bejeweled-chains

A code solution of mine written in MokoM (invented programming language) to a popular game problem called "Bejeweled Chains" consisting of a square/rectangular grid of different-colored jewels. The game is played by swapping the positions of jewels that are horizontally or vertically adjacent to create chains of three or more jewels of the same color.

You can refer to wikipedia for further gameplay details.

Background

We can represent a bejeweled game grid in MokoM with an array of array of numbers, with different numbers representing different colors. Consider the following board for example:

array(
  array( 3 , 3 , 3 , 4 , 1 , 2 ),
  array( 0 , 0 , 3 , 2 , 4 , 4 ),
  array( 4 , 0 , 0 , 2 , 2 , 1 ),
  array( 1 , 3 , 4 , 2 , 4 , 0 ),
  array( 0 , 1 , 0 , 2 , 0 , 1 ),
  array( 2 , 2 , 2 , 4 , 1 , 3 ),
)

As previously noted, the objective of the game is to form chains consisting of three or more jewels of the same color that are either horizontally or vertically adjacent. The example game board above contains 3 such chains, highlighted in red below:

array(
  array( 3 , 3 , 3 , 4 , 1 , 2 ),
  array( 0 , 0 , 3 , 2 , 4 , 4 ),
  array( 4 , 0 , 0 , 2 , 2 , 1 ),
  array( 1 , 3 , 4 , 2 , 4 , 0 ),
  array( 0 , 1 , 0 , 2 , 0 , 1 ),
  array( 2 , 2 , 2 , 4 , 1 , 3 ),
)

Problem - Bejeweled Chains

My task in this problem was to write a function count_chains that accepts a bejeweled game board as described above (an array of array of numbers) and then counts and returns the number of chains in the given game board. Each continuous horizontal or vertical chain counts as a single chain, regardless of length. However if a horizontal chain intersects a vertical chain, they still count as separate chains.

Some examples:

array(
  array( 0 , 3 , 3 ),
  array( 0 , 0 , 0 ),
  array( 0 , 1 , 4 )
)

The above has 2 chains (first column and second row).

array(
  array( 3 , 3 , 3 , 3 , 3 ),
  array( 0 , 0 , 1 , 2 , 1 ),
  array( 0 , 1 , 4 , 3 , 2 )
)

The above has one chain (first row)

I was to assume that I'll always be given a proper game board: ie. an array of equal-length arrays of only integer values. The length and width of the board can however be arbitrary, the number of different colors of jewels is also arbitrary.

My final solution was just to define the count_chains function. I didn't need to call it in any way. I was also permitted to define any number of helper functions to break down the task in a sensible way.

Example code the evaluator would run:

print count_chains(array(array( 0 , 3 , 3 ),array( 0 , 0 , 0 ),array( 0 , 1 , 4 )))

Example expected output:

Answer - Bejeweled Chains

To solve this type of complex problem, we would have to use 2 custom helper functions - countCompleteChains() and changeArrayOrientation(). The first helper function helps us count complete chains while the second helper function helps us swap the array structure from "m x n" array to an "n x m" array arrangment.

var count_chains = function(array) {
	var i = 0
	var numberOfChains = 0
	
	//Count Chains for original orientation
	while(i < length(array)) {
		numberOfChains = numberOfChains + countCompleteChains(get(array,i))
		i = i + 1
	}
	
	//Change Array's Orientation from Horizontal to Vertical
	array = changeArrayOrientation(array)
	
	//Count Chains for new array orientation
	i = 0
	while(i < length(array)) {
		numberOfChains = numberOfChains + countCompleteChains(get(array,i))
		i = i + 1
	}
	return numberOfChains
}

var countCompleteChains = function(array) {
	var i = 0
	var j = 0
	var currentNumber = 0
	var numberOfSameItems = 1
	var numberOfChains = 0
	while(i < length(array)) {
		currentNumber = get(array,i)
		j = i + 1
		if(j != length(array)) {
			if(currentNumber == get(array,j)) {
				numberOfSameItems = numberOfSameItems + 1
				if(numberOfSameItems == 3) {
					numberOfChains = numberOfChains + 1
				}
			} else {
                //Once the number is not the same reset numberOfSameItems
				numberOfSameItems = 1
			}
		}
		i = i + 1
	}
	return numberOfChains
}

var changeArrayOrientation = function(array) {
	var i = 0
	var k = 0
	var currentArray = array()
	var intermediateArray = array()
	var childArray = get(array, 0)
	var childArray_length = length(childArray)

        //Transfer item of each subarray into intermediate array for swapping
	while(i < length(array)) {
		currentArray = get(array, i)
		var j = 0
		while(j < length(currentArray)) {
			intermediateArray = insert(intermediateArray, k, get(currentArray, j))
			k = k + 1
			j = j + 1
		}
		i = i + 1
	}
	

        //Perform the swap 
	var l = 0
	var swappedArray = array()
	while(l < childArray_length) {
		var dynamicArray = array()
		var m = 0
		var n = l
		while(n < length(intermediateArray)) {
			dynamicArray = insert(dynamicArray, m, get(intermediateArray, n))
			n = n + childArray_length
			m = m + 1
		}
		swappedArray = insert(swappedArray, l, dynamicArray)
		l = l + 1
	}
	return swappedArray
}

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