CS3102 Theory of Computation - - PowerPoint PPT Presentation

CS3102 Theory of Computation

www.cs.virginia.edu/~njb2b/cstheory/s2020 Warm up: Software can be reconfigured, hardware cannot. When we build hardware, how do we decide what to implement? If you found the fitbit, the owner is down front, and is upset about all her wasted steps from the past 2 days.

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Logistics

• Exercise 2 due Tuesday
• Quiz 3 released Friday
• Exercise 3 is out this weekend

– Last “regular-sized” exercise before midterm – There will be a “tiny” exercise 4 due the Tuesday before the midterm

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Last Time

• Boolean Circuits as a model of computing
• Straightline Programs as a model of

computing

• Proved NAND-Straightline = NAND-Circ =

AON-Circ = AON-straightline

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Lookup

• Indexing into a bitstring
• The 𝑀𝑝𝑝𝑙𝑣𝑞 function of order 𝑙:

𝑀𝑃𝑃𝐿𝑉𝑄𝑙: 0,1 2𝑙+𝑙 → 0,1 Defined such that for 𝑦 ∈ 0,1 2𝑙, 𝑗 ∈ 0,1 𝑙: 𝑀𝑃𝑃𝐿𝑉𝑄𝑙 𝑦, 𝑗 = 𝑦𝑗

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𝑀𝑃𝑃𝐿𝑉𝑄𝑙

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1 1 1 1 𝑙 = 3 1 First 2𝑙 bits of input Considered as a bitstring Last 𝑙 bits of input Considered as an index

x: i:

Theorem

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There is a NAND-Cricuit that computes 𝑀𝑃𝑃𝐿𝑉𝑄𝑙: 0,1 2𝑙+𝑙 → {0,1} Moreover, the number of gates required is at most 4 ⋅ 2𝑙

Proof idea

• Consider index 𝑗
• If the first bit of 𝑗 is 0, then the bit we're

looking for is in the first half of 𝑦

• Do lookup for 𝑙 − 1

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1 1 1 1 1

𝑦: 𝑗: 𝑦: 𝑗:

1 1

Defining 𝑀𝑃𝑃𝐿𝑉𝑄𝑙

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For 𝑙 ≥ 2, 𝑀𝑃𝑃𝐿𝑉𝑄𝑙(𝑦0, … , 𝑦2𝑙−1, 𝑗0, … , 𝑗𝑙−1) is equal to: 𝐽𝐺(𝑗0, 𝑀𝑃𝑃𝐿𝑉𝑄𝑙−1 𝑦2𝑙−1, … , 𝑦2𝑙−1, 𝑗1, … , 𝑗𝑙−1 , 𝑀𝑃𝑃𝐿𝑉𝑄𝑙−1(𝑦0, … , 𝑦2𝑙−1−1, 𝑗1, … , 𝑗𝑙−1)

Base Case

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Next Step

LOOKUP2

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LOOKUP3 and 4

Counting Gates

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Show this uses at most 4 ⋅ 2𝑙 − 4 gates (lines of code)

Counting Gates

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Show this uses at most 4 ⋅ 2𝑙 − 4 gates (lines of code)

Computing Every Finite Function

• Next we'll show that NAND is “universal”
• Any finite function can be computed by some

NAND-straightline program (equivalently, a NAND-circuit)

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Idea

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Input Output 000 001 010 1 011 100 1 101 1 110 111

We will have one variable to represent each possible input. We'll do a lookup with the actual input to select the proper output

Consider the function 𝑔: 0,1 3 → {0,1}

Straightline Code for F

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Input Output 000 001 010 1 011 100 1 101 1 110 111

Getting 0 and 1

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Getting 0 and 1

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Computing any function

• Make a variable to represent each possible

input

• Assign its value to match the correct output
• Use LOOKUP to select the proper output for

the given input

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Straightline Code for F

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Input Output 000 001 010 1 011 100 1 101 1 110 111

How many gates?

• How many gates does this construction take?

You can compute any finite function 𝑔: 0,1 𝑜 → 0,1 𝑛 with a NAND Circuit using no more than 𝑑 ⋅ 𝑛 ⋅ 2𝑜 gates Note: This can be improved to 𝑑 ⋅ 𝑛 ⋅

2𝑜 𝑜 (theorem

4.16 in TCS)

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Counting gates

• 1. Create variables for each input
• 2. Assign 0,1 to each input
• 3. Do the LOOKUP

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What does this mean?

• Your laptop is a 64-bit machine. Given enough

transistors, it can compute any function 𝑔: 0,1 64 → 0,1 64

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Any to Every

• Previous theorem:

– We can compute ANY 𝑜-bit function using circuits/straightline programs

• What we want:

– A machine that can compute EVERY 𝑜-bit function

• How do we do this?:

– Define a function that "simulates" programs – Write a program that gives the same answer as a given program of 𝑜 inputs, 𝑛 outputs, and 𝑡 lines

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How are programs run?

