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

I made a not so fast but working version of the 'Narcissistic decimal number' problem. That is: find the first 25 non-negative integers $n$, with number of digits $m$, where the sum of the digits to the power $m$ is the number itself; for example: $153 = 1 + 125 + 27 = 1^3 + 5^3 + 3^3$.

powers = {};
found = 0; number = 0; totDigits = 0;
addpower[p_] := AppendTo[powers, Power[Range[0, 9], p]];
check[number_] := With[{digits = IntegerDigits[number]},
    If[Length[digits] > totDigits, addpower[Length[digits]]; totDigits++];
    If[Total[Map[powers[[Length[digits]]][[# + 1]] &, digits]] == number, number]];
While[found < 25, If[IntegerQ[check[number]], found++; Print[number]]; number++]

This works and only calculates the powers once, but still takes 250 s. I tried another approach but have some trouble finishing it. The idea is that instead of looping over the natural numbers, this loops over combinations of integers.

sumpow[digits_, len_] := Total[Map[Power[#, len] &, digits]];
checkN[digits_] := 
    Block[{len = Length[digits], 
    numbers = Map[FromDigits[#] &, Permutations[digits]]},
    If[MemberQ[numbers, number = sumpow[digits, len]], Print[number]]];
checklist = Subsets[{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}][[2;;]];
ParallelMap[checkN[#] &, checklist]

What this needs is (i) checklist also consisting of subsets with double numbers (9474 for examples is a narcissistic number) and (ii) a while loop that 'constructs' the subsets and stops after finding the 25th narcissistic number.

On the way I found that http://rosettacode.org/wiki/Combinations#Mathematica contains:

combinations[n_Integer, m_Integer]/;m>= 0:=Union[Sort /@ Permutations[Range[0, n - 1], {m}]]

While this one is shorter and seems to be an order of magnitude faster:

combinations[n_Integer, m_Integer] := Subsets[Range[0, n - 1], {m}]

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.

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

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:

I like the very short code! Thanks! great job.

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

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.

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 some of Mathematica's power.

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

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:

I added this code

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).

That looks a lot better! Sorry forgot about that; filenamejoin can not be used to add an extension; that was my fault.

I modified it slightly to make it more readable and added it too! Thanks!

Mathematica is quite unusual compared to most programming languages, so it's difficult if you don't know what kinds of functions to look for in the first place. A very vague rule of thumb I would recommend is looking for functions that work in "broad strokes." And generally you should try to write things in terms of functions, because of their composability and how they are used in those "broad strokes" functions. The quintessential example here is For vs Map:

newList = Range[Length[list]];
For[i = 1, i <= Length[list], i++,
    newList[[i]] = someF[list[[i]]]]

vs

Map[someF, list]

The real key though is that you have to have the impulse to look for these things. You ain't gonna find them if you aren't looking for them!

BTW in the decoder part for the move-to-front algorithm, remember that you get the index stream through the Fold function. Here's a more explicit version to make it easier to read:

f2[{output_, symList_}, nextIndex_] := Module[{char},
   char = symList[[nextIndex + 1]];
   {output~Append~char, Prepend[Delete[symList, nextIndex + 1], char]}];

Also:

{output_, symList_} ~f2~ nextIndex_ := Module[{char}...

I wouldn't recommend this infix form for a Rosetta submission, but it could help when thinking of how to write a function for Fold. Also:

Fold["f", initial, Range[4]] // TreeForm

Note that in this tree, the left branches/subtrees are where the 'state' is coming from.

Here is my suggestion for a complete algorithm, based on your encoding algorithm.

mtf[word_] :=
 Module[{f, f2},
  f[{output_, symList_}, next_] := 
   Module[{index}, index = Position[symList, next][[1, 1]] - 1;
    {output~Append~index, 
     Prepend[Delete[symList, index + 1], next]}];
  p = Fold[f, {{}, CharacterRange["a", "z"]}, 
     Characters[ToString[word]]][[1]];
  Print["'" word, "' encodes to: ", p];
  f2[{output_, symList_}, next_] := 
   Module[{index}, index = symList[[next + 1]];
    {output~Append~index, 
     Prepend[DeleteCases[symList, ToString[index]], index]}];
  q = Fold[f2, {{}, CharacterRange["a", "z"]}, p][[1]];
  Print[p, " decodes to: '", StringJoin@q, "'"]
  ]

Interestingly, I find the "old" algorithm to run a bit faster, tested with some random long string, with 1000 repetitions using AbsoluteTiming. Undoubtedly, though, this algorithm is much more elegant.

That would be excellent!

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'

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]]

Oh, that is very nice! Thanks for sharing, I need to keep my eyes out for more clever solutions instead of hard-coding some massive piece of code.

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.

I implemented Vampire number.

I implemented Rep-string, which was quite easy in Mathematica with smart usage of the pattern-matching abilities of Mathematica.

Hi Marcus, looks great!

I think:

Print@StringJoin[{"f[", ToString[i], "]= Overflow"}]

will be (nearly) identical to:

Print["f[",i, "]= Overflow"];

It will output it using row, and you don't have to think about ToString et cetera. Of course it is not the same, but as you 'print' it anyhow, it doesn't really matter. I will add it later.

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!

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.

Here is my very crude and probably extremely inefficient suggestion for a word-wrap.

string = "In olden times when wishing still helped one, there lived a \
king whose daughters were all beautiful, but the youngest was so \
beautiful that the sun itself, which has seen so much, was astonished \
whenever it shone in her face.  Close by the king's castle lay a \
great dark forest, and under an old lime-tree in the forest was a \
well, and when the day was very warm, the king's child went out into \
the forest and sat down by the side of the cool fountain, and when \
she was bored she took a golden ball, and threw it up on high and \
caught it, and this ball was her favorite plaything.";

wordWrap[textWidth_, spaceWidth_, string_] :=

  Module[{start, spaceLeft, masterString},
   spaceLeft = textWidth;
   start = 1;
   masterString = {};
   Do[
    If[i + 1 > Length@StringSplit@string,
     p = StringSplit[string][[start ;; i]]; 
     AppendTo[
      masterString, {StringJoin @@ 
        Riffle[p, StringJoin@ConstantArray[" ", spaceWidth]]}],
     If[StringLength[StringSplit@string][[i + 1]] + spaceWidth > 
       spaceLeft,
      spaceLeft = textWidth - StringLength[StringSplit@string][[i]]; 
      start = i; 
      AppendTo[
       masterString, {StringJoin @@ 
         Riffle[p, StringJoin@ConstantArray[" ", spaceWidth]]}],
      spaceLeft -= StringLength[StringSplit@string][[i]]; 
      spaceLeft -= spaceWidth; 
      p = StringSplit[string][[start ;; i]]]], {i, 1, 
     Length@StringSplit@string}];
   StringJoin @@ Riffle[masterString, "\n"]
   ];

wordWrap[72, 1, string]

I've hidden the output, looks like crap in the coding preview.

Edit: Added actual dependence on the space width in the code. Edit 2: Changed output style from n discrete lines to one single string output. Edit 3: Removed a lazy "Break[]" and "Clear[p]" left behind from building the code.

Please submit it in my place, I do not have an account and it seems you are already familiar with the submission process with regards to formatting and whatnot. I have now updated the code with your suggestion of outputting a single string, thank you for your input!

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