THE NEW VARIATION ON TOWERS OF HANOI 汉诺塔新变形的研究
|
|
- Prosper Day
- 7 years ago
- Views:
Transcription
1 THE NEW VARIATION ON TOWERS OF HANOI 汉诺塔新变形的研究 Team members: Hong Xuechen, Wang Zhiqi Tutor: Wang Kai School: Jiangsu Tianyi High School, Jiangsu, China
2 THE NEW VARIATION ON TOWERS OF HANOI 汉诺塔新变形的研究 Abstract French mathematician Edouard Lucas introduced Towers of Hanoi puzzle in Nine years later, W.R.Ball used a recursive algorithm to solve it perfectly. It has been more than one hundred years since the classical Towers of Hanoi was proposed, but modern mathematicians still put great emphasis on this problem and have suggested different variations on the classical Towers of Hanoi. We have seen a lot of variations on Towers of Hanoi. These different variations sparkle our interest and urge us to propose another new variation on the classical Towers of Hanoi puzzle. Many of these variations reduce the strict rules of the classical puzzle to yield more possible moving methods. Thus, we become more curious about this topic and have also been thinking to reduce the restrictions on the classical Towers of Hanoi. We introduce a new variation: changing the maximum number of disks that can be placed on one disk in this paper. After putting relentless effort into the research on our new variation, we can find out a method of moving disks that requires fewer steps and makes the recreational mathematic puzzle more interesting. By using mathematical induction, we prove the recursion formulas and the formulas for general terms of our new variation as following (suppose that f(n) represents the minimum steps of moving disks to complete this variation on Towers of Hanoi which has n disks): n = 1, f(1) = 1 n is a positive even integer recursion formula: f(n + ) = f(n) + formula for general term: f(n) = n+ n is a positive odd integer except 1 recursion formula: f(n + ) = f(n) + 1 formula for general term: f(n) = n+1 + n 3 1 Furthermore, we offer general optimal patterns of moving disks and a new algorithm to obtain the formulas for general terms. Keywords: Towers of Hanoi; variation; recursion formulas; formulas for general terms; optimality; mathematical induction; patterns of moving disks
3 1 Introduction 1.1 The classical puzzle and different variations Towers of Hanoi puzzle originally derives from a mysterious legend in India. This legend delineates an Indian temple in Kashi Vishwanath 1 which contained a hall with three time-worn posts surrounded by 64 golden disks. Towers of Hanoi, a recreational math puzzle, was proposed by the French mathematician Edouard Lucas. The initial configuration consists of three posts A, B, C and n disks of different sizes which are placed on post A in increasing order of size from top to bottom (i.e. the smallest on the top, the biggest on the bottom). The challenge is to move all the disks from post A to post C with the help of intermediate post B. At the same time the following four rules must be obeyed: 1) Only one disk can be moved at a time. ) Moving one disk from a post to another one is defined as one step. 3) A bigger disk can never be placed on the top of a smaller one. 4) Final configuration: n disks should be placed in the same sequence as that in the initial configuration. In 189, W.R.Ball utilized an optimal recursive algorithm and concluded that the minimum number of steps is n 1 [1]. Many variations of this puzzle have been proposed, in which the set of allowable moves has been extended or restricted. For example, Jeremy Barbay changed the insertion and removal point in each tower to invent Bouncing Tower []. Jonathan Chappelon, Urban Larsson and Akihiro Matsuura studied extension of the one-player setting to two players of Hanoi Towers, invoking classical winning conditions in combinatorial game theory [3].The Apprentices' Towers of Hanoi proposed by Cory B.H.Ball and Robert A.Beeler allows placing a larger disk on the top of a smaller one as long as all the other disks on the post are in increasing order of their diameter [4]. In addition, John McCarthy from the University of Stanford proposed the Towers of Stanford puzzle in which a smaller disk can be placed on the top of a bigger one provided that the disk on the bottom of that stack is the largest one on the post [5]. Furthermore, Lee Badger and Marc Williams from Weber State University offered the solution n n + 1 for the minimum steps to complete the Towers of Stanford with n disks [6]. Additionally, The super towers of Hanoi problem by Carole S.Klein and Steven Minsker stipulated that the distribution of disks was completely random in the initial configuration and the movement of disks should obey the first three rules of the classical Towers of Hanoi [7]. 1. Our research and process We try to claim a new variation and can get interesting results. The classical Towers of Hanoi puzzle has strict restrictions. We reduce the restrictions on the method of moving steps in this variation, which yields more feasible approaches. Therefore, we need to consider the optimality of more possibilities to find out the minimum steps and have a deeper analysis on finding out the optimal method of moving disks. Furthermore, it is essential for us to prove it, which is the core in this passage. Additionally, we find out the algorithm of the new variation is similar to the original one and there is a special relationship between the optimal patterns of moving disks in the classical puzzle and our new variation. The following is our research process: Firstly, we redefine the maximum number of disks that can be put on a disk. Secondly, we prove the optimality of a possibility and then apply mathematical induction to obtain the recursion 1 The name of an Indian temple. François Édouard Anatole Lucas was a French mathematician. He proposed Towers of Hanoi puzzle in 1883.
