How to resolve the algorithm Word ladder step by step in the Wren programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Word ladder step by step in the Wren programming language
Table of Contents
Problem Statement
Yet another shortest path problem. Given two words of equal length the task is to transpose the first into the second. Only one letter may be changed at a time and the change must result in a word in unixdict, the minimum number of intermediate words should be used. Demonstrate the following: A boy can be made into a man: boy -> bay -> ban -> man With a little more difficulty a girl can be made into a lady: girl -> gill -> gall -> gale -> gaze -> laze -> lazy -> lady A john can be made into a jane: john -> cohn -> conn -> cone -> cane -> jane A child can not be turned into an adult. Optional transpositions of your choice.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Word ladder step by step in the Wren programming language
Source code in the wren programming language
import "io" for File
import "/sort" for Find
var words = File.read("unixdict.txt").trim().split("\n")
var oneAway = Fn.new { |a, b|
var sum = 0
for (i in 0...a.count) if (a[i] != b[i]) sum = sum + 1
return sum == 1
}
var wordLadder = Fn.new { |a, b|
var l = a.count
var poss = words.where { |w| w.count == l }.toList
var todo = [[a]]
while (todo.count > 0) {
var curr = todo[0]
todo = todo[1..-1]
var next = poss.where { |w| oneAway.call(w, curr[-1]) }.toList
if (Find.first(next, b) != -1) {
curr.add(b)
System.print(curr.join(" -> "))
return
}
poss = poss.where { |p| !next.contains(p) }.toList
for (i in 0...next.count) {
var temp = curr.toList
temp.add(next[i])
todo.add(temp)
}
}
System.print("%(a) into %(b) cannot be done.")
}
var pairs = [
["boy", "man"],
["girl", "lady"],
["john", "jane"],
["child", "adult"]
]
for (pair in pairs) wordLadder.call(pair[0], pair[1])
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