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Integ TRS Inner 43557 pair #381740558
details
property
value
status
complete
benchmark
indirect.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n090.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.121539831161 seconds
cpu usage
0.11853828
max memory
8609792.0
stage attributes
key
value
output-size
1569
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = if2#(false, x, y) -> f#(x, y) if1#(true, I4, I5) -> h#(I4, I5) h#(I6, I7) -> if2#(I6 > I7, I6, I7) f#(I8, I9) -> if1#(I8 > I9, I8, I9) R = if2(false, x, y) -> f(x, y) if2(true, I0, I1) -> 0 if1(false, I2, I3) -> 0 if1(true, I4, I5) -> h(I4, I5) h(I6, I7) -> if2(I6 > I7, I6, I7) f(I8, I9) -> if1(I8 > I9, I8, I9) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = if2#(false, x, y) -> f#(x, y) if1#(true, I4, I5) -> h#(I4, I5) h#(I6, I7) -> if2#(I6 > I7, I6, I7) f#(I8, I9) -> if1#(I8 > I9, I8, I9) f#(I8, I9) -> h#(I8, I9) [I8 > I9] h#(I6, I7) -> f#(I6, I7) [not(I6 > I7)] R = if2(false, x, y) -> f(x, y) if2(true, I0, I1) -> 0 if1(false, I2, I3) -> 0 if1(true, I4, I5) -> h(I4, I5) h(I6, I7) -> if2(I6 > I7, I6, I7) f(I8, I9) -> if1(I8 > I9, I8, I9) The dependency graph for this problem is: 0 -> 4, 3 1 -> 5, 2 2 -> 3 -> 4 -> 2 5 -> 3 Where: 0) if2#(false, x, y) -> f#(x, y) 1) if1#(true, I4, I5) -> h#(I4, I5) 2) h#(I6, I7) -> if2#(I6 > I7, I6, I7) 3) f#(I8, I9) -> if1#(I8 > I9, I8, I9) 4) f#(I8, I9) -> h#(I8, I9) [I8 > I9] 5) h#(I6, I7) -> f#(I6, I7) [not(I6 > I7)] We have the following SCCs.
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