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I'm still doing the exercise from the book and the one exercise says that I should create a function: value :: Hand -> Int, which returns the value of the hand.
My code so far looks like this:
data Hand = PairOf Rank | ThreeOf1 Rank | ThreeOf2 Suit
value: Hand -> Int
otherwise = 0
--
I now have another problem, because I don't know how to describe "Nothing".
Nothing is described here as a possible hand combination in which Pair, ThreeOf1 and ThreeOf2 do not occur.
Would otherwise = 0 work or doesn't that make much sense?
Thanks again for the suggestions, I corrected them! Thanks also in advance for explanations and help.
otherwise won't work here. What you want is an irrefutable pattern that will match anything, and further since you don't care what gets matched, you specifically want the pattern _:
value :: Hand -> Int
value (PairOf r1) = 1
value (ThreeOf r1) = 2
value (Flush s1) = 3
value _ = 0
Since you don't care about what kind of pair, etc you get, you can also use _ in the other cases:
value :: Hand -> Int
value (PairOf _) = 1
value (ThreeOf _) = 2
value (Flush _) = 3
value _ = 0
If you wanted to match Nothing (or whatever name you come up with that doesn't conflict with the Maybe constructor) specifically, it's just
value Nothing = 0
One of the problems you might be having is that there is no Nothing.
Here's the full type of Hands for holdem:
data HandRank
= HighCard Rank Rank Rank Rank Rank
| OnePair Rank Rank Rank Rank
| TwoPair Rank Rank Rank
| ThreeOfAKind Rank Rank Rank
| Straight Rank
| Flush Rank Rank Rank Rank Rank
| FullHouse Rank Rank
| FourOfAKind Rank Rank
| StraightFlush Rank
deriving (Eq, Ord, Show, Generic)
The extra ranks in the data type are there to distinguish hands, so a pair of aces with a King & Queen kicker beats a pair of aces with a King & Jack kicker. But otherwise, it's the same as your setup.
Once you have a complete transformation from a set of cards to a hand value, there is no nothing. What you might be thinking of as nothing is the HighHand line in the sum type. If your context is that's it for possible hands, then I would add a NoValue to the Hand sum type, as better than outputting a Maybe Hand.
Using the wild cards otherwise _ or value _ introduces a chance of a bug because you might forget about coding up a full house, say, and your existing function would work. If you forget to code one of the sum types, and you don't have a match-any pattern, the compiler will complain. Providing the compiler with all branches of a sum type is also a hot development area of GHC, and it will be fast.
I'm new to Haskell and code in general and while trying to write a script that shows which of two numbers is smaller to practise using guards i got stuck with the "parse error on input ‘|’" error.
I read about using "let" somehow but I haven't been able to figure it out. How do I get past this error?
smallest a b = | a >= b = b
| a <= b = a
I'm posting this answer to clarify that you do need the equal sign; but you need one for each case, whereas you essentially duplicated it:
-- +--- this is the equal sign (one per case)
-- |
-- +----------|--- which you are duplicating here
-- | |
-- v v
smallest a b = | a >= b = b
| a <= b = a
that's the reason why the suggested code works.
By the way, the indentation is important, but you can write that snippet on 2 lines rather than 3,
smallest a b | a >= b = b
| otherwise = a
though it does not improve readability, and actually decreseas it if the part before the first | is long.
By the way, as far as I remember, otherwise is truly just a synonym for True (well, the former is a function, and the latter a value constructor, but I guess this makes little difference, as the former is like a 0-arguments function that generates the latter, so they are fundamentally synonyms).
When using guards, you don't need the equals sign. You can simply do this:
smallest a b
| a >= b = b
| otherwise = a
Note the indentation here. Also, since your first condition takes care of both the greater than and equal cases, you can simply use "otherwise" instead of explicitly stating the remaining case.
I am reading a haskell book (page 412). In this book, there is an explanation about normal form for sum of products:
All the existing algebraic rules for products and sums apply in type systems, and that includes the distributive property. Let’s take a look at how that works in arithmetic:
2 * (3 + 4)
2 * (7)
14
We can rewrite this with the multiplication distributed over the addition and obtain the same result:
2 * 3 + 2 * 4
(6) + (8)
14
This is known as a “sum of products.” In normal arithmetic, the expression is in normal form when it’s been reduced to a final result. However, if you think of the numerals in the above expressions as representations of set cardinality, then the sum of products expression is in normal form, as there is no computation to perform.
I've known that a normal form indiciates that an expression is fully reduced. In the above description, author of the book explains that sum of products can be seen as in normal form when we think of expression as representations of set cardinality. I don't understand that.
Cardinality of types means how many different values can be included in that type (like set). For example, Bool type in haskell has cardinality of 2, which is addition of 1 for False and 1 for True each.
Is sum of products (2 * 3 + 2 * 4) a normal form? This expression can be reduced furthre more because fully reduced expression would be 14. I don't understand why sum of products and cardinality of it are related to be normal form.
Let's declare ourselves some types to represent the numbers:
data Two = OneOfTwo | TwoOfTwo
data Three = OneOfThree | TwoOfThree | ThreeOfThree
data Four = ... (similar)
Now we can see that the number of possible values of type Two is, in fact, 2. Same for Three, Four, and Seven.
