Make one word into another - Word и Excel - помощь в работе с программами

How do I make one word into another?

Find and replace text

Go to Home > Replace or press Ctrl+H.
Enter the word or phrase you want to locate in the Find box.
Enter your new text in the Replace box.
Select Find Next until you come to the word you want to update.
Choose Replace. To update all instances at once, choose Replace All.

How do you solve a word ladder?

Puzzle: Word Ladders

Players get a starting word and an ending word.
Starting and ending words must be the same length (PIG and HOG, or CAT and DOG)
Players change one letter at a time, attempting to move from the starting word to the ending word.
Each intermediate step must be a valid word, and no proper nouns allowed!

What is the 9 letter word that still remains a word?

STARTLING

What is a word ladder example?

Replacing one letter at a time, the ladder for cat can become: cat – cot – dot – dog. This is a word ladder that starts at “cat” and ends at “dog.” For example, the word ladder for CAT can become: cat – scat – slat – slit – flit. This is a word ladder that starts with “cat” and ends with “flit.”

Who invented word ladders?

Lewis Carroll

What is meant by word ladder?

noun A kind of puzzle in which one word must be transformed into another specified word of the same length by changing one letter at a time, each step yielding a valid intermediate word, as in lead → load → goad → gold.

What are the 3 types of ladders?

Choosing the Right Ladder for the Job

Single Pole Ladders (maximum length 9 metres)
Extension Ladders (maximum length 15 metres)
Step Ladders (maximum height 6.1 metres)
Dual Purpose Ladders (stepladder hinged to provide an extension)
Platform (podium) Ladders.

What is ladder used for?

A ladder is a piece of equipment consisting of repeated bars or steps (rungs) between two upright lengths of metal, wood, or rope, used for climbing up or down something… So why are there so many different types of ladders?

What is another word for ladder?

In this page you can discover 18 synonyms, antonyms, idiomatic expressions, and related words for ladder, like: stair, step-stool, stairway, safety-rail, rung, belay, ledge, stepladder, handhold, steps and shot-line.

What is the opposite to the ladder?

What is the opposite of ladder?

closing	closure
join	juncture
perfection

What does it mean to ladder up?

First up is “ladder up,” a relatively new verb form. Those are two uses of the verb “ladder up.” The first means to refer to a higher level, where components need to relate to the “broader initiative.” In the second usage, “ladder up” means to climb the “career ladder,” a more familiar term.

What word rhymes with ladder?

Word	Rhyme rating	Meter
adder	100	[/x]
sadder	100	[/x]
madder	100	[/x]
gladder	100	[/x]

Does climb rhyme with time?

Word	Rhyme rating	Meter
time	100	[/]
crime	100	[/]
Prime	100	[/]
lime	100	[/]

What rhymes shouts?

Word	Rhyme rating	Categories
sprout	100	Verb, Noun
spout	100	Noun
pout	100	Noun, Verb
lout	100	Noun

What word rhymes with KISS?

Word	Rhyme rating	Categories
sis	100	Noun
hiss	100	Noun, Verb
amiss	100	Adverb
piss	100	Verb, Noun

What is the opposite of a kiss?

What are the antonyms for KISS? ram, knock, punch, whack.

What rhymes with kissing me?

Word	Rhyme rating	Categories
cris me	100	Phrase
dis me	100	Phrase
diss me	100	Phrase
dismiss me	100	Phrase

What rhymes with eternal life?

syllable: cruyff, fife, fyfe, fyffe, glyphe, greiff, knife, pfeife, pfeiff, phyfe, reife, rife, saif, schleif, schleife, slife, streife, streiff, strife, tgiff, wife.
syllables:
syllables:

What is another word for eternal life?

immortality	deathlessness
perpetuity	endlessness
eternity	indestructibility
timelessness	everlasting life
everlastingness	imperishability

What rhymes with the word New?

Word	Rhyme rating	♫
pursue	100	♫
flew	100	♫
jew	100	♫
screw	100	♫

Источник

dCode

Search for a tool

Word Mixer

Tools for mixing words (names, first names, etc.). The word mixer makes new or existing words (suitcase words, longest word, anagrams, etc.)

