Assuming that A = True, B = False and C = True.
Give the results of the following Boolean operations
AB + BC’ (A + B’C) + B
(b) ABC’ + (B + AC’)B + A’
in boolean multiply operation is AND (&&) operation
and plus + operation is OR(||) operation
and ' operator measn NOT(!)
so,
AB + BC’ (A + B’C) + B
can be written as
(A && B) || [(B && (!C)) && (A || ((!B) && C))] || B
PUTTING THE VALUE IN THE ABOVE EQUATION
= (1 && 0) || [(0 && (!0)) && (1 || ((!1) && 1))] || 1
= 0 || [(0 && 1 )&& (1 || (0 && 1)] ||1
= 0 || [(0 && 1) && (1 || 0)] ||1
= 0 || [0 && 1] ||1
= 0 || 0 || 1
= 1
I KNOW IT IS HARD TO FOLLOW BUT THE ANSWER WILL BE 1
AND FOLLOWING THE SAME LOGIC THE ANSWER TO THE 2ND STATEMENT WILL ALSO BE 1
IF YOU WANT TO UNDERSTAND HOW THIS WORKS PLEASE READ THE DOCS
I need such a condition in ts:
([treeLevel = 0] && [globalVar = GP:R > 0]) || [PIDinRootline = {$pages.2018}]
I wanna show block if page has treelevel=0 and the get var R > 0, or if page id = $pages.2018
It looks like the similar code in php:
if(($treeLevel == 0 && $r > 0) || (pid == number))
The all expression in first brackets should be right, or in second.
Is it exist the method to group it like the previous record or I can only use userfunc?
There is no grouping in TS conditions, but if you need this particular condition from your post I think it is not needed because brackets around && are useless in this case.
(p && q) || r
is exactly the same as
p && q || r
Did you tested it?
if ((status & 0x3F) == 1 ){ }..
the status is variable in swift language.
what is mean about this condition, & mean and (status & 0x3F) value return
& is the bitwise AND operator. It compares the bits of the two operands and sets the corresponding bit to 1 if it is 1 in both operands, or to 0 if either or both are 0.
So this statement:
((status & 0x3F) == 1)
is combining status with 0b111111 (the binary equivalent of 0x3F and checking if the result is exactly 1. This will only be true if the last 6 bits of status are 0b000001.
In this if:
if( (dtc24_state[2] & 0x8) == 0x8 ) {
self.haldexABCDTC24State.text = status_str + " - UNKNOWN"
self.haldexABCDTC24State.textColor = text_color
active_or_stored_dtc = true
}
dct24_state is an array of values. The value of dct24_state[2] is combined with 0x8 or 0b1000 and checked against 0x8. This is checking if the 4th bit from the right is set. Nothing else matters. If the 4th bit from the right is set, the if is true and the code block is executed.
0x3F is 111111. So, it means this:
for each bit of yourNumber in binary system presentation use and method.
This way truncates the left part of the number. and the result compares with 1.
e.g.
7777 is 1111001100001 after executing and this number converts into
100001. So the result is false.
But for 7745 (1111001000001) the result is 1. The result is true.
The rule for 'and' function: 0 & 0 = 0 ; 0 & 1 = 0; 1 & 0 = 1; 1 & 1 = 1.
I am under the impression that || is the "or" statement in MATLAB. Perhaps someone can explain the confusing behaviour I am seeing:
a = 2;
a == 2 %returns ans = 1 (true)
a == 2 || 3 %returns ans = 1 (true)
a == 3 || 4 %returns ans = 1 (true)??!!
What am I missing here? 'a' is neither 3 or 4, so shouldn't
a == 3 || 4
return ans = 0 (false)?
The expression
a == 3 || 4
is evaluated that way :
a == 3 => false
then
false || 4 => true
if you want to check whether a is equal to 3 or 4 you should write
(a == 3) || (a == 4)
which is evaluated that way
a == 3 => false
then
a == 4 => false
then
false || false => false
Thomas's answer is an excellent explanation of what's going on here; another way that you can compare a variable to multiple answers is using the any() function.
solutions = [3 4];
any(a==solutions);
The a==solutions line creates a matrix the same size as solutions, which contains 1's in indecies which where the conditional is true, and 0's where it is false.
A few more examples:
any(isprime([17:24])); %returns true; 17, 19 and 23 are prime
any(isinteger(sqrt([17:24]))); %(test for square number) returns false; no square numbers in this range
any(mod(magic(3)(:),6)==3); %returns true; 9 mod 6 == 3. Note (:) inserted so that any is evaluated against all entries of the square matrix created by magic
a == 3 || 4 %returns ans = 1 (true)??!!
The reason for the above behaviour is due to the fact that any real number other than '0' in MATLAB is always evaluated to true.
So what is happening here is
The expression a == 3 is evaluated first and found to be false.
Next, expression false || 4 is evaluated.
Since '4' is a real number other than zero, the resulting expression is false || true which is evaluated to true.
