Patent Application
20230221927
The present invention is related to a k-cluster residue number system, and more particularly, to a memory-based k-cluster residue number system capable of performing complement conversion, sign detection, magnitude comparison and division.
Edge artificial intelligence (AI) computing is an area of rapid growth, which integrates neural networks with Internet of Things (IoT) together for computer vision, natural language processing, and self-driving car applications, it quantizes the floating-point number to fixed-point integer for inference operations. In-memory architecture is one of the important Edge AI computing platforms, which stacks the memory over the top of the logic circuits for Memory Centric Neural Computing (MCNC). The data is directly loaded from stacked memory to Processing Elements (PEs) for computation, it avoids loading the data from the external memory and minimizes data transfer. It significantly reduces the latency and speeds up the operations. The performance is further enhanced using Residue Number System (RNS), it fully utilizes the internal memory to store the data for integer operations.
Residue Number System (RNS) is a number system, which first defines the moduli set and transforms the numbers to their integer remainders (also called residue) through modulo division, then performs the arithmetic operations (addition and multiplication) on the remainders only. For examples, the moduli set is defined as (7, 8, 9) with the number 13 and 17. The dynamic range is defined by the product of moduli set with the range 504. It first transforms the numbers to its residue through the modulo operations 13→(6, 5, 4) and 17→(3, 1, 8), then performs addition and multiplication on residues only, (6, 5, 4)+(3, 1, 8)=(9, 6, 12)→(2, 6, 3), which is equal to 30. (6, 5, 4)*(3, 1, 8)=(18, 5, 32)→(4, 5, 5), which is equal to 221. Since the remainder magnitude is much smaller, it only requires the simple logic for parallel computations. The drawback of RNS is sign detection, magnitude comparison and division support. The residues are required to convert back to the binary number domain for those operations.
In an embodiment, a method for generating a k-cluster residue number system comprises generating a modular set composed of p coprime integers, generating a dynamic range by taking a product of the p coprime integers, generating row indices for all integers in the dynamic range, and generating column indices for all integers in the dynamic range. The p coprime integers include 2.
In another embodiment, a k-cluster residue number system comprises a processor, and a memory coupled to the processor. The processor is used to generate a modular set composed of p coprime integers, generate a dynamic range by taking a product of the p coprime integers, generate row indices for all integers in the dynamic range, generate column indices for all integers in the dynamic range, and generate a look-up table according to the row indices, the column indices and all integers in the dynamic set. The memory is used to store the look-up table. The p coprime integers include 2.
These and other objectives of the present invention will no doubt become obvious to those of ordinary skill in the art after reading the following detailed description of the preferred embodiment that is illustrated in the various figures and drawings.
To represent an n-bit integer and its negative using a k-cluster residue number system (k-RNS), it first defines a modular set of p coprime integers as (m1, . . . , 2, . . . , mp) where a dynamic range is generated according to the product of the modular set (m1, . . . , 2, . . . , mp). When a modular set of 3 coprime integers is chosen to be (2n/2−1, 2, 2n/2+1), the dynamic range is set to [−(2n−1), (2n−2)]. The modular set is not limited to 3 coprime integers, the number of coprime integers in the modular set can be increased to increase the dynamic range and keep the moduli small. In this case, the k-RNS converts each integer in the dynamic range to its row indices and column index formed by remainders through modulo division such as Equation 1.
r
i
=I mod mi Equation 1:
where:
ri is a row index or a column index;
I is an integer in the dynamic range; and
mi is a coprime integer of the modular set.
The look-up table 8 may include 4 columns: cluster index, row indices (r1,r3), positive integer column, and negative integer column. In this example, since the modular set has 3 coprime integers, each integer has 2 row indices and a column index. The positive integer column may list positive integers from 0 to 14 in an ascending order. The negative integer column may list negative integers from −15 to −1 in an ascending order. The integers are grouped according to the first row index modulo behavior. The integers 0 to 2, and −15 to −13 may be grouped to cluster 1. The integers 3 to 5, and −12 to −10 may be grouped to cluster 2. The integers 6 to 8, and −9 to −7 may be grouped to cluster 3. The integers 9 to 11, and −6 to −4 may be grouped to cluster 4. The integers 12 to 14, and −3 to −1 may be grouped to cluster 5. This grouping approach is only for an illustrative purpose, not for limiting the scope of the embodiment.