• Have a table of variables
• Execute code in sequence
• Update values in table
• Return a value from the table

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Simulating XOR

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Variable Value

Simulating XOR

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Variable Value a b 1 u 1 v 1 w return 1

Defining EVAL

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Input: bit string representing a program (first 𝑇(𝑡) bits) plus input values (remaining 𝑜 bits) Output: the result of running the represented program

• n the provided input, or 𝑛 0's if there's a "compile

error"

𝐹𝑊𝐵𝑀𝑡,𝑜,𝑛: 0,1 𝑇 𝑡 +𝑜 → 0,1 𝑛

Programs as Bits

• To evaluate a program with another program, we need to

convert the first program into bits 1. Number each variable (first n go to input, last m to outputs) 2. Represent each line as 3 numbers (outvar, in1, in2) 3. Represent program as (n,m,[Lines])

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Variable Number a b 1 temp1 2 temp2 3 return 4 (2,0,0) (3,1,1) (4,2,3) (2,1,[(2,0,0),(3,1,1),(4,2,3)])

XOR to bits

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Variable Number 𝑜 = 𝑛 = 𝑡 = Total bits =

XOR to bits

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Variable Number a b 1 u 2 v 3 w 4 return 5 𝑜 = 2 𝑛 = 1 𝑡 = 4 Total bits = 3 [numbers per line] ⋅ 3 [bits per number] ⋅ 4[lines] + 6 [length of 𝑜 + 𝑛]

How big is this?

1. Number each variable 2. Represent each line as 3 numbers (outvar, in1, in2) 3. Represent program as (n,m,[Lines])

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⌈log2 3𝑡⌉ bits each

𝑇 𝑡 ≤ 4𝑡⌈log2 3𝑡⌉ ℓ = ⌈log2 3𝑡⌉

3𝑡 ⋅ ⌈log2 3𝑡⌉ bits 2⌈log2 𝑡⌉ bits

Defining EVAL

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Input: bit string representing a program (first 𝑇(𝑡) bits) plus input values (remaining 𝑜 bits) Output: the result of running the represented program

• n the provided input, or 𝑛 0's if there's a "compile

error"

𝐹𝑊𝐵𝑀𝑡,𝑜,𝑛: 0,1 𝑇 𝑡 +𝑜 → 0,1 𝑛

Defining the EVAL function

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Variable Value 1 2 3 4 5 Representation: (2, 0, 1), (3, 0, 2), (4, 1, 2), (5, 3, 4) Input: 0, 1

𝑜 = 2 𝑛 = 1

Psuedocode for EVAL

• Table 𝑈:

– holds variables and their values

• 𝐻𝐹𝑈(𝑈, 𝑗)

– Returns the bit of 𝑈 associated with variable 𝑗

• 𝑉𝑄𝐸𝐵𝑈𝐹(𝑈, 𝑗, 𝑐)

– Returns a new table such that variable 𝑗's value has been changed to 𝑐

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𝑈

Psuedocode for EVAL

• Input:

– Numbers 𝑜, 𝑛, 𝑡, 𝑢 representing the number of inputs, outputs, variables, and lines respectively – 𝑀, a list of triples representing the program – A string 𝑦 to be given as input to the program

• Output:

– Evaluation of the program represented by 𝑀 when run

• n input 𝑦

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Let 𝑈 be table of size 𝑢 For 𝑗 in range(𝑜): 𝑈 = UPDATE(𝑈, 𝑗, 𝑦[𝑗]) For (𝑗,𝑘,𝑙) in 𝑀: 𝑏 = GET(𝑈, 𝑘) 𝑐 = GET(𝑈, 𝑙) 𝑈 = UPDATE(𝑈, 𝑗, NAND(𝑏,𝑐)) For 𝑗 in range(𝑛): 𝑍[𝑗] = GET(𝑈, 𝑢 − 𝑛 + 𝑗) Return 𝑍

EVAL in NAND

• Next we implement 𝐹𝑊𝐵𝑀𝑡,𝑜,𝑛 using NAND

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𝐻𝐹𝑈(𝑈, 𝑗)

• Get the bit at “row” 𝑗 of 𝑈
• Look familiar?
• How many gates to implement?

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UPDATE

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• To change index 𝑗 of table 𝑈 to bit 𝑐
• For every index except 𝑗, return the same value
• For index 𝑗, return b instead
• Define 𝐹𝑅𝑉𝐵𝑀𝑘: 0,1 ℓ → {0,1} which returns 1 if

the input binary number is equal to j

Note: 𝐹𝑅𝑉𝐵𝑀𝑘 can be done in 𝑑 ⋅ ℓ gates 𝑉𝑄𝐸𝐵𝑈𝐹ℓ: 0,1 𝑡ℓ+ℓ+1 → 0,1 2ℓ

UPDATE pseudocode

For 𝑘 in range(2ℓ): 𝑏 = 𝐹𝑅𝑉𝐵𝑀𝑇

𝑘(𝑗)

𝑜𝑓𝑥𝑈[𝑘] = 𝐽𝐺(𝑏, 𝑐, 𝑈[𝑘]) Return 𝑜𝑓𝑥𝑈

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Runs 2ℓ times

Conclusion

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