Could built in function TuringMachine be used somehow to shorten implementation?

I think your method can be simplified:

ClearAll[NarcissiticQ]
NarcissiticQ[num_Integer] := Module[{dig = IntegerDigits[num], m},
  m = Length[dig];
  Total[dig^m] === num
  ]
start = AbsoluteTime[];
n = 25;
m = 0;

i = 0;
Dynamic[{i, m} // Column]
a = Reap[
   While[m < n,
    If[NarcissiticQ2[i],
     Sow[i];
     m++;
     ];
    i++;
    ]
   ];
a = a[[2, 1]]
AbsoluteTime[] - start

this takes 121 seconds on my machine.

Alternatively we can memorize (and memoize) 0..9^p and use replace instead of calculating the powers:

ClearAll[NarcissiticQ2, NarcissiticRules]
NarcissiticRules[n_] := NarcissiticRules[n] = Thread[Range[0, 9] -> Range[0, 9]^n]
NarcissiticQ2[num_Integer] := Module[{dig = IntegerDigits[num], rul},
  rul = NarcissiticRules[Length[dig]];
  Total[Replace[dig, rul, {1}]] === num
 ]

But this is slower; around 168 seconds on my machine.

Could you explain your method? It looks really complex!

Thanks Ian for adding :) looks like a very clean implementation!

The small confusion (aside from the transposition of values that Jason points out) is that the initial conditions are for a[0] and b[0] rather than a[1] and b[1].

So, with that said, this direct approach looks like

Clear[a, b, piCalc, $numericalPrecision]

$numericalPrecision = 150;

piCalc[n_] := (4*a[n]^2)/(1 - 
    Sum[2^(1 + k)*(a[k]^2 - b[k]^2), {k, 1, n}])

a[h_] := (a[h] = N[(a[h - 1] + b[h - 1])/2, $numericalPrecision])

b[h_] := (b[h] = N[Sqrt[a[h - 1] b[h - 1]], $numericalPrecision])

a[0] = 1;

b[0] = 1/Sqrt[2];

And here is an example of a calculation:

In[134]:= piCalc[100]

Out[134]= \ 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066

Some work needs to be done on this approach to adjust the value of $numericalPrecision and the argument of piCalc so that the desired number of digits are obtained.

Note that I have added memoization to the calculation to speed it up.

Hi all, So I have been trying rather unsuccessfully to implement the Arithmetic-geometric mean/Calculate Pi problem in the Wolfram Language, and I have been getting some rather disappointing and confusing results. Firstly, after studying the literature, it would seem to me that the method required by the Rosetta Code challenge is as follows,

Where a[n] is the arithmetic mean of a[n-1] and b[n-1], while b[n] is the geometric mean of a[n-1] and b[n-1].

But the problem is that I cannot find this exact formula anywhere else on the internet, and not even in the paper (pdf) that the Rosetta page references. Regardless, my implementation in the Wolfram language follows,

piCalc[n_]:=(4*(a[n])^2)/(1-Sum[(2^(k+1))*((a[k])^2-(b[k])^2),{k,1,n}])
a[h_]:=Mean[{a[h-1],b[h-1]}]
b[l_]:=GeometricMean[{a[l-1],b[l-1]}]
a[1]=Sqrt[2];
b[1]=1;

I have been unable to find what initial conditions for a and b will give Pi. I chose Sqrt[2] and 1 here because those are what is used in one of the papers mentioned below that use a similar algorithm, but not only do these number result in a negative result, it is nowhere near Pi.

Now there is some ambiguity from what I have seen concerning what a1] and what b[1] should be, as the Rosetta Code page makes no mention of these values, and I cannot determine from the other languages presentations what the other implementations use. I found another paper, [here, which makes mention of using simply the arithmetic and geometric mean formulas, to calculate Pi with the initial conditions, a2]=1/2, and a[2]=1/4, which doesn't make sense to me, but also makes Mathematica reach its recursion limit. Additionally, the [Wikipedia page appears to have an error on their page with respect to how c[n] is defined, as well as not providing initial conditions. Another site also makes mention of using the AGM to calculate Pi, and while it includes the exact formulation for Pi, it doesn't include the formula listed on the Rosetta Code site. It also mentions some of the other formulas I have seen, which seem both simpler than the Rosetta code's formulation, as well as quicker (though I believe in one or multiple of the papers it is proven that they converge quadratically). I am wondering if anyone here knows if there is anything wrong with my code, or if the formula from the Rosetta Code page is incorrect, or even if anyone know what initial conditions I should use for this, any help would be much appreciated.