Results

Word Mixer —

Tag(s) : Fun/Miscellaneous, Word Games

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dCode is free and its tools are a valuable help in games, maths, geocaching, puzzles and problems to solve every day!
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Fun/Miscellaneous
Word Mixer

Word/Name Mixer

Answers to Questions (FAQ)

How to mix word or names? (Definition)

There are several ways to mix words (or more precisely from words’ letters).

Generating a contraction word (portmanteau)

Two words can be contracted/fused into one (which exists or not)

Example: BRITAIN+EXIT=BREXIT

This method is popular on social networks to fusion two words and create hashtags

Generating an anagram

Letters can be mixed/scrambled and swapped together to get 1 or more words (but sometimes none exists in the dictionary).

Example: DOG <=> GOD

This method can also generate pseudonyms.

Example: SALVADOR DALI <=> AVIDA DOLLARS

Combining only some letters

It is sometimes impossible to generate anagrams but using some of the letters may be enough (similar to the longest word problem).

Example: The letters TWO+WORDS can create the words ROOT, DOTS, etc.

This also works with first names:

Example: TWO+FIRSTNAMES can give SIMONETTA, RAMSES, STEFANO, etc.

Why mixing word or names?

Mixing words makes it possible to create new concepts, the generator/mixer brings new ideas combining words, surnames or first names that have a meaning, both in the mechanics of mixing/combining and in the result (the generated word can / must remain comprehensible)

How to shuffle letters in a word?

Source code

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Источник

I came across this variation of edit-distance problem:

Design an algorithm which transforms a source word to a target word. for example: from head to tail, in each step, you just can replace one character, and the word must be valid. You’ll be given a dictionary.

It clearly is a variation of the edit distance problem, but in edit distance I do not care about if the word is valid or not. So how do I add this requirement to edit distance.

asked Feb 5, 2010 at 7:02

This can be modelled as a graph problem. You can think of the words as nodes of the graph and two nodes are connected if and only if they are of same length and differ in one char.

You can preprocess the dictionary and create this graph, should look like:

   stack  jack
    |      |
    |      |
   smack  back -- pack -- pick

You can then have a mapping from the word to the node representing the word, for this you can use a hash table, height balanced BST …

Once you have the above mapping in place, all you have to do is see if there exists a path between the two graph nodes, which can easily be done using BFS or DFS.

So you can summarize the algorithm as:

preprocess the dictionary and create the graph.
Given the two inputs words w1 and w2
if length(w1) != length(w2)
 Not possible to convert
else
 n1 = get_node(w1)
 n2 = get_node(w2)

 if(path_exists(n1,n2))
   Possible and nodes in the path represent intermediary words
 else
   Not possible

answered Feb 5, 2010 at 7:05

codaddictcodaddict

442k81 gold badges490 silver badges528 bronze badges

codaddict’s graph approach is valid, though it takes O(n^2) time to build each graph, where n is the number of words of a given length. If that’s a problem, you can build a bk-tree much more efficiently, which makes it possible to find all words with a given edit distance (in this case, 1) of a target word.

answered Feb 8, 2010 at 10:38

Nick JohnsonNick Johnson

101k16 gold badges128 silver badges198 bronze badges

Create a graph with each node representing word in the dictionary. Add an edge between two word nodes, if their corresponding words are at edit distance of 1. Then minimum number of transformations required would length of shortest path between source node and destination node.

answered Apr 16, 2013 at 0:30

prasadvkprasadvk

1,5682 gold badges13 silver badges13 bronze badges

You could simply use recursive back-tracking but this is far from the most optimal solution.

# Given two words of equal length that are in a dictionary, write a method to transform one word into another word by changing only
# one letter at a time.  The new word you get in each step must be in the
# dictionary.

