-
Notifications
You must be signed in to change notification settings - Fork 345
/
chapter10.tex
849 lines (749 loc) · 25.5 KB
/
chapter10.tex
1
2
3
4
5
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
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
\chapter{Bit manipulation}
All data in computer programs is internally stored as bits,
i.e., as numbers 0 and 1.
This chapter discusses the bit representation
of integers, and shows examples
of how to use bit operations.
It turns out that there are many uses for
bit manipulation in algorithm programming.
\section{Bit representation}
\index{bit representation}
In programming, an $n$ bit integer is internally
stored as a binary number that consists of $n$ bits.
For example, the C++ type \texttt{int} is
a 32-bit type, which means that every \texttt{int}
number consists of 32 bits.
Here is the bit representation of
the \texttt{int} number 43:
\[00000000000000000000000000101011\]
The bits in the representation are indexed from right to left.
To convert a bit representation $b_k \cdots b_2 b_1 b_0$ into a number,
we can use the formula
\[b_k 2^k + \ldots + b_2 2^2 + b_1 2^1 + b_0 2^0.\]
For example,
\[1 \cdot 2^5 + 1 \cdot 2^3 + 1 \cdot 2^1 + 1 \cdot 2^0 = 43.\]
The bit representation of a number is either
\key{signed} or \key{unsigned}.
Usually a signed representation is used,
which means that both negative and positive
numbers can be represented.
A signed variable of $n$ bits can contain any
integer between $-2^{n-1}$ and $2^{n-1}-1$.
For example, the \texttt{int} type in C++ is
a signed type, so an \texttt{int} variable can contain any
integer between $-2^{31}$ and $2^{31}-1$.
The first bit in a signed representation
is the sign of the number (0 for nonnegative numbers
and 1 for negative numbers), and
the remaining $n-1$ bits contain the magnitude of the number.
\key{Two's complement} is used, which means that the
opposite number of a number is calculated by first
inverting all the bits in the number,
and then increasing the number by one.
For example, the bit representation of
the \texttt{int} number $-43$ is
\[11111111111111111111111111010101.\]
In an unsigned representation, only nonnegative
numbers can be used, but the upper bound for the values is larger.
An unsigned variable of $n$ bits can contain any
integer between $0$ and $2^n-1$.
For example, in C++, an \texttt{unsigned int} variable
can contain any integer between $0$ and $2^{32}-1$.
There is a connection between the
representations:
a signed number $-x$ equals an unsigned number $2^n-x$.
For example, the following code shows that
the signed number $x=-43$ equals the unsigned
number $y=2^{32}-43$:
\begin{lstlisting}
int x = -43;
unsigned int y = x;
cout << x << "\n"; // -43
cout << y << "\n"; // 4294967253
\end{lstlisting}
If a number is larger than the upper bound
of the bit representation, the number will overflow.
In a signed representation,
the next number after $2^{n-1}-1$ is $-2^{n-1}$,
and in an unsigned representation,
the next number after $2^n-1$ is $0$.
For example, consider the following code:
\begin{lstlisting}
int x = 2147483647
cout << x << "\n"; // 2147483647
x++;
cout << x << "\n"; // -2147483648
\end{lstlisting}
Initially, the value of $x$ is $2^{31}-1$.
This is the largest value that can be stored
in an \texttt{int} variable,
so the next number after $2^{31}-1$ is $-2^{31}$.
\section{Bit operations}
\newcommand\XOR{\mathbin{\char`\^}}
\subsubsection{And operation}
\index{and operation}
The \key{and} operation $x$ \& $y$ produces a number
that has one bits in positions where both
$x$ and $y$ have one bits.
For example, $22$ \& $26$ = 18, because
\begin{center}
\begin{tabular}{rrr}
& 10110 & (22)\\
\& & 11010 & (26) \\
\hline
= & 10010 & (18) \\
\end{tabular}
\end{center}
Using the and operation, we can check if a number
$x$ is even because
$x$ \& $1$ = 0 if $x$ is even, and
$x$ \& $1$ = 1 if $x$ is odd.
More generally, $x$ is divisible by $2^k$
exactly when $x$ \& $(2^k-1)$ = 0.
