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From: | Jay Foad |
Subject: | Re: [Bug-apl] Problem with modulo arithmetic on Gaussian integers redux |
Date: | Thu, 15 Jun 2017 09:38:32 +0100 |
Jürgen, JayWith svn 958, I'm see the followinga ← ¯1J11b ← ¯12J10a | b¯11J¯1b ÷ a1J11J1 × a¯12J10a ← 1J¯11b ← 12J¯10a ∣ b11J1b ÷ a1J11J1 × a12J¯10So the error changed but persists on my platform.Regards,FredOn Wed, 2017-06-14 at 17:20 +0200, Juergen Sauermann wrote:Hi Jay, Fred,
thanks a lot, Jay, for figuring this out.
Fred, I made the change proposed by Jay. Please let me know if the
problem persists. SVN 958.
/// Jürgen
On 06/14/2017 03:13 PM, Jay Foad wrote:
It got broken in r939. The problem is in Cell.icc:
//---------------------------------------------------------- ------------------- inline boolCell::integral_within(APL_Complex A, APL_Float B) {const APL_Complex C = -A;return tolerant_floor(C, B) == tolerant_floor(A, B);}//---------------------------------------------------------- -------------------
You're missing a minus sign before the second "tolerant_floor".
Jay.
On 14 June 2017 at 13:04, Juergen Sauermann <address@hidden> wrote:
Hi Fred,
not sure what to do about that. On my machine I am getting:
1J3 ∣ 8J4
0
and your TGI0.apl program seems not to output anything.
Best Regqrds,
Jürgen Sauermann
On 06/14/2017 05:52 AM, Frederick Pitts wrote:
Juergen, I'm seeing errors with the mod (∣) operator applied to Gaussian integers again. With svn 896, the mod operator yields a nonzero residual result while the division operator yields an exact Gaussian integer quotient result as follows 1J3 ∣ 8J4 1J3 8J4 ÷ 1J3 2J¯2 1J3 × 2J¯2 8J4 I'm running Fedora 25, 64 bit, on an Intel Core i7-6700 4 core CPU with 16 Gbyte memory. The attached TGI0.apl generates many more failure examples if needed. Regards, Fred
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