I have a simple function which returns a table:
F[("A";"B");(1,-1)]
I would like to apply this function passing vectors as inputs:
a:((`A;`B);(`B;`C);(`C;`D))
b:((1;-1);(1;-1);(1;-1))
I have tried:
F each a,b
F each a cross b
but this doesn't work or combines the vectors rather than keeping the 2 components separate. In addition when I do get it to work how do I row bind the resulting list of tables? I am coming from a python background.
You need to use ' each-both :
q)F:{ ([] enlist x; enlist y)} /if F is simply creating a table
q)F[("A";"B");(1,-1)]
x y
---------
"AB" 1 -1
q)a:((`A;`B);(`B;`C);(`C;`D))
q)b:((1;-1);(1;-1);(1;-1))
q)F'[a;b] /each-both
+`x`y!(,`A`B;,1 -1)
+`x`y!(,`B`C;,1 -1)
+`x`y!(,`C`D;,1 -1)
raze will format it to a table (i think row binding means appending the tables together)
q)raze F'[a;b]
x y
--------
A B 1 -1
B C 1 -1
C D 1 -1
I'm newby in Matlab. I have took the work code with complex if-statement condition and need to rewrite it. This code should prepare some initial data to solve an optimization task. This if-statement condition looks like:
x=[784.8 959.2 468 572 279 341 139.5 170.5 76.5 93.5 45 55];
a=nchoosek(x,6); % all possible combinations from 6 elements of x
n=length(a);
q=[];
for i=1:n
if( ((a(i,1)==x(1)) & (a(i,2)==x(2))) |
((a(i,1)==x(3)) & (a(i,2)==x(4))) |
((a(i,1)==x(5)) & (a(i,2)==x(6))) |
((a(i,1)==x(7)) & (a(i,2)==x(8))) |
((a(i,2)==x(3)) & (a(i,3)==x(4))) |
((a(i,2)==x(5)) & (a(i,3)==x(6))) |
((a(i,2)==x(7)) & (a(i,3)==x(8))) |
((a(i,3)==x(3)) & (a(i,4)==x(4))) |
((a(i,3)==x(5)) & (a(i,4)==x(6))) |
((a(i,3)==x(7)) & (a(i,4)==x(8))) |
((a(i,3)==x(9)) & (a(i,4)==x(10)))|
((a(i,4)==x(5)) & (a(i,5)==x(6))) |
((a(i,4)==x(7)) & (a(i,5)==x(8))) |
((a(i,4)==x(9)) & (a(i,5)==x(10)))|
((a(i,5)==x(5)) & (a(i,6)==x(6))) |
((a(i,5)==x(7)) & (a(i,6)==x(8))) |
((a(i,5)==x(9)) & (a(i,6)==x(10)) |
((a(i,5)==x(11)) & (a(i,6)==x(12)))))
q(i,:)=a(i,:);
end;
end;
q;
R1=a-q;
R1(~any(R1,2),:) = [];
R1(:, ~any(R1)) = [];
Question: Could anyone give an idea how to rewrite if-statement to improve readability of code?
If I understood you correctly, what the convoluted if statement basically says
If "x(1) x(2)" or "x(3) x(4)" or ... "x(11) x(12)" appears anywhere consecutively in row i
Think about it:
((a(i,1)==x(1)) & (a(i,2)==x(2))) | ((a(i,1)==x(3)) & (a(i,2)==x(4))) |
((a(i,1)==x(5)) & (a(i,2)==x(6))) | ((a(i,1)==x(7)) & (a(i,2)==x(8)))
is no different from:
((a(i,1)==x(1)) & (a(i,2)==x(2))) | ((a(i,1)==x(3)) & (a(i,2)==x(4))) |
((a(i,1)==x(5)) & (a(i,2)==x(6))) | ((a(i,1)==x(7)) & (a(i,2)==x(8))) |
((a(i,1)==x(9)) & (a(i,2)==x(10))) | ((a(i,1)==x(11)) & (a(i,2)==x(12)))
since [x(9) x(10)] and [x(11) x(12)] will never appear at a(i, 1:2), so the line I added is always false and does not change the result of the chain of OR's. But if makes the logic much easier to understand. Same logic applies to a(i,2:3), a(i,3:4)..., complete those cases too and then you will get the first statement I made in this answer.
Then, instead of generating a directly from x, you should generate a from the INDEX of x, i.e. [1:12], as such:
a = nchoosek(1:length(x), 6);
Why? You said x consists of real numbers, and using == on real numbers does not guarantee success, and is a very bad practice in general.
Then, your target becomes:
find if sequence `[1 2]` or `[3 4]` or `[5 6]` ... exists in each line of `a`
which is equivalent to:
find if there is any odd number n followed by n+1
This logic can be represented as:
success = any (mod(a(:,1:end-1), 2) & diff(a,1,2)==1, 2)
Now success(i) will be true/false for the every a(i) that your statement evaluates to the same value. This method is better than your statement because it is very concise, automatically adapts to different sizes of x and does not need to run in a loop.
And if you want to get the actual combination of x values, just do
x(a(i)); % Get the ith permutation of x
x(a); % Get all permutation of x
x(a(success,:)); % Get all permutation of x that satisfy the requirement.
EDIT:
q = a; % q is basically a copy of a
q(~success,:) = 0; % except the `non-success` rows are zero
x(q) - x(a) % suppose q and a store index, this will give you the substraction.
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
Cause I've tried doing the truth table unfortunately one has 3 literals and the other has 4 so i got confused.
F = (A+B+C)(A+B+D')+B'C;
and this is the simplified version
F = A + B + C
http://www.belley.org/etc141/Boolean%20Sinplification%20Exercises/Boolean%20Simplification%20Exercise%20Questions.pdf
cause I think there's something wrong with this reviewer.. or is it accurate?
btw is simplification different from minimizing from Sum of Minterms to Sum of Products?
Yes, it is the same.
Draw the truth table for both expressions, assuming that there are four input variables in both. The value of D will not play into the second truth table: values in cells with D=1 will match values in cells with D=0. In other words, you can think of the second expression as
F = A +B + C + (0)(D)
You will see that both tables match: the (A+B+C)(A+B+D') subexpression has zeros in ABCD= {0000, 0001, 0011}; (A+B+C) has zeros only at {0000, 0001}. Adding B'C patches zero at 0011 in the first subexpressions, so the results are equivalent.
So I want to simplify z:=a+I*b; Im(z) where a, b are real variables So I try:
s:= 1+2*I
Im(s) // outputs 2
z:=a+I*b
Im(z) // outputs Im(a+I*b)
So I wonder is it any how possible to simplify Im(z) so to get b as output (here we look at general case meaning z could be any complex expression from real values (like a, b, c etc and complex I))?
You didn't tell Maple that a and b were real, so the simplification doesn't work because it doesn't necessarily hold. One way to get what you want is by using the assume command to let it know:
> s:=1+2*I;
s := 1 + 2 I
> Im(s);
2
> z:=a+I*b;
z := a + b I
> Im(z);
Im(a + b I)
> assume(a,real);
> assume(b,real);
> z;
a~ + b~ I
> Im(z);
b~
The evalc command works by considering unknowns as being real.
z:=a+I*b:
Im(z);
Im(a + I b)
evalc( Im(z) );
b
See its help-page, ?evalc.