How to resolve the algorithm 100 doors step by step in the Dc programming language
How to resolve the algorithm 100 doors step by step in the Dc programming language
Table of Contents
Problem Statement
There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm 100 doors step by step in the Dc programming language
Source code in the dc programming language
## NB: This code uses the dc command "r" via register "r".
## You may comment out the unwanted version.
[SxSyLxLy]sr # this should work with every "dc"
[r]sr # GNU dc can exchange top 2 stack values by "r"
## Now use "lrx" instead of "r" ...
0k # we work without decimal places
[q]sq # useful e.g. as loop termination
## (x)(y)>R == if (y)>(x) eval R
## isle x y --> (x <= y)
[
[1q]S. [ !<. 0 ]x s.L.
]sl
## l: isle
[
100 llx
]sL
## L: isle100
## for initcode condcode incrcode body
## [1] [2] [3] [4]
[
[q]S. 4:. 3:. 2:. 1:. 1;.x [2;.x 0=. 4;.x 3;.x 0;.x]d0:.x Os.L.o
]sf
## f: for
##----------------------------------------------------------------------------
## for( i=1 ; i<=100 ; ++i ) {
## door[i] = 0;
## }
#[init ...]P []ps-
[1si] [li lLx] [li1+si] [
li 0:d
]lfx
## for( s=1 ; s<=100 ; ++s ) {
## for( i=s ; i<=100 ; i+=s ) {
## door[i] = 1 - door[i]
## }
## }
[1ss] [ls lLx] [ls1+ss] [
#[step ]P lsn [ ...]ps-
[lssi] [li lLx] [lils+si] [
1 li;d - li:d
]lfx
]lfx
## long output:
## for( i=1 ; i<=100 ; ++i ) {
## print "door #", i, " is ", (door[i] ? "open" : "closed")), NL
## }
[
[1si] [li lLx] [li1+si] [
[door #]P
li n
[ is ]P
[closed]
[open]
li;d 0=r lrx s- n
[.]ps-
]lfx
]
## terse output:
## for( i=1 ; i<=100 ; ++i ) {
## if( door[i] ) {
## print i
## }
## print NL
## }
[
[1si] [li lLx] [li1+si] [
[] [ [ ]n lin ]
li;d 0=r lrx s- x
]lfx
[]ps-
]
lrx # comment out for the long output version
s- x
#[stack rest...]P []ps- f
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