# def transform(english_words, start, end):

# transform(english_words, 'damp', 'like')
# ['damp', 'lamp', 'limp', 'lime', 'like']
# ['damp', 'camp', 'came', 'lame', 'lime', 'like']


def is_diff_one(str1, str2):
    if len(str1) != len(str2):
        return False

    count = 0
    for i in range(0, len(str1)):
        if str1[i] != str2[i]:
            count = count + 1

    if count == 1:
        return True

    return False


potential_ans = []


def transform(english_words, start, end, count):
    global potential_ans
    if count == 0:
        count = count + 1
        potential_ans = [start]

    if start == end:
        print potential_ans
        return potential_ans

    for w in english_words:
        if is_diff_one(w, start) and w not in potential_ans:
            potential_ans.append(w)
            transform(english_words, w, end, count)
            potential_ans[:-1]

    return None


english_words = set(['damp', 'camp', 'came', 'lame', 'lime', 'like'])
transform(english_words, 'damp', 'lame', 0)

answered Oct 16, 2015 at 21:12

I don’t think this is edit distance.

I think this can be done using a graph. Just construct a graph from your dictionary, and attempt to navigate using your favorite graph traversal algorithm to the destination.

answered Feb 5, 2010 at 7:05

YuliyYuliy

17.3k6 gold badges41 silver badges47 bronze badges

@Codeaddict solution is correct but it misses on the opportunity to simplify and optimize the solution.

DFS vs BFS:

If we go with DFS, there are chances that we meet target string (or to_string) far deeper in the graph. Then we have to keep track of the levels at which it is found and the reference to that node, and finally find the minimum level possible and then trace it from root.

For example, consider this conversion from -> zoom:

               from
             /         
        fram            foom
        /              /   
    dram    drom     [zoom] food       << To traverse upto this level is enough
 ...         |           ...      
            doom                  
             |       
           [zoom]

Using BFS, we can simplify this process by a great deal. All we need to do is :

Start with from string at level 0. Add this string to visitedSetOfStrings.
Add non-visited valid strings to next level which are at edit distance +1 from the strings of current level.
Add all these strings to visitedSetOfStrings.
If this set contains the target string, stop further processing of nodes/strings. Otherwise continue to step 2.

To make the path tracing easier, we can add extra information of parent string in each node.

answered Sep 11, 2020 at 7:21

Saurav SahuSaurav Sahu

12.7k5 gold badges60 silver badges78 bronze badges

This is C# code to solve the problem using BFS:

//use a hash set for a fast check if a word is already in the dictionary
    static HashSet<string> Dictionary = new HashSet<string>();
    //dictionary used to find the parent in every node in the graph and to avoid traversing an already traversed node
    static Dictionary<string, string> parents = new Dictionary<string, string>();

    public static List<string> FindPath(List<string> input, string start, string end)
    {
        char[] allcharacters = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

        foreach (string s in input)
            Dictionary.Add(s);
        List<string> currentFrontier = new List<string>();
        List<string> nextFrontier = new List<string>();
        currentFrontier.Add(start);
        while (currentFrontier.Count > 0)
        {
            foreach (string s in currentFrontier)
            {
                for (int i = 0; i < s.Length; i++)
                {
                    foreach (char c in allcharacters)
                    {
                        StringBuilder newWordBuilder = new StringBuilder(s);
                        newWordBuilder[i] = c;
                        string newWord = newWordBuilder.ToString();
                        if (Dictionary.Contains(newWord))
                        {
                            //avoid traversing a previously traversed node
                            if (!parents.Keys.Contains(newWord))
                            {
                                parents.Add(newWord.ToString(), s);
                                nextFrontier.Add(newWord);
                            }

                        }
                        if (newWord.ToString() == end)
                        {
                            return ExtractPath(start, end);

                        }
                    }
                }
            }
            currentFrontier.Clear();
            currentFrontier.Concat(nextFrontier);
            nextFrontier.Clear();
        }
        throw new ArgumentException("The given dictionary cannot be used to get a path from start to end");
    }

    private static List<string> ExtractPath(string start,string end)
    {
        List<string> path = new List<string>();
        string current = end;
        path.Add(end);
        while (current != start)
        {
            current = parents[current];
            path.Add(current);
        }
         path.Reverse();
         return path;
    }

answered Apr 10, 2014 at 20:18

I don’t think we need graph or some other complicated data structure. My idea is to load the dictionary as a HashSet and use contains() method to find out if the word exists in the dictionary or not.

