algorithm - 2048 game: how many moves did I do? -


2048 used quite popular little while ago. played , lot of people posted nice screenshots accomplishments(myself among them). @ point began wonder if possible tell how long did play score. benchmarked , turns out that(at least on android application have) no more 1 move can made in 1 second. if play long enough(and fast enough) number of moves you've made quite approximation number of seconds you've played. question is: possible having screenshot of 2048 game compute how many moves made.

here example screenshot(actually best effort on game far):

from screenshot can see field layout @ current moment , number of points player has earned. so: information enough compute how many moves i've made , if so, algorithm that?

note: remind points scored when 2 tiles "combine" , number of points scored value of new tile(i.e. sum of values of tiles being combined).

the short answer is possible compute number of moves using information. explain algorithm , try post answer in steps. each step observation targeted @ helping solve problem. encourage reader try , solve problem alone after each tip.

  1. observation number one: after each move 1 tile appears. tile either 4 or 2. need count number of tiles appeared. @ least on version of game played game started 2 tiles 2 on them placed @ random.

  2. we don't care actual layout of field. care numbers present on it. become more obvious when explain algorithm.

  3. seeing values in cells on field can compute score if 2 had appeared after each move. call value twos_score.

  4. the number of fours have appeared equal difference of twos_score , actual_score divided 4. true because forming 4 2 2-s have scored 4 points, while if 4 appears straight away score 0. call number of fours fours.

  5. we can compute number of twos needed form numbers on field. after need subtract 2 * fours value single 4 replaces need of 2 2s. call twos.

using observations able solve problem. explain in more details how perform separate steps.

how compute score if 2 appeared?

i prove form number 2n, player score 2n*(n - 1) points(using induction).

  • the statements obvious 2 directly appears , therefor no points scored it.
  • let's assume fixed k number 2k user score 2k*(k - 1)
  • for k + 1: 2k + 1 can formed combining 2 numbers of value 2k. user score 2k*(k - 1) + 2k*(k - 1) + 2k+1(the score 2 numbers being combined plus score new number).
    equals: 2k + 1*(k - 1) + 2k+1= 2k+1 * (k - 1 + 1) = 2k+1 * k. completes induction.

therefor compute score if twos appeared need iterate on numbers on board , accumulate score them using formula above.

how compute number of twos needed form numbers on field?

it easier notice number of twos needed form 2n 2n - 1. strict proof can again done using induction, leave reader.

the code

i provide code solving problem in c++. not use language specific(appart vector dynamically expanding array) should easy port many other languages.

/**  * @param - vector containing values in field.   *   value of 0 means "empty cell".  * @param score - score player has.  * @return pair first number number of twos appeared  *    , second number number of fours appeared.  */ pair<int,int> solve(const vector<vector<int> >& a, int score) {   vector<int> counts(20, 0);   (int = 0; < (int)a.size(); ++i) {     (int j = 0; j < (int)a[0].size(); ++j) {       if (a[i][j] == 0) {         continue;       }       int num;       (int l = 1; l < 20; ++l) {         if (a[i][j] == 1 << l) {           num = l;           break;         }       }       counts[num]++;     }   }   // score if twos appeared every time   int twos_score = 0;   (int = 1; < 20; ++i) {     twos_score += counts[i] * (1 << i) * (i - 1);   }    // each 4 appears instead of 2 overall score decreases 4   int fours = (twos_score - score) / 4;    // how many twos needed numbers on field(ignoring score)   int twos = 0;   (int = 1; < 20; ++i) {     twos += counts[i] * (1 << (i - 1));   }   // each 4 replaces 2 2-s   twos -= fours * 2;    return make_pair(twos, fours); }

now answer how many moves we've made should add 2 values of pair returned function , subtract 2 because 2 tiles 2 appear straight away.


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