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Integer TRS Innermost pair #487529022
details
property
value
status
complete
benchmark
Ex09.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n146.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.24273109436 seconds
cpu usage
0.220906174
max memory
8835072.0
stage attributes
key
value
output-size
2140
starexec-result
MAYBE
output
/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = round#(x) -> if#(x % 2 = 0, x, x + 1) minusNat#(true, I2, y) -> minus#(I2, round(I2)) minusNat#(true, I2, y) -> round#(I2) minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) R = if(false, u, v) -> v if(true, I0, I1) -> I0 round(x) -> if(x % 2 = 0, x, x + 1) minusNat(true, I2, y) -> minus(I2, round(I2)) minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = round#(x) -> if#(x % 2 = 0, x, x + 1) minusNat#(true, I2, y) -> minus#(I2, round(I2)) minusNat#(true, I2, y) -> round#(I2) minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) minus#(I3, I4) -> minus#(I3, round(I3)) [I4 >= 0 && I3 = I4 + 1] minus#(I3, I4) -> round#(I3) [I4 >= 0 && I3 = I4 + 1] R = if(false, u, v) -> v if(true, I0, I1) -> I0 round(x) -> if(x % 2 = 0, x, x + 1) minusNat(true, I2, y) -> minus(I2, round(I2)) minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) The dependency graph for this problem is: 0 -> 1 -> 5, 4, 3 2 -> 0 3 -> 4 -> 5, 4, 3 5 -> 0 Where: 0) round#(x) -> if#(x % 2 = 0, x, x + 1) 1) minusNat#(true, I2, y) -> minus#(I2, round(I2)) 2) minusNat#(true, I2, y) -> round#(I2) 3) minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) 4) minus#(I3, I4) -> minus#(I3, round(I3)) [I4 >= 0 && I3 = I4 + 1] 5) minus#(I3, I4) -> round#(I3) [I4 >= 0 && I3 = I4 + 1] We have the following SCCs. { 4 } DP problem for innermost termination. P = minus#(I3, I4) -> minus#(I3, round(I3)) [I4 >= 0 && I3 = I4 + 1] R = if(false, u, v) -> v if(true, I0, I1) -> I0 round(x) -> if(x % 2 = 0, x, x + 1) minusNat(true, I2, y) -> minus(I2, round(I2)) minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4)
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