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Wordle: prefer common words via DreamDict frequency batch lookup

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Gather up to 5 candidate words from DreamDict RandomWord, then use
FrequencyBatch to pick the most commonly used one. Players get
recognizable everyday words instead of obscure ones that just clear
the min_freq floor.

Co-Authored-By: Claude Opus 4.6 <noreply@anthropic.com>
This commit is contained in:
prosolis
2026-04-10 17:15:16 -07:00
parent 76110f61ca
commit 08e9925d8f
1 changed files with 30 additions and 7 deletions
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37
internal/plugin/wordle.go
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@@ -397,22 +397,23 @@ func (p *WordlePlugin) createAndPostPuzzle(roomID id.RoomID, wordLength int, cat
return p.SendMessage(roomID, renderWordleStartAnnouncement(puzzleNumber, wordLength, hint))
}
// pickWord tries DreamDict's RandomWord API first, falling back to the local
// word pool if DreamDict is unavailable or only returns recently-used words.
// pickWord tries DreamDict's RandomWord API first, gathering multiple candidates
// and selecting the most common one by frequency. Falls back to the local word
// pool if DreamDict is unavailable.
func (p *WordlePlugin) pickWord(length int, category WordleCategory) string {
if p.dict != nil {
recent := loadRecentWordleAnswers(500)
lang := categoryLang(category)
// Try a few times to get a word that hasn't been used recently.
for range 10 {
// Gather candidate words, then pick the highest-frequency one.
var candidates []string
for range 15 {
word, err := p.dict.RandomWord(lang, "", length, length, 500)
if err != nil {
slog.Warn("wordle: DreamDict random word failed, falling back", "err", err)
break
}
word = strings.ToUpper(word)
// Verify the word is exactly the right length and contains only letters.
runes := []rune(word)
if len(runes) != length {
continue
@@ -427,9 +428,31 @@ func (p *WordlePlugin) pickWord(length int, category WordleCategory) string {
if !clean {
continue
}
if !recent[word] {
return word
if recent[word] {
continue
}
candidates = append(candidates, word)
if len(candidates) >= 5 {
break
}
}
if len(candidates) > 0 {
// Look up frequencies in batch and pick the most common word.
freqs, err := p.dict.FrequencyBatch(candidates, lang)
if err == nil && len(freqs) > 0 {
best := candidates[0]
bestFreq := 0
for _, w := range candidates {
if f, ok := freqs[strings.ToLower(w)]; ok && f > bestFreq {
best = w
bestFreq = f
}
}
return best
}
// FrequencyBatch failed — just use the first candidate.
return candidates[0]
}
}