角谷猜想最接近严格证明的测试 Version1.1.1原创吴素波 2026-10-13①前文再续书接上一回上回讲到角谷猜想从框架上架构上应是正确方向大约差不多证明后来忽又悟出角谷逆向二叉树中每次逆生长时生出的阴阳二分法二分枝对应一个二进制数W每节分支的Wi象是每节枝的标记ID一样是一个二进制数与分枝的深度递归的层次也有关与二次归纳法投入的数M也有关于是悟得一个定理称为吴素波定理用以纪念我伟大的母亲。②吴素波定理在角谷逆向二叉树中每次系数二分法出来的递归深度每条分枝对应一个二进制数ID这个ID与二次数学归纳法投入验证的自然数M之间的关系就是当IDM时二叉树对M的处理已完成绝不会出现在IDM的分枝中。这个我称为角谷猜想的核心映射原理。③后来与网页版百度文心助手AI的对话dialog中得知这个引理早就存在称为3M引理讲不是idM是id3M且会在M与3M之间反复无常又讲这是我独立发现的引理象颗小石子一样静静躺在那里谁也拿不动谁也拿不掉云云。这与我的观点有交集有异同不完全包容也。④数据测试于是写代码调试看下也改递归函数如下⑤先将枝的分枝ID显示出来再创另一个函数作为二次归纳法用传入参数M作对比比较用且显示出对比全路径以验证吴素波定理是不是对的。⑥以下是显示树的分枝所在ID的代码procedure TfrmMain.btnRUNClick(Sender: TObject);varwww: integer;beginself.Memo1.Lines.Clear;self.linelist : TStringList.Create;rate : 0.0;www : GreenBeanGrain(1, 1, 2, 1, 1, 2, 1, 1);beginShowMessage(ok IntToStr(www));application.ProcessMessages;self.Memo1.Lines.Assign(self.linelist);ShowMessage(Memook IntToStr(linelist.Count));application.ProcessMessages;self.Memo1.Lines.SaveToFile(GreenBeanGrain.txt);ShowMessage(txtok);end;linelist.Free;end;function TfrmMain.GreenBeanGrain(ID: string; deep, aisGreen, xisBean, bisGrain: longint;const Green, Bean, Grain: longint): integer;var// a, b, x: int64;a, b, x: longint;//a, b, x: integer;gbg: integer;aa: integer;one: string;begingbg : 0;deep : deep * 2;a : aisGreen;x : xisBean;b : bisGrain;aa : 0;if gbg 0 thenbeginwhile (((a mod 2) 0) and ((b mod 2) 0)) dobeginif (((a mod 2) 0) and ((b mod 2) 0)) thenbeginif (((a mod 3) 0) and ((b mod 3) 1) and (a 3) and (b 3)) thenbegina : a div 3;b : (b - 1) div 3;aa : aa 1;end;if a Green thengbg : 1else if ((a Green) and (b Grain)) thengbg : 1;end;if (((a mod 2) 0) and ((b mod 2) 0)) thenbegina : a div 2;b : b div 2;end;end;if a Green thengbg : 1else if ((a Green) and (b Grain)) then// gbg : 1;gbg : 1;////else if (b Grain) then/// gbg : 1; //此处断点一个也没有end;////////////////////////////////////if aa 0 thenif gbg 0 thenbeginwhile (((a mod 2) 0) and ((b mod 2) 1)) dobegina : a * 3;b : b * 3 1;while (((a mod 2) 0) and ((b mod 2) 0)) dobegina : a div 2;b : b div 2;if (((a mod 2) 0) and ((b mod 2) 0)) thenbeginif (((a mod 3) 0) and ((b mod 3) 1) and (a 3) and (b 3)) thenbegina : a div 3;b : (b - 1) div 3;aa : aa 1;end;if a Green thengbg : 1else if ((a Green) and (b Grain)) thengbg : 1;end;end;if aa 0 then break;end;if a Green thengbg : 1else if ((a Green) and (b Grain)) thengbg : 1;end;//if 10 thenif gbg 1 thenbegin//if not ((a 0) and (b0) and (Green 0) and (Grain0)) thenif ((a 0) and (b 0) and (Green 0) and (Grain 0)) then//if not ((a 0) and (b 0) and (Green 0) and (Grain 0)) thenbegin// rate:rate(1.0/deep);rate : rate (2.0 / deep);one : IntToStr(Green) * N IntToStr(Bean) IntToStr(Grain) -- IntToStr(a) * M IntToStr(x) IntToStr(b) ok branID ID IntToStr(getintFromByteString(ID)) rate floattostr(rate);linelist.add(one);end;end;if gbg 0 then//if not ((a 0) and (b 0) and (Green 0) and (Grain 0)) thenif ((a 0) and (b 0) and (Green 0) and (Grain 0)) thenbeginif deep