Re: Some list questions

*To*: mathgroup at smc.vnet.net*Subject*: [mg37150] Re: [mg37132] Some list questions*From*: BobHanlon at aol.com*Date*: Sun, 13 Oct 2002 05:56:38 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

In a message dated 10/12/2002 6:21:51 AM, flip_alpha at safebunch.com writes: >I have two seperate list questions that I was hoping to get help with. > >Question 1. > > I have a variable length list similar to that generated by FactorInteger, >that is {number, exponent} pairs. An example follows. > >lista = {{2,3},{3,1},{5,1}} ... this is the number 2^3 * 3^1 * 5^1 > >I want to generate a list of "all" the products of numbers from this list. > >I can tell that I get a total (3+1)*(1+1)*(1+1) = 4*2*2 = 16, products >and I >want a list showing all of those. > >These would be: > >2^3 can generate {2^0, 2^1, 2^2, 2^3} = {1, 2, 4 ,8} >3^1 can generate {3^0, 3^1} = {3} ... we dont care about the duplicate >"1" >5^1 can generate {5^0, 5^1} = {5} ... we dont care about the duplicate >"1" > >Hence the 4*2*2 = 16 (the product of one more of the exponents) above. > >Next we should get 16 products (from these lists), namely (I left them >as >products below to show what I am after): > >{1, 2, 4, 8, 1*3, 2*3, 4*3, 8*3, 1*5, 2 * 5, 4* 5, 8* 5, 1*3*5, >2*3*5, 4*3*5, 8*3*5} > >If the list were lista = {{2,4}, {3,2}, {5, 3},{7^5}}, we would have >(4+1)(2+1)(3+1)(5+1) = 360 products, for example and the return values >should be a single list showing all of those. > >Question 2. > >I have two lists and want to generate two new lists from them. These two >lists are {number, exponent} pairs. > >In the first list, I want the "minimum intersection" of {number, exponent} >pairs. > >In the second list, I want the "maximum union" of {number, exponent} pairs. > >Let me show an example: > >Input: > >list1 = {{2, 3}, {3, 4}, {5, 6}, {7, 2}, {17, 5}} > >list2 = {{2, 5}, {3, 2}, {5, 1}, {7, 3}} > >Output: > >minint = {{2, 3}, {3, 2}, {5, 1}, {7, 2}} > >Note: In this example we only kept those pairs where the intersection of >the >number exists and also keep the min power of those. > >maxint = {{2, 5}, {3, 4}, {5, 6}, {7, 3}, {17, 5}} > >Note: In this example we kept the union of lists and also keep the max >power >of each. > allProducts[x_] := Module[{sx = Sort[x, #2[[2]] < #1[[2]] &], f}, Union[Flatten[ Outer[f, Sequence @@ (PadRight[#, sx[[1, -1]] + 1, 1] & /@ (#[[1]]^ Range[0, #[[2]]] & /@ sx))]] /. f -> Times]]; lista = {{2, 3}, {3, 1}, {5, 1}}; allProducts[lista] {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120} minInt[x_, y_] := FactorInteger[GCD[ Times @@ (#[[1]]^#[[2]] & /@ x), Times @@ (#[[1]]^#[[2]] & /@ y)]]; maxInt[x_, y_] := Union[x, y] //. {s___, {b_, e1_}, {b_, e2_}, r___} :> {s, {b, Max[e1, e2]}, r} list1 = {{2, 3}, {3, 4}, {5, 6}, {7, 2}, {17, 5}}; list2 = {{2, 5}, {3, 2}, {5, 1}, {7, 3}}; minInt[list1, list2] {{2, 3}, {3, 2}, {5, 1}, {7, 2}} maxInt[list1, list2] {{2, 5}, {3, 4}, {5, 6}, {7, 3}, {17, 5}} Bob Hanlon