What is Effectively Computable?

Say that a problem can be solved by a computer --- how good is the solution (program)? How fast? How much storage space is needed? Is the solution the best possible for solving the problem?

These questions are studied within computational complexity theory. We learn about this topic by studying some simple algorithms that work with arrays.

Linear-time algorithms

Perhaps you have a database that is an array (sequence) of int keys connected to data records. The keys aren't arranged in any special way, so finding a data record by its key requires a linear search of an array of ints, e.g., find key 8 in array, [6, 10, 2, 4, 12, 16, 8, 14]:
Step 1:  V
        [6, 10, 2, 4, 12, 16, 8, 14]

Step 2:      V 
        [6, 10, 2, 4, 12, 16, 8, 14]

Step 7:                       V
        [6, 10, 2, 4, 12, 16, 8, 14]
We search the array from front to back. How fast is this algorithm? Without worrying about speeds of chips, we measure the speed of an array algorithm by how many array lookups/updates it must do.

Obviously, with an array of length 8, at most 8 lookups is needed; in general, if the array has length N, at most N lookups is needed. This is the worst-cast time complexity, and in this case is linearly proportional to the length of the array. We say that sequential search is a linear-time algorithm, that is, of order-N.

Here is a table of how a linear-time algorithm performs:

array size, N        worst-case time complexity, in terms of lookups
64                   64
512                  512
10,000               10,000
A linear-time algorithm looks "slow" to its user but is tolerable if the data structure is not "too large".

When you write a one-loop program that counts through the elements of an array, your algorithm runs in linear time.

Log-time algorithms

What if the array of keys was sorted so that the keys were in ascending order? In this case, we can use a "dictionary lookup" algorithm, called binary search, which jumps into the middle of the array, looks at the key there, and then jumps into the middle of the left half or right half, depending whether the desired key is less-than or greater-than the key in the middle:
Find 8 in  [2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22]:

Step 1: look in middle:     V
           [2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22]

Step 2: look in middle of left half:     V
                                  [2, 4, 6, 8, 10 ... ]

Step 3: look in middle of left's right subhalf:    V
                                            [ ..., 8, 10, ... ]
The subarray searched shrinks by half each time there is a lookup, and the key is quickly found --- in worst case, at most log2N lookups are needed to find the key. (Recall that log2N = M means that 2M = N.) This is log-time, that is, of order log-N. Here's a table:
array size, N   linear-time  log-time   
64               64          6
512              512         8
10,000           10,000      12
The speed-up is spectacular; this algorithm is markedly faster in theory and in practice. In practice, regardless of how a database is configured, the search operation must be order log-time or faster --- when you do a search on a web site, the computer program that does the searching is order log-time.

When you write an algorithm that "discards" half of a data structure during each processing step, you are writing a log-time algorithm.

Quadratic-time algorithms

How much time does it take to sort an array of keys into ascending order? A standard technique, called selection sort, works by repeatedly selecting and moving the least element from the unsorted part of the array into the sorted part:

Input array                     Output array
---------------------------    --------------------
[6, 10, 2, 4, 12, 16, 8, 14]    []

[6, 10, 4, 12, 16, 8, 14]       [2]

[6, 10, 12, 16, 8, 14]          [2, 4]

[10, 12, 16, 8, 14]             [2, 4, 6]

 . . .

[16]                            [2, 4, 6, 8, 10, 12, 14]

[]                              [2, 4, 6, 8, 10, 12, 14, 16]
The algorithm scans the N-element input array N-1 times, and 1/2(N2 - N) lookups are done; there are also N updates to the output array. This is a quadratic-time algorithm, order N2. Here is the comparison table:
array size, N   linear     log2N   N2          1/2(N2 + N)
64               64        6     4096         2064
512              512       8     262,144      131,200
10,000           10,000    12    100,000,000  50,002,500
(Note there isn't a gross difference between the numbers in the last two columns. That's why selection sort is called a quadratic-time algorithm.)

A quadratic-time algorithm runs slowly for even small data structures. It cannot be used for any interactive system --- you run this kind of algorithm while you go to lunch. When you write a loop-in-a-loop to process an array, you are writing a quadratic algorithm.

The standard sorting algorithms are quadratic-time, but there is a clever version of sorting, called "quicksort", that is based on a divide-and-conquer strategy, using a kind of binary search to replace one of the nested loops in the standard sorting algorithms. Quicksort runs in order N * log2N time:

N      log2N     N2          N*log2N
64     6       4096         384
512    8       262,144      4096
10,000 12      100,000,000  120,000

Intractable problems and their algorithms

Say you have a maze, and you want to calculate the shortest path of all the paths from the maze's entry to its exit. Or say you have a deck of cards, and you want to print all the shuffles of the deck. Or you have a roadmap of cities, and you want to calculate the shortest tour that lets you visit each city (this is the travelling salesman problem):

These problems require that you calculate all possible combinations of a data structure before you select the best one. Many important industrial and scientific problems, which ask for optimal solutions to complex constraints, are just the problems stated here.

The problems stated above are solvable --- indeed, here is a Python function that calculates all the permutations (shuffles) of a deck of size-many cards:

def permutations(size) :
    """computes a list of all permutations of [1,2,..upto..,size]
       param: size, a nonnegative int
       returns:  answer, a list of all the permutations defined above
    assert size >= 0
    if size == 0 :   
        answer = [[]]  # there is only the "empty permutation" of a size-0 deck
    else :
        sublist = permutations(size - 1)  # compute perms of [1,..upto..,size-1]
        answer = []   # build the answer for  size-many  cards in a deck
        for perm in sublist:  
            # insert  size  in all possible positions of each  perm:
            for i in range(len(perm)+1) :
                answer.append( perm[:i] + [size] + perm[i:] )
    return answer
It uses a recursion and a nested loop. For example, permutations(3) computes [[3, 2, 1], [2, 3, 1], [2, 1, 3], [3, 1, 2], [1, 3, 2], [1, 2, 3]]. Try this function on your computer and see how large an argument you can supply until the computer refuses to respond.

