I am making a CPLEX model with C++, and I need a function like:
IloConstraint f(...){
IloConstraint constr;
if(condition1){
constr = (x+y >= 1);
return constr;
}
if(condition2){
constr = false;
return constr;
}
constr = true;
return constr;
}
I think that I succeded in creating a true and false constraints by
constr = (x==x); and
constr = IloNot(x==x);
I assume that this approach is not very optimal because it adds extra conditions and variables. Is there a more optimal and more readable way to do this? Something like
constr = IloConstraint(IloFalse); ?
IloConstraint(IloFalse) will not work since this will be interpreted as IloConstraint((IloConstraintI*)0) (IloFalse just expands to the literal 0 (zero)), which will create a constraint without implementation.
There is no literal for a true or false constraint. You can go without extra variables if you do something like IloExpr(env, 1) == IloExpr(env, 1) (and != for the false constraint). Another option for a constant true constraint would be use an empty IloAnd or an empty IloOr.
However, just going with x == x, 1 >= 2 or things similar to that seems a lot more readable to me. Additional expressions should usually not pose a problem. The engine will remove those in preprocessing.
Another option would be to use IloCplex::ifThen() to create a conditional constraint. Maybe this is even more readable then your function that returns a constraint.
I have the following if else statement that created by myself in order to link to the if else statement given in second part:
m=4
if m==3
disp(true)
else
disp(false)
Second part ( this code is fix cannot be change):
if (true)
A=Hello World
else
A=Bye
If using the first part code, my output will be
A=Hello World
but my desire output is
A=Bye
Anyone one have idea to edit the first part, because now my return value in first part not able to link to my second part.
If you can't change the second part's code, I'm afraid your desire cannot be fulfilled. Or rather, I'm afraid your code won't run at all, because your perenthesis, quotes, end-statement (and arguably semicolons) are not in place.
if true
A = 'Hello World';
else
A = 'Bye';
end
This code will return A = 'Hello World', no matter what, since true is always true. If-else conditions work like this:
if (*what's in here evealuates to true*)
%do stuff
else (*if what's up there does not evaluate to true*)
%do other stuff
Clearly, true will always evaluate to true. So the above if-else condition will always return A = 'Hello World'.
You don't need two if statements in order to accomplish this task. One is more than enough to perform all what you need:
m = 4;
if (m == 3)
A = 'Hello World';
else
A = 'Bye';
end
disp(A);
A few comments concerning your code:
if statements need to be closed with an end
if (true) will always pass into the statement
the disp function doesn't assign a value, its only goal is to display it in the Command Window
in order to work with text, you have to enclose it within single quotes ' (char array) or double quotes " (string), more info here
If you posted only small excerpts of your code and you need to perform those two checks sequentially, in different parts of your script, then:
m = 4;
if (m == 3)
m_equals_3 = true;
disp('M == 3');
else
m_equals_3 = false;
disp('M ~= 3');
end
% then, somewhere else...
if (m_equals_3)
A = 'Hello World';
else
A = 'Bye';
end
% ...
I guess this is a homework exercise. You should disclose that if it’s the case.
The exercise requieres you to change the workspace such that the second bit of code evaluated the else case. This can be accomplished by changing the meaning of true. In your first bit of code, make it so that
true = flase;
Or equivalently,
true = 0;
Note that this is really bad form, if you ever do something like this outside of a homework exercise that explicitly asks you to do so, you’ll get fired or maybe even shot. You’ve been warned!
By the way, I assume that the missing quote characters around the strings and the missing ends are typos?
I have written a lot of Matlab code using structs and I have the following problem recurring.
I often need to test if a field exists and is not empty. So far, my solution is:
if isfield(var, 'field')
if ~isempty(var.field)
...
when one or the other test fails I need to perform the very same action. In these cases the solution is far from being elegant. For example:
if isfield(var, 'field')
if ~isempty(var.field)
fieldOk = true;
else
fieldOk = false;
end
else
fieldOk = false;
end
A better solution would be to do both tests at once. I could write a function that wraps all this but I am wondering if there is a native Matlab solution.
What about:
if isfield(var, 'field') && ~isempty(var.field)
fieldOk = true;
else
fieldOk = false;
end
The logical operator X && Y just evaluates Y if X is true.
Have a look here. Therefore it's exactly what you need. But you may have to turn it around:
if ~isempty(var.field) && isfield(var, 'field')
In Matlab, I've got a function that is called after some special points are found on an image. Depending on how the nearby pixels of that "special point" are, the function must return a structure with many parameters or return nothing.
Is it possible to have a function that by deafult will return something but, in some cases, nothing should be returned? How that "return nothing"'s code should be like? Thank you.
One common trick in matlab is to use the empty matrix [] to indicate nothing. You could write your function something like (untested code):
function result = analyze(image, special_point)
% your code here
if pixels_are_ok
result.a = 1;
result.b = 2;
else
result = [];
end
If you call this function from your other code, you can use isempty to see if you got a result or not:
result = analyze(image, special_point)
if isempty(result)
display('did not find anything')
else
display('found some interesting results')
display(result)
end
Given an arbitrary list of booleans, what is the most elegant way of determining that exactly one of them is true?