To get a desire result use (a == 3) || (a == 4) which is evaluated as false || false which returns false.
I've got three boolean values A, B and C. I need to write an IF statement which will execute if and only if no more than one of these values is True. In other words, here is the truth table:
A | B | C | Result
---+---+---+--------
0 | 0 | 0 | 1
0 | 0 | 1 | 1
0 | 1 | 0 | 1
0 | 1 | 1 | 0
1 | 0 | 0 | 1
1 | 0 | 1 | 0
1 | 1 | 0 | 0
1 | 1 | 1 | 0
What is the best way to write this? I know I can enumerate all possibilities, but that seems... too verbose. :P
Added: Just had one idea:
!(A && B) && !(B && C) && !(A && C)
This checks that no two values are set. The suggestion about sums is OK as well. Even more readable maybe...
(A?1:0) + (B?1:0) + (C?1:0) <= 1
P.S. This is for production code, so I'm going more for code readability than performance.
Added 2: Already accepted answer, but for the curious ones - it's C#. :) The question is pretty much language-agnostic though.
how about treating them as integer 1's and 0's, and checking that their sum equals 1?
EDIT:
now that we know that it's c#.net, i think the most readable solution would look somewhat like
public static class Extensions
{
public static int ToInt(this bool b)
{
return b ? 1 : 0;
}
}
the above tucked away in a class library (appcode?) where we don't have to see it, yet can easily access it (ctrl+click in r#, for instance) and then the implementation will simply be:
public bool noMoreThanOne(params bool[] bools)
{
return bools.ToList().Sum(b => b.ToInt()) <= 1;
}
...
bool check = noMoreThanOne(true, true, false, any, amount, of, bools);
You shold familiarize yourself with Karnaugh maps. Concept is most often applied to electronics but is very useful here too. It's very easy (thought Wikipedia explanation does look long -- it's thorough).
(A XOR B XOR C) OR NOT (A OR B OR C)
Edit: As pointed out by Vilx, this isn't right.
If A and B are both 1, and C is 0, A XOR B will be 0, the overall result will be 0.
How about:
NOT (A AND B) AND NOT (A AND C) AND NOT (B AND C)
If you turn the logic around, you want the condition to be false if you have any pair of booleans that are both true:
if (! ((a && b) || (a && c) || (b && c))) { ... }
For something completely different, you can put the booleans in an array and count how many true values there are:
if ((new bool[] { a, b, c }).Where(x => x).Count() <= 1) { ... }
I'd go for maximum maintainability and readability.
static bool ZeroOrOneAreTrue(params bool[] bools)
{
return NumThatAreTrue(bools) <= 1;
}
static int NumThatAreTrue(params bool[] bools)
{
return bools.Where(b => b).Count();
}
There are many answers here, but I have another one!
a ^ b ^ c ^ (a == b && b == c)
A general way of finding a minimal boolean expression for a given truth table is to use a Karnaugh map:
http://babbage.cs.qc.edu/courses/Minimize/
There are several online minimizers on the web. The one here (linked to from the article, it's in German, though) finds the following expression:
(!A && !B) || (!A && !C) || (!B && !C)
If you're going for code readability, though, I would probably go with the idea of "sum<=1". Take care that not all languages guarantee that false==0 and true==1 -- but you're probably aware of this since you've taken care of it in your own solution.
Good ol' logic:
+ = OR
. = AND
R = Abar.Bbar.Cbar + Abar.Bbar.C + Abar.B.Cbar + A.Bbar.Cbar
= Abar.Bbar.(Cbar + C) + Abar.B.Cbar + A.Bbar.Cbar
= Abar.Bbar + Abar.B.Cbar + A.Bbar.Cbar
= Abar.Bbar + CBar(A XOR B)
= NOT(A OR B) OR (NOT C AND (A XOR B))
Take the hint and simplify further if you want.
And yeah, get your self familiar with Karnaugh Maps
Depends whether you want something where it's easy to understand what you're trying to do, or something that's as logically simple as can be. Other people are posting logically simple answers, so here's one where it's more clear what's going on (and what the outcome will be for different inputs):
def only1st(a, b, c):
return a and not b and not c
if only1st(a, b, c) or only1st(b, a, c) or only1st(c, a, b):
print "Yes"
else:
print "No"
I like the addition solution, but here's a hack to do that with bit fields as well.
inline bool OnlyOneBitSet(int x)
{
// removes the leftmost bit, if zero, there was only one set.
return x & (x-1) == 0;
}
// macro for int conversion
#define BOOLASINT(x) ((x)?1:0)
// turn bools a, b, c into the bit field cba
int i = (BOOLASINT(a) << 0) | BOOLASINT(b) << 1 | BOOLASINT(c) << 2;
if (OnlyOneBitSet(i)) { /* tada */ }
Code demonstration of d's solution:
int total=0;
if (A) total++;
if (B) total++;
if (C) total++;
if (total<=1) // iff no more than one is true.
{
// execute
}