Sorting 1 million 8-decimal-digit numbers with 1 MB of RAM

Question

I have a computer with 1 MB of RAM and no other local storage  I must use it to accept 1 million 8-digit decimal numbers over a TCP connection  sort them  and then send the sorted list out over another TCP connection    The list of numbers may contain duplicates  which I must not discard  The code will be placed in ROM  so I need not subtract the size of my code from the 1 nbsp MB  I already have code to drive the Ethernet port and handle TCP IP connections  and it requires 2 nbsp KB for its state data  including a 1 nbsp KB buffer via which the code will read and write data  Is there a solution to this problem   Sources Of Question And Answer   slashdot org  cleaton net

User · Answer

Google s  bad  approach  from HN thread  Store RLE-style counts      Your initial data structure is  99999999 0   all zeros  haven t seen any numbers  and then lets say you see the number 3 866 344 so your data structure becomes  3866343 0 1 1 96133654 0  as you can see the numbers will always alternate between number of zero bits and number of  1  bits so you can just assume the odd numbers represent 0 bits and the even numbers 1 bits  This becomes  3866343 1 96133654    Their problem doesn t seem to cover duplicates  but let s say they use  0 1  for duplicates   Big problem  1  insertions for 1M integers would take ages   Big problem  2  like all plain delta encoding solutions  some distributions can t be covered this way  For example  1m integers with distances 0 99  e g   99 each one   Now think the same but with random distance in the range of 0 99   Note  99999999 1000000   99 99   Google s approach is both unworthy  slow  and incorrect  But to their defense  their problem might have been slightly different

User · Answer

Here is a generalized solution to this kind of problem   General procedure  The taken approach is as follows  The algorithm operates on a single buffer of 32-bit words  It performs the following procedure in a loop    We start with a buffer filled with compressed data from the last iteration  The buffer looks like this    compressed sorted empty  Calculate the maximum amount of numbers that can be stored in this buffer  both compressed and uncompressed  Split the buffer into these two sections  beginning with the space for compressed data  ending with the uncompressed data  The buffer looks like    compressed sorted empty empty  Fill the uncompressed section with numbers to be sorted  The buffer looks like    compressed sorted empty uncompressed unsorted  Sort the new numbers with an in-place sort  The buffer looks like    compressed sorted empty uncompressed sorted  Right-align any already compressed data from the previous iteration in the compressed section  At this point the buffer is partitioned    empty compressed sorted uncompressed sorted  Perform a streaming decompression-recompression on the compressed section  merging in the sorted data in the uncompressed section  The old compressed section is consumed as the new compressed section grows  The buffer looks like    compressed sorted empty    This procedure is performed until all numbers have been sorted   Compression  This algorithm of course only works when it s possible to calculate the final compressed size of the new sorting buffer before actually knowing what will actually be compressed  Next to that  the compression algorithm needs to be good enough to solve the actual problem   The used approach uses three steps  First  the algorithm will always store sorted sequences  therefore we can instead store purely the differences between consecutive entries  Each difference is in the range  0  99999999    These differences are then encoded as a unary bitstream  A 1 in this stream means  Add 1 to the accumulator  A 0 means  Emit the accumulator as an entry  and reset   So difference N will be represented by N 1 s and one 0   The sum of all differences will approach the maximum value that the algorithm supports  and the count of all differences will approach the amount of values inserted in the algorithm  This means we expect the stream to  at the end  contain max value 1 s and count 0 s  This allows us to calculate the expected probability of a 0 and 1 in the stream  Namely  the probability of a 0 is count  count maxval  and the probability of a 1 is maxval  count maxval    We use these probabilities to define an arithmetic coding model over this bitstream  This arithmetic code will encode exactly this amounts of 1 s and 0 s in optimal space  We can calculate the space used by this model for any intermediate bitstream as  bits   encoded   log2 1   amount   maxval    maxval   log2 1   maxval   amount   To calculate the total required space for the algorithm  set encoded equal to amount   To not require a ridiculous amount of iterations  a small overhead can be added to the buffer  This will ensure that the algorithm will at least operate on the amount of numbers that fit in this overhead  as by far the largest time cost of the algorithm is the arithmetic coding compression and decompression each cycle   Next to that  some overhead is necessary to store bookkeeping data and to handle slight inaccuracies in the fixed-point approximation of the arithmetic coding algorithm  but in total the algorithm is able to fit in 1MiB of space even with an extra buffer that can contain 8000 numbers  for a total of 1043916 bytes of space   Optimality  Outside of reducing the  small  overhead of the algorithm it should be theoretically impossible to get a smaller result  To just contain the entropy of the final result  1011717 bytes would be necessary  If we subtract the extra buffer added for efficiency this algorithm used 1011916 bytes to store the final result   overhead

User · Answer

I would try a Radix Tree  If you could store the data in a tree  you could then do an in-order traverse to transmit the data   I m not sure you could fit this into 1MB  but I think it s worth a try

User · Answer

My suggestions here owe a lot to Dan s solution  First off I assume the solution must handle all possible input lists  I think the popular answers do not make this assumption  which IMO is a huge mistake    It is known that no form of lossless compression will reduce the size of all inputs   All the popular answers assume they will be able to apply compression effective enough to allow them extra space  In fact  a chunk of extra space large enough to hold some portion of their partially completed list in an uncompressed form and allow them to perform their sorting operations  This is just a bad assumption   For such a solution  anyone with knowledge of how they do their compression will be able to design some input data that does not compress well for this scheme  and the  solution  will most likely then break due to running out of space   Instead I take a mathematical approach  Our possible outputs are all the lists of length LEN consisting of elements in the range 0  MAX  Here the LEN is 1 000 000 and our MAX is 100 000 000   For arbitrary LEN and MAX  the amount of bits needed to encode this state is   Log2 MAX Multichoose LEN   So for our numbers  once we have completed recieving and sorting  we will need at least Log2 100 000 000 MC 1 000 000  bits to store our result in a way that can uniquely distinguish all possible outputs   This is    988kb  So we actually have enough space to hold our result  From this point of view  it is possible    Deleted pointless rambling now that better examples exist      Best answer is here   Another good answer is here and basically uses insertion sort as the function to expand the list by one element  buffers a few elements and pre-sorts  to allow insertion of more than one at a time  saves a bit of time   uses a nice compact state encoding too  buckets of seven bit deltas