\subsubsection{Or operation}
\index{or operation}
The \key{or} operation $x$ | $y$ produces a number
that has one bits in positions where at least one
of $x$ and $y$ have one bits.
For example, $22$ | $26$ = 30, because
\begin{center}
\begin{tabular}{rrr}
& 10110 & (22)\\
| & 11010 & (26) \\
\hline
= & 11110 & (30) \\
\end{tabular}
\end{center}
\subsubsection{Xor operation}
\index{xor operation}
The \key{xor} operation $x$ $\XOR$ $y$ produces a number
that has one bits in positions where exactly one
of $x$ and $y$ have one bits.
For example, $22$ $\XOR$ $26$ = 12, because
\begin{center}
\begin{tabular}{rrr}
& 10110 & (22)\\
$\XOR$ & 11010 & (26) \\
\hline
= & 01100 & (12) \\
\end{tabular}
\end{center}
\subsubsection{Not operation}
\index{not operation}
The \key{not} operation \textasciitilde$x$
produces a number where all the bits of $x$
have been inverted.
The formula \textasciitilde$x = -x-1$ holds,
for example, \textasciitilde$29 = -30$.
The result of the not operation at the bit level
depends on the length of the bit representation,
because the operation inverts all bits.
For example, if the numbers are 32-bit
\texttt{int} numbers, the result is as follows:
\begin{center}
\begin{tabular}{rrrr}
$x$ & = & 29 & 00000000000000000000000000011101 \\
\textasciitilde$x$ & = & $-30$ & 11111111111111111111111111100010 \\
\end{tabular}
\end{center}
\subsubsection{Bit shifts}
\index{bit shift}
The left bit shift $x < < k$ appends $k$
zero bits to the number,
and the right bit shift $x > > k$
removes the $k$ last bits from the number.
For example, $14 < < 2 = 56$,
because $14$ and $56$ correspond to 1110 and 111000.
Similarly, $49 > > 3 = 6$,
because $49$ and $6$ correspond to 110001 and 110.
Note that $x < < k$
corresponds to multiplying $x$ by $2^k$,
and $x > > k$
corresponds to dividing $x$ by $2^k$
rounded down to an integer.
\subsubsection{Applications}
A number of the form $1 < < k$ has a one bit
in position $k$ and all other bits are zero,
so we can use such numbers to access single bits of numbers.
In particular, the $k$th bit of a number is one
exactly when $x$ \& $(1 < < k)$ is not zero.
The following code prints the bit representation
of an \texttt{int} number $x$:
\begin{lstlisting}
for (int i = 31; i >= 0; i--) {
if (x&(1<<i)) cout << "1";
else cout << "0";
}
\end{lstlisting}
It is also possible to modify single bits
of numbers using similar ideas.
For example, the formula $x$ | $(1 < < k)$
sets the $k$th bit of $x$ to one,
the formula
$x$ \& \textasciitilde $(1 < < k)$
sets the $k$th bit of $x$ to zero,
and the formula
$x$ $\XOR$ $(1 < < k)$
inverts the $k$th bit of $x$.
The formula $x$ \& $(x-1)$ sets the last
one bit of $x$ to zero,
and the formula $x$ \& $-x$ sets all the
one bits to zero, except for the last one bit.
The formula $x$ | $(x-1)$
inverts all the bits after the last one bit.
Also note that a positive number $x$ is
a power of two exactly when $x$ \& $(x-1) = 0$.
\subsubsection*{Additional functions}
The g++ compiler provides the following
functions for counting bits:
\begin{itemize}
\item
$\texttt{\_\_builtin\_clz}(x)$:
the number of zeros at the beginning of the number
\item
$\texttt{\_\_builtin\_ctz}(x)$:
the number of zeros at the end of the number
\item
$\texttt{\_\_builtin\_popcount}(x)$:
the number of ones in the number
\item
$\texttt{\_\_builtin\_parity}(x)$:
the parity (even or odd) of the number of ones
\end{itemize}
\begin{samepage}
The functions can be used as follows:
\begin{lstlisting}
int x = 5328; // 00000000000000000001010011010000
cout << __builtin_clz(x) << "\n"; // 19
cout << __builtin_ctz(x) << "\n"; // 4
cout << __builtin_popcount(x) << "\n"; // 5
cout << __builtin_parity(x) << "\n"; // 1
\end{lstlisting}
\end{samepage}
While the above functions only support \texttt{int} numbers,
there are also \texttt{long long} versions of
the functions available with the suffix \texttt{ll}.
