details

property	value
status	complete
benchmark	sqrt_patrs.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n094.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.205773830414 seconds
cpu usage	0.213617388
max memory	8998912.0

stage attributes

key	value
output-size	1376
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 = f#(I0, I1, I2, I3) -> f#(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0] sqrt#(I4) -> f#(I4, 0, 1, 1) R = f(x, z, u, w) -> z [w > x] f(I0, I1, I2, I3) -> f(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0] sqrt(I4) -> f(I4, 0, 1, 1) The dependency graph for this problem is: 0 -> 0 1 -> 0 Where: 0) f#(I0, I1, I2, I3) -> f#(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0] 1) sqrt#(I4) -> f#(I4, 0, 1, 1) We have the following SCCs. { 0 } DP problem for innermost termination. P = f#(I0, I1, I2, I3) -> f#(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0] R = f(x, z, u, w) -> z [w > x] f(I0, I1, I2, I3) -> f(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0] sqrt(I4) -> f(I4, 0, 1, 1) We use the reverse value criterion with the projection function NU: NU[f#(z1,z2,z3,z4)] = z1 + -1 * z4 This gives the following inequalities: I0 >= I3 && I2 >= 0 ==> I0 + -1 * I3 > I0 + -1 * (I3 + I2 + 2) with I0 + -1 * I3 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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

job log

popout

actions

all output return to Integ TRS Inner 43557