User · Answer

I think the solution is to combine techniques from video encoding  namely the discrete cosine transformation   In digital video  rather recording the changing the brightness or colour of video as regular values such as 110 112 115 116  each is subtracted from the last  similar to run length encoding   110 112 115 116 becomes 110 2 3 1   The values  2 3 1 require less bits than the originals  So lets say we create a list of the input values as they arrive on the socket  We are storing in each element  not the value  but the offset of the one before it   We sort as we go  so the offsets are only going to be positive   But the offset could be 8 decimal digits wide which this fits in 3 bytes  Each element can t be 3 bytes  so we need to pack these  We could use the top bit of each byte as a  quot continue bit quot   indicating that the next byte is part of the number and the lower 7 bits of each byte need to be combined  zero is valid for duplicates  As the list fills up  the numbers should be get closer together  meaning on average only 1 byte is used to determine the distance to the next value   7 bits of value and 1 bit of offset if convenient  but there may be a sweet spot that requires less than 8 bits for a  quot continue quot  value  Anyway  I did some experiment  I use a random number generator and I can fit a million sorted 8 digit decimal numbers into about 1279000 bytes  The average space between each number is consistently 99    public class Test       public static void main String   args  throws IOException              1 million values         int   values   new int 1000000               create random values up to 8 digits lrong         Random random   new Random            for  int x 0 x lt values length x                  values x    random nextInt 100000000                     Arrays sort values            ByteArrayOutputStream baos   new ByteArrayOutputStream             int av   0              writeCompact baos  values 0           first value         for  int x 1 x lt values length x                  int v   values x  - values x-1       difference             av    v              System out println values x     quot  diff  quot    v               writeCompact baos  v                      System out println  quot Average offset  quot     av values length            System out println  quot Fits in  quot    baos toByteArray   length              public static void writeCompact OutputStream os  long value  throws IOException           do               int b    int  value  amp  0x7f              value    value  amp  0x7fffffffffffffffl   gt  gt  7              os write value    0   b    b   0x80              while  value    0

User · Answer

If the input stream could be received few times this would be much easier  no info about that  idea and time-performance problem   Then  we could count the decimal values  With counted values it would be easy to make the output stream  Compress by counting the values  It depends what would be in the input stream

User · Answer

Sorting is a secondary problem here  As other said  just storing the integers is hard  and cannot work on all inputs  since 27 bits per number would be necessary   My take on this is  store only the differences between the consecutive  sorted  integers  as they will be most likely small  Then use a compression scheme  e g  with 2 additional bits per input number  to encode how many bits the number is stored on  Something like   00 - gt  5 bits 01 - gt  11 bits 10 - gt  19 bits 11 - gt  27 bits   It should be possible to store a fair number of possible input lists within the given memory constraint  The maths of how to pick the compression scheme to have it work on the maximum number of inputs  are beyond me   I hope you may be able to exploit domain-specific knowledge of your input to find a good enough integer compression scheme based on this   Oh and then  you do an insertion sort on that sorted list as you receive data

User · Answer

Suppose this task is possible   Just prior to output  there will be an in-memory representation of the million sorted numbers   How many different such representations are there   Since there may be repeated numbers we can t use nCr  choose   but there is an operation called multichoose that works on multisets    There are 2 2e2436455 ways to choose a million numbers in range 0  99 999 999  That requires 8 093 730 bits to represent every possible combination  or 1 011 717 bytes      So theoretically it may be possible  if you can come up with a sane  enough  representation of the sorted list of numbers   For example  an insane representation might require a 10MB lookup table or thousands of lines of code   However  if  1M RAM  means one million bytes  then clearly there is not enough space   The fact that 5  more memory makes it theoretically possible suggests to me that the representation will have to be VERY efficient and probably not sane

User · Answer

There is one rather sneaky trick not mentioned here so far  We assume that you have no extra way to store data  but that is not strictly true    One way around your problem is to do the following horrible thing  which should not be attempted by anyone under any circumstances  Use the network traffic to store data  And no  I don t mean NAS   You can sort the numbers with only a few bytes of RAM in the following way    First take 2 variables  COUNTER and VALUE  First set all registers to 0  Every time you receive an integer I  increment COUNTER and set VALUE to max VALUE  I   Then send an ICMP echo request packet with data set to I to the router  Erase I and repeat  Every time you receive the returned ICMP packet  you simply extract the integer and send it back out again in another echo request  This produces a huge number of ICMP requests scuttling backward and forward containing the integers    Once COUNTER reaches 1000000  you have all of the values stored in the incessant stream of ICMP requests  and VALUE now contains the maximum integer  Pick some threshold T  gt  gt  1000000  Set COUNTER to zero  Every time you receive an ICMP packet  increment COUNTER and send the contained integer I back out in another echo request  unless I VALUE  in which case transmit it to the destination for the sorted integers  Once COUNTER T  decrement VALUE by 1  reset COUNTER to zero and repeat  Once VALUE reaches zero you should have transmitted all integers in order from largest to smallest to the destination  and have only used about 47 bits of RAM for the two persistent variables  and whatever small amount you need for the temporary values    I know this is horrible  and I know there can be all sorts of practical issues  but I thought it might give some of you a laugh or at least horrify you

User · Answer

If the numbers are evenly distributed we can use Counting sort  We should keep the number of times that each number is repeated in an array   Available space is  1 MB - 3 KB   1045504 B or  8364032 bits Number of bits per number  8364032 1000000   8 Therefore  we can store the number of times each number is repeated to the maximum of 2 8-1 255  Using this approach we have an extra  364032 bits unused that can be used to handle cases where a number is repeated more than 255 times  For example we can say a number 255 indicates a repetition greater than or equal to 255  In this case we should store a sequence of numbers repetitions  We can handle 7745 special cases as shown bellow   364032   bits need to represent each number   bits needed to represent 1 million   364032    27 20  7745

User · Answer

If it is possible to read the input file more than once  your problem statement doesn t say it can t   the following should work  It is described in Benchley s book  Programming Perls   If we store each number in 8 bytes we can store 250 000 numbers in one megabyte  Use a program that makes 40 passes over the input file  On the first pass it reads into memory any integer between 0 and 249 999  sorts the  at most  250 000 integers and writes them to the output file  The second pass sorts the integers from 250 000 to 499 999 and so on to the 40th pass  which sorts 9 750 000 to 9 999 999

User · Answer

We could play with the networking stack to send the numbers in sorted order before we have all the numbers   If you send 1M of data  TCP IP will break it into 1500 byte packets and stream them in order to the target   Each packet will be given a sequence number   We can do this by hand   Just before we fill our RAM we can sort what we have and send the list to our target but leave holes in our sequence around each number   Then process the 2nd 1 2 of the numbers the same way using those holes in the sequence   The networking stack on the far end will assemble the resulting data stream in order of sequence before handing it up to the application   It s using the network to perform a merge sort   This is a total hack  but I was inspired by the other networking hack listed previously

