bash – 减少排列

#!/bin/bash

pool=( {a..d} )
for((i=0;i<${#pool[@]}-1;++i)); do
    for((j=i+1;j<${#pool[@]};++j)); do
        printf '%sn' "${pool[i]}${pool[j]}"
    done
 done

另一个：

#!/bin/bash

pool=( {a..d} )
while ((${#pool[@]}>1)); do
    h=${pool[0]}
    pool=("${pool[@]:1}")
    printf '%sn' "${pool[@]/#/$h}"
done

它们可以写成函数(或脚本)：

get_perms_ordered() {
    local i j
    for((i=1;i<"$#";++i)); do
        for((j=i+1;j<="$#";++j)); do
            printf '%sn' "${!i}${!j}"
        done
     done
}

要么

get_perms_ordered() {
    local h
    while (("$#">1)); do
        h=$1; shift
        printf '%sn' "${@/#/$h}"
    done
}

用于：

$get_perms_ordered {a..d}
ab
ac
ad
bc
bd
cd

最后一个可以很容易地转换成递归函数以获得给定长度的有序排列(没有替换 – 我使用愚蠢的球 – 瓮概率词汇表),例如,

get_withdraws_without_replacement() {
    # $1=number of balls to withdraw
    # $2,... are the ball "colors"
    # return is in array gwwr_ret
    local n=$1 h r=()
    shift
    ((n>0)) || return
    ((n==1)) && { gwwr_ret=( "$@" ); return; }
    while (("$#">=n)); do
        h=$1; shift
        get_withdraws_without_replacement "$((n-1))" "$@"
        r+=( "${gwwr_ret[@]/#/$h}" )
    done
    gwwr_ret=( "${r[@]}" )
}

然后：

$get_withdraws_without_replacement 3 {a..d}
$echo "${gwwr_ret[@]}"
abc abd acd bcd