In AIMMS, how do I do the following:
This is my constraint:
Which I'm trying to put into an AIMMS constraint:
Clearly, there is a contradiction:
In the mathematical summation, we sum (e.g. in the second summation expression) from j=1,...,p-1 and we have p=1,...,P of such constraints.
In AIMMS this translates to: sum[p | ord(p) <= ord(p) - 1, ... ] which clearly doesn't make any sense.
How can I make p=1,...,P constraints (predefined set) where I have to use the value of p in the definition (equation) as well?
In your AIMMS formulation I don't see the j index used anywhere that is used in the mathematical notation.
As far as I can see, it is just a matter of declaring an additional index j in the same set for which p is already an index and use this index j the same way it is used in the mathematical notation.
You can see the AIMMS blog (http://blog.aimms.com) for some information about using multiple indices in the same set.
Related
Total Julia noob here (with basic knowledge of Python). I am trying to do linear regression and things I read suggest the GLM package. Here is some sample code I found here:
using DataFrames, GLM
y = 1:10
df = DataFrame(y = y, x1 = y.^2, x2 = y.^3)
sm = GLM.lm( #formula(y ~ x1 + x2), df )
coef(sm)
Can someone explain the syntax here? What does #formula mean? Docs here say #foo means a
macro which I guess is basically just a function, but where do I find the function/macro formula? Just looking at the use here though, I would have thought it is maybe passing y ~ x1 + x2 (whatever that is) as the formula argument to lm? (similar to keyword arguments = in python?)
Next, what is ~ here? General docs say ~ means negation but I'm not seeing how that makes here.
Is there a place in the GLM docs where all of this is explained? I'm not seeing that. Only seeing a few examples but not a full breakdown of each function and all of its arguments.
You have stumbled upon the #formula language that is defined in the StatsModels.jl package and implemented in many statistics/econometrics related packages across the Julia ecosystem.
As you say, #formula is a macro, which transforms the expression given to it (here y ~ x1 + x2) into some other Julia expression. If you want to find out what happens when a macro gets called in Julia - which I admit can often look like magic to new (and sometimes experienced!) users - the #macroexpand macro can help you. In this case:
julia> #macroexpand #formula(y ~ x1 + x2)
:(StatsModels.Term(:y) ~ StatsModels.Term(:x1) + StatsModels.Term(:x2))
The result above is the expression constructed by the #formula macro. We see that the variables in our formula macro are transformed into StatsModels.Term objects. If we were to use StatsModels directly, we could construct this ourselves by doing:
julia> Term(:y) ~ Term(:x1) + Term(:x2)
FormulaTerm
Response:
y(unknown)
Predictors:
x1(unknown)
x2(unknown)
julia> (Term(:y) ~ Term(:x1) + Term(:x2)) == #formula(y ~ x1 + x2)
true
Now what is going on with ~, which as you say can be used for negation in Julia? What has happened here is that StatsModels has defined methods for ~ (which in Julia is and infix operator, that means essentially it is a function that can be written in between its arguments rather than having to be called with its arguments in brackets:
julia> (Term(:y) ~ Term(:x)) == ~(Term(:y), Term(:x))
true
So writing y::Term ~ x::Term is the same as calling ~(y::Term, x::Term), and this method for calling ~ with terms on the left and right hand side is defined by StatsModels (see method no. 6 below):
julia> methods(~)
# 6 methods for generic function "~":
[1] ~(x::BigInt) in Base.GMP at gmp.jl:542
[2] ~(::Missing) in Base at missing.jl:100
[3] ~(x::Bool) in Base at bool.jl:39
[4] ~(x::Union{Int128, Int16, Int32, Int64, Int8, UInt128, UInt16, UInt32, UInt64, UInt8}) in Base at int.jl:254
[5] ~(n::Integer) in Base at int.jl:138
[6] ~(lhs::Union{AbstractTerm, Tuple{Vararg{AbstractTerm,N}} where N}, rhs::Union{AbstractTerm, Tuple{Vararg{AbstractTerm,N}} where N}) in StatsModels at /home/nils/.julia/packages/StatsModels/pMxlJ/src/terms.jl:397
Note that you also find the general negation meaning here (method 3 above, which defines the behaviour for calling ~ on a boolean argument and is in Base Julia).
I agree that the GLM.jl docs maybe aren't the most comprehensive in the world, but one of the reasons for that is that the whole machinery behind #formula actually isn't a GLM.jl thing - so do check out the StatsModels docs linked above which are quite good I think.
I am trying to plot a function test(a) which returns the root of another (in this case x^3-2*x^2+a):
Why does this not work? Why does it work when I change a to x in the last line?