4 formulas and formulas for general terms. Thirdly, we state the optimal patterns of moving disks and discover the relationship of optimal moves between our variation and the classical puzzle. The new variation We change part of the rules and consider another variation on the Towers of Hanoi puzzle. The initial configuration is the same as classical Towers of Hanoi. Rules 1) and ) in the classical Towers of Hanoi puzzle should also be obeyed in this new variation. However, we redefine rule 3) and 4): 1) Only one disk can be moved at a time. ) Moving one disk from a post to another one is defined as one step. 3) n disks are marked as disks 1,, 3 n from the smallest disk to the biggest one. At most i disks can be put on the top of the disk which is marked as i ( 1 i n and i is an integer ) in any intermediate configurations. So in some configurations, a larger disk can be placed on a smaller one. 4) Final configuration: n disks are all placed on post C and their sequence only needs to obey the new rule 3). 3 Proof: mathematical induction 3.1 Proof for optimality In this passage we define the optimal way as the approach which requires the fewest steps. Additionally, function f(n) in terms of n stands for the minimum number of steps transporting n disks from post A to post C. From the statistics in the Appendix, we discover that for a certain f(n + ), we deal with the move of two disks at a time. Additionally, disk n + 1 is always on the bottom of post C. Disk n + is always adjacent to disk n + 1 and only can be placed on the top of disk n + 1 in the optimal final configuration (a vertical line stands for a post in all following figures): the first n disks disk n+ disk n+1 Figure 3-1 We first prove the optimality of the final configuration above: According to the new rule 3), for a certain f(n + ), only disk n + 1 and disk n + can be placed on the bottom of post C in the final configuration. Since disk n + 1 and disk n + are placed on the bottom of the post A in the initial configuration, we must make sure all of the first n disks are moved to post B and guarantee that post C is completely empty in order to move disk n + 1 or disk n + to the bottom of post C. As a result, the following three intermediate situations (X,Y and Z) are possible after moving disk n + 1 (the sequence of the first n disks is certain, optimal and the same in situation X and Y):
5 X disk n+1 disk n+ the first n disks A B Figure 3- C Y disk n+ the first n disks disk n+1 A B Figure 3-3 C Z part 1 disk n+1 disk n+ part A B (Part 1 and part are combinations of some disks, part 1+ part = the first n disks) Figure 3-4 C It is clear to see that disk n + 1 and disk n + can be moved only after all of the other n disks are removed from post A. So in order to attain situation Z, we need to first extract disk n + 1 from post A and then plug it back into the middle of the first n disks. However, we only need to place disk n + 1 on the top of the first n disks. Therefore, situation X is better than situation Z. Then, we need to compare the optimality of situation X and Y. These two situations can attain the following two intermediate situations *X and *Y respectively:
6 *X the first n disks disk n+1 disk n+ A B Figure 3-5 C *Y the first n disks disk n+ disk n+1 A B Figure 3-6 C Both situations *X and *Y need to move the first n disks from post B to post C to attain different final configurations which meet the requirements. Since the optimal way to move the first n disks is certain, the two situations need the same number of steps to attain the viable final configuration. We need to move two more steps from X to *X: move disk n + from post A to post C and then move disk n + 1 from post B to post C. However, we only need to move one more step from Y to *Y: move disk n + from post A to post C. As a result, situation Y is better than situation X. We obtain the conclusion that situation Y is the optimal one. 3. Proof for even terms Then we use mathematical induction to prove the recursion formula for even terms: f() = According to the Figure 3-3, 3-6 and 3-1, we can find the relationship between f(n + ) and f(n). For a certain even integer n, the minimum steps of moving n + disks from post A to post C can be represented by the sum of the following three parts: 1) the steps of moving the first n disks (except the two disks which are marked as n + 1 and n + ) from post A to B ) the steps of moving the two disks which are marked as n + 1 and n + from post A to the bottom of post C 3) the steps of moving the first n disks (except the two disks which are marked as n + 1 and n + ) from B to C The initial configurations of the first n disks in the above part 1) and f(n) are the same: disks 1,, 3 n are
7 placed on post A in increasing order of size from the top to the bottom. Since the optimal move for n disks in the same sequence is certain, the methods of moving disks in part 1) and f(n) are completely identical and they require the same number of steps. In addition, the steps required in the above part ) is. Then we need to figure out the minimum number of steps required in part 3). It is obvious that the steps of part 1) and 3) are continuous manipulations when we only pay attention to the position change of the first n disks. As a result, the initial permutation of disks in part 3) is the same as the ultimate one in part 1). Since the optimal move is certain for n disks, moving the first n disks back into the permutation in the beginning of part 1) is also the optimal way. Therefore, part 3) and part 1) are exactly inverse manipulations, which indicates the number of steps in part 3) equals the number of steps required to move n disks in both part 1) and f(n). From the induction above we can conclude that f(n + ) = f(n) + where n is a positive even integer. According to the proven recursion formula f(n + ) = f(n) + for even integers, we can estimate the function: f(n) = n+ We also apply mathematical induction to prove this general formula for positive even integers. Proof: 1. n = f() = + = This case satisfies the general formula for positive even integers.. We suppose that the case n = k (k can be any positive even integer except ) satisfies the general formula above. Then f(k) = k+. 3. n = k + According to the case n = k and the recursion formula, we can obtain that f(k + ) = f(k) + = ( k+ ) + = (k+)+. It satisfies the general formula f(n) = n+ Finally, we can elicit the conclusion that the recursion formula is f(n + ) = f(n) + and the general formula is f(n) = n+ (n is a positive even integer). 3.3 Proof for odd terms After excluding f(1) = 1, we utilize mathematical induction to prove the recursion formula for odd terms in a similar way. f(3) = 4 According to the Figure 3-3, 3-6 and 3-1, we can find the relationship between f(n + ) and f(n). For a certain odd integer n, the minimum steps of moving n + disks from post A to post C can also be represented by the sum of the following three parts: 1) the steps of moving the first n disks (except the two disks which are marked as n + 1 and n + ) from post A to B
8 ) the steps of moving the two disks which are marked as n + 1 and n + from post A to the bottom of post C 3) the steps of moving the first n disks (except the two disks which are marked as n + 1 and n + ) from B to C The same procedure in Chapter 3. (even integer terms) can be easily adapted to obtain that the number of steps in part 1) equals f(n) and part ) needs two steps to complete. Then we compare part 1) with part 3). We first neglect the existence of the smallest three disks among the first n disks (disks 1, and 3) since their positions are more changeable. The same method in Chapter 3. (inverse manipulations) can be utilized to obtain that the number of steps to move the n-3 disks in the middle are identical in part 1) and part 3). Then we focus on disks 1, and 3 which are ignored in the previous discussion. The first time to move these three disks from post A requires four steps according to the optimal situation when n = 3 that has been listed as f(3) = 4 in the Appendix. By listing all the possible methods of moving three disks, there are four different viable final configurations of the smallest three disks (from top to bottom) after switching their position for the first time: a) 1 3 b) 1 3 c) 3 1 d) 1 3 At least 5, 4, 4, 6 steps are required to attain a), b), c) and d) configurations respectively. We first consider the position of disk 1. If disk 1 is on the top, it will require more than three steps to move all the three disks to another post. Otherwise disk 1 will be on the bottom among the three disks, which contradicts the new rule. Therefore, placing disk 1 in the middle of disk and disk 3 is the most optimal one. As a result, situation c) and d) are more