The most obvious hack is type conversion: converting them to 0 for false and 1 for true and then summing them, and returning sum == 1.
I'd like to know if there is a way to do this without converting them to ints, actually using boolean logic.
(This seems like it should be trivial, idk, long week)
Edit: In case it wasn't obvious, this is more of a code-golf / theoretical question. I'm not fussed about using type conversion / int addition in PROD code, I'm just interested if there is way of doing it without that.
Edit2: Sorry folks it's a long week and I'm not explaining myself well. Let me try this:
In boolean logic, ANDing a collection of booleans is true if all of the booleans are true, ORing the collection is true if least one of them is true. Is there a logical construct that will be true if exactly one boolean is true? XOR is this for a collection of two booleans for example, but any more than that and it falls over.
You can actually accomplish this using only boolean logic, although there's perhaps no practical value of that in your example. The boolean version is much more involved than simply counting the number of true values.
Anyway, for the sake of satisfying intellectual curiosity, here goes. First, the idea of using a series of XORs is good, but it only gets us half way. For any two variables x and y,
x ⊻ y
is true whenever exactly one of them is true. However, this does not continue to be true if you add a third variable z,
x ⊻ y ⊻ z
The first part, x ⊻ y, is still true if exactly one of x and y is true. If either x or y is true, then z needs to be false for the whole expression to be true, which is what we want. But consider what happens if both x and y are true. Then x ⊻ y is false, yet the whole expression can become true if z is true as well. So either one variable or all three must be true. In general, if you have a statement that is a chain of XORs, it will be true if an uneven number of variables are true.
Since one is an uneven number, this might prove useful. Of course, checking for an uneven number of truths is not enough. We additionally need to ensure that no more than one variable is true. This can be done in a pairwise fashion by taking all pairs of two variables and checking that they are not both true. Taken together these two conditions ensure that exactly one if the variables are true.
Below is a small Python script to illustrate the approach.
from itertools import product
print("x|y|z|only_one_is_true")
print("======================")
for x, y, z in product([True, False], repeat=3):
uneven_number_is_true = x ^ y ^ z
max_one_is_true = (not (x and y)) and (not (x and z)) and (not (y and z))
only_one_is_true = uneven_number_is_true and max_one_is_true
print(int(x), int(y), int(z), only_one_is_true)
And here's the output.
x|y|z|only_one_is_true
======================
1 1 1 False
1 1 0 False
1 0 1 False
1 0 0 True
0 1 1 False
0 1 0 True
0 0 1 True
0 0 0 False
Sure, you could do something like this (pseudocode, since you didn't mention language):
found = false;
alreadyFound = false;
for (boolean in booleans):
if (boolean):
found = true;
if (alreadyFound):
found = false;
break;
else:
alreadyFound = true;
return found;
After your clarification, here it is with no integers.
bool IsExactlyOneBooleanTrue( bool *boolAry, int size )
{
bool areAnyTrue = false;
bool areTwoTrue = false;
for(int i = 0; (!areTwoTrue) && (i < size); i++) {
areTwoTrue = (areAnyTrue && boolAry[i]);
areAnyTrue |= boolAry[i];
}
return ((areAnyTrue) && (!areTwoTrue));
}
No-one mentioned that this "operation" we're looking for is shortcut-able similarly to boolean AND and OR in most languages. Here's an implementation in Java:
public static boolean exactlyOneOf(boolean... inputs) {
boolean foundAtLeastOne = false;
for (boolean bool : inputs) {
if (bool) {
if (foundAtLeastOne) {
// found a second one that's also true, shortcut like && and ||
return false;
}
foundAtLeastOne = true;
}
}
// we're happy if we found one, but if none found that's less than one
return foundAtLeastOne;
}
With plain boolean logic, it may not be possible to achieve what you want. Because what you are asking for is a truth evaluation not just based on the truth values but also on additional information(count in this case). But boolean evaluation is binary logic, it cannot depend on anything else but on the operands themselves. And there is no way to reverse engineer to find the operands given a truth value because there can be four possible combinations of operands but only two results. Given a false, can you tell if it is because of F ^ F or T ^ T in your case, so that the next evaluation can be determined based on that?.
booleanList.Where(y => y).Count() == 1;
Due to the large number of reads by now, here comes a quick clean up and additional information.
Option 1:
Ask if only the first variable is true, or only the second one, ..., or only the n-th variable.
x1 & !x2 & ... & !xn |
!x1 & x2 & ... & !xn |
...
!x1 & !x2 & ... & xn
This approach scales in O(n^2), the evaluation stops after the first positive match is found. Hence, preferred if it is likely that there is a positive match.
Option 2:
Ask if there is at least one variable true in total. Additionally check every pair to contain at most one true variable (Anders Johannsen's answer)
(x1 | x2 | ... | xn) &
(!x1 | !x2) &
...
