We combute hi th number to elemnts in the i-th equidistant subdivision of the interval [min L, max L] into n parts
i1 : M=(randomChainComplex({20,20},{20},ZeroMean=>true)).dd_1; 40 40 o1 : Matrix ZZ <--- ZZ |
i2 : (svds,U,Vt)=SVD(M**RR_53); |
i3 : (entries matrix {svds})_0/log o3 = {6.37106, 6.31472, 6.27245, 6.10348, 6.02102, 5.98252, 5.92934, 5.83927, ------------------------------------------------------------------------ 5.72509, 5.63923, 5.51957, 5.51441, 5.45378, 5.3237, 5.14787, 5.11063, ------------------------------------------------------------------------ 4.80679, 4.71988, 4.56427, 3.9834, -29.8017, -30.0022, -30.0403, ------------------------------------------------------------------------ -30.1295, -30.1765, -30.3182, -30.4807, -30.6149, -30.6391, -30.8203, ------------------------------------------------------------------------ -30.8781, -30.8935, -31.1443, -31.2944, -31.4568, -31.7739, -32.0257, ------------------------------------------------------------------------ -32.3668, -32.8797, -33.6968} o3 : List |
i4 : maximalEntry M o4 = 138 o4 : RR (of precision 53) |
i5 : histogram(svds/log,10) o5 = {20, 0, 0, 0, 0, 0, 0, 0, 0, 20} o5 : List |
i6 : histogram(svds_{0..19}/log,10) o6 = {1, 0, 1, 2, 2, 1, 4, 2, 4, 3} o6 : List |
i7 : histogram(svds_{20..39}/log,10) o7 = {1, 0, 1, 1, 2, 1, 2, 5, 2, 5} o7 : List |