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/*
* Copyright 2018 the original author or authors.
* Licensed under the Apache License, Version 2.0 (the "License"); you may not
* use this file except in compliance with the License. You may obtain a copy of
* the License at http://www.apache.org/licenses/LICENSE-2.0 Unless required by
* applicable law or agreed to in writing, software distributed under the
* License is distributed on an "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS
* OF ANY KIND, either express or implied. See the License for the specific
* language governing permissions and limitations under the License.
*/
package com.gitee.drinkjava2.frog.temp;
/**
*
* 临时类，待删，测试一个细胞有三个输入，并且每个输入是双通道（奖罚）情况下是否可以实现所有的模式识别
* 处理逻辑为 (信号x正权重)取1为饱和值 - 信号x负权重，当结果大于0.5时输出1。 这个逻辑是模拟网上找到的“大脑只需单个神经元就可进行XOR异或运算”一文
* 原理。只不过这里扩展到三个（或以上)输入的情况。
*
* 三个输入有8种排列组合，如一组参数能实现任意8种组合，则有2的8次方=256种排列组合，去除全为0的输入必须输出0，只计算有1信号的输入，则有2的7次方=128种组合）
* （经实测，有31种组合条件不，不能找到符合要求的正负权重）
*
* （4个通道有16种排列组合，如一组参数能实现任意16种组合，则有2的16次方=65536种排列组合，去除全为0的输入，则有2的15次方32768种组合）
* （经实测，32768组合中有29987种不能找到符合要求的正负权重，如只测前1024个组合，则有603种情况不能找到解，如只测前128组合，则有31种情况下找不到解）
*
*
*
*/
@SuppressWarnings("all")
public class TestInput3 {
// public static void main(String[] args) {
// testInput3();
// //testInput4();
// }
public static void testInput3() { //这里测试一个细胞有三个树突输入，每个输入有正负两种信号，且信号以1为饱和，测试结果发现有31种情况无法找到解
long notFoundCont = 0;
boolean pass = false;
float p = 0.1f;
for (int i = 0; i < 128; i++) {
found: //
for (float a = 0; a < 1.001; a += p) {
for (float b = 0; b < 1.001; b += p) {
for (float c = 0; c < 1.001; c += p) {
for (float d = 0; d < 1.001; d += p) {
for (float e = 0; e < 1.001; e += p) {
for (float f = 0; f < 1.001; f += p) {
pass = true;
int bt = 1;
int x, y, z;
for (int n = 1; n <= 7; n++) {
x = ((n & 4) > 0) ? 1 : 0;
y = ((n & 2) > 0) ? 1 : 0;
z = ((n & 1) > 0) ? 1 : 0;
int shouldbe = (i & bt) > 0 ? 1 : 0;
float v = x * a + y * b + z * c; //正信号累加
if (v > 1) //如饱和取1
v = 1f;
float v1 = x * d + y * e + z * f; //负信号累加
if (v1 > 1)
v1 = 1f;
v = v - v1;
int result = v >= 0.5 ? 1 : 0;
if (result != shouldbe) {
pass = false;
break;
}
bt = bt << 1;
}
if (pass) {
System.out.print("i=" + i + " found, i=" + bin(i));
System.out.println(" " + r(a) + ", " + r(b) + ", " + r(c) + " " + r(d) + ", " + r(e) + ", " + r(f));
bt = 1;
for (int n = 1; n <= 7 && false; n++) {
x = ((n & 4) > 0) ? 1 : 0;
y = ((n & 2) > 0) ? 1 : 0;
z = ((n & 1) > 0) ? 1 : 0;
int shouldbe = (i & bt) > 0 ? 1 : 0;
System.out.println(" " + x + "*" + r(a) + " + " + y + "*" + r(b) + " + " + z + "*" + r(c) + " - " + x + "*" + r(d) + " - " + y + "*" + r(e) + " - " + z
+ "*" + r(f) + " = " + shouldbe);
bt = bt << 1;
}
break found;
}
}
}
}
}
}
}
if (!pass) {
System.out.println("i=" + i + " not found, i=" + bin(i));
notFoundCont++;
}
}
System.out.println("notFoundCont=" + notFoundCont);
}
public static void testInput4() {//这里测试一个细胞有4个树突输入，每个输入有正负两种信号，且信号以1为饱和，测试结果发现有603种情况无法找到解
long notFoundCont = 0;
boolean pass = false;
float p = 0.2f;
for (int i = 0; i < 1024; i++) {
found: //
for (float a = 0; a < 1.001; a += p) {
for (float b = 0; b < 1.001; b += p) {
for (float c = 0; c < 1.001; c += p) {
for (float d = 0; d < 1.001; d += p) {
for (float e = 0; e < 1.001; e += p) {
for (float f = 0; f < 1.001; f += p) {
for (float g = 0; g < 1.001; g += p) {
for (float h = 0; h < 1.001; h += p) {
pass = true;
int bt = 1;
int x, y, z, m;
for (int n = 1; n <= 15; n++) {
x = ((n & 8) > 0) ? 1 : 0;
y = ((n & 4) > 0) ? 1 : 0;
z = ((n & 2) > 0) ? 1 : 0;
m = ((n & 1) > 0) ? 1 : 0;
int shouldbe = (i & bt) > 0 ? 1 : 0;
float v = x * a + y * b + z * c + m * d; //正信号累加
if (v > 1) //如饱和取1
v = 1f;
float v1 = x * e + y * f + z * g + m * h; //负信号累加
if (v1 > 1)
v1 = 1f;
v = v - v1;
int result = v >= 0.5 ? 1 : 0;
if (result != shouldbe) {
pass = false;
break;
}
bt = bt << 1;
}
if (pass) {
System.out.print("i=" + i + " found, i=" + bin(i));
System.out.println(" " + r(a) + ", " + r(b) + ", " + r(c) + ", " + r(d) + " " + r(e) + ", " + r(f) + ", " + r(g) + ", " + r(h));
bt = 1;
for (int n = 1; n <= 15; n++) {
x = ((n & 8) > 0) ? 1 : 0;
y = ((n & 4) > 0) ? 1 : 0;
z = ((n & 2) > 0) ? 1 : 0;
m = ((n & 1) > 0) ? 1 : 0;
int shouldbe = (i & bt) > 0 ? 1 : 0;
System.out.println(" " + x + "*" + r(a) + " + " + y + "*" + r(b) + " + " + z + "*" + r(c) + " + " + m + "*" + r(d) //
+ " - " + x + "*" + r(e) + " - " + y + "*" + r(f) + " - " + z + "*" + r(g) + " - " + m + "*" + r(h) + " = " + shouldbe);
bt = bt << 1;
}
break found;
}
}
}
}
}
}
}
}
}
if (!pass) {
System.out.println("i=" + i + " not found, i=" + bin(i));
notFoundCont++;
}
}
System.out.println("notFoundCont=" + notFoundCont);
}
static float r(float f) { //取小数后2位
return Math.round(f * 100) * 1.0f / 100;
}
static String bin(int i) { //转二进制
String ibin = Integer.toBinaryString(i);
while (ibin.length() < 7)
ibin = "0" + ibin;
return ibin;
}
}