Box Stacking Problem | DP-22 - GeeksforGeeks

30 Jun, 2021
You are given a set of n types of rectangular 3-D boxes, where the i^th box has height h(i), width w(i) and depth d(i) (all real numbers). You want to create a stack of boxes which is as tall as possible, but you can only stack a box on top of another box if the dimensions of the 2-D base of the lower box are each strictly larger than those of the 2-D base of the higher box. Of course, you can rotate a box so that any side functions as its base. It is also allowable to use multiple instances of the same type of box. Source: http://people.csail.mit.edu/bdean/6.046/dp/. The link also has a video for an explanation of the solution. 
 
The Box Stacking problem is a variation of LIS problem. We need to build a maximum height stack. Following are the key points to note in the problem statement: 1) A box can be placed on top of another box only if both width and depth of the upper placed box are smaller than width and depth of the lower box respectively. 2) We can rotate boxes such that width is smaller than depth. For example, if there is a box with dimensions {1x2x3} where 1 is height, 2×3 is base, then there can be three possibilities, {1x2x3}, {2x1x3} and {3x1x2} 3) We can use multiple instances of boxes. What it means is, we can have two different rotations of a box as part of our maximum height stack.Following is the solution based on DP solution of LIS problem.
Method 1 : dynamic programming using tabulation
1) Generate all 3 rotations of all boxes. The size of rotation array becomes 3 times the size of the original array. For simplicity, we consider width as always smaller than or equal to depth. 2) Sort the above generated 3n boxes in decreasing order of base area.3) After sorting the boxes, the problem is same as LIS with following optimal substructure property. MSH(i) = Maximum possible Stack Height with box i at top of stack MSH(i) = { Max ( MSH(j) ) + height(i) } where j < i and width(j) > width(i) and depth(j) > depth(i). If there is no such j then MSH(i) = height(i)4) To get overall maximum height, we return max(MSH(i)) where 0 < i < nFollowing is the implementation of the above solution.  
C++
Java
Python3
/* Dynamic Programming implementation of Box Stacking problem */#include<stdio.h>#include<stdlib.h> /* Representation of a box */struct Box{  // h --> height, w --> width, d --> depth  int h, w, d;  // for simplicity of solution, always keep w <= d}; // A utility function to get minimum of two integersint min (int x, int y){ return (x < y)? x : y; } // A utility function to get maximum of two integersint max (int x, int y){ return (x > y)? x : y; } /* Following function is needed for library function qsort(). We   use qsort() to sort boxes in decreasing order of base area.   Refer following link for help of qsort() and compare()   http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */int compare (const void *a, const void * b){    return ( (*(Box *)b).d * (*(Box *)b).w ) -           ( (*(Box *)a).d * (*(Box *)a).w );} /* Returns the height of the tallest stack that can be   formed with give type of boxes */int maxStackHeight( Box arr[], int n ){   /* Create an array of all rotations of given boxes      For example, for a box {1, 2, 3}, we consider three      instances{{1, 2, 3}, {2, 1, 3}, {3, 1, 2}} */   Box rot[3*n];   int index = 0;   for (int i = 0; i < n; i++)   {      // Copy the original box      rot[index].h = arr[i].h;      rot[index].d = max(arr[i].d, arr[i].w);      rot[index].w = min(arr[i].d, arr[i].w);      index++;       // First rotation of box      rot[index].h = arr[i].w;      rot[index].d = max(arr[i].h, arr[i].d);      rot[index].w = min(arr[i].h, arr[i].d);      index++;       // Second rotation of box      rot[index].h = arr[i].d;      rot[index].d = max(arr[i].h, arr[i].w);      rot[index].w = min(arr[i].h, arr[i].w);      index++;   }    // Now the number of boxes is 3n   n = 3*n;    /* Sort the array 'rot[]' in non-increasing order      of base area */   qsort (rot, n, sizeof(rot[0]), compare);    // Uncomment following two lines to print all rotations   // for (int i = 0; i < n; i++ )   //    printf("%d x %d x %d\n", rot[i].h, rot[i].w, rot[i].d);    /* Initialize msh values for all indexes      msh[i] --> Maximum possible Stack Height with box i on top */   int msh[n];   for (int i = 0; i < n; i++ )      msh[i] = rot[i].h;    /* Compute optimized msh values in bottom up manner */   for (int i = 1; i < n; i++ )      for (int j = 0; j < i; j++ )         if ( rot[i].w < rot[j].w &&              rot[i].d < rot[j].d &&              msh[i] < msh[j] + rot[i].h            )         {              msh[i] = msh[j] + rot[i].h;         }     /* Pick maximum of all msh values */   int max = -1;   for ( int i = 0; i < n; i++ )      if ( max < msh[i] )         max = msh[i];    return max;} /* Driver program to test above function */int main(){  Box arr[] = { {4, 6, 7}, {1, 2, 3}, {4, 5, 6}, {10, 12, 32} };  int n = sizeof(arr)/sizeof(arr[0]);   printf("The maximum possible height of stack is %d\n",         maxStackHeight (arr, n) );   return 0;}