Please, check this pseudocode to see my idea:

Two words are given: START and STOP. 
//List is our "way" from words START to STOP, so, we add the original word to it first.
    list.add(START);
//Finish to change the word when START equals STOP.
    while(!START.equals(STOP))
//Change each letter at START to the letter to STOP one by one and check if such word exists.
    for (int i = 0, i<STOP.length, i++){
        char temp = START[i];
        START[i] = STOP[i];
//If the word exists add a new word to the list of results. 
//And change another letter in the new word with the next pass of the loop.
        if dictionary.contains(START)
           list.add(START)
//If the word doesn't exist, leave it like it was and try to change another letter with the next pass of the loop.
        else START[i] = temp;}
    return list;

As I understand my code should work like that:

Input: DAMP, LIKE

Output: DAMP, LAMP, LIMP, LIME, LIKE

Input: BACK, PICK

Output: BACK, PACK, PICK

Regolith

2,9349 gold badges33 silver badges50 bronze badges

answered Dec 5, 2017 at 7:38

Boris Boris

3841 gold badge6 silver badges20 bronze badges

class Solution {
    //static int ans=Integer.MAX_VALUE;
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        HashMap<String,Integer> h=new HashMap<String,Integer>();
        HashMap<String,Integer> h1=new HashMap<String,Integer>();
        for(int i=0;i<wordList.size();i++)
        {
            h1.put(wordList.get(i),1);
        }
        int count=0;
        Queue<String> q=new LinkedList<String>();
        q.add(beginWord);
        q.add("-1");
        h.put(beginWord,1);
        int ans=ladderLengthUtil(beginWord,endWord,wordList,h,count,q,h1);
        return ans;
    }
    public int ladderLengthUtil(String beginWord, String endWord, List<String> wordList,HashMap<String,Integer> h,int count,Queue<String> q,HashMap<String,Integer> h1)
    {  
        int ans=1;
        while(!q.isEmpty()) 
        {
            String s=q.peek();
            q.poll();
            if(s.equals(endWord))
            {
                return ans;   
            }
            else if(s.equals("-1"))
            {
                if(q.isEmpty())
                {                    
                    break;
                }
                ans++;                
                q.add("-1");
            }
            else
            {
                for(int i=0;i<s.length();i++)
                {
                        for(int j=0;j<26;j++)
                        {
                            char a=(char)('a'+j);
                            String s1=s.substring(0,i)+a+s.substring(i+1);
                            //System.out.println("s1 is "+s1);
                            if(h1.containsKey(s1)&&!h.containsKey(s1))
                            {
                                h.put(s1,1);
                                q.add(s1);
                            }
                        }
                }
            }
        }
        return 0;    
    }
}

answered May 11, 2019 at 8:48

This is clearly a permutation problem. Using a graph is overkill. The problem statment is missing one important constraint; that you can change each position only once. This then makes it implicit that the solution is within 4 steps. Now all that needs to be decided is the sequence of the replace operations:

Operation1 = change «H» to «T»
Operation2 = change «E» to «A»
Operation3 = change «A» to «I»
Operation4 = change «D to «L»

The solution, the sequence of operations, is some permutation of the string «1234», where each digit represents the position of the character being replaced. e.g. «3124» indicates that first we apply operation3, then operation1, then operation2, then operation 4. At each step, if resulting word is not in dictionary, skip to next permutation. Reasonably trivial. Code anyone?

answered Feb 14, 2012 at 19:52

Источник

Combine up to 4 words into one unique word. You can make an unlimited amount of word combinations by putting in or taking out words.

Here are example combined words: Knuff + Archiater

Advertisements:

karchiater
kater
kchiater
ker
khiater
kiater
knarchiater
knater
knchiater
kner
knhiater
kniater
knr
knrchiater
knter
knuarchiater
knuater
knuchiater
knuer
knufarchiater
knufater
knufchiater
knufer
knuffarchiater
knuffater
knuffchiater
knuffer
knuffhiater
knuffiater
knuffr
knuffrchiater
knuffter
knufhiater
knufiater
knufr
knufrchiater
knufter
knuhiater
knuiater
knur
knurchiater
knuter
kr
krchiater
kter

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Источник

Anagram Solver is a tool used to help players rearrange letters to generate all the possible words from them. You input the letters, and Anagram Maker gives you the edge to win Scrabble, Words With Friends, or any other word game. No matter the length or difficulty of the word, Anagram Solver provides all available word options.