User · Answer

A radix tree representation would come close to handling this problem  since the radix tree takes advantage of  prefix compression    But it s hard to conceive of a radix tree representation that could represent a single node in one byte -- two is probably about the limit   But  regardless of how the data is represented  once it is sorted it can be stored in prefix-compressed form  where the numbers 10  11  and 12 would be represented by  say 001b  001b  001b  indicating an increment of 1 from the previous number   Perhaps  then  10101b would represent an increment of 5  1101001b an increment of 9  etc

User · Answer

Gilmanov s answer is very wrong in its assumptions  It starts speculating based in a pointless measure of a million consecutive integers  That means no gaps  Those random gaps  however small  really makes it a poor idea   Try it yourself  Get 1 million random 27 bit integers  sort them  compress with 7-Zip  xz  whatever LZMA you want  The result is over 1 5 nbsp MB  The premise on top is compression of sequential numbers  Even delta encoding of that is over 1 1 nbsp MB  And never mind this is using over 100 nbsp MB of RAM for compression  So even the compressed integers don t fit the problem and never mind run time RAM usage   It s saddens me how people just upvote pretty graphics and rationalization    include  lt stdint h gt   include  lt stdlib h gt   include  lt time h gt   int32 t ints 1000000      Random 27-bit integers  int cmpi32 const void  a  const void  b        return     int32 t   a -   int32 t   b       int main         int32 t  pi   ints     Pointer to input ints  REPLACE W  read from net          Fill pseudo-random integers of 27 bits     srand time NULL        for  int i   0  i  lt  1000000  i            ints i    rand    amp    1 lt  lt 27  - 1      Random 32 bits masked to 27 bits      qsort ints  1000000  sizeof  ints 0    cmpi32      Sort 1000000 int32s         Now delta encode  optional  store differences to previous int     for  int i   1  prev   ints 0   i  lt  1000000  i              ints i  -  prev          prev       ints i              FILE  f   fopen  ints bin    w        fwrite ints  4  1000000  f       fclose f       exit 0        Now compress ints bin with LZMA       xz -f --keep ints bin         100 MB RAM   7z a ints bin 7z ints bin     130 MB RAM   ls -lh ints bin      3 8M ints bin     1 1M ints bin 7z     1 2M ints bin xz

