PROGRAMMER's CHALLENGE

Volume Number: 18 (2002)
Issue Number: 10
Column Tag: PROGRAMMER's CHALLENGE

# PROGRAMMER's CHALLENGE

by Bob Boonstra

### Area

As those of you who are regular readers know, the Programmer's Challenge problems have become more difficult over time. Just as all scientific discoveries worth making had been made by the mid-twentieth century, so it is that all simple Challenge problems have been posed and solved by this time. Then again, as they say, maybe not. This month's problem is borrowed from http://www.polymathlove.com/, where Gary Smith posts software he uses in teaching mathematics to elementary and middle school students. One of his programs is called Area Puzzles, where students create rectangles with specified areas to cover a grid subject to certain constraints.

The prototype for the code you should write is:

```void Area(
const short *cells,
/* rectangle to be covered with smaller rectangles */
/* index [row][col] as cells[row*rectWidth + col[ */
/* value N>0 means this cell must be covered by a rectangle of area N */
short rectWidth,
short rectHeight,
Rect yourRects[]
);```

Your Area routine will be called with a rectangle of cells of width rectWidth and height rectHeight. Your task is to create a set of smaller rectangles (yourRect) that cover these cells. In doing so, you need to satisfy some constraints. Certain of the cells will have a nonzero value, and those cells must be covered by a rectangle with an area equal to that value. As an example, if the input cells were configured as follows ...

```   0  0  3  0  6  0  0  0  0  8
0  6  0  0  0  0  0  0  0  0
0  0  0  0  0  0  4  0  0  0
0  0  3  0  0  0  0  0  0  0
0  0  0  0  0  0  0  0  0  0
6  0  0  0  0  0  0  0 15  0
0  0  0  0  0  0  0  0  0  0
0 10  0  0 24  0  0  4  0  0
0  0  0  0  0  0  0  0  0  0
0  0  0  4  0  0  4  0  0  3```

... you might create a set of rectangles like this, where each cell is shown with the number of the rectangle including that cell.

```   1  1  1  2  2  2  3  3  3  3
4  4  5  2  2  2  3  3  3  3
4  4  5  6  6  6  6  7  7  7
4  4  5  8  8  8  8  7  7  7
9 10 10  8  8  8  8  7  7  7
9 10 10  8  8  8  8  7  7  7
9 10 10  8  8  8  8  7  7  7
9 10 10  8  8  8  8 13 13 14
9 10 10  8  8  8  8 13 13 14
9 11 11 11 11 12 12 12 12 14```

You should return the rectangles that cover the cell array and satisfy the constraints as yourRects. Each cell may be included in only one rectangle. If the cell has a nonzero value when Area is called, it must be included in a rectangle with an area equal to that value. Memory for the rectangles you create will be allocated for you, and there will be as many of those rectangles as there are nonzero values in the cells array. Any solution that covers the entire cells array and satisfies the constraints will be considered correct.

Scoring will be based on execution time - the winner will be the solution that correctly solves the puzzles with the smallest execution time.

This will be a native PowerPC Carbon C++ Challenge, using the Metrowerks CodeWarrior Pro 7.0 development environment. Please be certain that your code is carbonized, as I may evaluate this Challenge using Mac OS X. Also, when submitting you solution, please include the project file and the code you used to test your solution. Occasionally I receive a solution that will not compile and, while I always try to correct these problems, it is easier to do so if I have your entire project available.

### Winner of the July, 2002 Challenge

Congratulations to Alan Hart (United Kingdom) for winning the July One Time Pad Challenge. Recall that this Challenge required readers to decrypt a sequence of messages using a "one time pad". I place the term in quotation marks because the pad was neither "one time", as it was used multiple times, nor was it random, as a true one-time pad would be. Contestants had the advantage of possessing a dictionary of all the possible words in the communication.

Alan's solution tries each possible offset until the decoding attempt results in a sequence of words found in the dictionary. The speed of Alan's solution is due in part to his decision to test the decoding of the first four characters of the message for each offset against the dictionary before proceeding with the rest of the decoding. Another factor is Alan's reuse (with acknowledgement) of ideas from Ernst Munter's solution to the PlayFair Challenge, specifically the dictionary indexing approach. That approach creates an index for each word based on the first three characters that points to the first word in the dictionary beginning with those three letters. I'm pleased to see past Challenge code reused successfully.

Ernst Munter's second place entry also uses a brute force method. His approach is to select a possible offset from the pad, decrypt the message using that offset, verify that the decrypted message contains only words from the dictionary, and repeat with a new offset until successful. Ernst's solution also uses a modified version of the SpellTree dictionary class he developed for the PlayFair Challenge.

Jonny Taylor's third-place solution also examined the first three characters of the decoded message to determine whether an offset was promising enough to continue decoding. As noted by others, because the message may contain special characters not found in the dictionary, offsets rejected by this approach must be revisited to skip potential special characters if the message is not successfully decoded. Moses Hall takes a different approach, creating a finite state machine encoding the dictionary. Rounding out the remainder of the five top-scoring entries, Jan Schotsman used the Altivec programming model and reports achieving a 5% increase in speed over the non-vectorized version.

The table below lists, for each of the solutions submitted, the number of test cases processed correctly, the execution time in seconds, the bonus awarded for code clarity and commentary, and the total score for each solution. It also lists the programming language of each entry. As usual, the number in parentheses after the entrant's name is the total number of Challenge points earned in all Challenges prior to this one.

