LeetCode for Hardware Engineering, ASIC/FPGA/VLSI Digital Design and Verification Engineer
Synchronous reset has the following characteristics:
Continue reading → What are the differences between sync and async reset?
We discussed true LRU implementations, but they are usually costly. In this post, we will explain how to implement pseudo LRU.
An example is shown below. The data structure is similar to a binary tree. For N ways in a cache set, we need to keep (N – 1) bits. If there are 8 sets in a way, then we will need 7 bits.
Upon a cache access, all tree nodes point to that cache way will be flipped. For example, if set 3 is accessed, then h0, h1 and h4 will be flipped in next cycle.
To find LRU, we can perform a depth-first-search starting from h0, and traverse nodes in lower levels. If the node is 0, then we traverse the left sub-tree; otherwise, we traverse the right sub-tree. In the diagram above, the LRU is set 3.
Once we understand the basic idea of pseudo LRU, it is easy to write code for pseudo LRU.
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module lru_pseudo ( input clk, input rst_n, input acc_en, // access indication input [2:0] acc_idx, // index to accessed entry output logic [2:0] lru_idx // index to LRU entry ); // tree node logic h0, h1, h2, h3, h4, h5, h6; always_ff @(posedge clk or negedge rst_n) if (~rst_n) h0 <= '0; else if (acc_en) h0 <= ~h0; // always flip h0 if there's an access always_ff @(posedge clk or negedge rst_n) if (~rst_n) h1 <= '0; else if (acc_en & ~acc_idx[2]) h1 <= ~h1; // flip h1 if acc_idx == 0/1/2/3 always_ff @(posedge clk or negedge rst_n) if (~rst_n) h2 <= '0; else if (acc_en & acc_idx[2]) h2 <= ~h2; // flip h2 if acc_idx == 4/5/6/7 always_ff @(posedge clk or negedge rst_n) if (~rst_n) h3 <= '0; else if (acc_en & (acc_idx[2:1] == 2'b00)) h3 <= ~h3; // flip h3 if acc_idx == 0/1 always_ff @(posedge clk or negedge rst_n) if (~rst_n) h4 <= '0; else if (acc_en & (acc_idx[2:1] == 2'b01)) h4 <= ~h4; // flip h4 if acc_idx == 2/3 always_ff @(posedge clk or negedge rst_n) if (~rst_n) h5 <= '0; else if (acc_en & (acc_idx[2:1] == 2'b10)) h5 <= ~h5; // flip h5 if acc_idx == 4/5 always_ff @(posedge clk or negedge rst_n) if (~rst_n) h6 <= '0; else if (acc_en & (acc_idx[2:1] == 2'b11)) h6 <= ~h6; // flip h6 if acc_idx == 6/7 // lru_idx assignment assign lru_idx[2] = h0; assign lru_idx[1] = ~h0 & h1 | h0 & h2; assign lru_idx[0] = (~h0 & ~h1 & ~h3) | (~h0 & h1 & ~h4) | (h0 & ~h2 & ~h5) | (~h0 & h2 & ~h6); endmodule: lru_pseudo |
We covered 2 true LRU implementations: square matrix implementation and counter based implementation in previous post. We will discuss 2 more true LRU implementations in this post.
In the following discussion, we assume the number of ways in a cache set is N.
Linked list is probably the most straightforward LRU implementation. Nodes are linked from head to tail. Head points to the most recently used item, while tail points to the least recently used item.
Upon a cache access, the corresponding node “jumps” to the head, and the previous head becomes the next node of new head.
Least Recently Used, or LRU, is the optimal cache replacement policy, and it is used in various applications where arbitration is required. In this post, we will discuss different true LRU implementations, based on the paper “Highly Efficient LRU Implementations for High Associativity Cache Memory”.
In the following discussion, we assume the number of ways in a cache set is N.
If the sequence is not predefined, then we can no longer use traditional FSM based sequence detector. Assuming we know the sequence to detect is 5-bit, then we can use the following circuit to detect the sequence.
The circuit consists of a 5-stage shift register, and a 5-bit configuration register. The sequence to detect is programmed in the configuration register, and the input sequence is compared with the configuration register every cycle.
Assuming incoming bit stream is one bit per cycle, design a circuit that detects whether the integer number formed by the bit stream is a multiple of 5.
The idea is to have an FSM consisting of 5 states, S0, S1, S2, S3, S4. Each state represents divided by 5 remainder in previous cycle. If FSM stays at S0, then it means the number could be divided by 5 in previous cycle; otherwise, the number was not a multiple of 5.
Continue reading → Determine Whether An Infinite Sequence Is A Multiple of 5
Assuming the incoming bit stream is one bit per cycle, design a 3-b palindrome sequence detector.
Palindrome code is a sequence of characters which reads the same backward as forward. For example, 101, 010, 111, 000, etc.
The following diagram shows an example solution. We use Mealy Machine instead of Moore Machine since Moore Machine requires more states to detect 3b palindrome sequence.