Thanks,

Ian

Attachments:

This looks really nice! I'm not sure how it exactly works, but it works!

I have a program for Last letter-first letter. It simply enumerates all lists of every length. The main work is done with one line of Mathematica and two short functions. On an i5 desktop it takes 65 seconds. Printing takes a few more lines. The word list is an attached file. I have not yet submitted to the web site, brute force but workable. (updated for a little more speed)

pokemon = 
  Import[FileNameJoin[{NotebookDirectory[], "pokemon.txt"}]] // 
   StringSplit;
pindex[x_] := 
  First[ToCharacterCode[x]] - First[ToCharacterCode["a"]] + 1;
pdx = {pindex[StringTake[#, 1]], pindex[StringTake[#, -1]], 
     Position[pokemon, #][[1]]} & /@ pokemon;
pslice = Table[Cases[pdx, {i, _, _}], {i, pindex["z"]}];

merge[m_, n_] := Sow[{m[[1]], n[[2]], Join[m[[3]], n[[3]]]}];
reaper[p_ ] := 
  Reap[Do[If[(n[[2]] > 0 && pslice[[n[[2]]]] != {}), 
      Do[If[Not[MemberQ[n[[3]], q[[3, 1]]]], merge[n, q]], {q, 
        pslice[[n[[2]]]]}], Null], {n, First[p]}]][[2]];

 (* heart of the processing *)
 answers = NestWhileList[reaper, {pdx}, Length[#] > 0 &][[1 ;; -2]];

(* Summary of answers *)
Print["Number of lists of each length:"];
i = 0; Print[i = i + 1, " ", Length[#[[1]]]] & /@ answers;
Print["Length of longest list is ", i, ", examples follow"];
pokemon[[answers[[i, 1, 1, 3]]]]
pokemon[[answers[[i, 1, 2, 3]]]]

Output - Number of lists of each length:

1 70

2 172

3 494

4 1288

5 3235

6 7731

7 17628

8 37629

9 75122

10 139091

11 236679

12 367405

13 516210

14 650916

15 733915

16 727566

17 621835

18 446666

19 260862

20 119908

21 40296

22 10112

23 1248

Length of longest list is 23, examples follow

{"machamp", "petilil", "landorus", "scrafty", "yamask", "kricketune", \ "emboar", "registeel", "loudred", "darmanitan", "nosepass", \ "simisear", "relicanth", "heatmor", "rufflet", "trapinch", "haxorus", \ "seaking", "girafarig", "gabite", "exeggcute", "emolga", "audino"}

{"machamp", "petilil", "landorus", "scrafty", "yamask", "kricketune", \ "emboar", "registeel", "loudred", "darmanitan", "nosepass", \ "simisear", "rufflet", "trapinch", "heatmor", "relicanth", "haxorus", \ "seaking", "girafarig", "gabite", "exeggcute", "emolga", "audino"}

Attachments:

Thanks for the inspiration Luis!

Hi Luis,

I solved the Radix problem. We simply add a number to all numbers (such that all of them are positive), do the algorithm, and then subtract a number again.

ClearAll[SortByPos, RadixSort]
SortByPos[data : {_List ..}, pos_Integer] := Module[{digs, order},
  digs = data[[All, pos]];
  order = Ordering[digs];
  data[[order]]
 ]
RadixSort[x : {_Integer ..}] := Module[{y, digs, maxlen, offset},
  offset = Min[x];
  y = x - offset;
  digs = IntegerDigits /@ y;
  maxlen = Max[Length /@ digs];
  digs = IntegerDigits[#, 10, maxlen] & /@ y;
  digs = Fold[SortByPos, digs, -Range[maxlen]];
  digs = FromDigits /@ digs;
  digs += offset;
  digs
 ]

