Encoding/Decoding. Russell Impagliazzo and Miles Jones Thanks to Janine Tiefenbruck. May 9, 2016
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1 Encoding/Decoding Russell Impagliazzo and Miles Jones Thanks to Janine Tiefenbruck May 9, 2016
2 Review: Terminology A permutation of r elements from a set of n distinct objects is an ordered arrangement of them. There are Rosen p many of these. P(n,r) = n(n-1) (n-2) (n-r+1) A combination of r elements from a set of n distinct objects is an unordered slection of them. There are many of these. C(n,r) = n!/ ( r! (n-r)! ) Binomial coefficient "n choose r"
3 Fixed-density Binary Strings Rosen p. 413 How many length n binary strings contain k ones? Density is number of ones Objects: all strings made up of 0 1, 0 2, 1 1, 1 2, 1 3, 1 4 n! Categories: strings that agree except subscripts Size of each category: k!(n-k)! # categories = (# objects) / (size of each category) = n!/ ( k! (n-k)! ) = C(n,k) =
4 Encoding Fixed-density Binary Strings Rosen p. 413 What's the smallest number of bits that we need to specify a binary string if we know it has k ones and n-k zeros? A. n B. k C. log 2 ( C(n,k) ) D.??
5 Data Compression Store / transmit information in as little space as possible
6 Data Compression: Video Video: stored as sequence of still frames. Idea: instead of storing each frame fully, record change from previous frame.
7 Data Compression: Run-Length Encoding Image: described as grid of pixels, each with RED, GREEN, BLUE values. Idea: instead of storing RGB value of each pixel, store run-length of run of same color. When is this a good coding mechanism? Will there be any loss in this compression?
8 Lossy Compression: Singular Value Decomposition Image: described as grid of pixels, each with RED, GREEN, BLUE values. Idea: use Linear Algebra to compress data to a fraction of its size, with minimal loss.
9 Data Compression: Trade-off Complicated compression scheme save storage space may take a long time to encode / decode Data Encoding Algorithm Decoding Algorithm Data Stored in Computer
10 Encoding: Binary Palindromes Palindrome: string that reads the same forward and backward. Which of these are binary palindromes? A. The empty string. B C D E. All but one of the above.
11 Encoding: Binary Palindromes Palindrome: string that reads the same forward and backward. How many length n binary palindromes are there? A. 2 n B. n C. n/2 D. log 2 n E. None of the above
12 Encoding: Binary Palindromes Palindrome: string that reads the same forward and backward. How many bits are (optimally) required to encode length n binary palindromes? A. n B. n-1 C. n/2 D. log 2 n E. None of the above. Is there an algorithm that achieves this?
13 Goal: encode a length n binary string that we know has k ones (and n-k zeros). How would you represent such a string with n-1 bits?
14 Goal: encode a length n binary string that we know has k ones (and n-k zeros). How would you represent such a string with n-1 bits? Can we do better?
15 Goal: encode a length n binary string that we know has k ones (and n-k zeros). How would you represent such a string with n-1 bits? Can we do better? Idea: give positions of 1s in the string within some smaller window. - Fix window size. - If there is a 1 in the current "window" in the string, record its position and move the window over. - Otherwise, record a 0 and move the window over.
16 Idea: give positions of 1s in the string within some smaller window. - Fix window size. - If there is a 1 in the current "window" in the string, record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ?
17 Idea: give positions of 1s in the string within some smaller window. - Fix window size. - If there is a 1 in the current "window" in the string, record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: There's a 1! What's its position?
18 Idea: give positions of 1s in the string within some smaller window. - Fix window size. - If there is a 1 in the current "window" in the string, record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: 01 There's a 1! What's its position?
19 Idea: give positions of 1s in the string within some smaller window. - Fix window size. - If there is a 1 in the current "window" in the string, record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: 01 There's a 1! What's its position?