\section{Representing sets}
Every subset of a set
$\{0,1,2,\ldots,n-1\}$
can be represented as an $n$ bit integer
whose one bits indicate which
elements belong to the subset.
This is an efficient way to represent sets,
because every element requires only one bit of memory,
and set operations can be implemented as bit operations.
For example, since \texttt{int} is a 32-bit type,
an \texttt{int} number can represent any subset
of the set $\{0,1,2,\ldots,31\}$.
The bit representation of the set $\{1,3,4,8\}$ is
\[00000000000000000000000100011010,\]
which corresponds to the number $2^8+2^4+2^3+2^1=282$.
\subsubsection{Set implementation}
The following code declares an \texttt{int}
variable $x$ that can contain
a subset of $\{0,1,2,\ldots,31\}$.
After this, the code adds the elements 1, 3, 4 and 8
to the set and prints the size of the set.
\begin{lstlisting}
int x = 0;
x |= (1<<1);
x |= (1<<3);
x |= (1<<4);
x |= (1<<8);
cout << __builtin_popcount(x) << "\n"; // 4
\end{lstlisting}
Then, the following code prints all
elements that belong to the set:
\begin{lstlisting}
for (int i = 0; i < 32; i++) {
if (x&(1<<i)) cout << i << " ";
}
// output: 1 3 4 8
\end{lstlisting}
\subsubsection{Set operations}
Set operations can be implemented as follows as bit operations:
\begin{center}
\begin{tabular}{lll}
& set syntax & bit syntax \\
\hline
intersection & $a \cap b$ & $a$ \& $b$ \\
union & $a \cup b$ & $a$ | $b$ \\
complement & $\bar a$ & \textasciitilde$a$ \\
difference & $a \setminus b$ & $a$ \& (\textasciitilde$b$) \\
\end{tabular}
\end{center}
For example, the following code first constructs
the sets $x=\{1,3,4,8\}$ and $y=\{3,6,8,9\}$,
and then constructs the set $z = x \cup y = \{1,3,4,6,8,9\}$:
\begin{lstlisting}
int x = (1<<1)|(1<<3)|(1<<4)|(1<<8);
int y = (1<<3)|(1<<6)|(1<<8)|(1<<9);
int z = x|y;
cout << __builtin_popcount(z) << "\n"; // 6
\end{lstlisting}
\subsubsection{Iterating through subsets}
The following code goes through
the subsets of $\{0,1,\ldots,n-1\}$:
\begin{lstlisting}
for (int b = 0; b < (1<<n); b++) {
// process subset b
}
\end{lstlisting}
The following code goes through
the subsets with exactly $k$ elements:
\begin{lstlisting}
for (int b = 0; b < (1<<n); b++) {
if (__builtin_popcount(b) == k) {
// process subset b
}
}
\end{lstlisting}
The following code goes through the subsets
of a set $x$:
\begin{lstlisting}
int b = 0;
do {
// process subset b
} while (b=(b-x)&x);
\end{lstlisting}
\section{Bit optimizations}
Many algorithms can be optimized using
bit operations.
Such optimizations do not change the
time complexity of the algorithm,
but they may have a large impact
on the actual running time of the code.
In this section we discuss examples
of such situations.
\subsubsection{Hamming distances}
\index{Hamming distance}
The \key{Hamming distance}
$\texttt{hamming}(a,b)$ between two
strings $a$ and $b$ of equal length is
the number of positions where the strings differ.
For example,
\[\texttt{hamming}(01101,11001)=2.\]
Consider the following problem: Given
a list of $n$ bit strings, each of length $k$,
calculate the minimum Hamming distance
between two strings in the list.
For example, the answer for $[00111,01101,11110]$
is 2, because
\begin{itemize}[noitemsep]
\item $\texttt{hamming}(00111,01101)=2$,
\item $\texttt{hamming}(00111,11110)=3$, and
\item $\texttt{hamming}(01101,11110)=3$.
\end{itemize}
A straightforward way to solve the problem is
to go through all pairs of strings and calculate
their Hamming distances,
which yields an $O(n^2 k)$ time algorithm.
