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Can anyone help me with taking Wreath Products of Groups in Sagemath?
I haven't been able to find a online reference and it doesn't appear to be built in as far as I can tell.
As far as I know, you would have to use GAP to compute them within Sage (and then can manipulate them from within Sage as well). See e.g. this discussion from 2012. This question has information about it, here is the documentation, and here it is within Sage:
F = AbelianGroup(3,[2]*3)
G = PermutationGroup([[(1,2,3),(4,5)],[(3,4)]])
Gp = gap.StandardWreathProduct(gap(F),gap(G))
print Gp
However, if you try to get this back into Sage, you will get a NotImplementedError because Sage doesn't understand what GAP returns in this wacky case (which I hope even is legitimate). Presumably if a recognized group is returned then one could eventually get it back to Sage for further processing. In this case, you might be better off doing some GAP computations and then putting them back in Sage after doing all your group stuff (which isn't always the case).
(Misleading title: it's only one of a plethora of inter-related similar questions below: these sound like asking for a full reference manual but keep in mind for this topic there is no reference manual other than the entirety of GHC's source-codes of its STG pipeline stage, and the collective accumulated experience of others/"insiders"..)
I'm exploring "transpiling" Haskell (from scratch for fun/learning, ignoring existing projects; target language/s similarly high-level / "already-fit-for-STG-machine" with existing GC + lambdas/func-values + closures) and so I'm trying to become ever more familiar with GHC's STG IR. Having repeatedly gone through the dozen-or-two online articles/videos of varying age, depth, detail that actually deal with the topic (plus the original paper, plus StgSyn.hs), and understanding many-perhaps-most basic principles, seeing -ddump-stged output still baffles me in various parts (I won't manually parse it but reuse GHC API's in-memory AST later on of course) --- mostly I think I'm stuck mapping my "roughly known" concepts to the "still-foreign" abbreviated/codified identifiers of that IR. If you know your way around STG a bit, mind looking at the following mini-sample to clarify a few open questions and help further solidify my (and future searchers') grasp?
From a most simple .hs module, I have -ddump-stged twice, first (on the left) with -O0 and then (on the right) with -O2, both captured in this diff.
Walking through everything def-by-def..
Lines L_|R5-11: so in O2, testX1 and testX2 seem to be global constants/literals for the integers 4 and 5 --- O0 doesn't have them. Curious!
Is Str=DmdType something about strictness? "Strictness is of type on-demand" or some such? But then a top-level/heap-ish/"global" constant literal can't be "lazy" can it.. (one of the things where I can't just casually Ctrl+F in StgSyn.hs --- it's not in there! which is odd in its own way, how come there's STG syntax not in StgSyn.hs)
Caf have a rough idea about constant-applicative-forms, but Unf=OtherCon? "Other constructor" (unboxed/native Type.S#-related?) ..
Line L6|R14: Surprised to still see type-class information in there (Num), is that "just info/annotation" or is this crucial for any of the built-in code-gens to set up some "dictionary" lookup machinery at runtime? (I'd sure hope by the late STG / pre-CMM stage that would be resolved and inlined already where possible at least in O2. After all GHC has also decided to type-default 4 and 5 to Integer). Generally speaking I understand STG is "untyped" other than denoting prim types, saturated cons, perhaps strings (looks like it later on at the bottom), so such "typeclass" annotations can only be.. I guess for readers to find their way around the ddump-ed *.stg. But correct me if not.
GblId probably just "global identifier" aka top-level CAF right? Arity clear.
Line L7|R18: now Str=DmdType for testX is, only in O2, followed by a freakish <S(LLC(C(S))LLLL),U(1*C1(C1(U)),A,1*C1(C1(U)),A,A,A,C(U))><L,U>! What's that, SKI calculus? ;D no seriously, LLC.. LLLL.. stack or other memory layout hints for CMM? Any idea? Must be some optimization, would like to understand which-and-how..
Line L8|R20: $dNum_sGM (left) and $dNum_sIx (right) have me a bit concerned, they don't seem to be "defined at the module level" here anywhere. Typeclass "method dispatch dictionary lookup" kind of thing? Would eg. CMM take this together with the above Num annotation to set things up? It always appears together with the input func arg.
The whole function "body" for both left and right can be seen here essentially as "3 lets with a lambda-ish form for 3 atoms, 2 of which are statically known literal-constants" --- I suppose this is standard and to be expected in the STG IR AST? For the first of these, funnily enough we could say that O0 has "inlined the global (what is testX1 or testX2 in O2) and O2 hasn't" (making the latter much shorter as that applies to both these constant literals).
I've only ever seen Occ=Once, what are the others and how to interpret? Once for one isn't even in StgSyn.hs..