optimal beginnings for further moves. Since three steps are required to move the first three disks from one post to another each time in both permutations (3 1 ) and ( 1 3), we claim that we need at least three steps every time to move all the three disks from one post to another except the first time. Compared to d), situation c) is more optimal since it can be obtained more easily. (According to the case n=3 in the Appendix) Then, we can compare the steps of moving disks 1, and 3 in part 1) and 3). Since we need to move n disks in both part 1) and part 3) and the optimal method of moving is certain (for odd terms, the optimal method of moving is to move two disks at a time except the smallest three disks), we need to switch the position of disks 1, and 3 for the same number of times. Here we define the number of times to switch the position of the smallest three disks in both part 1) and 3) as m. From the poof above, we know that the first time to move the smallest three disks from post A in the initial configuration needs four steps and every following move of the three disks requires three steps. Hence, we can calculate the total steps of moving disks 1, and 3 in part 1) and 3). The number of steps in part 1) equals 4 + 3(m 1) = 3m + 1 And the number of steps in part 3) equals 3m As a result, the total steps of moving three smallest disks in part 1) require one more step than those in part 3). Since we have proved that the number of steps required to move the n-3 disks in the middle are identical in part 1) and part 3), we can conclude that the total number of steps in part 1) equals the total number of steps in part 3) plus1 step. As we claimed before in this Chapter, the methods of moving disks in part 1) and f(n) are the same. Therefore, the total number of steps in part 3) is f(n) 1. Finally, we can draw out our conclusion that f(n + ) = f(n) + 1 According to the proven recursion formula, we can also estimate the formula for general odd terms except n = 1: f(n) = n+1 + n 3 1 (n is a positive odd integer and n 3)
9 We exploit mathematical induction to prove the general formula for odd terms except n=1. Proof: 1. n = 3 f(3) = = 4 This case satisfies the general formula for positive odd integers.. We suppose that the case n = k (k can be any positive odd integer except 1 and 3) satisfies the general formula above. Then f(n) = k+1 + k n = k + According to the case n = k and the recursion formula, we can obtain that f(k + ) = f(k) + 1 = ( k+1 + k 3 1) + 1 = (k+1)+ + (k 3)+ 1 It satisfies the general formula f(n) = n+1 + n 3 1. Ultimately, we concluded that the recursion formula is f(n + ) = f(n) + 1 and the general formula is f(n) = n+1 + n 3 1 (n is a positive odd integer except 1). 3.4 Results of Proof Here we demonstrate our results of proof: Theorem: n = 1, f(1) = 1 n is a positive even integer recursion formula: f(n + ) = f(n) + formula for general term: f(n) = n+ n is a positive odd integer except 1 recursion formula: f(n + ) = f(n) + 1 formula for general term: f(n) = n+1 + n 3 1 We compare the results in our new variation with those in classical Towers of Hanoi puzzle. The two results are similar in some ways: both are related to the exponent of. There is a special relationship between the optimal patterns of moving disks for classical puzzle and for our new variation. This relationship will be stated in Chapter 4. However, unlike the results on classical Towers of Hanoi, our results are different for even and odd positive integers since the optimal methods of moving are a bit different in even and odd cases. In addition, the numerical value in our results for every term is smaller than that in the classical case, which indicates the new variation needs fewer steps. These differences are rational because the restrictions in our new variation are fewer.
10 4 The optimal patterns of moving disks 4.1 Definition of dividing group We define a new term called group. A group can contain j continuous adjacent disks in the initial configuration ( j n and j is a positive integer). It is not necessary that every group contains the same number of disks. The permutation of the disks in one group can be changed in intermediate positions, but it is on the basis of obeying the four rules in the new variation. Additionally, the disks in one group cannot be separated in intermediate positions and they are viewed as a whole. 4. Pattern for even terms We offer a general pattern for even terms in which n disks can be moved from post A to post C in the optimal way: In the initial configuration for even number of disks, we divide every two adjacent disks as a group from the top to the bottom and we mark them as groups 1,, 3 n (group 1 consists of disk 1 and disk, group n consists of disk n 1 and disk n). Here we prove a lemma. Lemma: The n groups in even cases should be placed in increasing order of size (from group 1 on the top to group n on the bottom) in any intermediate and final configurations. Proof: It is obvious that group n is on the bottom, otherwise the configuration will oppose to the rule 3)3 in the new variation. Since the disks in one group can't be separated, group n can only be placed above group n. Then we consider group n 4. It can be placed either in the middle of group n and group n or above group n. When group n 4 is middle of group n n and group, there can't be any groups in the middle of group n 4 and group n since the largest disk in the other groups is disk n 6 and the maximum number of disks on this disk is only n 6. As a result, the other groups are all above group n 4. In this case, we know that there are at least n 4 disks above disk n 5 in group n 4, which contradict the rule 3) in the new variation. Consequently, group n 4 can only be placed above group n n 6. In the same manner, group must be placed above group n 4 n 8, group result, we have proved the sequence of the groups. must be placed above group n 6 group 1 must be placed above group. As a Since the sequence of the groups is certain and requirements of the groups are the same as those in the classical Towers of Hanoi puzzle, the groups must obey the rule of the classical puzzle. According to the minimum steps of 3 The rule 3) was stated in Chapter.
11 moving n disks in the classical Towers of Hanoi puzzle 4, the minimum steps of moving n groups equals n 1 Since two steps are required to move the disks in a group when one group is moved once, the minimum steps of moving groups equals ( n 1) = n+ = n+ The result is the same as the formula for even terms from Theorem in Chapter 3.4.We also know the optimal pattern of moving even number of disks: move n groups in the same way as the optimal one in the classical Towers of Hanoi puzzle after defining every two adjacent disks in the initial configuration from top to bottom as a group. Furthermore, we find out there is a relationship between our algorithm and that in the classical Towers of Hanoi puzzle, whose optimal pattern is also related to ours: the results of our new variation can be calculated by utilizing W.R.Ball's recursive algorithm; the optimal pattern in the classical puzzle can be applied to move the disks in our optimal way in our the new variation. 4.3 Pattern for odd terms We also offer a general pattern to move odd number of disks in the fewest steps (n 5): In the initial configuration for odd number of disks, we divide every two adjacent disks except the three disks on the top (disks 1, and 3) as a group from the top to the bottom and we mark them as groups 1,, 3 n 3 (group 1 consists of disk 4 and disk 5, group n-3 consists of disk n 1 and disk n). According to the lemma in Chapter 4., the first n 3 groups can only be placed in increasing sequence of size in any configuration. Like Chapter 4., we can also employ the recursive algorithm presented by W.R.Ball from the classical puzzle's solution to calculate the fewest number of steps to move these n 3 groups which is n 3 1 The number of steps required is also two when we move one group once. Hence, the minimum steps of moving these n-3 disks can be represented as ( n 3 1) = n 3 Now we consider the moves of the smallest three disks (disks 1, and 3). According to the optimal way to move the three disks in Chapter 3.3, the first time to move these three disks require four steps and we require three steps to move the three disks every time except the first time. From the statistics listed in the Appendix, the numbers of times to move all the three smallest disks from one post to another for cases n = 5, 7, 9, 11 are, 4, 8, 16 respectively. Thus, we can conclude that the number of times to move the three disks is for odd integers (n 5). The total number of steps to move disks 1, and 3 is n 3 4 The result from W.R.Ball in 189 was stated in Chapter 1.1.