```
Name                 Cases     Time   Bonus   Score   Lang
Correct   (secs)
Alan Hart (39)        20       0.73    25%    5469.59   C++
Ernst Munter (872)    20       1.30    25%    9721.75   C++
Jonny Taylor (83)     20       1.67    25%   12499.44   C++
Moses Hall            20       2.54    15%   21572.92   C
Jan Schotsman (16)    20       3.05    5%    28959.52   C++
Tom Saxton (230)      20       3.08    5%    29268.66   C++
Damien Bobillot       15       12.11   15%  102918.96   C```

### Top Contestants ...

Listed here are the Top Contestants for the Programmer's Challenge, including everyone who has accumulated 20 or more points during the past two years. The numbers below include points awarded over the 24 most recent contests, including points earned by this month's entrants.

```
Rank    Name                Points      Wins     Total
(24 mo)    (24 mo)   Points
1.   Munter, Ernst         251        8         882
2.   Saxton, Tom            65        2         230
3.   Taylor, Jonathan       64        2          90
4.   Stenger, Allen         53        1         118
5.   Wihlborg, Claes        40        2          49
6.   Hart, Alan             34        1          59
7.   Rieken, Willeke        22        1         134
8.   Landsbert, Robin       22        1          22
9.   Gregg, Xan             20        1         140
10.   Mallett, Jeff         20        1         114
11.   Cooper, Tony          20        1          20
12.   Truskier, Peter       20        1          20```

Here is Alan's winning One Time Pad solution.

Alan Hart

```/******************************************\
Problem definition:
----------------
Decode multiple encrypted messages created using a known one time pad.
Each message is encrypted using an unknown offset in the one time pad.
Each character in the encrypted message is the sum of the corresponding  clear text
The sum is adjusted to remain in the valid character set range.
The character set is 62 upper/lower case alpha-numerics
that appear in words in a case-insensitive dictionary, plus 33 other punctuation and special
characters

("delimiters") that can appear in any locations between words.

Total time is to be minimized.

Assumptions:
-----------
We cannot make any assumptions about the number of delimiter characters that
may preceed the message or separate the words within it. In an ulikely extreme case
the message could be a sea of delimiters with a few short words distributed
anywhere within it. It is assumed that the test cases will not be pathological, and the
majority of characters in the message will form dictionary words. In particular, the
solution is optimized for messages with no leading delimiters before the first
dictionary word. It decrypts messges with leading delimiters during a second scan

Solution Summary:
-----------------
The external interface calls are passed to a Decoder class which does the work.

The gDecoder instance is dynamically assigned by InitOneTimePad () and allocates
a fixed sze array of 256 KBytes for the main dictionary index. Further index Branch
records are allocated dynamically during indexing.
This adds a further 100KBytes to the space required in the case of the dictionry
supplied with the test data.
The solution should fit comfortably in less than 500 KBytes heap space not
including the dictionary and message strings.

The decoder creates a small static lookup table containing two concatenated copies
of the character set to allow for wrap-around when subtracting the pad and cipher
characters, allowing a simple lookup for decoding, and avoiding the need for range
limiting.

