I am given three variables having finite values ( all are integers) m,n, r.
Now I need to do m<-r and n<-r ( assign m and n the value of r ) and I have read in "The Art of Computer Programming vol. 1 " that the operations can be combined as
m<-n<-r
But will the above statement not mean "assign m the value of n and then n the value of r".
Thanks in advance.
The order of assignment is from right to left. Thus, m<-n<-r will be interpreted as: n<-r and then m<-n.
Since n equals r after the first assignment, m<-n and m<-r are identical.
Assignment = operator is like assigning the right side value to left side. For eg
int a = 1 + 2;
Here first 1+2 is evaluated and assigned to a because it follows right to left associativity.
Now if you have something like this
int a=b=2;
It again follows right to left associativity. From right first b=2 is evaluated and assign 2 to b then b is assigned to a. It works like this a=(b=2)
Know in your question you have m<-n<-r . This will work like this m<-(n<-r)
You can see reference Operator Associativity
Related
I'm trying to implement the 'Sport Scheduling Problem' (with a Round-Robin approach to break symmetries). The actual problem is of no importance. I simply want to declare the value at x[1,1] to be the set {1,2} and base the sets in the same column upon the first set. This is modelled as in the code below. The output is included in a screenshot below it. The problem is that the first set is not printed as a set but rather some sort of range while the values at x[2,1] and x[3,1] are indeed printed as sets and x[4,1] again as a range. Why is this? I assume that in the declaration of x that set of 1..n is treated as an integer but if it is not, how to declare it as integers?
EDIT: ONLY the first column of the output is of importance.
int: n = 8;
int: nw = n-1;
int: np = n div 2;
array[1..np, 1..nw] of var set of 1..n: x;
% BEGIN FIX FIRST WEEK $
constraint(
x[1,1] = {1, 2}
);
constraint(
forall(t in 2..np) (x[t,1] = {t+1, n+2-t} )
);
solve satisfy;
output[
"\(x[p,w])" ++ if w == nw then "\n" else "\t" endif | p in 1..np, w in 1..nw
]
Backend solver: Gecode
(Here's a summarize of my comments above.)
The range syntax is simply a shorthand for contiguous values in a set: 1..8 is a shorthand of the set {1,2,3,4,5,6,7,8}, and 5..6 is a shorthand for the set {5,6}.
The reason for this shorthand is probably since it's often - and arguably - easier to read the shorthand version than the full list, especially if it's a long list of integers, e.g. 1..1024. It also save space in the output of solutions.
For the two set versions, e.g. {1,2}, this explicit enumeration might be clearer to read than 1..2, though I tend to prefer the shorthand version in all cases.
I have an if statement something like
t(2,5)
P(6,1)
if x:length(t)
t(2,3)>P
P=0
elseif t(2,3)<P
P=1
end
basically what I want to achieve is a loop that runs through t and compares it to p, if the value of t is smaller than the value of P a new matrix should be created and record the value as "1", if the value of t is bigger than the value of P record value as "0", I can't get it to work unfortunately.
If you just want to check if the values in array/matrix t is less than a certain constant value, say P, then better run t<P. That will do the job.
If t and P are both matrices, of equal length, then also you can use t<P. No need for loops or if statements.
How do i passed 2 variables to a lambda function, where x is a number and y is a symbol.
I have written this, but it wouldn't process
{[x;y]
// some calculation with x and y
}
each ((til 5) ,\:/: `a`b`c`d`f)
It seems to be complaining that i am missing another arg.
Here's an example that I think does what you're looking for:
q){string[x],string y}./: raze (til 5) ,\:/: `a`b`c`d`f
The issue with your example is that you need to raze the output of ((til 5) ,\:/: `a`b`c`d`f) to get your list of 2 inputs.
Passing a list of variables into a function is accomplished using "." (dot apply) http://code.kx.com/q/ref/unclassified/#apply
.e.g
q){x+y} . 10 2
12
In my example, I've then used an "each right" to then apply to each pair. http://code.kx.com/q/ref/adverbs/#each-right
Alternatively, you could use the each instead if you wrapped the function in another lamda
q){{string[x],string y} . x} each raze (til 5) ,\:/: `a`b`c`d`f
Instead of generating a list of arguments using cross or ",/:\:" and passing each of these into your function, modify your function with each left each right ("/:\:") to give you all combination. his should take the format;
x f/:\: y
Where x and y are both lists. Reusing the example {string[x],string y};
til[5] {string[x], string y}/:\:`a`b`c`d
This will give you a matrix of all combinations of x and y. If you want to flatten that list add a 'raze'
I have a problem with If statement in OpenScad.
I have 4 variables
a=20;
b=14;
w=1;
c=16;
I want to check witch number is bigger a or b.
And after depending who is smaller to take the value of smaller variable(in our case b < a) and to make a simple operation with c variable ( c=b-w).
I tried like this but it doesn't work.
a=20;
b=14;
w=1;
c=16;
if(a>b)
{
c=b-w;
}
if (a<b)
{
c=a-w;
}
if (a==b)
{
c=a-w;
}
It seems logic, but in openscad as I understood you can't change the value of variable inside a If statement. What trick can I use in order to get my goal.
Thank you!
in the 3. leg you confused the assignment-operator „=“ with the equal-operator „==“ (correct if (a==b)).
in your 3. leg you do the same as in the 2., so you could handle both as an „else“-leg.
Correct: assignment is not allowed in if-statement. In openscad you can use the ? operator instead:
c = a > b ? b-w : a-w;
After = follows the condition. The statement after the ? becomes the value if the condition is true, and the statement after the : becomes the value if the condition is false. Nested conditions are possible, e.g. your conditions:
c = a > b ? b-w : (a < b ? a-w : a-w);
More information in the documentation.
OpenSCAD's variable assignment is different. You can only assign variables inside a bracket. So c = b - w will only be assigned inside the if bracket. Outside if this bracket it will still be 16. Don't ask me why. You can read more in the Documentation of OpenSCAD.
c = min(c,min(a,b)/2-w);
this also solve the problem )
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.