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Joined 2026-09-21 Message
Incoming links: Bit mask
Bitwise AND 2026-10-20
Bitwise OR 2026-10-20
Byte-table population count 2026-10-25
Store the Hamming weight of each possible byte in a 256-entry table. Extract each byte of a word by a logical right shift and a bit mask, then sum its table entries. The byte positions partition the word's bits, proving the formula. A 32-bit word needs four lookups, which can be unrolled. Logical extraction does not depend on the endianness of a stored word. The speed advantage depends on table-access costs as well as instruction counts.
Digit dynamic programming 2026-10-20
Count or sum integers by processing their numerical digits and merging prefixes that have the same possible continuations. Typical state records include the current position, whether the prefix still equals an upper limit, a remainder, or a bit mask of digits already used. The final answer comes from accepting exactly those states satisfying the desired digit constraints.
Exclusive or Created 2026-09-28 Updated 2026-10-20
Exclusive or returns one exactly when its two input bits differ. Its truth table gives , , , and . The identitiesexpress exclusive or using logical conjunction, logical disjunction, and logical negation. The first expression accepts at least one input and then excludes both; the second explicitly lists the two accepted input pairs.
For numerical bits, exclusive or is addition in modular arithmetic modulo two:The binary operation is commutative and associative, because addition modulo two has those properties. It satisfies , , and . These identities explain why a bit can be toggled and restored with the same operation.
On binary strings of equal length, bitwise exclusive or applies the one-bit rule separately at each position. For example,A bit mask of toggles exactly the middle two bits. Applying the mask twice restores the original binary string: each selected bit changes twice, and each unselected bit stays fixed. This positional operation is distinct from taking the exclusive or of the overall truth values of two integers.
Paging 2026-10-13
Paging divides an address space into fixed-size memory pages and main computer memory into equally sized page frames. Here assume byte addressing, one memory page size bytes, and an unsigned virtual address . Its virtual page number and offset areThe brackets denote the floor function, and gives the remainder in Euclidean division. If the page table maps to physical frame number and permits the access, the physical address isThe low bits are unchanged; the other bits select the memory page or page frame. A bit shift and bit mask can extract these fields. [ref]
https://pages.cs.wisc.edu/~remzi/OSTEP/vm-paging.pdf. Arpaci-Dusseau and Arpaci-Dusseau, OSTEP chapter 18: Paging, Introduction, sections 18.1–18.2
For a 16-bit virtual address and bytes, take in hexadecimal. Then and . A mapping to givesThus the eight high bits choose a mapping; the eight low bits choose the same position in the resulting page frame. The 256-byte size is for illustration, not a claim about an actual architecture's supported memory page sizes.
Translating 0x1234 through an illustrative page table, with a separate offset and access check
. Details. 
The useful freedom is that adjacent memory pages can occupy unrelated page frames. In the diagram, virtual pages
0x12 and 0x13 occupy page frames 0x07 and 0x02; virtual addresses 0x12ff and 0x1300 are consecutive, but translate to physical addresses 0x07ff and 0x0200. Uniform units simplify placement and page replacement, at the cost of page table storage and unused bytes within partly filled memory pages. For example, placing 300 bytes in exclusive 256-byte page frames requires two page frames, leaving bytes unused. Past exam of the mathematics course of the University of Cambridge 2006 ia Paper 5 10 e Solution Created 2026-09-24 Updated 2026-10-14
Choose bit zero of each integer to represent the leftmost cell in its group of 32 columns. Then use a quotient to choose the word and a remainder to choose the bit:For valid ,
java
static boolean getBit(int[][] board, int i, int j) {
return ((board[i][j >>> 5] >>> (j & 31)) & 1) != 0;
}j >>> 5 is and ranges from zero to 31. The bit mask j & 31 is the offset . The logical right bit shift brings the desired bit to position zero; the final mask selects only that bit. Even bit 31 of a negative stored word is extracted correctly, because >>> inserts zeros. A different convention, with the leftmost cell in the most significant bit, would instead shift by 31 - (j & 31) and must be used consistently. Past exam of the mathematics course of the University of Cambridge 2006 ia Paper 5 3 Solution Created 2026-09-24 Updated 2026-10-14