Decoding is done by trying each possible pad offset in turn until the decode yields a
sequence of words that are found in the dictionary.

During the first pass candidate pad offsets are found by testing the first 4 characters
at each pad offset with the first 4 message characters. If this fails all pad offsets are
retested on the whole message in case the first word does not start at the first
character of the message. Decoding with each candidate pad is aborted if any
invalid sequence of consecutive dictionary characters is detected or the end of the

The validity of a character sequence is tested in three stages:

1. A sequence of three or more characters must have a "header" index value
matched by one or more dictionary words.
A shorter word must have an index value that matches a bit map of 1 and 2
character words.

2. Characters following a valid header index must form 4-character sequences that
exist somewhere in the dictionary.

3. Finally the word is compared with the dictionary entries using a case insensitive
character match and length comparison.

The dictionary index uses a 32 character (5 bit) enumeration to allow quick
calculation of compact indices, and to enable bit mapping in a single 32 bit word.
The 10 numeric characters are enumerated using 1 to 5, with pairs of digits sharing
the same index value. 6 to 31 enumerate the alphabetic characters, ignoring case.
Zero is used to denote any non-dictionary character.
This enumeration means that words in the supplied dictionary containing numerals
in the indexed characters may not be in correct index order. To avoid having to re-
sort the dictionary this is handled in the index building and searching procedures.

This mapping and indexing system allows a quick confidence check to be carried
out during decoding so that non-indexed final dictionary searching is only done on
longer words, and only when the word is likely to exist.

Words of one or two characters are recorded in a 32 x 32 bit map for a fast lookup.
The index for a word of three or more characters is calculated by concatenating the
index values of the first three characters.
The index selects one of an array of 32x32x32 (32K) Index records used for initial
and final validation of decoded words.
If the Index value is the header for words longer than three characters it has a
pointer to a dynamically allocated 32 entry lookup table. Each entry in the table
points to the first word in the dictionary whose 4th character matches
one character index value.
The Index record also has a bit map of the valid fourth character indices that can
follow this sequence when it appears at character positions 3, 6, 9 ... in the body of
any word in the dictionary.

Optimizations applied included using unsigned types to remove compiled sign
extend instructions, longs to remove compiled byte mask instructions, and the
addition of the first pass decoder to limit the tested pad offsets to those yielding
valid indices for the first 4 message characters. Instruction sequencing was also

Acknowledgement
----------------
The dictionary indexing system uses some ideas gleaned by revisiting
Ernst Munter's winning solution to the 1999 Playfair challenge.

\******************************************/
//   use unsigned types wherever possible to minimize the insertion of sign extend
// instructions by the compiler
typedef   unsigned char   uchar;
typedef   unsigned long   ulong;
typedef   unsigned long   blong;   //   used instead of bool to avoid character masking instructions

static const ulong IndexOffset [128] = {
//   lookup table to convert character codes to index offset values
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 0, 0, 0, 0, 0, 0,
0, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 0, 0, 0, 0, 0,
0, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 0, 0, 0, 0, 0
};
IndexValue
static inline ulong
IndexValue (const uchar* inWord) {
//   Return the index for the first three characters of a word
ulong   index = IndexOffset [*(inWord ++)];
index = (index << 5) | IndexOffset [*(inWord ++)];
return (index << 5) | IndexOffset [*inWord];
}
struct Branch
struct Branch {
const uchar**   fFirstWord;
const uchar**   fLastWord;
Branch () { fFirstWord = fLastWord = NULL; }

blong   ValidateWord (uchar* inWord, ulong inWordLength) {
//   Compare inWord with each word in this branch
//   The indices are known to match, so inWord already points to the 4th character of
// the test word
//   Note that the ambiguous indexing of numeric characters means it is possible to
// return a false positive.
//   This is not considered a problem, as we are only trying to verify that the one time
// pad is correctly aligned to produce a string of credible words. However, it means
// that shorter dictionary words can appear within within the indexed range, and
// must be ignored.
const   uchar**   wordPtr =fFirstWord;
do {
//   3 or more characters in this dictionary word
const uchar*   w1 = *wordPtr + 4;
uchar*      w2 = inWord;
long         difference =  0;
long         len = inWordLength;
while ( ! difference && *w1) {
//   compare the words until the end of the dictionary word
//   converting upper case characters in inWord to lower case
difference = (*(w1++) - (*(w2++) | 0x20));
len --;
}
if (difference == 0 && len == 0) return true;
// words match
if (difference > 0) return false;
//   this and subsequent words are greater than inWord
}
} while (wordPtr ++ < fLastWord);
return false;   //   no match found
}
};

