Merge branch 'master' of git://factorcode.org/git/factor

db4
Slava Pestov 2008-03-07 21:30:21 -06:00
commit 948898193c
5 changed files with 198 additions and 7 deletions

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@ -1,6 +1,7 @@
! Copyright (c) 2007 Aaron Schaefer.
! See http://factorcode.org/license.txt for BSD license.
USING: ascii io.files kernel math project-euler.common sequences sorting splitting ;
USING: ascii io.encodings.ascii io.files kernel math project-euler.common
sequences sequences.lib sorting splitting ;
IN: project-euler.022
! http://projecteuler.net/index.php?section=problems&id=22
@ -28,10 +29,10 @@ IN: project-euler.022
: source-022 ( -- seq )
"extra/project-euler/022/names.txt" resource-path
file-contents [ quotable? ] subset "," split ;
ascii file-contents [ quotable? ] subset "," split ;
: name-scores ( seq -- seq )
dup length [ 1+ swap alpha-value * ] 2map ;
[ 1+ swap alpha-value * ] map-index ;
PRIVATE>

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@ -0,0 +1,96 @@
! Copyright (c) 2008 Aaron Schaefer.
! See http://factorcode.org/license.txt for BSD license.
USING: arrays combinators.lib kernel math math.primes math.primes.factors
math.ranges namespaces sequences ;
IN: project-euler.047
! http://projecteuler.net/index.php?section=problems&id=47
! DESCRIPTION
! -----------
! The first two consecutive numbers to have two distinct prime factors are:
! 14 = 2 * 7
! 15 = 3 * 5
! The first three consecutive numbers to have three distinct prime factors are:
! 644 = 2² * 7 * 23
! 645 = 3 * 5 * 43
! 646 = 2 * 17 * 19.
! Find the first four consecutive integers to have four distinct primes
! factors. What is the first of these numbers?
! SOLUTION
! --------
! Brute force, not sure why it's incredibly slow compared to other languages
<PRIVATE
: (consecutive) ( count goal test -- n )
pick pick = [
swap - nip
] [
dup prime? [ [ drop 0 ] dipd ] [
2dup unique-factors length = [ [ 1+ ] dipd ] [ [ drop 0 ] dipd ] if
] if 1+ (consecutive)
] if ;
: consecutive ( goal test -- n )
0 -rot (consecutive) ;
PRIVATE>
: euler047 ( -- answer )
4 646 consecutive ;
! [ euler047 ] time
! 542708 ms run / 60548 ms GC time
! ALTERNATE SOLUTIONS
! -------------------
! Use a sieve to generate prime factor counts up to an arbitrary limit, then
! look for a repetition of the specified number of factors.
<PRIVATE
SYMBOL: sieve
: initialize-sieve ( n -- )
0 <repetition> >array sieve set ;
: is-prime? ( index -- ? )
sieve get nth zero? ;
: multiples ( n -- seq )
sieve get length 1- over <range> ;
: increment-counts ( n -- )
multiples [ sieve get [ 1+ ] change-nth ] each ;
: prime-tau-upto ( limit -- seq )
dup initialize-sieve 2 swap [a,b) [
dup is-prime? [ increment-counts ] [ drop ] if
] each sieve get ;
: consecutive-under ( m limit -- n/f )
prime-tau-upto [ dup <repetition> ] dip start ;
PRIVATE>
: euler047a ( -- answer )
4 200000 consecutive-under ;
! [ euler047a ] 100 ave-time
! 503 ms run / 5 ms GC ave time - 100 trials
! TODO: I don't like that you have to specify the upper bound, maybe try making
! this lazy so it could also short-circuit when it finds the answer?
MAIN: euler047a

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! Copyright (c) 2008 Aaron Schaefer, Slava Pestov.
! See http://factorcode.org/license.txt for BSD license.
USING: arrays ascii assocs hashtables io.encodings.ascii io.files kernel math
math.parser namespaces sequences sequences.lib sequences.private sorting
splitting strings ;
IN: project-euler.059
! http://projecteuler.net/index.php?section=problems&id=59
! DESCRIPTION
! -----------
! Each character on a computer is assigned a unique code and the preferred
! standard is ASCII (American Standard Code for Information Interchange). For
! example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.
! A modern encryption method is to take a text file, convert the bytes to
! ASCII, then XOR each byte with a given value, taken from a secret key. The
! advantage with the XOR function is that using the same encryption key on the
! cipher text, restores the plain text; for example, 65 XOR 42 = 107, then 107
! XOR 42 = 65.
! For unbreakable encryption, the key is the same length as the plain text
! message, and the key is made up of random bytes. The user would keep the
! encrypted message and the encryption key in different locations, and without
! both "halves", it is impossible to decrypt the message.
! Unfortunately, this method is impractical for most users, so the modified
! method is to use a password as a key. If the password is shorter than the
! message, which is likely, the key is repeated cyclically throughout the
! message. The balance for this method is using a sufficiently long password
! key for security, but short enough to be memorable.
! Your task has been made easy, as the encryption key consists of three lower
! case characters. Using cipher1.txt (right click and 'Save Link/Target
! As...'), a file containing the encrypted ASCII codes, and the knowledge that
! the plain text must contain common English words, decrypt the message and
! find the sum of the ASCII values in the original text.
! SOLUTION
! --------
! Assume that the space character will be the most common, so XOR the input
! text with a space character then group the text into three "columns" since
! that's how long our key is. Then do frequency analysis on each column to
! find out what the most likely candidate is for the key.
! NOTE: This technique would probably not work well in all cases, but luckily
! it did for this particular problem.
<PRIVATE
: source-059 ( -- seq )
"extra/project-euler/059/cipher1.txt" resource-path
ascii file-contents [ blank? ] right-trim "," split
[ string>number ] map ;
TUPLE: rollover seq n ;
C: <rollover> rollover
M: rollover length rollover-n ;
M: rollover nth-unsafe rollover-seq [ length mod ] keep nth-unsafe ;
INSTANCE: rollover immutable-sequence
: decrypt ( seq key -- seq )
over length <rollover> swap [ bitxor ] 2map ;
: frequency-analysis ( seq -- seq )
dup prune [
[ 2dup [ = ] curry count 2array , ] each
] { } make nip ; inline
: most-frequent ( seq -- elt )
frequency-analysis sort-values keys peek ;
: crack-key ( seq key-length -- key )
[ " " decrypt ] dip group 1 head-slice*
flip [ most-frequent ] map ;
PRIVATE>
: euler059 ( -- answer )
source-059 dup 3 crack-key decrypt sum ;
! [ euler059 ] 100 ave-time
! 13 ms run / 0 ms GC ave time - 100 trials
MAIN: euler059

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@ -13,10 +13,11 @@ USING: definitions io io.files kernel math math.parser project-euler.ave-time
project-euler.033 project-euler.034 project-euler.035 project-euler.036
project-euler.037 project-euler.038 project-euler.039 project-euler.040
project-euler.041 project-euler.042 project-euler.043 project-euler.044
project-euler.045 project-euler.046 project-euler.048 project-euler.052
project-euler.053 project-euler.056 project-euler.067 project-euler.075
project-euler.079 project-euler.092 project-euler.097 project-euler.134
project-euler.169 project-euler.173 project-euler.175 ;
project-euler.045 project-euler.046 project-euler.047 project-euler.048
project-euler.052 project-euler.053 project-euler.056 project-euler.059
project-euler.067 project-euler.075 project-euler.079 project-euler.092
project-euler.097 project-euler.134 project-euler.169 project-euler.173
project-euler.175 ;
IN: project-euler
<PRIVATE