Anagrams — Definition and Examples

Have you ever heard of an anagram? Maybe you recognize the term, but you’re not exactly sure what it means. On the other hand, you might be an expert at using anagrams and have fun with them when playing various word games and board games.

What is an Anagram?

Anagrams are words or phrases you spell by rearranging the letters of another word or phrase. For instance, fans of the Harry Potter series know that Lord Voldemort’s full name is actually an anagram of his birth name, and some people even play games challenging one another to make anagrams still relevant to the original term. For example, «schoolmaster» can be turned into «the classroom», «punishments» becomes «nine thumps», and «debit card» turns into «bad credit».

The only rule is that all the letters from the original word or phrase must be used when they’re reordered to say something entirely different.

History of Anagrams

Historians suggest that anagrams actually originated in the 4th century BC, but weren’t commonly used until the 13th century AD when they were sometimes thought of as mystical. Imagine that!

20 Cool Anagram Examples

Whatever your level of knowledge, Word Finder can be a great tool to assist you to unscramble letters and identify anagrams when playing online and offline games. Here are some examples to help you become more familiar with anagrams ─ starting with the word “anagram” itself.

anagram = nag a ram
below = elbow
study = dusty
night = thing
act = cat
dessert = stressed
bad credit = debit card
gainly = laying
conversation = voice rants on
eleven plus two = twelve plus one
they see = the eyes
funeral = real fun
meteor = remote
the classroom = schoolmaster
meal for one = for me alone
sweep the floor = too few helpers
older and wiser = I learned words
video game = give a demo
coins kept = in pockets
young lady = an old guy

Anagram Solver for Scrabble, Words with Friends and Crosswords

How does anagramming help with word games? Easily, it forces you to start reimagining your tiles in a less confusing way. You’ll start looking at how to make any phrase or word instead of simply struggling with what appears on the board and the rack.

Some people are naturals at coming up with anagrams. However, it’s a rare person who can look at language and expertly rearrange the consonants and vowels to arrive at interesting or entertaining new compositions.

What is an Anagram Solver?

An anagram solver is a terrific tool that many people like to rely on to create different letter combinations.

How to Use an Anagram Solver in 3 Simple Steps

Step #1: Recognize prefixes and suffixes.

Following are common ones:

Some prefixes that start words ─ ab, ad, dis, de, ex, re, sub, un

Some suffixes that end words ─ ed, er, ing, ism, ly, ment, ness, tion

Step #2: Pick them out.
Step #3: Reorder the letters into new words.

Anagramming Example

One example is that the word “painter” could become “repaint” by moving the suffix to the beginning so that it becomes a prefix. Alternatively, the letters could be rearranged to make the word “pertain”.

Using Anagram Maker

Now, you may not see how anagramming can really help you win at games such as Scrabble or Words with Friends. However, just think about it for a moment. If you have the board in front of you, and it is loaded with an array of pre-existing words and open spaces, your strategy demands you consider the most lucrative moves. It is not just about making the longest word, but more about the words that give the most points. Anagram generators, like ours, give you solutions with anything from two to six or more letters. You can then use them to plug into the available spaces, finding the highest points possible.

Scrabble Anagram Maker

Seasoned Scrabble players will already know the value of using an anagram generator. After pulling seven tiles from the Scrabble bag and laying them out on their rack, the first player must use a sufficient number to make a complete word to get the game going. There can be a lot riding on this initial play. So, it’s not an uncommon practice for participants to take a little time moving the letters around to see what arrangement will give them the highest score. After all, if they can keep their early advantage, they may eventually win the game!

What’s more, as the game progresses, players will sometimes become stumped about how to display the tiles that they have on the board to gain the most points for the play. In short, having an anagram creator can assist Scrabble players to use their tile points to make words with the best possible score quickly so that the game remains exciting.