class    Index
class    Index {
//   Dictionary index entry for a three-character sequence
public:
//   Data encapsulation is not used, so that the decoder can access Index members
// directly for higher performance
Branch   *fBranches;
//   Dynamically allocated list of 32 pointers to words that start with this index
ulong      fMap;
//   Map of valid fourth character indices that can follow this index in the body
// of a word
//   fMap bit 0 is used to register a valid three letter word for this index
Index () { fBranches = NULL; fMap = 0; }

~Index () { delete [] fBranches; }

ulong      //   Return the number of words registered for this index value
Register (
const uchar** inWordList,   // Pointer to the first word to register
ulong inIndex,            //   The index value for words to register
const uchar** inLastWord,   // The last word in the dictionary, for
//range checking
Index* inIndexArray
//   The array if Index records, used to registerword body sequences
) {
//   This is the dictionary indexing procedure
//   - Registers the existence of one or more 3 character words for this index in fMap
// bit zero
//   - If there are words of 4 or more characters with this index value, allocates a
// Branch array   and records the first and last words of the list for each 4th character.
//   - Registers the existence of each 4-letter sequence in the body of each word in the
// fMap for its index
const uchar**   wordPtr = inWordList;
ulong         index = inIndex;
const uchar*   word;
Branch*      branch = NULL;
do {
uchar   c = IndexOffset [*((*wordPtr) + 3)];
if (c) {
//   4 or more characters
if (fBranches == NULL) {
//   allocate a new branch list to index words on the 4th character
fBranches = new Branch [32];
if (fBranches == 0)
return 0;   //   bail out and signal allocation failure
}
//   record the start and end words in each branch
if (branch != fBranches + c) {   // end of previous branch
if (branch)
//   update the last word pointer for the old branch
branch->fLastWord = wordPtr - 1;
//   move to the next branch
branch = fBranches + c;
if (branch->fFirstWord == NULL)
//   this is the first word to be registered in this branch
//   insert the first word pointer for this branch
branch->fFirstWord = wordPtr;
}
//   register the remaining map bits for the body of this word
word = (*wordPtr) + 3;   //   skip the index
while (*word && *(word+1) && *(word+2)) {
//   while there are three or more characters left
//   get the 4th character index
c = IndexOffset [*(word + 3)];
if (c == 0) break;   // no 4th character
//   register the 4th character in the index bit map
inIndexArray [ IndexValue (word) ].fMap |= ( 1 <<  c);
word += 3;
}
} else
//   register the 3 character word in bit 0 of the index map
fMap |= 1;
wordPtr ++;   //   next word

//   until end of dictionary or index value changes
} while (wordPtr < inLastWord &&
index == IndexValue (*wordPtr));

if (branch)   //   update the last branch
branch->fLastWord = wordPtr - 1;
//   return the number of words registered
return wordPtr - inWordList;
}
blong   ValidateWord (uchar* inWord, ulong inWordLength)
{
//   The dictionary lookup procedure.
//   - Returns true if the word exists in the dictionary
if (inWordLength == 3)
//   3 character word. Check map bit 0
return (fMap & 0x01);
else if (fBranches) {
//   Compare the 4th and subsequent characters with the words in the appropriate
// branch
Branch*   branch = fBranches + IndexOffset [*(inWord + 3)];
if (branch->fFirstWord)
return branch->ValidateWord (inWord + 4, inWordLength - 4);
}
return false;
}
};
class Decoder
class    Decoder {
private:
//   pointer to the current first pad character
//   pointer to the last pad character to try
ulong         fMessageLength;      //   length of the encrypted string
//   Dictionary infomation
Index      fIndexArray [ 32*32*32 ];   //   the dictionary index array
ulong         fShortWordMap [32];
//   a bit map for one and two character words
//   The decoder table
ulong         fDecodeTable [190];
//   lookup table to convert a pair of cipher/pad characters to a clear character
public
{
//   Constructor initializes the data members
//   The dictionary index is built separately to allow Branch allocation failure to be
//  handled gracefully