12 4 + 3( n 3 1) = 3 n Hence, the minimum number of steps to move odd number of disks is ( n 3 ) + (3 n 3 + 1) = 5 n 3 1 The general formula for odd integers is also the same as that in Theorem (Chapter 3.4) since 5 n 3 1 = 4 n 3 + n 3 1 = n+1 + n 3 1 = n+1 + n 3 1 Additionally, we find out the optimal pattern of moving disks for odd terms: define every two adjacent disks as a group from top to bottom except disks 1, and 3 and move these n 3 groups in the same way as the optimal pattern in the classical Towers of Hanoi. Our algorithm and optimal pattern for odd terms are also related to W.R.Ball's algorithm and his optimal pattern: the algorithm of the new variation for odd terms is based on the original recursive algorithm presented by W.R.Ball and the optimal pattern of moving the n 3 as that in the classical Towers of Hanoi puzzle. groups is the same 5 Conclusion This new variation on Towers of Hanoi is originally proposed by us. We have also found out the minimum number of steps and proved it. Meanwhile, we offer general optimal patterns of moving disks. Compared to the results in the classical Towers of Hanoi puzzle, the steps required in our variation are fewer. In the future, we may do further research to look for applications. References: [1] W.R.Ball. Mathematical Recreations and Essays. McMillan, London, 189. [] Jeremy Barbay. Bouncing Towers move faster than Hanoi Towers, but still require exponential time. Computer Science, 016. [3] J.Chappelon, U.Larsson, A.Matsuura. Two-Player Tower of Hanoi, Computer Science, 015. [4] B.H.Ball, R.A.Beeler. The Apprentices' Tower of Hanoi. Journal of Mathematical Science, 016. [5] J.McCarthy. The Tower of Stanford, Problems and Solutions, American Mathematical Monthly, Vol.109, Number 7 (August-September 00), 664. [6] L.Badger, M.Williams.The Tower of Stanford, Problems and Solutions, American Mathematical Monthly, Vol.111, Number 4 (April 004), [7] C.S.Klein, S.Minsker. The super towers of Hanoi problem. Discrete Mathematics. 114 (1993),
13 Appendix: The patterns of the first eleven terms We search for general patterns about the new variation by listing the final configuration, the optimal way to attain the final configuration and the minimum numbers of steps for n = 1,, 3, 4, 5, 6, 7, 8, 9, 10, 11 respectively since we can find the patterns after 11 terms. (For convenience, we use i K J to denote every step. It means disk number i is moved from post K to post J, 1 i n and i is an integer, K and J can be post A, B or C): n = 1(1step) 1 A C Final configuration ( from top to bottom):1 n = ( steps ) 1 A C A C Final configuration (from top to bottom): 1 n = 3(4steps) 1 A B A C 1 B C 3 A C Final configuration (from top to bottom): 3 1 n = 4 ( 6 steps ) 1 A B A B 3 A C 4 A C B C 1 B C Final configuration (from top to bottom): n = 5 ( 9 steps ) 1 A C A B 1 C B 3 A B 4 A C 5 A C 3 B C 1 B C B C Final configuration (from top to bottom): n = 6 ( 14 steps ) 1 A C A C 3 A B 4 A B C B 1 C B 5 A C 6 A C 1 B A B A 4 B C 3 B C A C 1 A C Final configuration ( from top to bottom ): n = 7 ( 19 steps ) 1 A B A C 1 B C 3 A C 4 A B 5 A B 3 C B 1 C B C B 6 A C 7 A C B A 1 B A 3 B A 5 B C 4 B C 3 A C 1 A C A C Final configuration ( from top to bottom ): n = 8 ( 30 steps ) 1 A B A B 3 A C 4 A C A C 1 A C 5 A B 6 A B 1 C A C A 4 C B 3 C B A B 1 A B 7 A C 8 A C 1 B C B C 3 B A 4 B A C A 1 C A 6 B C 5 B C 1 A B A B 4 A C 3 A C B C 1 B C Final configuration ( from top to bottom ): n = 9 ( 39 steps ) 1 A C A B 1 C B 3 A B 4 A C 5 A C
14 3 B C 1 B C B C 6 A B 7 A B C A 1 C A 3 C A 5 C B 4 C B 3 A B 1 A B A B 8 A C 9 A C B C 1 B C 3 B C 4 B A 5 B A 3 C A 1 C A C A 7 B C 6 B C A B 1 A B 3 A B 5 A C 4 A C 3 B C 1 B C B C Final configuration ( from top to bottom ): n = 10 ( 6 steps ) 1 A C A C 3 A B 4 A B C B 1 C B 5 A C 6 A C 1 B A B A 4 B C 3 B C A C 1 A C 7 A B 8 A B 1 C B C B 3 C A 4 C A B A 1 B A 6 C B 5 C B 1 A C A C 4 A B 3 A B C B 1 C B 9 A C 10 A C 1 B A B A 3 B C 4 B C A C 1 A C 5 B A 6 B A 1 C B C B 4 C A 3 C A B A 1 B A 8 B C 7 B C 1 A C A C 3 A B 4 A B C B 1 C B 6 A C 5 A C 1 B A B A 4 B C 3 B C A C 1 A C Final configuration ( from top to bottom ): n = 11 ( 79 steps ) 1 A B A C 1 B C 3 A C 4 A B 5 A B 3 C B 1 C B C B 6 A C 7 A C B A 1 B A 3 B A 5 B C 4 B C 3 A C 1 A C A C 8 A B 9 A B C B 1 C B 3 C B 4 C A 5 C A 3 B A 1 B A B A 7 C B 6 C B A C 1 A C 3 A C 5 A B 4 A B 3 C B 1 C B C B 10 A C 11 A C B A 1 B A 3 B A 4 B C 5 B C 3 A C 1 A C A C 6 B A 7 B A C B 1 C B 3 C B 5 C A 4 C A 3 B A 1 B A B A 9 B C 8 B C A C 1 A C 3 A C 4 A B 5 A B 3 C B 1 C B C B 7 A C 6 A C B A 1 B A 3 B A 5 B C 4 B C 3 A C 1 A C A C Final configuration ( from top to bottom ): From the statistics above, we can obtain the following table: n Steps Table 1 According to the table, we can find out that there are different patterns for even and odd terms except n=1.