User · Answer

A solution is possible only because of the difference between 1 megabyte and 1 million bytes  There are about 2 to the power 8093729 5 different ways to choose 1 million 8-digit numbers with duplicates allowed and order unimportant  so a machine with only 1 million bytes of RAM doesn t have enough states to represent all the possibilities  But 1M  less 2k for TCP IP  is 1022 1024 8   8372224 bits  so a solution is possible   Part 1  initial solution  This approach needs a little more than 1M  I ll refine it to fit into 1M later   I ll store a compact sorted list of numbers in the range 0 to 99999999 as a sequence of sublists of 7-bit numbers  The first sublist holds numbers from 0 to 127  the second sublist holds numbers from 128 to 255  etc  100000000 128 is exactly 781250  so 781250 such sublists will be needed   Each sublist consists of a 2-bit sublist header followed by a sublist body  The sublist body takes up 7 bits per sublist entry  The sublists are all concatenated together  and the format makes it possible to tell where one sublist ends and the next begins  The total storage required for a fully populated list is 2 781250   7 1000000   8562500 bits  which is about 1 021 M-bytes   The 4 possible sublist header values are   00 Empty sublist  nothing follows   01 Singleton  there is only one entry in the sublist and and next 7 bits hold it   10 The sublist holds at least 2 distinct numbers  The entries are stored in non-decreasing order  except that the last entry is less than or equal to the first  This allows the end of the sublist to be identified  For example  the numbers 2 4 6 would be stored as  4 6 2   The numbers 2 2 3 4 4 would be stored as  2 3 4 4 2    11 The sublist holds 2 or more repetitions of a single number  The next 7 bits give the number  Then come zero or more 7-bit entries with the value 1  followed by a 7-bit entry with the value 0  The length of the sublist body dictates the number of repetitions  For example  the numbers 12 12 would be stored as  12 0   the numbers 12 12 12 would be stored as  12 1 0   12 12 12 12 would be  12 1 1 0  and so on   I start off with an empty list  read a bunch of numbers in and store them as 32 bit integers  sort the new numbers in place  using heapsort  probably  and then merge them into a new compact sorted list  Repeat until there are no more numbers to read  then walk the compact list once more to generate the output   The line below represents memory just before the start of the list merge operation  The  O s are the region that hold the sorted 32-bit integers  The  X s are the region that hold the old compact list  The     signs are the expansion room for the compact list  7 bits for each integer in the  O s  The  Z s are other random overhead   ZZZOOOOOOOOOOOOOOOOOOOOOOOOOO          XXXXXXXXXXXXXXXXXXXXXXXXXX   The merge routine starts reading at the leftmost  O  and at the leftmost  X   and starts writing at the leftmost      The write pointer doesn t catch the compact list read pointer until all of the new integers are merged  because both pointers advance 2 bits for each sublist and 7 bits for each entry in the old compact list  and there is enough extra room for the 7-bit entries for the new numbers   Part 2  cramming it into 1M  To Squeeze the solution above into 1M  I need to make the compact list format a bit more compact  I ll get rid of one of the sublist types  so that there will be just 3 different possible sublist header values  Then I can use  00    01  and  1  as the sublist header values and save a few bits  The sublist types are   A Empty sublist  nothing follows   B Singleton  there is only one entry in the sublist and and next 7 bits hold it   C The sublist holds at least 2 distinct numbers  The entries are stored in non-decreasing order  except that the last entry is less than or equal to the first  This allows the end of the sublist to be identified  For example  the numbers 2 4 6 would be stored as  4 6 2   The numbers 2 2 3 4 4 would be stored as  2 3 4 4 2    D The sublist consists of 2 or more repetitions of a single number   My 3 sublist header values will be  A    B  and  C   so I need a way to represent D-type sublists   Suppose I have the C-type sublist header followed by 3 entries  such as  C 17  101  58    This can t be part of a valid C-type sublist as described above  since the third entry is less than the second but more than the first  I can use this type of construct to represent a D-type sublist  In bit terms  anywhere I have  C 00       1        01        is an impossible C-type sublist  I ll use this to represent a sublist consisting of 3 or more repetitions of a single number  The first two 7-bit words encode the number  the  N  bits below  and are followed by zero or more  0100001  words followed by a  0100000  word   For example  3 repetitions   C 00NNNNN  1NN0000  0100000    4 repetitions   C 00NNNNN  1NN0000  0100001  0100000    and so on    That just leaves lists that hold exactly 2 repetitions of a single number  I ll represent those with another impossible C-type sublist pattern   C 0        11       10         There s plenty of room for the 7 bits of the number in the first 2 words  but this pattern is longer than the sublist that it represents  which makes things a bit more complex  The five question-marks at the end can be considered not part of the pattern  so I have   C 0NNNNNN  11N     10  as my pattern  with the number to be repeated stored in the  N s  That s 2 bits too long   I ll have to borrow 2 bits and pay them back from the 4 unused bits in this pattern  When reading  on encountering  C 0NNNNNN  11N00AB 10   output 2 instances of the number in the  N s  overwrite the  10  at the end with bits A and B  and rewind the read pointer by 2 bits  Destructive reads are ok for this algorithm  since each compact list gets walked only once   When writing a sublist of 2 repetitions of a single number  write  C 0NNNNNN 11N00  and set the borrowed bits counter to 2  At every write where the borrowed bits counter is non-zero  it is decremented for each bit written and  10  is written when the counter hits zero  So the next 2 bits written will go into slots A and B  and then the  10  will get dropped onto the end   With 3 sublist header values represented by  00    01  and  1   I can assign  1  to the most popular sublist type  I ll need a small table to map sublist header values to sublist types  and I ll need an occurrence counter for each sublist type so that I know what the best sublist header mapping is   The worst case minimal representation of a fully populated compact list occurs when all the sublist types are equally popular  In that case I save 1 bit for every 3 sublist headers  so the list size is 2 781250   7 1000000 - 781250 3   8302083 3 bits  Rounding up to a 32 bit word boundary  thats 8302112 bits  or 1037764 bytes   1M minus the 2k for TCP IP state and buffers is 1022 1024   1046528 bytes  leaving me 8764 bytes to play with   But what about the process of changing the sublist header mapping   In the memory map below   Z  is random overhead      is free space   X  is the compact list   ZZZ     XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX   Start reading at the leftmost  X  and start writing at the leftmost     and work right  When it s done the compact list will be a little shorter and it will be at the wrong end of memory   ZZZXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX          So then I ll need to shunt it to the right   ZZZ       XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX   In the header mapping change process  up to 1 3 of the sublist headers will be changing from 1-bit to 2-bit  In the worst case these will all be at the head of the list  so I ll need at least 781250 3 bits of free storage before I start  which takes me back to the memory requirements of the previous version of the compact list     To get around that  I ll split the 781250 sublists into 10 sublist groups of 78125 sublists each  Each group has its own independent sublist header mapping  Using the letters A to J for the groups   ZZZ     AAAAAABBCCCCDDDDDEEEFFFGGGGGGGGGGGHHIJJJJJJJJJJJJJJJJJJJJ   Each sublist group shrinks or stays the same during a sublist header mapping change   ZZZ     AAAAAABBCCCCDDDDDEEEFFFGGGGGGGGGGGHHIJJJJJJJJJJJJJJJJJJJJ ZZZAAAAAA     BBCCCCDDDDDEEEFFFGGGGGGGGGGGHHIJJJJJJJJJJJJJJJJJJJJ ZZZAAAAAABB     CCCCDDDDDEEEFFFGGGGGGGGGGGHHIJJJJJJJJJJJJJJJJJJJJ ZZZAAAAAABBCCC      DDDDDEEEFFFGGGGGGGGGGGHHIJJJJJJJJJJJJJJJJJJJJ ZZZAAAAAABBCCCDDDDD      EEEFFFGGGGGGGGGGGHHIJJJJJJJJJJJJJJJJJJJJ ZZZAAAAAABBCCCDDDDDEEE      FFFGGGGGGGGGGGHHIJJJJJJJJJJJJJJJJJJJJ ZZZAAAAAABBCCCDDDDDEEEFFF      GGGGGGGGGGGHHIJJJJJJJJJJJJJJJJJJJJ ZZZAAAAAABBCCCDDDDDEEEFFFGGGGGGGGGG       HHIJJJJJJJJJJJJJJJJJJJJ ZZZAAAAAABBCCCDDDDDEEEFFFGGGGGGGGGGHH       IJJJJJJJJJJJJJJJJJJJJ ZZZAAAAAABBCCCDDDDDEEEFFFGGGGGGGGGGHHI       JJJJJJJJJJJJJJJJJJJJ ZZZAAAAAABBCCCDDDDDEEEFFFGGGGGGGGGGHHIJJJJJJJJJJJJJJJJJJJJ        ZZZ       AAAAAABBCCCDDDDDEEEFFFGGGGGGGGGGHHIJJJJJJJJJJJJJJJJJJJJ   The worst case temporary expansion of a sublist group during a mapping change is 78125 3   26042 bits  under 4k  If I allow 4k plus the 1037764 bytes for a fully populated compact list  that leaves me 8764 - 4096   4668 bytes for the  Z s in the memory map   That should be plenty for the 10 sublist header mapping tables  30 sublist header occurrence counts and the other few counters  pointers and small buffers I ll need  and space I ve used without noticing  like stack space for function call return addresses and local variables   Part 3  how long would it take to run   With an empty compact list the 1-bit list header will be used for an empty sublist  and the starting size of the list will be 781250 bits  In the worst case the list grows 8 bits for each number added  so 32   8   40 bits of free space are needed for each of the 32-bit numbers to be placed at the top of the list buffer and then sorted and merged  In the worst case  changing the sublist header mapping results in a space usage of 2 781250   7 entries - 781250 3 bits   With a policy of changing the sublist header mapping after every fifth merge once there are at least 800000 numbers in the list  a worst case run would involve a total of about 30M of compact list reading and writing activity   Source   http   nick cleaton net ramsortsol html

User · Answer

My original answer was wrong  sorry for the bad math  see below the break    How about this   The first 27 bits store the lowest number you have seen  then the difference to the next number seen  encoded as follows  5 bits to store the number of bits used in storing the difference  then the difference  Use 00000 to indicate that you saw that number again   This works because as more numbers are inserted  the average difference between numbers goes down  so you use less bits to store the difference as you add more numbers  I believe this is called a delta list   The worst case I can think of is all numbers evenly spaced  by 100   e g  Assuming 0 is the first number   000000000000000000000000000 00111 1100100                                                                       a million times  27   1 000 000    5 7  bits     427k     Reddit to the rescue   If all you had to do was sort them  this problem would be easy  It takes 122k  1 million bits  to store which numbers you have seen  0th bit on if 0 was seen  2300th bit on if 2300 was seen  etc   You read the numbers  store them in the bit field  and then shift the bits out while keeping a count   BUT  you have to remember how many you have seen  I was inspired by the sublist answer above to come up with this scheme   Instead of using one bit  use either 2 or 27 bits    00 means you did not see the number  01 means you saw it once 1 means you saw it  and the next 26 bits are the count of how many times    I think this works  if there are no duplicates  you have a 244k list  In the worst case you see each number twice  if you see one number three times  it shortens the rest of the list for you   that means you have seen 50 000 more than once  and you have seen 950 000 items 0 or 1 times   50 000   27   950 000   2   396 7k   You can make further improvements if you use the following encoding   0 means you did not see the number 10 means you saw it once 11 is how you keep count   Which will  on average  result in 280 7k of storage   EDIT  my Sunday morning math was wrong   The worst case is we see 500 000 numbers twice  so the math becomes   500 000  27   500 000  2   1 77M  The alternate encoding results in an average storage of  500 000   27   500 000   1 70M

