(a) Tower of Hanoi: Assuming you have n bricks and 3 towers denoted by 0,1,2. Denote the current state by a n trinary numbers, for example (in the case n=9): 987654321 001102020 (current state) meaning that brick 9,8,5,3 and 1 are in the 0:th tower. Brick 7 and 6 in the 1:th tower and brick 4 and 2 in the 2nd tower. This would give you a state space of size 3^n, which is not too large. (This is only a partial answer. But every state-string will correspond to a legal state. That is to say, in each tower the size of the bricks will decrease from bottom to top,

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Answered 8 days After Sep 12, 2021

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Swapnil answered on Sep 20 2021

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(a) Tower of Hanoi:
    Assuming you have n
icks and 3 towers denoted by 0,1,2. Denote the cu
ent state by a n trinary numbers, for example (in the case n=9):
987654321
001102020 (cu
ent state)
meaning that
ick 9,8,5,3 and 1 are in the 0:th tower. Brick 7 and 6 in the 1:th tower and
ick 4 and 2 in the 2nd tower.
This would give you a state space of size 3^n, which is not too large.
(This is only a partial answer. But every state-string will co
espond to a legal state. That is to say,
in each tower the size of the
icks will decrease from bottom to top,
no
ick will appear in two different towers.
I therefore think that the suggested state space is minimal.
    (b) Easy-wristband:
    Initially there are 4 pairs of chains that are locked to each other. Every chains that are locked to each other. Every chains are identical. The goal is to lock and unlock chains such that it can form a wristband (shown in the right figure of Fig with minimal costs. It costs 30 cent to unlock and 50 cents to lock a pair of chains.
    (C) TSP
    Travelling Salesman Problem (TSP): Suppose the given set of the cities and the distance in between each pair of the cities and we have to return the starting point. Basically we need to find the shortest path from all nodes.
So the given set of the vertices {1,2,3,4…n} can be considered to the 1 as the starting point and the end point of the output. So every other vertex excluding 1 we need to find the minimum cost path with the 1 as the starting point. And i is the ending point for the all other vertices that will be appeared exact once.
Basically the cost of this path can be uses the co
esponding cycle that would be cost (i) + distance (i, 1) where we can get the distance (i, 1) is the distance from i to 1. Finally, we can return the minimum of all [cost(i) + dist (i, 1)] values.
    (d) Jumping-Counters:
    The jumping counters are one of the admissible heuristic and that is...

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