Sum and product puzzle: Difference between revisions
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show=. 'X=' , ": o X ,' Y=' , ": o Y , ' X+Y=' , ": o (X+Y) , ' X*Y=' , ": o (X*Y)
solve=. show :: (''"_) o filter3 o filter2 o filter1 o (] filter0 decompositions) f.</syntaxhighlight>
Example use:
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X=4 Y=13 X+Y=17 X*Y=52
solve
X=4 Y=13 X+Y=17 X*Y=52▼
solve 1685
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17567 candidates
2 candidates
X=4 4 Y=13 61 X+Y=17 65 X*Y=52 244
solve 1970
967272 candidates
70475 candidates
23985 candidates
3 candidates
▲X=4 4 16 Y=13 61 73 X+Y=17 65 89 X*Y=52 244 1168
solve 2522
1586340 candidates
116238 candidates
37748 candidates
4 candidates
X=4 4 16 16 Y=13 61 73 111 X+Y=17 65 89 127 X*Y=52 244 1168 1776</syntaxhighlight>
The code is tacit and fixed (in other words, it is point-free):
<syntaxhighlight lang="j"> _80 [\ (5!:5)<'solve'
('X=' , ":@:(0&{"1) , ' Y=' , ":@:(1&{"1) , ' X+Y=' , ":@:(0&{"1 + 1&{"1) , ' X*
Y=' , ":@:(0&{"1 * 1&{"1)) ::(''"_)@:(([ 0 0&$@(1!:2&2)@:(' candidates' ,~ ":@:#
))@:;@:(((0&{"1 + 1&{"1) </. ]) #~ 1 = #&>@:((0&{"1 + 1&{"1) </. ])))@:(([ 0 0&$
@(1!:2&2)@:(' candidates' ,~ ":@:#))@:;@:(((0&{"1 * 1&{"1) </. ]) #~ 1 = #&>@:((
0&{"1 * 1&{"1) </. ])))@:(([ 0 0&$@(1!:2&2)@:(' candidates' ,~ ":@:#))@:((((0&{"
1 + 1&{"1) <@:(0&{"1 * 1&{"1)/. ]) ('' -: -.)&>"0 _ <@:((0&{"1 * 1&{"1)@:;@:(((0
&{"1 * 1&{"1) </. ]) #~ 1 < (0&{"1 * 1&{"1) #/. ]))) ;@:# (0&{"1 + 1&{"1) </. ])
)@:(] ([ 0 0&$@(1!:2&2)@:(' candidates' ,~ ":@:#))@:((([ >: (0&{"1 + 1&{"1)@:])
*. ((1 < 0&{"1) *. (1 < 1&{"1) *. 0&{"1 < 1&{"1)@:]) # ]) >@:,@:{@:(;~)@:i.)</syntaxhighlight>
=={{header|Java}}==
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