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Mutual recursion: Difference between revisions
m
→{{header|REXX}}: used clear version of parse (for arg), broke some long line up, added whitespace.
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m (→{{header|REXX}}: used clear version of parse (for arg), broke some long line up, added whitespace.) |
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Line 1,730:
This version uses vertical formatting.
<lang rexx>/*REXX program shows mutual recursion (via Hofstadter Male & Female seq)*/
parse arg lim .; if lim
do j=0 to lim;
say pad
end /*j*/
exit /*stick a fork in it, we're done.*/
/*─────────────────────────────────────F, M, Jw subroutines────────────*/
F: procedure; parse arg n; if n==0 then return 1; return n-M(F(n-1))
M: procedure; parse arg n; if n==0 then return 0; return n-F(M(n-1))
Jw: return right(arg(1),length(lim)) /*right justifies # for nice look*/
</lang>
'''output''' (using the default of 40):
<pre style="height:30ex
F( 0) = 1 M( 0) = 0
F( 1) = 1 M( 1) = 0
Line 1,788 ⟶ 1,789:
<br><br>The optimization due to memoization is faster by many orders of magnitude.
<lang rexx>/*REXX program shows mutual recursion (via Hofstadter Male & Female seq)*/
parse arg lim .; if lim=='' then lim=99
hm.=; hm.0=0; hf.=; hf.0=1; Js=; Fs=; Ms=
do j=0 to lim;
end /*j*/
say 'Js=' Js
say 'Fs=' Fs
say 'Ms=' Ms
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────one─liner subroutines──────────────────────────────*/
F: procedure expose hm. hf.; parse arg n; if hf.n=='' then hf.n=n-M(F(n-1)); return hf.n
M: procedure expose hm. hf.; parse arg n; if hm.n=='' then hm.n=n-F(M(n-1)); return hm.n
Jw: return right(arg(1),length(lim))
'''output''' (using the default of 99):
<pre>
Js= 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99
Fs= 1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12 13 13 14 14 15 16 16 17 17 18 19 19 20 21 21 22 22 23 24 24 25 25 26 27 27 28 29 29 30 30 31 32 32 33 34 34 35 35 36 37 37 38 38 39 40 40 41 42 42 43 43 44 45 45 46 46 47 48 48 49 50 50 51 51 52 53 53 54 55 55 56 56 57 58 58 59 59 60 61 61
Line 1,814 ⟶ 1,816:
/*If LIM is negative, only show a single result for the abs(lim) entry.*/
parse arg lim .;
parse var lim . hm. hf. Js Fs Ms; hm.0=0; hf.0=1
do j=0 to Alim;
end
if lim>0 then say 'Js=' Js; else say 'J('aLim")=" word(Js,aLim+1)
Line 1,825 ⟶ 1,827:
if lim>0 then say 'Ms=' Ms; else say 'M('aLim")=" word(Ms,aLim+1)
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────one─liner subroutines──────────────────────────────*/
F: procedure expose hm. hf.; parse arg n; if hf.n=='' then hf.n=n-M(F(n-1)); return hf.n
M: procedure expose hm. hf.; parse arg n; if hm.n=='' then hm.n=n-F(M(n-1)); return hm.n
Jw: return right(arg(1),length(lim))
'''output''' using the input of: <tt> -70000 </tt>
<pre>
J(70000)= 70000
F(70000)= 43262
Line 1,836 ⟶ 1,838:
</pre>
'''output''' using the input of a ¼ million: <tt> -250000 </tt>
<pre>
J(250000)= 250000
F(250000)= 154509
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