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	n146.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.297606945038 seconds
cpu usage	0.220320937
max memory	8876032.0

stage attributes

key	value
output-size	1373
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/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

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job log

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actions

all output return to Integer TRS Innermost