Re: Suggestions for translating a Do[] loop ...

*To*: mathgroup at smc.vnet.net*Subject*: [mg69791] Re: Suggestions for translating a Do[] loop ...*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Sat, 23 Sep 2006 04:44:40 -0400 (EDT)*Organization*: The Open University, Milton Keynes, UK*References*: <eevtpj$hfr$1@smc.vnet.net>

David Annetts wrote: > Hi, > > Given that a, b & r are lists of the same length, can anyone suggest a > translation of the following Do[] loop to something functional? It looks > like an ideal application for Fold, but for me, the loop going backwards is > really obscuring things. > > The loop is > > Do[ > r[[j]] = a[[j]] * (b[[j]] + r[[j + 1]]) / (1 + b[[j]] * r[[j > + 1]]), > {j, Length@a - 1, 1, -1} > ]; > > with r[[Length@a]] = 0. > > Many thanks, > > Dave. > Dave, First, notice that if we reverse the lists a, b, and r, we can write the Do loop "from the beginning" as Do[r[[j]] = a[[j]]*((b[[j]] + r[[j - 1]])/(1 + b[[j]]*r[[j - 1]])), {j, 2, Length[a], 1}]; Second, the initial values of the list r -- except 0 -- are not used during the computation. Third, what we need is FoldList and a starting value of 0 to build the list r. In[1]:= a = Range[11]; b = Range[5, 15]; {a, b} = Rest /@ Reverse /@ {a, b}; Reverse[FoldList[#2[[1]]*((#2[[2]] + #1)/(1 + #2[[2]]*#1)) & , 0, Transpose[{a, b}]]] Out[4]= {25357030627/34045828151, 6036337922/3864138541, 576190351/406996435, 498170924/175771447, 20343315/13024927, 1414446/294913, 904463/687499, 61944/6115, 459/607, 140, 0} Result we can compare to the original code In[5]:= a = Range[11]; b = Range[5, 15]; r = Range[20, 30]; r[[Length[a]]] = 0; Do[r[[j]] = a[[j]]*((b[[j]] + r[[j + 1]])/(1 + b[[j]]*r[[j + 1]])), {j, Length[a] - 1, 1, -1}]; r Out[10]= {25357030627/34045828151, 6036337922/3864138541, 576190351/406996435, 498170924/175771447, 20343315/13024927, 1414446/294913, 904463/687499, 61944/6115, 459/607, 140, 0} Regards, Jean-Marc