By Susam Pal on (Sep)
Tower of Hanoi
A few weeks ago, I watched Rise of the Planet of the Apes . The movie showed a genetically engineered chimpanzee trying to solve a puzzle involving four discs, initially stacked in ascending order of size on one of three pegs. The chimpanzee was supposed to transfer the entire stack to one of the other pegs, moving only one disc at a time, and never placing a larger disc on a smaller one.
The problem was called the Lucas’ Tower in the movie. I have always known this problem as the Tower of Hanoi puzzle. The minimum number of moves required to solve the problem is (2 ^ n – 1 ) where (n ) is the number of discs. In the movie, the chimpanzee solved the problem in 15 moves, the minimum number of moves required when there are 4 discs.
Referring to the problem as the Lucas’ Tower made me wonder why it was called so instead of calling it the Tower of Hanoi. I guessed it was probably because the puzzle was invented by the French mathematician Édouard Lucas. Later when I checked the
Animated solution of the Tower of Hanoi puzzle created by André Karwath ( original source I’ll not discuss the solution of this puzzle in this blog post. There are plenty of articles on the web including the Wikpedia article that describes why it takes a minimum of (2 ^ n – 1 ) moves to solve the puzzle when there are (n ) discs involved. In this post, I’ll talk about an interesting result I discovered while playing with this puzzle one afternoon. Binary Numbers
Here is an interesting thing to note: If we denote the minimum number of moves required to solve the Tower of Hanoi puzzle as (T_n, ) then ( T_n ) when expressed in binary is the largest possible (n ) – bit integer. For example, (T_4=(_ {) }=_ {2}. ) That makes sense because (T_n=2 ^ n – 1 ) indeed represents the maximum possible (n ) – bit integer where all (n ) bits are set to (1. )
While playing with different values of (T_n ) for different values of (n, ) I stumbled upon an interesting result which I will pose as a problem in a later section below.
Before proceeding to the problem, let us define a few things to eliminate any possibility of ambiguity:
A positive integer (x ) is said to be an (n ) – bit integer if and only if the minimum number of bits required to express the integer is (n, ) or equivalently, ( lfloor log_2 x rfloor 1 =n. )
We will be dealing with arbitrary precision integers (bignums) in the problem, so let us also make a few assumptions:
Addition or subtraction of an (m ) – bit integer and an (n ) – bit integer ( (m le n )) takes (O (n) ) time. Counting the number of (1 ) – bits in an (n ) – bit integer takes (O (n) ) time.
The definition along with the assumptions lead to the following conclusions:
Adding or subtracting two integers (a ) and (b ) takes ( O ( log (max (a, b))) ) time. Counting the number of (1 ) – bits in an integer (a ) takes (O ( log (a)) ) time.
What is the most efficient way to compute the number of (1 ) – bits in [ T_1 T_2 dots T_n ] where (n ) is a positive integer, each (T_i=2 ^ i – 1 ) for integers (1 le i le n, ) and efficiency is measured in terms of time and space complexity?
The naive approach involves adding all the (n ) integers and counting the number of (1 ) – bits in the sum. It takes (O (n ^ 2) ) time to add the (n ) integers. The sum is an ((n 1) ) – bit integer, so it takes (O (n) ) time to count the number of (1 ) – bits in the sum. Since the sum is ((n 1) ) – bit long, it takes (O (n) ) memory to store the sum. If (n ) is as large as, say, (2 ^ { }, ) it takes 2 exbibytes plus one more byte of memory to store the sum.
We can arrive at a much more efficient solution if we look at what the binary representation of the sum looks like. We first arrive at a closed-form expression for the sum: begin {align *} T_1 T_2 dots T_n &=(2 – 1) (2 ^ 2 – 1) dots (2 ^ n – 1) \ &=(2 2 ^ 2 dots 2 ^ n) – n \ &=(2 ^ {n 1} – 2) – n \ &=(2 ^ {n 1} – 1) – (n – 1). end {align *}
Now (2 ^ {n 1} – 1 ) is an ((n 1) ) – bit number with all its bits set to (1. ) Subtracting (n – 1 ) from it is equivalent to performing the following operation with their binary representations: for each (1 ) – bit in ((n – 1), ) set the corresponding bit in ( (2 ^ {n 1} – 1) ) to (0. )
If we use the notation ( text {bitcount} (n) ) to represent the number of (1 ) – bits in the binary representation of a positive integer (n, ) then we get [text{bitcount}(T_1 T_2 dots T_n)=(n 1) – text{bitcount}(n – 1).]
Now the computation involves counting the number of (1 ) – bits in (n – 1 ) which takes (O ( log n) ) and subtracting this count from (n 1 ) which also takes (O ( log n) ) time. Further, the largest numbers that we keep in memory are (n 1 ) and (n – 1 ) which occupy (O ( log n) ) space each. Therefore, the entire problem can be solved in (O ( log n) ) time with (O ( log n) ) space.
What would have taken over 2 exbibytes of memory with the naive approach requires a little over 8 bytes of memory now.
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