Question

I’ve got an assignment I can’t figure out, any pointers will be much appreciated, it goes like so:

There’s a series of light bulbs represented as an array of true/false, there’s a switch for every light bulb, by clicking it for any light bulb, you toggle it as well as 2 adjacent ones (1 from left & another 1 from right; if clicked switch’s bulb on edge — only 1 adjacent toggled of course).

What is needed to accomplish is a method that accepts an array of a series of turned on/off light bulbs and another one representing another state of supposedly the 1st array after some switches have been clicked..!
So recursion must be used to find out whether there’s a combination of switch clicks that will transform array 1 to array 2.

Here’s the signature of the method:

public static boolean disco(boolean[] init, boolean[] target)

Will return true if array init can be transformed to target, false otherwise.
Method must be static and not use loops & any other static and global variables, only local.

Example:

boolean[] init = {true, false, true, false, true, false};
boolean[] target = {false, true, false, true, false, true};

For above 2 arrays, disco(init, target) will return true because toggling the 1st and 4th bulbs would yield the target state (remember adjacent bulbs being toggled as well).

Answer 1

new version

public static boolean disco(boolean[] init, boolean[] target)  
{  
  return recurse(init,boolean,0);  
}  

public static boolean recurse(boolean[] init, boolean[] target, int min)  
{  
  if (min == init.length)  
    if (init == target)  
      return true;  
    else  
      return false;  
  boolean[] temp = "init with a change at min";  
  boolean a = recurse(init, target, min+1);  
  boolean b =   recurse(temp, target, min+1);  
  return a||b;
}  

new new version

I’ve broken it down to three cases:

Case 1: length %3 = 0
By changing the first bulb and the second bulb, you can affect a change on only the 3rd.
Then a change to 4 and 5 will make the 6th the only one changed. We see that we can change every bulb with index divisible by 3.
Working backwards (starting at the end) we can do the same but this time it shows us we can change bulbs with indices (i+1) divisible by 3.
Combining the two, we see that if we want to change an index 0,1 mod 3 we can. But then to change a 2, we simply have to change a neighboring 0,1 pair and then do a change on the middle one. So in all cases these are solvable.

Case 2: length %3 = 1
Just like the first case, but we can affect single changes at 0,2 mod 3, and thus squeeze the 1 mod 3 cases.

Case 3: length %3 = 2
Working forward and backwards only yields cases 0 mod 3. The only remaining moves we have make two changes to the bulbs (since we can ignore any changes to positions 0). Changing a 1 or 2 will reverse the position surrounded by two 0’s whereas changing a 0 will swap the parity in adjacent blocks of 1,2 which have a 0 between them (it’s easier if you draw it). But what I’ve shown so far is that the 0’s can all be corrected and in any adjacent 1,2 you can fix both if they are wrong without changing anything else. If one is wrong then you propagate a change to an adjacent pair of 1,2. This change can be moved if necessary. What this means is that any even number of errors in 1,2 positions can be fixed, but an odd number cannot. All errors at position 0’s can be fixed.

public static boolean disco(boolean[] init, boolean[] target)  
{  
  if (init.length%3 == 0 || init.length%3 == 1)
    return true;
  return recurse(init,target,0, true);
}

public static boolean recurse(boolean[] init, boolean[] target, int index, boolean even)
{
  if (index = init.length)
    return even;
  if (init[index] != target[index] && index%3 != 0)
    even = !even;
  return recurse(init, target, index+1, even);
}