CONDITIONAL PROBABILITY

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1 Probability-based solution to N-Queen problem Madhusudan 1, Rachana Rangra 2 Abstract-This paper proposes the novel solution to N-Queen using CONDITIONAL PROBABILITY and BAYES THEOREM. N-Queen problem is a type of NP-hard problem, which aims at placing the queen in n*n chessboard in such a manner that no queen should kill another diagonally, horizontally and vertically. Earlier solutions have been provided using ACO (ant colony optimization), Genetic algorithm and DNA Sticker algorithm. The BAYES THEOREM updates the prior probability of theory being true to arrive at the posterior probability on the basis of given evidence, that is to find that present position is the safe position. By applying CONDITIONAL PROBABILITY the number of safe s in the next row with respect to the present position will be calculated. This algorithm will increase the reliability of the solution obtained. Index items-aco, Bayes Theorem, DNA Sticker, Genetic Algorithm, NP Hard Problem. I. INTRODUCTION N-Queen problem is a polynomial time problem defined as placing n queens on nxn chessboard, so that no queen attacks the other n-1 queens. This problem is categorized as 4 queen,8 queen and 16 queen problems. In 8 queen problem we are given with an 8x8 chessboard and the problem definition is placing all the 8 queens on the board such that no queen attacks each other [1].The solution is achieved by placing these queens on the board by different mechanism like backtracking [2],Anti Colony Optimization, DNASticker, and Genetic Algorithm etc. [3]. In Backtracking we start with placing a given queen on the board and then check the safes tates and repeat the algorithm till no safe s are left for the rest of the queens. When there are no safe s we backtrack by removing the last placed queen and again try to find the safe s by altering its position. If no safe is discovered then we move back to the queen who resulted in the unsafe for the rest of the queens. This algorithm generally works on a set of safe s and unsafe s. A safe is defined as a position on the board where a queen can be placed without any danger, and an unsafe is apposition where no queen can be placed since it is attacked by some other queen [4].Similarly the 4 queen and 16 queen problems are defined with 4x4chessboard and 16x16 chessboards respectively with the problem definition and solution definition being the same. Solution to n queen problem by ACO is such that we need to find the shortest path from source to destination with the condition that no two positions should be marked on one row. The idea is derived from the behavior of ant colony [5]. To apply ACO on n queen problem there are slight modifications. The vertices in search space are organized as rows * n column grid. Every vertex in a column is connected to all vertices in the next column through directed edges, except vertices in the nth column. An ant during a tour visits only n nodes (8 for 8-queen).The label of nodes in the tour cannot be the same. The node in a tour represents a cell of chessboard where a queen is to be placed. This restriction is added as if two nodes have same labels then 02 queens will be placed in same cell which is illegal. This essentially means that no two nodes in a tour will be in the same row of the search space [5]. We are proposing a solution to the given n queen problem using conditional probability and Bayes theorem. II. OBJECTIVES OF THE STUDY To study the existing solutions to the n queen problem and implement the probability theory to find a feasible solution that is reliable and consumes less time and space. III. PROPOSED SOLUTION In mathematical science probability is defined as the measure of occurrence of any event. The value of probability lies between 0 and 1 where 0 designates no occurrence of that event and 1 designates the total occurrence of that event. For example in a throw of die, the probability of occurrence of any of the number is 1/6. We are interested in the likelihood of an event which is defined as the favorable occurrence. Greater the value of likelihood better will be the solution [6]. The second concept used is that of conditional probability. In this case we find the probability of a given event on the basis of previous event. This is generally used when the events are dependent on the occurrence of some 426

2 other events. For example, if there is head then a die will be thrown else a coin will be tossed. So the probability of throwing a die is ½ and that of toss is also ½.The Bayes theorem defines the occurrence of an event A based on the occurrence of the events E1, E2, E3and so on [7]. P (A E1) = We proposed a solution using Conditional Probability and Bayes Theorem, for the given nxn chessboard. We find the probability of the safe s on the basis of placing a queen in the previous row/column. In the first row, since there is no danger, we can place the queen at any position. Then on the basis of the placement, we find the no of safe s and probability of safe s for the next row along with the maximum number of safe s for (n-2) queens. We discuss it with the help of 4x4. Let us assume that,,, Q4 be the positions in which the queens will be placed in future after finding the complete solution. Procedure:-Starting from the Row 1, since there is no danger and all the columns have equal probability that is 0.25, we can assume that queen is placed in any position which is Column 1, Column 2, Column 3,and Column 4. Step A: Assuming that in future the queen is placed in position (Column1 of Row 1) as shown in Fig.1.Now we calculate the number of safe s in the Row 2 corresponding to the position. S1 Fig 1. Number of safe s for position The shaded portion indicates the unsafe s for the next position with respect to position.as the Fig.1, indicates only two unshaded cells for Row 2. The number of safe s left are 2(S1 and ) Probability of each safe s = 1/4= 0.25 Since the probability of safe is same for both Column 3(S1) and Column 4() in Row 2 with respect to position. A.1 We can move further by assuming that the second queen will be placed in the position () (Column 2 of Row 2) as shown in Fig.2, which is the safe (S1) in Row 2 with respect to position.now we calculate the number of safe s in Row 3 with respect to position in Row