//   Fill in the decoder table with two concatenated copies of the character set
int   c;
ulong*   t = fDecodeTable;
for (c = 0x20; c < 0x7f; c ++, t ++)
*t = *(t + 0x5f) = c;

//   Clear the short word map
for (c = 0; c < 32; c ++)
fShortWordMap [c] = 0;
}

blong      IndexDictionary (const uchar** inDictionary,
ulong inNumWords)
{
//   Build the dictionary index and short word map
//   Return false if Branch allocation falis
const uchar**   wordPtr = inDictionary;
const uchar**   lastWord = inDictionary + inNumWords;
ulong      numWords;
do {
//   Process the next index value
ulong      index = *(ulong*)(*wordPtr);
numWords = 1;
if ((index & 0x0ff0000) == 0)
//   Register a 1 character word in short word map zero
*fShortWordMap |= (1 <<
IndexOffset [(index >> 24) & 0xff]);
else if ((index & 0x0ff00) == 0)
//   Register a 2 character word in the appropriate short word map
fShortWordMap[ IndexOffset[(index >> 24) & 0xff] ]
|= (1 << IndexOffset [(index >> 16) & 0xff]);
else {
//   3 or more characters
//   Pass the word list to the appropriate Index record for registration
index = IndexValue (*wordPtr);
numWords = fIndexArray [index].Register (wordPtr, index,
lastWord, fIndexArray);
if (numWords == 0)
//   branch list allocation failure - bail out
break;
}
//   Next word list
wordPtr += numWords;
} while (wordPtr < lastWord);

return (numWords > 0);
}

uchar* outDecryptedMessage, ulong *outOffset)
{
//   The main routine called to decode messages

//   Use trial and error to find a pad offset that successfully decodes the message
//   Return the offset found in *outOffset

//   Firat Pass
//   Optimized search for candidate pad offsets that decode a valid word index at the
// first encrypted character
//   Build lookup tables to decode each of the first 4 message characters to its index
// value for any pad character
//   Shift the pad through a 4-character buffer and apply this to the 4 message
// characters

ulong         c0, c1, c2;
ulong*       mapPtr;
blong         validOffset;
ulong         map, i;
const uchar*   cipher = inEncryptedMessage;

//   Measure the encrypted message length
while (*(++cipher)) {;}
fMessageLength = cipher - inEncryptedMessage;

cipher = inEncryptedMessage;
for (map = 0; map < 4; map ++, cipher ++) {
//   create the decode table for the next cipher character
//   clear the entries for invalid pad characters
*(mapPtr + 0x7f) = 0;
i = 0x20; while (i --) { *(mapPtr++) = 0; }
//   set the decoded index values for each valid pad character
//   calculate the first offset in the decode table for this cipher character
ulong*   c = fDecodeTable + 0x3f + *cipher;
i = 0x5f; while (i --) { *(mapPtr++) = I
ndexOffset [*(c --)]; }
}
//   Start at the beginning of the one time pad.
//   Decode these 4 characters with each pad offset and look for a valid word or index

validOffset = false;
//   prime a sequence of 4 characters with the first 3 valid pad characters
for (i = 0; i < 3; i ++) {
if (*pad == 0) return;   //   Abort if the pad is less than 4 characters
}
}
do {
//   Shift the next valid pad character into the set

//   1st character decodes to a dictionary character
if ( (c1 = padMapArray[1] [(padBuffer >> 16) & 0xff] ) ) {
//   2nd character decodes to a dictionary character
if ( (c2 = padMapArray [2] [(padBuffer >> 8) & 0xff]) ) {
//   select the appropriate Index record for the 3 characters
Index*   header =fIndexArray + ((((c0 << 5) | c1) << 5) |
c2);
//   decode the 4th character
//   check that the Branch index exists for the 4th character
else
//   check for a valid 3 character word
} else
//   check for a valid 2 character word
validOffset = fShortWordMap [c0] & (1 << c1);
} else
//   check for a valid single character word
validOffset = (*fShortWordMap) & (1 << c0);

if (validOffset) {
//   This pad offset decodes a valid word or index
//   try to decode the whole message using this candidate pad
outDecryptedMessage);
}
}
do {
else
//   We've run out of pad characters - end of pass 1
goto SecondPass;