Now if we construct a sum type:
data A = A Two
This type just straight up wraps a value of Two, so the number of possible values of A is also 2. So far so good?
Now let's construct a more complex one:
data B = B1 Three | B2 Four
Now, this type wraps either a value of type Three or a value of type Four (but not both at the same time!) This means that the number of possible values would be 3 + 4. Following so far?
Now, going further:
data C = C Two B
This type wraps two values at the same time - one value of type Two and one value of type B. This means that the number of possible values of C is the number of possible combinations of Two and B, which, as we know from middle-school mathematics, would be their product, or 2 * (3 + 4) = 2 * (7) = 14.
But here's the trick: we can write down an equivalent type in a different way:
data CNew = C1 Two Three | C2 Two Four
See what I did there? For CNew, the set of all possible combinations between values of Two, Three, and Four is the same as for C. Look: in both cases it's either a value of Two combined with a value of Three, or it's a value of Two combined with a value of Four. Except in CNew they're combined directly, but in C they're combined via B.
But the formula for CNew would be different: 2 * 3 + 2 * 4 = (6) + (8) = 14. This is what the book means.
Now to answer this bit more directly:
Is sum of products (2 * 3 + 2 * 4) a normal form? This expression can be reduced further more because fully reduced expression would be 14
This would be true if we were dealing with integer numbers, but we're not. We can rewrite C in the form of CNew, because that gives us all the same possible combinations of values. But we cannot rewrite them as a type that has straight up 14 possible values without combining 2, 3, and 4. That would be a completely new, unrelated type, as opposed to a combination of Two, Three, and Four.
And a possible terminology misunderstanding:
Is sum of products (2 * 3 + 2 * 4) a normal form?
The term "normal form" doesn't mean "the shortest". This term is usually used to denote a form that is very regular, and therefore easier to work with, and, crucially, that can represent all possible cases in the domain. In this case, normal form is defined as a "sum of products".
Could it be a "product of sums" instead? No, it couldn't, because, while a product of sums can always be converted to a sum of products, the reverse is not always possible, and this means that not every possible type would be representable in the normal form defined as "product of sums".
Could it be "just the number of possible values", like 14? Again no, because converting to such form loses some information (see above).
Assume I have an input variable x and three parameters a,b,c such that:
Given b we have c = f(x,a,b) for some (known) function f
Given c we have b = g(x,a,c) for some (known, different) function g.
I want to model this in a spreadsheet (Excel for instance). More precisely, if the user provides x,a and b then c will be evaluated and if c is given then b will be evaluated. It seems like this cannot be achieved directly, since a cell can hold either a value or a formula.
Is there a canonical way to do this? If not, what would be a best-practice workaround (probably some VBA magic)?
You can separate input fields from the calculated values and add some validation that only one of the mutually exclusive field is used, e.g.:
in my example, I used following conditional formatting to highlight invalid input:
=AND($B$4<>"", $B$5<>"")
and I used following the formulas for calculated values:
=B2
=B3
=IF(AND($B$4<>"", $B$5<>""), "#ERROR: only 1 value can be specified",
IF($B$4<>"", $B$4, $B$5-1))
=IF(AND($B$4<>"", $B$5<>""), "#ERROR: only 1 value can be specified",
IF($B$5<>"", $B$5, $B$4+1))
more generally:
=if(error_condition, error_message, if(b_is_not_empty, b, g(x,a,c)))
Suppose I have a sig A in my module that is related to a sig B.
Imagine now that we have the several instances :
A$1 -> B$1 , A$2 -> B$2
and
A$1 -> B$2 , A$2 -> B$1
I would like to express that B$1 and B$2 are equivalent (under certain conditions) such that only this instance would be generated A$1 -> B , A$2 -> B.
One of the solution might be to use the keyword "one" while declaring the sig B, but it won't work in my case, because B have several fields, making B atoms not necessarily equivalent . In short 2 atoms are equivalent only if they have their fields of same values.
Ideally I would like to remove the numbering for B. but still be able to have several atoms B .
The Alloy Analyzer doesn't give you much control over how the subsequent instances are generated (or how to break symmetries), so you typically have to work around those issues at the model level.
For your example, maybe this is the right model
sig B{}
one sig BA extends B {}
sig A {
b: one BA
}
run { #A=2 }
First you say that the two instances you describe are equivalent, although at a superficial level they appear to have distinct values for A$1 and A$2. Then you say "2 atoms are equivalent only if ... their fields [have the] same values". If that's the definition of equivalence, then your two instances are not equivalent; if your two instances are equivalent, then your definition is not capturing what you want.
It sounds as if you mean (1) that B atoms are either (1a) always equivalent, or (1b) equivalent under certain circumstances, and (2) that A atoms are equivalent if their fields all have values which are either identical or equivalent. And as if you want to prohibit equivalent A atoms. If that's so, then your jobs are:
Define a predicate that is true for two B elements if and only if they are equivalent.
Define a predicate for two A elements that is true if and only if they are equivalent.
Define a predicate (or a fact) that specifies that no two A atoms in the instance are equivalent.
If your problem is just that the Analyzer is showing you instances of the model which are not interestingly distinct from each other, then see Aleksandar Milicevic's response.