details

property	value
status	complete
benchmark	countUpNo.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.0796720981598 seconds
cpu usage	0.076067707
max memory	8757248.0

stage attributes

key	value
output-size	1145
starexec-result	MAYBE

output

/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = cu#(true, I0) -> cu#(I0 < exp(I0), I0 + 1) cu#(true, I0) -> exp#(I0) R = exp(x) -> 2 * x cu(true, I0) -> cu(I0 < exp(I0), I0 + 1) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = cu#(true, I0) -> cu_1#(I0) cu#(true, I0) -> exp#(I0) cu_1#(I0) -> cu#(I0 < exp(I0), I0 + 1) R = exp(x) -> 2 * x cu(true, I0) -> cu(I0 < exp(I0), I0 + 1) The dependency graph for this problem is: 0 -> 2 1 -> 2 -> 1, 0 Where: 0) cu#(true, I0) -> cu_1#(I0) 1) cu#(true, I0) -> exp#(I0) 2) cu_1#(I0) -> cu#(I0 < exp(I0), I0 + 1) We have the following SCCs. { 0, 2 } DP problem for innermost termination. P = cu#(true, I0) -> cu_1#(I0) cu_1#(I0) -> cu#(I0 < exp(I0), I0 + 1) R = exp(x) -> 2 * x cu(true, I0) -> cu(I0 < exp(I0), I0 + 1) popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to Integer TRS Innermost