(!x1 | !xn) &
(!x2 | !x3) &
...
(!x2 | !xn) &
...
This option also scales in O(n^2) due to the number of possible pairs. Lazy evaluation stops the formula after the first counter example. Hence, it is preferred if its likely there is a negative match.
(Option 3):
This option involves a subtraction and is thus no valid answer for the restricted setting. Nevertheless, it argues how looping the values might not be the most beneficial solution in an unrestricted stetting.
Treat x1 ... xn as a binary number x. Subtract one, then AND the results. The output is zero <=> x1 ... xn contains at most one true value. (the old "check power of two" algorithm)
x 00010000
x-1 00001111
AND 00000000
If the bits are already stored in such a bitboard, this might be beneficial over looping. Though, keep in mind this kills the readability and is limited by the available board length.
A last note to raise awareness: by now there exists a stack exchange called computer science which is exactly intended for this type of algorithmic questions
It can be done quite nicely with recursion, e.g. in Haskell
-- there isn't exactly one true element in the empty list
oneTrue [] = False
-- if the list starts with False, discard it
oneTrue (False : xs) = oneTrue xs
-- if the list starts with True, all other elements must be False
oneTrue (True : xs) = not (or xs)
// Javascript
Use .filter() on array and check the length of the new array.
// Example using array
isExactly1BooleanTrue(boolean:boolean[]) {
return booleans.filter(value => value === true).length === 1;
}
// Example using ...booleans
isExactly1BooleanTrue(...booleans) {
return booleans.filter(value => value === true).length === 1;
}
One way to do it is to perform pairwise AND and then check if any of the pairwise comparisons returned true with chained OR. In python I would implement it using
from itertools import combinations
def one_true(bools):
pairwise_comp = [comb[0] and comb[1] for comb in combinations(bools, 2)]
return not any(pairwise_comp)
This approach easily generalizes to lists of arbitrary length, although for very long lists, the number of possible pairs grows very quickly.
Python:
boolean_list.count(True) == 1
OK, another try. Call the different booleans b[i], and call a slice of them (a range of the array) b[i .. j]. Define functions none(b[i .. j]) and just_one(b[i .. j]) (can substitute the recursive definitions to get explicit formulas if required). We have, using C notation for logical operations (&& is and, || is or, ^ for xor (not really in C), ! is not):
none(b[i .. i + 1]) ~~> !b[i] && !b[i + 1]
just_one(b[i .. i + 1]) ~~> b[i] ^ b[i + 1]
And then recursively:
none(b[i .. j + 1]) ~~> none(b[i .. j]) && !b[j + 1]
just_one(b[i .. j + 1] ~~> (just_one(b[i .. j]) && !b[j + 1]) ^ (none(b[i .. j]) && b[j + 1])
And you are interested in just_one(b[1 .. n]).
The expressions will turn out horrible.
Have fun!
That python script does the job nicely. Here's the one-liner it uses:
((x ∨ (y ∨ z)) ∧ (¬(x ∧ y) ∧ (¬(z ∧ x) ∧ ¬(y ∧ z))))
Retracted for Privacy and Anders Johannsen provided already correct and simple answers. But both solutions do not scale very well (O(n^2)). If performance is important you can stick to the following solution, which performs in O(n):
def exact_one_of(array_of_bool):
exact_one = more_than_one = False
for array_elem in array_of_bool:
more_than_one = (exact_one and array_elem) or more_than_one
exact_one = (exact_one ^ array_elem) and (not more_than_one)
return exact_one
(I used python and a for loop for simplicity. But of course this loop could be unrolled to a sequence of NOT, AND, OR and XOR operations)
It works by tracking two states per boolean variable/list entry:
is there exactly one "True" from the beginning of the list until this entry?
are there more than one "True" from the beginning of the list until this entry?
The states of a list entry can be simply derived from the previous states and corresponding list entry/boolean variable.
Python:
let see using example...
steps:
below function exactly_one_topping takes three parameter
stores their values in the list as True, False
Check whether there exists only one true value by checking the count to be exact 1.
def exactly_one_topping(ketchup, mustard, onion):
args = [ketchup,mustard,onion]
if args.count(True) == 1: # check if Exactly one value is True
return True
else:
return False
How do you want to count how many are true without, you know, counting? Sure, you could do something messy like (C syntax, my Python is horrible):
for(i = 0; i < last && !booleans[i]; i++)
;
if(i == last)
return 0; /* No true one found */
/* We have a true one, check there isn't another */
for(i++; i < last && !booleans[i]; i++)
;
if(i == last)
return 1; /* No more true ones */
else
return 0; /* Found another true */
I'm sure you'll agree that the win (if any) is slight, and the readability is bad.
It is not possible without looping. Check BitSet cardinality() in java implementation.
http://fuseyism.com/classpath/doc/java/util/BitSet-source.html
We can do it this way:-
if (A=true or B=true)and(not(A=true and B=true)) then
<enter statements>
end if