Extract one four-bit digit at a time, beginning at the most significant end of the Java The logical right bit shift
int. A lookup string supplies characters, but no library performs the conversion:java
static String toHex(int value) {
String digits = "0123456789abcdef";
String answer = "";
boolean started = false;
for (int shift = 28; shift >= 0; shift -= 4) {
int digit = (value >>> shift) & 15;
if (started || digit != 0 || shift == 0) {
answer = answer + digits.charAt(digit);
started = true;
}
}
return answer;
}>>> inserts zeros, and the bit mask 15 retains exactly the low four bits of the shifted word. Each emitted character is therefore the corresponding hexadecimal digit. The started flag suppresses leading zero digits; the shift == 0 exception makes zero produce "0", rather than an empty string.A negative Java integer uses two's complement. Its top bit is set, so the first extracted digit is between eight and fifteen and all eight digits are emitted. We interpret its original 32-bit pattern, without taking an absolute value or negating a potentially unnegatable minimum integer. In particular,
toHex(19) returns "13", toHex(-1) returns "ffffffff", and toHex(Integer.MIN_VALUE) returns "80000000". There are exactly eight iterations, so for this fixed word size the time complexity is constant.For a 32-bit machine word and , use a left bit shift followed by a logical right shift:The left shift discards all but the lowest bits and moves those retained bits to the top. The logical right shift returns them to their original positions and fills the rest with zero. Hence
lsls r1, r0, #(32-k)
lsrs r1, r1, #(32-k)r1 contains , the remainder identified in (b). For the remainder is zero; for it is , so these endpoints can be handled directly.On a reduced instruction set computer, a bit mask such as may not fit an instruction's immediate field. Masking may therefore require materializing the constant in another processor register, possibly using several instructions or a memory load. Small shift counts are directly encodable, so the two shifts can save code space, a temporary processor register, or a load. This is an architecture-dependent advantage, rather than a universal speed claim. The source's compact instruction forms support these immediate shifts, while their bitwise AND form takes its other operand from a processor register.[ref]
Thumb 16-bit instruction set quick reference by Arm. page 1, Logical and Shift/rotate tables
Profile dynamic programming 2026-10-20
A dynamic programming method that advances a boundary across a structured object and records only the information needed on that boundary. For tilings by pieces of bounded height, a bit mask recording occupied cells in a few frontier rows can determine every legal continuation. Its state space is typically exponential in the width but independent of the total height.
Reduced instruction set computer 2026-10-24
A processor design using a relatively regular instruction repertoire and explicit restrictions on operand forms. Immediate constants may have limited encodings, so extracting low bits with two bit shifts can avoid constructing a large bit mask in a temporary processor register. The benefit depends on the actual instruction set. A compact example appears in the Thumb 16-bit instruction set quick reference by Arm, whose logical bitwise AND instruction uses a processor register operand while shifts accept small immediate counts.
Signed power-of-two division by biasing 2026-10-25
For signed two's complement integers, a conditional positive bias converts the rounding behaviour of an arithmetic right shift into C integer division. It can be formed without a conditional branch by turning the sign bit into either zero or a low--bit bit mask.
Proof. Nonnegative needs no adjustment. If , write , . Then , whose quotient rounded down is , exactly the quotient truncated towards zero. The bias fits safely for in an -bit signed word: it is added only to negative inputs, and the largest possible adjusted input is . The corresponding remainder is , where .
Step number 2026-10-20
A positive integer whose adjacent decimal numerical digits differ by exactly 1. Its digits form a walk on the path graph , with a nonzero starting digit. Counting pandigital numbers of this type can use digit dynamic programming with the final digit and a bit mask of visited digits as the state.
Page 1 lists the compact logical and shift instruction forms. Left bit shifts have immediate counts –, while logical right shifts and arithmetic right shifts have immediate counts –. The logical bitwise AND entry uses a processor register operand. These encoding details explain why extracting low bits with two immediate shifts can avoid materializing a bit mask.