3 Fig 2. Number of safe s for position The shaded portion indicates the unsafe s for next position with respect to position.as the Fig.2, indicates no unshaded cells for row 3. The number of safe s left is 0. Probability of safe s =0. A.2 If we assume that the second queen will be placed in position 1(Column 4 of Row 2) as shown in Fig.3,which is the safe () in Row 2 with respect to position.calculating the number of safe s in Row 3 with respect to position 1 in Row 2. S3 1 Fig 3. Number of safe s for 1 position The shaded portion indicates the unsafe s for next position with respect to position 1. The Fig.3, indicates one unshaded cell in row 3 (S3). The Number of safe left for Third queen=1(s3). Probability of safe =1/4=0.25 A.2.1 Since we are left with only one safe (S3) for third queen, we assume that next position of queen will be (Column 2 of Row 3) as shown in Fig.4. Fig 4. Number of safe s for position 428

4 The shaded portion indicates the unsafe s for next position with respect to position. The Fig.4 indicates no unshaded cell in row 4 which indicates no safe. The number of safe s left Fourth Queen=0 Probability of safe sate=0. Step B: Assuming that in future the queen is placed in position ( Column 2 of Row 1) as shown in Fig.5. Now we calculate the number of safe s in the Row 2 corresponding to the position. S1 Fig 5. Number of safe s for position The shaded portion indicates the unsafe s for the position of next queen in Row 2 with respect to position.from the Fig.5, we can see that only one unshaded cell is left (S1). The number of safe s left for Second queen =1(S1) Probability of safe s = ¼=0.25 B.1 Since we are left with only one safe () for second queen, we assume that next safe position of queen will be (Column 4 of Row 2) as shown in Fig.6. Calculating the number of safe sates for next queen in Row 3 corresponding to position.. Fig 6. Number of safe s for position The shaded portion indicates the unsafe s for the placement of next queen in Row 3 with respect to position.the above Fig.6 shows only one unshaded cell remaining (). 429

5 The number of safe s left for Third queen =1() Probability of safe s = ¼=0.25 B.2 We move further by assuming that the third queen will be placed in the position (Column 1of Row 3) as shown in Fig.7, which is the safe () with respect to position.calculating the number of safe sates for next queen in Row 4 corresponding to position in Row 3. S3 Fig 7. Number of safe s for position The shaded portion indicates the unsafe s for the placement of next queen in Row 4 with respect to position.the above Fig.7 shows only one unshaded cell remaining (S3). The no of safe s left for Fourth queen =1(S3) Probability of safe s = ¼=0.25 B.3 Since we are left with only one safe S3 for the fourth queen, hence we assume that the last queen will be placed in position Q4 (Column 3 of Row 4) as shown in Fig.8, which is the safe (S3) with respect to position in Row 3. Place the queen Q4 in S3 (Column 3, row4). Q4 Fig 8. Position of 4 queens Step C: Assuming that in future the queen is placed in position (Column 3 of Row 1) as shown in Fig.9. Now we calculate the number of safe s in the Row 2 corresponding to the position. 430

6 S1 Fig 9. Number of safe s for position The shaded portion indicates the unsafe s for the position of next queen in Row 2 with respectto position.from the Fig.9. We can see that only one unshaded cell is left (S1). The number of safe s left for Second queen =1(S1) Probability of safe s = ¼=0.25 C.1 Since we are left with only one safe (S1) for second queen, we assume that next safe position of queen will be (Column 1 of Row 2) as shown in Fig.10. Calculating the number of safe sates for next queen in Row 3 corresponding to position Fig 10. Number of safe s for position The shaded portion indicates the unsafe s for the placement of next queen in Row 3 with respect to position.the above Fig. 10 shows only one unshaded cell remaining (). The number of safe s left for Third queen =1() Probability of safe s = ¼=