} while ( ! validOffset);
if ( ! validOffset ) {
SecondPass:
//   Second Pass
//   The search for the first index may have failed due to leading delimiters
//   Try to decode the message using every pad offset in turn
do {
outDecryptedMessage);
} while ( ! validOffset && *(fFirstPadChar +
fMessageLength));
}
//   return the offset
}
uchar* outDecryptedMessage)
{
//   Called by FindPadOffset() with fFirstPadChar pointing to the start of a candidate
//   Attempt to decode the message using this pad, and return success/failure

const uchar*   cipher;   //   pointer to the next encrypted character to decode
uchar*      clear;   //   pointer to the next clear character to decode
ulong         index;   //   index value of current word
const ulong*   origin = fDecodeTable + 0x5f;
//   pointer to the origin in the DecodeTable
uchar*      word;   //   pointer to start of current word being processed
Index      *body, *header;   //   pointers to the Index records for the
//   current word
ulong         c0, c1, c2;
//   index values of the first three characters of teh current word
ulong         state = 0;   //   controls progress of the word decode process

clear = outDecryptedMessage;
cipher = inEncryptedMessage;

do {
//   for each character in the message

else goto Exit;   //   end of pad
}
//   decode the character
*clear = *(origin + *cipher - *pad);
switch (state) {
case 0:   //   looking for 1st character of a word
if ( (c0 = IndexOffset [*clear]) ) {
word = clear;   //   1st character of a word
state = 1;
}
break;
case 1:   //   looking for 2nd character of a word
if ( (c1 = IndexOffset [*clear]) ) state = 2;
//   2nd dictionary char
else if ((*fShortWordMap) & (1 << c0))
//   valid 1 character word
state = 0;
else goto Exit;
break;
case 2:   //   looking for 3rd character of a word
if ( (c2 = IndexOffset [*clear]) ) {
header =fIndexArray + ((((c0 << 5) | c1) << 5) | c2);
state = 3;   //   3rd dictionary char
} else if (fShortWordMap [c0] & (1 << c1))
//   valid 2 character word
state = 0;
else goto Exit;
break;
case 3:   //   looking for 4th character of a word
if ( (c0 = IndexOffset [*clear]) ) {
index = c0;   //   valid 4th dictionary char
body = NULL;
state = 4;   //   check body sequences
} else goto Exit;
} else if (header->fMap & 0x01)
//   valid 3 character word
state = 0;
else goto Exit;
break;
case 4:
if ( (c0 = IndexOffset [*clear]) ) {
if (body == NULL) {
index = (index << 5) | c0; // accumulate the body index
if (index > 1024)
//   third body index character. select the Index record
body = fIndexArray + index;
} else {   //   look up this character in the current body Index record
if (body->fMap & (1 << c0)) {
//   valid sequence - start building the next body index
index = c0;
body = NULL;
} else goto Exit;
}
} else {
//   end of the word, confirm it is in the dictionary
if (header->ValidateWord (word, clear - word))
state = 0;
else goto Exit;
}
default:
break;
}
clear ++;
} while (*(++cipher));

Exit:
*clear = 0;   //   terminate the decoded string
return ((*cipher) == 0);   //   successful if we reached the end of the
//   message
}

} *gDecoder;   //   global pointer to a dynamically allocated instance
//   Create the decoder
gDecoder = new Decoder ((const uchar*) oneTimePad);
if ( gDecoder) {
//   Index the dictionary
if ( ! gDecoder->IndexDictionary ((const uchar**) dictionary,
(ulong) numDictionaryWords)) {
//   Allocation failure
delete gDecoder;
gDecoder = NULL;
}
}
}
DecryptMessage
void DecryptMessage(const char *encryptedMessage, char *decryptedMessage, long *offset) {
if (gDecoder)
(uchar*) decryptedMessage, (ulong*)offset);
}
delete gDecoder;
}```

