Better way of finding combination for 3 boolean variables - boolean-logic

I have 3 bool variables x,y,z. Now at any given moment I can have one out of 2^3=8 combinations as below.
e.g. x=true, y=false and z=false or
x=false, y=true and z=true and so on.
If I see from programming perspective there are 8 cases or may be 8 or greater if else statement to determine what is the combination at that moment.
At any given moment if I want to know what combination is present(given the values of x,y,z) How can I know without using if-else ladder, which makes code logic little bulky. Is there any better/simple logic/way to do it.

If you must handle 8 situations separately. You could encode the value of x, y, z in a variable and then do a switch case on that variable. Pseudo code below -
v = 0
if (x) { v += 4 }
if (y) { v += 2 }
if (z) { v += 1 }
switch (v)
{
case 0 : // all false
case 1 : // z is true
case 2 : // y is true
case 3 : // z and y are true
case 4 : // x is true
...
}

It might be worth using bitwise operators, rather than the numeric value to determine which boolean variables are on or off.
// Assign the bitwise value of each variable
X = 4
Y = 2
Z = 1
// Setting X and Z as true using the bitwise OR operator.
v = X | Z // v = 4 + 1 = 5
// Checking if any of the variables are true using the bitwise OR operator
if (v | X+Y+Z) // v = 4 + 2 + 1 = 7
// Checking if ALL of the variables are true using the bitwise AND operator
if (v & X+Y+Z)
// Checking if variable Y is true using the bitwise OR operator
if (v | Y)
// Checking if variable Y is false using the bitwise OR operator
if (v | Y == false)
// Checking if ONLY variable Y is true using the bitwise AND operator
if (v & Y)
// Checking if ONLY variable Y is false using the bitwise AND operator
if (v & Y == false)
This saves you from messing up the resulting number of a combination of values X, Y, Z. It is also more readable.

Related

Matlab - Undefined function 'int_f_1' for input arguments of type 'double'

Currently, I am working on a program that integrates x + x^2 + e^x + 2cos(x/2) - 1 with three input variables, a, b, and n. What I need returned is the numerical integral from a to b with n increments. The function also has to return trapezoids for each n as a column vector. Thus, the integral value as a scalar, and a vector of values.
I've gotten to a point where the function int_f_1 is undefined for some reason, and I have no idea why. I thought by nesting that function under the test function, it would help. But it does not, and I don't know why that is. Any suggestions?
function [y] = test_function_1(x);
y = x + x.^2 + exp(x) + 2*cos(x/2) - 1
end
function [int_f, increment] = int_f_1 (a, b, n);
f = #test_function_1;
h = a + b ./ n
increments = h
int_f = integral(h, f)
end