And submitted it to the Rosetta stone

Bump! Added code for Set Puzzle:

colors = {Red, Green, Purple};
symbols = {"0", "\[TildeTilde]", "\[Diamond]"};
numbers = {1, 2, 3};
shadings = {"\[FilledSquare]", "\[Square]", "\[DoublePrime]"};

validTripleQ[l_List] := Entropy[l] != Entropy[{1, 1, 2}];
validSetQ[cards_List] := And @@ (validTripleQ /@ Transpose[cards]);

allCards = Tuples[{colors, symbols, numbers, shadings}];

deal[{numDeal_, setNum_}] := Module[{cards, count = 0}, 
   While[count != setNum,
    cards = RandomSample[allCards, numDeal];
    count = Count[Subsets[cards, {3}], _?validSetQ]];
   cards];

Row[{Style[#2, #1], #3, #4}] & @@@ deal[{9, 4}]

Fairly lazy approach but it shows off a lot of Mathematica's power.

Hi friend, now I leave my attempt to solve the sorting radix,I must say that is intended for positive integers,I do not know if you can also make for negative integers, please see it and correct what you think is wrong. Estamos comunicacion permanente

Attachments:

Hola Luis,

Muy bien code! Yo cambio el code un poco; por el legibilidad, y para hacer el code conciso. La tarea se compone dos partes: el algoritmo Gale-Shapley (incluido en la función StableMarriage) y perturbar los pares y prueba la estabilidad. Parte uno:

Quiet[Needs["Combinatorica`"]];
mals = {"abe", "bob", "col", "dan", "ed", "fred", "gav", "hal", "ian", "jon"};
fems = {"abi", "bea", "cath", "dee", "eve", "fay", "gay", "hope", "ivy", "jan"};
menstrg = {{"abi", "eve", "cath", "ivy", "jan", "dee", "fay", "bea", "hope", "gay"}, {"cath", "hope", "abi", "dee", "eve", "fay", "bea", "jan", "ivy", "gay"}, {"hope", "eve", "abi", "dee", "bea", "fay", "ivy", "gay", "cath", "jan"}, {"ivy", "fay", "dee", "gay", "hope", "eve", "jan", "bea", "cath", "abi"}, {"jan", "dee", "bea","cath", "fay", "eve", "abi", "ivy", "hope", "gay"}, {"bea", "abi", "dee", "gay", "eve", "ivy", "cath", "jan", "hope", "fay"}, {"gay", "eve", "ivy", "bea", "cath", "abi", "dee", "hope","jan", "fay"}, {"abi", "eve", "hope", "fay", "ivy", "cath", "jan", "bea", "gay", "dee"}, {"hope", "cath", "dee", "gay", "bea","abi", "fay", "ivy", "jan", "eve"}, {"abi", "fay", "jan", "gay", "eve", "bea", "dee", "cath", "ivy", "hope"}};
womstrg = {{"bob", "fred", "jon", "gav", "ian", "abe", "dan", "ed", "col", "hal"}, {"bob", "abe", "col", "fred", "gav", "dan", "ian", "ed", "jon", "hal"}, {"fred", "bob", "ed", "gav", "hal", "col", "ian", "abe", "dan", "jon"}, {"fred", "jon", "col", "abe", "ian", "hal", "gav", "dan", "bob", "ed"}, {"jon", "hal", "fred", "dan",  "abe", "gav", "col", "ed", "ian", "bob"}, {"bob", "abe", "ed", "ian", "jon", "dan", "fred", "gav", "col", "hal"}, {"jon", "gav", "hal", "fred", "bob", "abe", "col", "ed", "dan", "ian"}, {"gav", "jon", "bob", "abe", "ian", "dan", "hal", "ed", "col", "fred"}, {"ian", "col", "hal", "gav", "fred", "bob", "abe", "ed",  "jon", "dan"}, {"ed", "hal", "gav", "abe", "bob", "jon", "col", "ian", "fred", "dan"}};
male = menstrg /. Thread[fems -> Range[10]];
female = womstrg /. Thread[mals -> Range[10]];
lks = StableMarriage[male, female] /. Thread[ Range[10] -> fems];
MapThread[{#1, " fiance ", #2} &, {mals, lks}] // Grid

Output:

abe   fiance   ivy
bob   fiance   cath
col   fiance   dee
dan   fiance   fay
ed   fiance   jan
fred   fiance   bea
gav   fiance   gay
hal   fiance   eve
ian   fiance   hope
jon   fiance   abi

Replace ( /. ) funciona en todos los niveles (Levels en Mathematica). Table no es necesario.

