Towers of Hanoi: Difference between revisions
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m (Recursion moved to Towers of Hanoi: Towers of Hanoi is an excellent application of Recursion, but it does not necessarily exemplify it.) |
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return moves |
return moves |
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end hanoi |
end hanoi |
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==[[Java]]== |
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[[Category:Java]] |
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public void move(int n, int from, int to, int via) { |
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if (n == 1) { |
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System.out.println("Move disk from pole " + from + " to pole " + to); |
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} else { |
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move(n - 1, from, via, to); |
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move(1, from, to, via); |
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move(n - 1, via, to, from); |
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} |
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} |
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==[[Python]]== |
==[[Python]]== |
Revision as of 17:16, 30 January 2007
![Task](http://static.miraheze.org/rosettacodewiki/thumb/b/ba/Rcode-button-task-crushed.png/64px-Rcode-button-task-crushed.png)
You are encouraged to solve this task according to the task description, using any language you may know.
In this task, the goal is to solve the Towers of Hanoi problem with recursivity.
AppleScript
global moves --this is so the handler 'hanoi' can see the 'moves' variable set moves to "" hanoi(4, "peg A", "peg C", "peg B") on hanoi(ndisks, fromPeg, toPeg, withPeg) if ndisks is greater than 0 then hanoi(ndisks - 1, fromPeg, withPeg, toPeg) set moves to moves & "Move disk " & ndisks & " from " & fromPeg & " to " & toPeg & return hanoi(ndisks - 1, withPeg, toPeg, fromPeg) end if return moves end hanoi
Java
public void move(int n, int from, int to, int via) { if (n == 1) { System.out.println("Move disk from pole " + from + " to pole " + to); } else { move(n - 1, from, via, to); move(1, from, to, via); move(n - 1, via, to, from); } }
Python
def hanoi(ndisks, startPeg=1, endPeg=3): if ndisks: hanoi(ndisks-1, startPeg, 6-startPeg-endPeg) print "Move disk %d from peg %d to peg %d" % (ndisks, startPeg, endPeg) hanoi(ndisks-1, 6-startPeg-endPeg, endPeg) hanoi(ndisks=4)