The following function can be used to
calculate distances:
\begin{lstlisting}
int hamming(string a, string b) {
int d = 0;
for (int i = 0; i < k; i++) {
if (a[i] != b[i]) d++;
}
return d;
}
\end{lstlisting}
However, if $k$ is small, we can optimize the code
by storing the bit strings as integers and
calculating the Hamming distances using bit operations.
In particular, if $k \le 32$, we can just store
the strings as \texttt{int} values and use the
following function to calculate distances:
\begin{lstlisting}
int hamming(int a, int b) {
return __builtin_popcount(a^b);
}
\end{lstlisting}
In the above function, the xor operation constructs
a bit string that has one bits in positions
where $a$ and $b$ differ.
Then, the number of bits is calculated using
the \texttt{\_\_builtin\_popcount} function.
To compare the implementations, we generated
a list of 10000 random bit strings of length 30.
Using the first approach, the search took
13.5 seconds, and after the bit optimization,
it only took 0.5 seconds.
Thus, the bit optimized code was almost
30 times faster than the original code.
\subsubsection{Counting subgrids}
As another example, consider the
following problem:
Given an $n \times n$ grid whose
each square is either black (1) or white (0),
calculate the number of subgrids
whose all corners are black.
For example, the grid
\begin{center}
\begin{tikzpicture}[scale=0.5]
\fill[black] (1,1) rectangle (2,2);
\fill[black] (1,4) rectangle (2,5);
\fill[black] (4,1) rectangle (5,2);
\fill[black] (4,4) rectangle (5,5);
\fill[black] (1,3) rectangle (2,4);
\fill[black] (2,3) rectangle (3,4);
\fill[black] (2,1) rectangle (3,2);
\fill[black] (0,2) rectangle (1,3);
\draw (0,0) grid (5,5);
\end{tikzpicture}
\end{center}
contains two such subgrids:
\begin{center}
\begin{tikzpicture}[scale=0.5]
\fill[black] (1,1) rectangle (2,2);
\fill[black] (1,4) rectangle (2,5);
\fill[black] (4,1) rectangle (5,2);
\fill[black] (4,4) rectangle (5,5);
\fill[black] (1,3) rectangle (2,4);
\fill[black] (2,3) rectangle (3,4);
\fill[black] (2,1) rectangle (3,2);
\fill[black] (0,2) rectangle (1,3);
\draw (0,0) grid (5,5);
\fill[black] (7+1,1) rectangle (7+2,2);
\fill[black] (7+1,4) rectangle (7+2,5);
\fill[black] (7+4,1) rectangle (7+5,2);
\fill[black] (7+4,4) rectangle (7+5,5);
\fill[black] (7+1,3) rectangle (7+2,4);
\fill[black] (7+2,3) rectangle (7+3,4);
\fill[black] (7+2,1) rectangle (7+3,2);
\fill[black] (7+0,2) rectangle (7+1,3);
\draw (7+0,0) grid (7+5,5);
\draw[color=red,line width=1mm] (1,1) rectangle (3,4);
\draw[color=red,line width=1mm] (7+1,1) rectangle (7+5,5);
\end{tikzpicture}
\end{center}
There is an $O(n^3)$ time algorithm for solving the problem:
go through all $O(n^2)$ pairs of rows and for each pair
$(a,b)$ calculate the number of columns that contain a black
square in both rows in $O(n)$ time.
The following code assumes that $\texttt{color}[y][x]$
denotes the color in row $y$ and column $x$:
\begin{lstlisting}
int count = 0;
for (int i = 0; i < n; i++) {
if (color[a][i] == 1 && color[b][i] == 1) count++;
}
\end{lstlisting}
Then, those columns
account for $\texttt{count}(\texttt{count}-1)/2$ subgrids with black corners,
because we can choose any two of them to form a subgrid.
To optimize this algorithm, we divide the grid into blocks
of columns such that each block consists of $N$
consecutive columns. Then, each row is stored as
a list of $N$-bit numbers that describe the colors
of the squares. Now we can process $N$ columns at the same time
using bit operations. In the following code,
$\texttt{color}[y][k]$ represents
a block of $N$ colors as bits.
\begin{lstlisting}
int count = 0;
for (int i = 0; i <= n/N; i++) {
count += __builtin_popcount(color[a][i]&color[b][i]);
}
\end{lstlisting}
The resulting algorithm works in $O(n^3/N)$ time.
