Shifting the stack as it is

Question

There are $3$ saucers. And there are $5$ disks, say A,B,C,D,E, having respective areas $1,2,3,4,5$. That is, E is the largest disk and A is the smallest one.

They are placed in the order E,D,C,B,A in the leftmost saucer, that is, largest at the bottom, smallest at the top. The rest two saucers are empty.

You are supposed to transfer the entire stack in the same order in the rightmost saucer. That is, in the end, the first two saucers will remain empty and the third saucer will contain the disks in the order E,D,C,B,A, that is, largest disk at the bottom and the smallest one on the top.

You are allowed to use the middle saucer temporarily during the transfers, you can move only one disk at a time, you cannot place a bigger disk on a smaller one.

What is the least possible ways in which you can do this transfer? Can you devise an algorithm for any number of disks?

Answer 1

This is the famous Tower of Hanoi problem. This describes the optimal algorithm:

For an even number of disks:

make the legal move between pegs A and B
make the legal move between pegs A and C
make the legal move between pegs B and C
repeat until complete

For an odd number of disks:

make the legal move between pegs A and C
make the legal move between pegs A and B
make the legal move between pegs C and B
repeat until complete

In each case, a total of $2^n-1$ moves are made.

Answer 2

As has been stated, it's the Tower of Hanoi puzzle - but personally I didn't find that description of how to solve it particularly useful. I must be reading it wrong because 'repeat until complete' immediately fails on the second step (which attempts to move A to B, specifically disk C onto the middle saucer which already has BA on it and isn't a legal move).

So anyway, here is the long hand. Saucers are labelled 1,2,3 from left to right:

1: EDCBA
2:
3:

1: EDCB
2:
3: A

1: EDC
2: B
3: A

1: EDC
2: BA
3: 

1: ED
2: BA
3: C

1: EDA
2: B
3: C

1: EDA
2:
3: CB

1: ED
2: 
3: CBA

1: E
2: D
3: CBA

1: E
2: DA
3: CB

1: EB
2: DA
3: C

1: EBA
2: D
3: C

1: EBA
2: DC
3: 

1: EB
2: DC
3: A

1: E
2: DCB
3: A

1: E
2: DCBA
3: 

1: 
2: DCBA
3: E

1: A
2: DCB
3: E

1: A
2: DC
3: EB

1: 
2: DC
3: EBA

1: C
2: D
3: EBA

1: C
2: DA
3: EB

1: CB
2: DA
3: E

1: CBA
2: D
3: E

1: CBA
2: 
3: ED

1: CB
2: 
3: EDA

1: C
2: B
3: EDA

1: C
2: BA
3: ED

1: 
2: BA
3: EDC

1: A
2: B
3: EDC

1: A
2: 
3: EDCB

1: 
2: 
3: EDCBA