OCaml : recursive function dealing with parity between list elements and an int

This function should take two arguments a list and an int. if an element of the list and the number “a” parity is equal then they’d have to be summed, else the two numbers should be subtracted.
The calculation should be done in this order :
At the beginning, the residual value r is the value of a,
Each element e of lst (taken in the order given by the list) affects the residual value: if e and r are of the same parity (both odd or both even) then the new r’ is equal to the sum of r + e, if not then it should be equal to the subtraction of r - e,
The last r is the result expected.
To put this into an example:
par [4;7;3;6] 5
should return -1, it would work as follows :
5 and 4 have a different parity so we subtract -> 5 - 4 = 1
1 and 7 are both odd, so we add them together -> 1 + 7 = 8
8 and 3 have a different parity -> 8 - 3 = 5
Finally, 5 and 6 have different parity -> 5 - 6 = -1
I have thought of something like this below :
let rec par lst a =
match lst with
| [] -> 0
| h::t -> if (h mod 2 == 0 && a mod 2 == 0) || (h mod 2 == 1 && a mod 2 == 1) then a + h
| h::t -> if (h mod 2 == 0 && a mod 2 == 1) || (h mod 2 == 1 && a mod 2 == 0) then a - h :: par t a ;;
EDIT1 : Here is the error I get from the compiler :
Line 4, characters 83-88: Error: This expression has type int but an
expression was expected of type unit because it is in the result of a
conditional with no else branch
The idea is to build this function using no more than the following predefined functions List.hd, List.tl et List.length.
What is disturbing in my proposition above and how to remediate it? Anyone can help me resolve this, please?
EDIT 2:
I was able to do what is needed with if...then... else syntax (not the best I know for OCaml) but I personally have more difficulties sometimes understanding the pattern matching. Anyhow here's what I got :
let rec par lst a = (* Sorry it might hurt some sensible eyes *)
if List.length lst = 0 then a
else
let r = if (List.hd lst + a) mod 2 == 0 then (a + (List.hd lst))
else (a - (List.hd lst)) in
par (List.tl lst) r ;;
val par : int list -> int -> int = <fun>
Suggestions and help to put it into a pattern-matching syntax are welcomed.
Your code doesn't compile. Did you try compiling it? Did you read the errors and warnings produced by the compiler? Could you please add them to your question?
A few comments about your code:
| h::t -> if ... then ... should be | h::t when ... -> ...;
(h mod 2 == 0 && a mod 2 == 0) || (h mod 2 == 1 && a mod 2 == 1) can be simplified to (h - a) mod 2 == 0;
The compiler likes to know that the matching was exhaustive; in particular, you don't need to repeat the test in the third line of the matching (the third line will only be read if the test was false in the second line);
You are missing the recursive call in the second line of the matching;
In the third line of the matching, you are returning a list rather than a number (the compiler should have explicitly told you about that type mismatch!! did you not read the compiler error message?);
In the first line of the matching, in case the list is empty, you return 0. Are you sure that 0 is the value you want to return, when you've reached the end of the list? What about the residual value that you have calculated?
Once you have fixed this version of your code as a recursive function, I recommend trying to write a code solving the same problem using List.fold_left, rather than List.hd and List.tl as you are suggesting.
When I first wrote my answer, I included a fixed version of your code, but I think I'd be doing you a disservice by handing out the solution rather than letting you figure it out.

Function with vector as argument in Octave

How can I make a function with a vector as input and a matrix as an output?
I have to write a function that will convert cubic meters to liters and English gallons. The input should be a vector containing volume values ​​in m ^ 3 to be converted. The result should be a matrix in which the first column contains the result in m ^ 3, the second liter, the third English gallon.
I tried this:
function [liter, gallon] = function1 (x=[a, b, c, d]);
liter= a-10+d-c;
gallon= b+15+c;
endfunction
You're almost there.
The x=[a,b,c,d] part is superfluous, your argument should be just x.
function [liter, gallon] = function1 (x);
a = x(1); b = x(2); c = x(3); d = x(4);
liter = a - 10 + d - c;
gallon = b + 15 + c;
endfunction
If you want your code to be safe and guard against improper inputs, you can perform such checks manually inside the function, e.g.
assert( nargin < 1 || nargin > 4, "Wrong number of inputs supplied");
The syntax x=[a,b,c,d] does not apply to octave; this is reserved for setting up default arguments, in which case a, b, c, and d should be given specific values that you'd want as the defaults. if you had said something like x = [1,2,3,4], then this would be fine, and it would mean that if you called the function without an argument, it would set x up to this default value.

