details

property	value
status	complete
benchmark	A01.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n151.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.468897 seconds
cpu usage	0.170861
user time	0.087489
system time	0.083372
max virtual memory	113188.0
max residence set size	8504.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = cond#(true, x, y) -> minus#(x, y + 1) minus#(I2, I3) -> cond#(I2 > I3, I2, I3) R = cond(true, x, y) -> 1 + minus(x, y + 1) cond(false, I0, I1) -> 0 minus(I2, I3) -> cond(I2 > I3, I2, I3) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = cond#(true, x, y) -> minus#(x, y + 1) minus#(I2, I3) -> cond#(I2 > I3, I2, I3) minus#(I2, I3) -> minus#(I2, I3 + 1) [I2 > I3] R = cond(true, x, y) -> 1 + minus(x, y + 1) cond(false, I0, I1) -> 0 minus(I2, I3) -> cond(I2 > I3, I2, I3) The dependency graph for this problem is: 0 -> 2, 1 1 -> 2 -> 2, 1 Where: 0) cond#(true, x, y) -> minus#(x, y + 1) 1) minus#(I2, I3) -> cond#(I2 > I3, I2, I3) 2) minus#(I2, I3) -> minus#(I2, I3 + 1) [I2 > I3] We have the following SCCs. { 2 } DP problem for innermost termination. P = minus#(I2, I3) -> minus#(I2, I3 + 1) [I2 > I3] R = cond(true, x, y) -> 1 + minus(x, y + 1) cond(false, I0, I1) -> 0 minus(I2, I3) -> cond(I2 > I3, I2, I3) We use the reverse value criterion with the projection function NU: NU[minus#(z1,z2)] = z1 + -1 * z2 This gives the following inequalities: I2 > I3 ==> I2 + -1 * I3 > I2 + -1 * (I3 + 1) with I2 + -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