So if a problem is more difficult the language it represents is smaller?
I'm reading the definition of polynomial time reducible:
Let $L_1, L_2$ be two language. If $L_1$ is polynomial time reducible to $L_2$ then exists $f:{0,1}^*$ s.t. $forall xin{0,1}^*$ $$xin L_1iff f(x)in L_2$$
For me this means the $L_1$ may be bigger (in cardinality) than $L_2$, but $L_2$ is more difficult since $L_1$ can be solved after reduced to $L_2$?
np-complete reductions decision-problem
asked Dec 31 '18 at 14:32
Bit_hcAlgorithmBit_hcAlgorithm
1628
add a comment |
I'm reading the definition of polynomial time reducible:
Let $L_1, L_2$ be two language. If $L_1$ is polynomial time reducible to $L_2$ then exists $f:{0,1}^*$ s.t. $forall xin{0,1}^*$ $$xin L_1iff f(x)in L_2$$
For me this means the $L_1$ may be bigger (in cardinality) than $L_2$, but $L_2$ is more difficult since $L_1$ can be solved after reduced to $L_2$?
np-complete reductions decision-problem
asked Dec 31 '18 at 14:32
Bit_hcAlgorithmBit_hcAlgorithm
1628
How would define the cardinality of an infinite set being larger than another infinite set (both being countable sets)?
– dkaeae
Dec 31 '18 at 14:53
You can find examples in which $L_1$ is a strict subset of $L_2$ (and vice versa).
– Yuval Filmus
Dec 31 '18 at 15:17
add a comment |
I'm reading the definition of polynomial time reducible:
Let $L_1, L_2$ be two language. If $L_1$ is polynomial time reducible to $L_2$ then exists $f:{0,1}^*$ s.t. $forall xin{0,1}^*$ $$xin L_1iff f(x)in L_2$$
For me this means the $L_1$ may be bigger (in cardinality) than $L_2$, but $L_2$ is more difficult since $L_1$ can be solved after reduced to $L_2$?
np-complete reductions decision-problem
asked Dec 31 '18 at 14:32
Bit_hcAlgorithmBit_hcAlgorithm
1628
I'm reading the definition of polynomial time reducible:
Let $L_1, L_2$ be two language. If $L_1$ is polynomial time reducible to $L_2$ then exists $f:{0,1}^*$ s.t. $forall xin{0,1}^*$ $$xin L_1iff f(x)in L_2$$
For me this means the $L_1$ may be bigger (in cardinality) than $L_2$, but $L_2$ is more difficult since $L_1$ can be solved after reduced to $L_2$?
np-complete reductions decision-problem
np-complete reductions decision-problem
asked Dec 31 '18 at 14:32
Bit_hcAlgorithmBit_hcAlgorithm
1628
asked Dec 31 '18 at 14:32
Bit_hcAlgorithmBit_hcAlgorithm
1628
asked Dec 31 '18 at 14:32
Bit_hcAlgorithmBit_hcAlgorithm
1628
asked Dec 31 '18 at 14:32
Bit_hcAlgorithmBit_hcAlgorithm
1628
asked Dec 31 '18 at 14:32
Bit_hcAlgorithmBit_hcAlgorithm
1628
1628
How would define the cardinality of an infinite set being larger than another infinite set (both being countable sets)?
– dkaeae
Dec 31 '18 at 14:53
You can find examples in which $L_1$ is a strict subset of $L_2$ (and vice versa).
– Yuval Filmus
Dec 31 '18 at 15:17
add a comment |
How would define the cardinality of an infinite set being larger than another infinite set (both being countable sets)?
– dkaeae
Dec 31 '18 at 14:53
You can find examples in which $L_1$ is a strict subset of $L_2$ (and vice versa).
– Yuval Filmus
Dec 31 '18 at 15:17
How would define the cardinality of an infinite set being larger than another infinite set (both being countable sets)?
– dkaeae
Dec 31 '18 at 14:53
How would define the cardinality of an infinite set being larger than another infinite set (both being countable sets)?
– dkaeae
Dec 31 '18 at 14:53
You can find examples in which $L_1$ is a strict subset of $L_2$ (and vice versa).
– Yuval Filmus
Dec 31 '18 at 15:17
You can find examples in which $L_1$ is a strict subset of $L_2$ (and vice versa).
– Yuval Filmus
Dec 31 '18 at 15:17
add a comment |
1 Answer
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$L_1$ and $L_2$ are always countably infinite, and thus "equally big".
If any language is finite, then it is "constant time" recognizable.
answered Dec 31 '18 at 14:56
Pål GDPål GD
6,6502241
I forgot this fact that they're both infinite... Thanks!
– Bit_hcAlgorithm
Dec 31 '18 at 17:31
add a comment |
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1 Answer
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1 Answer
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$L_1$ and $L_2$ are always countably infinite, and thus "equally big".
If any language is finite, then it is "constant time" recognizable.
answered Dec 31 '18 at 14:56
Pål GDPål GD
6,6502241
I forgot this fact that they're both infinite... Thanks!
– Bit_hcAlgorithm
Dec 31 '18 at 17:31
add a comment |
$L_1$ and $L_2$ are always countably infinite, and thus "equally big".
If any language is finite, then it is "constant time" recognizable.
answered Dec 31 '18 at 14:56
Pål GDPål GD
6,6502241
I forgot this fact that they're both infinite... Thanks!
– Bit_hcAlgorithm
Dec 31 '18 at 17:31
add a comment |
$L_1$ and $L_2$ are always countably infinite, and thus "equally big".
If any language is finite, then it is "constant time" recognizable.
answered Dec 31 '18 at 14:56
Pål GDPål GD
6,6502241
$L_1$ and $L_2$ are always countably infinite, and thus "equally big".
If any language is finite, then it is "constant time" recognizable.
answered Dec 31 '18 at 14:56
Pål GDPål GD
6,6502241
answered Dec 31 '18 at 14:56
Pål GDPål GD
6,6502241
answered Dec 31 '18 at 14:56
Pål GDPål GD
6,6502241
answered Dec 31 '18 at 14:56
Pål GDPål GD
6,6502241
6,6502241
I forgot this fact that they're both infinite... Thanks!
– Bit_hcAlgorithm
Dec 31 '18 at 17:31
add a comment |
I forgot this fact that they're both infinite... Thanks!
– Bit_hcAlgorithm
Dec 31 '18 at 17:31
I forgot this fact that they're both infinite... Thanks!
– Bit_hcAlgorithm
Dec 31 '18 at 17:31
I forgot this fact that they're both infinite... Thanks!
– Bit_hcAlgorithm
Dec 31 '18 at 17:31
add a comment |
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How would define the cardinality of an infinite set being larger than another infinite set (both being countable sets)?
– dkaeae
Dec 31 '18 at 14:53
You can find examples in which $L_1$ is a strict subset of $L_2$ (and vice versa).
– Yuval Filmus
Dec 31 '18 at 15:17