20 Idea: give positions of 1s in the string within some smaller window. - Fix window size. - If there is a 1 in the current "window" in the string, record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: 0100 There's a 1! What's its position?
21 Idea: give positions of 1s in the string within some smaller window. - Fix window size. - If there is a 1 in the current "window" in the string, record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: 0100 No 1s in this window.
22 Idea: give positions of 1s in the string within some smaller window. - Fix window size. - If there is a 1 in the current "window" in the string, record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: No 1s in this window.
23 Idea: give positions of 1s in the string within some smaller window. - Fix window size. - If there is a 1 in the current "window" in the string, record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: There's a 1! What's its position?
24 Idea: give positions of 1s in the string within some smaller window. - Fix window size. - If there is a 1 in the current "window" in the string, record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: There's a 1! What's its position?
25 Idea: give positions of 1s in the string within some smaller window. - Fix window size. - If there is a 1 in the current "window" in the string, record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: No 1s in this window.
26 Idea: give positions of 1s in the string within some smaller window. - Fix window size. - If there is a 1 in the current "window" in the string, record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: No 1s in this window.
27 Idea: give positions of 1s in the string within some smaller window. - Fix window size. - If there is a 1 in the current "window" in the string, record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: Compressed to 8 bits! But can we recover the original string? Decoding
28 With n=12, k=3, window size n/k = 4. Output: Can be parsed as the (intended) input: s = ? But also: 01: one in position 1 0: no ones 00: one in position 0 11: one in position 3 0: no ones s' = Problem: two different inputs with same output. Can't uniquely decode.
29 Compression Algorithm A valid compression algorithm must: - Have outputs of shorter (or same) length as input. - Be uniquely decodable.
30 Can we modify this algorithm to get unique decodability? Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over.
31 Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output:
32 Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output:
33 Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: What output corresponds to these first few bits? A. 0 C. 01 E. None of the above. B. 1 D. 101
34 Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: 101 Interpret next bits as position of 1; this position is 01
35 Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: 101
36 Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: Interpret next bits as position of 1; this position is 00
37 Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output:
38 Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: No 1s in this window.
39 Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output:
40 Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: Interpret next bits as position of 1; this position is 11
41 Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output:
42 Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: No 1s in this window.
43 Output uses more bits than last time. Any redundancies? Encoding: Fixed Density Strings Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: Compare to previous output:
44 * After see the last 1, don't need to add 0s to indicate empty windows. * Encoding: Fixed Density Strings Idea: use marker bit to indicate when to interpret output as a position. - Fix window size. - If there is a 1 in the current "window" in the string, record a 1 to interpret next bits as position, then record its position and move the window over. - Otherwise, record a 0 and move the window over. Example n=12, k=3, window size n/k = 4. How do we encode s = ? Output: Compare to previous output:
45 procedure WindowEncode (input: b 1 b 2 b n, with exactly k ones and n-k zeros) 1. w := floor (n/k) 2. count := 0 3. location := 1 4. While count < k: 5. If there is a 1 in the window starting at current location 6. Output 1 as a marker, then output position of first 1 in window. 7. Increment count. 8. Update location to immediately after first 1 in this window. 9. Else 10. Output Update location to next index after current window. Uniquely decodable?
46 Decoding: Fixed Density Strings procedure WindowDecode (input: x 1 x 2 x m, target is exactly k ones and n-k zeros) 1. w := floor ( n/k ) 2. b := floor ( log 2 (w)) 3. s := empty string 4. i := 0 5. While i < m 6. If x i = 0 7. s += 0 0 (j times) 8. i += 1 9. Else 10. p := decimal value of the bits x i+1 x i+b 11. s += 0 0 (p times) 12. s += i := i+b If length(s) < n 15. s += 0 0 ( n-length(s) times ) 16. Output s.
47 Encoding/Decoding: Fixed Density Strings Correctness? E(s) = result of encoding string s of length n with k 1s, using WindowEncode. D(t) = result of decoding string t to create a string of length n with k 1s, using WindowDecode. Well-defined functions? Inverses? Goal: For each s, D(E(s)) = s. Strong Induction!
