details

property	value
status	complete
benchmark	countUpExp.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n013.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.246259 seconds
cpu usage	0.250888
user time	0.134851
system time	0.116037
max virtual memory	113176.0
max residence set size	8292.0

stage attributes

key	value
starexec-result	MAYBE

output

0.00/0.24 MAYBE 0.00/0.24 0.00/0.24 DP problem for innermost termination. 0.00/0.24 P = 0.00/0.24 if#(true, I0, I1) -> exp#(I0, I1 - 1) 0.00/0.24 exp#(I2, I3) -> if#(I3 > 0, I2, I3) 0.00/0.24 cu#(true, I4) -> cu#(I4 < exp(10, 2), I4 + 1) 0.00/0.24 cu#(true, I4) -> exp#(10, 2) 0.00/0.24 R = 0.00/0.24 if(false, x, y) -> 1 0.00/0.24 if(true, I0, I1) -> I0 * exp(I0, I1 - 1) 0.00/0.24 exp(I2, I3) -> if(I3 > 0, I2, I3) 0.00/0.24 cu(true, I4) -> cu(I4 < exp(10, 2), I4 + 1) 0.00/0.24 0.00/0.24 This problem is converted using chaining, where edges between chained DPs are removed. 0.00/0.24 0.00/0.24 DP problem for innermost termination. 0.00/0.24 P = 0.00/0.24 if#(true, I0, I1) -> exp#(I0, I1 - 1) 0.00/0.24 exp#(I2, I3) -> if#(I3 > 0, I2, I3) 0.00/0.24 cu#(true, I4) -> cu_1#(I4) 0.00/0.24 cu#(true, I4) -> exp#(10, 2) 0.00/0.24 cu_1#(I4) -> cu#(I4 < exp(10, 2), I4 + 1) 0.00/0.24 exp#(I2, I3) -> exp#(I2, I3 - 1) [I3 > 0] 0.00/0.24 R = 0.00/0.24 if(false, x, y) -> 1 0.00/0.24 if(true, I0, I1) -> I0 * exp(I0, I1 - 1) 0.00/0.24 exp(I2, I3) -> if(I3 > 0, I2, I3) 0.00/0.24 cu(true, I4) -> cu(I4 < exp(10, 2), I4 + 1) 0.00/0.24 0.00/0.24 The dependency graph for this problem is: 0.00/0.24 0 -> 5, 1 0.00/0.24 1 -> 0.00/0.24 2 -> 4 0.00/0.24 3 -> 5, 1 0.00/0.24 4 -> 3, 2 0.00/0.24 5 -> 5, 1 0.00/0.24 Where: 0.00/0.24 0) if#(true, I0, I1) -> exp#(I0, I1 - 1) 0.00/0.24 1) exp#(I2, I3) -> if#(I3 > 0, I2, I3) 0.00/0.24 2) cu#(true, I4) -> cu_1#(I4) 0.00/0.24 3) cu#(true, I4) -> exp#(10, 2) 0.00/0.24 4) cu_1#(I4) -> cu#(I4 < exp(10, 2), I4 + 1) 0.00/0.24 5) exp#(I2, I3) -> exp#(I2, I3 - 1) [I3 > 0] 0.00/0.24 0.00/0.24 We have the following SCCs. 0.00/0.24 { 2, 4 } 0.00/0.24 { 5 } 0.00/0.24 0.00/0.24 DP problem for innermost termination. 0.00/0.24 P = 0.00/0.24 exp#(I2, I3) -> exp#(I2, I3 - 1) [I3 > 0] 0.00/0.24 R = 0.00/0.24 if(false, x, y) -> 1 0.00/0.24 if(true, I0, I1) -> I0 * exp(I0, I1 - 1) 0.00/0.24 exp(I2, I3) -> if(I3 > 0, I2, I3) 0.00/0.24 cu(true, I4) -> cu(I4 < exp(10, 2), I4 + 1) 0.00/0.24 0.00/0.24 We use the reverse value criterion with the projection function NU: 0.00/0.24 NU[exp#(z1,z2)] = z2 + -1 * 0 0.00/0.24 0.00/0.24 This gives the following inequalities: 0.00/0.24 I3 > 0 ==> I3 + -1 * 0 > I3 - 1 + -1 * 0 with I3 + -1 * 0 >= 0 0.00/0.24 0.00/0.24 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/0.24 0.00/0.24 DP problem for innermost termination. 0.00/0.24 P = 0.00/0.24 cu#(true, I4) -> cu_1#(I4) 0.00/0.24 cu_1#(I4) -> cu#(I4 < exp(10, 2), I4 + 1) 0.00/0.24 R = 0.00/0.24 if(false, x, y) -> 1 0.00/0.24 if(true, I0, I1) -> I0 * exp(I0, I1 - 1) 0.00/0.24 exp(I2, I3) -> if(I3 > 0, I2, I3) 0.00/0.24 cu(true, I4) -> cu(I4 < exp(10, 2), I4 + 1) 0.00/0.24 0.00/3.22 EOF popout

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

job log

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actions

all output return to Integer_TRS_Innermost 2019-03-21 04.53