This is the solution I came up with in Go for Set 1 / Challenge 3 of the
matasano crypto challenges.
This one was a little more difficult than the previous two, on the off-chance
that you’re reading this post shortly after it was published; I plan to come
back tomorrow and detail my thought process / how I came to a solution so stay
tuned for an update.
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package main
import hex "encoding/hex"
import "fmt"
import "sort"
func main() {
sourceString := "1b37373331363f78151b7f2b783431333d78397828372d363c78373" +
"e783a393b3736"
sourceHex, _ := hex.DecodeString(sourceString)
// This is an array of every 'character' that may be the key used to
// decode the source string
decoderArray := make([]byte, 94)
// This is the array where we'll store a 'score' for each key used
// in the previous array
scoreArray := make([]int, 94)
scoreMap := make(map[int]string)
for i, _ := range decoderArray {
finalHex := make([]byte, 0)
for _, v := range sourceHex {
finalHex = append(finalHex, byte(i + 32) ^ v)
}
finalString := string(finalHex)
scoreArray[i] = scoreString(finalString)
scoreMap[i] = finalString
}
sortedScoreArray := make([]int, 94)
copy(sortedScoreArray, scoreArray)
sort.Ints(sortedScoreArray)
sort.Sort(sort.Reverse(sort.IntSlice(sortedScoreArray)))
highestScore := sortedScoreArray[0]
var decoderIndex int
for i, v := range scoreArray {
if v == highestScore {
decoderIndex = i
}
}
fmt.Println(scoreMap[decoderIndex])
}
func scoreString(target string) int {
score := 0
for _, v := range target {
switch {
case byte(v) < 0x65:
score = score - 1
}
switch string(v) {
case "e":
score = score + 5
case "t":
score = score + 5
case "a":
score = score + 4
case "o":
score = score + 4
case "i":
score = score + 4
}
}
return score
}
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Running this will output the expected result, like so:
1
2
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Pu:~ nate$ go run Set1Challenge3.go
Cooking MC's like a pound of bacon
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http://cryptopals.com/sets/1/challenges/3/
http://en.wikipedia.org/wiki/Damerau–Levenshtein_distance
http://en.wikipedia.org/wiki/Letter_frequency
http://en.wikipedia.org/wiki/ASCII
