九宫格密码有多少种组合.docx

九宫格密码有多少种组合九宫格密码有多少种组合通常设置密码至少4个点，最多9个点，规则通常是两点之间有一点，必须要过中间这个点，比如从1开始，必须要经过2才可以到3。1是可以直接到6的，但通常这种设置比较少。123456789运行附录的程序得到如下的数据：size: 4 count0: 144 count1: 96 count2:40size: 5 count0: 600 count1: 336 count2:152size: 6 count0: 2880 count1: 1344 count2:304size: 7 count0: 15120 count1: 4272 count2:496size: 8 count0: 80640 count1: 18432 count2:1024size: 9 count0: 362880 count1: 32256 count2:784sum: count0: 462264 count1: 56736 count2:2800use time: 453ms例如：4个点组合可能有144种可能，排除飞点的情况，剩下96种可能，如果不考虑跨点的情况，仅剩下40种可能。所以得出的结论是：4-9个点任意组合的九宫格有56736种组合，排除跨点的情况，有2800种可能。一般人设置在5-7个点，通常形状并不复杂的可能性仅为1000种左右，试出来的可能性还是很大的，而使用4位数字密码的组合则为10000种左右。源码附录：package string;import java.security.InvalidParameterException;public class HowMany public static void main(String args) String except1 = new String"46", "64", "28", "82", "19", "91", "37", "73","13", "31", "39", "93", "97", "79", "17", "71"String except2 = new String"16", "61", "18", "81", "34", "43", "38", "83","72", "27", "76", "67", "92", "29", "94", "49"long lastTime = System.currentTimeMillis;long sum0 = 0;long sum1 = 0;long sum2 = 0;for (int i = 4; i &lt= 9; i+) long count0 = 0;long count1 = 0;long count2 = 0;for (int j = 1; j &lt= 10 - i; j+) Combines num = new Combines(j, i);String str;boolean flag1 = false;boolean flag2 = false;do flag1 = false;flag2 = false;str = num.getOne;for (String except : except1) if (str.contains(except) flag1 = true;break;for (String except : except2) if (str.contains(except) flag2 = true;break;count0+;if (!flag1) count1+;if (!flag1 && !flag2) count2+;/System.out.println(str); while (num.moveNext);sum0 += count0;sum1 += count1;sum2 += count2;System.out.println("size: " + i + " count0: " + count0 + " count1: " + count1 + " count2:" + count2);System.out.println("sum: " + " count0: " + sum0 + " count1: " + sum1 + " count2:" + sum2);System.out.println("use time: " + (System.currentTimeMillis - lastTime) + "ms");private static class Combines private int base;private int number;private int poll;private long amount;private long count;public Combines(int start, int size) if (size &lt= 1) throw new InvalidParameterException;base = start;number = new intsize;poll = new intsize;amount = this.jieCheng(size);this.reset;public void reset initNumber;initPoll;count = 0;public String getOne StringBuffer strBuf = new StringBuffer;for (int i = 0; i &lt poll.length; i+) strBuf.append(getNumber(polli);return strBuf.toString;public boolean moveNext if (+count &lt amount) this.stepNext(1);this.initNumber;return true; else return false;private long jieCheng(int x) long y = 1;for (int i = 1; i &lt= x; i+) y = y * i;return y;private int getNumber(int index) int num;for (int i = 0; i &lt number.length; i+) if (numberi != 0) if (index = 0) num = numberi;numberi = 0;return num;index-;return 0;private void stepNext(int index) if (pollpoll.length - 1 - index = index) this.stepNext(index + 1);pollpoll.length - 1 - index = 0; else pollpoll.length - 1 - index+;return;private void initNumber for (int i = 0; i &lt number.length; i+) numberi = i + base;private void initPoll for (int i = 0; i &lt poll.length; i+) polli = 0;