User · Answer

As ROM size does not count one does not need any additional RAM besides the TCP buffers  Just implement a big finite-state machine  Each state represents a multi-set of numbers read in  After reading a million numbers one has just to print the numbers corresponding to the reached state

User · Answer

Have you tried converting to hex   I can see a big reduction on filesize after and before  then  work by part with the free space  Maybe  converting to dec again  order  hex  another chunk convert to dec  order     Sorry   I don t know if could work    for i in  1  10000  do echo   od -N1 -An -i  dev urandom    done  gt  10000numbers   for i in   cat 10000numbers    do printf   x n   i  done  gt  10000numbers hex   ls -lah total 100K drwxr-xr-x  2 diego diego 4 0K oct 22 22 32   drwx------ 39 diego diego  12K oct 22 22 31    -rw-r--r--  1 diego diego  29K oct 22 22 33 10000numbers hex -rw-r--r--  1 diego diego  35K oct 22 22 31 10000numbers

User · Answer

There is one solution to this problem across all possible inputs  Cheat    Read m values over TCP  where m is near the max that can be sorted in memory  maybe n 4  Sort the 250 000  or so  numbers and output them  Repeat for the other 3 quarters  Let the receiver merge the 4 lists of numbers it has received as it processes them   It s not much slower than using a single list

User · Answer

I would exploit the retransmission behaviour of TCP    Make the TCP component create a large receive window  Receive some amount of packets without sending an ACK for them    Process those in passes creating some  prefix  compressed data structure Send duplicate ack for last packet that is not needed anymore wait for retransmission timeout Goto 2  All packets were accepted   This assumes some kind of benefit of buckets or multiple passes   Probably by sorting the batches buckets and merging them  -  radix trees  Use this technique to accept and sort the first 80  then read the last 20   verify that the last 20  do not contain numbers that would land in the first 20  of the lowest numbers  Then send the 20  lowest numbers  remove from memory  accept the remaining 20  of new numbers and merge

User · Answer

We have 1 nbsp MB - 3 nbsp KB RAM   2 23 - 3 2 13 bits   8388608 - 24576   8364032 bits available   We are given 10 6 numbers in a 10 8 range  This gives an average gap of  100  lt  2 7   128  Let s first consider the simpler problem of fairly evenly spaced numbers when all gaps are  lt  128  This is easy  Just store the first number and the 7-bit gaps    27 bits    10 6 7-bit gap numbers   7000027 bits required  Note repeated numbers have gaps of 0   But what if we have gaps larger than 127   OK  let s say a gap size  lt  127 is represented directly  but a gap size of 127 is followed by a continuous 8-bit encoding for the actual gap length    10xxxxxx xxxxxxxx                         127    16 383  110xxxxx xxxxxxxx xxxxxxxx                16384    2 097 151   etc   Note this number representation describes its own length so we know when the next gap number starts   With just small gaps  lt  127  this still requires 7000027 bits   There can be up to  10 8   2 7    781250 23-bit gap number  requiring an extra 16 781 250   12 500 000 bits which is too much  We need a more compact and slowly increasing representation of gaps   The average gap size is 100 so if we reorder them as  100  99  101  98  102       2  198  1  199  0  200  201  202       and index this with a dense binary Fibonacci base encoding with no pairs of zeros  for example  11011 8 5 2 1 16  with numbers delimited by  00  then I think we can keep the gap representation short enough  but it needs more analysis

User · Answer

To represent the sorted array one can just store the first element and the difference between adjacent elements  In this way we are concerned with encoding 10 6 elements that can sum up to at most 10 8  Let s call this D  To encode the elements of D one can use a Huffman code  The dictionary for the Huffman code can be created on the go and the array updated every time a new item is inserted in the sorted array  insertion sort   Note that when the dictionary changes because of a new item the whole array should be updated to match the new encoding    The average number of bits for encoding each element of D is maximized if we have equal number of each unique element   Say elements d1  d2       dN in D each appear F times  In that case  in worst case we have both 0 and 10 8 in input sequence  we have  sum 1 lt  i lt  N  F  di   10 8   where   sum 1 lt  i lt  N  F   10 6  or F 10 6 N and the normalized frequency will be p  F 10  1 N  The average number of bits will be -log2 1 P    log2 N   Under these circumstances we should find a case that maximizes N  This happens if we have consecutive numbers for di starting from 0  or  di  i-1  therefore  10 8 sum 1 lt  i lt  N  F  di   sum 1 lt  i lt  N   10 6 N   i-1     10 6 N  N  N-1  2  i e   N  lt   201  And for this case average number of bits is log2 201  7 6511 which means we will need around 1 byte per input element for saving the sorted array  Note that this doesn t mean D in general cannot have more than 201 elements  It just sows that if elements of D are uniformly distributed  it cannot have more than 201 unique values

User · Answer

If we don t know anything about those numbers  we are limited by the following constraints    we need to load all numbers before we can sort them them  the set of numbers is not compressible    If these assumptions hold  there is no way to carry out your task  as you will need at least 26 575 425 bits of storage  3 321 929 bytes    What can you tell us about your data

User · Answer

What kind of computer are you using  It may not have any other  normal  local storage  but does it have video RAM  for example  1 megapixel x 32 bits per pixel  say  is pretty close to your required data input size    I largely ask in memory of the old Acorn RISC PC  which could  borrow  VRAM to expand the available system RAM  if you chose a low resolution or low colour-depth screen mode    This was rather useful on a machine with only a few MB of normal RAM

