Find longest word with letters

The longest word is a part of the Countdown TV program, whose purpose is to find the longest word by using only some selected letters (e.g. to rearrange letters in order to make a word from letters).

There are many letter games whose purpose is to make a word from letters (Scrabble, Wordox, Words with Friends, etc.). Most are similar to the longest word game, for example if the goal is to use all letters, it is an anagram.

In the original rules, a word list (dictionary reference) tells which word is an accepted solution or not (no proper noun). The program here is not limited and allows all kind of words, including conjugated verbs and sometimes some proper nouns.

In its original version, the player has to try to make an anagram of the letters, or remove some of them to get the longest/biggest word possible.

Example: ABCDEFGHIJ gives JIGHEAD (7 letters)

There are variants where letters can be used multiple times (repeating letters).

Example: ABCDEFGHIJ giving CHIFFCHAFF (10 letters)

It is also possible to search a word without scrambling the letters

Example: ABCDEFGHIJ allows A_C____HI_ (ACHI) (4 letters)

Finally, it is possible to mix the two options

Example: ABCDEFGHIJ gives BEEF (4 letters)

See also dCode solvers: Scrabble, Boggle, Words containing… etc.

In 1965, in a French TV Show by Armand Jammot, completed in 1972 by countdown numbers rounds.

The longest word varies according to the dictionary used:

— pneumonoultramicroscopicsilicovolcanoconiosis, but technical

— hippopotomonstrosesquipedaliophobia, a word that has been created to describe the fear of long words.

— antidisestablishmentarianism, found in all major dictionaries

I think this is related to the longest path (LP) problem that you mentioned, but it’s a little different. The primary difference is that the LP problem has a higher degree of connectivity than what your suggested problem does. By restricting your connections to the last and first letters, you remove a large number of potential combinations.

Here’s how I would recommend tackling this one:

1. For each word in the list, count the possible connections in and connections out.
2. Discard any words that have 0 ins and 0 outs.
3. Identify an initial set of «starter words» with the lowest numbers of ins and outs, and the outs must be greater than 0.
4. Each starter word receives its own working copy of the ins / outs connection count. This forms the head of the chain.
5. For each chain, identify a list of «next words» based upon:
   • last letter of starter or previous word
   • lowest number of of ins and outs connections (again, the outs must be greater than 0)
6. For each next word, repeat step 5 until the chain terminates.

Keep in mind that:

• You’ll need to keep track of the length of the chains and have some global mechanism to identify the longest chain.

• You’ll also need to remove each word from the working copy of the connection counts in order to avoid a recursive loop.

• At some point, your chain will terminate and you have to select a word with a 0 connection out count.

• You may have to recalculate ins / outs as words are removed from the working lists. At first glance, I don’t think this will be necessary as the overall sets will be relatively small. If you scaled out to 1000 words, then having static counts may slow down the algorithm from converging.

I kind of saw this as a packing problem. To me, the connections in and out identify the shape to be packed. The lower the connections, the more odd the shape. The more odd the shape, the sooner I want to pack it as I perceived having decreasing odds of being able to pack an odd shape the later I got into the chain.

As an example:

{dog, gopher, alpha, cube, elegant, this, that, bart}

dog     0, 1
gopher  1, 0
alpha   0, 0
cube    0, 1
elegant 1, 2
this    3, 0
that    2, 1
bart    0, 2

//alpha is dropped with 0 in and 0 out.
//two candidates found: dog, cube

//chain 1
dog => gopher
//chain 2
cube => elegant => that => this

//Note 1: the following chain won't occur due to selection rules
//that takes priority over this because of output count
cube => elegant => this

//Note 2: this chain won't occur either due to selection rules
bart => that => this

answered Aug 14, 2013 at 17:54

If you make 26X26 matrix to represent directed graph of vertex as each alphabet and words as edge.
For example word — APPLE connect vertex A and E with edge directed from A to E.
Now the problem reduces to finding largest Eulerian trail (path which includes maximum number of edges, visiting each edge once wit possible repetition of vertices) in the graph.
One of the O(E) algorithm would be to start randomly from a pair of vertices.
Find a path between them. Than keep relaxing the path till it is possible.

update
@GlenH7 I solved a similar question on www.hackerearth/jda recently, there was relative marks with respect to best solution and I scored the highest marks with the following approch-

Given list of words. Find the longest chain that can be formed by them. A chain is valid if every word begin with a letter *ending at the ending of last word.

Approch =

1)make the graph of alphabets as vertices and words as edges.
In place of using multiple edges use one with weight equal to
number of edges.

2)find the strongly connected component of graph with
maximum edges. Temporarily discard other edges.

3)For each vertex make its indegree equal to its outdegree.

4)Now their exists eulerian circuit in graph. Find it.

5)Now in remaining graph(w.r.t orignal graph find the longest
trail with first vertex in chosen strongly connected
component. I think this is NP hard.

6)Include the above trail in Elerian circuit converting eulerian
circuit into trail.

Why — I accept that this question is most probably NP hard(guess, not mathematically speaking). But the above approach works best when there is a long list(1000+) of uniformly distributed words(i.e. not intended to be w.c. for above approach).
Let us assume that after converting given list to graph mentioned above, it luckily turns out to be a eulerian graph(see http://en.wikipedia.org/wiki/Eulerian_path for conditions), then without any doubt we can say that answer to above question is P and is actually the eulerian path in the graph(see http://www.graph-magics.com/articles/euler.php for a very simple approch to do so and see this to verify that your graph has single http://www.geeksforgeeks.org/strongly-connected-components/ and if not temporarily clean other small scc because eulerian path exists for single scc). Thus for not lucky cases(which are almost all cases) I try to convert them to lucky cases(i.e eulerian trail condition are fulfilled). How to do this? I tried do increasing depth search for irrelevant edges(the set of edges in a path staring from vertex with outdegree greater than indegree and ending at vertex with indegree greater than outdegree). Increasing depth search means that first I searched for all such set of one edge in path than two edges in path and so on. It may seem at first look that ith depth search would take O(nodes^i) thus total time complexity of O(nodes + nodes^2 + nodes^3 + ….) till it is a lucky case. But amortized analysis will revel it is O(edges). Once it is reduced lucky case find eulerian circuit.

Till here it was all polynomial time. This would give almost the best solution. But to further increase your solution(perfect solution is NP hard) try some greedy approach in remaining graph to find a long trail staring with one of vertices in chosen scc. Now add this to above found eulerian trail to further increase it.

answered Jun 25, 2014 at 20:25

YSharpYSharp

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