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Runtime Complexity: TRS pair #487110646
details
property
value
status
complete
benchmark
minsort.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n143.star.cs.uiowa.edu
space
Mixed_TRS
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
1.57774 seconds
cpu usage
3.41045
user time
3.26012
system time
0.15033
max virtual memory
1.8275312E7
max residence set size
220516.0
stage attributes
key
value
starexec-result
WORST_CASE(NON_POLY, ?)
output
WORST_CASE(NON_POLY, ?) proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). (0) CpxTRS (1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] (2) TRS for Loop Detection (3) DecreasingLoopProof [FINISHED, 0 ms] (4) BOUNDS(EXP, INF) ---------------------------------------- (0) Obligation: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). The TRS R consists of the following rules: le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) eq(0, 0) -> true eq(0, s(y)) -> false eq(s(x), 0) -> false eq(s(x), s(y)) -> eq(x, y) if(true, x, y) -> x if(false, x, y) -> y minsort(nil) -> nil minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y)))) min(x, nil) -> x min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z)) del(x, nil) -> nil del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z))) S is empty. Rewrite Strategy: FULL ---------------------------------------- (1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID)) Transformed a relative TRS into a decreasing-loop problem. ---------------------------------------- (2) Obligation: Analyzing the following TRS for decreasing loops: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). The TRS R consists of the following rules: le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) eq(0, 0) -> true eq(0, s(y)) -> false eq(s(x), 0) -> false eq(s(x), s(y)) -> eq(x, y) if(true, x, y) -> x if(false, x, y) -> y minsort(nil) -> nil minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y)))) min(x, nil) -> x min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z)) del(x, nil) -> nil del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z))) S is empty. Rewrite Strategy: FULL ---------------------------------------- (3) DecreasingLoopProof (FINISHED) The following loop(s) give(s) rise to the lower bound EXP: The rewrite sequence min(x, cons(y, z)) ->^+ if(le(x, y), min(x, z), min(y, z)) gives rise to a decreasing loop by considering the right hand sides subterm at position [1]. The pumping substitution is [z / cons(y, z)]. The result substitution is [ ]. The rewrite sequence min(x, cons(y, z)) ->^+ if(le(x, y), min(x, z), min(y, z)) gives rise to a decreasing loop by considering the right hand sides subterm at position [2]. The pumping substitution is [z / cons(y, z)]. The result substitution is [x / y].
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