details

property	value
status	complete
benchmark	divMinus.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n109.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.159348964691 seconds
cpu usage	0.164672981
max memory	8769536.0

stage attributes

key	value
output-size	1586
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 = if#(true, x, y) -> div#(x - y, y) div#(I0, I1) -> if#(I0 >= I1 && I1 > 0, I0, I1) R = if(true, x, y) -> div(x - y, y) + 1 div(I0, I1) -> if(I0 >= I1 && I1 > 0, I0, I1) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = if#(true, x, y) -> div#(x - y, y) div#(I0, I1) -> if#(I0 >= I1 && I1 > 0, I0, I1) div#(I0, I1) -> div#(I0 - I1, I1) [I0 >= I1 && I1 > 0] R = if(true, x, y) -> div(x - y, y) + 1 div(I0, I1) -> if(I0 >= I1 && I1 > 0, I0, I1) The dependency graph for this problem is: 0 -> 2, 1 1 -> 2 -> 2, 1 Where: 0) if#(true, x, y) -> div#(x - y, y) 1) div#(I0, I1) -> if#(I0 >= I1 && I1 > 0, I0, I1) 2) div#(I0, I1) -> div#(I0 - I1, I1) [I0 >= I1 && I1 > 0] We have the following SCCs. { 2 } DP problem for innermost termination. P = div#(I0, I1) -> div#(I0 - I1, I1) [I0 >= I1 && I1 > 0] R = if(true, x, y) -> div(x - y, y) + 1 div(I0, I1) -> if(I0 >= I1 && I1 > 0, I0, I1) We use the reverse value criterion with the projection function NU: NU[div#(z1,z2)] = z1 This gives the following inequalities: I0 >= I1 && I1 > 0 ==> I0 > I0 - I1 with I0 >= 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 Integ TRS Inner 43557