Grus

• Working
• Examples
• Installation
  • Pre-built binaries
    • v0.1.0
• History

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 unveil)

Working

• Grus takes a word as input from the user
• Input is ordered in lexical order
• Ordered input is searched in grus's database

It returns unjumbled word along with all the anagrams.

Examples

Run grus help to get the usage printed.

grus tinesl # will unjumble tinesl

You can also use the grus-add python script to add words to grus.

Installation

Pre-built binaries

Pre-built binaries are available for OpenBSD.

v0.1.0

curl -s \
     https://tildegit.org/andinus/grus/raw/tag/v0.1.0/scripts/install.sh | sh

# Download the initialization scripts.
curl -o grus-add \
     https://tildegit.org/andinus/grus/raw/tag/v0.1.0/scripts/grus-add

curl -o init \
     https://tildegit.org/andinus/grus/raw/tag/v0.1.0/scripts/init

# Initialize the database.
chmod +x init && \
    ./init

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.