Words with Friends Anagram Finder

Similarly, an anagram word finder can be an invaluable device when enjoying Words with Friends. Faced with a jumble of letters, some players may be tempted to cheat or may try out words that they’re not very sure of. Would you believe that the English language has over 171,400 words? In addition, new words are added all the time. Therefore, it’s no wonder that game participants will sometimes become confused or perplexed when they’re attempting to solve multiple words and figure out where to make their next move.

Since Words with Friends is a digital game, you may be engaging with people anywhere in the world unless, of course, you choose to play solo. The game has the potential to be quite fast-paced, and you certainly don’t want to contemplate over your next move to slow things down ─ particularly when you may just be getting to know your opponent! This is where having a word anagram aid to use can be indispensable.

2 Tips to Solve Anagrams for Word Games Players

Are you ready for some final tips about solving anagrams? We’re sure that you can put the following information to good use!

Tip 1: Word Unscrambler

By employing Word Unscrambler, participants in word games are able to search for anagrams by entering the letters and wildcards that they have. Not only that, but they can use an advanced filter to discover words that start or end with particular letters and for other inquiries.

Here are a few examples:

The word “listen” is made up of letters EILNTS. When the word itself if entered in the Word Unscrambler, it quickly finds “silent”.

Along the same lines, “save”, comprised of AESV, reveals the word “vase” in the Word Unscrambler.

Tip 2: Phrase Unscrambler

When we study a phrase on its own, we can become quite stuck on its meaning and it can be difficult to see just how the words and letters can make something new. Hence, Phrase Unscrambler can be very valuable when players are looking to change the letters around in phrases to pinpoint anagrams.

Take a look at these examples:

“Dirty room” contains the following letters ─ DIMOORRTY. Putting the phrase into the Phrase Unscrambler uncovers the word “dormitory”.

By entering the phrase “moon starer” that has these letters ─ AEMNOORRST, the Phrase Unscrambler locates the word “Astronomer”.

Start playing with our anagram finder and discover the surprising number of options just a single collection of tiles can yield. Become an anagram creator today!

Источник

Everything You Need to Know about Word Unscramblers

Love playing Scrabble®? You know how difficult it is to find words among a bunch of letters. Sure, seeing vowels and consonants is everything some people need to win over any jumble.

However, figuring out a letter combination that forms an anagram isn’t a skill everyone possesses. If you’re one of those requiring word scramble help, I’ve got good news for you. It’s easy to figure out the missing word, even if you aren’t sure about it, especially if you are playing your favorite board game online.

You can discover new ways to make playing the game easy. Read on and discover your way to mastering any jumble.

What is a Word Unscramble Tool?

A word unscramble tool also goes by the name of «letter unscrambler» or «jumble solver.» It’s a tool that finds words hidden within jumbled letters.

An anagram solver lets you find all the words made from a list of letters presented in any order. You only need to locate the online tool and, in the search bar, enter any letters you can think of, including wild cards.

Many word solvers also let you choose a game dictionary. It gives you extra leeway to search with advanced options if you want to cheat with specific rules.

You don’t have to think of them as some unscramble cheat. Instead, using a scramble solver can help you study and practice your next Scrabble® or Words With Friends® match.

How to Unscramble Words and How to Use Advanced Options

Steps and Examples

The first thing you need to do is to find the best tool. Then, the steps are straightforward. Even more so, most tools follow the same steps; you’ll have a hard time getting lost with any scramble solver.

Step 1: Enter each of your current letter tiles in the search box. The maximum is fifteen. You can use two blank tiles («?» or SPACE).
Step 2: Hit the Search button. You will get to see different words coming up from the generator. Click on any word to see its definition.

Want to get even better at the popular word game? Alternatively, you can also use Advanced Options to add in more complexity to your favorite word game. So, you can decide what letter or letter pairs the word should start with, or the letter you will find at the end. A wildcard letter can generate many letter ideas.

You can also decide how many letters the word will contain, or the word pattern. For instance, you can search for high-scoring words that have the letter ‘n’ in a specific position. When you are done, all you need to do is hit the search button again.

Then, you can see the words database categorized by the number of letters.

Unscramble Words Methods

There are two approaches when it comes to word scramble help. Each method sets itself apart depending on how you’re solving the anagram.

