Let's assume you're a baseball manager. And you have N pitchers in your bullpen (N<=14) and they have to face M batters (M<=100). Also to mention you know the strength of each of the pitchers and each of the batters. For those who are not familiar to baseball once you brought in a relief pitcher he can pitch to k consecutive batters, but once he's taken out ofthe game he cannot come back.
For each pitcher the probability that he's gonna lose his match-ups is given by (sum of all batter he will face)/(his strength). Try to minimize these probabilities, i.e. try to maximize your chances of winning the game.
For example we have 3 pitchers and they have to face 3 batters. The batters' stregnths are:
10 40 30
While the strength of your pitchers is:
40 30 3
The most optimal solution would be to bring the strongest pitcher to face the first 2 batters and the second to face the third batter. Then the probability of every pitcher losing his game will be:
50/40 = 1.25 and 30/30 = 1
So the probability of losing the game would be 1.25 (This number can be bigger than 100).
How can you find the optimal number? I was thinking to take a greedy approach, but I suspect whether it will always hold. Also the fact that the pitcher can face unlimited number of batters (I mean it's only limited by M) poses the major problem for me.
Probabilities must be in the range [0.0, 1.0] so what you call a probability can't be a probability. I'm just going to call it a score and minimize it.
I'm going to assume for now that you somehow know the order in which the pitchers should play.
Given the order, what is left to decide is how long each pitcher plays. I think you can find this out using dynamic programming. Consider the batters to be faced in order. Build an NxM table best[pitchers, batter] where best[i, j] is the best score you can make considering just the first j batters using the first i pitchers, or HUGE if it does not make sense.
best[1,1] is just the score for the best pitcher against the first batter, and best[1,j] doesn't make sense for any other values of j.
For larger values of i you work out best[i,j] by considering when the last change of pitcher could be, considering all possibilities (so 1, 2, 3...i). If the last change of pitcher was at time t, then look up best[t, j-1] to get the score up to the time just before that change, and then calculate the a/b value to take account of the sum of batter strengths between time t+1 and time i. When you have considered all possible times, take the best score and use it as the value for best[i, j]. Note down enough info (such as the last time of pitcher change that turned out to be best) so that once you have calculated best[N, M], you can trace back to find the best schedule.
You don't actually know the order, and because the final score is the maximum of the a/b value for each pitcher, the order does matter. However, given a separation of players into groups, the best way to assign pitchers to groups is to assign the best pitcher to the group with the highest total score, the next best pitcher to the group with the next best total score, and so on. So you could alternate between dividing batters into groups, as described above, and then assigning pitchers to groups to work out the order the pitchers really should be in - keep doing this until the answer stops changing and hope the result is a global optimum. Unfortunately there is no guarantee of this.
I'm not convinced that your score is a good model for baseball, especially since it started out as a probability but can't be. Perhaps you should work out a few examples (maybe even solving small examples by brute force) and see if the results look reasonable.
Another way to approach this problem is via http://en.wikipedia.org/wiki/Branch_and_bound.
With branch and bound you need some way to describe partial answers, and you need some way to work out a value V for a given partial answer, such that no way of extending that partial answer can possibly produce a better answer than V. Then you run a tree search, extending partial answers in every possible way, but discarding partial answers which can't possibly be any better than the best answer found so far. It is good if you can start off with at least a guess at the best answer, because then you can discard poor partial answers from the start. My other answer might provide a way of getting this.
Here a partial answer is a selection of pitchers, in the order they should play, together with the number of batters they should pitch to. The first partial answer would have 0 pitchers, and you could extend this by choosing each possible pitcher, pitching to each possible number of batters, giving a list of partial answers each mentioning just one pitcher, most of which you could hopefully discard.
Given a partial answer, you can compute the (total batter strength)/(Pitcher strength) for each pitcher in its selection. The maximum found here is one possible way of working out V. There is another calculation you can do. Sum up the total strengths of all the batters left and divide by the total strengths of all the pitchers left. This would be the best possible result you could get for the pitchers left, because it is the result you get if you somehow manage to allocate pitchers to batters as evenly as possible. If this value is greater than the V you have calculated so far, use this instead of V to get a less optimistic (but more accurate) measure of how good any descendant of that partial answer could possibly be.
