Set consolidation: Difference between revisions
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:Given the two sets <tt>{A,B}</tt> and <tt>{B,D}</tt> then there is a common element <tt>B</tt> between the sets and the result is the single set <tt>{B,D,A}</tt>. (Note that order of items in a set is immaterial: <tt>{A,B,D}</tt> is the same as <tt>{B,D,A}</tt> and <tt>{D,A,B}</tt>, etc). |
:Given the two sets <tt>{A,B}</tt> and <tt>{B,D}</tt> then there is a common element <tt>B</tt> between the sets and the result is the single set <tt>{B,D,A}</tt>. (Note that order of items in a set is immaterial: <tt>{A,B,D}</tt> is the same as <tt>{B,D,A}</tt> and <tt>{D,A,B}</tt>, etc). |
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;'''Example 3:''' |
;'''Example 3:''' |
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:Given the three sets <tt>{A,B}</tt> and <tt>{C,D}</tt> and <tt>{D,B}</tt> then there is no common element between the sets <tt>{A,B}</tt> and <tt>{C,D}</tt> but the sets <tt>{A,B}</tt> and <tt>{D,B}</tt> do share a common element that consolidates to produce the result <tt>{B,D,A}</tt>. |
:Given the three sets <tt>{A,B}</tt> and <tt>{C,D}</tt> and <tt>{D,B}</tt> then there is no common element between the sets <tt>{A,B}</tt> and <tt>{C,D}</tt> but the sets <tt>{A,B}</tt> and <tt>{D,B}</tt> do share a common element that consolidates to produce the result <tt>{B,D,A}</tt>. |
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On examining this result with the remaining set, <tt>{C,D}</tt>, they share a common element and so consolidate to the final output of the single set <tt>{A,B,C,D}</tt> |
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;'''Example 4:''' |
;'''Example 4:''' |
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:The consolidation of the five sets: |
:The consolidation of the five sets: |
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end Set_Consolidation;</lang> |
end Set_Consolidation;</lang> |
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{{out}} |
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This generates the following output: |
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<pre>{A,B}{C,D} |
<pre>{A,B}{C,D} |
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{A,B,D} |
{A,B,D} |
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& test$(H+I+K A+B C+D D+B F+G+H) |
& test$(H+I+K A+B C+D D+B F+G+H) |
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);</lang> |
);</lang> |
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{{out}} |
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Output: |
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<pre>A+B C+D ==> A+B C+D |
<pre>A+B C+D ==> A+B C+D |
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A+B B+D ==> A+B+D |
A+B B+D ==> A+B+D |
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(test (consolidate {{'H' 'I' 'K'} {'A' 'B'} {'C' 'D'} {'D' 'B'} {'F' 'G' 'H'}})) |
(test (consolidate {{'H' 'I' 'K'} {'A' 'B'} {'C' 'D'} {'D' 'B'} {'F' 'G' 'H'}})) |
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</lang> |
</lang> |
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{{out}} |
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'''Output:''' |
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<lang egison> |
<lang egison> |
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{"DBAC" "HIKFG"} |
{"DBAC" "HIKFG"} |
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] |
] |
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0</lang> |
0</lang> |
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{{out}} |
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Output |
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<pre>seq [set ["C"; "D"]; set ["A"; "B"]] |
<pre>seq [set ["C"; "D"]; set ["A"; "B"]] |
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seq [set ["A"; "B"; "C"]] |
seq [set ["A"; "B"; "C"]] |
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;;</lang> |
;;</lang> |
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{{out}} |
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Output: |
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<pre>{ {A B} {C D} } |
<pre>{ {A B} {C D} } |
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{ {A B C} } |
{ {A B C} } |
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End |
End |
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Return strip(ol)</lang> |
Return strip(ol)</lang> |
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{{out}} |
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Output: |
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<pre> |
<pre> |
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Input 1 (B,A) (C,D) |
Input 1 (B,A) (C,D) |
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(consolidate (list (set 'h 'i 'k) (set 'a 'b) (set 'c 'd) (set 'd 'b) (set 'f 'g 'h))) |
(consolidate (list (set 'h 'i 'k) (set 'a 'b) (set 'c 'd) (set 'd 'b) (set 'f 'g 'h))) |
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</lang> |
</lang> |
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{{out}} |
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Output: |
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<lang racket> |
<lang racket> |
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(list (set 'b 'a) (set 'd 'c)) |
(list (set 'b 'a) (set 'd 'c)) |
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/*──────────────────────────────────iSIN subroutine─────────────────────*/ |
/*──────────────────────────────────iSIN subroutine─────────────────────*/ |
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isIn: return wordpos(arg(1), arg(2))\==0 /*is (word) arg1 in set arg2? */</lang> |
isIn: return wordpos(arg(1), arg(2))\==0 /*is (word) arg1 in set arg2? */</lang> |
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{{out}} when using the default supplied sample sets: |
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<pre> |
<pre> |
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the old sets= {A,B} {C,D} |
the old sets= {A,B} {C,D} |
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(format t "~s -> ~s\n" test (consoli test))))</lang> |
(format t "~s -> ~s\n" test (consoli test))))</lang> |
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{{out}} |
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Output: |
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<pre>((a b) (c d)) -> ((d c) (b a)) |
<pre>((a b) (c d)) -> ((d c) (b a)) |
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((a b) (b d)) -> ((d b a)) |
((a b) (b d)) -> ((d b a)) |
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[mapcar hash-keys (consoli [mapcar mkset test])])))</lang> |
[mapcar hash-keys (consoli [mapcar mkset test])])))</lang> |
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{{out}} |
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Output: |
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<pre>((a b) (c d)) -> ((b a) (d c)) |
<pre>((a b) (c d)) -> ((b a) (d c)) |
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((a b) (b d)) -> ((b a d)) |
((a b) (b d)) -> ((b a d)) |
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