The k-RNS converts 0 to (0,0,0) through dividing (3,2,5), the coprime integers of the modular set, since (0,0,0) are remainders of 0 over (3,2,5); and converts −15 to (0,1,0) through dividing (3,2,5) since (0,1,0) are remainders of −15 over (3,2,5). The k-RNS converts 1 to (1,1,1) through dividing (3,2,5) since (1,1,1) are remainders of 1 over (3,2,5) and converts −14 to (1,0,1) through dividing (3,2,5) since (1,0,1) are remainders of −14 over (3,2,5). The k-RNS converts 2 to (2,0,2) through dividing (3,2,5) since (2,0,2) are remainders of 2 over (3,2,5) and converts −13 to (2,1,2) through dividing (3,2,5) since (2,1,2) are remainders of −13 over (3,2,5). The same approach can be applied to other numbers, and is thus not elaborated herein.
Because 0 and −15 have the same row numbers (0,0), 0 and −15 are listed in the same row. Their difference is that 0 has a column number of 0, and −15 has a column number of 1. Because 1 and −14 have the same row numbers (1,1), 1 and −14 are listed in the same row. Their difference is that 1 has a column number of 1, and −14 has a column number of 0. Because 2 and −13 have the same row numbers (2,2), 2 and −13 are listed in the same row. Their difference is that 2 has a column number of 0, and −13 has a column number of 1.
When an unknown integer is provided with row indices and a column index, the complement of the unknown integer can be determined.
When an unknown integer is provided with row indices and a column index, the sign of the unknown integer can be determined.
Step S22: check a column index of a negative integer corresponding to row indices of an unknown integer;
Step S24: input a column index of the unknown integer to a first input of an XOR gate 14, and input the column index of the negative integer corresponding to the row indices of the unknown integer to a second input of the XOR gate 14 to generate an output;
Step S26: determine a sign of the unknown integer according to the output. If the output is 0, the unknown integer is negative; if the output is 1, the unknown integer is positive.
The method of determining the sign of the unknown integer can be exemplified as follows: When an unknown integer has row indices of (0,0) and a column index of 0, the column index of the negative integer with the row indices of (0,0) is 1, 0 and 1 are input to the XOR gate 14, the XOR gate 14 will output 1. Since the output is 1, the unknown integer is positive. When referring to the look-up table 8, the unknown integer having row indices of (0,0) and a column index of 0 is 0, which is a positive integer. This agrees with the output of the XOR gate 14.
When an unknown integer has row indices of (0,0) and a column index of 1, the column index of the negative integer with the row indices of (0,0) is 1, 1 and 1 are input to the XOR gate 14, the XOR gate 14 will output 0. Since the output is 0, the unknown integer is negative. When referring to the look-up table 8, the unknown integer having row indices of (0,0) and a column index of 1 is −15, which is a negative integer. This agrees with the output of the XOR gate 14.
When an unknown integer has row indices of (1,1) and a column index of 1, the column index of the negative integer with the row indices of (1,1) is 0, 1 and 0 are input to the XOR gate 14, the XOR gate 14 will output 1. Since the output is 1, the unknown integer is positive. When referring to the look-up table 8, the unknown integer having row indices of (1,1) and a column index of 1 is 1, which is a positive integer. This agrees with the output of the XOR gate 14.
When an unknown integer has row indices of (1,1) and a column index of 0, the column index of the negative integer with the row indices of (1,1) is 0, 0 and 0 are input to the XOR gate 14, the XOR gate 14 will output 0. Since the output is 0, the unknown integer is negative. When referring to the look-up table 8, the unknown integer having row indices of (1,1) and a column index of 0 is −14, which is a negative integer. This agrees with the output of the XOR gate 14.
When two unknown integers are each provided with row indices and a column index, the magnitude comparator 120 can be used to compare their magnitudes. First, the method 20 can be used to determine signs of two unknown integers. If the signs are different, the unknown integer with a positive sign is larger than the unknown integer with a negative sign. If the signs are the same, the unknown integer with a higher cluster index is larger. If the signs are the same and the cluster indices are also the same, the unknown integer with a higher entry position, that is, a higher row index ri−1 is larger. If the signs are the same, the cluster indices are the same, and the entry positions are also the same, the two unknown integers are equal.
