#Mathematical puzzle with movable disks series#

From this, we have the recurrence relation given by Mn = 2Mn+1(3.1). Number of moves of (n - 1) disks and one move of the Now for n >1 applying (3. Then we notice that we have to make two minimumđ Transfer all n disks from Peg 1 to either Peg 2 or Peg 3, Hence we get M = 1 (3.2) If we assume to be minimum number of moves to to transfer one disk from Peg 1 to either Peg 2 or Peg 3. Thus there will be only 1 minimum number of moves Required to move all n disks from Peg 1 to Peg 3.

Using the procedure briefed through Figures 2 toğurther, if we consider just one disk at Peg 1 ini-Ħ above, we can derive the minimum number of moves tially, then we can either keep it at Peg 2 or at Peg 3. Once this is done, we notice that all the n disks from Peg 1, has been transferred to Peg 3, thereby completingįigure 6: All the n disks transferred from Peg 1 to Peg 3 To accomplish our task of moving all the n disks from Peg 1 to say Peg 3, note that we have to first move the top (n - 1) disks from Peg 1 to Peg 2 as shown in Figure 3.įigure 3: Moving top (n - 1) disks from Peg 1 to Peg 2 Now we can move the largest disk at the bottom of Peg 1 to Peg 3 as shown in Figure 4.įigure 4: Moving remaining 1 disk from Peg 1 to Peg 3 We can move (n - 1) disks from Peg 2 to Peg 3 as shown in Figure 5.įigure 5: Moving (n - 1) disks from Peg 2 to Peg 3

Let us consider the case when there are say n disks at Peg 1 as shown in Figure 2.įigure 2: Tower of Hanoi Problem with initially n disks at Peg 1 With these two restrictions, what would be the minimum number of moves that we should make if there are initially n circular disks at the first peg? I will discuss about the solution to this fascinating problem and proceed further. (ii)ĝuring any move, only smaller disk can be placed above larger disk whereas, no larger disk can be placed above smaller. (i) Only one peg can be taken at a time which is located at top of that peg. A move refers to shifting of disk from a peg to another under the following two conditions: For doing this, we need to move the disks. The task of this mathematical puzzle is to transfer all the disks from the first peg to either second or third. Circular disks of different radius were stacked in one of the pegs in such a way that the larger disk lies at the bottom, the next larger lies above the largest one, until the smallest disk lies at the top as shown in Figure 1. Tower of Hanoi Problem In this interesting mathematical game problem, three identical pegs were placed in a stand.

#Mathematical puzzle with movable disks series#

Let £ a„ represent a divergent series of real Finally a geometrical interpretation was provided to have better understanding of the result derived. After doing this, I will determine the Ramanujan summation for the solution obtained. This problem has fascinating solution which I will derive in this paper. In 1883, French mathematician Edward Lucas introduced a wonderful mathematical combinatorial game called Tower of Hanoi Problem (Tower of Brahma in India), named after the Hanoi, the capital city of Vietnam. This idea created plenty of new results and generalizations in analytic number theory. Srinivasa Ramanujan introduced the concept of Ramanujan summation in connection with Bernoulli numbers and Riemann zeta function.

Keywords: Ramanujan Summation, Tower of Hanoi Problem, Definite Integral, Recurrence Relation, Mersenne Numbers.

In this paper, after briefly discussing Tower of Hanoi problem and deriving its solution, I had determined Ramanujan summation for the divergent series whose terms represent the solution of Tower of Hanoi problem. Similarly, one of the fascinating combinatorial mathematical games was called Tower of Hanoi Problem. The concept of Ramanujan Summation which is a novel way of assigning particular value to given divergent series was introduced by one of the great Indian mathematicians Srinivasa Ramanujan in early part of 20th Century. Independent Research Scholar African Moon University, South West Africa and USA RAMANUJAN SUMMATION FOR TOWER OF HANOI PROBLEM