For any Regular Language(RL) a DFA is always possible.
Let's see what information stored in the DFA (refer my colorful figure).
(note: In my explanation any number means zero or more times and Λ is null symbol)
State-1: is START state and information stored in it is even number of a has been come. And ZERO b.
Regular Expression(RE) for this state is = (aa)*.
State-4: Odd number of a has been come. And ZERO b.
Regular Expression for this state is = (aa)*a.
Figure: a BLUE states = EVEN number of a, and RED states = ODD number of a has been come.
NOTICE: Once first b has been come, move can't back to state-1 and state-4.
State-5: comes after Yellow b. Yellow b means b after odd numbers of a.
Once you gets b after odd numbers of a(at state-5) every thing is acceptable because there is self a loop for (b,a) at state-5.
You can write for state-5 : Yellow-b followed-by any string of a, b that is = Yellow-b (a + b)*
State-6: Just to differentiate whether odd a or even.
State-2: comes after even a then b then any number of b. = (aa)* bb*
State-3: comes after state-2 then first a then there is a loop via state-6.
We can write for state-3 comes = state-2 a (aa)* = (aa)*bb* a (aa)*
Because in our DFA, we have three final states so language accepted by DFA is union (+ in RE) of three RL (or three RE).
So the language accepted by the DFA is corresponding to three accepting states-2,3,5, And we can write like:
State-2 + state-3 + state-5
(aa)*bb* + (aa)*bb* a (aa)* + Yellow-b (a + b)*
I forgot to explain how Yellow-b comes?
ANSWER: Yellow-b is a b after state-4 or state-3. And we can write like:
Yellow-b = ( state-4 + state-3 ) b = ( (aa)*a + (aa)*bb* a (aa)* ) b
[ANSWER]
(aa)*bb* + (aa)*bb* a (aa)* + ( (aa)*a + (aa)*bb* a (aa)* ) b (a + b)*
English Description of Language: DFA accepts union of three languages
EVEN NUMBERs OF a's, FOLLOWED BY ONE OR MORE b's,
EVEN NUMBERs OF a's, FOLLOWED BY ONE OR MORE b's, FOLLOWED BY ODD NUMBERs OF a's.
A PREFIX STRING OF a AND b WITH ODD NUMBER OF a's, FOLLOWED BY b, FOLLOWED BY ANY STRING OF a AND b AND Λ.
English Description is complex but this the only way to describe the language. You can improve it by first convert given DFA into minimized DFA then write RE and description.
Also, there is a Derivative Method to find RE from a given Transition Graph using Arden's Theorem. I have explained here how to write a regular expression for a DFA using Arden's theorem. The transition graph must first be converted into a standard form without the null-move and single start state. But I prefer to learn Theory of computation by analysis instead of using the Mathematical derivation approach.
I guess this question isn't relevant anymore :) and it's probably better to guide you through it then just stating the answer, but I think I got a basic expression that covers it (it's probably minimizable), so i'll just write it down for future searchers
(aa)*b(b)* // for stoping at 2
U
(aa)*b(b)*a(aa)* // for stoping at 3
U
(aa)*b(b)*a(aa)*b((a)*(b)*)* // for stoping at 5 via 3
U
a(aa)*b((a)*(b)*)* // for stoping at 5 via 4
The examples (1 - 4) that you give there are not the language accepted by the DFA. They are merely strings that belong to the language that the DFA accepts. Therefore, they all fall in the same language.
If you want to figure out the regular expression that defines that DFA, you will need to do something called k-path induction, and you can read up on it here.