Mathematical Induction
Mathematical Induction (Handout March 8, 01) The Principle of Mathematical Induction provides a means to prove infinitely many statements all at once The principle is logical rather than strictly mathematical,
More informationThe Taxman Game. Robert K. Moniot September 5, 2003
The Taxman Game Robert K. Moniot September 5, 2003 1 Introduction Want to know how to beat the taxman? Legally, that is? Read on, and we will explore this cute little mathematical game. The taxman game
More information3. Mathematical Induction
3. MATHEMATICAL INDUCTION 83 3. Mathematical Induction 3.1. First Principle of Mathematical Induction. Let P (n) be a predicate with domain of discourse (over) the natural numbers N = {0, 1,,...}. If (1)
More informationcsci 210: Data Structures Recursion
csci 210: Data Structures Recursion Summary Topics recursion overview simple examples Sierpinski gasket Hanoi towers Blob check READING: GT textbook chapter 3.5 Recursion In general, a method of defining
More information8 Primes and Modular Arithmetic
8 Primes and Modular Arithmetic 8.1 Primes and Factors Over two millennia ago already, people all over the world were considering the properties of numbers. One of the simplest concepts is prime numbers.
More informationMidterm Practice Problems
6.042/8.062J Mathematics for Computer Science October 2, 200 Tom Leighton, Marten van Dijk, and Brooke Cowan Midterm Practice Problems Problem. [0 points] In problem set you showed that the nand operator
More informationSowing Games JEFF ERICKSON
Games of No Chance MSRI Publications Volume 29, 1996 Sowing Games JEFF ERICKSON Abstract. At the Workshop, John Conway and Richard Guy proposed the class of sowing games, loosely based on the ancient African
More informationMath 55: Discrete Mathematics
Math 55: Discrete Mathematics UC Berkeley, Fall 2011 Homework # 5, due Wednesday, February 22 5.1.4 Let P (n) be the statement that 1 3 + 2 3 + + n 3 = (n(n + 1)/2) 2 for the positive integer n. a) What
More informationThe positive minimum degree game on sparse graphs
The positive minimum degree game on sparse graphs József Balogh Department of Mathematical Sciences University of Illinois, USA jobal@math.uiuc.edu András Pluhár Department of Computer Science University
More informationIB Math Research Problem
Vincent Chu Block F IB Math Research Problem The product of all factors of 2000 can be found using several methods. One of the methods I employed in the beginning is a primitive one I wrote a computer
More informationA Study on the Necessary Conditions for Odd Perfect Numbers
A Study on the Necessary Conditions for Odd Perfect Numbers Ben Stevens U63750064 Abstract A collection of all of the known necessary conditions for an odd perfect number to exist, along with brief descriptions
More informationMath 55: Discrete Mathematics
Math 55: Discrete Mathematics UC Berkeley, Spring 2012 Homework # 9, due Wednesday, April 11 8.1.5 How many ways are there to pay a bill of 17 pesos using a currency with coins of values of 1 peso, 2 pesos,
More informationWRITING PROOFS. Christopher Heil Georgia Institute of Technology
WRITING PROOFS Christopher Heil Georgia Institute of Technology A theorem is just a statement of fact A proof of the theorem is a logical explanation of why the theorem is true Many theorems have this
More informationLAMC Beginners Circle: Parity of a Permutation Problems from Handout by Oleg Gleizer Solutions by James Newton
LAMC Beginners Circle: Parity of a Permutation Problems from Handout by Oleg Gleizer Solutions by James Newton 1. Take a two-digit number and write it down three times to form a six-digit number. For example,
More informationSection IV.1: Recursive Algorithms and Recursion Trees
Section IV.1: Recursive Algorithms and Recursion Trees Definition IV.1.1: A recursive algorithm is an algorithm that solves a problem by (1) reducing it to an instance of the same problem with smaller
More information6.2 Permutations continued
6.2 Permutations continued Theorem A permutation on a finite set A is either a cycle or can be expressed as a product (composition of disjoint cycles. Proof is by (strong induction on the number, r, of
More informationMath Workshop October 2010 Fractions and Repeating Decimals
Math Workshop October 2010 Fractions and Repeating Decimals This evening we will investigate the patterns that arise when converting fractions to decimals. As an example of what we will be looking at,
More informationarxiv:1112.0829v1 [math.pr] 5 Dec 2011
How Not to Win a Million Dollars: A Counterexample to a Conjecture of L. Breiman Thomas P. Hayes arxiv:1112.0829v1 [math.pr] 5 Dec 2011 Abstract Consider a gambling game in which we are allowed to repeatedly
More informationI. GROUPS: BASIC DEFINITIONS AND EXAMPLES
I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called
More informationMATHEMATICAL INDUCTION. Mathematical Induction. This is a powerful method to prove properties of positive integers.
MATHEMATICAL INDUCTION MIGUEL A LERMA (Last updated: February 8, 003) Mathematical Induction This is a powerful method to prove properties of positive integers Principle of Mathematical Induction Let P
More informationContinued Fractions and the Euclidean Algorithm
Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction
More informationTAKE-AWAY GAMES. ALLEN J. SCHWENK California Institute of Technology, Pasadena, California INTRODUCTION
TAKE-AWAY GAMES ALLEN J. SCHWENK California Institute of Technology, Pasadena, California L INTRODUCTION Several games of Tf take-away?f have become popular. The purpose of this paper is to determine the
More informationSHARP BOUNDS FOR THE SUM OF THE SQUARES OF THE DEGREES OF A GRAPH
31 Kragujevac J. Math. 25 (2003) 31 49. SHARP BOUNDS FOR THE SUM OF THE SQUARES OF THE DEGREES OF A GRAPH Kinkar Ch. Das Department of Mathematics, Indian Institute of Technology, Kharagpur 721302, W.B.,
More informationDiscrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 10
CS 70 Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 10 Introduction to Discrete Probability Probability theory has its origins in gambling analyzing card games, dice,
More informationSECTION 10-2 Mathematical Induction
73 0 Sequences and Series 6. Approximate e 0. using the first five terms of the series. Compare this approximation with your calculator evaluation of e 0.. 6. Approximate e 0.5 using the first five terms
More informationDiscrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 2
CS 70 Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 2 Proofs Intuitively, the concept of proof should already be familiar We all like to assert things, and few of us
More informationLottery Combinatorics
Published by the Applied Probability Trust Applied Probability Trust 2009 110 Lottery Combinatorics IAN MCPHERSON and DEREK HODSON The chance of landing the National Lottery jackpot (or a share of it)
More informationComputing exponents modulo a number: Repeated squaring
Computing exponents modulo a number: Repeated squaring How do you compute (1415) 13 mod 2537 = 2182 using just a calculator? Or how do you check that 2 340 mod 341 = 1? You can do this using the method
More informationNear Optimal Solutions
Near Optimal Solutions Many important optimization problems are lacking efficient solutions. NP-Complete problems unlikely to have polynomial time solutions. Good heuristics important for such problems.