We generated a random grid of size $2500 \times 2500$
and compared the original and bit optimized implementation.
While the original code took $29.6$ seconds,
the bit optimized version only took $3.1$ seconds
with $N=32$ (\texttt{int} numbers) and $1.7$ seconds
with $N=64$ (\texttt{long long} numbers).
\section{Dynamic programming}
Bit operations provide an efficient and convenient
way to implement dynamic programming algorithms
whose states contain subsets of elements,
because such states can be stored as integers.
Next we discuss examples of combining
bit operations and dynamic programming.
\subsubsection{Optimal selection}
As a first example, consider the following problem:
We are given the prices of $k$ products
over $n$ days, and we want to buy each product
exactly once.
However, we are allowed to buy at most one product
in a day.
What is the minimum total price?
For example, consider the following scenario ($k=3$ and $n=8$):
\begin{center}
\begin{tikzpicture}[scale=.65]
\draw (0, 0) grid (8,3);
\node at (-2.5,2.5) {product 0};
\node at (-2.5,1.5) {product 1};
\node at (-2.5,0.5) {product 2};
\foreach \x in {0,...,7}
{\node at (\x+0.5,3.5) {\x};}
\foreach \x/\v in {0/6,1/9,2/5,3/2,4/8,5/9,6/1,7/6}
{\node at (\x+0.5,2.5) {\v};}
\foreach \x/\v in {0/8,1/2,2/6,3/2,4/7,5/5,6/7,7/2}
{\node at (\x+0.5,1.5) {\v};}
\foreach \x/\v in {0/5,1/3,2/9,3/7,4/3,5/5,6/1,7/4}
{\node at (\x+0.5,0.5) {\v};}
\end{tikzpicture}
\end{center}
In this scenario, the minimum total price is $5$:
\begin{center}
\begin{tikzpicture}[scale=.65]
\fill [color=lightgray] (1, 1) rectangle (2, 2);
\fill [color=lightgray] (3, 2) rectangle (4, 3);
\fill [color=lightgray] (6, 0) rectangle (7, 1);
\draw (0, 0) grid (8,3);
\node at (-2.5,2.5) {product 0};
\node at (-2.5,1.5) {product 1};
\node at (-2.5,0.5) {product 2};
\foreach \x in {0,...,7}
{\node at (\x+0.5,3.5) {\x};}
\foreach \x/\v in {0/6,1/9,2/5,3/2,4/8,5/9,6/1,7/6}
{\node at (\x+0.5,2.5) {\v};}
\foreach \x/\v in {0/8,1/2,2/6,3/2,4/7,5/5,6/7,7/2}
{\node at (\x+0.5,1.5) {\v};}
\foreach \x/\v in {0/5,1/3,2/9,3/7,4/3,5/5,6/1,7/4}
{\node at (\x+0.5,0.5) {\v};}
\end{tikzpicture}
\end{center}
Let $\texttt{price}[x][d]$ denote the price of product $x$
on day $d$.
For example, in the above scenario $\texttt{price}[2][3] = 7$.
Then, let $\texttt{total}(S,d)$ denote the minimum total
price for buying a subset $S$ of products by day $d$.
Using this function, the solution to the problem is
$\texttt{total}(\{0 \ldots k-1\},n-1)$.
First, $\texttt{total}(\emptyset,d) = 0$,
because it does not cost anything to buy an empty set,
and $\texttt{total}(\{x\},0) = \texttt{price}[x][0]$,
because there is one way to buy one product on the first day.
Then, the following recurrence can be used:
\begin{equation*}
\begin{split}
\texttt{total}(S,d) = \min( & \texttt{total}(S,d-1), \\
& \min_{x \in S} (\texttt{total}(S \setminus x,d-1)+\texttt{price}[x][d]))
\end{split}
\end{equation*}
This means that we either do not buy any product on day $d$
or buy a product $x$ that belongs to $S$.
In the latter case, we remove $x$ from $S$ and add the
price of $x$ to the total price.