Perdona que mi Español sea malo, aún estoy aprendiendo.

For the english-only speaking:

The code above does part one of the task. Part two of the task is to perturb the 'best matches' and test stability. Anyone?

I added it! Thanks!

hola Sander, aqui esta mi propuesta para the Stable marriage problem, aunque es muy rustica cumple con el requisito de aplicar el algoritmo Gale/Shapley . Por favor hazle las modificaciones que cosideres necesarias, saludos y cuidate.

Attachments:

While I look over your suggestions on the remove line from file code, here is a start for reading a configuration file. I have ran into a brick wall with the "other family" list, but I might be attacking this problem from the wrong angle.

SetDirectory@NotebookDirectory[];
p = {}; m = {};
q = Fold[DeleteCases, Import["config.txt", "Data"], {{""}, {"#"}}];
Do[If[StringFreeQ[q[[i]], {"#"}][[1]],
   AppendTo[p, q[[i]]], ""], {i, Length@q}];
name = DeleteCases[
    StringCases[Flatten[p], "FULLNAME " ~~ x__ -> x], {}][[1, 1]];
m~AppendTo~{"name", name};
favf = DeleteCases[
    StringCases[Flatten[p], "FAVOURITEFRUIT " ~~ x__ -> x], {}][[1, 
    1]];
m~AppendTo~{"favouritefruit", favf};
If[StringMatchQ[
    Flatten[StringCases[Flatten[p], ___ ~~ "NEEDSPEELING"]], 
    "NEEDSPEELING"][[1]], needspeeling = "true", 
  needspeeling = "false"];
m~AppendTo~{"needspeeling", needspeeling};
If[StringMatchQ[
    Flatten[StringCases[Flatten[p], ___ ~~ "SEEDSREMOVED"]], 
    "SEEDSREMOVED"][[1]], seedsremoved = "true", 
  seedsremoved = "false"];
m~AppendTo~{"seedsremoved", seedsremoved};
Grid[m, Frame -> All]

Adding the config file as well.

http://rosettacode.org/wiki/Read_a_configuration_file

Attachments:

Here are some improvements that we should consider:

• document_ should be document_String to make it more foolproof, same for start_ and n_ Both should have head Integer.
• I would omit StringJoin[ToString@document, ".txt"] and just replace it by document; so the user can provide the filename directly.
• Maybe import it as "List", so you are sure that we have each 'line' of the file as an item.
• Dimensions[p][[1]] can be replaced by Length right?
• Table[{i}, {i, start, start + n - 1}] could be replaced by List/@Range[start,start+n-1] BUT:
• Delete[p, Table[{i}, {i, start, start + n - 1}]] could be replaced by Drop[p,{start,start+n-1}] directly. faster, and more concise.
• If the import is used using List export can be exported too using List: Export[FileNameJoin[{document,"_removed.txt"}],q,"List"]
• the variable q could be omitted: just overwrite p. (minor change of course...)

In general you should use the functions: FileNameSplit, FileNameJoin,FileNameTake,FileNameDrop, FileBaseName et cetera for combining filenames and folders. These are platform independent and should work on mac/linux/windows (the main problem is the "\" and "/" difference).

Remove lines from a file:

f[document_, start_, n_] :=
 Module[{p, q},
  SetDirectory@NotebookDirectory[];
  p = Import[StringJoin[ToString@document, ".txt"], "Data"];
  If[start + n - 1 <= Dimensions[p][[1]],
   q = Delete[p, Table[{i}, {i, start, start + n - 1}]];
   Export[StringJoin[ToString@document, "Removed.txt"], q];,
   Print["Too few lines in document. Operation aborted."]]]

Removes n lines including the designated starting line from the assigned txt file. Exports to a new file with suffix to indicate that the document has had lines removed, to not destroy the original file if the operation is run multiple times by mistake.

I would have used StringReplace:

StringReplace["abcdefghijkl","d"->""]

will output:

abcefghijkl

Well, that was pretty compact compared to my clunkstrosity! I guess it's quite obvious I am not familiar with a ton of the functions built in to Mathematica.

