Chopping words

One more challenge from Programming Praxis’ Word Games today (there are only a few left!). This time we have the challenge of cutting off bits of words, one letter at a time, such that each step is still a word.

The example given in their post is planet → plane → plan → pan → an → a, although surely many such examples exist.

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Dodgson’s Doublets

Today we have doublets source code, dictionary source code, queue source code. Using the same source code as the previous two posts (here and here, described originally here) for the dictionary, the code is a pretty straight forward case of using recursion to do backtracking. Basically, try all of the possible next words one letter different. Whenever you find a dead end, back up and try a different path. Something like this:

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Squaring the Bishop

Okay, this one was just neat. Based on word-squares source. I’ve only tested it in Racket 5.3+, but newer versions should work as well. Racket 5.2 won’t work without some tweaking as (at the very least) it’s missing a definition for string-trim.


Dictionary tries in Racket

For the next few posts, we’re going to need a way to represent a dictionary. You could go with just a flat list containing all of the words in the dictionary, but the runtime doesn’t seem optimal. Instead, we want a data structure that lets you easily get all possible words that start with a given prefix. We want a trie.

. Source: dictionary source code


Two Word Games

Another day, another post from Programming Praxis. Today they posted a word game that seems simple enough: first find all words in a given dictionary that contain all five vowels (a, e, i, o, u) in ascending order and then find any words (at least six letters long) where the letters are all in ascending alphabetical order.

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n-queens in 18 lines of code

One of the rites of passage for computer scientists it seems is to solve the Eight Queens Problem–where you must place 8 queens on a chessboard so that no pair of queens is attacking each other. Even better is when you can expand that to the n-queens problem with n queens on an n by n chessboard. After finding it again in older posts on both Programming Praxis and DataGenetics, I decided to go ahead and take a crack at it and I think the solution is pretty straight forward.

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A piece of the abc conjecture

There’s been a bit of hubbub in the in the math world the last few weeks with Shinichi Mochizuki's 500 page proof that of the ABC conjecture. Basically, the conjecture states that given three positive coprime integers a, b, and c such that a + b = c, the product of the distinct prime factors of a, b, and c is rarely much smaller than c. While this may sound strange, there are a number of interesting consequences that you can read about here.

To make a long story shorter, there was a challenge on Programming Praxis that intrigued me, which was to write code that given a upper bound on c would generate a list of all of the triples (a, b, c) such that the product is larger.

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A needle in a Pi-stack

Recently I’ve been watching a lot of find-in-pi source code (require racket/generator) (define (make-pi-spigot) (generator () (let loop ([q 1] [r 0] [t 1] [k 1] [n 3] [l 3]) (if (< (- (+ (* 4 q) r) t) (* n t)) (begin (yield n) (loop (* 10 q) (* 10 (- r (* n t))) t k (- (quotient (* 10 (+ (* 3 q) r)) t) (* 10 n)) l)) (loop (* q k) (* (+ (* 2 q) r) l) (* t l) (+ k 1) (quotient (+ (* q (+ (* 7 k) 2)) (* r l)) (* t l)) (+ l 2)))))) Simple enough to use, we can use Racket’s for to generate a list of n digits of pi or to convert the same to a string:

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