java - StackOverflow error while using recursion -

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• bfs arithmetic operations 3 answers

i trying number (n) (m) , me algorithm makes sense yet crashes stackoverflow error. help? here's code:

private static int solve(int n, int m, int steps) {      if (n > m || n <= 0) {         //--steps;         return 0;     } else if (n == m)         return steps;      return math.min(solve(n * 2, m, steps++), solve(n - 1, m, steps++)); } 

update::: code solve problem elegantly

private static int solve(int n, int m) {      int steps = 0;     int cur = n;     arraylist<integer> arr = new arraylist<>();     //setting list of best minimum track.     arr.add(m);     (int = 0; !arr.contains(1); ++i)         arr.add((int) math.round((double) arr.get(i) / 2));     //aligning number n track , applying best track.(imagine stair of steps , have reach top of it, you'll align 1 of steps , voooom :) )     while (cur != m) {         if (arr.contains(cur))             cur *= 2;         else             cur--;          steps++;     }     return steps; } 

this algorithm traverses all possible paths in tree 1 branch represents multiplication , 1 decrement 1. problem "all", since there path this:

n = 2 m = 5 branch selection  *2 -1 -1 *2 -1 -1 *2 ... n after branch     4  3  2  4  3  2  4 ... 

which never abort.

and doubt algorithm will, though makes perfect sense , apart obvious issue stackoverflowexception solve problem in cases. guess, problem find minimum-number of multiplications , decrements required transform n m. algorithm cases return 0.

a correct solution require bfs, unlike (incorrect) dfs-implementation solution uses. there additional set visited nodes required implement correct dfs-traversal, invalid solutions have marked explicitly. bfs on other hand aswell require store depth @ value visited:

public int solve(int n , int m , int step , map<integer , integer> visitedat){     //no valid solution       if(n > m * 2 || n <= 0)         return -1;      //a solution found @ step     if(n == m)         return step;      //the node visited less steps     if(visitedat.containskey(n) && visitedat.get(n) <= step)         return -1;      //store visited value , step @ visited     visitedat.put(n , step);     //calculate number of steps required reach m if next step     //either decrement or multiply 2     int dec = solve(n - 1 , m , step + 1 , visitedat);     int mul = solve(n * 2 , m , step + 1 , visitedat);      if(dec == -1)//no possibility reach m decrementing n         return mul;     else if(mul == -1)//cant reach m multiplying n 2         return dec;     else //both lead valid solutions, find minimum number of steps         return math.min(dec , mul); } 

though solution isn't elegant implementation of bfs-algorithm. fulfills requirement of being recursive.