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Here is a suitable set of rules to solve the first part of the problem:
[ seq(n, c) := seq(n/2, c+1) :: n%2 = 0, seq(n, c) := seq(3n+1, c+1) :: n%2 = 1 :: n > 1 ] |
Given the initial formula `seq(6, 0)', application of these rules produces the following sequence of formulas:
seq( 3, 1) seq(10, 2) seq( 5, 3) seq(16, 4) seq( 8, 5) seq( 4, 6) seq( 2, 7) seq( 1, 8) |
whereupon neither of the rules match, and rewriting stops.
We can pretty this up a bit with a couple more rules:
[ seq(n) := seq(n, 0), seq(1, c) := c, ... ] |
Now, given `seq(6)' as the starting configuration, we get 8 as the result.
The change to return a vector is quite simple:
[ seq(n) := seq(n, []) :: integer(n) :: n > 0, seq(1, v) := v | 1, seq(n, v) := seq(n/2, v | n) :: n%2 = 0, seq(n, v) := seq(3n+1, v | n) :: n%2 = 1 ] |
Given `seq(6)', the result is `[6, 3, 10, 5, 16, 8, 4, 2, 1]'.
Notice that the n > 1 guard is no longer necessary on the last rule since the n = 1 case is now detected by another rule. But a guard has been added to the initial rule to make sure the initial value is suitable before the computation begins.
While still a good idea, this guard is not as vitally important as it
was for the fib
function, since calling, say, `seq(x, [])'
will not get into an infinite loop. Calc will not be able to prove
the symbol `x' is either even or odd, so none of the rules will
apply and the rewrites will stop right away.
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