User · Answer

The trick is to represent the algorithms state  which is an integer multi-set  as a compressed stream of  increment counter      and  output counter      characters  For example  the set  0 3 3 4  would be represented as             followed by any number of     characters  To modify the multi-set you stream out the characters  keeping only a constant amount decompressed at a time  and make changes inplace before streaming them back in compressed form   Details  We know there are exactly 10 6 numbers in the final set  so there are at most 10 6     characters  We also know that our range has size 10 8  meaning there are at most 10 8     characters  The number of ways we can arrange 10 6    s amongst 10 8    s is  10 8   10 6  choose 10 6  and so specifying some particular arrangement takes  0 965 MiB  of data  That ll be a tight fit   We can treat each character as independent without exceeding our quota  There are exactly 100 times more     characters than     characters  which simplifies to 100 1 odds of each character being a     if we forget that they are dependent  Odds of 100 101 corresponds to  0 08 bits per character  for an almost identical total of  0 965 MiB  ignoring the dependency has a cost of only  12 bits in this case     The simplest technique for storing independent characters with known prior probability is Huffman coding  Note that we need an impractically large tree  A huffman tree for blocks of 10 characters has an average cost per block of about 2 4 bits  for a total of  2 9 Mib  A huffman tree for blocks of 20 characters has an average cost per block of about 3 bits  which is a total of  1 8 MiB  We re probably going to need a block of size on the order of a hundred  implying more nodes in our tree than all the computer equipment that has ever existed can store    However  ROM is technically  free  according to the problem and practical solutions that take advantage of the regularity in the tree will look essentially the same   Pseudo-code   Have a sufficiently large huffman tree  or similar block-by-block compression data  stored in ROM Start with a compressed string of 10 8     characters  To insert the number N  stream out the compressed string until N     characters have gone past then insert a      Stream the recompressed string back over the previous one as you go  keeping a constant amount of buffered blocks to avoid over under-runs  Repeat one million times   input  stream decompress insert compress   then decompress to output

User · Answer

There are 10 6 values in a range of 10 8  so there s one value per hundred code points on average  Store the distance from the Nth point to the  N 1 th  Duplicate values have a skip of 0  This means that the skip needs an average of just under 7 bits to store  so a million of them will happily fit into our 8 million bits of storage   These skips need to be encoded into a bitstream  say by Huffman encoding  Insertion is by iterating through the bitstream and rewriting after the new value  Output by iterating through   and writing out the implied values  For practicality  it probably wants to be done as  say  10 4 lists covering 10 4 code points  and an average of 100 values  each   A good Huffman tree for random data can be built a priori by assuming a Poisson distribution  mean variance 100  on the length of the skips  but real statistics can be kept on the input and used to generate an optimal tree to deal with pathological cases

User · Answer

Here s some working C   code which solves the problem   Proof that the memory constraints are satisfied   Editor  There is no proof of the maximum memory requirements offered by the author either in this post or in his blogs  Since the number of bits necessary to encode a value depends on the values previously encoded  such a proof is likely non-trivial  The author notes that the largest encoded size he could stumble upon empirically was 1011732  and chose the buffer size 1013000 arbitrarily   typedef unsigned int u32   namespace WorkArea       static const u32 circularSize   253250      u32 circular circularSize      0               consumes 1013000 bytes      static const u32 stageSize   8000      u32 stage stageSize                            consumes 32000 bytes            Together  these two arrays take 1045000 bytes of storage  That leaves 1048576 - 1045000 - 2 times 1024   1528 bytes for remaining variables and stack space   It runs in about 23 seconds on my Xeon W3520  You can verify that the program works using the following Python script  assuming a program name of sort1mb exe   from subprocess import   import random  sequence    random randint 0  99999999  for i in xrange 1000000    sorter   Popen  sort1mb exe   stdin PIPE  stdout PIPE  for value in sequence      sorter stdin write   08d n    value  sorter stdin close    result    int line  for line in sorter stdout  print  OK   if result    sorted sequence  else  Error      A detailed explanation of the algorithm can be found in the following series of posts    1MB Sorting Explained Arithmetic Coding and the 1MB Sorting Problem Arithmetic Encoding Using Fixed-Point Math

User · Answer

You have to count up to at most 99 999 999 and indicate 1 000 000 stops along the way   So a bitstream can be used that is interpreted such that at 1 indicates in increment a counter and a 0 indicates to output a number   If the first 8 bits in the stream are 00110010  we would have 0  0  2  2  3 so far   log 99 999 999   1 000 000    log 2    26 59   You have 2 28 bits in your memory   You only need to use half

User · Answer

Now aiming to an actual solution  covering all possible cases of input in the 8 digit range with only 1MB of RAM  NOTE  work in progress  tomorrow will continue  Using arithmetic coding of deltas of the sorted ints  worst case for 1M sorted ints would cost about 7bits per entry  since 99999999 1000000 is 99  and log2 99  is almost 7 bits    But you need the 1m integers sorted to get to 7 or 8 bits  Shorter series would have bigger deltas  therefore more bits per element   I m working on taking as many as possible and compressing  almost  in-place  First batch of close to 250K ints would need about 9 bits each at best  So result would take about 275KB  Repeat with remaining free memory a few times  Then decompress-merge-in-place-compress those compressed chunks  This is quite hard  but possible  I think   The merged lists would get closer and closer to the 7bit per integer target  But I don t know how many iterations it would take of the merge loop  Perhaps 3   But the imprecision of the arithmetic coding implementation might make it impossible  If this problem is possible at all  it would be extremely tight   Any volunteers

User · Answer

I think one way to think about this is from a combinatorics viewpoint  how many possible combinations of sorted number orderings are there  If we give the combination 0 0 0      0 the code 0  and 0 0 0     1 the code 1  and 99999999  99999999      99999999 the code N  what is N  In other words  how big is the result space   Well  one way to think about this is noticing that this is a bijection of the problem of finding the number of monotonic paths in an N x M grid  where N   1 000 000 and M   100 000 000  In other words  if you have a grid that is 1 000 000 wide and 100 000 000 tall  how many shortest paths from the bottom left to the top right are there  Shortest paths of course require you only ever either move right or up  if you were to move down or left you would be undoing previously accomplished progress   To see how this is a bijection of our number sorting problem  observe the following   You can imagine any horizontal leg in our path as a number in our ordering  where the Y location of the leg represents the value     So if the path simply moves to the right all the way to the end  then jumps all the way to the top  that is equivalent to the ordering 0 0 0     0  if it instead begins by jumping all the way to the top and then moves to the right 1 000 000 times  that is equivalent to 99999999 99999999      99999999  A path where it moves right once  then up once  then right one  then up once  etc to the very end  then necessarily jumps all the way to the top   is equivalent to 0 1 2 3     999999   Luckily for us this problem has already been solved  such a grid has  N   M  Choose  M  paths    1 000 000   100 000 000  Choose  100 000 000     2 27   10 2436455  N thus equals 2 27   10 2436455  and so the code 0 represents 0 0 0     0 and the code 2 27   10 2436455 and some change represents 99999999 99999999      99999999   In order to store all the numbers from 0 to 2 27   10 2436455 you need lg2  2 27   10 2436455    8 0937   10 6 bits   1 megabyte   8388608 bits   8093700 bits  So it appears that we at least actually have enough room to store the result  Now of course the interesting bit is doing the sorting as the numbers stream in  Not sure the best approach to this is given we have 294908 bits remaining  I imagine an interesting technique would be to at each point assume that that is is the entire ordering  finding the code for that ordering  and then as you receive a new number going back and updating the previous code  Hand wave hand wave

