So if a problem is more difficult the language it represents is smaller?












3














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







share|cite|improve this question






















  • How would define the cardinality of an infinite set being larger than another infinite set (both being countable sets)?
    – dkaeae
    10 hours ago










  • You can find examples in which $L_1$ is a strict subset of $L_2$ (and vice versa).
    – Yuval Filmus
    10 hours ago
















3














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







share|cite|improve this question






















  • How would define the cardinality of an infinite set being larger than another infinite set (both being countable sets)?
    – dkaeae
    10 hours ago










  • You can find examples in which $L_1$ is a strict subset of $L_2$ (and vice versa).
    – Yuval Filmus
    10 hours ago














3












3








3







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 11 hours ago









Bit_hcAlgorithm

1408




1408












  • How would define the cardinality of an infinite set being larger than another infinite set (both being countable sets)?
    – dkaeae
    10 hours ago










  • You can find examples in which $L_1$ is a strict subset of $L_2$ (and vice versa).
    – Yuval Filmus
    10 hours ago


















  • How would define the cardinality of an infinite set being larger than another infinite set (both being countable sets)?
    – dkaeae
    10 hours ago










  • You can find examples in which $L_1$ is a strict subset of $L_2$ (and vice versa).
    – Yuval Filmus
    10 hours ago
















How would define the cardinality of an infinite set being larger than another infinite set (both being countable sets)?
– dkaeae
10 hours ago




How would define the cardinality of an infinite set being larger than another infinite set (both being countable sets)?
– dkaeae
10 hours ago












You can find examples in which $L_1$ is a strict subset of $L_2$ (and vice versa).
– Yuval Filmus
10 hours ago




You can find examples in which $L_1$ is a strict subset of $L_2$ (and vice versa).
– Yuval Filmus
10 hours ago










1 Answer
1






active

oldest

votes


















5














$L_1$ and $L_2$ are always countably infinite, and thus "equally big".



If any language is finite, then it is "constant time" recognizable.






share|cite|improve this answer





















  • I forgot this fact that they're both infinite... Thanks!
    – Bit_hcAlgorithm
    8 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5














$L_1$ and $L_2$ are always countably infinite, and thus "equally big".



If any language is finite, then it is "constant time" recognizable.






share|cite|improve this answer





















  • I forgot this fact that they're both infinite... Thanks!
    – Bit_hcAlgorithm
    8 hours ago
















5














$L_1$ and $L_2$ are always countably infinite, and thus "equally big".



If any language is finite, then it is "constant time" recognizable.






share|cite|improve this answer





















  • I forgot this fact that they're both infinite... Thanks!
    – Bit_hcAlgorithm
    8 hours ago














5












5








5






$L_1$ and $L_2$ are always countably infinite, and thus "equally big".



If any language is finite, then it is "constant time" recognizable.






share|cite|improve this answer












$L_1$ and $L_2$ are always countably infinite, and thus "equally big".



If any language is finite, then it is "constant time" recognizable.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 10 hours ago









Pål GD

5,9621939




5,9621939












  • I forgot this fact that they're both infinite... Thanks!
    – Bit_hcAlgorithm
    8 hours ago


















  • I forgot this fact that they're both infinite... Thanks!
    – Bit_hcAlgorithm
    8 hours ago
















I forgot this fact that they're both infinite... Thanks!
– Bit_hcAlgorithm
8 hours ago




I forgot this fact that they're both infinite... Thanks!
– Bit_hcAlgorithm
8 hours ago


















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