Grus
Table of Contents
Grus is a simple word unjumbler written in Go.
Project Home | Grus |
Source Code | Andinus / Grus |
GitHub (Mirror) | Grus - GitHub |
Tested on:
- OpenBSD 6.6 (with pledge & unveil)
Demo Video | System Information |
---|---|
Grus v0.2.0 | OpenBSD 6.6 (with pledge & unveil) |
Installation
Pre-built binaries
Pre-built binaries are available for OpenBSD, FreeBSD, NetBSD, DragonFly BSD, Linux & macOS.
This will just print the steps to install grus & you have to run those commands
manually. Piping directly to sh
is not a good idea, don't run this unless you
understand what you're doing.
v0.3.0
curl -s https://tildegit.org/andinus/grus/raw/tag/v0.3.0/scripts/install.sh | sh
v0.2.0
curl -s https://tildegit.org/andinus/grus/raw/tag/v0.2.0/scripts/install.sh | sh
Post install
You need to have a dictionary for grus to work, if you don't have one then you can download the Webster's Second International Dictionary, all 234,936 words worth. The 1934 copyright has lapsed.
curl -L -o /usr/local/share/dict/web2 \
https://archive.org/download/grus-v0.2.0/web2
There is also another big dictionary with around half a million english words. I'm not allowed to distribute it, you can get it directly from GitHub.
curl -o /usr/local/share/dict/words \
https://raw.githubusercontent.com/dwyl/english-words/master/words.txt
Documentation
Note: This is documentation for latest release, releases are tagged & so previous documentation can be checked by browsing source at tags.
Environment Variables
GRUS_SEARCH_ALL
Search in all dictionaries, by default Grus will exit after searching in first dictionary.
GRUS_ANAGRAMS
Prints all anagrams if set to true, by default Grus will print all anagrams.
GRUS_PRINT_PATH
Prints path to dictionary if set to true, this is set to false by default.
GRUS_STRICT_UNJUMBLE
Overrides everything & will try to print at least one match, if it doesn't find
any then it will exit the program with a non-zero exit code. This will ignore
GRUS_SEARCH_ALL
till it finds at least one match.
Default Dictionaries
These files will be checked by default (in order).
/usr/local/share/dict/words
/usr/local/share/dict/web2
/usr/share/dict/words
/usr/share/dict/web2
/usr/share/dict/special/4bsd
/usr/share/dict/special/math
Examples
# print grus version grus version # print grus env grus env # unjumble word grus word # don't print all anagrams GRUS_ANAGRAMS=false grus word # search for word in custom dictionaries too grus word /path/to/dict1 /path/to/dict2 # search for word in all dictionaries GRUS_SEARCH_ALL=true grus word /path/to/dict1 /path/to/dict2 # print path to dictionary GRUS_PRINT_PATH=1 grus word # find at least one match GRUS_STRICT_UNJUMBLE=1 grus word
History
Initial version of Grus was just a simple shell script that used the slowest method of unjumbling words, it checked every permutation of the word with all words in the file with same length.
Later I rewrote the above logic in python, I wanted to use a better method. Next version used logic similar to the current one. It still had to iterate through all the words in the file but it eliminated lots of cases very quickly so it was faster. It first used the length check then it used this little thing to match the words.
import collections match = lambda s1, s2: collections.Counter(s1) == collections.Counter(s2)
I don't understand how it works but it's fast, faster than convert the string to list & sorting the list. Actually I did that initially & you'll still find it in grus-add script.
lexical = ''.join(sorted(word)) if word == lexical: print(word)
This is equivalent to lexical.SlowSort in current version.
package lexical import ( "sort" "strings" ) // SlowSort returns string in lexical order. This function is slower // than Lexical. func SlowSort(word string) (sorted string) { // Convert word to a slice, sort the slice. t := strings.Split(word, "") sort.Strings(t) sorted = strings.Join(t, "") return }
Next version was also in python & it was stupid, for some reason using a database didn't cross my mind then. It sorted the word & then created a file with name as lexical order of that word (if word is "test" then filename would be "estt"), and it appended the word to that file.
It took user input & sorted the word, then it just had to print the file (if word is "test" then it had to print "estt"). This was a lot faster than iterating through all the words but we had to prepare the files before we could do this.
This was very stupid because the dictionary I was using had around 1/2 million words so this meant we got around half a million files, actually less than that because anagrams got appended into a single file but it was still a lot of small files. Handling that many small files is stupid.
I don't have previous versions of this program. I decided to rewrite this in Go,
this version does things differently & is faster than all previous versions.
Currently we first sort the word in lexical order, we do that by converting the
string to []rune
& sorting it, this is faster than lexical.SlowSort.
lexical.SlowSort converts the string to []string
& sorts it.
package lexical import "sort" // Sort takes a string as input and returns the lexical order. func Sort(word string) (sorted string) { // Convert the string to []rune. var r []rune for _, char := range word { r = append(r, char) } sort.Slice(r, func(i, j int) bool { return r[i] < r[j] }) sorted = string(r) return }
Instead of creating lots of small files, entries are stored in a sqlite3 database.
This was true till v0.1.0, v0.2.0 was rewritten & it dropped the use of database or any form of pre-parsing the dictionary. Instead it would look through each line of dictionary & unjumble the word, while this may be slower than previous version but this is simpler.