The problem is that mnewton requires an expression which evaluates to a number after assigning values to the specified variables (just x in the example). To make this work, you need to prevent mnewton from being called until a has a value. For example, something like:
plot2d (lambda ([a], rhs(mnewton(p(x, a) = 0, x, 20)[1][1])), [a, -1, 1]);
where p(x, a) is the polynomial expression.
I've made the range of a smaller, since I get an error from mnewton for larger values of a.
You can also try find_root to solve equations of 1 variable.
Since the function is question is a cubic polynomial, there is an exact solution and Maxima can find it via: algsys([p(x, a)], [x]);
Is there a typed programming language where I can constrain types like the following two examples?
A Probability is a floating point number with minimum value 0.0 and maximum value 1.0.
type Probability subtype of float
where
max_value = 0.0
min_value = 1.0
A Discrete Probability Distribution is a map, where: the keys should all be the same type, the values are all Probabilities, and the sum of the values = 1.0.
type DPD<K> subtype of map<K, Probability>
where
sum(values) = 1.0
As far as I understand, this is not possible with Haskell or Agda.
What you want is called refinement types.
It's possible to define Probability in Agda: Prob.agda
The probability mass function type, with sum condition is defined at line 264.
There are languages with more direct refinement types than in Agda, for example ATS
You can do this in Haskell with Liquid Haskell which extends Haskell with refinement types. The predicates are managed by an SMT solver at compile time which means that the proofs are fully automatic but the logic you can use is limited by what the SMT solver handles. (Happily, modern SMT solvers are reasonably versatile!)
One problem is that I don't think Liquid Haskell currently supports floats. If it doesn't though, it should be possible to rectify because there are theories of floating point numbers for SMT solvers. You could also pretend floating point numbers were actually rational (or even use Rational in Haskell!). With this in mind, your first type could look like this:
{p : Float | p >= 0 && p <= 1}
Your second type would be a bit harder to encode, especially because maps are an abstract type that's hard to reason about. If you used a list of pairs instead of a map, you could write a "measure" like this:
measure total :: [(a, Float)] -> Float
total [] = 0
total ((_, p):ps) = p + probDist ps
(You might want to wrap [] in a newtype too.)
Now you can use total in a refinement to constrain a list:
{dist: [(a, Float)] | total dist == 1}
The neat trick with Liquid Haskell is that all the reasoning is automated for you at compile time, in return for using a somewhat constrained logic. (Measures like total are also very constrained in how they can be written—it's a small subset of Haskell with rules like "exactly one case per constructor".) This means that refinement types in this style are less powerful but much easier to use than full-on dependent types, making them more practical.
Perl6 has a notion of "type subsets" which can add arbitrary conditions to create a "sub type."
For your question specifically:
subset Probability of Real where 0 .. 1;
and
role DPD[::T] {
has Map[T, Probability] $.map
where [+](.values) == 1; # calls `.values` on Map
}
(note: in current implementations, the "where" part is checked at run-time, but since "real types" are checked at compile-time (that includes your classes), and since there are pure annotations (is pure) inside the std (which is mostly perl6) (those are also on operators like *, etc), it's only a matter of effort put into it (and it shouldn't be much more).
More generally:
# (%% is the "divisible by", which we can negate, becoming "!%%")
subset Even of Int where * %% 2; # * creates a closure around its expression
subset Odd of Int where -> $n { $n !%% 2 } # using a real "closure" ("pointy block")
Then you can check if a number matches with the Smart Matching operator ~~:
say 4 ~~ Even; # True
say 4 ~~ Odd; # False
say 5 ~~ Odd; # True
And, thanks to multi subs (or multi whatever, really – multi methods or others), we can dispatch based on that:
multi say-parity(Odd $n) { say "Number $n is odd" }
multi say-parity(Even) { say "This number is even" } # we don't name the argument, we just put its type
#Also, the last semicolon in a block is optional
Nimrod is a new language that supports this concept. They are called Subranges. Here is an example. You can learn more about the language here link
type
TSubrange = range[0..5]
For the first part, yes, that would be Pascal, which has integer subranges.
The Whiley language supports something very much like what you are saying. For example:
type natural is (int x) where x >= 0
type probability is (real x) where 0.0 <= x && x <= 1.0
These types can also be implemented as pre-/post-conditions like so:
function abs(int x) => (int r)
ensures r >= 0:
//
if x >= 0:
return x
else:
return -x
The language is very expressive. These invariants and pre-/post-conditions are verified statically using an SMT solver. This handles examples like the above very well, but currently struggles with more complex examples involving arrays and loop invariants.
For anyone interested, I thought I'd add an example of how you might solve this in Nim as of 2019.