1. Unscramble Letters

The first approach is to unscramble letter combinations to make words. This way tends to be the most commonly sought-after because it’s easier to score more points and win when you’re not focusing on a specific word.

When we talk about having to unscramble letters to make words, the possibilities are more extensive.

This word scramble help consists of what you learned earlier. The unscrambler tool receives combinations of letters and proceeds to unscramble them into different words.

If your objective is to rely less on that random wildcard and increase your vocabulary, this way is the best.

2. Unscramble Words

This type of word solver is much more restrictive. If you go with it, you’re choosing to unscramble jumbled words. It’s the closest you can get to a literal anagram.

To unscramble this anagram is much more difficult. You’re going after an individual result instead of many possibilities.

Online tools to unscramble jumbled words are usually more difficult to find. Often, the easiest way to unscramble a specific word with online help is to use filters. This way, you can limit the results and narrow them down to what you want.

Tips and Tricks to Unscramble Long Words

Words longer than five letters can be a nightmare. However, there are a few tips we can give you to make your life easier.

Tip 1: Focus on Syllables

Firstly, you can exploit the mighty syllable. People make words from syllables, not letters. You can merge vowels and consonants and form letter combinations (like suffixes and prefixes) that often go together. This way makes it easier to visualize possible words.

Tip 2: Vowels vs Consonants

Another way is to separate consonants and vowels. It often makes answers more noticeable than having everything jumbled.

Tip 3: Separate the Letter S

Lastly, the chances are that your language pluralizes words by adding an S in the end. If you’re playing Scrabble® and have a noisy S, taking up space, you probably can place it as adjacent letters at the end of your next word.

Most Popular Unscrambling Examples

There are ways to make the next puzzle game more exciting. Additionally, you can use these «rules» to focus on particular vocabularies you want to improve.

A. Three Word Finding Examples by Length

The first example is to unscramble anagrams into a set number of random letters using advanced options.

Make 7 letter words with these letters: AHSJFTSIKATL
Fajitas
Saltish
Khalifa
Make 6 letter words with these letters: OKLIYNCMZHOF
Colony
Flinch
Kimono
Make 5 letter words with these letters: MGJDUHSIAOET
Audio,
Amuse
Guest

B. Two Word Solving Examples by Topic

The other way to solve a letter scramble puzzle is to focus on a topic. You can choose specific categories for your anagram, or you can limit your jumble to a certain language like German or French to make things harder!

Find home utilities with these letters: KSIETNCHOFRK
Kitchen
Fork
Knife
Find food-related words with these letters: AJDOQIUESHNM
Quinoa
Queso
Squid

If you are looking to get better in the board game faster, this Word Unscrambler is the one you need to check out – for sure! For Crossword Puzzles lovers, we have a different tool. Try it here when you are stuck in solving any clue.

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Paste (Ctrl + V) your article below then click Submit to watch this article rewriter do it’s thing!

Select Language:

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There are four other answers here, and they all have very sensible advice. But none of them, as far as I can see, points out the terrible runtime complexity of the chosen algorithm, so I think an additional answer is necessary.

1. Bug

>>> english_words = set('bar bat cat war'.split())
>>> transform(english_words, 'bar', 'war', 0)
['bar', 'bat', 'cat', 'war']

As pointed out by Barry, ths is because the line

potential_ans[:-1]

is a mistake for:

potential_ans.pop()

In what follows, I’m going to assume that the bug is fixed.

2. Abstracting the problem

The problem being solved here can be thought of as a problem on a graph, in which each word is a node of the graph, and two words are connected if they differ by one letter. Supposing that the words are cake, came, camp, dame, damp, dike, dime, lake, lame, lamp, like, lime, and limp, then the graph looks like this:

A transformation of one word to another corresponds to a path in this graph. For example one possible transformation of cake to limp corresponds to the highlighted path here:

Finding the shortest transformation between two words corresponds to finding the shortest path between the two nodes in the graph.

There are several benefits of thinking about a problem in terms of abstract data structures like graphs, including:

There are standard terms for describing the problem: here nodes, edges, paths.
There are concrete representations of the data structure that support efficient operations.
You can look up the best known algorithms for solving the problem, instead of having to invent your own.