7 C.2 We move further by assuming that the third queen will be placed in the position (Column 4 of Row 3) as shown in Fig.11, which is the safe () with respect to position.calculating the number of safe sates for next queen in Row 4 corresponding to position in Row 3. S3 Fig 11. Number of safe s for position The shaded portion indicates the unsafe s for the placement of next queen in Row 4 with respect to position.the above Fig.11 shows only one unshaded cell remaining (S3). The number of safe s left for Fourth queen =1(S3) Probability of safe s = ¼=0.25 C.3 Since we are left with only one safe S3 for fourth queen, hence we assume that the last queen will be placed in position Q4 (Column 2 of Row 4) as shown in Fig.12, which is the safe (S3) with respect to position in Row 3 Place the queen Q4 in S3(Column 3, row4). Q4 Fig 12. Position of 4 queen Step D: Assuming that in future the queen is placed in position (Column 4 of Row 1) as shown in Fig.13.Now we calculate the number of safe s in the Row 2 corresponding to the position. 432

8 S1 Fig 13. Number of safe s for position The shaded portion indicates the unsafe s for the placement of next queen with respect to position.fig.13 indicates only two unshaded cells remaining. The number of safe s left are 2(S1 and ) Probability of each safe =1/4=0.25 Since the probability of safe is same for both Column 1(S1) and Column 2() in Row 2 with respect to position D.1. We can move further by assuming that the second queen will be placed in position (Column 1 of Row 2) as shown in Fig.14, which is the safe (S1) in Row 2 with respect to position.now Calculate the number of safe s in Row 3 with respect to position in Row 2. S3 Fig 14. Number of safe s for position The shaded portion indicates the unsafe s for next position with respect to position.as we can see from the Fig.14 only one unshaded cell is left in row 3. The number of safe s left for Third Queen is 1(S3) Probability of safe s =1/4=0.25. D.1.1. Since we are left with only one safe (S3) for third queen, we assume that next position of queen will be (Column 3 for Row 3) as shown in Fig

9 Fig 15. Number of safe s for position The shaded portion indicates the unsafe s for next position with respect to position.as we can see from the Fig.15 there is no unshaded portion left in row 4 which indicates no safe. S1 1 The number of safe s left Fourth Queen=0 Probability of safe sate=0. Fig 16. Number of safe s for 1 position D.2. If we assume that the second queen will be placed in position 1(Column 2 of Row 2) as shown in Fig.16, which is the safe () in Row 2 with respect to position.calculating the number of safe s in Row 3 with respect to position 1 in Row 2 The shaded portion indicates the unsafe s for next position with respect to position 1.As we can see from the Fig.16 there is no unshaded cellleft in row 3. The Number of safe left for Third queen=0 Probability of safe =0 I. Probability table for Row 1 Column 1 IV.RESULTS Row Column No of safe Probability Of safe 434

10 II. Probability table for Row 1 Column 1 ` Row Column III. Probability table for Row 1 Column No of safe Probability Of safe Row Column No of safe Probability Of safe IV. Probability table for Row 1 Column 3 Row Column No of safe Probability Of safe V. Probability table for Row 1 Column 4 Row Column No of safe Probability Of safe VI. Probability table for Row 1 Column 4 Row Column No of safe Probability Of safe VII. Total probability of safe s for each column in Row 1 Table I Table II Table III Table IV Table V Table VI

11 Feasible solution:-we will choose the set of columns which maximizes the total probability of safe s. From the above table we can find that the maximum probability of safe s is given by Column 2 and Column 3 that is The solution set is given by following combination of columns i)2, 4, 1, 3 ii)3, 1, 4, 2 V. CONCLUSION & FUTURE SCOPE The solution we proposed provides a reliable and time efficient results to the 4 queen problem, and the solution can be spanned for any values of n except n=2 and n=3.calculating the total probability of all the safe s for the present, and then selecting the solution set with the maximum total probability, provides the set of safe s for the N-Queen problem. Contrary to placing the queen and then checking for the safe, we first search for all the safe s with their probability and then chooses the set of safe s which has highest probability. REFERENCES [1] Prabhakar. Gupta, Vineet.Agarwal and Manish.Varshney, Design and Analysis of Algorithms, Second Edition, PHI Learning Private Ltd., [2] Elaine. Rich, Kevin. Knight and Shivashankar.Nair, Artificial Intelligence, McGraw Hill Education, Third Edition, [3] S.Pothumani, Solving N Queen Problem Using Various Algorithms-A Survey, IJARCSSE, Vol. 3, Issue 2, Feb [4] Prabhakar. Gupta, Vineet. Agarwal and Manish.Varshney, Design and Analysis of Algorithms, Second Edition, PHI Learning Private Ltd., [5] Salabat.Khan, Mohsin.Bilal, M.Sharif, Malik. Sajid and Rauf.Baig, Solution of n-queen Problem Using ACO, In Proceedings of Multitopic Conference, INMIC 2009, IEEE 13th International Source: IEEE Xplore. [6] Dan W. Patterson, Introduction to Artificial Intelligence and Expert Systems, PHI Learning Private Ltd., Reprint [7] Dan W. Patterson, Introduction to Artificial Intelligence and Expert Systems, PHI Learning Private Ltd., Reprint

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