Suppose you have a computer that require 1 minute to solve problem instances of size=1,000.Suppose you buy a new computer that runs 1,000 times faster than the old one. What instance sizes can be run in 1 minute,assuming the following time complexities T(n) for our algorithm?\n\na) T(n)=n.\nb) T(n)=n^3.\nc) T(n)=10^n.\n\n************************************************************
122903
In case of T(n) = n, say O(n), and n is 1000, then on the old machine, 1 minute and the new one, 1/1000 minute!\nfor n^3, size will be 1000^3 = 1000,000,000 which takes 1000,000 minutes on the old machine and 1000 on the new one, and for 10^n, of course you don't expect me to write 1000 ZEROs here?! but takes 10^997 minutes on the old one, and 10^994 minutes on the new one!
229478
The links below have good info. The first briefly addresses your first question, and the second is a good brief history of the problem which provides some insight as to the difficulties encountered. I actually played around with this problem a little on an elementary level and it is interesting. Cheers.\n\nhttp://www.sciam.com/print_version.cfm?articleID=000DBA55-B698-1C71-9EB7809EC588F2D7\n\nhttp://www.geocities.com/fermatnow/flt/
626056
You have to measure the approximate number or operations for each algorithm.  Say one is O(n) and another is O(2n).  \n\ncheck out some sites about algorithm complexity\n\nhttp://www.cs.ucd.ie/staff/jcarthy/home/complexity.html\nhttp://en.wikipedia.org/wiki/Big_O_notation
180457
Consider the point with the smallest number of neighbors, say k.  You can disconnect this point from the graph in k moves by choosing each of its neighbors in turn.\n\nIn a graph of n points and 100 edges, the point with the smallest number of neighbors can have no more than:\n\nk = Floor(200 / n)\n\nneighbors.
99017
15 people  8 hours       100 items\n6 people   x hours        50 items\n\n\n1 men  would made 100 items in 15*8=120 hours,\n6 men would need  120/6 = 20 hours to make 100 items; \nhence, for 50 items they need 20/2 hours which is 10 hours.
486712
1.  The number of infected computers is\n1 + 20 + 20^2 + 20^3 + ... + 20^n\nwhere n is the number of 5 minute intervals that have elapsed.\n\nThis is a geometric series with n+1 terms, whose sum is\n(20^(n+1)-1)/(20-1)\n\nPut this equal to 10^8 and solve\n[after you get 20^(n+1) = 19*10^8, take logs of both sides.]\nAlternatively you could just evaluate the sum formula for n = 4, 5, .. until the result exceeds 100 million, and multiply your value of n by 5 minutes.\n\n2.  Use your calculator or a spreadsheet to evaluate these functions.  Then just plot the points you got and draw a curve (or in one case a line) through them. e.g.\n\nFor the first one you get -3, -7, 9, 305, 3297.  Not a good example, but you plot the points\n(1, -3) (2, -7) (4, 9) and the others are off the page.  Not a good question, the others will be easier to graph.  For this one, you also need to work out the value for x = 0 [f(0) = 1, right?]\nand x = -1  ?[f(-1) = -1?  You get that?]\nThen you  draw a curve that goes up to a peak just before coming down through (0,1), then goes through a trough between (2, -7) and (4, 9).\n\nAs for classifications, some of them are too obvious to be able to give a reason for.  The first one is polynomial, you should find one of them is linear "because its graph is a straight line"; one is quadratic because its highest power is x^2, or its graph is a parabola; the one with e is exponential; the one with log is logarithmic.  Do the last part yourself by looking at the graphs and just making commonsense judgements.
410953
C.  Statement 1 tells us nothing about set T.  For Statement 2, consider the sums, given that Sm is the median  element of set S and Tm is the median element of set T.  We have 5 Sm  = 7 Tm, but there are obviously numerous values of Sm and Tm that would work other than the trivial case Sm = Tm = 0.  However, Statement 1 excludes all cases except the trivial case, so the bottom line is that both are needed.
252985
Hopefully, I'm not doing your homework for you, and you are just curious about how to find this answer. Follow along on a piece of paper so it is easier to understand.\n\nIn a 12 hour period of time, reguardless of what time you start the count, only 1:00 through 6:00 can possibly be used in order to stay within the "6" limit. Let's see how many numbers will add to 6 in 1:00. \n\n1:05, 1:50, 1:14, 1:41, 1:23, 1:32. (6 Total)\n\nNow for the rest of the hours:\n\n2:04, 2:40, 2:13, 2:31, 2:22. (5 Total)\n3:03, 3:30, 3:12, 3:21. (4 Total)\n\nSee a pattern developing?\n\n4:02, 4:20, 4:11. (3 Total)\n5:01, 5:10. (2 Total)\n6:00. (1 Total)\n\nNow add the totals to get 21 times that the digits add up to 6!\nCongratulations.