/* Dynamic Programming implementationof Box Stacking problem in Java*/import java.util.*; public class GFG {         /* Representation of a box */    static class Box implements Comparable<Box>{             // h --> height, w --> width,        // d --> depth        int h, w, d, area;                 // for simplicity of solution,        // always keep w <= d         /*Constructor to initialise object*/        public Box(int h, int w, int d) {            this.h = h;            this.w = w;            this.d = d;        }                 /*To sort the box array on the basis        of area in decreasing order of area */        @Override        public int compareTo(Box o) {            return o.area-this.area;        }    }     /* Returns the height of the tallest    stack that can be formed with give    type of boxes */    static int maxStackHeight( Box arr[], int n){                 Box[] rot = new Box[n*3];                 /* New Array of boxes is created -        considering all 3 possible rotations,        with width always greater than equal        to width */        for(int i = 0;i < n;i++){            Box box = arr[i];                         /* Original Box*/            rot[3*i] = new Box(box.h, Math.max(box.w,box.d),                                    Math.min(box.w,box.d));                         /* First rotation of box*/            rot[3*i + 1] = new Box(box.w, Math.max(box.h,box.d),                                       Math.min(box.h,box.d));                         /* Second rotation of box*/            rot[3*i + 2] = new Box(box.d, Math.max(box.w,box.h),                                       Math.min(box.w,box.h));        }                 /* Calculating base area of        each of the boxes.*/        for(int i = 0; i < rot.length; i++)            rot[i].area = rot[i].w * rot[i].d;                 /* Sorting the Boxes on the bases        of Area in non Increasing order.*/        Arrays.sort(rot);                 int count = 3 * n;                 /* Initialize msh values for all        indexes        msh[i] --> Maximum possible Stack Height                   with box i on top */        int[]msh = new int[count];        for (int i = 0; i < count; i++ )            msh[i] = rot[i].h;                 /* Computing optimized msh[]        values in bottom up manner */        for(int i = 0; i < count; i++){            msh[i] = 0;            Box box = rot[i];            int val = 0;                         for(int j = 0; j < i; j++){                Box prevBox = rot[j];                if(box.w < prevBox.w && box.d < prevBox.d){                    val = Math.max(val, msh[j]);                }            }            msh[i] = val + box.h;        }                 int max = -1;                 /* Pick maximum of all msh values */        for(int i = 0; i < count; i++){            max = Math.max(max, msh[i]);        }                 return max;    }         /* Driver program to test above function */    public static void main(String[] args) {                 Box[] arr = new Box[4];        arr[0] = new Box(4, 6, 7);        arr[1] = new Box(1, 2, 3);        arr[2] = new Box(4, 5, 6);        arr[3] = new Box(10, 12, 32);                 System.out.println("The maximum possible "+                           "height of stack is " +                           maxStackHeight(arr,4));    }} // This code is contributed by Divyam
# Dynamic Programming implementation# of Box Stacking problemclass Box:         # Representation of a box    def __init__(self, h, w, d):        self.h = h        self.w = w        self.d = d     def __lt__(self, other):        return self.d * self.w < other.d * other.w def maxStackHeight(arr, n):     # Create an array of all rotations of    # given boxes. For example, for a box {1, 2, 3},    # we consider three instances{{1, 2, 3},    # {2, 1, 3}, {3, 1, 2}}    rot = [Box(0, 0, 0) for _ in range(3 * n)]    index = 0     for i in range(n):         # Copy the original box        rot[index].h = arr[i].h        rot[index].d = max(arr[i].d, arr[i].w)        rot[index].w = min(arr[i].d, arr[i].w)        index += 1         # First rotation of the box        rot[index].h = arr[i].w        rot[index].d = max(arr[i].h, arr[i].d)        rot[index].w = min(arr[i].h, arr[i].d)        index += 1         # Second rotation of the box        rot[index].h = arr[i].d        rot[index].d = max(arr[i].h, arr[i].w)        rot[index].w = min(arr[i].h, arr[i].w)        index += 1     # Now the number of boxes is 3n    n *= 3     # Sort the array 'rot[]' in non-increasing    # order of base area    rot.sort(reverse = True)     # Uncomment following two lines to print    # all rotations    # for i in range(n):    #     print(rot[i].h, 'x', rot[i].w, 'x', rot[i].d)     # Initialize msh values for all indexes    # msh[i] --> Maximum possible Stack Height    # with box i on top    msh = [0] * n     for i in range(n):        msh[i] = rot[i].h     # Compute optimized msh values    # in bottom up manner    for i in range(1, n):        for j in range(0, i):            if (rot[i].w < rot[j].w and                rot[i].d < rot[j].d):                if msh[i] < msh[j] + rot[i].h:                    msh[i] = msh[j] + rot[i].h     maxm = -1    for i in range(n):        maxm = max(maxm, msh[i])     return maxm # Driver Codeif __name__ == "__main__":    arr = [Box(4, 6, 7), Box(1, 2, 3),           Box(4, 5, 6), Box(10, 12, 32)]    n = len(arr)    print("The maximum possible height of stack is",           maxStackHeight(arr, n)) # This code is contributed by vibhu4agarwal
The maximum possible height of stack is 60
In the above program, given input boxes are {4, 6, 7}, {1, 2, 3}, {4, 5, 6}, {10, 12, 32}. Following are all rotations of the boxes in decreasing order of base area. 
