DouKaku?

import std.stdio;
import std.regexp;
import std.string;

void main(){
    string[] operators = ["and", "or", "not", "and"];
    string operatorsStr;  // 各演算子の頭文字からなる文字列 (e.g. "aona")
    foreach(operator; operators){
        operatorsStr ~= operator[0];
    }
    operatorsStr = operatorsStr.sub("nn", "", "g");  // 2 連続の not はないのと同じ

    int requiredOperandNum = operatorsStr.sub("n", "", "g").length + 1;
    for(int i = (1 << requiredOperandNum) - 1; i >= 0; i--){
        string operands = format("%0*b", requiredOperandNum, i);  // オペランドからなる文字列 (e.g. "1011")

        // 式を生成
        string expression;
        int j;
        foreach(operator; operatorsStr){
            if(operator == 'n'){
                expression ~= operator;
            }
            else{
                expression ~= [operands[j++], operator];
            }
        }
        expression ~= operands[j];  // e.g. "1a0on1a1"

        // eval
        expression = expression.sub("n0", "1", "g").sub("n1", "0", "g");  // 先に not を処理 (e.g. "1a0o0a1")
        while(expression.length > 1){  // 式左端の 2 項演算だけを順に
            expression = expression.sub("^1a1", "1").sub("^.a.", "0")
                                   .sub("^0o0", "0").sub("^.o.", "1");
        }  // e.g. "1a0o0a1"=>"0o0a1"=>"0a1"=>"0"

        if(expression == "1"){
            bool[] operandsArray;
            foreach(operand; operands){
                operandsArray ~= (operand == '1');
            }
            writefln(operandsArray);
        }
    }
}