The problem seems to arise whenever the last letter is the last of the symbol string. I am posting a fix in a very short while. Notice that the second and third "If" argument in the decoding "Do" are slightly different. I have introduced a different string of symbols in one, and forgot to change the other. I will post a fix right after I have written out this post.

The code is first deleting the symbol from the list, which reduces the string length by 1, and then it tries to prepend the same symbol which should be at position 26, however the value has been deleted and the string is only 25 symbols long, which spawns the errors. By introducing a list number 2 which is deleted from, we can prepend the value to list 2 by funneling the symbol from list 1.

The line of code you pasted removes the input letter from the symbol string, the following code would prepend to the symbol string. The reason it's so clunky is that you can't do:

symTable = "abc";
StringDrop[symTable,"b"]

But rather

symTable = "abc";
StringDrop[symTable, {2}]

This means that we have to know where the "b" occurs;

StringPosition[symTable, StringTake[string , {i}]]

Outputs a list of the start- and stop-position of the string you are looking for, of the form

{{2,2}}

For which we only need one of the numbers to ge the correct placement. We can then replace the 2 in

StringDrop[symTable, {2}]

By the following code:

StringPosition[symTable, StringTake[string , {i}]][[1,1]]

Move-to-front algorithm:

mtf[string_] :=
 Module[{word, p, q, symTable, symTable2},
  p = {};
  q = {};
  symTable = StringJoin @@ CharacterRange["a", "z"];
  Do[
   If[
    i == StringLength@string,
    AppendTo[p, 
     StringPosition[symTable, StringTake[string, {i}]][[1, 1]] - 1];
    symTable = 
     StringDrop[
      symTable, {StringPosition[symTable, StringTake[string, {i}]][[1,
         1]]}];
    symTable = StringJoin[StringTake[string, {i}], symTable];
    Print["'", string, "' encodes to: ", p],
    AppendTo[p, 
     StringPosition[symTable, StringTake[string, {i}]][[1, 1]] - 1];
    symTable = 
     StringDrop[
      symTable, {StringPosition[symTable, StringTake[string, {i}]][[1,
         1]]}];
    symTable = StringJoin[StringTake[string, {i}], symTable];
    ]
   , {i, StringLength@string}];
  symTable = StringJoin @@ CharacterRange["a", "z"];
  Do[
   If[
    j == StringLength@string,
    q = StringJoin[q, StringTake[symTable, {p[[j]] + 1}]];
    symTable2 = 
     StringDrop[
      symTable, {StringPosition[symTable, 
         StringTake[symTable, {p[[j]] + 1}]][[1, 1]]}];
    symTable = 
     StringJoin[StringTake[symTable, {p[[j]] + 1}], symTable2];
    Print[p, " decodes to: '", q, "'"],
    q = StringJoin[q, StringTake[symTable, {p[[j]] + 1}]];
    symTable2 = 
     StringDrop[
      symTable, {StringPosition[symTable, 
         StringTake[symTable, {p[[j]] + 1}]][[1, 1]]}];
    symTable = 
     StringJoin[StringTake[symTable, {p[[j]] + 1}], symTable2]
    ]
   , {j, StringLength@string}];
  ]

Any suggestions? Clunky code, but it gets the job done. Admittedly, I have cheated to get the problem "zero-indexed". This can apparently be done with the Notation package, but I haven't gotten a chance to look at it. Now back to my thesis!

In[347]:= Timing[mtf["broood"]]

During evaluation of In[347]:= 'broood' encodes to: {1,17,15,0,0,5}

During evaluation of In[347]:= {1,17,15,0,0,5} decodes to: 'broood'

Out[347]= {0.000612, Null}

Edit: I guess it's not a symbol table array either, it's just a string of the whole alphabet. Oh well. Edit 2: Fixed the code. Good bug-catching!

mtf["zbmcfkukwmeogkhgkspansqgytihgwudpevgaelx"]

'zbmcfkukwmeogkhgkspansqgytihgwudpevgaelx' encodes to: {25,2,13,4,7,12,21,1,23,5,10,17,12,5,13,2,2,21,19,14,19,3,20,6,25,23,20,10,4,14,15,20,12,15,24,6,14,3,23,25}

{25,2,13,4,7,12,21,1,23,5,10,17,12,5,13,2,2,21,19,14,19,3,20,6,25,23,20,10,4,14,15,20,12,15,24,6,14,3,23,25} decodes to: 'zbmcfkukwmeogkhgkspansqgytihgwudpevgaelx'

I implemented Zebra puzzle, which was great fun!

