details

property	value
status	complete
benchmark	indirect.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n074.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.115714 seconds
cpu usage	0.117809
user time	0.064018
system time	0.053791
max virtual memory	113176.0
max residence set size	7760.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.11 YES 0.00/0.11 0.00/0.11 DP problem for innermost termination. 0.00/0.11 P = 0.00/0.11 if2#(false, x, y) -> f#(x, y) 0.00/0.11 if1#(true, I4, I5) -> h#(I4, I5) 0.00/0.11 h#(I6, I7) -> if2#(I6 > I7, I6, I7) 0.00/0.11 f#(I8, I9) -> if1#(I8 > I9, I8, I9) 0.00/0.11 R = 0.00/0.11 if2(false, x, y) -> f(x, y) 0.00/0.11 if2(true, I0, I1) -> 0 0.00/0.11 if1(false, I2, I3) -> 0 0.00/0.11 if1(true, I4, I5) -> h(I4, I5) 0.00/0.11 h(I6, I7) -> if2(I6 > I7, I6, I7) 0.00/0.11 f(I8, I9) -> if1(I8 > I9, I8, I9) 0.00/0.11 0.00/0.11 This problem is converted using chaining, where edges between chained DPs are removed. 0.00/0.11 0.00/0.11 DP problem for innermost termination. 0.00/0.11 P = 0.00/0.11 if2#(false, x, y) -> f#(x, y) 0.00/0.11 if1#(true, I4, I5) -> h#(I4, I5) 0.00/0.11 h#(I6, I7) -> if2#(I6 > I7, I6, I7) 0.00/0.11 f#(I8, I9) -> if1#(I8 > I9, I8, I9) 0.00/0.11 f#(I8, I9) -> h#(I8, I9) [I8 > I9] 0.00/0.11 h#(I6, I7) -> f#(I6, I7) [not(I6 > I7)] 0.00/0.11 R = 0.00/0.11 if2(false, x, y) -> f(x, y) 0.00/0.11 if2(true, I0, I1) -> 0 0.00/0.11 if1(false, I2, I3) -> 0 0.00/0.11 if1(true, I4, I5) -> h(I4, I5) 0.00/0.11 h(I6, I7) -> if2(I6 > I7, I6, I7) 0.00/0.11 f(I8, I9) -> if1(I8 > I9, I8, I9) 0.00/0.11 0.00/0.11 The dependency graph for this problem is: 0.00/0.11 0 -> 4, 3 0.00/0.11 1 -> 5, 2 0.00/0.11 2 -> 0.00/0.11 3 -> 0.00/0.11 4 -> 2 0.00/0.11 5 -> 3 0.00/0.11 Where: 0.00/0.11 0) if2#(false, x, y) -> f#(x, y) 0.00/0.11 1) if1#(true, I4, I5) -> h#(I4, I5) 0.00/0.11 2) h#(I6, I7) -> if2#(I6 > I7, I6, I7) 0.00/0.11 3) f#(I8, I9) -> if1#(I8 > I9, I8, I9) 0.00/0.11 4) f#(I8, I9) -> h#(I8, I9) [I8 > I9] 0.00/0.11 5) h#(I6, I7) -> f#(I6, I7) [not(I6 > I7)] 0.00/0.11 0.00/0.11 We have the following SCCs. 0.00/0.11 0.00/3.09 EOF popout

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job log

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actions

all output return to Integer_TRS_Innermost 2019-03-21 04.53