514565
1. d=t, any form of d=t+b would be parallel. b = any number. That is a line.\n\n2. d=t would also work here.\n\n3. Well if you go 60 MPH and you wnat to know how many miles a minute you can go convert using (60miles/hour)*(1hour/60minutes) and you get 1mile/minute. If you need feet, 1 mile is 5280 feet, so 5280 feet/minute
229824
This is kinda like finding the LCM of the two numbers.  If Lou can run a lap in 24 seconds, then:\n\n2 laps - 48 seconds\n3 laps - 72 seconds\n4 laps - 96 seconds\n5 laps - 120 seconds\n\nLikewise for Ty:\n2 laps - 60 seconds\n3 laps - 90 seconds\n4 laps - 120 seconds\n\nSo, after Ty runs 4 laps, Lou will have run 5.  All this happens in 120 seconds, or 2 minutes.
60373
Here's the algorithm\n\n    solve ( current, final )\n    {\n 1      let max be the number of disks\n 2      let dest be the final place of max\n 3      let disk = max\n        repeat\n 4          while disk > 0 do\n 5              if disk is already on dest,\n 6              or, moving it succeeds then\n 7                  if disk = max then\n 8                      decrement max by 1\n 9                      if max = 0 then\n10                          return    // done\n                        end if\n11                      let dest be the final place of max\n                    end if\n                else\n12                  let dest be the alternative place between dest and\n                    the current place of disk\n                end if\n13              decrement disk by 1\n            end while\n14          let p and q be the places different of dest\n15          let disk be the smaller of the disks on top of p and q\n16          let dest be the place between p and q with greater disk on top\n        end repeat\n    } \n\n\nif u need more help..\n Just let me know...
128237
Basically this question is one of trial and error. You have to think of the many possibilities and try eliminating some to get the desired results. It would be mind-boggling to finish these questions off in half an hour.
430022
unfortunately as all files are not  of equal size, it is not possible to predict or calculate the answer.\n\nhowever otherwise it will be 300*100/40  seconds
516785
OK: 60 seconds in one minute, therefore 13 x 60 = 780 seconds.\n780 + 19 = 797 seconds or 13 minutes 19 seconds\n797 / 10 = 79.7, correct?\n\nOR did you mean 13 hours and 19 minutes???\n13 hours x 3600 seconds per hour = 46800 seconds\n19 minutes x 60 seconds per minute = 1140 seconds\n46800 + 1140 = 47940 seconds\n47940 / 10 = 4794
77544
Let x be the side of a square\nThen, area of the square is, A=x^2\nThe rate of change of area w.r.t side is dA/dx =2x\nIn the problem another parameter is given, that is time.It is given in the problem that there is a constant change in the length of the side at the rate 0.4cm /sec.Here we have to notice that corresponding change in area is not uniform throughout.Therefore we can only find instantaneous change in area or accumulated area in the given periodor interval.That means change in area at a given instant of time or interval.\nFor example, let initial area be 0.Now, to find the area between the interval a and b we have to reintegrate. ie. integration of dA between the interval a and b =integration of 2x in the interval a & b.\nie. [A]a&b= [x^2]a&b=b^2-a^2 square units\nFor instance, change in area between the interval 9th and 10th seconds is =[x^2]a&b=b^2-a^2 square units\nGiven, 0.4units/sec,considering intial length and time as 0,we have, length of the side on the 9th second is a=9*0.4=3.6units and in the same way length on the 10th second is b= 4units\nAccordingly change in area(accumulated area) at the 10th second is =b^2-a^2\n=16-12.96=3.04 square units [this is an instataneous change]\nPlease note that, this change cannot be written as units^2 /sec
340203
Whats the probability that the problem is *not* solved? Tom doesn't solve it (0.75) and Alice doesn't (0.5) so its 0.75*0.5 = 0.375. So the probability it is solved is 1-0.375 = 0.625.\n\nNow, we are given that it is solved. What is the probability that Tom solves it? 0.25. \nSo:\nP(Tom | solved) = P(Tom and solved) / P(solved) = P(Tom) / P(solved) = 0.25/0.625 = 0.4.\n\n(edit - note that the reason the below answer is not quite correct is because it is possible for *both* of them to solve it - so the probability Tom solves it when solved + probability Alice solves it when solved should *not* equal 100% - it should be more.)
470459
double min(double* s, int length)\n{\ndouble min = *s;\nint i;\nfor(i=1, i< length; ++i)\nif((*s) < min ) {\n    min = *s;\n}\nreturn min;\n}