Related
I am writing an AI to play 5 card poker, where you are allowed to discard a card from your hand and swap it for another randomly dealt one if you wish. My AI can value every possible poker hand as shown in the answer to my previous question. In short, it assigns a unique value to each possible hand where a higher value correlates to a better/winning hand.
My task is to now write a function, int getDiscardProbability(int cardNumber) that gives my AI a number from 0-100 relating to whether or not it should discard this card (0 = defintely do not discard, 100 = definitely discard).
The approach I have thought up of was to compute every possible hand by swapping this card for every other card in the deck (assume there are still 47 left, for now), then compare each of their values with the current hand, count how many are better and so (count / 47) * 100 is my probability.
However, this solution is simply looking for any better hand, and not distinguishing between how much better one hand is. For example, if my AI had the hand 23457, it could discard the 7 for an K, producing a very slightly better hand (better high card), or it could exchange the 7 for an A or a 6, completing the Straight - a much better hand (much higher value) than a High King.
So, when my AI is calculating this probability, it would be increased by the same amount when it sees that the hand could be improved by getting the K than it would when it sees that the hand could be improved by getting an A or 6. Because of this, I somehow need to factor in the difference in value from my hand and each of the possible hands when calculating this probability. What would be a good approach to achieve this with?
Games in general have a chicken-egg problem: you want to design an AI that can beat a good player, but you need a good AI to train your AI against. I'll assume you're making an AI for a 2-player version of poker that has antes but no betting.
First, I'd note that if I had a table of probabilities for win-rate for each possible poker hand (of which there are surprisingly few really different ones), one can write a function that tells you the expected value from discarding a set of cards from your hand: simply enumerate all possible replacement cards and average the probability of winning with the hands. There's not that many cards to evaluate -- even if you don't ignore suits, and you're replacing the maximum 3 cards, you have only 47 * 46 * 43 / 6 = 16215 possibilities. In practice, there's many fewer interesting possibilities -- for example, if the cards you don't discard aren't all of the same suit, you can ignore suits completely, and if they are of the same suit, you only need to distinguish "same suit" replacements with "different suits" replacement. This is slightly trickier than I describe it, since you've got to be careful to count possibilities right.
Then your AI can work by enumerating all the possible sets of cards to discard of which there are (5 choose 0) + (5 choose 1) + (5 choose 2) + (5 choose 3) = 1 + 5 + 10 + 10 = 26, and pick the one with the highest expectation, as computed above.
The chicken-egg problem is that you don't have a table of win-rate probabilities per hand. I describe an approach for a different poker-related game here, but the idea is the same: http://paulhankin.github.io/ChinesePoker/ . This approach is not my idea, and essentially the same idea is used for example in game-theory-optimal solvers for real poker variants like piosolver.
Here's the method.
Start with a table of probabilities made up somehow. Perhaps you just start assuming the highest rank hand (AKQJTs) wins 100% of the time and the worst hand (75432) wins 0% of the time, and that probabilities are linear in between. It won't matter much.
Now, simulate tens of thousands of hands with your AI and count how often each hand rank is played. You can use this to construct a new table of win-rate probabilities. This new table of win-rate probabilities is (ignoring some minor theoretical issues) an optimal counter-strategy to your AI in that an AI that uses this table knows how likely your original AI is to end up with each hand, and plays optimally against that.
The natural idea is now to repeat the process again, and hope this yields better and better AIs. However, the process will probably oscillate and not settle down. For example, if at one stage of your training your AI tends to draw to big hands, the counter AI will tend to play very conservatively, beating your AI when it misses its draw. And against a very conservative AI, a slightly less conservative AI will do better. So you'll tend to get a sequence of less and less conservative AIs, and then a tipping point where your AI is beaten again by an ultra-conservative one.
But the fix for this is relatively simple -- just blend the old table and the new table in some way (one standard way is to, at step i, replace the table with a weighted average of 1/i of the new table and (i-1)/i of the old table). This has the effect of not over-adjusting to the most recent iteration. And ignoring some minor details that occur because of assumptions (for example, ignoring replacement effects from the original cards in your hand), this approach will give you a game-theoretically optimal AI, as described in: "An iterative method of solving a game, Julia Robinson (1950)."
A simple (but not so simple) way would be to use some kind of database with the hand combination probabilities (maybe University of Alberta Computer Poker Research Group Database).