More informationMATH10040 Chapter 2: Prime and relatively prime numbers
MATH10040 Chapter 2: Prime and relatively prime numbers Recall the basic definition: 1. Prime numbers Definition 1.1. Recall that a positive integer is said to be prime if it has precisely two positive
More informationCOUNTING INDEPENDENT SETS IN SOME CLASSES OF (ALMOST) REGULAR GRAPHS
COUNTING INDEPENDENT SETS IN SOME CLASSES OF (ALMOST) REGULAR GRAPHS Alexander Burstein Department of Mathematics Howard University Washington, DC 259, USA aburstein@howard.edu Sergey Kitaev Mathematics
More informationCatalan Numbers. Thomas A. Dowling, Department of Mathematics, Ohio State Uni- versity.
7 Catalan Numbers Thomas A. Dowling, Department of Mathematics, Ohio State Uni- Author: versity. Prerequisites: The prerequisites for this chapter are recursive definitions, basic counting principles,
More informationWe can express this in decimal notation (in contrast to the underline notation we have been using) as follows: 9081 + 900b + 90c = 9001 + 100c + 10b
In this session, we ll learn how to solve problems related to place value. This is one of the fundamental concepts in arithmetic, something every elementary and middle school mathematics teacher should
More informationLemma 5.2. Let S be a set. (1) Let f and g be two permutations of S. Then the composition of f and g is a permutation of S.
Definition 51 Let S be a set bijection f : S S 5 Permutation groups A permutation of S is simply a Lemma 52 Let S be a set (1) Let f and g be two permutations of S Then the composition of f and g is a
More informationThe Mathematics of the RSA Public-Key Cryptosystem
The Mathematics of the RSA Public-Key Cryptosystem Burt Kaliski RSA Laboratories ABOUT THE AUTHOR: Dr Burt Kaliski is a computer scientist whose involvement with the security industry has been through
More informationMATH 289 PROBLEM SET 4: NUMBER THEORY
MATH 289 PROBLEM SET 4: NUMBER THEORY 1. The greatest common divisor If d and n are integers, then we say that d divides n if and only if there exists an integer q such that n = qd. Notice that if d divides
More informationBreaking The Code. Ryan Lowe. Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and
Breaking The Code Ryan Lowe Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and a minor in Applied Physics. As a sophomore, he took an independent study
More informationNIM with Cash. Abstract. loses. This game has been well studied. For example, it is known that for NIM(1, 2, 3; n)
NIM with Cash William Gasarch Univ. of MD at College Park John Purtilo Univ. of MD at College Park Abstract NIM(a 1,..., a k ; n) is a -player game where initially there are n stones on the board and the
More informationGREATEST COMMON DIVISOR
DEFINITION: GREATEST COMMON DIVISOR The greatest common divisor (gcd) of a and b, denoted by (a, b), is the largest common divisor of integers a and b. THEOREM: If a and b are nonzero integers, then their
More informationThe BBP Algorithm for Pi
The BBP Algorithm for Pi David H. Bailey September 17, 2006 1. Introduction The Bailey-Borwein-Plouffe (BBP) algorithm for π is based on the BBP formula for π, which was discovered in 1995 and published
More informationMathematical Induction
Mathematical Induction In logic, we often want to prove that every member of an infinite set has some feature. E.g., we would like to show: N 1 : is a number 1 : has the feature Φ ( x)(n 1 x! 1 x) How
More information5.1 Radical Notation and Rational Exponents
Section 5.1 Radical Notation and Rational Exponents 1 5.1 Radical Notation and Rational Exponents We now review how exponents can be used to describe not only powers (such as 5 2 and 2 3 ), but also roots
More informationProof of the conservation of momentum and kinetic energy
Experiment 04 Proof of the conservation of momentum and kinetic energy By Christian Redeker 27.10.2007 Contents 1.) Hypothesis...3 2.) Diagram...7 3.) Method...7 3.1) Apparatus...7 3.2) Procedure...7 4.)
More informationKenken For Teachers. Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles June 27, 2010. Abstract
Kenken For Teachers Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles June 7, 00 Abstract Kenken is a puzzle whose solution requires a combination of logic and simple arithmetic skills.
More informationCONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12
CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.
More informationCONTINUED FRACTIONS AND FACTORING. Niels Lauritzen
CONTINUED FRACTIONS AND FACTORING Niels Lauritzen ii NIELS LAURITZEN DEPARTMENT OF MATHEMATICAL SCIENCES UNIVERSITY OF AARHUS, DENMARK EMAIL: niels@imf.au.dk URL: http://home.imf.au.dk/niels/ Contents
More informationAbout the inverse football pool problem for 9 games 1
Seventh International Workshop on Optimal Codes and Related Topics September 6-1, 013, Albena, Bulgaria pp. 15-133 About the inverse football pool problem for 9 games 1 Emil Kolev Tsonka Baicheva Institute
More informationPigeonhole Principle Solutions
Pigeonhole Principle Solutions 1. Show that if we take n + 1 numbers from the set {1, 2,..., 2n}, then some pair of numbers will have no factors in common. Solution: Note that consecutive numbers (such
More informationALGEBRAIC APPROACH TO COMPOSITE INTEGER FACTORIZATION
ALGEBRAIC APPROACH TO COMPOSITE INTEGER FACTORIZATION Aldrin W. Wanambisi 1* School of Pure and Applied Science, Mount Kenya University, P.O box 553-50100, Kakamega, Kenya. Shem Aywa 2 Department of Mathematics,
More informationCHAPTER 5. Number Theory. 1. Integers and Division. Discussion
CHAPTER 5 Number Theory 1. Integers and Division 1.1. Divisibility. Definition 1.1.1. Given two integers a and b we say a divides b if there is an integer c such that b = ac. If a divides b, we write a
More informationAN ANALYSIS OF A WAR-LIKE CARD GAME. Introduction
AN ANALYSIS OF A WAR-LIKE CARD GAME BORIS ALEXEEV AND JACOB TSIMERMAN Abstract. In his book Mathematical Mind-Benders, Peter Winkler poses the following open problem, originally due to the first author:
More informationCombinatorial Proofs
Combinatorial Proofs Two Counting Principles Some proofs concerning finite sets involve counting the number of elements of the sets, so we will look at the basics of counting. Addition Principle: If A
More informationSample Induction Proofs
Math 3 Worksheet: Induction Proofs III, Sample Proofs A.J. Hildebrand Sample Induction Proofs Below are model solutions to some of the practice problems on the induction worksheets. The solutions given
More informationEXPONENTIAL FUNCTIONS 8.1.1 8.1.6
EXPONENTIAL FUNCTIONS 8.1.1 8.1.6 In these sections, students generalize what they have learned about geometric sequences to investigate exponential functions. Students study exponential functions of the
More informationSudoku Puzzles Generating: from Easy to Evil
Team # 3485 Page 1 of 20 Sudoku Puzzles Generating: from Easy to Evil Abstract As Sudoku puzzle becomes worldwide popular among many players in different intellectual levels, the task is to devise an algorithm
More informationSo let us begin our quest to find the holy grail of real analysis.