The first part of the questions is straightfoward, since in the interval since since this question was asked, Nim has gained the ability to generate subrange types on floats (as well as ordinal and enum types). The code below defines two new float subranges types, Probability and ProbOne.
The second part of the question is more tricky -- defining a type with constrains on a function of it's fields. My proposed solution doesn't directly define such a type but instead uses a macro (makePmf) to tie the creation of a constant Table[T,Probability] object to the ability to create a valid ProbOne object (thus ensuring that the PMF is valid). The makePmf macro is evaluated at compile time, ensuring that you can't create an invalid PMF table.
Note that I'm a relative newcomer to Nim so this may not be the most idiomatic way to write this macro:
import macros, tables
type
Probability = range[0.0 .. 1.0]
ProbOne = range[1.0..1.0]
macro makePmf(name: untyped, tbl: untyped): untyped =
## Construct a Table[T, Probability] ensuring
## Sum(Probabilities) == 1.0
# helper templates
template asTable(tc: untyped): untyped =
tc.toTable
template asProb(f: float): untyped =
Probability(f)
# ensure that passed value is already is already
# a table constructor
tbl.expectKind nnkTableConstr
var
totprob: Probability = 0.0
fval: float
newtbl = newTree(nnkTableConstr)
# create Table[T, Probability]
for child in tbl:
child.expectKind nnkExprColonExpr
child[1].expectKind nnkFloatLit
fval = floatVal(child[1])
totprob += Probability(fval)
newtbl.add(newColonExpr(child[0], getAst(asProb(fval))))
# this serves as the check that probs sum to 1.0
discard ProbOne(totprob)
result = newStmtList(newConstStmt(name, getAst(asTable(newtbl))))
makePmf(uniformpmf, {"A": 0.25, "B": 0.25, "C": 0.25, "D": 0.25})
# this static block will show that the macro was evaluated at compile time
static:
echo uniformpmf
# the following invalid PMF won't compile
# makePmf(invalidpmf, {"A": 0.25, "B": 0.25, "C": 0.25, "D": 0.15})
Note: A cool benefit of using a macro is that nimsuggest (as integrated into VS Code) will even highlight attempts to create an invalid Pmf table.
Modula 3 has subrange types. (Subranges of ordinals.) So for your Example 1, if you're willing to map probability to an integer range of some precision, you could use this:
TYPE PROBABILITY = [0..100]
Add significant digits as necessary.
Ref: More about subrange ordinals here.
I was looking through a programming question, when the following question suddenly seemed related.
How do you convert a string to another string using as few swaps as follows. The strings are guaranteed to be interconvertible (they have the same set of characters, this is given), but the characters can be repeated. I saw web results on the same question, without the characters being repeated though.
Any two characters in the string can be swapped.
For instance : "aabbccdd" can be converted to "ddbbccaa" in two swaps, and "abcc" can be converted to "accb" in one swap.
Thanks!
This is an expanded and corrected version of Subhasis's answer.
Formally, the problem is, given a n-letter alphabet V and two m-letter words, x and y, for which there exists a permutation p such that p(x) = y, determine the least number of swaps (permutations that fix all but two elements) whose composition q satisfies q(x) = y. Assuming that n-letter words are maps from the set {1, ..., m} to V and that p and q are permutations on {1, ..., m}, the action p(x) is defined as the composition p followed by x.
The least number of swaps whose composition is p can be expressed in terms of the cycle decomposition of p. When j1, ..., jk are pairwise distinct in {1, ..., m}, the cycle (j1 ... jk) is a permutation that maps ji to ji + 1 for i in {1, ..., k - 1}, maps jk to j1, and maps every other element to itself. The permutation p is the composition of every distinct cycle (j p(j) p(p(j)) ... j'), where j is arbitrary and p(j') = j. The order of composition does not matter, since each element appears in exactly one of the composed cycles. A k-element cycle (j1 ... jk) can be written as the product (j1 jk) (j1 jk - 1) ... (j1 j2) of k - 1 cycles. In general, every permutation can be written as a composition of m swaps minus the number of cycles comprising its cycle decomposition. A straightforward induction proof shows that this is optimal.
Now we get to the heart of Subhasis's answer. Instances of the asker's problem correspond one-to-one with Eulerian (for every vertex, in-degree equals out-degree) digraphs G with vertices V and m arcs labeled 1, ..., m. For j in {1, ..., n}, the arc labeled j goes from y(j) to x(j). The problem in terms of G is to determine how many parts a partition of the arcs of G into directed cycles can have. (Since G is Eulerian, such a partition always exists.) This is because the permutations q such that q(x) = y are in one-to-one correspondence with the partitions, as follows. For each cycle (j1 ... jk) of q, there is a part whose directed cycle is comprised of the arcs labeled j1, ..., jk.
