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
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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,
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5.72509, 5.63923, 5.51957, 5.51441, 5.45378, 5.3237, 5.14787, 5.11063,
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4.80679, 4.71988, 4.56427, 3.9834, -30.0577, -30.4488, -30.4836,
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-30.5326, -30.64, -30.7324, -30.8622, -30.9132, -31.0039, -31.0986,
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-31.1857, -31.278, -31.5279, -31.7783, -32.0976, -32.2601, -32.4935,
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-32.6476, -33.5527, -33.998}
o3 : List
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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
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i6 : histogram(svds_{0..19}/log,10)
o6 = {1, 0, 1, 2, 2, 1, 4, 2, 4, 3}
o6 : List
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i7 : histogram(svds_{20..39}/log,10)
o7 = {1, 1, 0, 2, 2, 1, 2, 5, 4, 2}
o7 : List
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