The idea is getting to know each combination how much percentage of winning has. And doing the combination and comparing that percentage of each possible hand.
For instance, you have 5 cards, AAAKJ, and it's time to discard (or not).
AAAKJ has a winning percentage (which I ignore, lets say 75)
AAAK (discarting J) has a 78 percentage (let's say).
AAAJ (discarting K) has x.
AAA (discarting KJ) has y.
AA (discarting AKJ) has z.
KJ (discarting AAA) has 11 (?)..
etc..
And the AI would keep the one from the combination which had a higher rate of success.
Instead of counting how many are better you might compute a sum of probabilities Pi that the new hand (with swapped card) will win, i = 1, ..., 47.
This might be a tough call because of other players as you don't know their cards, and thus, their current chances to win. To make it easier, maybe an approximation of some sort can be applied.
For example, Pi = N_lose / N where N_lose is the amount of hands that would lose to the new hand with ith card, and N is the total possible amount of hands without the 5 that the AI is holding. Finally, you use the sum of Pi instead of count.
I’m having trouble solving a deceptively simple problem. My girlfriend and I are trying to formulate weekly meal plans and I had this brilliant idea that I could optimize what we buy in order to maximize the things that we could make from it. The trouble is, the problem is not as easy as it appears. Here’s the problem statement in a nutshell:
The problem:
Given a list of 100 ingredients and a list of 50 dishes that are composed of one or more of the 100 ingredients, find a list of 32 ingredients that can produce the maximum number of dishes.
This problem seems simple, but I’m finding that computing the answer is not trivial. The approach that I’ve taken is that I’ve computed a combination of the 32 ingredients as a 100 bit string with 32 of the bits set. Then I do a check of what dishes can be made with that ingredient number. If the number of dishes is greater than the current maximum, I save off the list. Then I compute the next valid ingredient combination and repeat, repeat, and repeat.
The number of combinations of the 32 ingredients is staggering! The way that I see it, it would take about 300 trillion years to calculate using my method. I’ve optimized the code so that each combination takes a mere 75 microseconds to figure out. Assuming that I can optimize the code, I might be able to reduce the run time to a mere trillion years.
I’m thinking that a completely new approach is in order. I'm currently coding this in XOJO (REALbasic), but I think the real problem is with approach rather than specific implementation. Anybody have an idea for an approach that has a chance of completion during this century?
Thanks,
Ron
mcdowella's branch and bound solution will be a big improvement over exhaustive enumeration, but it might still take a few thousand years. This is the kind of problem that is really best solved by an ILP solver.
Assuming that the set of ingredients for meal i is given by R[i] = { R[i][1], R[i][2], ..., R[i][|R[i]|] }, you can encode the problem as follows:
Create an integer variable x[i] for each ingredient 1 <= i <= 100. Each of these variables should be constrained to the range [0, 1].
Create an integer variable y[i] for each meal 1 <= i <= 50. Each of these variables should be constrained to the range [0, 1].
For each meal i, create |R[i]| additional constraints of the form y[i] <= x[R[i][j]] for 1 <= j <= |R[i]|. These will guarantee that we can only set y[i] to 1 if all of meal i's ingredients have been included.
Add a constraint that the sum of all x[i] must be <= 32.
Finally, the objective function should be the sum of all y[i], and we should be trying to maximise this.
Solving this will produce assignments for all the variables x[i]: 1 means the ingredient should be included, 0 means it should not.
My feeling is that a commercial ILP solver like CPLEX or Gurobi will probably solve a 150-variable ILP problem like this in milliseconds; even freely available solvers like lp_solve, which as a rule are much slower, should have no problems. In the unlikely case that it seems to be taking forever, you can still solve the LP relaxation, which will be very fast (milliseconds) and will give you (a) an upper bound on the maximum number of meals that can be prepared and (b) "hints" in the variable values: although the x[i] will in general not be exactly 0 or 1, values close to 1 are suggestive of ingredients that should be included, while values close to 0 suggest unhelpful ingredients.
There will be a http://en.wikipedia.org/wiki/Branch_and_bound solution to this, but it may be too expensive to get the exact answer - ILP as suggested by j_random_hacker is probably better - the LP relaxation of that is probably a better heuristic than the relaxation proposed here, and the ILP solver will be heavily optimized.