From the look-up table 8, a cluster index table can be illustrated as follows:
(r
)
indicates data missing or illegible when filed
Step S32: determine signs of two unknown integers;
Step S34: check if the signs are different; if so, go to Step S36, else go to Step S38;
Step S36: the unknown integer with a positive sign is larger.
Step S38: determine the cluster indices of the two unknown integers;
Step S40: check if the cluster indices of the two unknown integers are different; if so, go to Step S42, else go to Step S44;
Step S42: the unknown integer with a higher cluster index is larger.
Step S44: determine entry positions ri−1 of the unknown integers;
Step S46: check if the entry positions of the two unknown integers are different; if so, go to Step S48, else go to Step S49;
Step S48: the unknown integer with a higher entry position is larger.
Step S49: the two unknown integers are equal.
In Step S38, according to the look-up table 8, if the first unknown integer has row indices of (2,2), its cluster index would be 1. If the second unknown integer has row indices of (2,1), its cluster index would be 4, thus regardless of whether the two unknown integers are both positive or both negative, Step S42 can determine that the second unknown integer is larger than the first unknown integer.
In Step S44, according to the look-up table 8, if the first unknown integer has row indices of (2,2), its cluster index would be 1. If the second unknown integer has row indices of (1,1), its cluster index would also be 1, since the entry position of the first unknown integer is higher than the second unknown integer, that is, the row index ri−1 of the first unknown integer which is 2 is larger than the row index ri−1 of the second unknown integer which is 1, regardless of whether the two unknown integers are both positive or both negative. Step S48 can determine that the first unknown integer is larger than the second unknown integer.
In another method to compare magnitudes of two unknown integers, like the first method 30, the method 20 can be used to determine signs of the unknown integers. If the signs are different, the unknown integer with a positive sign is larger than the unknown integer with a negative sign. If the signs are the same, the unknown integers can be subtracted with each other, then the method 20 can be applied again on the difference to determine the sign of the difference. If the difference is a positive integer, then the minuend is larger.
Step S52: determine signs of two unknown integers;
Step S54: check if the signs are different; if so, go to Step S56, else go to Step S58;
Step S56: the unknown integer with a positive sign is larger.
Step S58: subtract the two unknown integers with each other;
Step S60: check if the difference of the two unknown integers is 0; if so, go to Step S62, else go to Step S64;
Step S62: the two unknown integers are equal.
Step S64: determine the sign of the difference of the two unknown integers;
Step S66: check if the difference of the two unknown integers is positive; if so, go to Step S68, else go to Step S70;
Step S68: the minuend is larger.
Step S70: the subtrahend is larger.
The magnitude comparison can be done using the subtraction approach. Suppose the first unknown integer has row indices of (2,2) and a column index of 0, and is thus represented as (2,0,2), the second unknown integer has row indices of (2,1) and a column index of 1, and is thus represented as (2,1,1), and Step S54 determines that their signs are the same since both are positive, then Step S58 would subtract the first unknown integer (2,0,2) by the second unknown integer (2,1,1) to obtain (0,1,1). In Step S60, since the difference of the two unknown integers is not 0, Step S64 is performed to determine the sign of the difference of the two unknown integers. By applying the method 20 in Step S66, it can be seen that the difference of the two unknown integers is negative, thus Step S70 can determine that the subtrahend is larger. By verifying with the look-up table 8, the integer represented as (2,0,2) is 2, the integer represented as (2,1,1) is 11, and the integer represented as (0,1,1) is −9. Thus the subtrahend 11 is indeed larger than the minuend 2.
The k-RNS 2 can also perform iterative subtraction to implement division of two unknown integers if the two unknown integers have the same sign. The division is to look for quotient Q and remainder R of dividend X and divisor Y. Let initial dividend X0=Xi initial quotient Q0=0, and iterative subtraction X′=Xi−Y.
Step S82: X0=X, Q0=0;
Step S84: X′=Xi−Y;
Step S86: check if X′≥0; if so, go to Step S90, else go to step S88;
Step S88: Q=Qi, R=Xi.
