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.
OneTimePad.cp
Copyright (c) 2002
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
and pad characters.
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
of the pad.
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
pad is reached.
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
adjusted to minimize the time spent on processing failed pad offsets.
Acknowledgement
----------------
The dictionary indexing system uses some ideas gleaned by revisiting
Ernst Munter's winning solution to the 1999 Playfair challenge.
\******************************************/
#include "OneTimePad.h"
// 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 {
ulong header = *(ulong*)(*wordPtr);
if ((header & 0x0ff0000) && (header & 0x0ff00)) {
// 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:
// Pad information
const uchar* fPad; // local copy of the pad pointer
const uchar* fFirstPadChar;
// pointer to the current first pad character
const uchar* fLastPadChar;
// 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
Decoder (const uchar* inPad)
{
// Constructor initializes the data members
// The dictionary index is built separately to allow Branch allocation failure to be
// handled gracefully
// initialize pad pointers and pad offset
fPad =inPad;
fFirstPadChar = fPad;
// 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);
}
void FindPadOffset (const uchar* inEncryptedMessage,
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
typedef ulong PadMap [128];
ulong c0, c1, c2;
ulong* mapPtr;
const uchar* pad;
blong validOffset;
ulong map, i;
const uchar* cipher = inEncryptedMessage;
PadMap padMapArray [4];
// 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
mapPtr = padMapArray [map];
// 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
fFirstPadChar = fPad;
validOffset = false;
pad = fPad;
// prime a sequence of 4 characters with the first 3 valid pad characters
ulong padBuffer = 0;
for (i = 0; i < 3; i ++) {
while (*pad < 0x20 || *pad > 0x7e) {
if (*pad == 0) return; // Abort if the pad is less than 4 characters
pad ++;
}
padBuffer = (padBuffer << 8) | *(pad ++);
}
do {
// Shift the next valid pad character into the set
while (*pad < 0x20 || *pad > 0x7e) { pad ++; }
padBuffer = (padBuffer << 8) | *(pad ++);
if ( (c0 = padMapArray[0] [padBuffer >> 24]) ) {
// 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
if ( (c0 = padMapArray [3] [padBuffer & 0xff]) )
// check that the Branch index exists for the 4th character
validOffset = header->fBranches &&
header->fBranches [c0].fFirstWord;
else
// check for a valid 3 character word
validOffset = header->fMap & 0x01;
} 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
validOffset = ApplyPad (inEncryptedMessage,
outDecryptedMessage);
}
}
// move the pad offset to the next valid pad character
do {
if (*(fFirstPadChar + fMessageLength))
fFirstPadChar ++;
else
// We've run out of pad characters - end of pass 1
goto SecondPass;
} while (*fFirstPadChar < 0x20 || *fFirstPadChar > 0x7e);
} 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
fFirstPadChar = fPad;
do {
validOffset = ApplyPad (inEncryptedMessage,
outDecryptedMessage);
fFirstPadChar ++;
} while ( ! validOffset && *(fFirstPadChar +
fMessageLength));
}
// return the offset
*outOffset = fFirstPadChar - fPad - 1;
}
blong ApplyPad (const uchar* inEncryptedMessage,
uchar* outDecryptedMessage)
{
// Called by FindPadOffset() with fFirstPadChar pointing to the start of a candidate
// pad
// Attempt to decode the message using this pad, and return success/failure
const uchar* pad; // pointer to the current pad character
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;
pad = fFirstPadChar;
do {
// for each character in the message
// next valid pad character
while (*pad < 0x20 || *pad > 0x7e) {
if (*pad) pad++;
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]) ) {
if (header->fBranches &&
header->fBranches [c0].fFirstWord) {
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;
}
pad ++;
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
InitOneTimePad
void InitOneTimePad (const char *oneTimePad, const char *dictionary[], long numDictionaryWords) {
// 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)
gDecoder->FindPadOffset ((uchar*) encryptedMessage,
(uchar*) decryptedMessage, (ulong*)offset);
}
TermOneTimePad
void TermOneTimePad(void) {
delete gDecoder;
}
