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The Power and the Limits of Computation. Elaine Rich. The Earliest Digital Computers. 1945 ENIAC. Stored 20 10-digit decimal numbers. The IBM 7090. A dual 7090 system at NASA in about 1962. Could store 32,768 36-bit words. That’s about .00015 gigabytes. Cost: about $3,000,000. - PowerPoint PPT Presentation
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The Power and the Limits of Computation
Elaine Rich
The Earliest Digital Computers
1945 ENIAC
Stored 20 10-digit decimal numbers
The IBM 7090A dual 7090 system at NASA in about 1962.
Could store 32,768 36-bit words.
That’s about .00015 gigabytes.
Cost:
about $3,000,000.
or $19,794,000 2005 dollars
The Earth SimulatorThe Earth Simulator (ES) is a project of Japanese agencies to develop
a 40 TFLOPS system for climate modeling. The ES site is a new location in an industrial area of Yokohama, an
hour drive west of Tokyo. The facility became operational in late 2001.
Hardware
The ES is based on: 5,120 (640 8-way nodes) 500 MHz NEC CPUs
8 GFLOPS per CPU (41 TFLOPS total) 2 GB (4 512 MB FPLRAM modules) per CPU (10 TB total)
shared memory inside the node 640 × 640 crossbar switch between the nodes
16 GB/s inter-node bandwidth 20 kVA power consumption per node
The iPod Nano
2005
Can store 4 gigabytes (1000 songs).
Cost: about $250.
Compute Power Increases Over Time
From Hans Moravec, Robot Mere Machine to Transcendent Mind 1998.
Moore’s Law
What Can We Do With All that Power?
Overheard in ACES last week: “Genomics has turned biology into an information science.”
Is there anything we can’t do?
The Traveling Salesman Problem
Finding a Solution to the TSP
Given n cities:
•n choices for a starting point.•n-1 for the next city•n-2 for the next city
For a total of n! paths to be considered.
Finding a Solution to the TSP
Given n cities:
•n choices for a starting point.•n-1 for the next city•n-2 for the next city
For a total of n! paths to be considered.
We notice that it doesn’t matter where we start (since we need to make a loop).
And the cost is the same forward or backward. So we can cut the number of paths down to:
(n – 1)!/2
The Growth Rate of n!n! = n(n-1)(n-2)…(1)
2 2 11 4790016003 6 12 62270208004 24 13 871782912005 120 14 13076743680006 720 15 209227898880007 5040 16 3556874280960008 40320 17 64023737057280009 362880 18 12164510040883200010 3628800 19 243290200817664000011 39916800 36 3.6 1041
Putting that Rate into Perspective
Speed of light: 3 108 m/sec
Width of a proton: 10-15 m
If we perform one operation in the time light crosses a proton, we can perform: 3 1023 ops/sec
Seconds since the big bang: 3 1017
Operations since the big bang: 9 1040
Compared to 36! 3.6 1041
The Post Correspondence Problem
i X Y1 b bbb2 babbb ba3 ba a4 bbbaa babbb
The Post Correspondence Problem
i X Y1 b bbb2 babbb ba3 ba a4 bbbaa babbb
2
X b a b b b
Y b a
The Post Correspondence Problem
i X Y1 b bbb2 babbb ba3 ba a4 bbbaa babbb
2 1
X b a b b b b
Y b a b b b
The Post Correspondence Problem
i X Y1 b bbb2 babbb ba3 ba a4 bbbaa babbb
2 1 1
X b a b b b b b
Y b a b b b b b b
The Post Correspondence Problem
i X Y1 b bbb2 babbb ba3 ba a4 bbbaa babbb
2 1 1 3
X b a b b b b b b a
Y b a b b b b b b a
The Post Correspondence Problem
i X Y1 10 1012 011 113 101 011
i X Y1 1101 12 0110 113 1 110
The Post Correspondence ProblemA program to solve this problem:
Until a solution is found do:Generate the next candidate solution.Test it. If it is a solution, halt and report yes.
So, if there are say 4 rows in the table, we’ll try:1 2 3 4 1,1 1,2 1,3 1,4 1,5 2,1 …… 1,1,1 ….
The Post Correspondence ProblemA program to solve this problem:
Until a solution is found do:Generate the next candidate solution.Test it. If it is a solution, halt and report yes.
So, if there are say 4 rows in the table, we’ll try:1 2 3 4 1,1 1,2 1,3 1,4 1,5 2,1 …… 1,1,1 ….
But what if there is no solution?
A Tiling Problem
A Tiling Problem
A Tiling Problem
A Tiling Problem
A Tiling Problem
A Tiling Problem
A Tiling Problem
A Tiling Problem
A Tiling Problem
A Tiling Problem
Programs Debug Programs
read nif 2*int(n/2) = n then print “even” else print “odd”
read nresult = 1for i = 2 to n do
result = result * iprint result
Given an arbitrary program, can it be guaranteed to halt?
A Problem That Cannot be Solved in Any Amount of Time
read nif 2*int(n/2) = n then print “even” else print “odd”
read nresult = 1for i = 2 to n do
result = result * iprint result
Given an arbitrary program, can it be guaranteed to halt?
result1
i= 2 2 i= 3 6i= 4 24i= 5 120
Other Kinds of Loops
result = 0count = 0until result > 100 do
read n count = count + 1 result = result+nprint count
Other Kinds of Loops
result = 0count = 0until result > 100 do
read n count = count + 1 result = result+nprint count
Suppose all the inputs are positive integers.
Other Kinds of Loops
result = 0count = 0until result > 100 do
read n count = count + 1 result = result+nprint count
Suppose some of the integers are negative.
The Halting Problem Is Not Solvable
Suppose that the following program existed:
Halts(program, string) returns:•True if program halts on string•False otherwise
The Halting Problem is Not Solvable
Consider the following program:
Trouble(string) = If Halts(string, string) then loop forever else halt.
Now we invoke Trouble(<Trouble>).
What should Halts(<Trouble>, <Trouble>) say? If it:•Returns True (Trouble will halt): Trouble loops•Returns False (Trouble will not halt): Trouble halts
So there is no answer that Halts can give that does not lead to a contradiction.
Other Unsolvable Problems
PCP:
•We can encode a <program>,<input> pair as an instance of PCP so that the PCP problem has a solution iff <program> halts on <input>.
•So if PCP were solvable then Halting would be.
•But Halting isn’t. So neither is PCP.
Other Unsolvable ProblemsTiling:
•We can encode a <program>,<input> pair as an instance of a tiling problem so that there is an infinite tiling iff <program> halts on <input>.
00010000111000000111110000000000000 0001000011101000011111000000000000000010000111011000111110000000000000 …
•So if the tiling problem were solvable then Halting would be.
•But Halting isn’t. So neither is the tiling problem.
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