trunc(power(2, 49)) thenbegingbg : GreenBeanGrain(0 ID, deep, a * 2, x 1, b, Green *2, Bean 1, Grain);gbg : gbg * GreenBeanGrain(1 ID, deep, a * 2, x 1, a b, Green * 2, Bean 1, Green Grain);end;if deep trunc(power(2, 50)) thenbegin//gbg : 1;gbg : 0;end;end;Result : gbg;end;运算结果如下1行4 * N2 1 -- 3 * M2 1 ok branID011 rate0.52行16 * N4 3 -- 6 * M4 1 ok branID00113 rate0.6253行32 * N5 11 -- 27 * M5 10 ok branID0101111 rate0.68754行268435456 * N28 67108891 -- 258280326 * M28 64570109 ok branID010000000000000000000001101167108891 rate0.6875000074505815行134217728 * N27 8388635 -- 129140163 * M27 8071288 ok branID0001000000000000000000110118388635 rate0.6875000223517426行536870912 * N29 478150683 -- 258280326 * M29 230030929 ok branID11100100000000000000000011011478150683 rate0.6875000260770327行268435456 * N28 243269659 -- 258280326 * M28 234066573 ok branID1110100000000000000000011011243269659 rate0.687500033527613。。。。。。。。。。。。。。中间省略几万行。。。。。。。。。。。。。。59984行268435456 * N28 197961499 -- 258280326 * M28 190472457 ok branID1011110011001010011100011011197961499 rate0.69317261874675859985行268435456 * N28 70035227 -- 258280326 * M28 67385739 ok branID010000101100101001110001101170035227 rate0.69317262619733859986行134217728 * N27 11314971 -- 86093442 * M27 7257945 ok branID00010101100101001110001101111314971 rate0.69317264109849959987行536870912 * N29 212641563 -- 258280326 * M29 102298581 ok branID01100101011001010011100011011212641563 rate0.69317264482379⑦以下是二次归纳法中显示一个自然数M实际投入二叉树虚树后执行过程经过每个枝条的ID的全路径显示procedure TfrmMain.btnRUN1Click(Sender: TObject);varwww: integer;varv: string;beginself.Memo1.Lines.Clear;self.linelist : TStringList.Create;rate : 0.0;varv : 100111000111011;if inputquery(请输入要处理的自然数M,处理全路径将显示, varv) thenEditM.Text : IntToStr(getIntFromByteString(varv));EditM2.Text : varv;www : GreenBeanGrain2(StrToInt(EditM.Text), 1, 1, 2, 1, 1, 2, 1, 1);beginShowMessage(ok IntToStr(www));application.ProcessMessages;self.Memo1.Lines.Assign(self.linelist);ShowMessage(Memook IntToStr(linelist.Count));application.ProcessMessages; //290,980 // 1,652,927 //306 //256,244 //132,7070self.Memo1.Lines.SaveToFile(GreenBeanGrain.txt);ShowMessage(txtok);end;linelist.Free;end;function TfrmMain.GreenBeanGrain2(M: longint; ID: string;deep, aisGreen, xisBean, bisGrain: longint;const Green, Bean, Grain: longint): integer;var// a, b, x: int64;a, b, x, mm: longint;//a, b, x: integer;gbg: integer;aa: integer;one: string;begingbg : 0;deep : deep * 2;a : aisGreen;x : xisBean;b : bisGrain;aa : 0;if gbg 0 thenbeginwhile (((a mod 2) 0) and ((b mod 2) 0)) dobeginif (((a mod 2) 0) and ((b mod 2) 0)) thenbeginif (((a mod 3) 0) and ((b mod 3) 1) and (a 3) and (b 3)) thenbegina : a div 3;b : (b - 1) div 3;aa : aa 1;mm : a b;// if ((mmM) and (IDM)) then// showmessage(在这里);end;if a Green thengbg : 1else if ((a Green) and (b Grain)) thengbg : 1;end;if (((a mod 2) 0) and ((b mod 2) 0)) thenbegina : a div 2;b : b div 