The next step is to calculate the values of the function
using dynamic programming.
To store the function values, we declare an array
\begin{lstlisting}
int total[1<<K][N];
\end{lstlisting}
where $K$ and $N$ are suitably large constants.
The first dimension of the array corresponds to a bit
representation of a subset.
First, the cases where $d=0$ can be processed as follows:
\begin{lstlisting}
for (int x = 0; x < k; x++) {
total[1<<x][0] = price[x][0];
}
\end{lstlisting}
Then, the recurrence translates into the following code:
\begin{lstlisting}
for (int d = 1; d < n; d++) {
for (int s = 0; s < (1<<k); s++) {
total[s][d] = total[s][d-1];
for (int x = 0; x < k; x++) {
if (s&(1<<x)) {
total[s][d] = min(total[s][d],
total[s^(1<<x)][d-1]+price[x][d]);
}
}
}
}
\end{lstlisting}
The time complexity of the algorithm is $O(n 2^k k)$.
\subsubsection{From permutations to subsets}
Using dynamic programming, it is often possible
to change an iteration over permutations into
an iteration over subsets\footnote{This technique was introduced in 1962
by M. Held and R. M. Karp \cite{hel62}.}.
The benefit of this is that
$n!$, the number of permutations,
is much larger than $2^n$, the number of subsets.
For example, if $n=20$, then
$n! \approx 2.4 \cdot 10^{18}$ and $2^n \approx 10^6$.
Thus, for certain values of $n$,
we can efficiently go through the subsets but not through the permutations.
As an example, consider the following problem:
There is an elevator with maximum weight $x$,
and $n$ people with known weights
who want to get from the ground floor
to the top floor.
What is the minimum number of rides needed
if the people enter the elevator in an optimal order?
For example, suppose that $x=10$, $n=5$
and the weights are as follows:
\begin{center}
\begin{tabular}{ll}
person & weight \\
\hline
0 & 2 \\
1 & 3 \\
2 & 3 \\
3 & 5 \\
4 & 6 \\
\end{tabular}
\end{center}
In this case, the minimum number of rides is 2.
One optimal order is $\{0,2,3,1,4\}$,
which partitions the people into two rides:
first $\{0,2,3\}$ (total weight 10),
and then $\{1,4\}$ (total weight 9).
The problem can be easily solved in $O(n! n)$ time
by testing all possible permutations of $n$ people.
However, we can use dynamic programming to get
a more efficient $O(2^n n)$ time algorithm.
The idea is to calculate for each subset of people
two values: the minimum number of rides needed and
the minimum weight of people who ride in the last group.
Let $\texttt{weight}[p]$ denote the weight of
person $p$.
We define two functions:
$\texttt{rides}(S)$ is the minimum number of
rides for a subset $S$,
and $\texttt{last}(S)$ is the minimum weight
of the last ride.
For example, in the above scenario
\[ \texttt{rides}(\{1,3,4\})=2 \hspace{10px} \textrm{and}
\hspace{10px} \texttt{last}(\{1,3,4\})=5,\]
because the optimal rides are $\{1,4\}$ and $\{3\}$,
and the second ride has weight 5.
Of course, our final goal is to calculate the value
of $\texttt{rides}(\{0 \ldots n-1\})$.
We can calculate the values
of the functions recursively and then apply
dynamic programming.
The idea is to go through all people
who belong to $S$ and optimally
choose the last person $p$ who enters the elevator.
Each such choice yields a subproblem
for a smaller subset of people.
If $\texttt{last}(S \setminus p)+\texttt{weight}[p] \le x$,
we can add $p$ to the last ride.
Otherwise, we have to reserve a new ride
that initially only contains $p$.
To implement dynamic programming,
we declare an array
\begin{lstlisting}
pair<int,int> best[1<<N];
\end{lstlisting}
that contains for each subset $S$
a pair $(\texttt{rides}(S),\texttt{last}(S))$.