The basic idea is that you do a recursive depth first search of a tree of partial solutions, extending them one at a time. Once you recurse far enough down to reach a fully populated solution you can start keeping track of the best solution found so far. If I label your ingredients A, B, C, D... a partial solution is a list of ingredients of length <= 32. You start with the zero-length solution, then when you visit a partial solution e.g. ABC you consider ABCD, ABCE, ... and so on, and may visit some of these.
For each partial solution you work out the maximum score that any descendant of that solution could achieve. Getting an accurate idea of this is important. Here is one suggestion - suppose you have a partial solution of length 20. This leaves 12 ingredients to be chosen, so the best you could possibly do is to make all dishes which require no more than 12 ingredients not already in the 20 you have chosen so far work out how many of those there are and this is one example of a best possible score to any descendant of the partial solution.
Now when you consider extending the partial solution ABC to ABCD or ABCE or ABCF... if you have a best solution found so far you can ignore all extensions that cannot possibly score more than the best solution so far - this means that you don't need to consider all possible combinations of your 32 ingredients.
Once you have worked out which of the possible extensions might contain a new best answer, your recursive search should continue with the most promising of these possible extensions, because this is the one most likely to survive finding a better best solution so far.
One way to make this fast is to code it cleverly so that recursing up and down means only small changes to the existing data structure which you typically make on the way down and reverse on the way up.
Another way is to cut corners. One obvious way is to stop when you run out of time and go for the best solution found so far at that stage. Another way is to discard partial solutions more aggressively. If you have a score so far of e.g. 100 you could discard partial solutions that couldn't score any better than 110. This speeds up the search, and you know that although you might have best a better answer than 100 whatever you missed could not have been better than 110.
Solving some discrete mathematics huh? Well here is the wiki.
You also have not factored in anything about quantity. For example, flour would be used in a lot of fried recipes but buying 10 pounds of flour might not be great. And cost might be prohibitive for some ingredients that your solution wants. Not to mention a lot of ingredients are in everything. (milk, water, salt, pepper, sugar things like that)
In reality, optimization to this degree is probably not necessary. But I will not provide relationship advice on SO.
As for a new solution:
I would suggest identifying a lot of what you want to make and with what, and then writing a program to suggest things to make with the rest.
Why not just order the list of ingredients by the number of dishes they are used in?
This would be more like a greedy solution, of course, but it should give you some clues about what ingredients are most often used. From that you can compile a list of dishes that can be cooked already with the top 30 (or whatever) ingredients.
Also you could order the list of remaining (non-cookable) dishes by number of missing ingredients and maybe try to optimize on that to maximize the number of cookable dishes.
To be more "algorithmic", I think a local search is most promising here. Start with a candidate solution (random assignments to the 32 ingredients) and calculate as a fitness function the number of cookable dishes. Then check the neighboring states (switching one ingredient) and move to the state with the highest value. Repeat until a maximum is reached. Do this veeeery often and you should find a good solution. (This would be a simple greedy hill-climbing algorithm)
There are a lot of local search algorithms, you should be able to find more than enough information on the net. Most often you won't find the optimal solution (of course that depends on the problem), but a very good one nonetheless.
People visit my website, and I have an algorithm that produces a score between 1 and 0. The higher the score, the greater the probability that this person will buy something, but the score isn't a probability, and it may not be a linear relationship with the purchase probability.
I have a bunch of data about what scores I gave people in the past, and whether or not those people actually make a purchase.
Using this data about what happened with scores in the past, I want to be able to take a score and translate it into the corresponding probability based on this past data.
Any ideas?
edit: A few people are suggesting bucketing, and I should have mentioned that I had considered this approach, but I'm sure there must be a way to do it "smoothly". A while ago I asked a question about a different but possibly related problem here, I have a feeling that something similar may be applicable but I'm not sure.
edit2: Let's say I told you that of the 100 customers with a score above 0.5, 12 of them purchased, and of the 25 customers with a score below 0.5, 2 of them purchased. What can I conclude, if anything, about the estimated purchase probability of someone with a score of 0.5?
Draw a chart - plot the ratio of buyers to non buyers on the Y axis and the score on the X axis - fit a curve - then for a given score you can get the probability by the hieght of the curve.
(you don't need to phyically create a chart - but the algorithm should be evident from the exercise)
Simples.