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Dreamweaver CC 2018 is available as part of Adobe Creative Cloud for as little as \$19.99/month (or \$9.99/month if you're a previous Dreamweaver customer). Adobe Dreamweaver CC 2018 allows you to... Read more
Adobe Flash Player 29.0.0.113 - Plug-in...
Adobe Flash Player is a cross-platform, browser-based application runtime that provides uncompromised viewing of expressive applications, content, and videos across browsers and operating systems.... Read more
Drive Genius 5.2.0 - \$79.00
Drive Genius features a comprehensive Malware Scan. Automate your malware protection. Protect your investment from any threat. The Malware Scan is part of the automated DrivePulse utility. DrivePulse... Read more
MegaSeg 6.0.6 - Professional DJ and radi...
MegaSeg is a complete solution for pro audio/video DJ mixing, radio automation, and music scheduling with rock-solid performance and an easy-to-use design. Mix with visual waveforms and Magic... Read more
ffWorks 1.0.7 - Convert multimedia files...
ffWorks (was iFFmpeg), focused on simplicity, brings a fresh approach to the use of FFmpeg, allowing you to create ultra-high-quality movies without the need to write a single line of code on the... Read more
Dash 4.1.5 - Instant search and offline...
Dash is an API documentation browser and code snippet manager. Dash helps you store snippets of code, as well as instantly search and browse documentation for almost any API you might use (for a full... Read more
Evernote 7.0.3 - Create searchable notes...
Evernote allows you to easily capture information in any environment using whatever device or platform you find most convenient, and makes this information accessible and searchable at anytime, from... Read more
jAlbum Pro 15.3 - Organize your digital...
jAlbum Pro has all the features you love in jAlbum, but comes with a commercial license. You can create gorgeous custom photo galleries for the Web without writing a line of code! Beginner-friendly... Read more

## Latest Forum Discussions

Around the Empire: What have you missed...
Oh hi nice reader, and thanks for popping in to check out our weekly round-up of all the stuff that you might have missed across the Steel Media network. Yeah, that's right, it's a big ol' network. Obviously 148Apps is the best, but there are some... | Read more »
All the best games on sale for iPhone an...
It might not have been the greatest week for new releases on the App Store, but don't let that get you down, because there are some truly incredible games on sale for iPhone and iPad right now. Seriously, you could buy anything on this list and I... | Read more »
Everything You Need to Know About The Fo...
In just over a week, Epic Games has made a flurry of announcements. First, they revealed that Fortnite—their ultra-popular PUBG competitor—is coming to mobile. This was followed by brief sign-up period for interested beta testers before sending out... | Read more »
The best games that came out for iPhone...
It's not been the best week for games on the App Store. There are a few decent ones here and there, but nothing that's really going to make you throw down what you're doing and run to the nearest WiFi hotspot in order to download it. That's not to... | Read more »
Death Coming (Games)
Death Coming 1.1.1.536 Device: iOS Universal Category: Games Price: \$1.99, Version: 1.1.1.536 (iTunes) Description: --- Background Story ---You Died. Pure and simple, but death was not the end. You have become an agent of Death: a... | Read more »
Hints, tips, and tricks for Empires and...
Empires and Puzzles is a slick match-stuff RPG that mixes in a bunch of city-building aspects to keep things fresh. And it's currently the Game of the Day over on the App Store. So, if you're picking it up for the first time today, we thought it'd... | Read more »
What You Need to Know About Sam Barlow’s...
Sam Barlow’s follow up to Her Story is #WarGames, an interactive video series that reimagines the 1983 film WarGames in a more present day context. It’s not exactly a game, but it’s definitely still interesting. Here are the top things you should... | Read more »
Pixel Plex Guide - How to Build Better T...
Pixel Plex is the latest city builder that has come to the App Store, and it takes a pretty different tact than the ones that came before it. Instead of being in charge of your own city by yourself, you have to work together with other players to... | Read more »
Fortnite Will Be Better Than PUBG on Mob...
Before last week, if you asked me which game I prefer between Fortnite Battle Royale and PlayerUnknown’s Battlegrounds (PUBG), I’d choose the latter just about 100% of the time. Now that we know that both games are primed to hit our mobile screens... | Read more »
Siege of Dragonspear (Games)
Siege of Dragonspear 2.5.12 Device: iOS Universal Category: Games Price: \$9.99, Version: 2.5.12 (iTunes) Description: Experience the Siege of Dragonspear, an epic Baldur’s Gate tale, filled with with intrigue, magic, and monsters.... | Read more »