We set the value for an empty group as follows:
\begin{lstlisting}
best[0] = {1,0};
\end{lstlisting}
Then, we can fill the array as follows:
\begin{lstlisting}
for (int s = 1; s < (1<<n); s++) {
// initial value: n+1 rides are needed
best[s] = {n+1,0};
for (int p = 0; p < n; p++) {
if (s&(1<<p)) {
auto option = best[s^(1<<p)];
if (option.second+weight[p] <= x) {
// add p to an existing ride
option.second += weight[p];
} else {
// reserve a new ride for p
option.first++;
option.second = weight[p];
}
best[s] = min(best[s], option);
}
}
}
\end{lstlisting}
Note that the above loop guarantees that
for any two subsets $S_1$ and $S_2$
such that $S_1 \subset S_2$, we process $S_1$ before $S_2$.
Thus, the dynamic programming values are calculated in the
correct order.
\subsubsection{Counting subsets}
Our last problem in this chapter is as follows:
Let $X=\{0 \ldots n-1\}$, and each subset $S \subset X$
is assigned an integer $\texttt{value}[S]$.
Our task is to calculate for each $S$
\[\texttt{sum}(S) = \sum_{A \subset S} \texttt{value}[A],\]
i.e., the sum of values of subsets of $S$.
For example, suppose that $n=3$ and the values are as follows:
\begin{multicols}{2}
\begin{itemize}
\item $\texttt{value}[\emptyset] = 3$
\item $\texttt{value}[\{0\}] = 1$
\item $\texttt{value}[\{1\}] = 4$
\item $\texttt{value}[\{0,1\}] = 5$
\item $\texttt{value}[\{2\}] = 5$
\item $\texttt{value}[\{0,2\}] = 1$
\item $\texttt{value}[\{1,2\}] = 3$
\item $\texttt{value}[\{0,1,2\}] = 3$
\end{itemize}
\end{multicols}
In this case, for example,
\begin{equation*}
\begin{split}
\texttt{sum}(\{0,2\}) &= \texttt{value}[\emptyset]+\texttt{value}[\{0\}]+\texttt{value}[\{2\}]+\texttt{value}[\{0,2\}] \\
&= 3 + 1 + 5 + 1 = 10.
\end{split}
\end{equation*}
Because there are a total of $2^n$ subsets,
one possible solution is to go through all
pairs of subsets in $O(2^{2n})$ time.
However, using dynamic programming, we
can solve the problem in $O(2^n n)$ time.
The idea is to focus on sums where the
elements that may be removed from $S$ are restricted.
Let $\texttt{partial}(S,k)$ denote the sum of
values of subsets of $S$ with the restriction
that only elements $0 \ldots k$
may be removed from $S$.
For example,
\[\texttt{partial}(\{0,2\},1)=\texttt{value}[\{2\}]+\texttt{value}[\{0,2\}],\]
because we may only remove elements $0 \ldots 1$.
We can calculate values of \texttt{sum} using
values of \texttt{partial}, because
\[\texttt{sum}(S) = \texttt{partial}(S,n-1).\]
The base cases for the function are
\[\texttt{partial}(S,-1)=\texttt{value}[S],\]
because in this case no elements can be removed from $S$.
Then, in the general case we can use the following recurrence:
\begin{equation*}
\texttt{partial}(S,k) = \begin{cases}
\texttt{partial}(S,k-1) & k \notin S \\
\texttt{partial}(S,k-1) + \texttt{partial}(S \setminus \{k\},k-1) & k \in S
\end{cases}
\end{equation*}
Here we focus on the element $k$.
If $k \in S$, we have two options: we may either keep $k$ in $S$
or remove it from $S$.
There is a particularly clever way to implement the
calculation of sums. We can declare an array
\begin{lstlisting}
int sum[1<<N];
\end{lstlisting}
that will contain the sum of each subset.
The array is initialized as follows:
\begin{lstlisting}
for (int s = 0; s < (1<<n); s++) {
sum[s] = value[s];
}
\end{lstlisting}
Then, we can fill the array as follows:
\begin{lstlisting}
for (int k = 0; k < n; k++) {
for (int s = 0; s < (1<<n); s++) {
if (s&(1<<k)) sum[s] += sum[s^(1<<k)];
}
}
\end{lstlisting}
This code calculates the values of $\texttt{partial}(S,k)$
for $k=0 \ldots n-1$ to the array \texttt{sum}.
Since $\texttt{partial}(S,k)$ is always based on
$\texttt{partial}(S,k-1)$, we can reuse the array
\texttt{sum}, which yields a very efficient implementation.