   10 x 12 x 32
   12 x 10 x 32
   32 x 10 x 12
   4 x 6 x 7
   4 x 5 x 6
   6 x 4 x 7
   5 x 4 x 6
   7 x 4 x 6
   6 x 4 x 5
   1 x 2 x 3
   2 x 1 x 3
   3 x 1 x 2
The height 60 is obtained by boxes { {3, 1, 2}, {1, 2, 3}, {6, 4, 5}, {4, 5, 6}, {4, 6, 7}, {32, 10, 12}, {10, 12, 32}}Time Complexity: O(n^2) Auxiliary Space: O(n)
Method 2 : dynamic programming using memoization (top-down)
C++
/* Dynamic Programming top-down implementation of Box * Stacking problem */#include <bits/stdc++.h>using namespace std; /* Representation of a box */class Box {public:    int length;    int width;    int height;}; // dp arrayint dp[303]; /*    boxes -> vector of Box    bottom_box_index -> index of the bottom box    index -> index of current box*//* NOTE: we can use only one variable in place of bottom_box_index and index     but it has been avoided to make it simple */int findMaxHeight(vector<Box>& boxes, int bottom_box_index, int index){     // base case    if (index < 0)        return 0;     if (dp[index] != -1)        return dp[index];     int maximumHeight = 0;     // recurse    for (int i = index; i >= 0; i--) {         // if there is no bottom box        if (bottom_box_index == -1             // or if length & width of new box is < that of            // bottom box            || (boxes[i].length                    < boxes[bottom_box_index].length                && boxes[i].width                       < boxes[bottom_box_index].width))             maximumHeight                = max(maximumHeight,                      findMaxHeight(boxes, i, i - 1)                          + boxes[i].height);    }     return dp[index] = maximumHeight;} /* wrapper function for recursive calls whichReturns the height of the tallest stack that can beformed with give type of boxes */int maxStackHeight(int height[], int width[], int length[],                   int types){    // creating a vector of type Box class    vector<Box> boxes;     // Initialize dp array with -1    memset(dp, -1, sizeof(dp));     Box box;     /* Create an array of all rotations of given boxes    For example, for a box {1, 2, 3}, we consider three    instances{{1, 2, 3}, {2, 1, 3}, {3, 1, 2}} */    for (int i = 0; i < types; i++) {         // copy original box        box.height = height[i];        box.length = max(length[i], width[i]);        box.width = min(length[i], width[i]);         boxes.push_back(box);         // First rotation of box        box.height = width[i];        box.length = max(length[i], height[i]);        box.width = min(length[i], height[i]);         boxes.push_back(box);         // Second rotation of box        box.height = length[i];        box.length = max(width[i], height[i]);        box.width = min(width[i], height[i]);         boxes.push_back(box);    }     // sort by area in ascending order .. because we will be dealing with this vector in reverse    sort(boxes.begin(), boxes.end(), [](Box b1, Box b2) {        // if area of box1 < area of box2        return (b1.length * b1.width)               < (b2.length * b2.width);    });      // Uncomment following two lines to print all rotations    //for (int i = boxes.size() - 1; i >= 0; i-- )   //   printf("%d x %d x %d\n", boxes[i].length, boxes[i].width, boxes[i].height);     return findMaxHeight(boxes, -1, boxes.size() - 1);} int main(){     // where length, width and height of a particular box    // are at ith index of the following arrays    int length[] = { 4, 1, 4, 10 };    int width[] = { 6, 2, 5, 12 };    int height[] = { 7, 3, 6, 32 };     int n = sizeof(length) / sizeof(length[0]);     printf("The maximum possible height of stack is %d\n",           maxStackHeight(height, length, width, n));     return 0;}
The maximum possible height of stack is 60
In the above program,  for boxes of dimensions of {4, 6, 7}, {1, 2, 3}, {4, 5, 6}, {10, 12, 32} on giving the input as {4, 1, 4, 10} for length, {6, 2, 5, 12} for width and {7, 3, 6, 32} for height. Following rotations are possible for the boxes in decreasing order of base area. 
32 x 12 x 10 <-
32 x 10 x 12
12 x 10 x 32 <-
7 x 6 x 4      <-
6 x 5 x 4      <-
7 x 4 x 6
6 x 4 x 5
6 x 4 x 7
5 x 4 x 6    <-
3 x 2 x 1    <-
3 x 1 x 2
2 x 1 x 3    <-
The maximum possible height of stack is 60
The height 60 is obtained by boxes { {2, 1, 3}, {3, 2, 1}, {5, 4, 6}, {6, 5, 4}, {7, 6, 4}, {12, 10, 32}, {32, 12, 10}}
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