The problem with Subhasis's NP-hardness reduction is that arc-disjoint cycle packing on Eulerian digraphs is a special case of arc-disjoint cycle packing on general digraphs, so an NP-hardness result for the latter has no direct implications for the complexity status of the former. In very recent work (see the citation below), however, it has been shown that, indeed, even the Eulerian special case is NP-hard. Thus, by the correspondence above, the asker's problem is as well.
As Subhasis hints, this problem can be solved in polynomial time when n, the size of the alphabet, is fixed (fixed-parameter tractable). Since there are O(n!) distinguishable cycles when the arcs are unlabeled, we can use dynamic programming on a state space of size O(mn), the number of distinguishable subgraphs. In practice, that might be sufficient for (let's say) a binary alphabet, but if I were to try to try to solve this problem exactly on instances with large alphabets, then I likely would try branch and bound, obtaining bounds by using linear programming with column generation to pack cycles fractionally.
#article{DBLP:journals/corr/GutinJSW14,
author = {Gregory Gutin and
Mark Jones and
Bin Sheng and
Magnus Wahlstr{\"o}m},
title = {Parameterized Directed \$k\$-Chinese Postman Problem and \$k\$
Arc-Disjoint Cycles Problem on Euler Digraphs},
journal = {CoRR},
volume = {abs/1402.2137},
year = {2014},
ee = {http://arxiv.org/abs/1402.2137},
bibsource = {DBLP, http://dblp.uni-trier.de}
}
You can construct the "difference" strings S and S', i.e. a string which contains the characters at the differing positions of the two strings, e.g. for acbacb and abcabc it will be cbcb and bcbc. Let us say this contains n characters.
You can now construct a "permutation graph" G which will have n nodes and an edge from i to j if S[i] == S'[j]. In the case of all unique characters, it is easy to see that the required number of swaps will be (n - number of cycles in G), which can be found out in O(n) time.
However, in the case where there are any number of duplicate characters, this reduces to the problem of finding out the largest number of cycles in a directed graph, which, I think, is NP-hard, (e.g. check out: http://www.math.ucsd.edu/~jverstra/dcig.pdf ).
In that paper a few greedy algorithms are pointed out, one of which is particularly simple:
At each step, find the minimum length cycle in the graph (e.g. Find cycle of shortest length in a directed graph with positive weights )
Delete it
Repeat until all vertexes have not been covered.
However, there may be efficient algorithms utilizing the properties of your case (the only one I can think of is that your graphs will be K-partite, where K is the number of unique characters in S). Good luck!
Edit:
Please refer to David's answer for a fuller and correct explanation of the problem.
Do an A* search (see http://en.wikipedia.org/wiki/A-star_search_algorithm for an explanation) for the shortest path through the graph of equivalent strings from one string to the other. Use the Levenshtein distance / 2 as your cost heuristic.
Haskell implementation of the familiar Fibonacci function
fibSlow n
| n == 0 = 1 --fib.1
| n == 1 = 1 --fib.2
| otherwise = fibSlow(n-1) + fibSlow(n-2) --fib.3
What is the induction proof of correctness for fibSlow?
To prove correctness of a function on the natural numbers by induction, you would show that it's correct for certain base cases, and then that it's correct for higher values of the parameter given the assumption that it's correct for lower ones. So you'd verify first that fibSlow 0 = 1, and then that fibSlow 1 = 1, and then that for n > 1, fibSlow n is equal to the (n-1)th fibonacci number plus the (n-2)th fibonacci number. Here you get to assume that those numbers are fibSlow (n-1) and fibSlow (n-2), since fibSlow is correct for all inputs less than n by the inductive hypothesis.
This might seem all rather trivial... because it is! The whole point of such an example in Haskell is that you can write code that's obviously correct. When you go to prove it correct, the proof just writes itself and amounts to looking at the code and noting that it clearly says exactly what you're trying to prove. This is one of the nice properties of a declarative language like Haskell.
Apologies I haven't formally seen this kind of material for a while, so you're probably best looking at other sources if this is homework.
I think you want to show the existence of a monotone function which describes the "progress" of the recursion. This case should be pretty simple: the argument itself is monotonically decreasing. For a nonnegative n, the recursive call will be made with a lesser n', and that n' will never be less than zero.
You can also use power induction to argue the function is defined on all n. You have declared it defined on 0 and 1, and it suffices to say that if it's defined on n and n+1, then it's defined on n+2. This is obvious by the definition of the recursive call.
I think you might be able to read up on some formalities in Jech's Set Theory book, in the Ordinals chapter.