User · Answer

If the range of the numbers is limited  there can be only mod 2 8 digit numbers  or only 10 different 8 digit numbers for example   then you could write an optimized sorting algorithm  But if you want to sort all possible 8 digit numbers  this is not possible with that low amount of memory

User · Answer

If the input stream could be received few times this would be much easier  no information about that  idea and time-performance problem     Then  we could count the decimal values  With counted values it would be easy to make the output stream  Compress by counting the values  It depends what would be in the input stream

User · Answer

While receiving the stream do these steps  1st set some reasonable chunk size Pseudo Code idea   The first step would be to find all the duplicates and stick them in a dictionary with its count and remove them  The third step would be to place number that exist in sequence of their algorithmic steps and place them in counters special dictionaries with the first number and their step like n  n 1     n 2  2n  2n 1  2n 2    Begin to compress in chunks some reasonable ranges of number like every 1000 or ever 10000 the remaining numbers that appear less often to repeat  Uncompress that range if a number is found and add it to the range and leave it uncompressed for a while longer  Otherwise just add that number to a byte chunkSize   Continue the first 4 steps while receiving the stream  The final step would be to either fail if you exceeded memory or start outputting the result once all the data is collected by beginning to sort the ranges and spit out the results in order and uncompressing those in order that need to be uncompressed and sort them when you get to them

User · Answer

I have a computer with 1M of RAM and no other local storage   Another way to cheat  you could use non-local  networked  storage instead  your question does not preclude this  and call a networked service that could use straightforward disk-based mergesort  or just enough RAM to sort in-memory  since you only need to accept 1M numbers   without needing the  admittedly extremely ingenious  solutions already given   This might be cheating  but it s not clear whether you are looking for a solution to a real-world problem  or a puzzle that invites bending of the rules    if the latter  then a simple cheat may get better results than a complex but  genuine  solution  which as others have pointed out  can only work for compressible inputs

User · Answer

You just need to store the differences between the numbers in sequence  and use an encoding to compress these sequence numbers  We have 2 23 bits  We shall divide it into 6bit chunks  and let the last bit indicate whether the number extends to another 6 bits  5bits plus extending chunk     Thus  000010 is 1  and 000100 is 2  000001100000 is 128  Now  we consider the worst cast in representing differences in sequence of a numbers up to 10 000 000  There can be 10 000 000 2 5 differences greater than 2 5  10 000 000 2 10 differences greater than 2 10  and 10 000 000 2 15 differences greater than 2 15  etc   So  we add how many bits it will take to represent our the sequence  We have 1 000 000 6   roundup 10 000 000 2 5  6 roundup 10 000 000 2 10  6 roundup 10 000 000 2 15  6 roundup 10 000 000 2 20  4 7935479   2 24   8388608  Since 8388608   7935479  we should easily have enough memory  We will probably need another little bit of memory to store the sum of where are when we insert new numbers  We then go through the sequence  and find where to insert our new number  decrease the next difference if necessary  and shift everything after it right