I used a 'sodoku kind of way' to solve it. Each house has for each kind, all the candidates. Then, by applying the rules you filter out candidates until you're left with 1 candidates per kind per house.

Sorry I also don't have access to this book. If no-one has access to it, we have to implement ourselves... Thanks for the hint! Hopefully someone has this book.

I implemented IBAN, quite easy to implement.

I replaced the Prints, and uploaded it to the website; thanks! If you forget the commas it will see it as multiplication, multiplications of a string and an number will (generally) be displayed as "numberstring". Thanks!

You are right, the output is virtually the same. I think your version is better as it carries only one third of the commands for printing the values and hence is more elegant.

I actually tried with a similar construction before resorting to printing strings, but it somehow displayed the value before the text strings somehow. I don't remember exactly how I formulated the command though.

Edit: I think I omitted the commas within the Print command. That doesn't work out so well apparently.

Trabb Pardo-Knuth algorithm;

numbers = RandomReal[{-2, 6}, 11]

tpk[numbers_, overflowVal_] :=
 Module[{revNumbers},
  revNumbers = Reverse[numbers];
  f[x_] := Abs[x]^0.5 + 5 x^3;
  Do[If[f[i] > overflowVal,
    Print@StringJoin[{"f[", ToString[i], "]= Overflow"}],
    Print@StringJoin[{"f[", ToString[i], "]= ", ToString[f[i]]}]], {i,revNumbers}]]

tpk[numbers, 400]

Had to resort to discrete printings of strings, not sure how to do it otherwise. Feel free to make improvement suggestions. If you would please submit this one as well, Sander, I would appreciate that.

hi Sander, looking at the page of the Rosetta Code I found the stable marriage problem and I remember that Stan Wagon proposes a solution to this problem in his book "Mathematica in action " third edition, but not if we have problems with Stan to post your code

Looks great! It works pretty neatly I think. Works good, but it might be nicer if it would return a string rather than print it.

Did you add it to the Rosetta website? If you want me to do it (because you e.g. don't have an account), let me know.

I added the code for the Combinations and permutation task.

Going by this video I added one for Huffman Coding, done in my usual potentially-too-fancy-for-its-own-good style:

huffman[s_String] := huffman[Characters[s]];
huffman[l_List] := Module[{merge, structure, rules},

   (*merge front two branches. list is assumed to be sorted*)
   merge[k_] := Replace[k, {{a_, aC_}, {b_, bC_}, rest___} :> {{{a, b}, aC + bC}, rest}];

   structure = FixedPoint[
      Composition[merge, SortBy[#, Last] &],
      Tally[l]][[1, 1]];

   rules = (# -> Flatten[Position[structure, #] - 1]) & /@ DeleteDuplicates[l];

   {Flatten[l /. rules], rules}];

It seems to be correct.

An addendum to Sander's FibonacciWord code above:

step["0", {_?EvenQ}] = RotationTransform[Pi/2];
step["0", {_?OddQ}] = RotationTransform[-Pi/2];
step[___] = Identity;

steps = MapIndexed[step, Characters[FibonacciWord[10]]];
dirs = ComposeList[steps, {0, 1}];
Graphics[Line[FoldList[Plus, {0, 0}, dirs]]]

Also in a playful programming style. I'm not certain about the correctness of this one, though.

Here is a start for the Parse an IP Address task, feel free to add/modify:

testcases={"127.0.0.1","127.0.0.1:80","::1","[::1]:80","2605:2700:0:3::4713:93e3","[2605:2700:0:3::4713:93e3]:80"};
ClearAll[ParseIP]
ParseIP[str_String]:=Module[{},
 Which[
  Count[Characters[str],"."]==3,
  Print["IPV4"];
  ,
  Count[Characters[str],":"]>=2,
  Print["IPV6"];
  ,
  True,
  Print["Not sure what this is!"]
 ]
]
ParseIP/@testcases