That is what logistic regression, probit regression, and company were invented for. Nowdays most people would use logistic regression, but fitting involves iterative algorithms - there are, of course, lots of implementations, but you might not want to write one yourself. Probit regression has an approximate explicit solution described at the link that might be good enough for your purposes.
A possible way to assess whether logistic regression would work for your data, would be to look at a plot of each score versus the logit of the probability of purchase (log(p/(1-p)), and see whether these form a straight line.
I eventually found exactly what I was looking for, an algorithm called “pair-adjacent violators”. I initially found it in this paper, however be warned that there is a flaw in their description of the implementation.
I describe the algorithm, this flaw, and the solution to it on my blog.
Well, the straightforward way to do this would be to calculate which percentage of people in a score interval purchased something and do this for all intervals (say, every .05 points).
Have you noticed an actual correlation between a higher score and an increased likelihood of purchases in your data?
I'm not an expert in statistics and there might be a better answer though.
You could divide the scores into a number of buckets, e.g. 0.0-0.1, 0.1-0.2,... and count the number of customers who purchased and did not purchase something for each bucket.
Alternatively, you may want to plot each score against the amount spent (as a scattergram) and see if there is any obvious relationship.
You could use exponential decay to produce a weighted average.
Take your users, arrange them in order of scores (break ties randomly).
Working from left to right, start with a running average of 0. Each user you get, change the average to average = (1-p) * average + p * (sale ? 1 : 0). Do the same thing from the right to the left, except start with 1.
The smaller you make p, the smoother your curve will become. Play around with your data until you have a value of p that gives you results that you like.
Incidentally this is the key idea behind how load averages get calculated by Unix systems.
Based upon your edit2 comment you would not have enough data to make a statement. Your overall purchase rate is 11.2% That is not statistically different from your 2 purchase rates which are above/below .5 Additionally to validate your score, you would have to insure that the purchase percentages were monotonically increasing as your score increased. You could bucket but you would need to check your results against a probability calculator to make sure they did not occur by chance.
http://stattrek.com/Tables/Binomial.aspx
Hey guys, I have a sort of speed dating type application (not used for dating, just a similar concept) that compares users and matches them in a round based event.
Currently I am storing each user to user comparison (using cosine similarity) and then finding a round in which both users are available. My current set up works fine for smaller scale but I seem to be missing a few matchings in larger data sets.
For example with a setup like so (assuming 6 users, 3 from each group)
Round (User1, User2)
----------------------------
1 (x1,y1) (x2,y2) (x3,y3)
2 (x1,y2) (x2,y3) (x3,y1)
3 (x1,y3) (x2,y1) (x3,y2)
My approach works well right now to ensure I have each user meeting the appropriate user without having overlaps so a user is left out, just not with larger data sets.
My current algorithm
I store a comparison of each user from x to each user from y like so
Round, user1, user2, similarity
And to build the event schedule I simply sort the comparisons by similarity and then iterate over the results, finding an open round for both users, like so:
event.user_maps.all(:order => 'similarity desc').each do |map|
(1..event.rounds).each do |round|
if user_free_in_round?(map.user1) and user_free_in_round?(map.user2)
#creates the pairing and breaks from the loop
end
end
end
This isn't exact code but the general algorithm to build the schedule. Does anyone know a better way of filling in a matrix of item pairings where no one item can be in more than one place in the same slot?
EDIT
For some clarification, the issue I am having is that in larger sets my algorithm of placing highest similarity matches first can sometimes result in collisions. What I mean by that is that the users are paired in such a way that they have no other user to meet with.
Like so:
Round (User1, User2)
----------------------------
1 (x1,y1) (x2,y2) (x3,y3)
2 (x1,y3) (x2,nil) (x3,y1)
3 (x1,y2) (x2,y1) (x3,y2)
I want to be able to prevent this from happening while preserving the need for higher similar users given higher priority in scheduling.
In real scenarios there are far more matches than there are available rounds and an uneven number of x users to y users and in my test cases instead of getting every round full I will only have about 90% or so of them filled while collisions like the above are causing problems.
I think the question still needs clarification even after edit, but I could be missing something.
As far as I can tell, what you want is that each new round should start with the best possible matching (defined as sum of the cosine similarities of all the matched pairs). After any pair (x_i,y_j) have been matched in a round, they are not eligible for the next round.