User · Answer

Please see the first correct answer or the later answer with arithmetic encoding  Below you may find some fun  but not a 100  bullet-proof solution   This is quite an interesting task and here is an another solution  I hope somebody would find the result useful  or at least interesting    Stage 1  Initial data structure  rough compression approach  basic results  Let s do some simple math  we have 1M  1048576 bytes  of RAM initially available to store 10 6 8 digit decimal numbers   0 99999999   So to store one number 27 bits are needed  taking the assumption that unsigned numbers will be used   Thus  to store a raw stream  3 5M of RAM will be needed  Somebody already said it doesn t seem to be feasible  but I would say the task can be solved if the input is  good enough   Basically  the idea is to compress the input data with compression factor 0 29 or higher and do sorting in a proper manner   Let s solve the compression issue first  There are some relevant tests already available   http   www theeggeadventure com wikimedia index php Java Data Compression      I ran a test to compress one million consecutive integers using   various forms of compression  The results are as follows     None     4000027 Deflate  2006803 Filtered 1391833 BZip2    427067 Lzma     255040   It looks like LZMA  Lempel   Ziv   Markov chain algorithm  is a good choice to continue with  I ve prepared a simple PoC  but there are still some details to be highlighted    Memory is limited so the idea is to presort numbers and use  compressed buckets  dynamic size  as temporary storage It is easier to achieve a better compression factor with presorted data  so there is a static buffer for each bucket  numbers from the buffer are to be sorted before LZMA  Each bucket holds a specific range  so the final sort can be done for each bucket separately  Bucket s size can be properly set  so there will be enough memory to decompress stored data and do the final sort for each bucket separately     Please note  attached code is a POC  it can t be used as a final solution  it just demonstrates the idea of using several smaller buffers to store presorted numbers in some optimal way  possibly compressed   LZMA is not proposed as a final solution  It is used as a fastest possible way to introduce a compression to this PoC   See the PoC code below  please note it just a demo  to compile it LZMA-Java will be needed    public class MemorySortDemo    static final int NUM COUNT   1000000  static final int NUM MAX     100000000   static final int BUCKETS        5  static final int DICT SIZE      16   1024     LZMA dictionary size static final int BUCKET SIZE    1024  static final int BUFFER SIZE    10   1024  static final int BUCKET RANGE   NUM MAX   BUCKETS   static class Producer       private Random random   new Random        public int produce     return random nextInt NUM MAX        static class Bucket       public int size  pointer      public int   buffer   new int BUFFER SIZE        public ByteArrayOutputStream tempOut   new ByteArrayOutputStream        public DataOutputStream tempDataOut   new DataOutputStream tempOut       public ByteArrayOutputStream compressedOut   new ByteArrayOutputStream         public void submitBuffer   throws IOException           Arrays sort buffer  0  pointer            for  int j   0  j  lt  pointer  j                  tempDataOut writeInt buffer j                size                                  pointer   0             public void write int value  throws IOException           if  isBufferFull                  submitBuffer                      buffer pointer      value             public boolean isBufferFull             return pointer    BUFFER SIZE             public byte   compressData   throws IOException           tempDataOut close            return compress tempOut toByteArray                        private byte   compress byte   input  throws IOException           final BufferedInputStream in   new BufferedInputStream new ByteArrayInputStream input            final DataOutputStream out   new DataOutputStream new BufferedOutputStream compressedOut             final Encoder encoder   new Encoder            encoder setEndMarkerMode true           encoder setNumFastBytes 0x20           encoder setDictionarySize DICT SIZE           encoder setMatchFinder Encoder EMatchFinderTypeBT4            ByteArrayOutputStream encoderPrperties   new ByteArrayOutputStream            encoder writeCoderProperties encoderPrperties           encoderPrperties flush            encoderPrperties close             encoder code in  out  -1  -1  null           out flush            out close            in close             return encoderPrperties toByteArray               public int   decompress byte   properties  throws IOException           InputStream in   new ByteArrayInputStream compressedOut toByteArray             ByteArrayOutputStream data   new ByteArrayOutputStream 10   1024           BufferedOutputStream out   new BufferedOutputStream data            Decoder decoder   new Decoder            decoder setDecoderProperties properties           decoder code in  out  4   size            out flush            out close            in close             DataInputStream input   new DataInputStream new ByteArrayInputStream data toByteArray              int   array   new int size           for  int k   0  k  lt  size  k                  array k    input readInt                       return array           static class Sorter       private Bucket   bucket   new Bucket BUCKETS        public void doSort Producer p  Consumer c  throws IOException            for  int i   0  i  lt  bucket length  i          allocate buckets             bucket i    new Bucket                       for int i 0  i lt  NUM COUNT  i                 produce some data             int value   p produce                int bucketId   value BUCKET RANGE              bucket bucketId  write value               c register value                      for  int i   0  i  lt  bucket length  i         submit non-empty buffers             bucket i  submitBuffer                       byte   compressProperties   null          for  int i   0  i  lt  bucket length  i         compress the data             compressProperties   bucket i  compressData                       printStatistics             for  int i   0  i  lt  bucket length  i         decode  amp  sort buckets one by one             int   array   bucket i  decompress compressProperties               Arrays sort array                for int v   array                    c consume v                                   c finalCheck               public void printStatistics             int size   0          int sizeCompressed   0           for  int i   0  i  lt  BUCKETS  i                  int bucketSize   4 bucket i  size              size    bucketSize              sizeCompressed    bucket i  compressedOut size                 System out println    bucket     i                          contains      bucket i  size                         numbers  compressed size      bucket i  compressedOut size                         String format   compression factor    2f     double bucket i  compressedOut size    bucketSize                       System out println String format  Data size    2fM   double size  1014 1024                     String format   compressed   2fM   double sizeCompressed  1014 1024                     String format   compression factor   2f   double sizeCompressed size             static class Consumer       private Set lt Integer gt  values   new HashSet lt  gt          int v   -1      public void consume int value            if v  lt  0  v   value           if v  gt  value                throw new IllegalArgumentException  Current value is greater than previous      v      gt      value            else              v   value              values remove value                        public void register int value            values add value              public void finalCheck             System out println values size    gt  0    NOT OK      values size      OK              public static void main String   args  throws IOException       Producer p   new Producer        Consumer c   new Consumer        Sorter sorter   new Sorter         sorter doSort p  c         With random numbers it produces the following   bucket 0  contains  200357 numbers  compressed size  353679 compression factor  0 44 bucket 1  contains  199465 numbers  compressed size  352127 compression factor  0 44 bucket 2  contains  199682 numbers  compressed size  352464 compression factor  0 44 bucket 3  contains  199949 numbers  compressed size  352947 compression factor  0 44 bucket 4  contains  200547 numbers  compressed size  353914 compression factor  0 44 Data size  3 85M compressed 1 70M compression factor 0 44   For a simple ascending sequence  one bucket is used  it produces   bucket 0  contains  1000000 numbers  compressed size  256700 compression factor  0 06 Data size  3 85M compressed 0 25M compression factor 0 06   EDIT  Conclusion    Don t try to fool the Nature Use simpler compression with lower memory footprint Some additional clues are really needed  Common bullet-proof solution does not seem to be feasible     Stage 2  Enhanced compression  final conclusion  As was already mentioned in the previous section  any suitable compression technique can be used  So let s get rid of LZMA in favor of simpler and better  if possible   approach  There are a lot of good solutions including Arithmetic coding  Radix tree etc    Anyway  simple but useful encoding scheme will be more illustrative than yet another external library  providing some nifty algorithm  The actual solution is pretty straightforward  since there are buckets with partially sorted data  deltas can be used instead of numbers      Random input test shows slightly better results   bucket 0  contains  10103 numbers  compressed size  13683 compression factor  0 34 bucket 1  contains  9885 numbers  compressed size  13479 compression factor  0 34     bucket 98  contains  10026 numbers  compressed size  13612 compression factor  0 34 bucket 99  contains  10058 numbers  compressed size  13701 compression factor  0 34 Data size  3 85M compressed 1 31M compression factor 0 34   Sample code    public static void encode int   buffer  int length  BinaryOut output        short size    short  length  amp  0x7FFF        output write size       output write buffer 0         for int i 1  i lt  size  i              int next   buffer i  - buffer i-1           int bits   getBinarySize next            int len   bits           if bits  gt  24              output write 3  2             len   bits - 24           else if bits  gt  16              output write 2  2             len   bits-16           else if bits  gt  8              output write 1  2             len   bits - 8           else            output write 0  2                      if  len  gt  0                if   len   2   gt  0                    len   len   2                  output write len  2                   output write false                 else                   len   len   2 - 1                  output write len  2                              output write next  bits                      public static short decode BinaryIn input  int   buffer  int offset        short length   input readShort        int value   input readInt        buffer offset    value       for  int i   1  i  lt  length  i              int flag   input readInt 2            int bits          int next   0          switch  flag                case 0                  bits   2   input readInt 2    2                  next   input readInt bits                   break              case 1                  bits   8   2   input readInt 2   2                  next   input readInt bits                   break              case 2                  bits   16   2   input readInt 2   2                  next   input readInt bits                   break              case 3                  bits   24   2   input readInt 2   2                  next   input readInt bits                   break                     buffer offset   i    buffer offset   i - 1    next            return length      Please note  this approach    does not consume a lot of memory works with streams provides not so bad results   Full code can be found here  BinaryInput and BinaryOutput implementations can be found here  Final conclusion  No final conclusion    Sometimes it is really good idea to move one level up and review the task from a meta-level point of view    It was fun to spend some time with this task  BTW  there are a lot of interesting answers below  Thank you for your attention and happy codding

[algorithm] Sorting 1 million 8-decimal-digit numbers with 1 MB of RAM

Examples related to algorithm

Examples related to sorting

Examples related to embedded

Examples related to ram