## Price Scanner via MacPrices.net

Sunday Sales: \$200 off 13″ Touch Bar MacBook...
Amazon has new 2017 13″ 3.1GHz Touch Bar MacBook Pros on sale this weekend for \$200 off MSRP, each including free shipping: – 13″ 3.1GHz/256GB Space Gray MacBook Pro (MPXV2LL/A): \$1599.99 \$200 off... Read more
B&H drops prices on 15″ MacBook Pros up t...
B&H Photo has dropped prices on new 2017 15″ MacBook Pros, now up to \$300 off MSRP and matching Adorama’s price drop yesterday. Shipping is free, and B&H charges sales tax for NY & NJ... Read more
Apple restocks Certified Refurbished 2017 13″...
Apple has restocked Certified Refurbished 2017 13″ 2.3GHz MacBook Pros for \$200-\$230 off MSRP. A standard Apple one-year warranty is included with each MacBook, models receive new outer cases, and... Read more
13″ Space Gray Touch Bar MacBook Pros on sale...
Adorama has new 2017 13″ Space Gray Touch Bar MacBook Pros on sale for \$150 off MSRP. Shipping is free, and Adorama charges sales tax in NY & NJ only: – 13″ 3.1GHz/256GB Space Gray MacBook Pro (... Read more
Best deal of the year on 15″ Apple MacBook Pr...
Adorama has New 2017 15″ MacBook Pros on sale for up to \$300 off MSRP. Shipping is free, and Adorama charges sales tax in NJ and NY only: – 15″ 2.8GHz Touch Bar MacBook Pro Space Gray (MPTR2LL/A): \$... Read more
Save \$100-\$150+ on 13″ Touch Bar MacBook Pros...
B&H Photo has 13″ Touch Bar MacBook Pros on sale for \$100-\$150 off MSRP. Shipping is free, and B&H charges sales tax for NY & NJ residents only: – 13″ 3.1GHz/256GB Space Gray MacBook Pro... Read more
Current deals on 27″ Apple iMacs, models up t...
B&H Photo has 27″ iMacs on sale for up to \$150 off MSRP. Shipping is free, and B&H charges sales tax for NY & NJ residents only: – 27″ 3.8GHz iMac (MNED2LL/A): \$2149 \$150 off MSRP – 27″ 3... Read more
Thursday Deal: 13″ 2.3GHz MacBook Pro for \$11...
B&H Photo has the 13″ 2.3GHz/128GB Space Gray MacBook Pro on sale for \$100 off MSRP. Shipping is free, and B&H charges sales tax for NY & NJ residents only: – 13-inch 2.3GHz/128GB Space... Read more
How to save \$100-\$190 on 10″ & 12″ iPad P...
Apple is now offering Certified Refurbished 2017 10″ and 12″ iPad Pros for \$100-\$190 off MSRP, depending on the model. An Apple one-year warranty is included with each model, and shipping is free: –... Read more
Silver 12″ 1.3GHz MacBook on sale at B&H...
B&H Photo has the 2017 12″ 1.3GHz Silver MacBook on sale for \$1399.99 including free shipping plus sales tax for NY & NJ residents only. Their price is \$200 off MSRP, and it’s the lowest... Read more

## Jobs Board

Firmware Engineer - *Apple* Accessories - A...
# Firmware Engineer - Apple Accessories Job Number: 113452350 Santa Clara Valley, California, United States Posted: 28-Feb-2018 Weekly Hours: 40.00 **Job Summary** Read more
Automation and Performance Engineer, *Apple*...
# Automation and Performance Engineer, Apple Pay Job Number: 113557967 Santa Clara Valley, California, United States Posted: 09-Mar-2018 Weekly Hours: 40.00 **Job Read more
Hardware Systems Architect - *Apple* Watch...
# Hardware Systems Architect - Apple Watch Job Number: 113565323 Santa Clara Valley, California, United States Posted: 05-Mar-2018 Weekly Hours: 40.00 **Job Read more
Lead *Apple* Solution Consultant - Apple (U...
# Lead Apple Solution Consultant Chicago IL Job Number: 113560644 Chicago, Illinois, United States Posted: 10-Mar-2018 Weekly Hours: 40.00 **Job Summary** As a Lead Read more
Art Director, *Apple* Music + Beats1 Market...
# Art Director, Apple Music + Beats1 Marketing Design Job Number: 113258081 Culver City, California, United States Posted: 07-Mar-2018 Weekly Hours: 40.00 **Job Read more