Dinesman's multiple-dwelling problem: Difference between revisions

===== Using Permutation Step Function =====

===== Using Amp function =====

<syntaxhighlight lang="m2000 interpreter">

Module Using_AmbFunction {

Enum Solution {First, Any=-1}

Function Amb(way as Solution, failure) {

read a

c1=lambda i=0, a, (&any, &ret) ->{

any=(array(a,i),)

ret=any

i++

ok=i=len(a)

if ok then i=0

=ok

}

m=stack.size

if m=0 then Error "At least two arrays needed"

c=c1

while m>0 {

read a

c1=lambda c2=c, i=0, a, (&any, &ret) ->{

any=(array(a,i),)

ret=(,) : ok=false : anyother=(,)

ok=c2(&anyother, &ret)

ret=cons(ret, any)

if ok then i++

ok=i=len(a)

if ok then i=0

=ok

}

c=c1 : m--

}

ok=false

any=(,)

flush

while not ok

ret=(,)

ok=c(&any, &ret)

s=stack(ret)

if not failure(! s) then data ret : if way>0 then ok=true

End While

if empty then

ret=(("",),)

else

ret=array([])

end if

=ret

}

Range=lambda (a, f) ->{

for i=a to f-1: data i: next

=array([])

}

Baker=range(1, 5)

Cooper=range(2, 6)

Fletcher=range(2, 5)

Miller=range(1,6)

Smith=range(1,6)

failure=lambda (Baker, Cooper, Fletcher, Miller, Smith)->{

if Baker=Cooper or Baker=Fletcher or Baker=Miller or Baker=Smith then =true:exit

if Cooper=Fletcher or Cooper =Miller or Cooper=Smith then =true:exit

if Fletcher=Miller or Fletcher=Smith or Miller=Smith then =true:exit

if Miller<Cooper or abs(Cooper-Fletcher)=1 or abs(Smith-Fletcher)=1 then =true:exit

}

all=amb(Any, failure, Baker, Cooper, Fletcher, Miller, Smith)

k=each(all)

s=("Baker", "Cooper", "Fletcher", "Miller", "Smith")

while k

z=array(k)

zz=each(z, , -2)

while zz

? s#val$(zz^)+" ("+array(zz)+"), ";

end while

zz=each(z, -1)

while zz

? s#val$(zz^)+" ("+array(zz)+") "

end while

end while

}

Using_AmbFunction

</syntaxhighlight>

{{out}}

<pre>

Baker (3), Cooper (2), Fletcher (4), Miller (5), Smith (1)

</pre>