Step S90: check if X′=0; if so, go to Step S94, else go to step S92;
Step S92: Qi+1=Qi+1, Xi+1=X′, go to Step S84;
Step S94: Q=Qi+1, R=0.
If the two unknown integers X and Y are determined to be positive integers, the method 80 can be performed directly. If the two unknown integers X and Y are determined to be negative integers, the complement converter 150 can be used to generate the complements of the two unknown integers X and Y, then the complements of the two unknown integers X and Y can be used to perform the method 80.
An example, supposing X=14 represented as (2,0,4), and Y=3 represented as (0,1,3), can be illustrated as follows: In Step S82, X0=(2,0,4), Q0=0, represented as (0,0,0). In Step S84, X′=X0−Y=(2,0,4)−(0,1,3)=(2,1,1) since the modular set is (3,2,5). In Step S86, the method 20 can determine (2,1,1) to be positive. In Step S90, (2,1,1) is not (0,0,0), Step S92 can determine) Q1=Q0+1=(0,0,0)+(1,1,1)=(1,1,1), and X1=(2,1,1).
Then Step S84 is again performed. In Step S84, X′=X1−Y=(2,1,1)−(0,1,3)=(2,0,3). In Step S86, the method 20 can determine (2,0,3) to be positive. In Step S90, (2,0,3) is not (0,0,0), Step S92 can determine Q2=Q1+1=(1,1,1)+(1,1,1)=(2,0,2), X2=(2,0,3).
Then Step S84 is again performed. In Step S84, X′=X2−Y=(2,0,3)−(0,1,3)=(2,1,0). In Step S86, the method 20 can determine (2,1,0) to be positive. In Step S90, (2,1,0) is not (0,0,0), Step S92 can determine Q3=Q2+1=(2,0,2)+(1,1,1)=(0,1,3), X3=(2,1,0).
Then Step S84 is again performed. In Step S84, X′=X3−Y=(2,1,0)−(0,1,3)=(2,0,2). In Step S86, the method 20 can determine (2,0,2) to be positive. In Step S90, (2,0,2) is not (0,0,0), Step S92 can determine Q4=Q3+1=(0,1,3)+(1,1,1)=(1,0,4), X4=(2,0,2).
Then Step S84 is again performed. In Step S84, X′=X4−Y=(2, 0,2)−(0,1,3)=(2,1,4). In Step S86, the method 20 can determine (2,1,4) to be negative. In Step S88, Q=Q4=(1,0,4) which in decimal is 4, R=X4=(2,0,2) which in decimal is 2. This result agrees with the decimal result of 14 divided by 3.
The k-RNS can also perform another iterative subtraction to implement division when two unknown integers have the same sign. The division is also to look for quotient Q and remainder R of dividend X and divisor Y. Let initial dividend X0=Xi initial quotient Q0=0, and iterative subtraction X′=Xi-qiY.
The multiplier 114 has a first input coupled to the output of the quotient factor generator 112 for receiving the quotient factor qi, a second input for receiving the divisor Y, and an output for outputting the product of the quotient factor qi and the divisor Y. The subtractor 102 has a first input for receiving the dividend Xi, a second input for receiving the product of the quotient factor qi and the divisor Y, and an output for outputting the difference between the dividend Xi, and the product of the quotient factor qi and the divisor Y. The sign detector 104 has an input coupled to the output of the subtractor 102 for receiving the difference between the dividend Xi, and the product of the quotient factor qi and the divisor Y, a first output, and a second output. The dividend register 106 has a first input coupled to the output of the subtractor 102 for receiving the difference between the dividend Xi and the product qiY, a second input coupled to the first output of the sign detector 104 for receiving the sign of the difference between the dividend Xi and the product qiY, and an output coupled to the first input of the quotient factor generator 112 and the first input of the subtractor 102. If the difference is a non-zero positive integer, the dividend register 106 will output the difference as an updated dividend Xi+1 to the first input of the quotient factor generator 112 and the first input of the subtractor 102. The adder 108 has a first input coupled to the output of the quotient factor generator 112 for receiving the quotient factor qi, a second input for receiving a quotient Qi, and an output for outputting the sum of the quotient factor qi and the quotient Qi. The quotient register 110 has a first input coupled to the output of the adder 108 for receiving the sum of the quotient factor qi and the quotient Qi as an updated quotient Qi+1, a second input coupled to the second output of the sign detector 104 for receiving the sign of the difference between the dividend Xi and the product qiY, a first output coupled to the second input of the adder 108 for outputting the updated quotient Qi+1 if the sign of the difference between the dividend Xi and the product qiY is positive, and a second output for outputting the quotient Qi if the sign of the difference between the dividend Xi and the divisor Y is negative. Compared to the division device 100, the division device 200 uses the quotient factor generator 112 to speed up the division process.