48 Encoding/Decoding: Fixed Density Strings Output size? Assume n/k is a power of two. Consider s a binary string of length n with k 1s. How long is E(s)? A. n-1 B. log 2 (n/k) C. Depends on where 1s are located in s
49 Encoding/Decoding: Fixed Density Strings Output size? Assume n/k is a power of two. Consider s a binary string of length n with k 1s. For which strings is E(s) shortest? A. More 1s toward the beginning. B. More 1s toward the end. C. 1s spread evenly throughout.
50 Encoding/Decoding: Fixed Density Strings Output size? Assume n/k is a power of two. Consider s a binary string of length n with k 1s. Best case : 1s toward the beginning of the string. E(s) has - One bit for each 1 in s to indicate that next bits denote positions in window. - log 2 (n/k) bits for each 1 in s to specify position of that 1 in a window. - k such ones. - No bits representing 0s because all 0s are "caught" in windows with 1s or after the last 1. Total E(s) = k log 2 (n/k) + k
51 Encoding/Decoding: Fixed Density Strings Output size? Assume n/k is a power of two. Consider s a binary string of length n with k 1s. Worst case : 1s toward the end of the string. E(s) has - Some bits representing 0s since there are no 1s in first several windows. - One bit for each 1 in s to indicate that next bits denote positions in window. - log 2 (n/k) bits for each 1 in s to specify position of that 1 in a window. - k such ones. What's an upper bound on the number of these bits? A. n D. 1 B. n-k E. None of the above. C. k
52 Encoding/Decoding: Fixed Density Strings Output size? Assume n/k is a power of two. Consider s a binary string of length n with k 1s. Worst case : 1s toward the end of the string. E(s) has - At most k bits representing 0s since there are no 1s in first several windows. - One bit for each 1 in s to indicate that next bits denote positions in window. - log 2 (n/k) bits for each 1 in s to specify position of that 1 in a window. - k such ones. Total E(s) <= k log 2 (n/k) + 2k
53 Encoding/Decoding: Fixed Density Strings Output size? Assume n/k is a power of two. Consider s a binary string of length n with k 1s. k log 2 (n/k) + k <= E(s) <= k log 2 (n/k) + 2k Using this inequality, there are at most length n strings with k 1s. A. 2 n D. (n/k) k B. n E. None of the above. C. (n/k) 2
54 Encoding/Decoding: Fixed Density Strings Output size? Assume n/k is a power of two. Consider s a binary string of length n with k 1s. Given E(s) <= k log 2 (n/k) + 2k, we need at most k log 2 (n/k) + 2k bits to represent all length n binary strings with k 1s. Hence, there are at most 2 many such strings.
55 Encoding/Decoding: Fixed Density Strings Output size? Assume n/k is a power of two. Consider s a binary string of length n with k 1s. Given E(s) <= k log 2 (n/k) + 2k, we need at most k log 2 (n/k) + 2k bits to represent all length n binary strings with k 1s. Hence, there are at most 2 many such strings.
56 Encoding/Decoding: Fixed Density Strings Output size? Assume n/k is a power of two. Consider s a binary string of length n with k 1s. Given E(s) <= k log 2 (n/k) + 2k, we need at most k log 2 (n/k) + 2k bits to represent all length n binary strings with k 1s. Hence, there are at most 2 many such strings. C(n,k) = # Length n binary strings with k 1s <= (4n/k) k
57 Bounds for Binomial Coefficients Using windowencode(): Lower bound? Idea: find a way to count a subset of the fixed density binary strings.. Some fixed density binary strings have one 1 in each of k chunks of size n/k. How many such strings are there? A. n n B. k! C. (n/k) k D. C(n,k) k E. None of the above.
58 Bounds for Binomial Coefficients Using windowencode(): Using evenly spread strings: Counting helps us analyze our compression algorithm. Compression algorithms help us count.
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