3. Description and complexity of algorithm

With the bug noted in §1 fixed, what your program does is to generate all the simple paths (paths with no repeated nodes) in the graph of words beginning at start, until it finds a path that leads to end.

But this can take a very very long time:

>>> from timeit import timeit
>>> english_words = set('bar bat cat eat fat hat mat oat pat rat sat tat vat war'.split())
>>> timeit(lambda:transform(english_words, 'bar', 'war', 0), number=1)
['bar', 'war']
2732.44891167

Why does the algorithm take three-quarters of an hour to find this result? Well, the graph being searched looks like this:

The algorithm starts at bar and keeps adding nodes to its path until it can’t go any further without repeating a node. For example (depending on the order in which the words are retrieved from the set), it might end up with this path:

After backtracking a couple of steps, it will find this path:

and so on. You’ll see that it will explore every path that starts bar—bat and then visits all the —at words. There are $ 11! = 39,916,800 $ such paths, and only when it has explored them all will it consider the path bar—war.

Checking to see if two words with $ m $ letters differ by one letter takes time $ O(m) $, so if there are $ n $ words of $ m $ letters each, the total runtime is $ O(n!m) $.

4. Better algorithm

It is possible to do much better than this if you use one of the standard graph search algorithms such as Dijkstra’s or A*. Here’s an implementation using the latter.

from collections import defaultdict, namedtuple
from heapq import heappush, heappop

class NotFound(Exception):
    pass

def word_ladder(words, start, end):
    """Return a word ladder (a list of words each of which differs from
    the last by one letter) linking start and end, using the given
    collection of words. Raise NotFound if there is no ladder.

    >>> words = 'card care cold cord core ward warm'.split()
    >>> ' '.join(word_ladder(words, 'cold', 'warm'))
    'cold cord card ward warm'

    """
    # Find the neighbourhood of each word.
    placeholder = object()
    matches = defaultdict(list)
    neighbours = defaultdict(list)
    for word in words:
        for i in range(len(word)):
            pattern = tuple(placeholder if i == j else c
                            for j, c in enumerate(word))
            m = matches[pattern]
            m.append(word)
            neighbours[word].append(m)

    # A* algorithm: see https://en.wikipedia.org/wiki/A*_search_algorithm

    # Admissible estimate of the steps to get from word to end.
    def h_score(word):
        return sum(a != b for a, b in zip(word, end))

    # Closed set: of words visited in the search.
    closed_set = set()

    # Open set: search nodes that have been found but not yet
    # processed. Accompanied by a min-heap of 4-tuples (f-score,
    # g-score, word, previous-node) so that we can efficiently find
    # the node with the smallest f-score.
    Node = namedtuple('Node', 'f g word previous')
    open_set = set([start])
    open_heap = [Node(h_score(start), 0, start, None)]
    while open_heap:
        node = heappop(open_heap)
        if node.word == end:
            result = []
            while node:
                result.append(node.word)
                node = node.previous
            return result[::-1]
        open_set.remove(node.word)
        closed_set.add(node.word)
        g = node.g + 1
        for neighbourhood in neighbours[node.word]:
            for w in neighbourhood:
                if w not in closed_set and w not in open_set:
                    next_node = Node(h_score(w) + g, g, w, node)
                    heappush(open_heap, next_node)
                    open_set.add(w)

    raise NotFound("No ladder from {} to {}".format(start, end))

Suppose that there are $ n $ words of $ m $ letters, and $ e $ edges in the graph of words. Then this algorithm takes $ O(m^2n) $ to find the edges and then $ O(e + mn + n log n) $ to find the shortest path.

So this finds shortest ladders on large sets of words in a reasonable amount of time:

>>> dictionary = [w.strip() for w in open('/usr/share/dict/words') if w == w.lower()]
>>> five_letter_words = [w for w in dictionary if len(w) == 5]
>>> len(five_letter_words)
8497
>>> from timeit import timeit
>>> timeit(lambda:print(' '.join(word_ladder(five_letter_words, 'above', 'below'))), number=1)
above amove amoke smoke smoky sooky booky booly bolly bally balli balai balao baloo balow below
0.3188382713124156

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