Now LclId a counterpart to the earlier encountered GblId. That's denoting the scope of the identifier? Could it also be anything else, in this expression context? As in: if traversing the AST I roughly know how deep I am, I can ignore this since if I'm at the top-level it must be GblId and otherwise LclId? Hm.. maybe better take what STG gives me but then I need to be sure about the semantics and possibilities.. guys, using StgSyn.hs I have the wrong source file, right? Nothing on this in there either.. (always hopeful as its comments are quite well-done)
the rest is just metadata as string constants, OK.. oh wait, look at O2, there's Str=DmdType m1 and Str=DmdType m, what's the m/m1 about, another thing I don't see "defined anywhere at the module level" here? And it's not in O0..
still going strong? Merely a bonus question (for now), tell us about srt:SRT:[] ;)
Just a few tidbits - a full answer is quite beyond my knowledge.
The type of your function is
testX :: GHC.Num.Num a => a -> a
It’s compiled to a function with two arguments: a dictionary of the Num type class, and the actual argument.
The $d… names stand for dictionaries of type class instances. The <S(LLC(C(S))LLLL),… annotations are strictness information about the function arguments. They basically say which part of the argument will be used by your function and which not. Looks a bit weird here because it contains information about all the class instance members.
Some of this is explained here:
https://ghc.haskell.org/trac/ghc/wiki/Commentary/Compiler/Demand
The str:STR: is the „Static reference table“, i.e. list of free variables of the expression - in your case, always [].
In the bottom i gave link to whole program (ciao) to make help easier.
I try to make in Prolog function that will be have list of questions like
questions([[[What, are,you,doing,?],[Where,am,I,Incorrect,?]]]).
answers([[Im,doing,exercise],[I,do,nothing]],[[You,are,incorrect,at,'..'],[i,dont,know]]]).
wordkeys([[[Incorrect,50],[doing,20]]]).
I know it look really messy but I really need help and will be grateful.
Main function is checking which answer is the best (having biggest sum of keywords points).
My problem is that all look fine(made some write() to see what happening) till it go to last function here :
count_pnt_keys()
Prolog checking all words if their equal but when is out from keywords should come back to function called it but its just 'no' . Maybe I should check if list is empty before I call again the same function with just Tail? How to do this?
rules:
count_pnt([],[],[]).
count_pnt([Ah|At],Keys,RList) :- %choose one answer from answer list and go further
count_pnt_word(Ah,Keys,Pnts), %return sum of points for one answer
count_ADD_POINT(RList,Pnts), %not important here
count_pnt(At,Keys,RList). %call again with next question
/*vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv*/
count_pnt_word([],[],0)
count_pnt_word([Ah|At],Keys,Pnts) :- %choose one WORD from answer and go further
count_pnt_keys(Ah,Keys,Pnts),
count_pnt_word(At,Keys,Pnts). %call recursion with next word from answer
/*vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv*/
count_pnt_keys([],[],0)
count_pnt_keys(AWord,[Kh|Kt],Pnts) :- %check word with all keys for first question
get_tail_h(Kh,KWORD,KPOINTS), %just return head and tail from first key
AWord==KWORD, %check if they are equal
/*counting points*/ !, %counting not important when end counting points go out to next
count_pnt_keys(AWord,Kt,Pnts). %call again if not equal and try with next key
and I call it:
test :-
write(AnswerFirst),
count_pnt(FirstPackOfAnswer,FirstPackofKeys,ReturnedListwithpoints),
write(ReturnedListwithpoints).
link to code (ciao)
http://wklej.org/id/754478/
You call count_pnt with three free arguments, which means that count_pnt will first unify all its arguments with an empty list. Upon backtracking the recursive count_pnt clause is called which leads to count_pnt_keys with again three free arguments, that will lead to Ah being unified with [] etc rather than a failure.
Do you really call count_pnt as suggested by the code for test?
count_pnts(_,_,[],_).
count_pnt_word(_,[],_).
count_pnt_keys([],_,_).
should look like this that was a problem
Original question:
I know Mathematica has a built in map(f, x), but what does this function look like? I know you need to look at every element in the list.
Any help or suggestions?
Edit (by Jefromi, pieced together from Mike's comments):
I am working on a program what needs to move through a list like the Map, but I am not allowed to use it. I'm not allowed to use Table either; I need to move through the list without help of another function. I'm working on a recursive version, I have an empty list one down, but moving through a list with items in it is not working out. Here is my first case: newMap[#, {}] = {} (the map of an empty list is just an empty list)
I posted a recursive solution but then decided to delete it, since from the comments this sounds like a homework problem, and I'm normally a teach-to-fish person.