Haskell Integer Odd Digits Checker

I seem to be stuck on a question and have no idea how to approach it or what Im doing wrong with my current code.
I have to write a function called oddDigits which takes a single integer argument and returns a boolean result. It should return True if and only if the argument is a positive integer with an odd number of digits. If the argument is zero or negative, the function should stop with an error message.
Also, cant convert the argument into a string. Have to use recursion.
I have a feeling each digit could be stored in a list recursively and then the length of the list could determine the answer.
So far, I have this:
oddDigits :: Integer -> Bool
lst = []
oddDigits x
| (x < 0) || (x == 0) = error
| x `mod` 10 ++ lst ++ oddDigits(x `div` 10)
| length(lst) `mod` 2 /= 0 = True
| otherwise = False
Sorry if the code looks horrible. I am new to Haskell and still learning. What exactly am I doing wrong and how could I correct it?
First off, this seems a pretty weird thing to check. Perhaps what you're doing wrong is to ever consider this problem...
But if you persist you want to know the property of an integer having an odd number of digits... oh well. There's a lot that could be improved. For starters, (x < 0) || (x == 0) doesn't need the parentheses – < and == (infix 4) bind more tightly than ||. If you're not sure about this, you can always ask GHCi:
Prelude> :i ==
class Eq a where
(==) :: a -> a -> Bool
...
-- Defined in ‘GHC.Classes’
infix 4 ==
Prelude> :i ||
(||) :: Bool -> Bool -> Bool -- Defined in ‘GHC.Classes’
infixr 2 ||
But here you don't need || anyway because there's a dedicated operator for less-than-or-equal. Hence you can just write
oddDigits x
| x <= 0 = error "bla bla"
| ...
Then, you can “convert” the number to a string. Converting to string is generally a really frowned-upon thing to do because it throws all structure, typechecking etc. out of the window; however “number of digits” basically is a property of a string (the decimal expansion), rather than a number itself, so this is not entirely unsensible for this specific task. This would work:
oddDigits x
| x <= 0 = error "blearg"
| length (show x)`mod`2 /= 0 = True
| otherwise = False
however it's a bit redundancy department redundant. You're checking if something is True, then give True as the result... why not just put it in one clause:
oddDigits x
| x <= 0 = error "blearg"
| otherwise = length (show x)`mod`2 /= 0
That's perhaps in fact the best implementation.
For any proper, sensible task, I would not recommend going the string route. Recursion is better. Here's what it could look like:
oddDigits 1 = True
oddDigits x
| x <= 0 = error "blearg"
| otherwise = not . oddDigits $ x`div`10
There's nothing wrong with your general approach of converting to a list of digits, then finding the length of the list. Really where you went wrong is trying to cram everything into one function. As you found out first hand, it makes it very difficult to debug. Functional programming works best with very small functions.
If you separate out the responsibility of converting an integer to a list of digits, using a digs function like the one from this answer, the rest of your algorithm simplifies to:
oddDigits x | x <= 0 = error
oddDigits x = odd . length $ digs x
leftaroundabout's eventual answer is very nice, however it fails for numbers like 2,3 and 23. Here's a fix.
oddDigits x
| x <= 0 = error "blearg"
| x < 10 = True
| otherwise = not . oddDigits $ x`div`10
Its much more elegant than my initial answer, below. I'm including it to introduce a common functional paradigm, a worker/wrapper transformation of the problem. Here the wrapper gives the interface and passes off the work to another function. Notice that the negativity check only needs to be done once now.
oddDigits :: Integer -> Bool
oddDigits x
| x <= 0 = False
| otherwise = oddDigits' True x
oddDigits' :: Bool -> Integer -> Bool
oddDigits' t x
| x < 10 = t
| otherwise = oddDigits' (not t) $ x `div` 10
oddDigits' carries a piece of internal data with it, the initial Bool. My first first thought was to have that Bool be a digit accumulator, counting the number of digits. In that case, an "unwrapper" needs to be supplied, in this case the standard "odd" function:
oddDigits x
| x <= 0 = False
| otherwise = odd . oddDigits'' 1 $ x
where oddDigits'' :: Integer -> Integer -> Integer.

Pattern matching on Ints

I am a beginner in learning Haskell, and I wanted to know if you could pattern match on Ints like so:
add x 0 = x
add x (1 + y) = 1 + x + add x y,
Or maybe in this way:
add x 0 = x
add x (successor y) = 1 + x + add x y
There is an extension that lets you do that, but instead you should simply pattern match on y, and subtract 1 manually:
add x y = 1 + x + add x (y - 1)
The extension is called NPlusKPatterns. If you really want to use it (keep in mind it's deprecated in haskell 2010), it can be enabled by either passing a -XNPlusKPatterns parameter to GHC, or putting a {-# LANGUAGE NPlusKPatterns #-} at the top of your file.
Pattern matching isn't arbitrary case analysis. It's a disciplined, but limited form of case analysis, where the cases are the constructors of a data type.
In the specific case of pattern matching integers, the constructors are taken to be the integer values. So you can use integer values as the cases for pattern-matching:
foo 0 = ...
foo 2 = ...
foo x = ...
But you can't use arbitrary expressions. The following code is illegal:
foo (2 * x) = ...
foo (2 * x + 1) = ...
You may know that ever integer is either of the form 2 * x or 2 * x + 1. But the type system doesn't know.
The formatting of your code is a bit off so it is difficult to know what you're asking but you can using pattern matching for input of type Int. An example would be
add x 0 = x
add x y = x + y