可以暴力破解。为了提高效率，可以丢弃所有具有重复字母的单词，并对单词进行预处理以使用它们具有哪些字母的位掩码（有 26 个字母，所以这适合 32 位无符号整数）。
然后只需进行深度优先搜索，维护与目前找到的单词不相交的单词列表（位掩码）。
我已经编写了一些执行此操作的 go 代码。它使用仅包含解决方案单词的简短单词列表（完整的单词列表太长，无法在此处包含），但即使使用完整列表，代码也会在几秒钟内运行。
由于它使用位掩码来表示单词，因此解决方案中可能存在多个具有相同字母的单词。该程序显示了介于两者之间的|。解决方案中只有一对：cylix|xylic：
bling treck waqfs jumpy vozhd
pling treck waqfs jumby vozhd
brick glent waqfs jumpy vozhd
kreng clipt waqfs jumby vozhd
fjord chunk vibex gymps waltz
fjord gucks vibex nymph waltz
prick glent waqfs jumby vozhd
kempt brung waqfs cylix|xylic vozhd
blunk waqfs cimex grypt vozhd
clunk waqfs bemix grypt vozhd
可以在这里运行：https://go.dev/play/p/wVEDjx3G1fE
package main
import (
"fmt"
"math/bits"
"sort"
"strings"
)
var allWords = []string{
"bemix", "bling", "blunk", "brick", "brung", "chunk", "cimex", "clipt", "clunk", "cylix", "fjord", "glent", "grypt", "gucks", "gymps", "jumby", "jumpy", "kempt", "kreng", "nymph", "pling", "prick", "treck", "vibex", "vozhd", "waltz", "waqfs", "xylic",
}
func printSol(res []uint32, masks map[uint32][]string) {
var b strings.Builder
for i, r := range res {
if i > 0 {
b.WriteString(" ")
}
b.WriteString(strings.Join(masks[r], "|"))
}
fmt.Println(b.String())
}
func find5(w []uint32, mask uint32, n int, res []uint32, masks map[uint32][]string) {
if n == 5 {
printSol(res, masks)
return
}
sub := []uint32{}
for _, x := range w {
if x&mask != 0 {
continue
}
sub = append(sub, x)
}
for i, x := range sub {
res[n] = x
find5(sub[i+1:], mask|x, n+1, res, masks)
}
}
func find5clique() {
masks := map[uint32][]string{}
for _, x := range allWords {
m := uint32(0)
for _, c := range x {
m |= 1 << (c - 'a')
}
if bits.OnesCount32(m) == 5 {
masks[m] = append(masks[m], x)
}
}
maskSlice := []uint32{}
for m := range masks {
maskSlice = append(maskSlice, m)
}
sort.Slice(maskSlice, func(i, j int) bool {
return maskSlice[i] < maskSlice[j]
})
find5(maskSlice, uint32(0), 0, make([]uint32, 5, 5), masks)
}
func main() {
find5clique()
}