# Difference between revisions of "Bit-String Flicking"

Bit strings (strings of binary digits) are frequently manipulated bit-by-bit using the logical operators NOT, AND, OR, and XOR. Bits strings are manipulated as a unit using SHIFT and CIRCULATE operators. The bits on the left are called the most significant bits and those on the right are the least significant bits.

Most high-level languages (e.g., Python, Java, C++), support bit-string operations. Programmers typically use bit strings to maintain a set of flags. Suppose that a program supports 8 options, each of which can be either “on” or “off”. One could maintain this information using an array of size 8, or one could use a single variable (if it is internally stored using at least 8 bits or 1 byte, which is usually the case) and represent each option with a single bit. In addition to saving space, the program is often cleaner if a single variable is involved rather than an array. Bits strings are often used to maintain a set where values are either in the set or not. Shifting of bits is also used to multiply or divide by powers of 2.

Mastering this topic is essential for systems programming, programming in assembly language, optimizing code, and hardware design.

## Operators

### Bitwise Operators

The logical operators are NOT (~ or \$\neg\$), AND (&), OR (|), and XOR (\$\oplus\$). These operators should be familiar to ACSL students from the Boolean Algebra and Digital Electronics categories.

• NOT is a unary operator that performs logical negation on each bit. Bits that are 0 become 1, and those that are 1 become 0. For example: ~101110 has a value of 010001.
• AND is a binary operator that performs the logical AND of each bit in each of its operands. The AND of two values is 1 only if both values are 1. For example, 1011011 and 011001 has a value of 001001. The AND function is often used to isolate the value of a bit in a bit-string or to clear the value of a bit in a bit-string.
• OR is a binary operator that performs the logical OR of each bit in each of its operands. The OR of two values is 1 only if one or both values are 1. For example, 1011011 or 0011001 has a value of 1011011. The OR function is often use to force the value of a bit in a bit-string to be 1, if it isn't already.
• XOR is a binary operator that performs the logical XOR of each bit in each of its operands. The XOR of two values is 1 if the values are different and 0 if they are the same. For example, 1011011 xor 011001 = 110010. The XOR function is often used to change the value of a particular bit.

All binary operators (AND, OR, or XOR) must operate on bit-strings that are of the same length. If the operands are not the same length, the shorter one is padded with 0's on the left as needed. For example, 11010 and 1110 would have value of 11010 and 01110 = 01010.

The following table summarizes the operators:

[math]x[/math] [math]y[/math] not [math]x[/math] [math]x[/math] and [math]y[/math] [math]x[/math] or [math]y[/math] [math]x[/math] xor [math]y[/math]
0 0 1 0 0 0
0 1 1 0 1 1
1 0 0 0 1 1
1 1 0 1 1 0

### Shift Operators

Logical shifts (LSHIFT-x and RSHIFT-x) “ripple” the bit-string x positions in the indicated direction, either to the left or to the right. Bits shifted out are lost; zeros are shifted in at the other end.

Circulates (RCIRC-x and LCIRC-x) “ripple” the bit string x positions in the indicated direction. As each bit is shifted out one end, it is shifted in at the other end. The effect of this is that the bits remain in the same order on the other side of the string.

The size of a bit-string does not change with shifts, or circulates. If any bit strings are initially of different lengths, all shorter ones are padded with zeros in the left bits until all strings are of the same length.

The following table gives some examples of these operations:

x (LSHIFT-2 x) (RSHIFT-3 x) (LCIRC-3 x) (RCIRC-1 x)
01101 10100 00001 01011 10110
10 00 00 01 01
1110 1000 0001 0111 0111
1011011 1101100 0001011 1011101 1101101

### Order of Precedence

The order of precedence (from highest to lowest) is: NOT; SHIFT and CIRC; AND; XOR; and finally, OR. In other words, all unary operators are performed on a single operator first. Operators with equal precedence are evaluated left to right; all unary operators bind from right to left.

## Sample Problems

### Problem 1

Evaluate the following expression:

(101110 AND NOT 110110 OR (LSHIFT-3 101010))

Solution: The expression evaluates as follows:

(101110 AND 001001 OR (LSHIFT-3 101010))
(001000 OR (LSHIFT-3 101010))
(001000 OR 010000)
011000

### Problem 2

Evaluate the following expression:

(RSHIFT-1 (LCIRC-4 (RCIRC-2 01101)))

Solution: The expression evaluates as follows, starting at the innermost parentheses:

(RCIRC-2 01101) => 01011
(LCIRC-4 01011) => 10101
(RSHIFT-1 10101) = 01010

### Problem 3

List all possible values of x (5 bits long) that solve the following equation.

(LSHIFT-1 (10110 XOR (RCIRC-3 x) AND 11011)) = 01100

Solution: Since x is a string 5 bits long, represent it by abcde.

(RCIRC-3 abcde) => cdeab
(cdeab AND 11011) => cd0ab
(10110 XOR cd0ab) => Cd1Ab (the capital letter is the NOT of its lower case)
(LSHIFT-1 Cd1Ab) => d1Ab0

So, d1Ab0 = 01100.

Meaning, we must have d=0, A=1 (hence a=0), b=0. Thus, the solution must be in the form 00*0*, where * is an “I-don’t-care”.

The four possible values of x are: 00000, 00001, 00100 and 00101.

### Problem 4

Evaluate the following expression:

((RCIRC-14 (LCIRC-23 01101)) | (LSHIFT-1 10011) & (RSHIFT-2 10111))

Solution: The problem can be rewritten as

A | B & C

The AND has higher precedence than the OR.

The evaluation of expression A can be done in a straightforward way: (LCIRC-23 01101) is the same as (LCIRC-3 01101) which has a value of 01011, and (RCIRC-14 01011) is the same as (RCIRC-4 01011) which has a value of 10110. Another strategy is to offset the left and right circulates. So, ((RCIRC-14 (LCIRC-23 01101)) has the same value as (LCIRC-9 01101), which has the same value as (LCIRC-4 01101) which is also 11010.

Expressions B and C are pretty easy to evaluate:

B = (LSHIFT-1 10011) = 00110
C = (RSHIFT-2 10111) = 00101

The expression becomes

A | B & C = 10110 | 00110 & 00101 = 10110 | 00100 = 10110

## Video Resources

The following YouTube videos show ACSL students and advisors working out some ACSL problems that have appeared in previous contests. Some of the videos contain ads; ACSL is not responsible for the ads and does not receive compensation in any form for those ads.

 "ACSL Math: Bit String Flicking" (Quick Coding Bytes) This video introduces the topic, then using an example problem, explains the methodology to solve problems that appear on ACSL contests. Bit String Flicking (Intro) (CalculusNguyenify) A great two-part tutorial on this ACSL category. Part 1 covers bitwise operations AND, OR, NOT, and XOR. Bit String Flicking Shifts and Circs (CalculusNguyenify) Part 2 covers logical shifts and circulate operations. Bit String Flicking (Tangerine Code) Shows the solution to the problem: (RSHIFT-3 (LCIRC-2 (NOT 10110))) Bit String Flicking by Ravi Yeluru (hemsra) Walks through two problems from the Junior Division. ACSL BitString Flicking Contest 2 Worksheet 1 (misterminich) Solves a handful of problems given in previous years at the Intermediate Division level.

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