The solution to the above is written in just a few lines of Prolog, which is an ideal language for calculating all combinations that satisfy a set of constraints:

/* permutation(Xs, Zs)  holds true if list Zs is a reordering of list Xs */
permutation([], []).
permutation(Xs, [Z|Zs]) :- select(Z, Xs, Ys),  permutation(Ys, Zs).

/* select(X, HasAnX, HasOneLessX)  "extracts" X from  HasAnX, giving HasOneLessX.  
   It's a built-in Prolog operation, but here's the code:  */
select(X, [X|Rest], Rest).
select(X, [Y|Ys], [Y|Zs]) :- select(X, Ys, Zs).

/* This query computes all the permutations and saves them in  Answer:  */
?- findall(Perm, permutations([0,1,2,...,size], Perm), Answer).

One way of sorting a deck of cards is by computing all the shuffles (permutations) and keeping the one that places the cards in order. But this strategy will be very slow! How slow?

Permutation problems, like all-N-shuffles and travelling-salesman-to-N-cities, require 1*2*3*..upto..*N data-structure lookups/updates to compute their answers. This number is abbreviated as N!, called factorial N, and it grows huge quickly:

N      log2N     N2          N*log2N   N!
64     6       4096         384       126886932185884164103433389335161480802865516174545192198801894375214704230400000000000000
512    8       262,144      4096      347728979313260536328304591754560471199225065564351457034247483155161041206635254347320985033950225364432243311021394545295001702070069013264153113260937941358711864044716186861040899557497361427588282356254968425012480396855239725120562512065555822121708786443620799246550959187232026838081415178588172535280020786313470076859739980965720873849904291373826841584712798618430387338042329771801724767691095019545758986942732515033551529595009876999279553931070378592917099002397061907147143424113252117585950817850896618433994140232823316432187410356341262386332496954319973130407342567282027398579382543048456876800862349928140411905431276197435674603281842530744177527365885721629512253872386613118821540847897493107398381956081763695236422795880296204301770808809477147632428639299038833046264585834888158847387737841843413664892833586209196366979775748895821826924040057845140287522238675082137570315954526727437094904914796782641000740777897919134093393530422760955140211387173650047358347353379234387609261306673773281412893026941927424000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
10,000 12      100,000,000  120,000   ????????????????????????????????????????????????????????????????????????????????
No computer, no matter how fast, can use an algorithm that requires factorial time. A problem that has no faster solution than this is called intractable --- unsolvable in practice.

There is a way to improve the performance of permutation problems: remember partial solutions and reuse them over and over. You can do this in the travelling-salesman problem, where the distances over subpaths in the graph can be saved in a table and reused as often as possible when computing longer paths. (This technique is called dynamic programming.) The resulting algorithm uses about 2N lookups --- exponential time:

N      log2N     N2          N*log2N   2N
64     6       4096         384       18446744073709551616
512    8       262,144      4096      13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084096
10,000 12      100,000,000  120,000   ????????????????????????????????????????????????????????????????????????????????
Exponential-time algorithms are still too slow to use in practice. At best, you start one and come back in a day. (I am not kidding; people do this.)

Computational complexity theory is the study of the fundamental time and space requirements for solving important problems.


  1. Run these tests with the permutations function given earlier:
    Notice that each answer is larger than the previous. How much larger? Well, the size of the answer for permutations(4) is 1*2*3*4 = 24, which is 4 times larger than the answer for permutations(3) (which is sized 1*2*3). And, the size of the answer to permutations(5) will be 5 times larger (and take 5 times as long to compute!) than the one for permutations(4).

    Finish this exercise as follows:

    Compute permutations(5), permutations(6), etc., as far as your computer lets you go. While you wait for each answer to compute and then print, calculate how large each answer will be. Remember that each next test case, permutations(i+1), will be i+1 times slower and its answer is i+i times larger than its predecessor, permutations(i).

    How far did you progress with your experiments until your computer failed to respond? Is the computer lost? tired? worn out? What if you bought a new computer that is twice as fast or even ten times as fast as the one you are now using --- would it help here? Is it realistic to compute all the permutations of an ordinary 52-card deck of cards so that you can prepare for your next trip to Las Vegas? Can't computers do things like this?

  2. If a list of ints is mixed up, we can sort it by finding a permutation of it that is ordered. An int list is ordered if its ints are arranged from smallest to highest, e.g., [1, 2, 4, 5, 6] is ordered, but [6, 5, 4, 1, 2] is not. Write this function in Python: def isOrdered(nums) : """isOrdered looks at the int list, nums, and returns True if nums is ordered. It returns False if nums is not ordered.""" Use your function with this one:
    sortList(nums) :
        """sortList  finds the sorted permutation of int list  nums  by brute force"""
        allindexpermutations = permutations(len(nums))
        for indexperm in allindexpermutations :
            possibleanswer = []
            for i in indexperm :
                possibleanswer = possibleanswer.append(nums[i-1])
            print "Possible sorted version is", possibleanswer
            if isOrdered(possibleanswer) :
                print "Found it!"
                return possibleanswer
    Find a better way of sorting a list of ints.


Herbert Enderton. A Mathematical Introduction to Logic. Academic Press, 1972.
Peter Suber. Gödel's Proof. Earlham College, http://legacy.earlham.edu/~peters/courses/logsys/g-proof.htm, 1997.
Nigel J. Cutland. Computability. Cambridge University Press, 1980.
Robert Sedgewick. Algorithms, 2d ed. Addison-Wesley, 1988.