You could do this by building a bipartite graph where your Xs are nodes in one side and Ys are nodes in another side, and the edge weight is cosine similarity. Then you find the max weighted match in this graph. For the next rounds, you eliminate the edges that have already been used in previous round and run the matching algorithm again. For details on how to code max weight matching in bipartite graph, see here.
BTW, this solution is not optimum since we are proceeding from one round to next in a greedy fashion. I have a feeling that getting the optimum solution would be NP hard, but I don't have a proof so can't be sure.
I agree that the question still needs clarification. As Amit expressed, I have a gut feeling that this is an NP hard problem, so I am assuming that you are looking for an approximate solution.
That said, I would need more information on the tradeoffs you would be willing to make (and perhaps I'm just missing something in your question). What are the explicit goals of the algorithm?
Is there a lower threshold for similarity below which you don't want a pairing to happen? I'm still a bit confused as to why there would be individuals which could not be paired up at all during a given round...
Essentially, you are performing a search over the space of possible pairings, correct? Maybe you could use backtracking or some form of constraint-based algorithm to make sure that you can obtain a complete solution for a given round...?
There is a picture book with 100 pages. If dice are rolled randomly to select one of the pages and subsequently rerolled in order to search for a certain picture in the book -- how do I determine the best, worst and average case complexity of this problem?
Proposed answer:
best case: picture is found on the first dice roll
worst case: picture is found on 100th dice roll or picture does not exist
average case: picture is found after 50 dice rolls (= 100 / 2)
Assumption: incorrect pictures are searched at most one time
Given your description of the problem, I don't think your assumption (that incorrect pictures are only "searched" once) sounds right. If you don't make that assumption, then the answer is as shown below. You'll see the answers are somewhat different from what you proposed.
It is possible that you could be successful on the first attempt. Therefore, the first answer is 1.
If you were unlucky, you could keep rolling the wrong number forever. So the second answer is infinity.
The third question is less obvious.
What is the average number of rolls? You need to be familiar with the Geometric Distribution: the number of trials needed to get a single success.
Define p as the probability for a successful trial; p=0.01.
Let Pr(x = k) be the probability that the first successful trial being the k th. Then we're going to have to have (k-1) failures and one success. So Pr(x=k) = (1-p)^(k-1) * p. Verify that this is the "probability mass function" on the wiki page (left column).
The mean of the geometric distribution is 1/p, which is therefore 100. This is the average number of rolls required to find the specific picture.
(Note: We need to consider 1 as the lowest possible value, rather than 0, so use the left hand column of the table on the Wikipedia page.)
To analyze this, think about what the best, worst and average cases actually are. You need to answer three questions to find those three cases:
What is the fewest number of rolls to find the desired page?
What is the largest number of rolls to find the desired page?
What is the average number of rolls to find the desired page?
Once you find the first two, the third should be less tricky. If you need asymptotic notation as opposed to just the number of rolls, think about how the answers to each question change if you change the number of pages in the book (e.g. 200 pages vs 100 pages vs 50 pages).
The worst case is not the page found after 100 dice rolls. That would be is your dice always returned different numbers. The worst case is that you never find the page (the way you stated the problem).
The average case is not average of the best and worst cases, fortunately.
The average case is:
1 * (probability of finding page on the first dice roll)
+ 2 * (probability of finding page on the second dice roll)
+ ...
And yes, the sum is infinite, since in thinking about the worst case we determined that you may have an arbitrarily large number of dice rolls. It doesn't mean that it can't be computed (it could mean that, but it doesn't have to).
The probability of finding the page on the first try is 1/100. What's the probability of finding it on the second dice roll?
You're almost there, but (1 + 2 + ... + 100)/100 isn't 50.
It might help to observe that your random selection method is equivalent to randomly shuffling the whole deck and then searching it in order for your target. Each position is equally likely, so the average is straightforward to calculate. Except of course that you aren't doing all that work up front, just as much as is needed to generate each random number and access the corresponding element.
Note that if your book were stored as a linked list, then the cost of moving from each randomly-selected page to the next selection depends on how far apart they are, which will complicate the analysis quite a lot. You don't actually state that you have constant-time access, and it's possibly debatable whether a "real book" provides that or not.
For that matter, there's more than one way to choose random numbers without repeats, and not all of them have the same running time.
So, you'd need more detail in order to analyse the algorithm in terms of anything other than "number of pages visited".