2;end;end;if a Green thengbg : 1else if ((a Green) and (b Grain)) then// gbg : 1;gbg : 1;////else if (b Grain) then/// gbg : 1; //此处断点一个也没有end;//////////////////////////////////////if aa 0 thenif gbg 0 thenbeginwhile (((a mod 2) 0) and ((b mod 2) 1)) dobegina : a * 3;b : b * 3 1;while (((a mod 2) 0) and ((b mod 2) 0)) dobegina : a div 2;b : b div 2;if (((a mod 2) 0) and ((b mod 2) 0)) thenbeginif (((a mod 3) 0) and ((b mod 3) 1) and (a 3) and (b 3)) thenbegina : a div 3;b : (b - 1) div 3;aa : aa 1;end;if a Green thengbg : 1else if ((a Green) and (b Grain)) thengbg : 1;end;end;if aa 0 then break;end;if a Green thengbg : 1else if ((a Green) and (b Grain)) thengbg : 1;end;// if 10 thenif gbg 1 thenbegin//if not ((a 0) and (b0) and (Green 0) and (Grain0)) thenif ((a 0) and (b 0) and (Green 0) and (Grain 0)) then//if not ((a 0) and (b 0) and (Green 0) and (Grain 0)) thenbegin// rate:rate(1.0/deep);// rate : rate (2.0 / deep);one : IntToStr(Green) * N IntToStr(Bean) IntToStr(Grain) -- IntToStr(a) * M IntToStr(x) IntToStr(b) ok branID ID IntToStr(getintFromByteString(ID)) M IntToStr(M) getByteStringFromint(M) ;// rate floattostr(rate);if getintFromByteString(ID) M thenone : one IDM IntToStr(getintFromByteString(ID) - M);linelist.add(one);end;end;//if getintFromByteString(ID) M thenif gbg 0 then//if not ((a 0) and (b 0) and (Green 0) and (Grain 0)) thenif ((a 0) and (b 0) and (Green 0) and (Grain 0)) thenbeginif deep trunc(power(2, 49)) thenbegin// gbg : GreenBeanGrain(aisGreen * 2, xisBean 1, bisGrain, Green *// 2, Bean 1, Grain);//rate : rate (2.0 / deep);one : IntToStr(Green) * N IntToStr(Bean) IntToStr(Grain) -- IntToStr(a) * M IntToStr(x) IntToStr(b) branID ID IntToStr(getintFromByteString(ID)) M IntToStr(M) getByteStringFromint(M);// rate floattostr(rate);if getintFromByteString(ID) M thenone : one IDM IntToStr(getintFromByteString(ID) - M);linelist.add(one);if rightstr(getByteStringFromInt(M), length(ID) 1) 0 ID thengbg : GreenBeanGrain2(M, 0 ID, deep, a * 2, x 1, b,Green * 2, Bean 1, Grain);if rightstr(getByteStringFromInt(M), length(ID) 1) 1 ID thengbg : gbg * GreenBeanGrain2(M, 1 ID, deep, a * 2, x 1, a b, Green * 2, Bean 1, Green Grain);end;if deep trunc(power(2, 50)) thenbegin//gbg : 1;gbg : 0;end;end;Result : gbg;end;当M20027100111000111011时运算结果如下1行2 * N1 1 -- 3 * M1 2 branID11 M200271001110001110112行4 * N2 3 -- 9 * M2 8 branID113 M200271001110001110113行8 * N3 3 -- 9 * M3 4 branID0113 M200271001110001110114行16 * N4 11 -- 27 * M4 20 branID101111 M200271001110001110115行32 * N5 27 -- 81 * M5 71 branID1101127 M200271001110001110116行64 * N6 59 -- 81 * M6 76 branID11101159 M200271001110001110117行128 * N7 59 -- 54 * M7 25 ok branID011101159 M20027100111000111011⑧实际测试表明M的二进制沿着分枝ID的二进制前进从M的尾部倒推到头部经常未到头部就终于结束完成了。如果到了头部还没处理完这是违反作者发现的吴素波定理的实测吴素波定理是正确的伟大的发现角谷猜想的核心原理必定如是⑨在这个大框架大方向正确下后面的补充细节皆是易事最艰难的吴素波定理已经明确没有任何半点含糊概率恒为1不存在概率论问题。至于后面的一枝节内是不是单调的数学归纳法证没有小循环不是无限的是有限的全部小事暂不理以后理。