Rosetta Code

Mutual recursion: Difference between revisions

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Mutual recursion (view source)

Revision as of 21:54, 28 November 2014

39 bytes removed ,  9 years ago
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{{task|recursion}}
Two functions are said to be mutually recursive if the first calls the second, and in turn the second calls the first.
and in turn the second calls the first.
 
Write two mutually recursive functions that compute members of the [[wp:Hofstadter sequence#Hofstadter Female and Male sequences|Hofstadter Female and Male sequences]] defined as:
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</math>
 
<br>(If a language does not allow for a solution using mutually recursive functions then state this rather than give a solution by other means).
then state this rather than give a solution by other means).
 
=={{header|ACL2}}==
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new line(stand out)
)</lang>
{{out}}
Output:
<pre>
1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12
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DEF FNm(n%) IF n% = 0 THEN = 0 ELSE = n% - FNf(FNm(n% - 1))</lang>
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'''Output:'''
<pre>
F sequence:
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</lang>
 
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output
 
<lang>
> coffee mutual_recurse.coffee
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println 'm(0..20): ' + (0..20).collect { m(it) }</lang>
 
{{out}}
Output:
<pre>f(0..20): [1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12, 13]
m(0..20): [0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12, 12]</pre>
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}
print(out.F + "\n" + out.M);</lang>
{{out}}
outputs:
<pre>1,1,2,2,3,3,4,5,5,6,6,7,8,8,9,9,10,11,11,12
0,0,1,2,2,3,4,4,5,6,6,7,7,8,9,9,10,11,11,12</pre>
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<lang julia>F(n) = n < 1 ? one(n) : n - M(F(n - 1))
M(n) = n < 1 ? zero(n) : n - F(M(n - 1))</lang>
{{out}}
Output:
<pre>
julia> [F(i) for i = 0:19], [M(i) for i = 0:19]
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}
}</lang>
{{out}}
Output:
<pre>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
0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 16, 16, 17, 17, 18, 19, 19, 20, 20, 21, 22, 22, 23, 24, 24, 25, 25, 26, 27, 27, 28, 29, 29, 30, 30, 31, 32, 32, 33, 33, 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, 54, 55, 56, 56, 57, 58, 58, 59, 59, 60, 61, 61</pre>
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end</lang>
 
{{out}}
Sample Output:
<lang MATLAB>>> n = (0:10);
>> arrayfun(@female,n)
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TRAP 0,Halt,0</lang>
 
{{out}}
Output:
~/MIX/MMIX/Rosetta> mmix mutualrecurs1
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
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CloseConsole()
EndIf</lang>
{{out}}
Sample output:
<pre>1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12
0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12</pre>
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print ([ M(n) for n in range(20) ])</lang>
 
{{out}}
Output:
<pre>[1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12]
[0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12]</pre>
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print ["F:" mold fs crlf "M:" mold ms]</lang>
 
{{out}}
Output:
 
<pre>F: [1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12]
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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'''{{out}} using the default input of: &nbsp; <tt> 40 </tt>
<pre style="height:30ex">
F( 0) = 1 M( 0) = 0
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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)) /*right justifies # for nice look*/</lang>
'''output'''{{out}} using the default input of: &nbsp; <tt> 99 </tt>
<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
Line 1,846 ⟶ 1,847:
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)) /*right justifies # for nice look*/</lang>
'''output'''{{out}} using the input of: &nbsp; <tt> -70000 </tt>
<pre>
J(70000)= 70000
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M(70000)= 43262
</pre>
'''output'''{{out}} using the input of a ¼ million: &nbsp; <tt> -250000 </tt>
<pre>
J(250000)= 250000
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p (Array.new(20) {|n| M(n) })</lang>
 
{{out}}
Output:
<pre>[1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12]
[0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12]</pre>
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if n <> 0 then m = n - f(m(n - 1))
end function</lang>
{{out}}
Output:
<pre>F sequence:1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12 13
M sequence:0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12 12</pre>
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println!("")
}</lang>
{{out}}
Output:
<pre>1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12
0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12</pre>
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println((0 until 20).map(M).mkString(", "))</lang>
 
{{out}}
Output:
<pre>1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12
0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12</pre>
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end</lang>
 
{{out}}
Output:
<pre>M: 0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12 12 13 14 14 15 16 16
F: 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</pre>
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test = ^(F*,M*) iota 20</lang>
{{out}}
output:
<pre>
(
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]</lang>
 
{{out}}
Output:
<pre>
1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12
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