You're on the way to a recursive solution with your definition newMap[f_, {}] := {}.
Mathematica's pattern-matching is your friend. Consider how you might implement the definition for newMap[f_, {e_}], and from there, newMap[f_, {e_, rest___}].
One last hint: once you can define that last function, you don't actually need the case for {e_}.
UPDATE:
Based on your comments, maybe this example will help you see how to apply an arbitrary function:
func[a_, b_] := a[b]
In[4]:= func[Abs, x]
Out[4]= Abs[x]
SOLUTION
Since the OP caught a fish, so to speak, (congrats!) here are two recursive solutions, to satisfy the curiosity of any onlookers. This first one is probably what I would consider "idiomatic" Mathematica:
map1[f_, {}] := {}
map1[f_, {e_, rest___}] := {f[e], Sequence##map1[f,{rest}]}
Here is the approach that does not leverage pattern matching quite as much, which is basically what the OP ended up with:
map2[f_, {}] := {}
map2[f_, lis_] := {f[First[lis]], Sequence##map2[f, Rest[lis]]}
The {f[e], Sequence##map[f,{rest}]} part can be expressed in a variety of equivalent ways, for example:
Prepend[map[f, {rest}], f[e]]
Join[{f[e]}, map[f, {rest}] (#Mike used this method)
Flatten[{{f[e]}, map[f, {rest}]}, 1]
I'll leave it to the reader to think of any more, and to ponder the performance implications of most of those =)
Finally, for fun, here's a procedural version, even though writing it made me a little nauseous: ;-)
map3[f_, lis_] :=
(* copy lis since it is read-only *)
Module[{ret = lis, i},
For[i = 1, i <= Length[lis], i++,
ret[[i]] = f[lis[[i]]]
];
ret
]
To answer the question you posed in the comments, the first argument in Map is a function that accepts a single argument. This can be a pure function, or the name of a function that already only accepts a single argument like
In[1]:=f[x_]:= x + 2
Map[f, {1,2,3}]
Out[1]:={3,4,5}
As to how to replace Map with a recursive function of your own devising ... Following Jefromi's example, I'm not going to give to much away, as this is homework. But, you'll obviously need some way of operating on a piece of the list while keeping the rest of the list intact for the recursive part of you map function. As he said, Part is a good starting place, but I'd look at some of the other functions it references and see if they are more useful, like First and Rest. Also, I can see where Flatten would be useful. Finally, you'll need a way to end the recursion, so learning how to constrain patterns may be useful. Incidentally, this can be done in one or two lines depending on if you create a second definition for your map (the easier way), or not.
Hint: Now that you have your end condition, you need to answer three questions:
how do I extract a single element from my list,
how do I reference the remaining elements of the list, and
how do I put it back together?
It helps to think of a single step in the process, and what do you need to accomplish in that step.
I have question that comes from a algorithms book I'm reading and I am stumped on how to solve it (it's been a long time since I've done log or exponent math). The problem is as follows:
Suppose we are comparing implementations of insertion sort and merge sort on the same
machine. For inputs of size n, insertion sort runs in 8n^2 steps, while merge sort runs in 64n log n steps. For which values of n does insertion sort beat merge sort?
Log is base 2. I've started out trying to solve for equality, but get stuck around n = 8 log n.
I would like the answer to discuss how to solve this mathematically (brute force with excel not admissible sorry ;) ). Any links to the description of log math would be very helpful in my understanding your answer as well.
Thank you in advance!
http://www.wolframalpha.com/input/?i=solve%288+log%282%2Cn%29%3Dn%2Cn%29
(edited since old link stopped working)
Your best bet is to use Newton;s method.
http://en.wikipedia.org/wiki/Newton%27s_method
One technique to solving this would be to simply grab a graphing calculator and graph both functions (see the Wolfram link in another answer). Find the intersection that interests you (in case there are multiple intersections, as there are in your example).
In any case, there isn't a simple expression to solve n = 8 log₂ n (as far as I know). It may be simpler to rephrase the question as: "Find a zero of f(n) = n - 8 log₂ n". First, find a region containing the intersection you're interested in, and keep shrinking that region. For instance, suppose you know your target n is greater than 42, but less than 44. f(42) is less than 0, and f(44) is greater than 0. Try f(43). It's less than 0, so try 43.5. It's still less than 0, so try 43.75. It's greater than 0, so try 43.625. It's greater than 0, so keep going down, and so on. This technique is called binary search.
Sorry, that's just a variation of "brute force with excel" :-)
Edit:
For the fun of it, I made a spreadsheet that solves this problem with binary search: binary‑search.xls . The binary search logic is in the second data column, and I just auto-extended that.