From 5089ecd4d18be9b2f50f4a85a1a331b058f444c2 Mon Sep 17 00:00:00 2001 From: 13766800364 <13766800364@qq.com> Date: Thu, 9 Oct 2025 16:08:50 +0800 Subject: [PATCH] Add File --- history/015_testinput3/TestInput3.java | 201 +++++++++++++++++++++++++ 1 file changed, 201 insertions(+) create mode 100644 history/015_testinput3/TestInput3.java diff --git a/history/015_testinput3/TestInput3.java b/history/015_testinput3/TestInput3.java new file mode 100644 index 0000000..e0c2539 --- /dev/null +++ b/history/015_testinput3/TestInput3.java @@ -0,0 +1,201 @@ +/* + * 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; + } + +}