1 Section 5.2 The Complete Ordered Field: Purpose of Section We present an axiomatic description of the real numbers as a complete ordered field. The axioms which describe the arithmetic of the real numbers
More informationFor additional information, see the Math Notes boxes in Lesson B.1.3 and B.2.3.
EXPONENTIAL FUNCTIONS B.1.1 B.1.6 In these sections, students generalize what they have learned about geometric sequences to investigate exponential functions. Students study exponential functions of the
More informationCOMBINATORIAL PROPERTIES OF THE HIGMAN-SIMS GRAPH. 1. Introduction
COMBINATORIAL PROPERTIES OF THE HIGMAN-SIMS GRAPH ZACHARY ABEL 1. Introduction In this survey we discuss properties of the Higman-Sims graph, which has 100 vertices, 1100 edges, and is 22 regular. In fact
More informationCollatz Sequence. Fibbonacci Sequence. n is even; Recurrence Relation: a n+1 = a n + a n 1.
Fibonacci Roulette In this game you will be constructing a recurrence relation, that is, a sequence of numbers where you find the next number by looking at the previous numbers in the sequence. Your job
More informationBasics of Counting. The product rule. Product rule example. 22C:19, Chapter 6 Hantao Zhang. Sample question. Total is 18 * 325 = 5850
Basics of Counting 22C:19, Chapter 6 Hantao Zhang 1 The product rule Also called the multiplication rule If there are n 1 ways to do task 1, and n 2 ways to do task 2 Then there are n 1 n 2 ways to do
More informationIntroduction. Appendix D Mathematical Induction D1
Appendix D Mathematical Induction D D Mathematical Induction Use mathematical induction to prove a formula. Find a sum of powers of integers. Find a formula for a finite sum. Use finite differences to
More informationJust the Factors, Ma am
1 Introduction Just the Factors, Ma am The purpose of this note is to find and study a method for determining and counting all the positive integer divisors of a positive integer Let N be a given positive
More informationIf A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C?
Problem 3 If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C? Suggested Questions to ask students about Problem 3 The key to this question
More informationReal Roots of Univariate Polynomials with Real Coefficients
Real Roots of Univariate Polynomials with Real Coefficients mostly written by Christina Hewitt March 22, 2012 1 Introduction Polynomial equations are used throughout mathematics. When solving polynomials
More informationCycles in a Graph Whose Lengths Differ by One or Two
Cycles in a Graph Whose Lengths Differ by One or Two J. A. Bondy 1 and A. Vince 2 1 LABORATOIRE DE MATHÉMATIQUES DISCRÉTES UNIVERSITÉ CLAUDE-BERNARD LYON 1 69622 VILLEURBANNE, FRANCE 2 DEPARTMENT OF MATHEMATICS
More informationLecture 1: Course overview, circuits, and formulas
Lecture 1: Course overview, circuits, and formulas Topics in Complexity Theory and Pseudorandomness (Spring 2013) Rutgers University Swastik Kopparty Scribes: John Kim, Ben Lund 1 Course Information Swastik
More informationSquaring, Cubing, and Cube Rooting
Squaring, Cubing, and Cube Rooting Arthur T. Benjamin Harvey Mudd College Claremont, CA 91711 benjamin@math.hmc.edu I still recall my thrill and disappointment when I read Mathematical Carnival [4], by
More informationChapter 3. Cartesian Products and Relations. 3.1 Cartesian Products
Chapter 3 Cartesian Products and Relations The material in this chapter is the first real encounter with abstraction. Relations are very general thing they are a special type of subset. After introducing
More informationMathematical Induction. Mary Barnes Sue Gordon
Mathematics Learning Centre Mathematical Induction Mary Barnes Sue Gordon c 1987 University of Sydney Contents 1 Mathematical Induction 1 1.1 Why do we need proof by induction?.... 1 1. What is proof by
More informationContinued Fractions. Darren C. Collins
Continued Fractions Darren C Collins Abstract In this paper, we discuss continued fractions First, we discuss the definition and notation Second, we discuss the development of the subject throughout history
More informationThe thing that started it 8.6 THE BINOMIAL THEOREM
476 Chapter 8 Discrete Mathematics: Functions on the Set of Natural Numbers (b) Based on your results for (a), guess the minimum number of moves required if you start with an arbitrary number of n disks.
More informationPUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.
PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include
More informationCartesian Products and Relations
Cartesian Products and Relations Definition (Cartesian product) If A and B are sets, the Cartesian product of A and B is the set A B = {(a, b) :(a A) and (b B)}. The following points are worth special
More informationAssignment 5 - Due Friday March 6
Assignment 5 - Due Friday March 6 (1) Discovering Fibonacci Relationships By experimenting with numerous examples in search of a pattern, determine a simple formula for (F n+1 ) 2 + (F n ) 2 that is, a
More informationNumber Theory. Proof. Suppose otherwise. Then there would be a finite number n of primes, which we may
Number Theory Divisibility and Primes Definition. If a and b are integers and there is some integer c such that a = b c, then we say that b divides a or is a factor or divisor of a and write b a. Definition
More informationMATH 4330/5330, Fourier Analysis Section 11, The Discrete Fourier Transform
MATH 433/533, Fourier Analysis Section 11, The Discrete Fourier Transform Now, instead of considering functions defined on a continuous domain, like the interval [, 1) or the whole real line R, we wish
More informationThe last three chapters introduced three major proof techniques: direct,
CHAPTER 7 Proving Non-Conditional Statements The last three chapters introduced three major proof techniques: direct, contrapositive and contradiction. These three techniques are used to prove statements
More informationCongruent Number Problem
University of Waterloo October 28th, 2015 Number Theory Number theory, can be described as the mathematics of discovering and explaining patterns in numbers. There is nothing in the world which pleases
More informationTower Of Hanoi & Reve Puzzles
McCann 1 (of 9) William McCann Professor Langston Discrete Mathematics August 29, 2007 Tower Of Hanoi & Reve Puzzles Abstract In this paper I will describe the Tower Of Hanoi puzzle along with an algorithm
More informationHandout NUMBER THEORY
Handout of NUMBER THEORY by Kus Prihantoso Krisnawan MATHEMATICS DEPARTMENT FACULTY OF MATHEMATICS AND NATURAL SCIENCES YOGYAKARTA STATE UNIVERSITY 2012 Contents Contents i 1 Some Preliminary Considerations
More information1(a). How many ways are there to rearrange the letters in the word COMPUTER?
CS 280 Solution Guide Homework 5 by Tze Kiat Tan 1(a). How many ways are there to rearrange the letters in the word COMPUTER? There are 8 distinct letters in the word COMPUTER. Therefore, the number of
More informationSYSTEMS OF PYTHAGOREAN TRIPLES. Acknowledgements. I would like to thank Professor Laura Schueller for advising and guiding me
SYSTEMS OF PYTHAGOREAN TRIPLES CHRISTOPHER TOBIN-CAMPBELL Abstract. This paper explores systems of Pythagorean triples. It describes the generating formulas for primitive Pythagorean triples, determines
More informationCardinality. The set of all finite strings over the alphabet of lowercase letters is countable. The set of real numbers R is an uncountable set.
Section 2.5 Cardinality (another) Definition: The cardinality of a set A is equal to the cardinality of a set B, denoted A = B, if and only if there is a bijection from A to B. If there is an injection
More information3 0 + 4 + 3 1 + 1 + 3 9 + 6 + 3 0 + 1 + 3 0 + 1 + 3 2 mod 10 = 4 + 3 + 1 + 27 + 6 + 1 + 1 + 6 mod 10 = 49 mod 10 = 9.
SOLUTIONS TO HOMEWORK 2 - MATH 170, SUMMER SESSION I (2012) (1) (Exercise 11, Page 107) Which of the following is the correct UPC for Progresso minestrone soup? Show why the other numbers are not valid
More information1.2. Successive Differences
1. An Application of Inductive Reasoning: Number Patterns In the previous section we introduced inductive reasoning, and we showed how it can be applied in predicting what comes next in a list of numbers
More information14.1 Rent-or-buy problem
CS787: Advanced Algorithms Lecture 14: Online algorithms We now shift focus to a different kind of algorithmic problem where we need to perform some optimization without knowing the input in advance. Algorithms
More informationCONTRIBUTIONS TO ZERO SUM PROBLEMS
CONTRIBUTIONS TO ZERO SUM PROBLEMS S. D. ADHIKARI, Y. G. CHEN, J. B. FRIEDLANDER, S. V. KONYAGIN AND F. PAPPALARDI Abstract. A prototype of zero sum theorems, the well known theorem of Erdős, Ginzburg
More informationSCORE SETS IN ORIENTED GRAPHS
Applicable Analysis and Discrete Mathematics, 2 (2008), 107 113. Available electronically at http://pefmath.etf.bg.ac.yu SCORE SETS IN ORIENTED GRAPHS S. Pirzada, T. A. Naikoo The score of a vertex v in
More information0.8 Rational Expressions and Equations
96 Prerequisites 0.8 Rational Expressions and Equations We now turn our attention to rational expressions - that is, algebraic fractions - and equations which contain them. The reader is encouraged to
More informationWinter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com
Polynomials Alexander Remorov alexanderrem@gmail.com Warm-up Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)).
More informationThe Pointless Machine and Escape of the Clones
MATH 64091 Jenya Soprunova, KSU The Pointless Machine and Escape of the Clones The Pointless Machine that operates on ordered pairs of positive integers (a, b) has three modes: In Mode 1 the machine adds
More informationA Short Guide to Significant Figures
A Short Guide to Significant Figures Quick Reference Section Here are the basic rules for significant figures - read the full text of this guide to gain a complete understanding of what these rules really
More informationLecture 13 - Basic Number Theory.
Lecture 13 - Basic Number Theory. Boaz Barak March 22, 2010 Divisibility and primes Unless mentioned otherwise throughout this lecture all numbers are non-negative integers. We say that A divides B, denoted
More information2 When is a 2-Digit Number the Sum of the Squares of its Digits?
When Does a Number Equal the Sum of the Squares or Cubes of its Digits? An Exposition and a Call for a More elegant Proof 1 Introduction We will look at theorems of the following form: by William Gasarch
More informationC H A P T E R Regular Expressions regular expression
7 CHAPTER Regular Expressions Most programmers and other power-users of computer systems have used tools that match text patterns. You may have used a Web search engine with a pattern like travel cancun
More informationCONTENTS 1. Peter Kahn. Spring 2007
CONTENTS 1 MATH 304: CONSTRUCTING THE REAL NUMBERS Peter Kahn Spring 2007 Contents 2 The Integers 1 2.1 The basic construction.......................... 1 2.2 Adding integers..............................
More informationLecture 3: Finding integer solutions to systems of linear equations
Lecture 3: Finding integer solutions to systems of linear equations Algorithmic Number Theory (Fall 2014) Rutgers University Swastik Kopparty Scribe: Abhishek Bhrushundi 1 Overview The goal of this lecture
More informationSequences. A sequence is a list of numbers, or a pattern, which obeys a rule.
Sequences A sequence is a list of numbers, or a pattern, which obeys a rule. Each number in a sequence is called a term. ie the fourth term of the sequence 2, 4, 6, 8, 10, 12... is 8, because it is the
More informationk, then n = p2α 1 1 pα k
Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square
More informationStudent Outcomes. Lesson Notes. Classwork. Discussion (10 minutes)
NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 5 8 Student Outcomes Students know the definition of a number raised to a negative exponent. Students simplify and write equivalent expressions that contain
More information