Step S132: X0=X, Q0=0;
Step S133: generate qi based on the cluster index of Xi and the cluster index of Y using the quotient factor look-up table A;
Step S134: X′=Xi−qiY;
Step S136: check if X′≥0; if so, go to Step S140, else go to step S138;
Step S138: Q=Qi, R=Xi.
Step S140: check if X′=0; if so, go to Step S144, else go to step S142;
Step S142: Qi+1=Qi+qi, Xi+1=X′, go to Step S133;
Step S144: Q=Qi+qi, R=0.
If the two unknown integers X and Y are determined to be positive integers, the method 130 can be performed directly. If the two unknown integers X and Y are determined to be negative integers, the complement converter 150 can be used to generate the complements of the two unknown integers X and Y, then the complements of the two unknown integers X and Y can be used to perform the method 130.
An example supposing X=14 represented as (2,0,4) and Y=2 represented as (2,0,2) can be illustrated as follows: In Step S132, X0=(2,0,4), Q0 represents as (0,0,0). In Step 133, the cluster index of (2,0,4) is 5, and the cluster index of (2,0,2) is 1, since the smallest positive integer in the cluster index of 5 is 12, the largest positive integer in the cluster index of 1 is 2, q0 is set to 6 which is the quotient of 12 and 2, and represented as (0,0,1). That is, the quotient factor qi is the quotient of the smallest positive integer in the cluster index of Xi and the largest positive integer in the cluster index of Y when the cluster index of Xi is larger than the cluster index of Y, otherwise the quotient factor qi is equal to 1. And the quotient factor qi can be determined by the cluster index of Xi and the cluster index of Y using the quotient factor look-up table A. In Step S134, X′=X0−q0Y=(2,0,4)−(0,0,1)×(2,0,2)=(2,0,2). In Step S136, the method 20 can determine (2,0,2) to be positive. In Step S140, (2,0,2) is not equal to 0. Step S142 can determine Q1=Q0+q0=(0,0,0)+(0,0,1)=(0,0,1), and X1=(2,0,2).
Step S133 is again performed. In Step S133, the cluster index of (2,0,2) is 1, thus q1 is set to (1,1,1). In Step S134, X′=X1−q1Y=(2,0,2)−(2,0,2)=(0,0,0). In Step S136, the method 20 can determine (0,0,0) to be 0. In Step S140, (0,0,0) is equal to 0, thus Step S144 is performed to determine Q=(0,0,1)+(1,1,1)=(1,1,2) and R=0. From the look-up table 8, (1,1,2) is the representation of the integer 7. Thus the quotient is 7, and the remainder is 0. The result agrees with the decimal result of 14 divided by 2.
The k-RNS 2 can also perform iterative addition to implement division of two unknown integers if the two unknown integers have different signs. The division is to look for quotient Q and remainder R of a negative dividend X and a positive divisor Y. Let initial dividend X0=Xi initial quotient Q0=0, and iterative addition X′=Xi+Y.
Step S162: X0=X, Q0=0;
Step S164: X′=Xi+Y;
Step S166: check if X′≤0; if so, go to Step S170, else go to step S168;
Step S168: Q=complement of Qi, R=Xi.
Step S170: check if X′=0; if so, go to Step S174, else go to step S172;
Step S172: Qi+1=Qi+1, Xi+1=X′, go to Step S164;
Step S174: Q=complement of (Qi+1), R=0.
The k-RNS can also perform another iterative addition to implement division when two unknown integers have different signs. The division is also to look for quotient Q and remainder R of a negative dividend X and a positive divisor Y. Let initial dividend X0=Xi initial quotient Q0=0, and iterative addition X′=Xi+qiY.
