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Runti Compl Inner Rewri 22807 pair #381904431
details
property
value
status
complete
benchmark
#4.32.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n071.star.cs.uiowa.edu
space
Strategy_removed_AG01
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
6.67447805405 seconds
cpu usage
20.136918871
max memory
3.259887616E9
stage attributes
key
value
output-size
8463
starexec-result
WORST_CASE(Omega(n^1), O(n^1))
output
/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^1), O(n^1)) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1). (0) CpxTRS (1) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms] (2) CdtProblem (3) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 44 ms] (4) CdtProblem (5) CdtInstantiationProof [BOTH BOUNDS(ID, ID), 0 ms] (6) CdtProblem (7) CdtRhsSimplificationProcessorProof [BOTH BOUNDS(ID, ID), 0 ms] (8) CdtProblem (9) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms] (10) CdtProblem (11) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 7 ms] (12) CdtProblem (13) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms] (14) BOUNDS(1, 1) (15) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] (16) TRS for Loop Detection (17) DecreasingLoopProof [LOWER BOUND(ID), 0 ms] (18) BEST (19) proven lower bound (20) LowerBoundPropagationProof [FINISHED, 0 ms] (21) BOUNDS(n^1, INF) (22) TRS for Loop Detection ---------------------------------------- (0) Obligation: The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1). The TRS R consists of the following rules: f(x, c(y)) -> f(x, s(f(y, y))) f(s(x), s(y)) -> f(x, s(c(s(y)))) S is empty. Rewrite Strategy: INNERMOST ---------------------------------------- (1) CpxTrsToCdtProof (UPPER BOUND(ID)) Converted Cpx (relative) TRS to CDT ---------------------------------------- (2) Obligation: Complexity Dependency Tuples Problem Rules: f(z0, c(z1)) -> f(z0, s(f(z1, z1))) f(s(z0), s(z1)) -> f(z0, s(c(s(z1)))) Tuples: F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1)) F(s(z0), s(z1)) -> c2(F(z0, s(c(s(z1))))) S tuples: F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1)) F(s(z0), s(z1)) -> c2(F(z0, s(c(s(z1))))) K tuples:none Defined Rule Symbols: f_2 Defined Pair Symbols: F_2 Compound Symbols: c1_2, c2_1 ---------------------------------------- (3) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1))) Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S. F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1)) We considered the (Usable) Rules:none And the Tuples: F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1)) F(s(z0), s(z1)) -> c2(F(z0, s(c(s(z1))))) The order we found is given by the following interpretation: Polynomial interpretation : POL(F(x_1, x_2)) = x_2 POL(c(x_1)) = [1] + x_1 POL(c1(x_1, x_2)) = x_1 + x_2 POL(c2(x_1)) = x_1 POL(f(x_1, x_2)) = [1] + x_2 POL(s(x_1)) = 0 ----------------------------------------
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