Towers of Hanoi: Difference between revisions
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Using it (5 is the number of disks.)
<lang joy>[source destination temp] 5 hanoi.</lang>
=={{header|jq}}==
{{works with|jq|1.4}}
The algorithm used here is used elsewhere on this page but it is worthwhile pointing out that it can also be read as a proof that:
(a) move(n;"A";"B";"C") will logically succeed for n>=0; and
(b) move(n;"A";"B";"C") will generate the sequence of moves, assuming sufficient computing resources.
The proof of (a) is by induction:
* As explained in the comments, the algorithm establishes that move(n;x;y;z) is possible for all n>=0 and distinct x,y,z if move(n-1;x;y;z)) is possible;
* Since move(0;x;y;z) evidently succeeds, (a) is established by induction.
The truth of (b) follows from the fact that the algorithm emits an instruction of what to do when moving a single disk.
<lang jq># n is the number of disks to move from From to To
def move(n; From; To; Via):
if n > 0 then
# move all but the largest at From to Via (according to the rules):
move(n-1; From; Via; To),
# ... so the largest disk at From is now free to move to its final destination:
"Move disk from \(From) to \(To)",
# Move the remaining disks at Via to To:
move(n-1; Via; To; From)
else empty
end;</lang>
'''Example''':
move(5; "A"; "B"; "C")
=={{header|K}}==
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