The multiplier 414 has a first input coupled to the output of the quotient factor generator 412 for receiving the quotient factor qi, a second input for receiving the divisor Y, and an output for outputting the product of the quotient factor qi and the divisor Y. The first adder 402 has a first input for receiving the dividend Xi, a second input for receiving the product of the quotient factor qi and the divisor Y, and an output for outputting the sum of the dividend Xi, and the product of the quotient factor qi and the divisor Y. The sign detector 404 has an input coupled to the output of the first adder 402 for receiving the sum of the dividend Xi, and the product of the quotient factor qi and the divisor Y, a first output, and a second output. The dividend register 406 has a first input coupled to the output of the first adder 402 for receiving the sum of the dividend Xi and the product qiY, a second input coupled to the first output of the sign detector 404 for receiving the sign of the sum of the dividend Xi and the product qiY, and an output coupled to the first input of the quotient factor generator 412 and the first input of the first adder 402. If the sum is a negative integer, the dividend register 406 will output the sum as an updated dividend Xi+1 to the first input of the quotient factor generator 412 and the first input of the first adder 402. The second adder 408 has a first input coupled to the output of the quotient factor generator 412 for receiving the quotient factor qi, a second input for receiving a quotient Qi, and an output for outputting the sum of the quotient factor qi and the quotient Qi. The quotient register 410 has a first input coupled to the output of the second adder 408 for receiving the sum of the quotient factor qi and the quotient Qi as an updated quotient Qi+1, a second input coupled to the second output of the sign detector 404 for receiving the sign of the sum of the dividend Xi and the product qiY, a first output coupled to the second input of the second adder 408 for outputting the updated quotient Qi+1 if the sign of the sum of the dividend Xi and the product qiY is negative, and coupled to the complement converter 150 so that the complement converter 150 can generate the complement of the updated quotient Qi+1 if the sum of the dividend Xi and the product qiY is zero, and a second output coupled to the complement converter 150 so that the complement converter 150 can generate the complement of the quotient Qi if the sum of the dividend Xi and the product qiY is a non-zero positive integer. Compared to the division device 300, the division device 400 uses the quotient factor generator 412 to speed up the division process.
Step S182: X0=X, Q0=0;
Step S184: generate qi based on the cluster index of Xi and the cluster index of Y using the quotient factor look-up table B;
Step S186: X′=Xi+qiY;
Step S188: check if X′≤0; if so, go to Step S192, else go to step S190;
Step S190: Q=complement of Qi, R=Xi.
Step S192: check if X′=0; if so, go to Step S196, else go to step S194;
Step S194: Qi+1=Qi+qi, Xi+1=X′, go to Step S184;
Step S196: Q=complement of (Qi+qi), R=0.
In the method 160, 180, if the dividend X is determined to be a negative integer, and the divisor Y is determined to be a positive integer, the method 160, 180 can be performed directly. If the dividend X is determined to be a positive integer, and the divisor Y is determined to be a negative integer, the complement converter 150 can be used to generate the complements of the two unknown integers X and Y, then the complements of the two unknown integers X and Y can be used to perform the method 160, 180.
In the k-cluster residue number system 2, since the modular set is composed of coprime integers, and each integer is represented by row indices and a column index, the memory space used to store the look-up table 8 in the memory 6 is minimized. Since 2 is among the coprime integers and is used as a basis for column indices, complement conversion and sign detection can be easily performed. Since the processor 4 can perform complement conversion, sign detection, magnitude comparison and division in residue numbers, the k-cluster residue number system 2 greatly reduces the amount of calculations and improves the performance of the processor 4. Once calculations are performed, row indices and a column index of an integer can be easily used to retrieve the integer in the dynamic range. The look-up table 8 can be extended to RNS/binary conversion without using Chinese Remainder Theorem (CRT) or Mixed Radix Conversion (MRC). Therefore, the k-cluster residue number system 2 can enhance the performance of edge artificial intelligence (AI) computing. The k-cluster residue number system 2 can also be applied to other signal processing applications.
Those skilled in the art will readily observe that numerous modifications and alterations of the device and method may be made while retaining the teachings of the invention. Accordingly, the above disclosure should be construed as limited only by the metes and bounds of the appended claims.
